Chapter 1
Practice Exercises 1.1
(a) SF6 contains 1 S and 6 F atoms per molecule (b) (C2H5)2N2H2 contains 4 C, 12 H, and 2 N per molecule (c) Ca3(PO4)2 contains 3 Ca, 2 P, and 8 O atoms per formula unit (d) Co(NO3)2·6H2O contains 1 Co, 2 N, 12 O, and 12 H per formula unit
1.2
(a) NH4NO3 contains 2 N nitrogen, 4 H hydrogen, 3 O oxygen atoms per formula unit (b) FeNH4(SO4)2 contains 1 Fe iron, 1 N nitrogen, 4 H hydrogen, 2 S sulfur, 8 O oxygen atoms per formula unit (c) Mo(NO3)2·5H2O contains 1 Mo molybdenum, 2 N nitrogen, 11 O oxygen, and 10 H hydrogen atoms per formula unit (d) C6H4ClNO2 contains 6 C carbon, 4 H hydrogen, 1 Cl chlorine, 1 N nitrogen, and 2 O oxygen atoms per molecule
1.3
C2H7N, CH3NHCH3.
1.4
Reactants: 4 N, 12 H, and 6 O; Products: 4 N, 12 H, and 6 O.
1.5
Reactants: 6 N, 42 H, 2 P, 20 O, 3 Ba, 12 C; Products: 3 Ba, 2 P, 20 O, 6 N, 42 H, 12 C; The reaction is balanced .
+
2
4
7
+
6
1.6
Review Questions 1.1
This answer will be student dependent.
1.2
Observation, testing and explanation.
1.3
(a) (b) (c)
A law is a description of behavior based on the results of many experiments which are true while a theory is a tested explanation of the results of many experiments. An observation is a statement that accurately describes something we see, hear, taste, feel or smell while a conclusion is a statement that is based on a series of observations. Data are the observations made while performing experiments.
1.4
A theory is valid as long as there is no experimental evidence to disprove it. Any experimental evidence that contradicts the theory therefore, disproves the theory.
1.5
Matter has mass and occupies space. All items in the question are examples of matter.
1.6
A physical change does not change the chemical composition of matter. Melting, boiling, change of shape, or mass, and the formation of a mixture are examples of physical changes to matter. A chemical change changes the chemical composition of matter. Formation of new compounds from the reaction of other substances is an example.
1
Chapter 1
A chemical changes involves the change in composition while a physical change does not change in the composition of matter. 1.7
The reaction of calcium metal with water is a chemical change resulting in the formation of new compounds, hydrogen gas and calcium hydroxide. It is not stated in the problem, but the water also increases in temperature, which is a physical change.
1.8
These are physical changes.
1.9
(a) (b) (c) (d) (e) (f) (g)
An element is a pure substance that cannot be decomposed into something simpler. A compound is a pure substance that is composed of two or more elements in some fixed or characteristic proportion. Mixtures result from combinations of pure substances in varying proportions. A homogeneous mixture has one phase. It has the same properties throughout the sample. A heterogeneous mixture has more than one phase. The different phases have different properties. A phase is a region of a mixture that has properties that are different from other regions of the mixture. A solution is a homogeneous mixture.
1.10
Changing a compound into its element is a chemical change.
1.11
(a) Cl (e) Na (i) Hg
1.12
(a) potassium (e) manganese (i) carbon
1.13
(a) This is a heterogeneous mixture. (b) This is a pure substance and is an element, such as H2, O2, N2 or a halogen. (c) This is a homogeneous mixture. (d) This is a pure substance and is a molecule such as H2O.
1.14
(a) Diagrams (a) and (d) contain pure elements (b) Diagram (c) contains a compound (c) Diagram (a) and (b) contain diatomic molecules
1.15
The first law of chemical combination is the law of conservation of mass: no detectable gain or loss of mass occurs in chemical reactions. The other law is the law of definite proportions: in a given chemical compound, the elements are always combined in the same proportions by mass.
1.16
Conservation of mass derives from the postulate that atoms are not destroyed in normal chemical reactions. The Law of Definite Proportions derives from the notion that compound substances are always composed of the same types and numbers of atoms of the various elements in the compound.
1.17
This is the Law of Definite Proportions, which guarantees that a single pure substance is always composed of the same ratio of masses of the elements that compose it.
1.18
The law of multiple proportions states that when elements combine to make a molecule the ratio of atoms is always a small, whole number. When elements combine to form more than one compound the ratio of one element in the compound compared to a fixed amount of a second element is always a small, whole number.
(b) S (f) P (j) Ca
(c) Fe (g) I
(d) Ag (h) Cu
(b) zinc (f) magnesium (j) nitrogen
(c) silicon (g) nickel
(d) tin (h) aluminum
The first compound in the diagram has two blue and three red atoms. The second compound has two blue and one red atom. Since each compound has two blue atoms we can compare the ratio of red atoms between the
2
Chapter 1
two compounds. Thus, the ratio of red atoms in the two compounds to two blue atoms in the two compounds, is three to one. 1.19
The ratio of red between the two compounds compared to fixed number of blue atoms is not one-to-one so they are different compounds. You could also compare the ratio of blue to red in each compound to demonstrate that these are two different compounds.
1.20
These two molecules do not have a different ratio of red in the two compounds compared to a fixed number of blue atoms. The ratio of red to one blue atom is two red to one blue. Therefore, they do not demonstrate the law of multiple proportions.
1.21
This may stand for the name of an element or for the name of one atom of an element.
1.22
The smallest particle that is representative of a particular element is the atom of that element. A molecule is a representative unit that is made up of two or more atoms linked together.
1.23
H2, hydrogen Cl2, chlorine
1.24
(a) nitrogen, N (b) bromine, Br (c) chlorine, Cl (d) oxygen, O
1.25
(a) carbon, C (b) hydrogen, H (c) sulfur, S (d) iodine, I
1.26
DNA contains hydrogen, carbon, nitrogen, oxygen, and phosphorus atoms.
1.27
A chemical reaction is balanced when there is the same number of each kind of atom on both the reactant and product side of the equation; and the total charges on both the reactant and product sides of the equation are the same. These conditions must be met due to the law of conservation of matter.
1.28
Reactants are the substances to the left of the arrow in a reaction that are present before the reaction begins. Products are the substances to the right of the arrow in a reaction and they are formed during the reaction and are present when the reaction is over.
1.29
(a) (b) (c)
N2, nitrogen, Br2. bromine
O2, oxygen I2, iodine
F2, fluorine
Magnesium reacts with oxygen to give (yield) magnesium oxide. The reactants are Mg and O2. The product is MgO.
Review Problems 1.30
3 Fe, 2 Al, 3 Si, 12 O
1.31
3 Ca, 5 Mg, 8 Si, 24 O, 2 H
1.32
KNaC4H4O6
1.33
MgSO4·7H2O
3
Chapter 1
1.34
CH3COOH or C2H4O2
1.35
CH3SCH3, or C2H6S; using parentheses we can write the formula as (CH3)2S
1.36
NH3
1.37
HOCH2CH2OH, or C2H6O2
1.38 b H
H N
H.
1.39
(a) is the proper representation for C2H6O2
1.40
(a) 1 Ca calcium, 2 N nitrogen, 6 O oxygen (b) 3 H hydrogen, 1 P phosphorus, 4 O oxygen (c) 6 C carbon, 14 H hydrogen (d) 1 H hydrogen, 1 C carbon, 1 N nitrogen (e) 1 Cu copper, 1 S sulfur, 4 O oxygen
1.41
(a) (b) (c) (d) (e)
1.42
(a) 1 Sr strontium, 1 Cr chromium, 4 O oxygen (b) 1 K potassium, 1 Mn manganese, 4 O oxygen (c) 2 N nitrogen, 8 H hydrogen, 2 S sulfur, 3 O oxygen (d) 1 Mg magnesium, 1 S sulfur, 11 O oxygen, 14 H hydrogen (e) 2 Fe iron, 3 S sulfur, 12 O oxygen
1.43
(a) (b) (c) (d) (e)
1.44
(a) 6 C, 18 H (b) 4 N, 16 H, 2 S, 8 O (c) 4 Cu, 8 Cl, 16 H, 8 O
1.45
(a) (b) (c)
1.46
C2H6O2, C3H6O These two compounds illustrate that different masses of carbon combine with the same mass of hydrogen and those masses are in small, whole number ratios. It is also true for the ratio of oxygen to a fixed mass of hydrogen.
1.47
NH3, N2H4
3 H, 1 P, 4 O, hydrogen, phosphorus, oxygen 1 Ca, 4 H, 2 P, 8 O, calcium, hydrogen, phosphorus, oxygen 4 C, 9 H, 1 Br, carbon, hydrogen, bromine 3 Fe, 2 As, 8 O, iron, arsenic, oxygen 3 C, 8 H, 3 O, carbon, hydrogen, oxygen
6 C, 12 H, 2 O, carbon, hydrogen, oxygen 1 Mg, 1 S, 14 H, 11 O, magnesium, sulfur, hydrogen, oxygen 1 K, 1 Al, 2 S, 20 O, 24 H, potassium, aluminum, sulfur, oxygen, hydrogen 1 Cu, 2 N, 6 O, copper, nitrogen, oxygen 4 C, 10 H, 1 O, carbon, hydrogen, oxygen
14 C, 28 H, 14 O 4 N, 8 H, 2 C, 2 O 15 C, 40 H, 15 O
The ratio of H to a fixed mass of N in the two compounds is 3 to 2. Also, the ratio of N to a fixed mass of H in the two compounds is 2 to 3. 4
Chapter 1
1.48
(a) 2 N
(b) 10 O
1.49
(a)
1.50
2 S(CH3)2, 9 O2, 2 SO2, 4 CO2, 6 H2O 2 S(CH3)2 + 9 O2(g) → 2 SO2(g) + 4 CO2(g) + 6 H2O(l)
1.51
1 CH3OH, 3 N2O, 1 CO2, 2 H2O, 3 N2
12 C
(b)
CH3OH(l) + 3 N2O(g) 1.52
(c) 2 Na
(d) 1 S 28 H
(c)
38 O
→ CO2(g) + 2 H2O(l) + 3 N2(g)
Reaction is not balanced as written. The proper balanced reaction is:
C3H8(l) + 5O2(g) → 3CO2(g) + 4H2O(l)
1.53
The reaction is not balanced as written. The proper balanced reaction is: 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
Additional Exercises 1.54
The ratio of N in N2O to N in NO2 compared to a fixed mass of O is 4 to 1. Examine the ratio of nitrogen to a fixes mass of oxygen: 2 N2O:NO2
1.55
Ethanol, C2H6O
CH3CH2OH
Diethyl ether, C2H2O (CH3)2O 1.56
A repeat experiment, using exactly the same conditions, though possibly different masses of reactants, would result in a sample having the same composition, namely 3 CO2 to 2 N2 molecules. Also, the molecules produced would always have the same ratio of atoms in the molecules, namely CO2 and N2. The law of definite composition states that elements combine in the same mass ratio, or proportion, for a given compound.
1.57
(a) A physical change will not separate the molecules into their respective constituent chemical elements. (b) A physical change could be used to separate the molecules. By cooling the gas mixture, one gas would liquefy before the other and thus would allow one to separate the mixture. (c) A chemical change would be required to separate the molecules into their component chemical elements. (d) You would need to do a chemical reaction on CO2 to reduce it to carbon and oxygen. N2 is already an element. (e) reducing CO2 would result in one element in molecular form, namely oxygen (O2).
5
Chapter 2
Practice Exercises
4 3 4 πr , the SI unit for radius, r, is meters, the numbers and π do not have units. Therefore, the SI 3 3 unit for volume is meter3 or m3.
2.1
V=
2.2
Force equals mass × acceleration (F = ma), and acceleration equals change in velocity divided by change in change in v d time (a = ), and velocity equals distance divided by time (v = ). Put the equations together: change in t t
change in v F = m change in t d change in d change in t F= m = m 2 change in t change in t The unit for mass is kilogram (kg); the unit for distance is meter (m) and the unit for time is second (s). Substitute the units into the equation above: m Unit for force in SI base units = kg or kg m s–2 s2
2.3
2.4
2.5
9 tF = 5
9 5
F 86 C C
(
)
+ 32 F = 187 F
5 C 5 C t C = t F − 32 F = 50 F − 32 F = 10 C 9 F 9 F To convert from °F to K we first convert to °C. In the 5 C 5 C t C = t F − 32 F = 68 F − 32 F = 20 C 9 F 9 F 1K 1K TK = (273 °C + tC) = (273 °C + 20 °C) 1 °C = 293 K 1 °C (a) (b) (c)
2.6
F t + 32 F = C C
(a) (b) (c) (d) (e)
(
)
(
)
(
)
(
)
21.0233 g + 21.0 g = 42.0233 g: rounded correctly to 42.0 g 10.0324 g / 11.7 mL = 0.8574 g / mL: rounded correctly to 0.857 g / mL 14.24 cm × 12.334 cm = 148.57 cm: rounded correctly to 149 cm ( 2.223 cm − 1.04 cm ) 32.02 mL – 2.0 mL = 30. mL 54.183 g – 0.0278 g = 54.155 g 10.0 g + 1.03 g + 0.243 g = 11.3 g 1 ft 43.4 in × = 3.62 ft (1 and 12 are exact numbers) 12 in.
1.03 m × 2.074 m × 3.9 m × = 0.48 m2 12.46 m + 4.778 m
6
Chapter 2
2
2
2.7
30.48 cm 1 m 2 m 2 = 124 ft 2 = 11.5 m 1 ft 100 cm
2.8
(a)
3 ft 12 in. in. = ( 3.00 yd ) = 108 in. 1 yd 1 ft
(b)
1000 m 100 cm cm = (1.25 km ) = 1.25 × 105 cm 1 km 1 m
(c)
1 m 100 cm 1 in. 1 ft ft = ( 3.27 mm ) = 0.0107 ft 1000 mm 1 m 2.54 cm 12 in.
(d)
km 20.2 mile 1.609 km 1 gal km = = 8.59 L L 1 gal 1 mile 3.785 L
2.9
(
)
Density =
mass volume
365 g
= 16.5 g/cm3 22.12 cm3 The object is not composed of pure gold since the density of gold is 19.3 g/cm3. Density of the object =
2.10
The density of the alloy is 12.6 g/cm3. To determine the mass of the 0.822 ft3 sample of the alloy, first convert the density from g/cm3 to lb/ft3, then find the weight. 3
12.6 g 1 lb 30.48 cm 3 = 787 lb/ft cm3 453.6 g 1 ft Mass of sample alloy = (0.822 ft3) (787 lb/ft3) = 647 lb Density in lb/ft3 =
2.11 . 2.12
density = mass/volume = (1.24 × 106 g)/(1.38 × 106 cm3) = 0.899 g/cm3
1 g 1 cm3 volume of one-carat diamond = 225 mg = 0.0639 cm3 1000 mg 3.52 g
Review Questions 2.1
Physical properties include boiling point, melting point, density, color, refractive index, mass and volume.
2.2
A chemical property describes a property that changes the chemical nature of a substance while physical properties describe properties that do not change the chemical nature of a substance. For example, boiling water does not change the chemical composition of water.
2.3
(a) Physical change. Copper does not change chemically when electricity flows through it: It remains copper. (b) Physical change. Gallium is changes its state, not its chemical composition when it melts. (c) Chemical change. This is an example of the Maillard reaction describing the chemical reaction of sugar molecules and amino acids. (d) Chemical change. Wine contains ethanol which can be converted to acetic acid. (e) Chemical change. Concrete is composed of many different substances that undergo a chemical process called hydration when water is added to it. 7
Chapter 2
2.4
(a) Physical change. When corn is popped water is turned into steam by heating the corn. The pressure of the steam caused the kernel to pop open resulting in popped corn. (b) Physical change. Generally alloys are mixtures of substances and no chemical change occurs. On occasion, a chemical change can occur during the production of an alloy. An example is when iron and carbon are mixed together to make steel. During this process compounds of iron and carbon such as cementite, Fe3C , are produced. (c) Physical change. During the whipping process air is mixed with cream to increase its volume. (d) Physical change. During the production of butter fat molecules aggregate, due to the agitation of whipping, and separate from the water. (e) Physical change. The aluminum is not chemically altered during recycling.
2.5
Extensive properties, such as volume, and size, are properties that depend on the amount of substance or mass of substance while intensive properties, such as density, are not dependent on the amount of substance. The density of a milliliter of water is the same as the density of a liter of water at the same temperature.
2.6
(a) Extensive
Obviously, mass is a mass dependent property.
(b) Intensive
The boiling point of a substance is the same for a mL as it is for a L of the compound so it is mass independent.
(c) Intensive
The color of a substance does not change when you change the amount of substance.
(d) Intensive
The physical state, gas, liquid, or solid, depends on temperature and pressure but not on the mass of the substance.
(a) Intensive
The melting point of 1.0 g of water is that same as 100.0 thus melting point is not mass Dependent.
(b) Intensive
The density of 1.0 g of water is the same as 100.0 g if both samples are at the same Temperature. Thus, density is not dependent on the mass of substance.
(c) Extensive
The volume occupied by a substance is dependent on the mass of substance.
(d) Extensive
Surface area depends on the amount of substance. It also depends on the nature of the Substance. A bar of metal has a smaller surface area than that of the same bar ground Into fine particles.
(a) Gas
Temperature, density, volume, viscosity
(b) Liquid
Temperature, density, volume, viscosity
(c) Solid
Temperature, density, volume
2.7
2.8
2.9
(a) Hydrogen is a gas at room temperature (b) Aluminum is a solid at room temperature (c) Nitrogen is a gas at room temperature (d) Mercury is a liquid at room temperature
8
Chapter 2
2.10
(a) Potassium chloride is a solid at room temperature. (b) Carbon dioxide is a gas at room temperature. (c) Ethyl alcohol is a liquid at room temperature. (d) Methane is a gas at room temperature. (e) Sucrose is a solid at room temperature.
2.11
(a) Sodium chloride is a solid at room temperature. (b) Ozone is a gas at room temperature. (c) Teflon is a solid at room temperature. (d) Cholesterol is a solid at room temperature. (e) Silicon dioxide is a solid at room temperature.
2.12
Measurements involve a comparison. The unit gives the number meaning.
2.13
Kilogram
2.14
(a) (b) (c) (d) (e) (f) (g)
2.15
(a) c (e) n
2.16
The melting points and boiling points of water at 1 atmosphere pressure. On the Celsius scale these points correspond to 0 °C and 100 °C respectively.
2.17
(a) (b) (c)
2.18
The digits that are significant figures in a quantity are those that are known (measured) with certainty plus the last digit, which contains some uncertainty.
2.19
The accuracy of a measured value is the closeness of that value to the true value of the quantity. The precision of a number of repeated measurements of the same quantity is the closeness of the measurements to one another. The minimum uncertainty that is implied in this measurement is ± 0.01 cm.
2.20
0.01 0.001 1000 0.000001 0.000000001 0.000000000001 1,000,000 (b) m (f) p
10–2 10–3 103 10–6 10–9 10–12 106 (c) k (g) M
(d) µ
1 Fahrenheit degree < 1 Celsius degree 1 Celsius degree = 1 Kelvin 1 Fahrenheit degree < 1 Kelvin
2.21
The problem with using the fraction 3 yd/1 ft as a conversion factor is that there are 3 feet in one yard. The conversion factor should be 1 yd/3 ft. For the second part of the question, it is not possible to construct a valid conversion factor relating centimeters to meters from the equation 1 cm = 1000 m, since 100 cm = 1 m.
2.22
To convert 250 seconds to hours multiply 250 by: 9
Chapter 2
1h 3600 s To convert 3.84 hours to seconds multiply 3.84 hours by: 3600 s 1h 2.23
Four significant figures would be correct because the conversion factor contains exact values. The measured value determines the number of significant figures.
2.24
d=
2.25
10.5 g silver = 1 cm3 silver
m : d = density; m = mass; v = volume v
10.5 g Ag 1 cm3
and
1 cm3 10.5 g Ag
Review Problems 2.26
(a) 100 cm (d) 10 dm
(b) 0.001 km (e) 1000 g
(c) 1 × 10 (f) 100 cg
2.27
(a) 10–9 (d) 106
(b) 10–6 (e) 10–3
(c) 103 (f) 0.1
2.28
(a)
(b) (c)
(d) (e)
(f)
2.29
−12
m
TC = (TF − 32)( 5 9 ) TC = (135 − 32)( 5 9 ) TC = (103)( 5 9 ) = 57.2°C TC = (61 − 32)( 5 9 ) TC = (29)( 5 9 ) = 16°C TF = ( 9 5 )TC + 32 TF = ( 9 5 )( −3.6) + 32 TF = − 6.48 + 32 = 26°F TF = ( 9 5 )(15) + 32 TF = 27 + 32 = 59°F TK = TC + 273.15 236 = TC + 273.15 −37°C = TC TK = 39 + 273.15 TK = 312 K
(a)
5 °C 5 °C tC = tF – 32 °F) = (96 °F – 32 °F) = 36 °C 9 °F 9 °F
(b)
5 °C 5 °C tC = (tF – 32 °F) = 9 °F (–6 °F –32 °F) = –21 °C 9 ° F
(c)
9 °F 9 °F tF = (tC) + 32 °F = 5 °C (–55 °C) +32 °F = –67 °F 5 °C
10
Chapter 2
2.30
(d)
1 °C 1 °C tC = (TK – 273 K) = (273 K – 273 K) = 0 °C 1K 1K
(e)
1 °C 1 °C tC = (TK – 273 K) = (299 K – 273 K) = 26 °C 1K 1K
(f)
1K 1K TK = (tC + 273 °C) = (40 °C + 273 °C) 1 °C = 313 K 1 ° C
5 C 5 °C t C = t F − 32 F = 104.5 F − 32 F = 40.3 C 9 °F 9 F This dog has a fever; the temperature is out of normal canine range.
(
)
(
)
2.31
Convert –96 °F to tc: 5 °C 5 °C tC = (tF – 32 °F) = 9 °F (–96 °F – 32 °F) = –71 °C 9 ° F
2.32
Range in Kelvins: 1 × 106 K K= (10 MK ) = 1.0 × 107 K 1 MK 1 × 106 K K= ( 25 MK ) = 2.5 × 107 K 1 MK Range in degrees Celsius: 1 °C 1 °C 7 7 tC = (TK – 273 K) = (1.0 × 10 K – 273 K) 1 K ≈ 1.0 × 10 °C 1 K
1 °C 1 °C 7 tC = (TK – 273 K) = (2.5 × 107 K – 273 K) ≈ 2.5 × 10 °C 1K 1K Range in degrees Fahrenheit: 9 °F 9 °F 7 7 tF = (°C) + 32 °F = 5 °C (1.0 × 10 °C) + 32 °F ≈ 1.8 × 10 °F 5 ° C
9 °F 9 °F 7 7 tF = (°C) + 32 °F = 5 °C (2.5 × 10 °C) + 32 °F ≈ 4.5 × 10 °F 5 ° C 2.33
Convert 111 K to tC: 1 °C 1 °C tC = (TK – 273 K) = (111 K – 273K) = –162 °C 1K 1K Convert –162 °C to tF 9 °F tF = (tC) + 32 °F = 5 °C
9 °F 5 °C (–162°C) +32 °F = –260 °F
2.34
1 °C 1 °C tC = (TK – 273 K) = (4 K – 273 K) = –269 °C 1K 1K
2.35
Convert 6000 K to tC: 11
Chapter 2
1 °C 1 °C tC = (TK – 273 K) = (6000 K – 273 K) = 5700 °C 1K 1K This is hot enough to melt concrete, since it is hotter than 2000 °C.
2.36
(a) 3 significant figures (b) 5 significant figures (c) 5 significant figures
(d) 3 significant figures (e) 4 significant figures (f) 2 significant figures
2.37
(a) 3 significant figures (b) 6 significant figures (c) 1 significant figures
(d) 5 significant figures (e) 1 significant figures (f) 5 significant figures
2.38
(a) 0.72 m2 (d) 19.42 g/mL (b) 84.24 kg (e) 858.0 cm2 3 (c) 4.19 g/cm (dividing a number with 4 sig. figs by one with 3 sig. figs)
2.39
(a) 2.06 g/mL (b) 4.02 mL (c) 12.4 g/mL
2.40
(a) (b) (c)
1 m 1 km 3600 s km/hr = ( 29.5 dm/s ) = 10.6 km/h 10 dm 1000 m 1 h 1 g 1 × 106 µg 1000 mL µg/L = (18.3 mg/mL ) = 1.83 × 107 µg/L 1g 1000 mg 1 L 1 g 1 kg −5 kg = ( 94.3 mg ) = 9.43 × 10 kg 1000 mg 1000 g
(d)
1L L = (105.8 mL ) = 0.1058 L 1000 mL
(e)
1000 mL mL = ( 0.075 L ) = 75 mL 1L
(f)
2.41
(d) 0.276 g/mL (e) 0.0006 m/s
(a)
(b)
(c)
2 1× 10
dm = ( 976 pm )
2
m 10 dm = 9.76 × 10−20 dm 2 1 pm 1 m −12
2
3 1 L 1000 cm 1 m µm = (92 dL) 100 cm 10 dL 1 L 3
3
3
106 µm = 9.2 × 1015 µm3 1m
1 g 106 µg µg = (22 ng) = 0.022 µg 109 ng 1 g 1 L 109 nL nL = (83 pL) = 0.083 nL 1012 pL 1 L 3
(d)
1000 m 11 3 m3 = (230 km3) = 2.3 × 10 m 1 km
(e)
1 m 1 km 3600 s 4 -2 km hr-2 = (87.3 cm s–2) = 1.13 × 10 km hr 100 cm 1000 m 1 hr
2
12
Chapter 2
2
2
2.42
2.43
2.44
(f)
1 m 1 nm 14 2 nm2 = (238 mm2) = 2.38 × 10 nm 1000 mm 10−9 m
(a)
2.54 cm cm = ( 27 in.) = 69 cm 1 in.
(b)
1 kg kg = ( 7.6 lb ) = 3.4 kg 2.205 lb
(c)
946.4 mL mL = ( 2.7 qt ) = 2600 mL 1 qt
(d)
29.6 mL mL = (12 oz ) = 360 mL 1 oz
(e)
1.609 km 2 km/hr = ( 65 mi/hr ) = 1.0 × 10 km/hr 1 mi
(f)
1.609 km km = ( 68.0 mi ) = 109 km 1 mi
(a)
1 qt qt = ( 250 mL ) = 0.26 qt 946.4 mL
(b)
12 in. 2.54 cm 1 m m = ( 3.0 ft ) = 0.91 m 1 ft 1 in. 100 cm
(c)
2.205 lb lb = (1.62 kg ) = 3.57 lb 1 kg
(d)
1000 mL 1 oz oz = (1.75 L ) = 59.1 oz 1 L 29.6 mL
(e)
1 mi mi/hr = ( 35 km/hr ) = 22 mi/hr 1.609 km
(f)
1 mi mi = ( 80.0 km ) = 49.7 mi 1.609 km
(a)
cm 2 =
(b)
km 2 =
(c)
30.48 cm cm3 = 176 ft 3 1 ft
(a)
0.9144 m 2 m2 = (2.4 yd2) = 2.0 m 1 yd
(b)
2.54 cm 10 mm 2 mm2 = (8.3 in2) 1 cm = 5400 mm 1 in
cm (9.8 ft ) 30.48 1 ft 2
( (
2
= 9100 cm 2 2
1.609 km 2 546 mi 2 = 1410 km 1 mi
)
)
3
= 4.98 × 106 cm3
2
2.45
2
2
13
Chapter 2
3
(c)
2.46
2.47
3
3
1 yd 0.9144 m 100 cm 1 mL 1 L L = (9.1 ft3) = 260 L 3ft 1 yd 1 m 1 cm3 1000 mL
946.35 mL 3 mL = ( 4.2 qt ) = 4.0 × 10 mL (stomach volume) 1 qt 4.0 × 103 mL ÷ 0.9 mL = 4,000 pistachios (don’t try this at home) To determine if 50 eggs will fit into 4.2 quarts, calculate the volume of fifty eggs, then compare the answer to the volume of the stomach: 53 mL 1 L 1.057 qt Volume of 50 eggs = (50 eggs) = 2.8 qt 1 egg 1000 mL 1 L 2.8 qt < 4.2 qt Luke can eat 50 eggs.
2.48
−2 m m 200 mi 5280 ft 30.48 cm 1×10 m 1 hr 1 min = = 90 s s 1 hr 1 mi 1 ft 1 cm 60 min 60 s
2.49
2435 ft 1 yd 0.9144 m 1 km 3600 s km/h = = 2672 km/h s 3 ft 1 yd 1000 m 1 h
2.50
mi mi 2230 ft 1 mi 60 s 60 min = 5280 ft 1 min 1 hr = 1520 hr h 1 s
2.51
tons/day = 2.05 × 105
2.52
2.53
2.54
2.55
2.56
ft 3 62.4 lb 1 ton 3600 s 24 h 8 = 5.53 × 10 tons/day s 1 ft 3 2000 lb 1 h 1 d
365.25 d 24 h 3600 s 3.00 × 108 m 1 light year = 1 y = 9.47 × 1015 m 1 h 1 y 1 d 1 s 15 9.47 × 10 m 1 km 1 mi 13 miles = 8.7 light years = 5.1 × 10 mi 1 light year 1000 m 1.609 km There are 360 degrees of latitude around the circumference of the earth. 60 nautical miles 1.151 statute miles 4 statute miles = 360 degree latitude = 2.49 × 10 statute miles 1 degree latitude 1 nautical mile
6 ft 1 yd 0.9144 m meters = 6033.5 fathoms = 11,034 m 1 fathom 3 ft 1 yd 14.7 lb/in 2 pounds/in2 = 11,034 m = 16,200 lb/in2 10 m 1 ton tons/in2 = 162,000 lb/in2 2000 lb = 8.10 ton/in2 density = mass/ volume = 27.7 g/34.8 mL = 0.796 g/mL 14
Chapter 2
2.57
density =
mass volume
d=
16.7 g 9.87 cm
2.58
1 mL mL = 37.0g = 46.8 mL 0.791 g
2.59
1 mL mL = (17.632 g ) = 17.684 mL 0.99704 g
2.60
1.492 g g = 235 mL = 351 g 1 mL
2.61
kg =
3
= 1.69 g/cm3
1000 mL 0.65 g 1 kg = 22 kg 1 L 1 mL 1000 g
( 34 L )
2.2 lbs lbs = ( 22 kg ) = 48 lbs 1 kg 2.62
mass of silver = 62.00 g – 27.35 g = 34.65 g volume of silver = 18.3 mL –15 mL = 3.3 mL or 3.3 cm3 density of silver = (mass of silver)/(volume of silver) = (34.65 g)/(3.3 cm3) = 11 g/cm3
2.63
volume of titanium = (2.36 cm)(1.24 cm)(2.12 cm) = 6.20 cm3 density of titanium = 28.2 g/6.20 cm3 = 4.55 g/cm3
2.64
227, 641 lb density = = 0.591 lb gal 385, 265 gal
sp. gr. of liquid hydrogen =
(
0.5909 lb gal = 0.0708 8.34 lb gal
)
density = 0.07085 1.00 g mL = 0.0709 g mL 3
2.65
10.1 ft x 32.3 ft x 4.00 in x
3.08 m3 x 0.686
1 ft 0.3048 m x = 3.08 m3 12 in ft
kg 1000 L x = 2112 kg L3 m3
Additional Exercises 2.66
0.959
CN$ 1 US$ 3.785 L x x = 3.178 US$ L 1.142 CN$ gal
15
Chapter 2
2.67
2.68
Hausberg Tarn
1 yd 4350 m = 4760 yd 0.9144 m
Mount Kenya
1 yd 4600 m = 5000 yd 0.9144 m
Temperature
9 °F ∆tF = (tC) = 5 °C
1 yd 4700 m = 5100 yd 0.9144 m
9 °F 5 °C (4.0 °C) = 7.2 °F
If the density is in metric tons… 4.93 mL 1 cm3 1×108 tons 1000 kg 1× 103 g g = 1 teaspoon 3 1 ton 1 kg 1 tsp 1 mL 1 cm = 4.93 × 1014 g If the density is in English tons…
4.93 mL 1 cm3 1× 108 tons 2000 lbs 453.59 g g = 1 teaspoon 3 1 ton 1 lb 1 tsp 1 mL 1 cm = 4.47 × 1014 g 2.69
3600 s 24 h 365 d 15 1 light year = 3.00 × 108 m/s = 9.46 × 10 m 1 h 1 d 1 y Distance to Arcturus: 1 h 1 d 1s 1000 m 4 days = 3.50 × 1014 km = 1.35 × 10 d 8 1 km 3.00 × 10 m 3600 s 24 h 1000 m 1 light year light years = 3.50 × 1014 km = 37.0 light years 1 km 9.46 × 1015 m
2.70
(a) In order to determine the volume of the pycnometer, we need to determine the volume of the water that fills it. We will do this using the mass of the water and its density. mass of water = mass of filled pycnometer – mass of empty pycnometer = 32.954 g – 23.426 g = 9.528 g
1 mL volume = (9.528 g) = 9.556 mL 0.99704 g (b) We know the volume of chloroform from part (a). The mass of chloroform is determined in the same way that we determined the mass of water. mass of chloroform = mass of filled pycnometer – mass of empty pycnometer = 37.540 g – 23.426 g = 14.114 g
2.71
14.114 g Density of chloroform = = 1.477 g/mL 9.556 mL For the message to get to the moon: 1s 1.609 km 1000 m s = ( 239, 000 miles ) = 1.28 s 8 1 mile 1 km 3.00 × 10 m The reply would take the same amount of time, so the total time would be: 1.28 s × 2 = 2.56 s
16
Chapter 2
2.72
2.73
$5.75 $0.19 = 30 min min
(a)
$5.75 = 30 min
(b)
$0.19 60 min $ = 2 hr × + 15 min min = $25.65 hr
(c)
1 min min = ( $29.75 ) = 157 min $0.19
1.025 g 1 lb 30.48 cm dsea water = cm3 453.59 g 1 ft
ft 3 =
30 min 1 min = $5.75 $0.19
3
= 64.0 lb
ft 3
2000 lbs 1 ft 3 = 1.183 × 105 ft 3 1 ton 64.0 lb
( 3785 tons )
3
2.74 2.75
2.76
1 in 0.00011 lbs 453.59 g g = 2510 cm3 = 7.6 g 2.54 cm 1 in 3 1 lb The experimental density most closely matches the known density of methanol (0.7914 g/mL). The density of ethanol is 0.7893 g/mL. Melting point and boiling point could also distinguish these two alcohols, but not color.
453.59 g 1 ft g/mL = 69.22 lb/ft 1 lb 30.48 cm 3
3
1 cm3 1 mL = 1.1088g/mL
Since the density closely matches the known value, we conclude that this is an authentic sample of ethylene glycol. 2.77 2.78
1 °C 1 °C tc = (TK – 273 K) = (5800 K – 273 K) = 5500 ºC 1K 1K We solve by combining two equations: 9 °C tF = (tC) + 32 °F 5 °F tF = tC If tF = tC, we can use the same variable for both temperatures:
9 °C tC = (tC) + 32 °F 5 °F
2.79
5 9 °C tc = (tC) + 32 °F 5 5 °F −4 t c = 32 5 −5 tc = 32 = –40, therefore the answer is –40 °C. 4 Both the Rankine and the Kelvin scales have the same temperature at absolute zero: 0 R = 0 K. For converting from tF to TR: 5 °C 1 °C tC = (tF – 32 °F) and tC = (TK – 273 K) 9 °F 1K
17
Chapter 2
therefore at TK = 0 K = 0 R
1 °C 5 °C (TK – 273 K) = (tF – 32 °F) 1 K 9 °F 1 °C 5 °C (0 K – 273 K) = (tF – 32 °F) 1 K 9 °F 5 °C –273 °C = (tF – 32 °F) 9 °F –491 °F = tF – 32 °F tF = –459 °F at absolute zero 1R TR = (tF + 459 °F) 1 °F
Also, TR at absolute zero is 0 R and So, the boiling point of water is 212 °F and in TR: 1R TR = (212 °F + 459 °F) = 671 R 1 °F 2.80
Sand d = 2.84 g/mL Gold d = 19.3 g/mL Mixture d = 3.10 g/mL 1000 g 3 1.55 kg mixture = 1.55 × 10 g of mixture 1 kg 1.55 × 103 g of mixture = msand + mgold msand = (dsand)(Vsand) mgold = (dgold)(Vgold) 1.55 × 103 g of mixture = (dsand)(Vsand) + (dgold)(Vgold) 1.55 × 103 g of mixture = (2.84 g/mL)(Vsand) + (19.3 g/mL)(Vgold) Vmixture = Vsand + Vgold m d= V
1.55 × 103 g = 425 mL 3.65 g mL Vsand + Vgold = 425 mL Vsand = 425 mL – Vgold 1.55 × 103 g of mixture = (2.84 g/mL)(425 mL – Vgold) + (19.3 g/mL)(Vgold) 1.55 × 103 g of mixture = 1207 g sand – (2.84 g/mL)(Vgold) + (19.3 g/mL)(Vgold) 1.55 × 103 g of mixture – 1207 g sand = (16.5 g/mL)(Vgold) 20.8 mL = Vgold 1.55 × 103 g of mixture – 1207 g sand = 343 g gold 343 g gold % mass of gold = × 100% = 22.1% gold 1550 g total 2
2.81
30.48 cm Area of gold in cm2 = 14.6 ft2 = 1.36 × 104 cm2 1 ft 1 cm 3 Volume of gold in cm3 = 1.36 × 104 cm2 × 2.50 µm × = 3.39 cm 1× 104 µm 1 mL 19.3 g 1 troy ounce $1125.10 Cost of gold = 3.39 cm3 × × = $2367 × × 1 cm3 1 mL 31.1035 g troy ounce
18
Chapter 2
2
2.82
1 Volume of cylindrical metal bar = π × r2 × h = π × ( 0.828 cm ) × 2.12 cm = 1.14 cm3 2 mass 9.276 g Density of cylindrical metal bar = = 8.14 g/mL = volume 1.14 cm3 3
8.14 g 1 lb 1 mL 30.48 cm 3 Density in lb/ft3 = 453.59 g = 508 lb/ft 3 1 mL 1 ft 1 cm
2.83
2.84
2.85
200 mg 1 g 1 mL Volume of diamond = 2.25 carat = 0.128 mL 1 carat 1000 mg 3.51 g 3.2 ×10−4 g 106 µg 1 L Concentration of lead in blood in µg/dL = = 32 µg/dL 1 g 10 dL L This person is in danger of exhibiting the effects of lead poisoning since the 32 µg/dL is above the threshold of 10 µg of lead/dL. Radius of ball bearing = 2.000 mm × (1/2) = 1.000 mm Volume of ball bearing= 4/3 × π × r3 = 4/3 × π × (1.000 mm)3 = 4.189 mm3 Radius of ball bearing + gold = 1.000 mm + 0.500 mm = 1.500 mm Volume of ball bearing + gold = 4/3 × π × r3 = 4/3 × π × (1.500 mm)3 = 14.14 mm3 Volume of gold = (volume of ball bearing + gold) – (volume of ball bearing) = 14.14 mm3 – 4.189 mm3 = 9.95 mm3 3
2.86
1 cm 19.31 g Mass of gold = 9.95 mm3 = 0.192 g 10 mm 1 cm The question is asking to calculate the number of mile/gallon/person for a jet airliner and a car. The answer is: 1 mile 5.0 gallons of jet fuel = 3.5 × 10–4 mile/gallon/person Rate of fuel consumption = 568 people
21.5 miles 1 gallon Rate of fuel consumption for car = = 11 mile/gallon/person 2 people But a more insightful answer would be to calculate the number of gallons/person/mile which would give the number of gallons each person uses per mile. 5.0 gallons of jet fuel 568 people = 0.0088 gallon/person/mile Rate of fuel consumption = 1 mile 1 gallon 2 people Rate of fuel consumption for car = = 0.75 gallon/person/mile 21.5 miles This would indicate that the jet airliner has better fuel consumption. 5.0 gallons 3.785 L 1000 mL 0.803 g 1 lb 5 Pounds of jet fuel = 3470 miles = 1.16 × 10 lb 1 mile 1 gallon 1 L 1 mL 453 g
19
Chapter 3 Practice Exercises 3.1
240 94 Pu
, 94 electrons
The bottom number is the atomic number, found on the periodic table (number of protons). The top number is the mass number (sum of the number of protons and the number of neutrons). Since it is a neutral atom, it has 94 electrons. 3.2
35 17 Cl
3.3
We can discard the 17 since the 17 tells the number of protons which is information that the symbol "Cl" also provides. In addition, the number of protons equals the number of electrons in a neutral atom, so the symbol "Cl" also indicates the number of electrons. The 35 is necessary to state which isotope of chlorine is in question and therefore the number of neutrons in the atom.
3.4
2.24845 × 12 u = 26.9814 u
3.5
Copper is 63.546 u ÷12 u = 5.2955 times as heavy as carbon
3.6
(0.199 × 10.0129 u) + (0.801 × 11.0093 u) = 10.8 u
3.7
(0.90483 x 19.992 u) + (0.00271 x 20.994 u) + (0.09253 x 21.991 u) = 20.2 u
3.8
The number of protons is equal to the atomic number of an element. The number of electrons is equal to the atomic number for a neutral atom. If the atom has a positive charge, the number of electrons is determined by subtracting the charge from the atomic number. If the atom has a negative charge the number of electrons is determined by adding the charge to the atomic number. (a) Fe has 26 protons, and 26 electrons (b) Fe3+ has 26 protons, and 23 electrons (c) N3- has 7 protons, and 10 electrons (d) N has 7 protons, and 7 electrons.
3.9
The number of protons is equal to the atomic number of an element. The number of electrons is equal to the atomic number for a neutral atom. If the atom has a positive charge, the number of electrons is determined by subtracting the charge from the atomic number. If the atom has a negative charge the number of electrons is determined by adding the charge to the atomic number. (a) O has 8 protons, and 8 electrons (b) O2- has 8 protons, and 10 electrons (c) Al3+ had 13 protons, and 10 electrons (d) Al has 13 protons, and 13 electrons
3.10
(a) NaF
(b) Na2O
(c) MgF2
(d) Al4C3
3.11
(a) Ca3N2
(b) AlBr3
(c) K2S
(d) CsCl
3.12
(a) (b)
CrCl3 and CrCl2, Cr2O3 and CrO CuCl, CuCl2, Cu2O and CuO
3.13
(a) (b)
Au2S and Au2S3, Au3N and AuN TiS, Ti2S3 and TiS2; Ti3N2,TiN and Ti3N4
3.14
(a) KC2H3O2
contains 17 protons, 17 electrons, and 18 neutrons.
(b) Sr(NO3)2
(c) Fe(C2H3O2)3
20
3.15
(a) Na2CO3
(b) (NH4)2SO4
3.16
(a) K2S (e) Ca3P2
(b) BaBr2
3.17
(a) (c) (e)
3.18
lithium sulfide, magnesium phosphide, nickel(II) chloride, titanium(II) chloride iron(III) oxide
3.19
(a) Al2S3 (e) Au2O3
3.20
(a) (c)
3.21
(a) KClO3
3.22
The term "octa' means eight, therefore there are 8 carbon atoms in octane. The formula for an alkane is CnH2n+2, so octane has 8 carbons and ((2 × 8) + 2) = 18 H. The molecular formula is C8H18 and the condensed structural formula is CH3CH2CH2CH2CH2CH2CH2CH3.
3.23
The molecular formula is C10H22 and the condensed structural formula is CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3.
3.24
(a) Propanol: C3H8O, CH3CH2CH2OH
(b) Butanol: C4H10O, CH3CH2CH2CH2OH
3.25
(a) phosphorus trichloride (d) hydrogen sulfide
(b) sulfur dioxide
3.26
(a) AsCl5
3.27
diodine pentoxide We drop the last letter of the prefix if the name starts with a vowel
3.28
chromium(III) acetate
3.29
potassium perchlorate
(c) NaCN
aluminum chloride sodium bromide potassium phosphide
(b) SrF2
(b) (d)
(b) SCl6
barium hydroxide calcium fluoride
(c) TiO2
lithium carbonate iron(III) hydroxide (b) NaOCl
(d) Al(OH)3
(d) CoO
(b) potassium permanganate
(c) Ni3(PO4)2
(c) S2Cl2
(d) H2Te
Review Questions 3.1
(c) dichlorine heptoxide
Protons, 11 p + , +1 charge Electron, 00 e− , –1 charge Neutron, 01 n , no charge
21
3.2
Nearly all of the mass is located in the nucleus, because this is the portion of the atom where the protons and the neutrons are located.
3.3
A nucleon is a subatomic particle found in the atomic nucleus. We have studied neutrons and protons.
3.4
The charge-to-mass ratio of the electron was determined by J.J. Thomson using a cathode ray tube. Electrons were generated and passed through the tube to a detection screen. The ratio of e/m was determined by applying an electric and magnetic field, at 900 to each other, to the electrons and observing to what degree they were bend when the fields were applied.
3.5
Millikan generated a fine mist of oil in a container. Electrons were then generated in the container and attached themselves to the oils drops. He applied an electric field to the oil drops and observed their motion on the electric field. From his observations of the effect of an electric field on the oil drops he was able to determine the charge of a single electron. Using the ratio, e/m, of an electron determined by Thompson, he was able to determine the mass of an electron.
3.6
The proton was discover by passing hydrogen gas through a mass spectrometer. The hydrogen was ionized leaving the nucleus. Three different species were observed corresponding to the mass of a proton, a proton with one neutron, and a proton with two neutrons.
3.7
Rutherford passes alpha particles through a thin sheet of metal foil and observed their deflection to determine the existence and nature of the nucleus.
3.8
The atomic number is equal to the number of protons in the nucleus of the atom, and the mass number is the sum of the number of neutrons and the number of protons. The atomic number (symbol Z) is designated by a subscript preceding the chemical symbol and the mass number (symbol A) is a superscript preceding the chemical symbol.
3.9
(a)
3.10
(a)
3.11
The statement does not indicate which isotope corresponds to the given mass.
3.12
For all group IA elements (the alkali metals), the formula is MX, that is one Cl per atom of metal. For all group IIA elements (the alkaline earth metals), the formula is MX2, that is two Cl atoms per atom of metal. The correspondence in formula and the similarities in chemical behavior allowed Mendeleev to locate theses two series into their separate groups on the periodic table.
3.13
Mendeleev constructed his periodic table by arranging the elements in order of increasing atomic weight, and grouping the elements by their recurring properties. The modern periodic table is arranged in order of increasing number of protons.
mass number
131 53 I
(b)
90 38 Sr
(b)
(c)
atomic number
137 55 Ce
(d)
22
18 9F
3.14
Strontium and calcium are in the same Group of the periodic table, so they are expected to have similar chemical properties. Strontium should therefore form compounds that are similar to those of calcium, including the sorts of compounds found in bone.
3.15
Silver and gold are in the same periodic table group as copper, so they might well be expected to occur together in nature, because of their similar properties and tendencies to form similar compounds.
3.16
Cadmium is in the same periodic table group as zinc, but silver is not. Therefore cadmium would be expected to have properties similar to those of zinc, whereas silver would not.
3.17
The superscript before the symbol indicates the mass number; the superscript after the symbol indicates the charge on the atom; the subscript before the symbol indicates the atomic number; and the subscript after the symbol indicates the number of atoms in the compound. For example:
mass number charge atomic number N number of atoms
14 7 N2
Nitrogen has 7 protons and 7 electrons for a neutral atom. The molecule has two atoms in it, and the isotope with 7 neutrons gives it a mass number of 14 3.18
See Figure in the margin of page 74.
3.19
(a)
Li
(b)
I
(c)
W
(d)
Xe
(e)
Sm
(f)
Pu
(g)
Mg
3.20
Luster, electrical conductivity, thermal conductivity, ductility, and malleability are the characteristic properties of metals.
3.21
Mercury is used in thermometers because it is a liquid, and tungsten is used in light bulbs because is has such a high melting point.
3.22
The noble gases: He, Ne, Ar, Kr, Xe, and Rn
3.23
Mercury and bromine
3.24
They are semiconductors.
3.25
See figure 3.8, page 75.
3.26
The heavy line separates the metals from the nonmetals, and the metalloids border the line.
23
3.27
Metals which are used to make jewelry are those that do not corrode, silver, gold, and platinum. Iron would be useless for jewelry because it is susceptible to rusting. Potassium reacts violently with water to form hydrogen and potassium hydroxide.
3.28
Luster, malleability, color, and brittleness are some trends that are mentioned in terms of moving from the metals to the nonmetals across the periodic table, or moving down a group from nonmetals to metals.
3.29
(a) (b) (c)
In general, melting points decrease from left to right across the periodic table and increases from top to bottom. In general, boiling points decrease from left to right across the periodic table and increases from top to bottom. In general, density has a maximum in the middle of the periodic table and falls off to the right and left. Also, the density increases moving down a group.
3.30
An ionic compound is formed by the transfer of electrons, and it is accompanied by the formation of ions of opposite charge.
3.31
Metals react with nonmetals.
3.32
An ion is a charged particle. It can be monatomic or polyatomic, and it can have either a positive or a negative charge. It is derived from an atom or a molecule by gain or loss of electrons. Atoms and molecules are neutral.
3.33
In ionic substances, no molecules exist. Rather we have a continuous array of cations and anions, which are present in a constant ratio. The ratio is given by the formula unit.
3.34
(a)
Na, Na+
(b)
These particles have the same number of nuclei.
(c)
These particles have the same number of protons
(d)
These particles could have different numbers of neutrons, if they are different isotopes.
(e)
These particles do not have the same number of electrons; Na+ has one less electron.
3.35
A cation is a positively charged ion with one or fewer electrons than its neutral atom. An anion is a negatively charged ion with one or more electrons than its neutral atom. A polyatomic ion is made up of more than one atom; the whole unit is the ion.
3.36
Titanium lost four electrons to form Ti4+; it has 22 protons and 18 electrons.
3.37
Negative
3.38
Nitrogen gained 3 electrons to form N3–; it has 7 protons and 10 electrons.
3.39
Rb forms a +1 cation (Rb+) and Cl forms a –1 anion (Cl–), so the formula should be RbCl. The cation is first in the formula; therefore the formula should be Na2S.
3.40
The formula should have the smallest whole numbers possible. The formula should be TiO2.
24
3.41
3.42
(a)
Fe2+, Fe3+
(b)
Co2+, Co3+
(c)
Hg2+, Hg22+
(d)
Cr2+, Cr3+
(e)
Sn2+, Sn4+
(f)
Mn2+, Mn4+
The incorrect ones are a, d, and e. (a) should be written as Na2O (d) should be written as AlCl3 (e) should be written as Mg3P2
3.43
3.44
3.45
3.46
3.47
3.48
(a)
CN–
(b)
NH4+
(c)
NO3–
(d)
SO32–
(e)
ClO3–
f)
SO42–
(a)
OCl–
(b)
HSO4–
(c)
PO43–
(d)
H2PO4–
(e)
MnO4–
(f)
C2O42–
(a)
dichromate ion
(b)
hydroxide ion
(c)
acetate ion
(d)
carbonate ion
(e)
cyanide ion
(f)
perchlorate ion
(a)
Ca(s) + Cl2(g) CaCl2(s)
(b)
2Mg(s) + O2(g) 2MgO(s)
(c)
4Al(s) + 3O2(g) 2Al2O3(s)
(d)
S(s) + 2Na(s) Na2S(s)
(a)
Fe(OH)3(s) + 3HCl(g) H2O + FeCl3(aq)
(b)
2AgNO3(aq) + BaCl2(aq) 2AgCl(s) + Ba(NO3)2(aq)
(a)
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
(b)
2Na(s) + 2H2O 2NaOH(aq) + H2(g)
or
S8(s) + 16Na(s) → 8Na2S(s)
3.49
Nonmetals react with metals, nonmetals, and metalloids.
3.50
The Noble gases, He, Ne, Ar, Kr, Xe, Rn
3.51
H2, N2, O2, F2, Cl2, Br2, I2
25
3.52
Nonmetals
3.53
Nonmetals are more frequently found in compounds because of the large variety of ways they may combine. A particularly illustrative example is the combination of carbon, a nonmetal, with other elements. So many compounds are possible that there is one entire area of chemistry devoted to the study of carbon compounds, organic chemistry.
3.54
Al2Cl6 is molecular because the smallest whole number ratio of elements is not used in the formula.
3.55
(a) CH4
3.56
PH3
3.57
HAt
3.58
SnH4
3.59
(a) CH4, component of natural gas
(b) CH3CH3, component of natural gas
(c) CH3CH2CH3, gas-fired barbecues
(d) CH3CH2CH2CH3, cigarette lighters
(b) NH3
CH3OH
(c) TeH2
(d) HI
(b)
CH3CH2OH
3.60
(a)
3.61
C10H22 or CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
3.62
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3 or CH3(CH2)21CH3 or C23H48
3.63
All of the elements are nonmetals, and the formula is not in the smallest whole number ratio.
3.64
methane:
ethane:
26
propane:
butane:
methanol:
ethanol:
27
decane (10 carbons):
Another display of decane:
23 carbon hydrocarbon
28
There are several forms of glucose, C6H12O6. Shown below are the chain form and one of the cyclic forms. Chain form
Cyclic form (α-D-glucopyranose)
3.65
Binary compounds, such as CCl4 contain two elements only. A diatomic substance is composed of molecules having two atoms, such as HCl or N2. In the latter, the two atoms may or may not be the same.
3.66
When naming the compound, molecular compounds need the prefixes to specify the number of atoms in the molecule. Ionic compounds made with transition metals or post-transition metals need to have the charge of the metal specified.
3.67
For naming ionic compounds of the transition elements it is essential to know the charge on the anion since that will help determine the charge on the transition element. Transition elements can have more than one charge.
3.68
(a)
Greek prefixes are used to specify the number of atoms of each element in a molecular compound (PCl5 is phosphorous pentachloride; specify the number of water molecules in a hydrate (CuSO4·5H2O is copper sulfate pentahydrate); specify the number of atoms in polyatomic ions; and organic compounds use Greek prefixes to specify the number of carbon atoms in the compound (pentane has five carbons).
(b)
Roman numerals are used in the name of transition metal compounds to specify the charge of the metal.
Review Problems 3.69
Since we know that the formula is CH4, we know that one fourth of the total mass due to the hydrogen atom constitutes the mass that may be compared to the carbon. Hence we have 0.33597 g H ÷ 4 = 0.083993 g H and 1.00 g assigned to the amount of C-12 in the compound. Then it is necessary to realize that the ratio 1.00 g C ÷ 12 for carbon is equal to the ratio 0.083993 g H ÷ X, where X equals the relative atomic mass of hydrogen.
1.000 g C 0.083993 g H 12 u C = = 1.008 u X
29
3.70
Since we know that the formula is CCl4, we know that one fourth of the total mass due to the chlorine atom constitutes the mass that may be compared to the carbon. Hence we have 11.818 g Cl ÷ 4 = 2.9545 g Cl and 1.00 g assigned to the amount of C-12 in the compound. Then it is necessary to realize that the ratio 1.00 g C ÷ 12 for carbon is equal to the ratio 2.9545 g Cl ÷ X, where X equals the relative atomic mass of hydrogen.
1.000 g C 2.9545 g Cl 12 u C = = 35.454 u X
3.71
Using the ratio of the number of atoms of O and X and the atomic mass of O, we can compare that to the ratio of the masses of O and X to calculate the atomic mass of X:
1.125 g X 2 atoms X uX = 1.000 g O 3 atoms O 15.9994 u O u X = 27.00 u The element is aluminum. 3.72
Since we know that the formula is M3N, we know that one third of the total mass due to the metal atom constitutes the mass that may be compared to the nitrogen. Hence we have 1.486 g M ÷ 3 = 0.4953 g M and 1.00 g assigned to the amount of N in the compound. Then it is necessary to realize that the ratio 1.00 g N ÷ 14.0067 for nitrogen (we use the weighted average atomic mass for N since no specific isotope was designated) is equal to the ratio 0.4953 g N ÷ X, where X equals the relative atomic mass of the metal.
1.000 g N 14.0067 u N = 0.4953 = 6.938 u The metal is lithium, Li 3.73
Regardless of the definition, the ratio of the mass of hydrogen to that of carbon would be the same. If C–12 were assigned a mass of 24 (twice its accepted value), then hydrogen would also have a mass twice its current value, or 2.01588 u.
3.74
Taking the mass ratio of 109Ag to 12C and multiplying it by 12 we see that the mass of Ag-109 is 108.90 u.
3.75
(0.51839 × 106.9051 u) + (0.48161 × 108.9048 u) = 107.87 u
3.76
(0.9223 × 27.9769 u) + (0.0467 × 28.9765 u) + (0.0310 × 29.9738 u) = 28.09 u
3.77
(a)
152
Sm has 62 protons, 62 electrons, and 90 neutrons
(b) 205Tl has 81 protons, 81 electrons, and 124 neutrons (c) 18O has 8 protons, 8 electrons, and 10 neutrons (d) 71Ga has 31 protons, 31 electrons, and 40 neutrons
30
3.78 electrons
protons
neutrons
(a)
Selenium-78
34
34
44
(b)
209
83
83
126
(c)
Neodymium-143
60
60
83
(d)
96
42
42
54
Bi
Mo
131
I has 53 protons, 53 electrons, and 78 neutrons.
3.80
99
Tc has 43 protons, 36 electrons in the +7 state, and 56 neutrons
3.81
(a)
Rb+
(b)
I–
(d)
Se2–
(e)
Ga3+
3.82
(a) (d)
3.83
3.79
(c)
Ca2+
Ba2+ Sr2+
(b) (e)
O2– Rb+
(c)
F–
(a) (d)
KCl BaCl2
(b) (e)
NaF CaI2
(c)
MgS
3.84
(a) (b) (c) (d) (e)
CrCl2 and CrCl3 FeCl2 and FeCl3 MnCl2 and MnCl3 CuCl and CuCl2 ZnCl2
3.85
(a) (d)
NaNO3 Cr2(CO3)3
(b) (e)
Ba(C2H3O2)2 Ca3(PO4)2
(c)
NH4Br
3.86
(a) (d)
Cu(OH)2 K2SO4
(b) (e)
Cu2CrO4 NaHCO3
(c)
MgSO3
3.87
(a)
PbO and PbO2
3.88
(d) (a) (d)
FeO and Fe2O3 CdCl2 NiCl2
3.89
(a) (c) (e)
magnesium oxide potassium nitride sodium sulfide
(b) (d)
gallium chloride calcium arsenide
3.90
(a) (c) (e)
sodium fluoride lithium nitride potassium selenide
(b) (d)
magnesium carbide aluminum oxide
(b)
SnO and SnO2 (e) (b)
(c)
Cu2O and CuO AgCl
31
MnO and Mn2O3
(c)
ZnCl2
3.91
(a) (c)
carbon tetrabromide diphosphorus pentoxide
(b) (d)
nitrogen dioxide phosphorus pentachloride
3.92
(a) (c)
chlorine trifluoride dinitrogen pentoxide
(b) (d)
disulfur dichloride arsenic pentachloride
3.93
(a) (c)
iron(II) sulfide tin(IV) oxide
(b) (d)
copper(II) oxide cobalt(II) chloride hexahydrate
3.94
(a) (c)
manganese(III) oxide lead(II) sulfide
(b) (d)
mercury(I) chloride chromium(III) chloride tetrahydrate
3.95
(a) (c)
lithium nitrite copper(II) chloride dihydrate
(b) (d)
strontium chromate sodium thiocyanate
3.96
(a) (c)
potassium phosphate iron(III) carbonate
(b) (d)
ammonium acetate sodium thiosulfate pentahydrate
3.97
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)
ionic molecular ionic molecular ionic molecular ionic ionic ionic molecular
chromium(II) chloride disulfur dichloride ammonium acetate sulfur trioxide potassium iodate tetraphosphorus hexoxide calcium sulfite silver cyanide zinc(II) bromide hydrogen selenide
3.98
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)
ionic ionic ionic ionic molecular ionic ionic molecular molecular molecular
vanadium(III) nitrate cobalt(II) acetate gold(III) sulfide gold(I) sulfide germanium tetrabromide potassium chromate iron(II) hydroxide diiodine tetroxide tetraiodine nonoxide tetraphosphorus triselenide
3.99
(a) (e)
K2HPO4 Cu(CN)2
(b) (f)
Na2Se MnO2
(c) (g)
Pb(C2H3O2)4 SbCl5
(d)
P5O10
3.100
(a) (e) (g)
Al2Cl6 NH4SCN I 2O 5
(b) (f)
As4O10 K 2 S2 O 3
(c)
Mg(OH)2
(d)
Cu(HSO4)2
32
3.101
(a) (e)
(NH4)2S SnCl4
(b) (f)
Cr2(SO4)3·6H2O (c) H2Se (g)
SiF4 P 4 S7
(d)
MoS2
3.102
(a) (e)
Hg(C2H3O2)2 Mg(H2PO4)2
(b) (f)
Ba(HSO3)2 CaC2O4
BCl3 XeF4
(d)
Ca3P2
3.103
diselenium hexasulfide and diselenium tetrasulfide
3.104
diphosphorous pentasulfide
(c) (g)
Additional Exercises 3.105
(a) (b) (c) (d) (e)
nonmetal 84 36 Kr 48 neutrons 36 electrons 83.80 = 6.983 times heavier than a C-12 atom 12.000
3.106
Group 1A elements lose electrons to become positive ions and group 7A elements gains electrons to become negative ions.
3.107
(0.68077 × 57.9353 u) + (0.26223 × 59.9308 u) + (0.01140 × 60.9311 u) + (0.03634 × 61.9283 u) + (0.00926 × 63.9280 u) = 58.69 u
3.108
Let x equal the abundance of 79Br and y equal the abundance of 81Br. We know that x + y = 1 and x(78.9183) + y(80.9163) = 79.904. Substituting y = 1 – x we get x(78.9183) + (1 – x)80.9163 = 79.904. Solving for x we get x = 0.5067 and y = 0.4933.
3.109
(40.08)1.6605389 × 10–24 g = 6.655 × 10–23 g (63.55)1.6605389 × 10–24 g = 1.055 × 10–22 g Comparing answers, we see that both numbers are on the order of 8 × 10–23. We would expect 8 × 1023 atoms of K in 39.10 g of K.
3.110
Hg2(NO3)2·2H2O
Hg(NO3)2·H2O
3.111
(a) (b) (c) (d) (e) (f) (g)
HCl SO2 HCl BaS FeBr2 HCl FeBr2
BaS FeBr2 F2 F2 BaS F2 BaS
FeBr2
PH3
PH3 HCl
SO2
PH3 LiNO3
SO2
33
SO2
3.112
3.113
(a)
Ca(s) + Br2(l) CaBr2(s)
Mass of Br 79.904 × 2 = = 3.99 Mass of Ca 40.078
(b)
C(s) + 2Cl2(g) CCl4(l)
Mass of Cl 35.453 × 4 = = 11.8 Mass of C 12.0107
(c)
2Al(s) + 3S(g) Al2S3(s)
Mass of S 32.065 × 2 = = 0.792 Mass of Al 26.9815 × 3
N2O5(g) + 3SO2(g) 3SO3(g) + 2NO(g) The small whole number ratio for oxygen in these oxides is: 5:2 for N2O5, 1:1 for NO, 2:1 for SO2, and 3:1 for SO3. The small whole number ratio for oxygen in the sulfur oxides is 3 to 2.
3.114
There are three combinations of the two isotopes:
3.115
CH3CH2OH
Figure (c)
NaCl
Figure (d)
SnCl4
Figure (b)
H 2O
Figure (a)
3.116
79
Br 79 Br ,
79
Br 81 Br ,
The protons are red. The neutrons are gray. The electrons are the gold cloud. There a three protons in the nucleus so the element is lithium, Li. There are three protons and four neutrons on the nucleus so the mass number is 7. 7 3
Li
34
81
Br 81 Br
Chapter 4 Practice Exercise 4.1
1 mol Al mol Al = 3.47 g Al = 0.129 mol Al 26.98 g Al
4.2
1 mol K 2SO 4 -5 Uncertainty in moles = ±0.002 g = ±1.15 x 10 mol K2SO4 174.25 g K 2SO 4
4.3
Find the mass of 5.64 × 1018 molecules of C18H38 (MW = 254.50 g/mol
1 mol C18 H38 g = 5.64 × 1018 23 6.022 × 10 molecules C18 H38
254.50 g C H 18 38 –3 1 mol C18 H38 = 2.38 × 10 g
g = 2.38 × 10–3 g = 0.00238 g Many laboratory balances can measure 1 mg (0.001 g); therefore, it is possible to weigh 5.64 × 1018 molecules of C18H38. 4.4
Formula mass of sucrose = (12 C)(12.011 g/mol) + (22 H)(1.0079 g/mol) + (11 O)(15.9994 g/mol) = 342.299 g/mol If the massing uncertainty is ±0.002 g what is the uncertainty of mol of sucrose?
1 mol sucrose –6 Uncertainty of mol of sucrose = ±0.002 g = ±5.8 × 10 mol sucrose 342.299 g
6.022 × 1023 molecules sucrose Uncertainty of molecules of sucrose = ±5.8 × 10–6 mol sucrose = 1 mol sucrose ±3.5× 1018 molecules of sucrose 4.5
Aluminum sulfate: Al2(SO4)3, the aluminum is Al3+ 3+
mole Al = 0.0774 mol SO4
4.6
2 mol Al3+ 3 mol SO 2− 4
2–
= 0.0516 mol Al3+
The formula of dinitrogen pentoxide is N2O5
2 mol N mol N = ( 8.60 mol O ) = 3.44 mol N atoms 5 mol O
4.7
1 mol O 2 mol Fe 55.85 g Fe g Fe = ( 25.6 g O ) = 59.6 g Fe 16.00 g O 3 mol O 1 mol Fe
35
4.8
1 mol Fe2 O3 2 mol Fe 55.85 g Fe g Fe = (15.0 g Fe2 O3 ) = 10.5 g Fe 159.7 g Fe2 O3 1 mol Fe2 O3 1 mol Fe
4.9
1 mol O 1 mol TiO 2 g Ti = (12.0 g O) 16.00 g O 2 mol O
4.10
4.10
1 mol Ti 1 mol TiO 2
47.87 g Ti = 18.0 g Ti 1 mol Ti
0.0870 g H mass H × 100% = % H = × 100% = 13.04% total mass 0.6672 g total
0.3481 g C mass C % C = × 100% = × 100% = 52.17% total mass 0.6672 g total It is likely that the compound contains another element since the percentages do not add up to 100%. 4.11
% N = 0.2012/0.5462 × 100% = 36.84% N % O = 0.3450/0.5462 × 100% = 63.16% O Since these two values constitute 100%, there are no other elements present.
4.12
We first determine the number of grams of each element that are present in one mol of sample: 2 mol N × 14.01 g/mol = 28.02 g N 4 mol O × 16.00 g/mol = 64.00 g O The percentages by mass are then obtained using the formula mass of N2O4 (92.02 g): % N = (28.02/92.02) × 100% = 30.45% N % O = (64.00/92.02) × 100% = 69.55% O
4.13
N2O:
NO:
NO2:
N 2O 3:
Formula mass = 44.02 g/mol 2 mol N × 14.01 g/mol = 28.02 g N
% N = (28.02/44.02) × 100% = 63.65% N
1 mol O × 16.00 g/mol = 16.00 g O
% O = (16.00/44.02) × 100% = 36.34% O
Formula mass = 30.01 g/mol 1 mol N × 14.01 g/mol = 14.01 g N
% N = (14.01/30.01) × 100% = 46.68% N
1 mol O × 16.00 g/mol = 16.00 g O
% O = (16.00/30.01) × 100% = 53.32% O
Formula mass = 46.01 g/mol 1 mol N × 14.01 g/mol = 14.01 g N
% N = (14.01/46.01) × 100% = 30.45% N
2 mol O × 16.00 g/mol = 32.00 g O
% O = (32.00/46.01) × 100% = 69.55% O
Formula mass = 76.02 g/mol
36
N 2O 4:
N 2O 5:
2 mol N × 14.01 g/mol = 28.02 g N
% N = (28.02/76.02) × 100% = 36.86% N
3 mol O × 16.00 g/mol = 48.00 g O
% O = (48.00/76.02) × 100% = 63.14% O
Formula mass = 92.02 g/mol 2 mol N × 14.01 g/mol = 28.02 g N
% N = (28.02/92.02) × 100% = 30.45% N
4 mol O × 16.00 g/mol = 64.00 g O
% O = (64.00/92.02) × 100% = 69.55% O
Formula mass = 108.02 g/mol 2 mol N × 14.01 g/mol = 28.02 g N
% N = (28.02/108.02) × 100% = 25.94% N
5 mol O × 16.00 g/mol = 80.00 g O
% O = (80.00/108.02) × 100% = 74.06% O
The compound N2O3 corresponds to the data in Practice Exercise 4.11.
4.14
We first determine the number of mol of each element as follows:
1 mol N mol N = ( 0.712 g N ) = 0.0508 mol N 14.01 g N We need to know the number of grams of O. Since there is a total of 1.525 g of compound and the only other element present is N, the mass of O = 1.525 g – 0.712 g = 0.813 g O.
1 mol O mol O = ( 0.813 g O ) = 0.0508 mol O 16.00 g O Since these two mole amounts are the same, the empirical formula is NO.
4.15
First, find the number of moles of each element, then determine the empirical formula by comparing the ratio of the number of moles of each element. Start with the number of moles of S:
1 mol S mol S = 0.7625 g S = 0.02378 mol S 32.066 g S Then find the number of moles of O: since there are only two elements in the compound, S and O, the remaining mass is O g O = 1.525 g compound – 0.7625 g S = 0.7625 g O
1 mol O mol O = 0.7625 g O = 0.04766 mol O 15.9994 g O The empirical formula is
37
S0.02378O0.4766 The empirical formula must be in whole numbers, so divide by the smaller subscript:
S 0.02378 O 0.04766 which becomes SO2 0.02378
4.16
0.02378
2000 lb Al 454 g Al 1 mol Al 5 mol Al = 5.68 tons Al = 1.91 × 10 mol Al 1 ton Al 1 lb Al 26.98 g Al 2000 lb O 454 g O 1 mol O 5 mol O = 5.04 tons O = 2.86 × 10 mol O 1 ton O 1 lb O 16.00 g O Empirical Formula: Al1.91×105 O 2.86×105 In whole numbers: Al1.91×105 O 2.86×105 which becomes AlO1.5 and multiply the subscripts by 2: Al2O3 1.91×105
4.17
1.91×105
We first determine the number of mol of each element as follows:
1 mol N mol N = ( 0.522 g N ) = 0.0373 mol N 14.01 g N We need to know the number of grams of O. Since there is a total of 2.012 g of compound and the only other element present is N, the mass of O = 2.012 g – 0.522 g = 1.490 g O.
1 mol O mol O = (1.490 g O ) = 0.0931 mol O 16.00 g O Since these two mole amounts are not the same, the empirical formula is N0.0373O0.0931; to have the empirical formula in whole numbers, first divide by the smaller number of moles:
N 0.0373 O 0.0931 which gives 0.0373
NO 2.5
0.0373
Now to have whole numbers, multiply the subscripts by 2: N2O5.
4.18
It is convenient to assume that we have 100 g of the sample, so that the % by mass values may be taken directly to represent masses. Thus there is 32.4 g of Na, 22.6 g of S and (100.00 – 32.4 – 22.6) = 45.0 g of O. Now, convert these masses to a number of mol:
1 mol Na mol Na = ( 32.4 g Na ) = 1.40 mol Na 23.00 g Na 1 mol S mol S = ( 22.6 g S) = 0.705 mol S 32.06 g S
38
1 mol O mol O = ( 45.0 g O ) = 2.81 mol O 16.00 g O Next, we divide each of these mol amounts by the smallest in order to deduce the simplest whole number ratio: For Na: 1.40 mol/0.705 mol = 1.99 For S: 0.705 mol/0.705 mol = 1.00 For O: 2.81 mol/0.705 mol = 3.99 The empirical formula is Na2SO4. 4.19
It is convenient to assume that we have 100 g of the sample, so that the % by mass values may be taken directly to represent masses. Thus there is 81.79 g of C, 6.10 g of H and (100.00 – 81.79 – 6.10) = 12.11 g of O. Now, convert these masses to a number of mol:
1 mol C mol C = ( 81.79 g C ) = 6.81 mol C 12.01 g C 1 mol H mol H = ( 6.10 g H ) = 6.05 mol H 1.008 g H 1 mol O mol O = (12.11 g O ) = 0.757 mol O 16.00 g O Next, we divide each of these mol amounts by the smallest in order to deduce the simplest whole number ratio: For C: 6.81 mol/0.757 mol = 9.00 For H: 6.05 mol/0.757 mol = 7.99 For O: 0.757 mol/0.757 mol = 1.00 The empirical formula is C9H8O.
4.20
Find the moles of S and C using the stoichiometric ratios, and then find the empirical formula from the ratio of moles of S and C. Molar mass of SO2 = 64.06 g mol–1
Molar mass of CO2 = 44.01 g mol–1
1 mol SO 2 1 mol S –3 mol S = 0.640 g SO2 = 9.99 × 10 mol 64.06 g SO 2 1 mol SO 2
1 mol CO2 1 mol C –3 mol C = 0.220 g CO2 = 5.00 × 10 mol 44.01 g CO 1 mol CO 2 2
39
Empirical Formula C5.00×10−3 S9.99×10−3 divide both subscripts by 5.00 × 10–3 to get CS2. 4.21
Since the entire amount of carbon that was present in the original sample appears among the products only as CO2, we calculate the amount of carbon in the sample as follows:
1 mol CO2 1 mol C g C = ( 7.406 g CO 2 ) 44.01 g CO 2 1 mol CO 2
12.01 g C = 2.021 g C 1 mol C
Similarly, the entire mass of hydrogen that was present in the original sample appears among the products only as H2O. Thus the mass of hydrogen in the sample is:
1 mol H 2 O 2 mol H 1.008 g H g H = ( 3.027 g H 2 O ) = 0.3386 g H 18.02 g H 2 O 1 mol H 2 O 1 mol H The mass of oxygen in the original sample is determined by difference: 5.048 g – 2.021 g – 0.3386 g = 2.688 g O Next, these mass amounts are converted to the corresponding mol amounts:
1 mol C mol C = ( 2.021 g C ) = 0.1683 mol C 12.01 g C 1 mol H mol H = ( 0.3386 g H ) = 0.3359 mol H 1.008 g H 1 mol O mol O = ( 2.688 g O ) = 0.1680 mol O 16.00 g O The simplest formula is obtained by dividing each of these mol amounts by the smallest: For C: 0.1683 mol/0.1680 mol= 1.002 for H: 0.3359 mol/0.1680 mol= 1.999 For O: 0.1680 mol/0.1680 mol = 1.000 These values give us the simplest formula directly, namely CH2O.
4.22
To find the molecular formula, divide the molecular mass by the formula mass of the empirical formula, then multiply the subscripts of the empirical formula by that value. Formula mass of CH2Cl: 49.48 g mol–1 Formula mass of CHCl: 48.47 g mol–1
40
For CH2Cl
100 289 = 2.02 and = 5.84 49.48 49.48
For CHCl:
100 289 = 2.06 and = 5.96 48.47 48.47
The CH2Cl rounds better using the molecular mass of 100, therefore multiply the subscripts by 2 and the formula is C2H4Cl2. For CHCl, the molecular mass of 289 gives a multiple of 6, therefore the formula is C6H6Cl6.
4.23
The formula mass of the empirical unit is 1 N + 2 H = 16.03. Since this is half of the molecular mass, the molecular formula is N2H4. 32.0 g/mol hydrazine x 1 mol NH2/16.03 g = 2 mol NH2/mol hydrazine
4.24
3CaCl2(aq) + 2K3PO4(aq) Ca3(PO4)2(s) + 6KCl(aq)
4.25
3Ca(NO3)2(aq) + 2(NH4)3PO4(aq) →
4.26
1 mol O 2 mol O 2 = ( 6.76 mol SO3 ) = 3.38 mol O 2 2 mol SO3
4.27
1 mol H 2SO 4 mol H 2SO 4 = ( 0.366 mol NaOH ) = 0.183 mol H 2SO 4 2 mol NaOH
4.28
Fe2O3(s) + 2Al(s)
→
Ca3(PO4)2 (s) + 6 NH4NO3(aq)
2Fe(l) + Al2O3(s)
1 mol Fe 1 mol Al2 O3 102.0 g Al2 O3 g Al2 O3 = ( 86.0 g Fe ) 55.85 g Fe 2 mol Fe 1 mol Al2 O3 mol Al2 O3 = 78.5 g Al2O3
4.29
4.30
1 mol CaO 1 mol CO 2 44.01 g CO 2 2 g CO2 = (1.50 × 102 g CaO) 1 mol CO = 1.18 × 10 g CO2 56.08 g CaO 1 mol CaO 2 First determine the number of grams of CaCO3 that would be required to react completely with the given amount of HCl:
1 mol HCl 1 mol CaCO3 100.088 g CaCO3 g CaCO3 = (125 g HCl) = 171.57 g CaCO3 36.461 g HCl 2 mol HCl 1 mol CaO3 Since this is more than the amount that is available, we conclude that CaCO3 is the limiting reactant. The rest of the calculation is therefore based on the available amount of CaCO3:
41
1 mol CaCO3 1 mol CO2 44.01 g CO2 g CO2 = (125 g CaCO3) 100.088 g CaCO3 1 mol CaCO3 1 mol CO 2 = 55.0 g CO2
For the number of grams of left over HCl, the excess reagent, find the amount of HCl used and then subtract that from the amount of HCl started with, 125 g.
1 mol CaCO3 2 mol HCl 36.461 g HCl g HCl used = (125 g CaCO3) 100.088 g CaCO3 1 mol CaCO3 1 mol HCl = 91.1 g HCl g HCl remaining = 125 g – 91.1 g = 34 g HCl remaining
4.31
First determine the number of grams of O2 that would be required to react completely with the given amount of ammonia:
1 mol NH3 5 mol O 2 32.00 g O 2 g O2 = ( 30.00 g NH3 ) 17.03 g NH3 4 mol NH3 1 mol O 2 = 70.46 g O 2 Since this is more than the amount that is available, we conclude that oxygen is the limiting reactant. The rest of the calculation is therefore based on the available amount of oxygen:
1 mol O2 4 mol NO 30.01 g NO g NO = ( 40.00 g O 2 ) 32.00 g O2 5 mol O2 1 mol NO = 30.01 g NO
4.32
First determine the number of grams of salicylic acid, HOOCC6H4OH that would be required to react completely with the given amount of acetic anhydride, C4H6O3:
g HOOCC6H4OH = (15.6 g C4H6O3) ×
1 mol C4 H 6 O3 2 mol HOOCC6 H 4 OH 138.12 g HOOCC6 H 4 OH 1 mol C4 H 6 O3 102.09 g C4 H 6 O3 1 mol HOOCC6 H 4 OH = 42.2 g HOOCC6H4OH Since more salicylic acid is required than is available, it is the limiting reagent. Once 28.2 g of salicylic acid is reacted the reaction will stop, even though there are 15.6 g of acetic anhydride present. Therefore the salicylic acid is the limiting reactant. The theoretical yield of aspirin HOOCC6H4O2C2H3 is therefore based on the amount of salicylic acid added. This is calculated below:
42
g HOOCC6H4O2C2H3 = (28.2 g HOOCC6H4OH) ×
1 mol HOOCC6 H 4 OH 2 mol HOOCC6 H 4 O 2 C2 H3 180.16 g HOOCC6 H 4 O2 C2 H3 138.12 g HOOCC6 H 4 OH 2 mol HOOCC6 H 4 OH 1 mol HOOCC6 H 4 O 2 C2 H3 = 36.78 g HOOCC6H4O2C2H3 Now the percentage yield can be calculated from the amount of acetyl salicylic acid actually produced, 30.7 g:
30.7 g HOOCC6 H 4 O 2 C2 H3 actual yield percent yield = × 100% × 100% = theoretical yield 36.78 g HOOCC6 H 4 O 2 C2 H3 = 83.5% 4.33
First determine the number of grams of C2H5OH that would be required to react completely with the given amount of sodium dichromate:
1 mol Na 2 Cr2 O7 3 mol C2 H5 OH 46.08 g C2 H5 OH g C2 H5 OH = ( 90.0 g Na 2 Cr2 O7 ) 262.0 g Na 2 Cr2 O7 2 mol Na 2 Cr2 O7 1 mol C2 H5 OH = 23.7 g C2 H5 OH Once this amount of C2 H5 OH is reacted the reaction will stop, even though there are 24.0 g C2H5OH present, because the Na 2 Cr2 O7 will be used up. Therefore Na 2 Cr2 O7 is the limiting reactant. The theoretical yield of acetic acid (HC2H3O2) is therefore based on the amount of Na 2 Cr2 O7 added. This is calculated below:
1 mol Na 2 Cr2 O7 3 mol HC2 H3O2 60.06 g HC2 H3O2 g HC2 H3O 2 = ( 90.0 g Na 2 Cr2 O7 ) 262.0 g Na 2 Cr2 O7 2 mol Na 2 Cr2 O7 1 mol HC2 H3O 2 = 30.9 g HC2 H3O 2 Now the percentage yield can be calculated from the amount of acetic acid actually produced, 26.6 g:
26.6 g HC2 H3O 2 actual yield percent yield = × 100 = theoretical yield 30.9 g HC2 H3O 2 4.34
× 100 = 86.1%
Three step synthesis overall yield = (0.872 x 0.911 x 0.863) x 100 = 68.6 % Two step synthesis overall yield = (0.855 x 0.843) x 100 = 72.1 % Therefore, the two step process is the preferred process.
43
Review Questions 4.1
To estimate the number of atoms in a gram of iron, using atomic mass units, u, convert g to kg, then use the relationship, 1.661 × 10–27 kg = 1 u, finally using the atomic mass of Fe (55.85 u) to find the number of atoms: 1 molecule 1 kg 1u 1 g Fe = 1.08 × 1022 atoms Fe − 27 kg 55.85 u 1000 g 1.661 × 10
4.2
The mole is the SI unit for the amount of a substance. A mole is equal in quantity to Avogadro’s number (6.022 × 1023) of particles, or the formula mass in grams of a substance.
4.3
Moles are used for calculations instead of atomic mass units because they have the right units for converting from grams to moles and vice versa.
4.4
There are the same number of molecules in 2.5 moles of H2O and 2.5 moles of H2.
4.5
There are 2 moles of iron atoms in 1 mole of Fe2O3. The stoichiometric equivalent between Fe and Fe2O is 2 mol Fe ≡ 1 mol Fe2O3. For the number of iron atoms in 1 mole of Fe2O3:
1 mol Fe2O3 2 mol Fe 1 mole Fe O 2 3 4.6
(a)
1 mol S 2 mol O
(b)
2 mol As 3 mol O
(c)
(d)
4.7
6.022 × 1023 Fe atoms 24 = 1.204 × 10 atoms Fe 1 mol Fe
1 mol S 2 mol O 1 mol SO 2 1 mol SO 2 3 mol O 2 mol As 3 mol O 2 mol As 1 mol As O 1 mol As O 2 3 2 3
2 mol O 1 mol S
2 mol K 1 mol S 4 mol O 1 mol K 2SO 4 1 mol K 2SO 4 1 mol K 2SO 4 1 mol S 1 mol S 4 mol O 4 mol O 2 mol K 4 mol O 2 mol K 1 mol S
2 mol K 1 mol S
2 mol K 4 mol O
2 mol Na 1 mol H 1 mol P 4 mol O 1 mol Na 2 HPO 4 1 mol Na 2 HPO 4 1 mol Na 2 HPO 4 1 mol Na 2 HPO 4 2 mol Na 2 mol Na 2 mol Na 1 mol H 1 mol H 1 mol H 1 mol H 1 mol P 4 mol O 2 mol Na 1 mol P 4 mol O 1 mol P 2 mol Na
1 mol P 1 mol H
(a)
3 mol Mn 4 mol O
4 mol O 3 mol Mn
(b)
2 mol Sb 5 mol S
5 mol S 2 mol Sb
1 mol P 4 mol O
4 mol O 2 mol Na
3 mol Mn 1 mol Mn 3O 4 2 mol Sb 1 mol Sb2S5
44
4 mol O 1 mol H
4 mol O 1 mol Mn 3O 4 5 mol S 1 mol Sb2S5
4 mol O 1 mol P
2 mol N 1 mol ( NH 4 ) SO4 2 4 mol O 1 mol ( NH 4 ) SO4 2 8 mol H 1 mol S 4 mol O 2 mol N
(c)
2 mol N 8 mol H
1 mol S 1 mol ( NH 4 ) SO4 2
2 mol N 1 mol S
1 mol S 8 mol H
2 mol Cl 2 mol Hg
2 mol Hg 2 mol Cl
(d)
8 mol H 1 mol ( NH 4 ) SO4 2
2 mol N 4 mol O
1 mol S 4 mol O
2 mol Hg 2 mol Hg 2 Cl2
8 mol H 2 mol N
4 mol O 2 mol N
4 mol O 8 mol H
8 mol H 1 mol S 4 mol O 1 mol S
2 mol Cl 2 mol Hg 2 Cl2
4.8
The molecular mass is required to convert grams of a substance to moles of that same substance.
4.9
The statement, 1 mol O, does not indicate whether this is atomic oxygen, O, or molecular oxygen, O2. The statement 64 g of oxygen is not ambiguous because the source of oxygen is not important.
4.10
1 mol Al 26.98 g Al 1 mol Al and 26.98 g Al
4.11
At a minimum, the identity and mass of each atomic element present must be known. If the total mass of the compound is known, then it is necessary to know all but one mass of the elements that compose the compound.
4.12
When balancing a chemical equation, changing the subscripts changes the identity of the substance.
4.13 4.14
The subscripts in a formula may not be changed unless one is determining the molecular formula from the empirical formula. There are three distinct empirical formulas represented AB2, AB3, and A3B8. There are two molecules with the empirical formula AB3; AB3 and A2B6. There is one A3B8, and there are two with the formula AB2; A6B12 and A3B6.
4.15
Avagadro's number would become 5 × 1023.
1000 g Avagadro's number = 2 × 10−27 kg × 1 kg
(
4.16
−1
)
= 5 × 1023
To convert grams of a substance to molecules of the same substance, the molecular mass of the substance, and Avagadro's number are needed.
4.17 H H
H H H
H H
+
3
5
45
+
4
4.18
O
+
2
2
O
S O
4.19
Student B is correct. Student A wrote a properly balanced equation. However, by changing the subscript for the product of the reaction from an implied one, NaCl, to a two, NaCl2, this student has changed the identity of the product. When balancing chemical equations, never change the values of the subscripts given in the unbalanced equation.
4.20
Convert moles of B to moles of compound, A5B2; then using the stoichiometric ratio of moles of A to moles of A5B2, determine the moles of A; and finally convert the moles of A to grams of A using the molecular mass of A. 1 mol A 5 B2 5 mol A 100.0 g A (10 mol B) 2 mol B 1 mol A 5 B2 1 mol A The pieces of information that were not needed were the molecular mass of B and the number of molecules of A in a mole of A
4.21
Their formula weights must be identical.
4.22
To determine the number of grams of sulfur that would react with a gram of arsenic, the stoichiometric ratio of the arsenic to the sulfur in the compound is needed, as well as the atomic masses of sulfur and arsenic.
4.23
(a)
The balanced equation describes the stoichiometry
(b)
The scale of the reaction is determined by the number of moles used as reactants in the experiment.
4.24
2H2O2 2H2O + O2
4.25
First write the balanced equation for the reaction of NH4NO3 as an explosive: 2NH4NO3 (s) 2N2 (g) + O2 (g) + 4H2O (g) Then find the molecular mass of NH4NO3 (80.04 g/mol). Then calculate the number of moles of NH4NO3 is in 1.00 kg of NH4NO3: Finally, using the stoichiometric ratio of N2 to NH4NO3 calculate the number of moles of N2 and then multiply by Avagadro's number:
46
1000 g NH 4 NO3 1 mol NH 4 NO3 molecules of N2 = (1.00 kg NH4NO3) × 1 kg NH 4 NO3 80.06 g NH 4 NO3 1 mol N 2 1 mol NH NO 4 3 4.26
6.022 × 1023 molecules N 2 1 mol N 2
Reaction 1
= 7.53 × 1024 molecules of N2
Reaction 2
Review Problems 4.27
1:2,
4.28
1:4,
4.29
4.30
2 mol N to 4 mol O or in the smallest whole number ratio 1 mol N to 2 mol O
1 mol C:4 mol H 1 mol Ta –3 1.56 × 1021 atoms Ta = 2.59 × 10 mole Ta 23 6.022 × 10 atoms Ta
1 mol I2 3.65 × 1024 molecules of I2 6.022 × 1023 molecules I 2
4.31
(a) (b) (c) (d)
6 atom C:11 atom H 12 mole C:11 mole O 2 atom H: 1 atom O 2 mole H: 1 mole O
4.32
(a) (b) (c) (d)
2 atom C: 1 atom O 2 mole C: 1 mole O 1 atom C:2 atom H 1 mole C:2 mole H
= 6.06 mole I2
4.33
2 mol Bi mol Bi = (2.24 mol O) = 1.49 mol Bi 3 mol O
4.34
2 mol V mol V = (0.565 mol O) = 0.226 mol V 5 mol O
4.35
2 mol Cr mol Cr = (3.64 mol Cr2O3) = 7.28 mol Cr 1 mol Cr2 O3
47
4.36
3 mol O mol O = (4.25 mol CaCO3) = 12.8 mol O 1 mol CaCO3
4.37
(a)
2 mol Al 3 mol S 3 mol S or 2 mol Al
(b)
3 mol S 1 mol Al2 (SO4 )3 or 3 mol S 1 mol Al2 (SO4 )3
(c)
mol Al =
(d)
mol S =
(a)
3 mol Fe 1 mol Fe3O 4 or 3 mol Fe 1 mol Fe3O 4
(b)
3 mol Fe 4 mol O 4 molO or 3 mol Fe
(c)
3 mol Fe mol Fe = (2.75 mol Fe3O4) = 8.25 mol Fe 1 mol Fe3O 4
(d)
mol Fe 2 O3 =
4.38
4.39
( 0.638 mol S)
2 mol Al = 0.425 mol Al 3 mol S
3 mol S = 7.62 mol S 1 mol Al2 (SO 4 )3
( 2.54 mol Al2 (SO4 )3 )
3 mol Fe 1 mol Fe3O 4
( 4.50 mol Fe3O4 )
1 mol Fe 2 O3 = 6.75 mol Fe2 O3 2 mol Fe
Based on the balanced equation: 2 NH3(g) N2(g) + 3H2(g) From this equation the conversion factors can be written:
1 mol N 2 3 mol H 2 and 2 mol NH3 2 mol NH3 To determine the moles produced, simply convert from starting moles to end moles:
1 mol N 2 mole N2 = 0.287 mol NH3 = 0.144 mol N 2 2 mol NH3 The moles of hydrogen are calculated similarly:
3 mol H 2 mole H2 = 0.287 mol NH3 = 0.431 mol H 2 2 mol NH3
48
4.40
Based on the balanced equation: 2 Al(s) + 3 S(g) Al2S3(s) From this equation the conversion factor can be written:
3 mol S 2 mol Al To determine the moles of S needed, simply convert from the moles of Al2S3 produced:
3 mol S mol S = (0.225 mol Al) = 0.338 mol S 2 mol Al
4.41
4 mol F mol UF6 = (1.25 mol CF4) 1 mol CF4
1 mol UF6 6 mol F
4.42
2 mol Fe 1 mol Fe3O 4 mol Fe3O4 = (0.395 mol Fe2O3) 1 mol Fe2 O3 3 mol Fe
4.43
1 mol C3 H8 atoms C = (4.13 mol H) 8 mol H
= 0.833 mol UF6 = 0.263 mol Fe3O4
23 6.022 × 10 molecules C3 H8 1 mol C3 H8
3 atoms C 1 molecule C3 H8
= 9.33 × 1023 atoms C 4.44
4.45
8 mol H atom H = (3.21 mol C3H8) 1 mol C3 H8
6.022 × 1023 atoms H = 1.55 × 1025 atoms H 1 mol H
Number of C, H and O atoms in glucose = 6 atoms C + 12 atoms H + 6 atoms O = 24 atoms 6.022 × 1023 molecules glucose 24 atoms Number of atoms = (0.260 mol glucose) 1 mol glucose 1 molecule glucose = 3.76 × 1024 atoms
4.46
Number of N, H and O atoms in glucose = 2 atoms N + 4 atoms H + 3 atoms O = 9 atoms 6.022 × 1023 molecules NH NO 9 atoms 4 3 Number of atoms = (0.648 mol NH4NO3) 1 mol NH NO 1 molecule glucose 4 3 = 3.51 × 1024 atoms
4.47
1 mol C-12 mol C–12 = 8.00 g × = 0.667 mol C–12 12.00 g C-12 6.022 × 1023 atoms C-12 Number of atoms C–12 = 0.667 mol = 4.01 × 1023 atoms C–12 1 mol C-12
49
4.48
6.022 × 1023 atoms C-12 Number of atoms of C–12 = 1.5 mol C–12 = 9.033 × 1023 atoms C–12 1 mol C-12 12.00 g C-12 g C–12 = 1.5 mol C–12 × = 18 g C–12 1 mol C-12
4.49
4.50
(a)
55.85 g Fe g Fe = (2.46 mol Fe) = 137 g Fe 1 mole Fe
(b)
16.0 g O g O = (13.8 mol O) = 221 g O 1 mole O
(c)
40.08 g Ca g Ca = (0.688 mol Ca) = 27.6 g Ca 1 mole Ca
(a)
32.07 g S g S = (0.546 mol S) = 17.5 g S 1 mole S
(b)
14.01 g N g N = (3.29 mol N) = 46.1 g N 1 mole N
(c)
26.98 g N g Al = (8.11 mol Al) = 219 g Al 1 mole N
4.51
39.10 g K 1 mol K –20 g K = 4 × 102 atoms K = 3 × 10 g K 23 1 mol K 6.022 × 10 atoms K
4.52
196.9665 g Au 1 mol Au –4 g Au = 4 × 1017 atoms Au = 1.31 × 10 g Au 23 1 mol Au 6.022 × 10 atoms Au
4.53
1 mol Ni mol Ni = 22.4 g Ni = 0.382 mol Ni 58.69 g Ni
4.54
1 mol Cr mol Cr = 85.7 g Cr = 1.65 mol Cr 52.00 g Cr
4.55
Note: all masses are in g/mole (a)
(b)
NaHCO3
(NH4)2CO3
=
1Na + 1H + 1C + 3O
=
(22.98977) + (1.00794) + (12.0107) + (3 × 15.9994)
=
84.00661 g/mole = 84.0066 g/mol
=
2N + 8H + C + 3O
50
(c)
(d)
(e)
4.56
CuSO4·5H2O
K2Cr2O7
Al2(SO4)3
=
(2 × 14.0067) + (8 × 1.00794) + (12.0107) + (3 × 15.9994)
=
96.08582 g/mole = 96.0858 g/mol
=
1Cu + 1S +9O + 10H
=
63.546 + 32.065 + (9 × 15.9994) + (10 × 1.00794)
=
249.685 g/mole
=
2K + 2Cr + 7O
=
(2 × 39.0983) + (2 × 51.9961) + (7 × 15.9994)
=
294.1846 g/mole
=
2Al + 3S + 12O
=
(2 × 26.98154) + (3 × 32.065) + (12 × 15.9994)
=
342.15088 g/mole = 342.151 g/mol
Note: all masses are in g/mole (a)
(b)
(c)
(d)
(e)
Ca(NO3)2
Pb(C2H5)4
Na2SO4·10H2O
Fe4[Fe(CN)6]3
Mg3(PO4)2
=
1Ca + 2N + 6O
=
(40.078) + (2 × 14.0067) + (6 × 15.9994)
=
164.0878 g/mole = 164.088 g/mol
=
1Pb + 8C + 20H
=
(207.2) + (8 × 12.0107) + (20 × 1.00794)
=
323.4 g/mole (Since the mass of Pb is known exactly.)
=
2Na + 1S +14O + 20H
= =
(2 × 22.98977) + 32.065 + (14 × 15.9994) + (20 × 1.00794) 322.19494 g/mole = 322.195 g/mol
=
7Fe + 18C + 18N
=
(7 × 55.845) + (18 × 12.0107) + (18 × 14.0067)
=
859.2282 g/mole = 859.228 g/mol
=
3Mg + 2P + 8O
=
(3 × 24.3050) + (2 × 30.97376) + (8 × 15.9994)
=
262.85772 g/mole = 262.8577 g/mol
51
4.57
(a)
310.18 g Ca 3 ( PO 4 ) 2 g Ca3(PO4)2 = (3.25 mol Ca3(PO4)2) 1 mol Ca 3 ( PO 4 ) 2
= 1010 g Ca3(PO4)2
(b)
241.86 mg Fe ( NO3 ) 3 mg Fe(NO3)3 = (0.975 mmol Fe(NO3)3) = 236 mg Fe(NO3)3 1 mmol Fe ( NO3 ) 3 = 0.236 g Fe(NO3)3
4.58
(c)
58.12 µg C4 H10 –5 µg C4H10 = (0.750 µmol C4H10) = 43.6 µg C4H10 = 4.36 × 10 g C4H10 1 µ mol C H 4 10
(d)
96.09 g ( NH 4 ) CO3 2 = 308 g (NH4)2CO3 g (NH4)2CO3 = (3.21 mol (NH4)2CO3) 1 mol ( NH 4 ) CO3 2
(a)
136.31 g ZnCl 2 g ZnCl2 = (0.754 mol ZnCl2) = 103 g ZnCl2 1 mol ZnCl2
(b)
1 mol KIO 3 214.00 g KIO3 g KIO3 = (0.194 µmol KIO3) 106 µmol KIO 1 mol KIO3 3 = 4.15 × 10–5 g KIO3
4.59
(c)
1 mol POCl 3 153.33 g POCl3 = 0.0494 g POCl3 g POCl3 = (0.322 mmol POCl3) 103 mmol POCl 1 mol POCl3 3
(d)
132.1 g ( NH 4 ) HPO 4 2 g (NH4)2HPO4 = (4.31 ×10–3 mol (NH4)2HPO4) 1 mol ( NH 4 ) HPO 4 2
(a)
1 mole CaCO3 moles CaCO3 = ( 42.6 g CaCO3 ) = 0.426 moles CaCO3 100.09 g CaCO3
(b)
1 g NH 1 mole NH 3 3 moles NH3 = ( 2.16 ng NH3 ) = 1.27 × 10−10 moles NH3 9 1×10 ng NH 17.03 g NH3 3
(c)
1 mole Sr ( NO3 ) 2 moles Sr ( NO3 )2 = 9.78 g Sr ( NO3 )2 211.6 g Sr ( NO3 ) 2
(d)
1 g Na CrO 2 4 moles Na 2 CrO 4 = ( 4.89 µg Na 2 CrO 4 ) 106 µg Na CrO 2 4
(
)
= 3.01 × 10−8 moles Na 2 CrO 4
52
= 0.569 g (NH4)2HPO4
= 4.62 × 10−2 moles Sr ( NO3 )2
1 mole Na CrO 2 4 162.0 g Na 2 CrO 4
4.60
(a)
1 mole Ca(OH)2 mol Ca(OH)2 = (9.36 g Ca(OH)2 = 0.126 mol Ca(OH)2 74.10 g Ca(OH)2
(b)
1000 g PbSO 4 1 mole PbSO4 mol PbSO4 = (38.2 kg PbSO4) = 126 mol PbSO4 1 kg PbSO 4 303.3 g PbSO 4
(c)
1 mole H 2 O 2 mol H2O2 = (4.29 g H2O2) = 0.126 mol H2O2 34.01 g H 2 O2
(d)
1 g NaAuCl4 1 mol NaAuCl4 mol NaAuCl4 = 4.65 mg NaAuCl4 1000 mg NaAuCl4 361.8 g NaAuCl4 = 1.29 × 10–5 mol NaAuCl4
4.61
The formula CaC2 indicates that there is 1 mole of Ca for every 2 moles of C. Therefore, if there are 0.278 moles of C there must be 0.139 moles of Ca.
40.078 g Ca g Ca = (0.139 mol Ca) = 5.57 g Ca 1 mole Ca
4.62
2 moles I mol I = 0.500 mol Ca ( IO3 )2 1 mole Ca ( IO3 ) 2
(
)
= 1.00 moles I
389.9 g Ca ( IO3 ) 2 g Ca ( IO3 )2 = 0.500 mol Ca ( IO3 )2 1 mole Ca ( IO3 ) 2
(
4.63
)
= 195 g Ca ( IO3 ) 2
2 moles N = 1.30 mol N mol N = (0.650 mol (NH4)2CO3) 1 mole ( NH 4 ) CO3 2 96.09 g (NH 4 )2 CO3 g (NH4)2CO3 = (0.650 mol (NH4)2CO3) = 62.5 g (NH4)2CO3 1 mole (NH 4 )2 CO3
4.64
2 moles N mol N = (0.549 mol NH4NO3) = 1.10 mol N 1 mole NH 4 NO3 80.04 g NH 4 NO3 g NH4NO3 = (0.549 mol NH4NO3) = 43.9 g NH4NO3 1 mole NH 4 NO3
4.65
1000 g N 1 mol N 1 mol ( NH 4 )2 CO3 kg fertilizer = (1 kg N) 2 mol N 1 kg N 14.01 g N 96.09 g ( NH 4 ) CO3 1 kg ( NH 4 ) CO3 2 2 = 3.43 kg fertilizer × 1 mol ( NH 4 ) CO3 1000 g ( NH 4 ) CO3 2 2
53
4.66
4.67
1000 g P 1 mol P 1 mol P2 O5 kg P2O5 = (1.5 kg P) 1 kg P 30.97 g P 2 mol P = 3.4 kg P2O5
141.94 g P2 O5 1 kg P2 O5 1 mol P2 O5 1000 g P2 O5
Assume one mole total for each of the following. (a)
The molar mass of NaH2PO4 is 119.98 g/mol.
% Na =
(b)
(c)
(d)
23.0 g Na × 100% = 19.2% 119.98 g NaH 2 PO 4
%H=
2.02 g H × 100% = 1.68% 119.98 g NaH 2 PO4
%P=
31.0 g P × 100% = 25.8% 119.98 g NaH 2 PO4
%O=
64.0 g O × 100 % = 53.3 % 119.98 g NaH 2 PO4
The molar mass of NH4H2PO4 is 115.05 g/mol.
%N=
14.0 g N × 100% = 12.2% 115.05 g NH 4 H 2 PO4
%H=
6.05 g H × 100% = 5.26% 115.05 g NH 4 H 2 PO4
%P=
31.0 g P × 100% = 26.9% 115.05 g NH 4 H 2 PO4
%O=
64.0 g O × 100 % = 55.6 % 115.05 g NH 4 H 2 PO4
The molar mass of (CH3)2CO is 58.08 g/mol
%C=
36.0 g C × 100% = 62.0% 58.08 g ( CH3 )2 CO
%H=
6.05 g H × 100% = 10.4% 58.08 g ( CH3 )2 CO
%O=
16.0 g O × 100% = 27.6% 58.08 g ( CH3 )2 CO
The molar mass of calcium sulfate dihydrate is 172.2 g/mol.
54
% Ca =
(e)
%S=
32.1 g S × 100% = 18.6% 172.2 g CaSO 4 • 2H 2 O
%O=
96.0 g O × 100% = 55.7% 172.2 g CaSO 4 • 2H 2 O
%H=
4.03 g H × 100 % = 2.34 % 172.2 g CaSO 4 • 2H 2 O
The molar mass of CaSO4•2H2O is 172.2 g/mol.
% Ca =
4.68
(a)
(b)
40.1 g Ca × 100% = 23.3% 172.2 g CaSO 4 • 2H 2 O
40.1 g Ca × 100% = 23.3% 172.2 g CaSO 4 • 2H 2 O
%S=
32.1 g S × 100% = 18.6% 172.2 g CaSO 4 • 2H 2 O
%O=
96.0 g O × 100% = 55.7% 172.2 g CaSO 4 • 2H 2 O
%H=
4.03 g H × 100 % = 2.34 % 172.2 g CaSO 4 • 2H 2 O
The molar mass of (CH3)2N2H2 is 60.12 g/mol.
%C=
24.02 g C × 100% = 40.0% 60.12 g (CH3 )2 N 2 H 2
%H=
8.06 g H × 100% = 13.4% 60.12 g (CH3 )2 N 2 H 2
%N=
28.0 g N × 100% = 46.6% 60.12 g (CH3 )2 N 2 H 2
The molar mass of CaCO3 is 100.1 g/mol.
% Ca =
40.08 g Ca × 100% = 40.0% 100.1 g CaCO3
%C=
12.01 g C × 100% = 12.0% 100.1 g CaCO3
%O=
48.00 g O × 100% = 48.0% 100.1 g CaCO3
55
(c)
(d)
(e)
The molar mass of Fe(NO3)3 is 241.9 g/mol.
% Fe =
55.85 g Fe × 100% = 23.1% 241.9 g Fe ( NO3 )3
%N =
42.03 g N × 100% = 17.4% 241.9 g Fe ( NO3 )3
%O =
144.00 g O × 100% = 59.5% 241.9 g Fe ( NO3 )3
The molar mass of C3H8 is 44.11 g/mol.
%C =
36.03 g C × 100% = 81.7% 44.11 g C3 H8
%H =
8.08 g H × 100% = 18.3% 44.11 g C3 H8
The molar mass of Al2(SO4)3 is 342.2 g/mol.
% Al =
4.69
54.0 g Al × 100% = 15.8% 342.2 g Al2 ( SO4 )3
%S =
96.2 g S × 100% = 28.1% 342.2 g Al2 ( SO 4 )3
%O =
192.0 g O × 100% = 56.1% 342.2 g Al2 ( SO 4 )3
% C in morphine =
% C in heroin =
204.17 g C × 100% = 71.556% C 285.36 g C17 H19 NO3
252.21 g C × 100% = 68.276% C 369.44 g C21H 23 NO5
Therefore morphine has a higher percentage carbon.
4.70
% N in carbamazepine =
28.02 g N × 100% = 11.9% N 236.29 g C15 H12 N 2 O
% N in carbetapentane =
14.01 g N × 100% = 4.20% N 333.52 g C20 H31 NO3
Therefore, carbamazepine has a higher percentage of nitrogen.
56
4.71
% Cl in Freon-12 =
70.90 g Cl × 100% = 58.63% Cl 120.92 g CCl2 F2
% Cl in Freon 141b =
70.9 g Cl × 100% = 60.62% Cl 116.95 g C2 H3Cl2 F
Therefore Freon 141b has a higher percentage chlorine. 4.72
% Cl in Freon-12 =
70.90 g Cl × 100% = 58.63% Cl 120.92 g CCl2 F2
% Cl in Freon 113 =
106.35 g Cl × 100% = 56.759% Cl 187.37 g C2 Cl3 F3
Therefore Freon-12 has a higher percentage chlorine. 4.73
%P =
0.976 g P × 100% = 22.9% 4.26 g compound
% Cl = 100% − 22.9% = 77.1% 4.74
%N =
0.896 g N × 100% = 25.9% 3.46 g compound
%O = 100% − 25.9% = 74.1% 4.75
For C17H25N, the molar mass (17C + 25H + 1N) equals 243.43 g/mole, and the three theoretical values for % by weight are calculated as follows:
%C=
204.2 g C × 100% = 83.89% 243.4 g C17 H 25 N
%H=
25.20 g H × 100% = 10.35% 243.4 g C17 H 25 N
%N=
14.01 g N × 100% = 5.76% 243.4 g C17 H 25 N
These data are consistent with the experimental values cited in the problem. 4.76
For C20H25N3O, the molar mass (20C + 25H + 3N + O) equals 323.44 g/mole, and the theoretical values for % by weight are calculated as follows:
57
%C=
240.22 g C × 100% = 74.27% 323.44 g C20 H 25 N3O
%H=
25.20 g H × 100% = 7.791% 323.44 g C20 H 25 N3O
%N=
42.02 g N × 100% = 12.99% 323.44 g C20 H 25 N3O
%O=
16.00 g O × 100% = 4.947% 323.44 g C20 H 25 N3O
(a)
The % by mass oxygen in the suspected sample may be determined by difference: 100% – (74.07 + 7.95 + 9.99)% = 7.99 %.
(b)
These data are not consistent with the theoretical formula for LSD.
4.77
5 mol O 16.0 g O 1 mol N g O = 1.28 × 1022 atoms N = 0.850 g O 23 6.02 × 10 atoms N 2 mol N 1 mol O
4.78
5 mol C 12.01 g C 1 mol H g C = (4.25 × 1023 atoms H) = 3.53 g C 23 6.02 × 10 atoms H 12 mol H 1 mol C
4.79
(
)
The molecular formula is some integer multiple of the empirical formula. This means that we can divide the molecular formula by the largest possible whole number that gives an integer ratio among the atoms in the empirical formula. (a)
SCl
(b)
CH2O
(c)
NH3
(d)
AsO3
(e)
HO
4.80
(a)
CH3O
(b)
HSO4
(c)
C2H5
(d)
BH3
(e)
C2H6O
4.81
We begin by realizing that the mass of oxygen in the compound may be determined by difference: 0.896 g total – (0.111 g Na + 0.477 g Tc) = 0.308 g O. Next we can convert each mass of an element into the corresponding number of moles of that element as follows:
1 mol Na −3 mol Na = ( 0.111 g Na ) = 4.83 × 10 mol Na 23.00 g Na 1 mol Tc −3 mol Tc = ( 0.477 g Tc ) = 4.82 × 10 mol Tc 98.9 g Tc 1 mol O −2 mol O = ( 0.308 g O ) = 1.93 × 10 mol O 16.0 g O Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: for Na, 4.83 × 10–3 moles / 4.82 × 10–3 moles = 1.00 for Tc, 4.82 × 10–3 moles / 4.82 × 10–3 moles = 1.00
58
for O, 1.93 × 10–2 moles / 4.82 × 10–3 moles = 4.00 These relative mole amounts give us the empirical formula: NaTcO4. 4.82
1 mol C mol C = (0.423 g C) = 0.0352 mol C 12.01 g C 1 mol Cl mol Cl = (2.50 g Cl) = 0.0705 mol Cl 35.45 g Cl 1 mol F mol F = (1.34 g F) = 0.0705 mol F 19.00 g F Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: for C, 0.0352 moles / 0.0352 moles = 1.00 for Cl, 0.0705 moles / 0.0352 moles = 2.000 for F, 0.0705 moles / 0.0352 moles = 2.00 These relative mole amounts give us the empirical formula CCl2F2
4.83
To solve this problem we will assume that we have a 100 g sample. This implies we have 14.5 g C,and 85.5 g Cl.:
1 mol C mol C = (14.5 g C) = 1.21 mol C 12.01 g C 1 mol Cl mol Cl = (85.5 g Cl) = 2.41 mol Cl 35.45 g Cl Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: for C, 1.21 moles / 1.21 moles = 1.00 for Cl, 2.41 moles /1.21 moles = 2.000 These relative mole amounts give us the empirical formula CCl2 4.84
To solve this problem we will assume that we have a 100 g sample. This implies that we have 77.26 g Hg, 9.25 g C, 1.17 g H and 12.32 g O. The amount of oxygen was determined by subtracting the total amounts of the other three elements from the total assumed mass of 100 g. Convert each of these masses into a number of moles:
1 mole Hg moles Hg = ( 77.26 g Hg ) = 0.3852 moles Hg 200.59 g Hg
59
1 mole C moles C = ( 9.25 g C ) = 0.770 moles C 12.011 g C 1 mole H moles H = (1.17 g H ) = 1.16 moles H 1.008 g H 1 mole O moles O = (12.32 g O ) = 0.7700 moles O 15.999 g O The relative mole amounts are determined as follows: for Hg, 0.3852 moles / 0.3852 moles = 1.000 for C, 0.770 moles / 0.3852 moles = 2.00 for H, 1.16 moles / 0.3852 moles = 3.01 for O, 0.7700 moles / 0.3852 moles = 1.999 and the empirical formula is HgC2H3O2. The empirical formula weight is 259.6 g/mole, which must be multiplied by 2 in order to obtain the molecular weight. This means that the molecular formula is twice the empirical formula, or Hg2C4H6O4. 4.85
To solve this problem we will assume that we have a 100 g sample. This implies we have 72.96 g C, 5.40 g H, and 21.64 g O. The grams of O are determined knowing that the total mass of all elements in the compound must add to 100 g.:
1 mol C mol C = (72.96 g C) = 6.074 mol C 12.011 g C 1 mol H mol H = (5.40 g H) = 5.36 mol H 1.008 g H To find the number of moles of O, first we have to find the number of grams of O: 100 g total = (72.96 g C) + (5.40 g H) + (x g O) g O = 21.64 g O
1 mol O mol O = (21.64 g O) = 1.353 mol O 15.999 g O Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: for C, 6.074 moles / 1.353 moles = 4.49 for H, 5.36 moles / 1.353 moles = 3.96 for O, 1.353 moles / 1.353 moles = 1.00
60
These relative mole amounts give us the empirical formula C4.5H4O Since we cannot have decimals as subscripts, multiply all of the subscripts by 2 to get the formula: C9H8O2 4.86
Assume a 100 g sample:
1 mol C mol C = (63.2 g C) = 5.26 mol C 12.01 g C 1 mol H mol H = (5.26 g H) = 5.22 mol H 1.008 g H 1 mol O mol O = (31.6 g O) = 1.98 mol O 16.00 g O Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: for C, 5.26 moles / 1.98 moles = 2.66 = (8/3) for H, 5.22 moles / 1.98 moles = 2.64 = (8/3) for O, 1.98 moles / 1.98 moles = 1.00 = (3/3) These relative mole amounts give us the empirical formula C8H8O3 4.87
All of the carbon is converted to carbon dioxide so,
1 mol CO 2 1 mol C 12.01 g C g C = (1.104 g CO2) = 0.3013 g C 44.01 g CO 2 1 mol CO 2 1 mol C 1 mol C –2 mol C = (0.3013 g C) = 2.509 × 10 mol C 12.01 g C All of the hydrogen is converted to H2O, so
1 mol H 2 O 2 mol H 1.008 g H g H = (0.678 g H2O) = 0.0758 g H 18.02 g H 2 O 1 mol H 2 O 1 mol H 1 mol H –2 mol H = (0.0758 g H) = 7.52 × 10 mol H 1.008 g H The amount of O in the compound is determined by subtracting the mass of C and the mass of H from the sample. g O = 0.578 g – 0.3013 g – 0.0758 g = 0.201 g O
1 mol O –2 mol O = (0.201 g O) = 1.26 × 10 mol O 16.00 g O
61
The relative mole ratios are: for C, 0.02509 moles / 0.0126 moles = 1.99 for H, 0.0758 moles/ 0.0126 moles = 6.02 for O, 0.0126 moles / 0.0126 moles = 1.00 The relative mole amounts give the empirical formula C2H6O 4.88
All of the carbon is converted to carbon dioxide so,
1 mol CO 2 1 mol C g C = (2.01 g CO2) 44.01 g CO 2 1 mol CO 2
12.01 g C = 0.549 g C 1 mol C
1 mol C mol C = (0.549 g C) = 0.0457 mol C 12.01 g C All of the hydrogen is converted to H2O, so
1 mol H 2 O 2 mol H 1.008 g H g H = (0.827 g H2O) = 0.0925 g H 18.02 g H 2 O 1 mol H 2 O 1 mol H 1 mol H mol H = (0.0925 g H) = 0.0918 mol H 1.008 g H The amount of O in the compound is determined by subtracting the mass of C and the mass of H from the sample. g O = 0.822 g – 0.549 g – 0.0925 g = 0.181 g
1 mol O mol O = (0.181 g O) = 0.0113 mol O 16.00 g O The relative mole ratios are: for C, 0.0457 moles / 0.0113 moles = 4.04 for H, 0.0918 moles/ 0.0113 moles = 8.12 for O, 0.0113 moles / 0.0113 moles = 1.00 The relative mole amounts give the empirical formula C4H8O 4.89
This type of combustion analysis takes advantage of the fact that the entire amount of carbon in the original sample appears as CO2 among the products. Hence the mass of carbon in the original sample must be equal to the mass of carbon that is found in the CO2.
62
1 mole CO 2 1 mole C g C = (19.73 × 10–3 g CO2) 44.01 g CO2 1 mole CO 2
12.011 g C –3 = 5.385 × 10 g C 1 mole C
Similarly, the entire mass of hydrogen that was present in the original sample ends up in the products as H2O:
1 mole H 2 O 2 mole H 1.008 g H g H = (6.391 × 10–3 g H2O) = 7.150 × 10–4 g H 18.02 g H 2 O 1 mole H 2 O 1 mole H The mass of oxygen is determined by subtracting the mass due to C and H from the total mass: 6.853 mg total – (5.385 mg C + 0.7150 mg H) = 0.753 mg O. Now, convert these masses to a number of moles:
1 mol C –4 mol C = (5.385 × 10–3 g C) = 4.484 × 10 mol C 12.011 g C 1 mol H –4 mol H = (7.150 × 10–4 g H) = 7.094 × 10 mol H 1.0079 g H 1 mol O –5 mol O = (7.53 × 10–4 g O) = 4.71 × 10 mol H 15.999 g O The relative mole amounts are: for C, 4.483 × 10–4 mol / 4.71 × 10–5 mol = 9.52 for H, 7.094 × 10–4 mol / 4.71 × 10–5 mol = 15.1 for O, 4.71 × 10–5 mol / 4.71 × 10–5 mol = 1.00 The relative mole amounts are not whole numbers as we would like. However, we see that if we double the relative number of moles of each compound, there are approximately 19 moles of C, 30 moles of H and 2 moles of O. If we assume these numbers are correct, the empirical formula is C19H30O2, for which the formula weight is 290 g/mole. In most problems where we attempt to determine an empirical formula, the relative mole amounts should work out to give a “nice” set of values for the formula. Rarely will a problem be designed that gives very odd coefficients. With experience and practice, you will recognize when a set of values is reasonable. 4.90
This type of combustion analysis takes advantage of the fact that the entire amount of carbon in the original sample appears as CO2 among the products. Hence the mass of carbon in the original sample must be equal to the mass of carbon that is found in the CO2.
1 mole CO 2 1 mole C 12.0107 g C –3 g C = (23.085 × 10–3 g CO2) = 6.3002 × 10 g C 44.0095 g CO 1 mole CO 1 mole C 2 2
63
Similarly, the entire mass of hydrogen that was present in the original sample ends up in the products as H2O:
1 mole H 2 O 2 mole H 1.0079 g H –4 g H = (7.764 × 10–3 g H2O) = 8.688 × 10 g H 18.015 g H O 1 mole H O 1 mole H 2 2 The mass of oxygen is determined by subtracting the mass due to C and H from the total mass: 7.468 mg total – (6.3002 mg C + 0.8688 mg H) = 0.299 mg O. Now, convert these masses to a number of moles:
1 mol C –4 mol C = (6.3002 × 10–3 g C) = 5.2455 × 10 mol C 12.0107 g C 1 mol H –4 mol H = (8.688 × 10–4 g H) = 8.620 × 10 mol H 1.0079 g H 1 mol O –5 mol O = (2.99 × 10–4 g O) = 1.87 × 10 mol H 15.999 g O The relative mole amounts are: for C, 5.2455 × 10–4 mol / 1.87 × 10–5 mol = 28.1 for H, 8.620 × 10–4 mol / 1.87 × 10–5 mol = 46.1 for O, 1.87 × 10–5 mol / 1.87 × 10–5 mol = 1.00 and the empirical formula is C28H46O. 4.91
(a)
Formula mass = 135.1 g
270.4 g/mol = 2.001 135.1 g/mol The molecular formula is Na2S4O6 (b)
Formula mass = 73.50 g
147.0 g/mol = 2.000 73.50 g/mol The molecular formula is C6H4Cl2 (c)
Formula mass = 60.48 g
181.4 g/mol = 2.999 60.48 g/mol The molecular formula is C6H3Cl3
64
4.92
(a)
Formula mass = 122.1 g
732.6 g/mol = 6.000 122.1 g/mol The molecular formula is Na12Si6O18 (b)
Formula mass = 102.0 g
305.9 g/mol = 2.999 102.0 g/mol The molecular formula is Na3P3O9 (c)
Formula mass = 31.03 g
62.1 g/mol = 2.00 31.03 g/mol The molecular formula is C2H6O2
4.93
The formula mass for the compound C19H30O2 is 290 g/mol. Thus, the empirical and molecular formulas are equivalent.
4.94
The formula mass for the compound C28H46O is 399 g/mol. Thus, the empirical and molecular formulas are equivalent.
4.95
From the information provided, we can determine the mass of mercury as the difference between the total mass and the mass of bromine: g Hg = 0.595 g compound – 0.170 g Br = 0.425 g Hg To determine the empirical formula, first convert the two masses to a number of moles.
1 mole Hg –3 mol Hg = (0.425 g Hg) = 2.12 × 10 mol Hg 200.59 g Hg 1 mole Br –3 mol Br = (0.170 g Br) = 2.13 × 10 mol Br 79.904 g Br Now, we would divide each of these values by the smaller quantity to determine the simplest mole ratio between the two elements. By inspection, though, we can see there are the same number of moles of Hg and Br. Consequently, the simplest mole ratio is 1:1 and the empirical formula is HgBr.
65
To determine the molecular formula, recall that the ratio of the molecular mass to the empirical mass is equivalent to the ratio of the molecular formula to the empirical formula. Thus, we need to calculate an empirical mass: (1 mole Hg)(200.59 g Hg/mole Hg) + (1 mole Br)(79.904 g Br/mole Br) = 280.49 g/mole HgBr. The molecular mass, as reported in the problem is 561 g/mole. The ratio of these is:
561 g/mole = 2.00 280.49 g/mole So, the molecular formula is two times the empirical formula or Hg2Br2. 4.96
From the information provided, the mass of sulfur is the difference between the total mass and the mass of antimony:
g S = 0.6662 g compound – 0.4017 g Sb = 0.2645 g S
To determine the empirical formula, first convert the two masses to a number of moles.
1 mole S –3 mol S = (0.2645 g S) = 8.249 × 10 mol S 32.065 g S 1 mole Sb –3 mol Sb = (0.4017 g Sb) = 3.299 × 10 mol Sb 121.76 g Sb
Now, divide each of these values by the smaller quantity to determine the simplest mole ratio between the two elements:
For Sb: 3.299 × 10–3 moles/3.299 × 10–3 moles = 1.000 mol Sb For S: 8.249 × 10–3 moles/3.299 × 10–3 moles = 2.500 mol S
Hence the empirical formula is Sb2S5, and the empirical mass is (2 × Sb) + (5 × S) = 403.85 g/mol. Since the molecular mass reported in the problem is the same as the calculated empirical mass, the empirical formula is the same as the molecular formula.
4.97
First, determine the amount of oxygen in the sample by subtracting the masses of the other elements from the total mass: 0.8306 g – (0.2318 g C + 0.01944 g H + 0.2705 g N) = 0.3089 g O. Now, convert these masses into a number of moles for each element:
66
1 mole C –2 mol C = (0.2318 g C) = 1.930 × 10 mol C 12.011 g C 1 mole H –2 mol H = (0.01944 g H) = 1.929 × 10 mol H 1.0079 g H 1 mole N –2 mol N = (0.2705 g N) = 1.931 × 10 mol N 14.007 g N 1 mole O –2 mol O = (0.3089 g O) = 1.931 × 10 mol O 15.999 g O These are clearly all the same mole amounts, and we deduce that the empirical formula is CHNO, which has a formula weight of 43. It can be seen that the number 43 must be multiplied by the integer 3 in order to obtain the molar mass (3 × 43 = 129), and this means that the empirical formula should similarly be multiplied by 3 in order to arrive at the molecular formula, C3H3N3O3. 4.98
To solve this problem we will assume that we have a 100 g sample. This implies that we have 75.42 g C, 6.63 g H, 8.38 g N and 9.57 g O. The amount of oxygen was determined by subtracting the total amounts of the other three elements from the total assumed mass of 100 g. Convert each of these masses into a number of moles:
1 mole C mol C = (75.42 g C) = 6.279 mol C 12.011 g C 1 mole H mol H = (6.63 g H) = 6.58 mol H 1.008 g H 1 mole N mol N = (8.38 g N) = 0.598 mol N 14.01 g N 1 mole O mol O = (9.57 g O) = 0.598 mol O 16.00 g O The relative mole amounts are determined as follows: for C, 6.279 mol / 0.598 mol = 10.5 for H, 6.58 mol / 0.598 mol = 11.0 for N, 0.598 mol / 0.598 mol = 1.00 for O, 0.598 mol / 0.598 mol = 1.00
In order to obtain whole numbers, each of these values is multiplied by 2 and we determine the empirical formula is C21H22N2O2. The empirical formula weight is 334 g/mole. This means that the molecular formula is the same as the empirical formula, or C21H22N2O2.
67
4.99
2 Ba(OH)2•8H2O contains: (a) 2 atoms of Ba, 20 atoms of O, and 36 atoms of H
(b) 2 moles of Ba, 20 moles of O, and 36 moles of H
4.100
3Ca3(PO4)2 contains: 9 atoms of Ca, 6 atoms of P, and 24 atoms of O. There are 9 moles of Ca, 6 moles of P, and 24 moles of O present.
4.101
4Fe(s) + 3O2(g) 2Fe2O3(s)
4.102
2NO(g) + O2(g) 2NO2(g)
4.103
(a)
Ca(OH)2 + 2HCl CaCl2 + 2H2O
(b)
2AgNO3 + CaCl2 Ca(NO3)2 + 2AgCl
(c)
Pb(NO3)2 + Na2SO4 PbSO4 + 2NaNO3
(d)
2Fe2O3 + 3C 4Fe + 3CO2
(e)
2C4H10 + 13O2 8CO2 + 10H2O
(a)
2SO2 + O2 2SO3
(b)
2NaHCO3 + H2SO4 Na2SO4 + 2H2O + 2CO2
(c)
P4O10 + 6H2O 4H3PO4
(d)
Fe2O3 + 3H2 2Fe + 3H2O
(e)
2Al + 3H2SO4 Al2(SO4)3 + 3H2
(a)
Mg(OH)2 + 2HBr MgBr2 + 2H2O
(b)
2HCl + Ca(OH)2 CaCl2 + 2H2O
(c)
Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O
(d)
2KHCO3 + H3PO4 K2HPO4 + 2H2O + 2CO2
(e)
C9H20 + 14O2 9CO2 + 10H2O
(a)
CaO + 2HNO3 → Ca(NO3)2 + H2O
(b)
Na2CO3 + Mg(NO3)2 MgCO3 + 2NaNO3
(c)
(NH4)3PO4 + 3NaOH Na3PO4 + 3NH3 + 3H2O
(d)
2LiHCO3 + H2SO4 Li2SO4 + 2H2O + 2CO2
4.104
4.105
4.106
68
(e)
C4H10O + 6O2 4CO2 + 5H2O
4.107
4NH2CHO + 5O2
→
4CO2 + 6H2O + 2N2
4.108
4NH2CHO + 5O2
→
4CO2 + 6H2O + 2N2
4.109
(a)
1 mole Na 2S2 O3 mol Na2S2O3 = (0.24 mol Cl2) = 0.060 mol Na2S2O3 4 mole Cl2
(b)
8 mole HCl mol HCl = (0.24 mol Cl2) = 0.48 mol HCl 4 mole Cl2
(c)
5 mole H 2 O mol H2O = (0.24 mol Cl2) = 0.30 mol H2O 4 mole Cl2
(d)
5 mole H 2 O mol H2O = (0.48 mol HCl) = 0.30 mol H2O 8 mole HCl
(a)
25 mole O2 mol O2 = (6.84 mol C8H18) = 85.5 mol O2 2 mole C8 H18
(b)
16 mole CO 2 mol CO2 = (0.511 mol C8H18) = 4.09 mol CO2 2 mole C8 H18
(c)
18 mole H 2 O mol H2O = (8.20 mol C8H18) 2 mole C8 H18
(d)
25 mole O 2 mol O2 = ( 6.00 mol CO2) = 9.38 mol O2 16 mole CO2
4.110
= 73.8 mol H2O
2 mole C8 H18 mol C8H18 = (6.00 mol CO2) = 0.750 mol C8H18 16 mole CO 2
4.111
(a)
1 mol Zn 0.23 mol Au(CN)2– − 2 mol Au(CN)2
–
2 mol Au
(b)
0.23 mol Au(CN)2 − 2 mol Au(CN)2
(c)
2 mol Au(CN) − 2 0.23 mol Zn 1 mol Zn
65.39 g Zn 1 mol Zn = 7.5 g Zn 197.0 g Au 1 mol Au = 45 g Au
249.0 g Au(CN) − 2 1 mol Au(CN) − 2
69
= 110 g Au(CN)2–
4.112
4.113
4.114
(a)
5 mol O 2 32.00 g O 2 3.45 mol C3H8 = 552 g O2 1 mol C3 H8 1 mol O 2
(b)
3 mol CO 2 44.01 g CO 2 0.177 mol C3H8 = 23 g CO2 1 mol C3 H8 1 mol CO 2
(c)
4 mol H 2 O 18.01 g H 2 O 4.86 mol C3H8 = 350 g H2O 1 mol C3 H8 1 mol H 2 O
(a)
4P + 5O2 P4O10
(b)
1 mol P 5 mol O 2 32.0 g O2 g O2 = (6.85 g P) = 8.85 g O2 30.97 g P 4 mol P 1 mol O 2
(c)
1 mol O2 1 mol P4 O10 283.9 g P4 O10 g P4O10 = (8.00 g O2) = 14.2 g P4O10 32.00 g O2 5 mol O 2 1 mol P4 O10
(d)
1 mol P4 O10 4 mol P g P = (7.46 g P4O10) 283.9 g P4 O10 1 mol P4 O10
(a)
2C4H10 + 13O2 8CO2 + 10H2O
(b)
1 mol H 2 O 2 mol C4 H10 58.12 g C4 H10 g C4H10 = (5.32 g H2O) = 3.43 g C4H10 18.02 g H 2 O 10 mol H 2 O 1 mol C4 H10
(c)
1 mol C4 H10 13 mol O2 32.0 g O 2 g O2 = (3.43 g C4H10) = 12.3 g O2 58.12 g C4 H10 2 mol C4 H10 1 mol O 2
(d)
8 mol CO 2 44.01 g CO 2 g CO2 = (0.0590 mol C4H10) = 10.4 g CO2 2 mol C4 H10 1 mol CO 2
30.97 g P = 3.26 g P 1 mol P
4.115
1 mol Cu 8 mol HNO3 63.013 g HNO3 g HNO3 = (9.705 g Cu) = 25.66 g HNO3 63.546 g Cu 3 mol Cu 1 mol HNO3
4.116
1 mol N 2 H 4 7 mol H 2 O 2 34.02 g H 2 O 2 g H2O2 = (852 g N2H4) = 6330 g H2O2 32.05 g N 2 H 4 1 mol N 2 H 4 1 mol H 2 O 2
4.117
1000 g H 2 O 2 1 mol H 2 O 2 1 mol O2 32.00 g O 2 1 kg O2 kg O2 = 1.0 kg H2O2 1 kg H 2 O 2 34.01 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 1000 g O 2 = 0.47 kg O2
70
4.118
1000 g KClO3 1 mol KClO3 3 mol O2 32.00 g O 2 1 kg O 2 kg O2 = 1.5 kg KClO2 1 kg KClO3 122.6 g KClO3 2 mol KClO3 1 mol O 2 1000 g O 2 = 0.59 kg O2
4.119
The picture shows 9 molecules of O2 and 3 molecules of C2H6S. The balanced reaction shows that 2 molecules of C2H6S react with 9 moles of O2. Therefore, O2 is the limiting reagent. Molecules of SO2 = (9 molecules O2)(2 molecules SO2/9 molecules O2) = 2 molecules of SO2
4.120
The product mixture consists of 2 SO2, 6 H2O, 4 CO2, and 1 O2 molecules. Since you have an unreacted oxygen molecule in the mixture, the reaction mixture had an excess of O2 and C2H6S was the limiting reagent.
4.121
(a)
First determine the amount of Fe2O3 that would be required to react completely with the given amount of Al:
1 mol Fe2 O3 mol Fe2O3 = (4.95 mol Al) = 2.48 mol Fe2O3 2 mol Al Since only 2.35 mol of Fe2O3 are supplied, it is the limiting reactant. This can be confirmed by calculating the amount of Al that would be required to react completely with all of the available Fe2O3: 2 mol Al mol Al = (2.35 mol Fe2O3) = 4.70 mol Al 1 mol Fe2 O3 Since an excess (4.95 mol – 4.70 mol = 0.25 mol) of Al is present, Fe2O3 must be the limiting reactant, as determined above. (b) 4.122
(a)
2 mol Fe 55.847 g Fe g Fe = (2.35 mol Fe2O3) = 262 g Fe 1 mol Fe2 O3 1 mol Fe First determine the amount of H2O that would be required to react completely with the given amount of C2H4: 1000 g C2 H 4 1 mol C2 H 4 1 mol H 2 O kg H2O = 1.62 kg C2H4 × 1 kg C H 28.05 g C H 1 mol C H 2 4 2 4 2 4 18.02 g H 2 O 1 kg H 2 O = 1.04 kg H2O 1 mol H 2 O 1000 g H 2 O Since only 0.0148 kg of H2O are supplied, it is the limiting reactant. This can be confirmed by calculating the amount of C2H4 that would be required to react completely with all of the available H2O:
1000 g H 2 O 1 mol H 2 O 1 mol C2 H 4 g C2H4 = 0.0148 kg H2O 1 kg H 2 O 18.02 g H 2 O 1 mol H 2 O
71
28.05 g C2 H 4 1 kg C2 H 4 = 0.024 kg C2H4 1 mol C2 H 4 1000 g C2 H 4 Thus, water is the limiting reagent/ The mass of ethanol produced from 0.0148 kg of water is: 1000 g H 2 O 1 mol H 2 O g C2H5OH = (0.0148kg H2O) 1 kg H 2 O 18.02 g H 2 O 1 mol C2 H5 OH 46.08 g C2 H5 OH × = 37.8 g C2H5OH 1 mol H 2 O 1 mol C2 H5 OH
4.123
3AgNO3 + FeCl3 → 3AgCl + Fe(NO3)3 Calculate the amount of FeCl3 that are required to react completely with all of the available silver nitrate:
1 mol AgNO3 1 mol FeCl3 162.21 g FeCl3 g FeCl3 = (18.0 g AgNO3) 169.87 g AgNO3 3 mol AgNO3 1 mol FeCl3 = 5.73 g FeCl3 Since more than this minimum amount is available, FeCl3 is present in excess, and AgNO3 must be the limiting reactant. We know that only 5.73 g FeCl3 will be used. Therefore, the amount left unused is: 32.4 g total – 5.73 g used = 26.7 g FeCl3 Notice that we decided in the beginning to calculate the amount of FeCl3 required to react with AgNO3. Instead, we could have calculated the grams of AgNO3 required to react with FeCl3. In that case we would have seen that AgNO3 was in excess.
4.124
First, calculate the amount of H2O needed to completely react with the available ClO2;
1 mol ClO 2 3 mol H 2 O 18.02 g H 2 O g H2O = 168.0 g ClO2 = 22.44 g H2O 67.45 g ClO 2 6 mol ClO2 1 mol H 2 O
So, there is excess H2O present. The amount that remains is 39.7 g – 22.44 g = 17.3 g H2O. 4.125
First calculate the number of moles of water that are needed to react completely with the given amount of NO2:
1 mol NO2 1 mol H 2 O 18.02 g H 2 O –4 g H2O = 0.0010 g NO2 = 1.3 × 10 g H2O 46.01 g NO 3 mol NO 1 mol H O 2 2 2 Since this is less than the amount of water that is supplied, the limiting reactant must be NO2. Therefore, to calculate the amount of HNO3:
72
1 mol NO 2 2 mol HNO3 63.02 g HNO3 g HNO3 = 0.0010 g NO2 = 0.913 mg HNO3 46.01 g NO 2 3 mol NO 2 1 mol HNO3 4.126
(a)
First calculate the number of moles of water that are needed to react completely with the given amount of PCl5:
4 mole H 2 O mol H2O = (0.450 mol PCl5) = 1.80 mol H2O 1 moles PCl5 Since this is less than the amount of water that is supplied, the limiting reactant must be PCl5. This can be confirmed by the following calculation: 1 mol PCl5 mol PCl5 = (3.80 mol H2O) = 0.950 mol PCl5 4 mol H 2 O which also demonstrates that the limiting reactant is PCl5. 5 mol HCl g HCl = (0.450 mol PCl5) 1 mol PCl5 First determine the theoretical yield:
(b) 4.127
36.46 g HCl = 82.0 g HCl 1 mol HCl
1 mol Ba(NO3 ) 2 1 mol BaSO 4 233.39 g BaSO 4 g BaSO4 = (79.25 g Ba(NO3)2 261.34 g Ba(NO3 )2 1 mol Ba(NO3 ) 2 1 mol BaSO 4 = 70.77 g BaSO4 Then calculate a % yield: actual yield 63.78 g × 100 = × 100 = 90.12% % yield = theoretical yield 70.77 g 4.128
The theoretical yield is
1 mole NaCl 1 mol NaHCO3 1 mol Na 2 CO3 105.99 g Na 2 CO3 g Na2CO3 = 120 g NaCl 58.44 g NaCl 1 mol NaCl 2 mol NaHCO3 1 mol Na 2 CO3 = 108.8 g Na2CO3 actual yield 85.4 g % yield = × 100 = × 100 = 78.5% theoretical yield 108.8 g 4.129
First, determine how much H2SO4 is needed to completely react with the AlCl3
1 mol AlCl3 3 mol H 2SO 4 g H2SO4 = 25.0 g AlCl3 133.34 g AlCl3 2 mol AlCl3 = 27.58 g H2SO4 There is an excess of H2SO4 present. Determine the theoretical yield:
73
98.08 g H 2SO 4 1 mol H 2SO 4
1 mol AlCl3 1 mol Al2 (SO 4 )3 342.17 g Al2 (SO 4 )3 g Al2(SO4)3 = 25.0 g AlCl3 133.33 g AlCl3 2 mol AlCl3 1 mol Al2 (SO4 )3 = 32.08 g Al2(SO4)3
% yield = 4.130
actual yield 28.46 g × 100 = × 100 = 88.72% theoretical yield 32.08 g
Assume there is excess oxygen present and determine the theoretical yield of carbon dioxide.
1 mol CH3OH 2 mol CO 2 44.01 g CO2 g CO 2 = ( 7.35 g CH3OH ) = 10.1 g CO 2 32.04 g CH3OH 2 mol CH3OH 1 mol CO 2 % yield =
4.131
8.46 g × 100% = 83.8% 10.1g
If the yield for this reaction is only 71% and we need to have 12.8 g of product, we will attempt to make 18.0 g of product. This is determined by dividing the actual yield by the percent yield. Recall actual yield × 100 . If we rearrange this equation we can see that; % yield = theoretical yield that theoretical yield =
actual yield × 100 . Substituting the values from this problem gives the 18.0 g of % yield
product mentioned above.
4.132
1 mol KC7 H5 O 2 1 mol C7 H8 92.14 g C7 H8 g C7 H8 = 18.0 g KC7 H5O 2 = 10.4 g C7 H8 160.21 g KC7 H5 O2 1 mol KC7 H5 O2 1 mol C7 H8 First, determine how much MnI2 is needed to completely react with the F2 1 mol F2 2 mol MnI 2 308.75 g MnI2 g MnI2 = 15.0 g F2 = 18.8 g MnI2 38.00 g F2 13 mol F2 1 mol MnI2
There is an excess of F2 present. Note that MnI2 is the limiting reactant and that MnI2 and MnF3 are in a 1:1 ratio, so the number of mole of MnI2 equals the number of moles of MnF3. According to the problem statement, we will only prepare 56% of this number of moles of MnF3.
1 mol MnI2 2 mol MnF3 111.93 g MnF3 g MnF3 = 15.0 g MnI2 ( 0.56 ) = 3.05 g MnF3 308.75 g MnI 2 2 mol MnI 2 1 mol MnF3 Additional Problems 4.133
2000 lb Hg 1 kg Hg 1000 g Hg 1 mol Hg lb (CH3)2Hg = (263 tons Hg)(0.010) 1 ton Hg 2.205 lb Hg 1 kg Hg 200.59 g Hg
74
1 mol (CH3 )2 Hg 230.66 g (CH3 ) 2 Hg 6 = 2.7 × 10 g (CH3)2Hg 1 mol Hg 1 mol (CH ) Hg 3 2 1k g (CH3 )2 Hg 2.205 lb (CH3 )2 Hg 2.7 × 106 g (CH3)2Hg = 5950 lb (CH3)2Hg 1000 g (CH3 ) 2 Hg 1 kg (CH3 )2 Hg 4.134
Only 27% of the paint is left in paint chip after 73% has evaporated. The mass of the wet paint is:
0.23 g = 0.85 g 0.27 g PbCr2O7 = 0.85 g sample × 14.5% PbCr2O7 = 0.124 g PbCr2O7
1 mol PbCr2 O7 1 mol Pb 207.2 g Pb g Pb = 0.124 g PbCr2O7 = 0.060 g Pb 423.2 g PbCr O 1 mol PbCr O 2 7 2 7 1 mol Pb
4.135
Assume 100 g of magnesium boron compound, therefore there are 52.9 g of Mg and 47.1 g of B.
1 mol Mg mol of Mg = 52.9 g Mg = 2.18 mol Mg 24.305 g Mg 1 mol B mol of B = 47.1g B = 4.36 mol Mg 10.811 g B Divide the number of moles of each element by the least number of moles:
2.18 mol Mg =1 2.18 mol Mg 4.26 mol B =2 2.18 mol Mg Formula is MgB2.
4.136
First, we determine the number of grams of chlorine in the original sample:
1 mol AgCl 1 mol Cl 35.453 g Cl g Cl = (0.4857 g AgCl) = 0.1201 g Cl 143.32 g AgCl 1 mol AgCl 1 mol Cl
The mass of Cr in the original sample is thus 0.1789 – 0.1201 g = 0.0588 g Cr. Converting to moles, we have:
75
1 mol Cl –3 for Cl: 0.1201 g = 3.388 × 10 mol Cl 35.453 g Cl 1 mol Cr –3 for Cr: 0.0588 g Cr = 1.13 × 10 mol Cr 51.996 g Cr The relative mole amounts are: for Cl: 3.388 × 10–3 mol / 1.13 × 10–3 mol = 3.00 for Cr: 1.13 × 10–3 mol / 1.13 × 10–3 mol = 1.00 The empirical formula is thus CrCl3.
4.137
First determine the percentage by weight of each element in the respective original samples. This is done by determining the mass of the element in question present in each of the original samples. The percentage by weight of each element in the unknown will be the same as the values we calculate.
1 mol CaCO3 1 mol Ca 40.1 g Ca g Ca = (0.240 g CaCO3) = 0.0962 g Ca 100.09 g CaCO3 1 mol CaCO3 1 mol Ca
% Ca = (0.0962/0.375) × 100% = 25.7% Ca
1 mol BaSO 4 1 mol S 32.07 g S g S = (0.374 g BaSO4) = 0.0513 g S 233.8 g BaSO 4 1 mol BaSO 4 1 mol S
% S = (0.0513/0.125) × 100% = 41.0% S
1 mol NH3 1 mol N 14.01 g N g N = (0.206 g NH3) = 0.169 g N 17.03 g NH3 1 mol NH3 1 mol N % N = (0.169/0.946) × 100% = 17.9% N % C = 100.0 – (25.7 + 41.0 + 17.9) = 15.4% C. Next, we assume 100 g of the compound, and convert these weight percentages into mole amounts:
76
1 mol Ca mol Ca = ( 25.7 g Ca ) = 0.641 mol Ca 40.08 g Ca 1 mol S mol S = ( 41.0 g S ) = 1.28 mol S 32.07 g S 1 mol N mol N = (17.9 g N ) = 1.27 mol N 14.07 g N 1 mole C moles C = (15.4 g C ) = 1.28 moles C 12.01 g C Dividing each of these mole amounts by the smallest, we have: For Ca: 0.641 mol / 0.641 mol = 1.00 For S: 1.28 mol / 0.641 mol = 2.00 For N: 1.27 mol / 0.641 mol = 1.98 For C: 1.28 mol / 0.641 mol = 2.00 The empirical formula is therefore CaC2S2N2, and the mass of the empirical unit is Ca + 2S + 2N + 2C = 156 g/mol. Since the molecular mass is the same as the empirical mass, the molecular formula is CaC2S2N2. 4.138
(a)
One mole of N2, 2 moles of H2O and 1/2 mole of O2 for a total of 3 1/2 moles of gases.
(b)
2000 lb 453.59 g mol of gases = (1.00 ton NH4NO3) 1 ton 1 lb 1 mol NH 4 NO3 3.5 mol gas 4 × = 3.97 × 10 mol gas 80.04 g NH 4 NO3 1 mol NH 4 NO3
4.139
4.140
6.00 g N 1 mol N 1 mol (NH 2 )2 CO 60.06 g (NH 2 )2 CO g (NH2)2CO = 150g fertilizer 1 mol (NH ) CO 2 mol N 100 g Fertilizer 14.007 g N 2 2 = 19.4 g (NH2)2CO
Calculate the cost of one mole nitrogen from each compound: (a)
(b)
1 kg NH 4 NO3 80.04 g NH 4 NO3 1 mol NH 4 NO3 $625 $ per mol N = 2 mol N 25 kg NH 4 NO3 1000 g NH 4 NO3 1 mol NH 4 NO3 = $1.00 per mol N $55 $ per mol N = 1 kg ( NH 4 ) HPO 4 2 1 mol ( NH 4 ) HPO4 2 2 mol N
1 kg ( NH 4 ) HPO 4 2 1000 g ( NH 4 ) HPO 4 2
= $3.66 per mol N
77
132.1 g ( NH 4 ) HPO 4 2 1 mol ( NH 4 ) HPO 4 2
(c)
(d)
1 kg CH 4 ON 2 60.06 g CH 4 ON 2 $60 $ per mol N = 5 kg CH 4 ON 2 1000 g CH 4 ON 2 1 mol CH 4 ON 2 = $0.36 per mol N
1 mol CH 4 ON 2 2 mol N
$128 1 kg NH3 17.03 g NH3 1 mol NH3 $ per mol N = 50 kg NH3 1000 g NH3 1 mol NH3 1 mol N = $0.04 per mol N
NH3 is the cheapest and could be the most economical. 4.141
4.142
1 mol C6 H 6 6 mol C g Na2C2O4 = (155 g C6H6) 78.11 g C6 H 6 1 mol C6 H 6 = 798 g NaC2O4
1 mol Na 2 C2 O 4 2 mol C
134.00 g Na 2 C2 O 4 1 mol Na C O 2 2 4
Assume the hydrogen is the limiting reactant.
1 lb O 2 453.59237 g 1 mol H 2 1 mol O 2 31.9988 g O 2 lb O2 = 227,641 lb H2 1 lb 2.01588 g H 2 2 mol H 2 1 mol O 2 453.59237 g O2 =1,806,714 lb O2 Since this is more than the amount of O2 that is supplied, the limiting reactant must be O2. Next calculate the amount of H2 needed to react completely with all of the available O2.
1 lb H 2 453.59237 g 1 mol O 2 2 mol H 2 2.01588 g H 2 lb H2 = 1,361,936 lb O2 1 lb 31.9988 g O 2 1 mol O 2 1 mol H 2 453.59237 g H 2 =171,600 lb H2
Since only 171,600 lb. of H2 reacted, there are 227,641 lb. – 171,600 lb. = 56,041 lb. of unreacted H2.
4.143
Since 6.00 g represents 86.0% of the required amount, we can solve for the amount that should be made: 6.00 g = 86.0 % of X; X = 6.98 g Pb(NO3)2.
1 mol Pb(NO3 )2 1 mol PbO g PbO = (6.98 g Pb(NO3)2) 331.21 g Pb(NO3 ) 2 1 mol Pb(NO3 )2
223.2 g PbO = 4.70 g PbO 1 mol PbO
4.144
1 mol Cl 1 mol F 18.998 g F –9 g F = (2.2 × 10–9 g Cl) = 1.2 × 10 g F 35.453 g Cl 1 mol Cl 1 mol F
4.145
Overall percentage yield = (0.835)(0.714) × 100% = 59.6%
78
Chapter 5
Practice Exercises 5.1
5.2
5.3
(a)
FeCl3(s) Fe3+(aq) + 3Cl–(aq)
(b)
K3PO4(s) 3K+(aq) + PO43–(aq)
(a)
MgCl2(s) Mg2+(aq) + 2Cl–(aq)
(b)
Al(NO3)3(s) Al3+(aq) + 3NO3–(aq)
(c)
Na2CO3(s) 2Na+(aq) + CO32–(aq)
molecular: (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) ionic: 2NH4+(aq) + SO42–(aq) + Ba2+(aq) + 2NO3–(aq) BaSO4(s) + 2NH4+(aq) + 2NO3–(aq) net ionic: Ba2+(aq) + SO42–(aq) BaSO4(s)
5.4
molecular: CdCl2(aq) + Na2S(aq) CdS(s) + 2NaCl(aq) ionic: Cd2+(aq) + 2Cl–(aq) + 2Na+(aq) + S2–(aq) CdS(s) + 2Na+(aq) + 2Cl–(aq) net ionic: Cd2+(aq) + S2–(aq) CdS(s)
5.5
HCHO2(aq) + H2O H3O+(aq) + CHO2–(aq)
5.6
Note that the drawing below is an anion. The negative charge has not been included in the drawing.
CH3CH2CH2CHOOH(l ) + H2O
5.7
→ CH3CH2CH2COO-(aq) + H3O+(aq)
H3C6H5O7(s) + H2O H3O+(aq) + H2C6H5O7–(aq) H2C6H5O7–(aq) + H2O H3O+(aq) + HC6H5O72–(aq) HC6H5O72–(aq) + H2O H3O+(aq) + C6H5O73–(aq)
5.8
(C2H5)3N(aq) + H2O (C2H5)3NH+(aq) + OH–(aq)
79
Chapter 5
5.9
HONH2(aq) + H2O HONH3+(aq) + OH–(aq)
5.10
The drawing below is the cation formed and does not include the positive charge of the ion.
CH3CH2NH2(aq) + H2O i CH3CH2NH3+(aq) + OH–(aq)
5.11
CH3NH2(aq) + H2O CH3NH3+(aq) + OH–(aq)
5.12
HNO2(aq) + H2O H3O+(aq) + NO2–(aq)
5.13
Sodium arsenate
5.14
Calcium formate, calcium methanoate
5.15
HF: Hydrofluoric acid, sodium salt = sodium fluoride (NaF) HBr: Hydrobromic acid, sodium salt = sodium bromide (NaBr)
5.16
Iodic acid
5.17
NaHSO3, sodium hydrogen sulfite
5.18
H3PO4(aq) + NaOH(aq) NaH2PO4(aq) + H2O
sodium dihydrogen phosphate
NaH2PO4(aq) + NaOH(aq) Na2HPO4(aq) + H2O
sodium hydrogen phosphate
Na2HPO4(aq) + NaOH(aq) Na3PO4(aq) + H2O
sodium phosphate
80
Chapter 5
5.19
molecular: Zn(NO3)2(aq) + Ca(C2H3O2)2(aq) Zn(C2H3O2)2(aq) + Ca(NO3)2(aq) ionic: Zn2+(aq) + 2NO3–(aq) + Ca2+(aq) + 2C2H3O2–(aq) Zn2+(aq) + 2C2H3O2–(aq) + Ca2+(aq) + 2NO3–(aq) net ionic: No reaction
5.20
(a)
molecular: AgNO3(aq) + NH4Cl(aq) AgCl(s) + NH4NO3(aq) ionic: Ag+(aq) + NO3–(aq) + NH4+(aq) + Cl–(aq) AgCl(s) + NH4+(aq) + NO3–(aq) net ionic: Ag+(aq) + Cl–(aq) AgCl(s)
(b)
molecular: Na2S(aq) + Pb(C2H3O2)2(aq) 2NaC2H3O2(aq) + PbS(s) ionic: 2Na+(aq) + S2–(aq) + Pb2+(aq) + 2C2H3O2–(aq) 2Na+(aq) + 2C2H3O2–(aq) + PbS(s) net ionic: S2–(aq) + Pb2+(aq) PbS(s)
5.21
molecular: 2HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2H2O ionic: 2H+(aq) + 2NO3–(aq) + Ca2+(aq) + 2OH–(aq) Ca2+(aq) + 2NO3–(aq) + 2H2O net ionic: H+(aq) + OH–(aq) H2O
5.22
(a)
molecular: HCl(aq) + KOH(aq) H2O + KCl(aq) ionic: H+(aq) + Cl–(aq) + K+(aq) + OH–(aq) H2O + K+(aq) + Cl–(aq) net ionic: H+(aq) + OH–(aq) H2O
(b)
molecular: HCHO2(aq) + LiOH(aq) H2O + LiCHO2(aq) ionic: HCHO2(aq) + Li+(aq) + OH–(aq) H2O + Li+(aq) + CHO2–(aq) net ionic: HCHO2(aq) + OH–(aq) H2O + CHO2–(aq)
(c)
molecular: N2H4(aq) + HCl(aq) N2H5Cl(aq) ionic: N2H4(aq) + H+(aq) + Cl–(aq) N2H5+(aq) + Cl–(aq) net ionic: N2H4(aq) + H+(aq) N2H5+(aq)
5.23
molecular: CH3NH2(aq) + HCHO2(aq) CH3NH3CHO2(aq) ionic: CH3NH2(aq) + HCHO2(aq) CH3NH3+(aq) + CHO2–(aq) net ionic: CH3NH2(aq) + HCHO2(aq) CH3NH3+(aq) + CHO2–(aq)
5.24
molecular: 2HCHO2(aq) + Co(OH)2(s) Co(CHO2)2(aq) + 2H2O ionic: 2HCHO2(aq) + Co(OH)2(s) 2CHO2–(aq) + Co2+(aq) + 2H2O
81
Chapter 5
net ionic: 2HCHO2(aq) + Co(OH)2(s) 2CHO2–(aq) + Co2+(aq) + 2H2O
5.25
(a)
Formic acid, a weak acid will form. molecular: KCHO2(aq) + HCl(aq) KCl(aq) + HCHO2(aq) ionic: K+(aq) + CHO2– (aq) + H+(aq) + Cl–(aq) K+(aq) + Cl–(aq) + HCHO2(aq) net ionic: CHO2–(aq) + H+(aq) HCHO2(aq)
(b)
Carbonic acid will form and it will further dissociate to water and carbon dioxide: CuCO3(s) + 2H+(aq) CO2(g) + H2O + Cu2+(aq) molecular: CuCO3(s) + 2HC2H3O2(aq) CO2(g) + H2O + Cu(C2H3O2)2(aq) ionic: CuCO3(s) + 2HC2H3O2(aq) CO2(g) + H2O + Cu2+ + 2C2H3O2–(aq) net ionic: CuCO3(s) + 2HC2H3O2(aq) CO2(g) + H2O + Cu2+ + 2C2H3O2–(aq)
(c)
No reaction will occur. All acetate salts and nitrate salts are soluble
(d)
Insoluble nickel hydroxide will precipitate. Ni2+(aq) + 2OH–(aq) Ni(OH)2(s) molecular: NiCl2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaCl(aq) ionic: Ni2+(aq) + 2Cl–(aq) + 2Na+(aq) + 2OH–(aq) Ni(OH)2(s) + 2Na+(aq) + 2Cl–(aq) net ionic: Ni2+(aq) + 2OH– (aq) Ni(OH)2(s)
5.26
CuO(s) + 2HNO3(aq)
→
Cu(NO3)2(aq) + H2O(l)
Or Cu(OH)2(s) + 2HNO3(aq) →
5.27
You want to use a metathesis reaction that produces CoS, which is insoluble, and a second product that is soluble. You may want the reactants to be soluble. CoCl2(aq) + Na2S(aq)
5.28
Cu(NO3)2(aq) + 2H2O(l)
→
CoS(s) + 2NaCl(aq)
1 mol HNO3 mol HNO3 = (16.9 g HNO3) = 0.268 mol HNO3 63.02 g HNO3
82
Chapter 5
0.268 mol HNO3 1000 mL solution M HNO3 = = 1.53 M HNO3 175 ml solution 1 L solution The amount of HNO3 does not change as the solution is diluted.
5.29
1 mol NaCl mol NaCl in 1.223 g NaCl = (1.223 g NaCl) = 0.02093 mol NaCl 58.443 g NaCl mol NaCl in 1.461 g NaCl = 0.02500 mol NaCl Total mol NaCl = 0.02093 mol NaCl + 0.02500 mol NaCl = 0.04593 mol NaCl
0.04593 mol NaCl 1000 mL solution Molarity NaCl = = 0.1837 M 250.0 mL solution 1 L solution
5.30
5.31
1 L solution 0.250 mol HCl mol HCl = 175 mL HCl solution = 0.0438 mol HCl 1000 mL solution 1 L solution
Determine the moles of HCl in 1.30 g. This is the number of moles of HCl in the 0.250 M solution so we need to determine what volume that is required to divide the moles by to get 0.250 M.
1 mol HCl / Vsoln = 0.250 M 36.46 g HCl
(1.30 g HCl)
Vsoln = 143 mL of 0.250 M HCl 5.32
If we were working with a full liter of this solution, it would contain 0.2 mol of Sr(NO3)2. The molar mass of the salt is 211.62 g mol–1, so 0.2 mol is slightly more than 40 g. However, we are working with just 50 mL, so the amount of Sr(NO3)2 needed is slightly more than a twentieth of 40 g, or 2 g. The answer, 2.11 g, is close to this, so it makes sense.
1 L solution 0.2 mol Sr ( NO3 )2 g Sr(NO3)2 = 50 mL 1000 mL solution 1 L solution
5.33
211.62 g Sr ( NO3 ) 2 1 mol Sr ( NO3 ) 2
= 2.11 g Sr(NO3)2
1 L sol’n 0.0125 mol AgNO3 169.9 g AgNO3 g AgNO3 = (250 mL sol'n) = 0.531 g AgNO3 1 L sol’n 1000 mL sol’n 1 mol AgNO3
83
Chapter 5
5.34
1 0.125 mol H 2SO 4 mL = (100.0 mL solution) = 250.0 mL 1 L solution ( 0.0500 M )
5.35
1 L solution 0.50 mol HCl mol HCl = (150 mL solution) = 0.075 mol HCl 1000 mL solution 1 L solution 1 L solution 1000 mL solution mL = (0.075 mol HCl) = 750 mL 0.10 mol HCl 1 L solution To find the number of mL of water to add to the solution subtract the number of mL of the concentrated solution from the total volume: 750 mL solution – 150 mL = 600 mL Add 600 mL of water.
5.36
1 L solution 0.100 mol KOH 1 mol H3 PO 4 mol H3PO4 = (45.0 mL KOH) 1000 mL solution 1 L solution 3 mol KOH = 1.5 × 10–3 mol H3PO4 1 L solution mL H3PO4 = (1.5 × 10–3 mol H3PO4) 0.0475 mol H3 PO 4
5.37
1 L H 2SO 4 0.108 mol H 2SO 4 mL NaOH = (15.4 mL H2SO4) 1 L H 2SO 4 1000 mL H 2SO 4 2 mol NaOH × 1 mol H 2SO4
5.38
1000 mL solution = 31.6 mL H3PO4 1 L solution
1 L NaOH 1000 mL NaOH = 26.8 mL NaOH 0.124 mol NaOH 1 L NaOH
FeCl3 Fe3+ + 3Cl–
0.40 mol FeCl3 1 mol Fe3+ M Fe3+ = = 0.40 M Fe3+ 1 L FeCl soln 1 mol FeCl 3 3 0.40 mol FeCl3 3 mol Cl− M Cl– = = 1.2 M Cl– 1 L FeCl soln 1 mol FeCl 3 3
5.39
0.250 mol PO 3− 3 mol Na + 4 M Na+ = 1 L Na 3 PO 4 soln 1 mol PO 3− 4
= 0.750 M Na+
5.40
1 L AgNO3 0.100 mol AgNO3 1 mol Ag + mol CaCl2 = 18.4 mL AgNO3 1 L AgNO3 1000 mL AgNO3 1 mol AgNO3
84
Chapter 5
1 mol Cl− × 1 mol Ag +
1 mol CaCl2 = 9.20 × 10–4 mol CaCl2 2 mol Cl−
9.20 × 10−4 mol CaCl 2 M CaCl2 = 20.5 mL CaCl 2
5.41
1000 mL CaCl 2 = 0.0449 M CaCl 2 1 L CaCl2
The balanced net ionic equation is: Fe2+(aq) + 2OH–(aq) Fe(OH)2(s). First determine the number of moles of Fe2+ present.
0.250 mol FeCl2 1 mol Fe2+ mol Fe2+ = (60.0 mL Fe2+) 1000 mL solution 1 mol FeCl2
= 1.50 × 10–2 mol Fe2+
Now, determine the amount of KOH needed to react with the Fe2+.
5.42
2 mol OH − mL KOH = (1.50× 10–2 mol Fe2+) 2+ 1 mol Fe
1 mol KOH 1 mol OH −
1000 mL solution = 60.0 mL KOH 0.500 mol KOH
0.150 mol BaCl2 g Na2SO4 = 28.40 mL BaCl2 1000 mL BaCl2
1 mol Ba 2+ 1 mol BaCl2
1 mol SO4 2− 1 mol Ba 2+
1 mol Na SO 2 4 × 1 mol SO 2− 4
5.43
(a) (b) (c)
5.44
142.05 g Na SO 2 4 = 0.605 g Na SO 2 4 1 mol Na 2SO 4
= 5.41 × 10–3 mol Ca2+ Since all of the Ca2+ is precipitated as CaSO4, there were originally 5.41 × 10–3 moles of Ca2+ in the sample. All of the Ca2+ comes from CaCl2, so there were 5.41 × 10–3 moles of CaCl2 in the sample. 1 mol CaSO 4 mol Ca2+ = (0.736 g CaSO4) 136.14 g CaSO 4
1 mol Ca 2+ 1 mol CaSO 4
(d)
110.98 g CaCl2 g CaCl2 = (5.41 × 10–3 mol CaCl2) = 0.600 g CaCl2 1 mol CaCl2
(e)
% CaCl2 =
(f)
% MgCl2 = 100 % - 30 % = 70 %
0.600 g CaCl2 × 100% = 30.0% CaCl2 2.000 g sample
Balanced equation: H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O
1 L NaOH soln 0.147 mol NaOH 1 mol H 2SO4 mol HsSO4 = (36.42 mL NaOH) 1000 mL NaOH soln 1 L NaOH soln 2 mol NaOH
85
Chapter 5
= 2.68 × 10–3 mol H2SO4
2.68 × 10−3 mol H SO 2 4 M H2SO4 = 15.00 mL H 2 SO4
5.45
1000 mL H SO 2 4 = 0.178 M H SO 2 4 1 L H 2 SO 4
1 L KOH soln 0.0100 mol KOH soln 1 mol HCl –4 mol HCl = (11.00 mL) 1 mol KOH = 1.1 × 10 mol HCl 1000 mL NaOH soln 1 L KOH soln 1.1× 10−4 mol HCl 1000 mL HCl soln M HCl = = 0.0220 M HCl 5.00 mL HCl soln 1 L HCl soln 0.0220 mol HCl 36.5 g HCl –3 g HCl = (5.00 mL HCl) = 4.02 × 10 g HCl 1000 mL HCl 1 mol HCl
weight % =
4.02 × 10 −3 g × 100% = 0.0803% 5.00 g
Review Questions 5.1 (a) Solvent – the medium into which something (a solute) is dissolved to make a solution
5.2
(b)
Solute – Something dissolved in a solvent to make a solution
(c)
Concentration – the ratio of the quantity of solute to the quantity of solution or quantity of solvent
(a)
Concentrated – a solution that has a large ratio of the amounts of solute to solvent
(b)
Dilute – a solution in which the ratio of the quantities of solute to solvent is small
(c)
Saturated – a solution that holds as much solute as it can at a given temperature
(d)
Unsaturated – Any solution with a concentration less than that of a saturated solution of the same solute and solvent
(e)
Supersaturated – a solution whose concentration of solute exceeds the equilibrium concentration
(f)
Solubility – the ratio of the quantity of solute to the quantity of solvent in a saturated solution
5.3
Chemical reactions are often carried out using solutions because this allows the reactants to move about and come in contact with each other. Furthermore, solutions can be made with a high enough concentration to allow the reaction to proceed at a reasonable rate.
5.4
When a sugar crystal is added to (a) a saturated sugar solution, the sugar crystal will not dissolve.
86
Chapter 5
5.5
(b)
a supersaturated sugar solution, the sugar crystal will cause the extra sugar in solution to precipitate forming more sugar crystals.
(c)
an unsaturated sugar solution, the added sugar crystal will dissolve.
Precipitate – a solid that separates from a solution usually as the result of a chemical reaction For a precipitate to form spontaneously in a solution, the equilibrium must be disrupted. A supersaturated solution may form a precipitate spontaneously, or if the temperature changes in the direction that will cause a precipitate to form.
5.6
Electrolytes are soluble, ionic compounds. The following are likely to be electrolytes: CuBr2, iron(II) chloride, and (NH4)2SO4. The following are molecular compounds and are not likely to be electrolytes: C12H22O11, and CH3OH.
5.7
nonelectrolyte does not have a charge, so it cannot allow ions to move. An ion is hydrated when it is surrounded by water molecules.
5.8
Dissociation – the dissolving of an ionic compound in water such that the individual ions that compose the ionic compound become separated from one another (via hydration), and move about freely in solution, acting more or less independently of one another.
5.9
(a)
CaCl2(aq) Ca2+(aq) + 2Cl–(aq)
(b)
(NH4)2SO4(aq) 2NH4+(aq) + SO42–(aq)
(c)
NaC2H3O2(aq) Na+(aq) + C2H3O2–(aq)
(d)
Cu(ClO4)2(aq) Cu2+(aq) + 2ClO4-(aq)
5.10
The spectator ions are Na+ and Cl–. The net ionic equation is: Co2+ + 2OH– Co(OH)2(s)
5.11
There are no counter ions for the Al3+ or the OH- so this must be a net ionic equation.
5.12
In a balanced ionic equation, both the mass and the electrical charge must be balanced. It must have the correct formulas of reactants and products. The product is not correct in the equation. Co3+(aq) + HPO42–(aq) CoPO4(s) + H+(aq)
5.13
Acid – sour taste, turns litmus red, corrode some metals, etc... Base – bitter taste, turns litmus blue, soapy “feel”, etc...
5.14
If a solution is believed to be basic, red litmus paper should be used so that it would turn blue. The blue litmus paper may not change color if the solution is neutral.
5.15
According to the definition of Arrhenius, an acid gives H+ ions in water, and a base gives OH– ions in water.
87
Chapter 5
5.16
(a)
NaOH
dissociation
(b)
HNO3
ionization
(c)
NH3
ionization
(d)
H2SO4 ionization
5.17
(a) (b) (c) (d)
P4O10 K 2O SeO3 Cl2O7
5.18
Dynamic equilibrium is a condition in which two opposing processes are occurring at equal rates. Acetic acid is not a strong acid, so that it forms an equilibrium between the molecular form, HC2H3O2, and the ionized form, H+ and C2H3O2–.
5.19
Double arrows are not used for the reaction of a strong acid with water because the reaction is not in equilibrium. These are not reversible reactions, i.e., the reverse reaction has practically no tendency to occur.
5.20
(a)
HCN: weak
(b)
HNO3: strong
(c)
H2SO3: weak
(d)
HCl: strong
(e)
HCHO2: weak
(f)
HNO2: weak
(a)
C5H5N: weak
(b)
Ba(OH)2: strong
(c)
KOH: strong
(d)
C6H5NH2: weak
(e)
Cs2O: strong
(f)
N2O5: acidic solution
5.21
acidic solutions basic solutions acidic solutions acidic solutions
5.22 H H 3C N H
H H 3C N H
+
H
88
Chapter 5
5.23
The student removed a hydrogen attached to the methyl group, CH3. Hydrogen attached to carbon atoms are not acidic protons and will not be removed in water. The correct structure of the ion is one where the hydrogen attached to the oxygen atom is removed. The structure should be:
5.24
The molecules is diethylamine, as base. In water, the molecule would add an H+ to the nitrogen atom. The structure of the resulting ion is:
The balanced chemical equation is given below.
(CH 3CH 2 ) 2 NH(l ) + H 2 O(l ) 5.25
(a) (b)
hydrogen selenide hydroselenic acid
5.26
(a) (b) (c) (d) (e)
periodic acid iodic acid iodous acid hypoiodous acid hydroiodic acid
5.27
(a) (b)
IO4– periodate
5.28
(a) (b) (c)
H2CrO4 H2CO3 H2C2O4
IO3– iodate
→ (CH 3CH 2 )2 NH 2+ (aq ) + OH − (aq )
IO2– iodite
89
IO– hypoiodite
I– iodide
Chapter 5
5.29
(a) (b) (c)
sodium bicarbonate or sodium hydrogen carbonate potassium dihydrogen phosphate ammonium hydrogen phosphate
5.30
NaH2PO4
5.31
(a) (b) (c) (d)
5.32
H3AsO3
5.33
sodium butyrate
5.34
propionic acid
5.35
Formation of a weak electrolyte, water, a gas, or an insoluble solid.
5.36
A metathesis reaction is also called a double replacement reaction.
5.37
Since AgBr is insoluble, the concentrations of Ag+ and Br– in a saturated solution of AgBr are very small. When solutions of AgNO3 and NaBr are mixed, the concentrations of Ag+ and Br– are momentarily larger than those in a saturated AgBr solution. Since this solution is immediately supersaturated in the moment of mixing, a precipitate of AgBr forms spontaneously.
5.38
3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
Na2HPO4
hypochlorous acid iodous acid bromic acid perchloric acid
Na3PO4
sodium hypochlorite sodium iodite sodium bromate sodium perchlorate
NaOCl NaIO2 NaBrO3 NaClO4
3Mg2+(aq) + 2PO43–(aq) Mg3(PO4)2(s) 5.39
The substance is an electrolyte that dissolves readily in water: Na2CO3·10H2O(s) 2Na+(aq) + CO32–(aq) + 10H2O The carbonate anion then serves to cause the precipitation of calcium cations: Ca2+(aq) + CO32–(aq) CaCO3(s)
5.40
HCHO2 will react with the following: (a)
KOH
HCHO2 + KOH KCHO2 + H2O
(b)
MgO
2HCHO2 + MgO Mg(CHO2)2 + 2H2O
(c)
NH3
HCHO2 + NH3 NH4+ + CHO2–
90
Chapter 5
5.41
Any solution containing ammonium ion will react with a strong base to yield ammonia. The presence of ammonia is easily detected by its odor.
5.42
(a)
HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O + CO2(g)
(b)
2HCl(aq) +Na2S(aq) 2NaCl(aq) + H2S(g)
(c)
2HCl(aq) + K2SO3(aq) 2KCl(aq) + H2O + SO2(g)
5.43
Molarity is the number of moles of solute per liter of solution, also known as molar concentration. mmol 1 mol 1000 mL 1000 mol mol mL 1000 mmol 1 L = 1000 L = L
5.44
The two conversion factors are:
0.25 mol HCl 1 L HCl
5.45
1 L HCl 0.25 mol HCl
mol M×L= × L = mol L
5.46
The number of moles of HNO3 in the solution has not changed because none of the original sample was removed. Instead, the concentration has decreased since more water was added.
5.47
The number of moles of CaCl2 is the same in both solutions, but A is 0.10 M CaCl2 and B is 0.20 M CaCl2. The volume of solution A is 50 mL, therefore the volume of B is 25 mL:
1 L 0.1 mol CaCl2 –3 mol CaCl2 present = 50 mL = 5 × 10 mol CaCl2 1000 mL 1 L CaCl2 1 L CaCl2 volume of solution B = (5 × 10–3 mol CaCl2) 0.2 mol CaCl2 5.48
1000 mL = 25 mL solution B 1 L
Qualitative analysis is the use of experimental procedures to determine what elements are present in a substance. Quantitative analysis determines the percentage composition of a compound or the percentage of a component in a mixture. Qualitative analysis answers the question, "what is in the sample?" Quantitative analysis answers the question, "how much is in the sample?"
5.49
(a)
Buret – a long glass tube fitted with a stopcock, graduated in mL, and used for the controlled, measured addition of a volume of a solution to a receiving flask.
(b)
Titration – a procedure for obtaining quantitative information about a reactant by a controlled addition of one substance to another until a signal (usually a color change of an indicator) shows that equivalent quantities have reacted. Titrant – the solution delivered from a buret during a titration. End point – that point during a titration when the indicator changes color, the titration is stopped, and the total added volume of the titrant is recorded.
(c) (d)
91
Chapter 5
5.50
The indicator provides a visible signal that the solution has changed from an acid to a base. (a) (b)
Phenolphthalein is colorless in acid solution. Phenolphthalein is pink in base solution.
Review Problems 5.51
5.52
(a)
ionic: 2NH4+(aq) + CO32–(aq) + Mg2+(aq) + 2Cl–(aq) 2NH4+(aq) + 2Cl–(aq) + MgCO3(s) net: Mg2+(aq) + CO32–(aq) MgCO3(s)
(b)
ionic: Cu2+(aq) + 2Cl–(aq) + 2Na+(aq) + 2OH–(aq) Cu(OH)2(s) + 2Na+(aq) + 2Cl–(aq) net: Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)
(c)
ionic: 3Fe2+(aq) + 3SO42–(aq) + 6Na+(aq) + 2PO43–(aq) Fe3(PO4)2(s) + 6Na+(aq) + 3SO42–(aq) Net: 3Fe2+(aq) + 2PO43–(aq) Fe3(PO4)2(s)
(d)
ionic: 2Ag+(aq) + 2C2H3O2–(aq) + Ni2+(aq) + 2Cl–(aq) 2AgCl(s) + Ni2+(aq) + 2C2H3O–(aq) Net: 2Ag+(aq) + 2Cl–(aq) 2AgCl(s)
(a)
ionic: Cu2+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) BaSO4(s) + Cu2+(aq) + 2Cl–(aq) net: Ba2+(aq) + SO42–(aq) BaSO4(s)
(b)
Fe3+(aq) + 3NO3–(aq) + 3Li+(aq) + 3OH–(aq) Fe(OH)3(s) + 3Li+(aq) + 3NO3–(aq) net: Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)
(c)
6Na+(aq) + 2PO43–(aq) + 3Ca2+(aq) + 6Cl–(aq) Ca3(PO4)2(s) + 6Na+(aq) + 6Cl–(aq) net: 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
(d)
2Na+(aq) + S2–(aq) + 2Ag+(aq) + 2C2H3O2–(aq) 2Na+(aq) + 2C2H3O2–(aq) + Ag2S(s) net: 2Ag+(aq) + S2–(aq) Ag2S(s)
5.53
This is an ionization reaction: HClO4(l) + H2O H3O+(aq) + ClO4–(aq)
5.54
HBr(l) + H2O H3O+(aq) + Br–(aq)
5.55
N2H4(aq) + H2O N2H5+(aq) + OH–(aq)
5.56
C5H5N(aq) + H2O C5H5NH+(aq) + OH–(aq)
5.57
HNO2(aq) + H2O H3O+(aq) + NO2–(aq)
5.58
HC5H9O2(aq) + H2O H3O+(aq) + C5H9O2–(aq)
5.59
H2CO3(aq) + H2O H3O+(aq) + HCO3–(aq)
92
Chapter 5
HCO3–(aq) + H2O H3O+(aq) + CO32–(aq) 5.60
H3PO4(aq) + H2O H3O+(aq) + H2PO4–(aq) H2PO4–(aq) + H2O H3O+(aq) + HPO42–(aq) HPO42–(aq) + H2O H3O+(aq) + PO43–(aq)
5.61
molecular: Na2S(aq) + Cu(NO3)2(aq) CuS(s) + 2NaNO3(aq) ionic: 2Na+(aq) + S2–(aq) + Cu2+ (aq) + 2NO3– (aq) CuS(s) + 2Na+(aq) + 2NO3–(aq) net: Cu2+(aq) + S2–(aq) CuS(s)
5.62
molecular: Fe2(SO4)3(aq) + 3BaCl2(aq) 3BaSO4(s) + 2FeCl3(aq) ionic: 2Fe3+(aq) + 3SO42–(aq) + 3Ba2+(aq) + 6Cl–(aq) 3BaSO4(s) + 2Fe3+(aq) + 6Cl–(aq) net: 3Ba2+(aq) + 3SO42–(aq) 3BaSO4(s)
5.63
The soluble ones are (a), (b), and (d).
5.64
The soluble ones are (a), (b), (d), and (f).
5.65
(a)
molecular equation: FeSO4(aq) + K2CO3(aq) K2SO4(aq) + FeCO3(s) ionic equation: Fe2+(aq) + SO42–(aq) + 2K+(aq) + CO32−(aq) 2K+(aq) + SO42−(aq)
+ FeCO3 (s) net ionic equation: Fe2+(aq) + CO32−(aq) FeCO3(s) (b)
molecular equation: 2AgC2H3O2(aq) + ZnCl2(aq) 2AgCl(s) + Zn(C2H3O2)2 (aq) ionic equation: 2Ag+(aq) + 2C2H3O2−(aq) + Zn2+(aq) + 2Cl− (aq) 2AgCl(s) + Zn2+ (aq) + 2C2H3O2−(aq)
net ionic equation: Ag+(aq) + Cl−(aq) AgCl(s) (c)
molecular equation: 2CrCl3(aq) + 3Ca(OH)2(aq) 3CaCl2(aq) + 2Cr (OH)3 (s) ionic equation: 2Cr3+(aq) + 6Cl−(aq) + 3Ca2+(aq) + 6OH−(aq) 3Ca2+(aq) + 6Cl−(aq) + 2Cr(OH)3 (s)
net ionic equation:Cr3+(aq) + 3OH−(aq) Cr(OH)3(s) 5.66
(a) (b) (c)
5.67
(a)
ionic: Fe3+(aq) + 3NO3–(aq) + 3K+(aq) + 3OH–(aq) Fe(OH)3(s) + 3K+(aq) + 3NO3–(aq) net: Fe3+(aq) + 3OH–(aq) Fe(OH)3(s) ionic: 6Na+(aq) + 2PO43–(aq) + 3Sr2+(aq) + 6Cl–(aq) Sr3(PO4)2(s) + 6Na+(aq) + 6Cl–(aq) net: 3Sr2+(aq) + 2PO43–(aq) Sr3(PO4)2(s) ionic: Pb2+(aq) + 2C2H3O2–(aq) + 2NH42+(aq) + SO42–(aq) PbSO4(s) + 2NH4+(aq) + 2C2H3O2–(aq) Net: Pb2+(aq) + SO42–(aq) PbSO4(s) molecular equation: Ba(OH)2(aq) + 2HNO3(aq) 2H2O(l) + Ba(NO3)2(aq) ionic equation: Ba2+(aq) + 2OH−(aq) + 2H+(aq) + 2NO3−(aq) 2H2O(l) + Ba2+(aq) + 2NO3−(aq) net ionic equation: OH−(aq) + H+(aq) H2O(l)
93
Chapter 5
5.68
5.69
(b)
molecular equation: Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l) ionic equation: Al2O3(s) + 6H+(aq) +6Cl–(aq) 2Al3+(aq) + 6Cl–(aq) + 3H2O (l) net ionic equation: Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O(l)
(c)
molecular equation: Cu(OH)2(s) + H2SO4(aq) 2H2O(l) + CuSO4(aq) ionic equation: Cu(OH)2(s) + 2H+(aq) + SO42−(aq) 2H2O(l) + Cu2+(aq) + SO42−(aq) net ionic equation: Cu(OH)2(s) + 2H+(aq) 2H2O(l) + Cu2+(aq)
(a)
molecular: 2HC2H3O2(aq) + Mg(OH)2(s) Mg(C2H3O2)2(aq) + 2H2O ionic: 2HC2H3O2(aq) + Mg(OH)2(s) Mg2+(aq) + 2C2H3O2–(aq) + 2H2O net: 2HC2H3O2(aq) + Mg(OH)2(s) Mg2+(aq) + 2C2H3O2–(aq) + 2H2O
(b)
molecular: HClO4(aq) + NH3(aq) NH4ClO4(aq) ionic: H+(aq) + ClO4–(aq) + NH3(aq) NH4+(aq) + ClO4–(aq) net: H+(aq) + NH3(aq) NH4+(aq)
(c)
molecular: H2CO3(aq) + 2NH3(aq) (NH4)2CO3(aq) ionic: H2CO3(aq) + 2NH3(aq) 2NH4+(aq) + CO32–(aq) net: H2CO3(aq) + 2NH3(aq) 2NH4+(aq) + CO32–(aq)
The electrical conductivity would decrease as the solution is neutralized because there are half the amount of ions as products of this reaction than there were to start with. Cu(OH)2(s) + 2H+(aq) 2H2O(l) + Cu2+(aq)
5.70
The electrical conductivity would increase since HC2C3O2 is a weak acid and is only partially dissociated, and as the NH3 is added two ions are formed, increasing the concentration of ions in solution. NH3(aq) + HC2H3O2(aq) NH4+(aq) + C2H3O2–(aq) Once the point of neutralization has been reached, adding more NH3 will not significantly change the amount of electrolytes in solution since it is a weak base.
5.71
(a) (b)
2H+(aq) + CO32–(aq) H2O(l) + CO2(g) NH4+(aq) +OH–(aq) NH3(g) + H2O
5.72
(a) (b)
H+(aq) + HSO3–(aq) H2O(l) + CO2(g) 2H+(aq) + SO32–(aq) → H2O(l) + SO2(g)
5.73
These reactions have the following "driving forces":
5.74
(a)
formation of insoluble Cr(OH)3
(b)
formation of water, a weak electrolyte
These reactions have the following "driving forces": (a) formation of a gas, CO2
94
Chapter 5
5.75
5.76
(b)
formation of a weak electrolyte, H2C2O4
(a)
molecular: 3HNO3(aq) + Cr(OH)3(s) Cr(NO3)3(aq) + 3H2O ionic: 3H+(aq) + 3NO3–(aq) + Cr(OH)3(s) Cr3+(aq) + 3NO3–(aq) + 3H2O net: 3H+(aq) + Cr(OH)3(s) Cr3+(aq) + 3H2O
(b)
molecular: HClO4(aq) + NaOH(aq) NaClO4(aq) + H2O ionic: H+(aq) + ClO4–(aq) + Na+(aq) + OH–(aq) Na+(aq) + ClO4–(aq) + H2O net: H+(aq) + OH–(aq) H2O
(c)
molecular: Cu(OH)2(s) + 2HC2H3O2(aq) Cu(C2H3O2)2(aq) + 2H2O ionic: Cu(OH)2(s) + 2H+(aq) + 2C2H3O2–(aq) Cu2+(aq) + 2C2H3O2–(aq) + 2H2O net: Cu(OH)2(s) + 2H+(aq) Cu2+(aq) + 2H2O
(d)
molecular: ZnO(s) + H2SO4(aq) ZnSO4(aq) + H2O ionic: ZnO(s) + 2H+(aq) + SO42–(aq) Zn2+(aq) + SO42–(aq) + H2O net: ZnO(s) + 2H+(aq) Zn2+(aq) + H2O
(a)
molecular: NaHSO3(aq) + HBr(aq) SO2(g) + NaBr(aq) + H2O ionic: Na+(aq) + HSO3–(aq) + H+(aq) + Br–(aq) SO2(g) + Na+(aq) + Br–(aq) + H2O net: HSO3–(aq) + H+(aq) SO2(g) + H2O
(b)
molecular: (NH4)2CO3(aq) + 2NaOH(aq) 2NH3(g) + Na2CO3(aq) + 2H2O ionic: 2NH4+(aq) + CO32–(aq) + 2Na+(aq) + 2OH–(aq) 2NH3(g) + 2Na+(aq) + CO32–(aq) + 2H2O net: NH4 (aq) + OH (aq) NH3(g) + H2O +
5.77
–
(c)
molecular: (NH4)2CO3(aq) + Ba(OH)2(aq) BaCO3(s) + 2NH3(g) + 2H2O ionic: 2NH4+(aq) + CO32–(aq) + Ba2+(aq) + 2OH–(aq) BaCO3(s) + 2NH3(g) + 2H2O net: 2NH4+(aq) + CO32–(aq) + Ba2+(aq) + 2OH–(aq) BaCO3(s) + 2NH3(g) + 2H2O
(d)
molecular: FeS(s) + 2HCl(aq) FeCl2(aq) + H2S(g) ionic: FeS(s) + 2H+(aq) + 2Cl–(aq) Fe2+(aq) + 2Cl–(aq) + H2S(g) net: FeS(s) + 2H+(aq) Fe2+(aq) + H2S(g)
(a)
molecular: Na2SO3(aq) + Ba(NO3)2(aq) BaSO3(s) + 2NaNO3(aq) ionic: 2Na+(aq) + SO32–(aq) + Ba2+(aq) + 2NO3–(aq) BaSO3(s) + 2Na+(aq) + 2NO3–(aq) net: Ba2+(aq) + SO32–(aq) BaSO3(s)
(b)
molecular: 2HCHO2(aq) + K2CO3(aq) CO2(g) + H2O + 2KCHO2(aq) ionic: 2H+(aq) + 2CHO2–(aq) + 2K+(aq) + CO32–(aq) CO2(g) + H2O + 2K+(aq) + 2CHO2–(aq) net: 2H+(aq) + CO32–(aq) CO2(g) + H2O
(c)
molecular: 2NH4Br(aq) + Pb(C2H3O2)2(aq) 2NH4C2H3O2(aq) + PbBr2(s) ionic: 2NH4+(aq) + 2Br–(aq) + Pb2+(aq) + 2C2H3O2–(aq) 2NH4+(aq) + 2C2H3O2–(aq) + PbBr2(s) net: Pb2+(aq) + 2Br–(aq) PbBr2(s)
95
Chapter 5
5.78
5.79
(d)
molecular: 2NH4ClO4(aq) + Cu(NO3)2(aq) Cu(ClO4)2(aq) + 2NH4NO3(aq) ionic: 2NH4+(aq) + 2ClO4–(aq) + Cu2+(aq) + 2NO3–(aq) Cu2+(aq) + 2ClO4–(aq) + 2NO3–(aq) + 2NH4+(aq) net: N.R.
(a)
molecular: (NH4)2S(aq) + 2NaOH(aq) 2NH3(g) + 2H2O + Na2S(aq) ionic: 2NH4+(aq) + S2–(aq) + 2Na+(aq) + 2OH–(aq) 2NH3(g) + 2H2O + 2Na+(aq) + S2–(aq) net: NH4+(aq) + OH–(aq) NH3(g) + H2O
(b)
molecular: Cr2(SO4)3(aq) + 3K2CO3(aq) Cr2(CO3)3(s) + 3K2SO4(aq) ionic: 2Cr3+(aq) + 3SO42–(aq) + 6K+(aq) + 3CO32–(aq) Cr2(CO3)3(s) + 6K+(aq) + 3SO42–(aq) net: 2Cr3+(aq) + 3CO32–(aq) Cr2(CO3)3(s)
(c)
molecular: 3AgNO3(aq) + Cr(C2H3O2)3(aq) 3AgC2H3O2(aq) + Cr(NO3)3(aq) ionic: 3Ag+(aq) + 3NO3(aq) + Cr3+(aq) + 3C2H3O2(aq) 3Ag+(aq) + 3C2H3O2(aq) + Cr3+(aq) +3NO3(aq) net: NR
(d)
molecular: Sr(OH)2(aq) + MgCl2(aq) SrCl2(aq) + Mg(OH)2(s) ionic: Sr2+(aq) + 2OH–(aq) + Mg2+(aq) + 2Cl–(aq) Mg(OH)2(s) + Sr2+(aq) + 2Cl–(aq) net: Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)
There are numerous possible answers. One of many possible sets of answers would be: (a) (b) (c) (d) (e)
NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O FeCl2(aq) + 2NaOH(aq) Fe(OH)2(s) + 2NaCl(aq) Ba(NO3)2(aq) + K2SO3(aq) BaSO3(s) + 2KNO3(aq) 2AgNO3(aq) + Na2S(aq) Ag2S(s) + 2NaNO3(aq) ZnO(s) + 2HCl(aq) ZnCl2(aq) + H2O
5.80
We need to choose a set of reactants that are both soluble and that react to yield only one solid product. Choose (b). It can form CuCO3, depending on the concentration of (NH4)2CO3. The solution needs to be kept acidic enough to prevent the formation of Cu(OH)2. Choices (a), (c), and (e) all have insoluble reactants, and for (d), the K2CO3 is basic enough to form Cu(OH)2.
5.81
(a)
NaOH Na+ + OH–
1 mol NaOH mol NaOH = (5.00 g NaOH) = 0.125 mol NaOH 40.00 g NaOH 0.125 mol NaOH 1000 mL NaOH M NaOH solution = = 0.500 M NaOH 250.0 mL NaOH 1 L NaOH (b)
CaCl2 Ca2+ + 2Cl–
1 mol CaCl2 mol CaCl2 = (14.0 g CaCl2) = 0.126 mol CaCl2 110.98 g CaCl2 0.126 mol CaCl2 1000 mL CaCl2 M CaCl2 solution = = 0.631 M CaCl2 200.0 mL CaCl2 1 L CaCl2
96
Chapter 5
5.82
(a)
H2SO4 2H+ + SO42–
1 mol H 2SO 4 mol H2SO4 = (3.60 g H2SO4) = 0.0367 mol H2SO4 98.08 g H 2SO 4 0.0367 mol H 2SO 4 1000 mL H 2SO4 M H2SO4 solution = = 0.0816 M H2SO4 450.0 mL H 2SO4 1 L H 2SO 4 (b)
Fe(NO3)2 Fe2+ + 2NO3–
2.00 ×10−3 mol Fe(NO ) 3 2 M Fe(NO3)2 solution = 12.0 mL Fe(NO ) 3 2
5.83
1000 mL Fe(NO ) 3 2 = 0.17 M Fe(NO3)2 1 L Fe(NO3 )2
1 mol NaC2 H3O 2 1 L NaC2 H3O 2 mL NaC2H3O2 = (12.6 g NaC2H3O2) 82.03 g NaC H O 0.265 mol NaC H O 2 3 2 2 3 2 1000 mL NaC2 H3O 2 1 L NaC2 H3O 2
= 580. mL NaC2H3O2
5.84
1 mol HNO3 1000 mL HNO3 1 L HNO3 mL HNO3 = (1.35 g HNO3) = 34.8 mL HNO3 63.02 g HNO 0.615 mol HNO 1 L HNO3 3 3
5.85
(a) (b)
(c)
5.86
(a) (b) (c)
5.87
1 L 0.225 mol NaCl 58.44 g NaCl g NaCl = 125 mL soln 1 mol NaCl = 1.64 g NaCl 1 L soln 1000 mL 1 L 0.320 mol C6 H12 O6 180.2 g C6 H12 O6 g C2H12O6 = 250 mL soln 1 L soln 1000 mL 1 mol C6 H12 O6 = 14.4 g C2H12O6 1 L 0.275 mol H 2SO 4 98.08 g H 2SO 4 g H2SO4 = 250 mL soln 1 mol H SO = 6.74 g H2SO4 1 L soln 1000 mL 2 4 1 L 0.100 mol K 2SO 4 174.3 g K 2SO 4 g K2SO4 = 150.0 mL soln 1 mol K SO = 2.61 g K2SO4 1 L soln 1000 mL 2 4 1 L 0.250 mol FeCl3 162.2 g FeCl3 g FeCl3 = 75.0 mL soln = 3.04 g FeCl3 1 L soln 1000 mL 1 mol FeCl3 g Ba(C2H3O2)2 = 250.0 mL soln × 1 L 0.400 mol Ba ( C2 H3O 2 )2 255.4 g Ba ( C2 H3O 2 )2 = 25.5 g Ba(C2H3O2)2 1 mol Ba ( C2 H3O 2 ) 1 L soln 1000 mL 2
0.86 mol H 2SO 4 mol of H2SO4 = 25.0 mL H2SO4 1 L soln = 0.0215 mol H2SO4 1 L soln 1000 mL soln 0.0215 mol H 2SO 4 1000 mL H 2SO 4 M of final solution = = 0.172 M H2SO4 125 mL H 2SO 4 1 L H 2SO4
97
Chapter 5
5.88
0.560 mol HNO3 mol of HNO3 = 150 mL HNO3 1 L soln = 0.084 mol HNO3 1000 mL soln 1 L soln 0.084 mol HNO3 1000 mL HNO3 M of final solution = = 0.187 M HNO3 450 mL HNO3 1 L HNO3
5.89
5.90
M V V2 = 1 1 M2 (18.0 M H 2SO4 )( 25.0 mL ) V2 = = 300 mL H2SO4 1.50 M H 2SO 4 The 25.0 mL of H2SO4 must be diluted to 300 mL. M1V1 = M2V2
M V V2 = 1 1 M2 (1.75 M HCl )( 25.0 mL ) = 219 mL HCl V2 = 0.200 M HCl The 25.0 mL of HCl must be diluted to 219 mL. M1V1 = M2V2
M V V2 = 1 1 M2 ( 3.00 M KOH )(150.0 mL ) = 360. mL KOH V2 = 1.25 M KOH The 150.0 mL of KOH must be diluted to 360. mL. The volume of water to be added is: 360. mL of V2 – 150 mL of V1 = 210 mL water
5.91
M1V1 = M2V2
5.92
M1V1 = M2V2
5.93
(a)
M V V2 = 1 1 M2 (1.50 M HCl )(120 mL ) = 180 mL HCl V2 = 1.00 M HCl The 120.0 mL of HCl must be diluted to 180 mL. The volume of water to be added is: 180 mL of V2 – 120 mL of V1 = 60. mL water
(b)
CaCl2 Ca2+ + 2Cl– mol CaCl2 = 0.625 mol/L × 0.0323 L = 0.0202 mol CaCl2 2+ 2+ 0.0202 mol CaCl2 1 mol Ca =0.0202 mol Ca 1 mol CaCl2 2 mol Cl− – 0.0202 mol CaCl2 =0.0404 mol Cl 1 mol CaCl2 AlCl3 Al3+ + 3Cl– mol AlCl3 = 0.380 mol/L × 0.0500 L = 0.0190 mol AlCl3 3+ 0.0190 mol AlCl3 1 mol Al =0.0190 mol Al3+ 1 mol AlCl3
98
Chapter 5
5.94
(a)
− 0.0190 mol AlCl3 3 mol Cl =0.0570 mol Cl– 1 mol AlCl3 (NH4)2CO3 2NH4+ + CO32–
mol (NH4)2CO3 = 0.40 mol/L × 0.0185 L = 0.0074 mol (NH4)2CO3 0.0074 mol (NH4)2CO3 × 2 mol NH4+/mol (NH4)2CO3 = 0.015 mol NH4+ 0.0074 mol (NH4)2CO3 × 1 mol CO32–/mol (NH4)2CO3 = 0.0074 mol CO32– (b)
Al2(SO4)3 2Al3+ + 3SO42– mol Al2(SO4)3 = 0.35 mol/L × 0.0300 L = 0.011 mol Al2(SO4)3 0.0105 mol Al2(SO4)3 × 2 mol Al3+/mol Al2(SO4)3 = 0.021 mol Al3+ 0.0105 mol Al2(SO4)3 × 3 mol SO42–/mol Al2(SO4)3 = 0.033 mol SO42–
5.95
(a)
Cr(NO3)2 Cr2+ + 2NO3–
0.45 mol Cr(NO3 )2 M Cr 2+ = 1 L Cr(NO3 )2 soln
1 mol Cr 2+ 1 mol Cr(NO3 )2
0.45 mol Cr(NO3 )2 M NO3− = 1 L Cr(NO3 )2 soln (b)
1 mol Cu 2+ 1 mol CuSO4
0.15 mol CuSO 4 M SO4 2− = 1 L CuSO 4 soln
= 0.90 M NO3−
= 0.15 M Cu 2+
1 mol SO 4 2− 1 mol CuSO 4
= 0.15 M SO 42 −
Na3PO4 3Na+ + PO43–
0.37 mol Na 3 PO 4 M Na + = 1 L Na 3 PO 4 soln
3 mol Na + 1 mol Na 3 PO 4
0.37 mol Na 3 PO 4 M PO43− = 1 L Na 3 PO 4 soln (d)
2 mol NO3− 1 mol Cr(NO3 )2
CuSO4 Cu2+ + SO42–
0.15 mol CuSO 4 M Cu 2+ = 1 L CuSO 4 soln
(c)
= 0.45 M Cr 2+
= 1.1 M Na +
1 mol PO 43− 1 mol Na 3 PO4
= 0.37 M PO 43−
Al2(SO4)3 2Al3+ + 3SO42–
0.050 mol Al2 (SO 4 )3 2 mol Al3+ M Al3+ = = 0.10 M Al3+ 1 L Al2 (SO 4 )3 soln 1 mol Al2 (SO4 )3 0.050 mol Al2 (SO 4 )3 3 mol SO 4 2− M SO4 2− = = 0.15 M SO 4 2− 1 L Al (SO ) soln 1 mol Al (SO ) 2 4 3 2 4 3
99
Chapter 5
5.96
(a)
(b)
Ca(OH)2 Ca2+ + 2OH–
0.060 mol Ca(OH)2 M Ca 2+ = 1 L Ca(OH) 2 soln
1 mol Ca 2+ 1 mol Ca(OH)2
= 0.060 M Ca 2+
0.060 mol Ca(OH)2 M OH − = 1 L Ca(OH) 2 soln
2 mol OH − 1 mol Ca(OH)2
= 0.12 M OH −
FeCl3 Fe3+ + 3Cl–
0.15 mol FeCl3 1 mol Fe3+ M Fe3+ = = 0.15 M Fe3+ 1 L FeCl soln 1 mol FeCl 3 3 0.15 mol FeCl3 3 mol Cl− M Cl− = = 0.45 M Cl− 1 L FeCl soln 1 mol FeCl 3 3 (c)
Cr2(SO4)3 2Cr3+ + 3SO42–
0.22 mol Cr2 (SO 4 )3 2 mol Cr3+ M Cr3+ = = 0.44 M Cr3+ 1 L Cr2 (SO 4 )3 soln 1 mol Cr2 (SO4 )3 0.22 mol Cr2 (SO4 )3 3 mol SO 42 − M SO4 2− = = 0.66 M SO 4 2− 1 L Cr (SO ) soln 1 mol Cr (SO ) 2 4 3 2 4 3 (d)
5.97
5.98
5.99
(NH4)2SO4 2NH4+ + SO42–
0.60 mol (NH 4 )2SO4 M NH 4+ = 1 L (NH 4 ) 2SO 4 soln
2 mol NH 4 + 1 mol (NH 4 ) 2SO 4
= 1.2 M NH 4+
0.60 mol (NH 4 ) 2SO 4 M SO4 2− = 1 L (NH 4 ) 2SO 4 soln
1 mol SO4 2 − 1 mol (NH 4 ) 2SO 4
= 0.60 M SO 4 2−
0.175 mol Al3+ g Al2(SO4)3 = (60.0 mL soln) 1000 mL soln = 1.80 g Al2(SO4)3
1 mol Al2 (SO 4 )3 342.14 g Al2 (SO 4 )3 2 mol Al3+ 1 mol Al2 (SO 4 )3
0.0726 mol Cl− 1 mol NiCl2 g NiCl2 = (175 mL solution) 1000 mL solution 2 mol Cl−
129.6 g NiCl2 = 0.823 g NiCl2 1 mole NiCl2
0.153 mol Na 2 CO3 1 mol NiCl2 1000 mL soln mL NiCl2 soln = 20.0 mL soln 1000 mL soln 1 mol Na 2 CO3 0.415 mol NiCl2 = 7.37 mL NiCl2 soln 0.415 mol NiCl2 1 mol NiCO3 118.7 g NiCO3 g NiCO3 = 7.37 mL NiCl2 soln = 0.363 g NiCO3 1000 mL soln 1 mol NiCl2 1 mol NiCO3
5.100
0.250 mol H 2 C4 H 4 O6 mL NaOH soln = 25.0 mL soln 1000 mL soln
100
2 mol NaOH 1 mol H 2 C4 H 4 O6
1000 mL soln 0.100 mol NaOH
Chapter 5
= 125 mL NaOH soln
( 20.78 mL HCl soln ) 5.101
M KOH =
1 L HCl soln 0.116 mol HCl 1 mol KOH 1 mol HCl 1000 mL HCl soln 1 L HCl 1 L KOH 21.34 mL KOH 1000 mL KOH
= 0.113 M KOH KOH(aq) + HCl(aq) KCl(aq) + H2O 5.102
1 L NaOH soln 0.1024 mol NaOH 1 mol H3 PO 4 mol H3PO4 = (39.43 mL NaOH) 3 mol NaOH 1 L NaOH 1000 mL NaOH soln –3 = 1.346 × 10 mol H3PO4 1.346 × 10−3 mol H PO 1000 mL H PO –2 3 4 3 4 M H3PO4 = = 6.729 × 10 M H3PO4 20.00 mL H PO 1 L H PO 3 4 3 4 H3PO4(aq) + 3NaOH(aq) 3H2O + Na3PO4(aq)
5.103
Al2(SO4)3(aq) + 3Ba(OH)2(aq) 2Al(OH)3(s) + 3BaSO4(s)
0.0500 mol Ba(OH)2 g Al2(SO4)3 = 85.0 mL soln 1000 mL soln
1 mol Al2 (SO 4 )3 342.2 g Al2 (SO 4 )3 3 mol Ba(OH) 1 mol Al (SO ) 2 2 4 3
= 0.485 g Al2(SO4)3 5.104
NaHCO3 + HCl NaCl + H2O + CO2
0.052 mol HCl 1 mol NaHCO3 84.01 g NaHCO3 g NaHCO3 = (185 mL HCl soln) 1000 mL HCl soln 1 mol HCl 1 mol NaHCO3 = 0.81 g NaHCO3
5.105
0.0625 mol AgNO3 1 mol Ag + mL FeCl3 soln = 26.45 mL AgNO3 1000 mL soln 1 mol AgNO3
1 mol Cl− 1 mol Ag +
1 mol FeCl3 1000 mL soln × = 3.67 mL FeCl3 soln 3 mol Cl− 0.150 mol FeCl3 1 L AgNO3 0.0625 mol AgNO3 3 mol AgCl g AgCl = (26.45 mL AgNO3) 1 L AgNO3 1000 mL AgNO3 3 mol AgNO3 = 0.237 g AgCl
5.106
− 2+ 0.200 mol KOH 1 mol OH 1 mol Co g CoCl2 = 60.0 mL KOH soln − 1000 mL soln 1 mol KOH 2 mol OH
1 mol CoCl2 129.8 g CoCl2 × 1 mol Co 2+ 1 mol CoCl2
101
= 0.779 g CoCl2
143.32 AgCl 1 mol AgCl
Chapter 5
5.107
Ag+ + Cl– → AgCl(s)
0.500 mol AgC2 H3O 2 mL AlCl3 = (25.0 mL AgC2H3O2) 1000 mL AgC2 H3O2
1 mol Ag + 1 mol AgC2 H3O 2
1 mol Cl− 1 mol Ag +
1 mol AlCl3 1000 mL AlCl3 × = 16.7 mL AlCl3 3 mol Cl− 0.250 moles AlCl3
5.108
− 1.00 mol NaOH 1 mol OH mL (NH4)2SO4 soln = (50.0 mL NaOH soln) 1000 mL NaOH soln 1 mol NaOH
1 mol NH + 4 × 1 mol OH − 5.109
1 mol (NH ) SO 4 2 4 2 mol NH + 4
1000 mL (NH ) SO soln 4 2 4 = 1.00 × 102 mL (NH4)2SO4 soln 0.250 mol (NH 4 )2SO 4
Fe2O3 + 6HCl 2FeCl3 + 3H2O 0.0250 L HCl × 0.500 mol/L = 1.25 × 10–2 mol HCl
1 mol Fe2 O3 2 mol Fe3+ mol Fe3+ = (1.25 × 10–2 mol HCl) = 4.17 × 10–3 mol Fe3+ 6 mol HCl 1 mol Fe 2 O3 M Fe3+ =
4.17 × 10−3 mol Fe3+ = 0.167 M Fe3+ 0.0250 L soln
1 mol Fe 2 O3 159.69 g Fe2 O3 g Fe2O3 = (4.17 × 10–3 mol Fe3+) = 0.333 g Fe2O3 2 mol Fe3+ 1 mol Fe2 O3 Therefore, the mass of Fe2O3 that remains unreacted is: (4.72 g – 0.333 g) = 4.39 g 5.110
First determine the moles of H2SO4 present
0.500 mol H 2SO 4 mol H2SO4 = (30.0 mL H2SO4) = 0.015 mol H2SO4 1000 mL H 2SO 4 Next determine the number of moles of Mg(OH)2 present: 1 mol Mg(OH) 2 mol Mg(OH)2 = (3.50 g Mg(OH)2) = 0.0600 mol Mg(OH)2 58.32 g Mg(OH)2 Thus, from the reaction, H2SO4(aq) + Mg(OH)2(s) 2H2O + MgSO4(aq), we see that 0.015 mol Mg(OH)2 will react with 0.015 mol H2SO4. This produces 0.015 mol of MgSO4(aq) in 30.0 mL of solution and leaves 0.0600 – 0.015 = 0.045 mol Mg(OH)2 unreacted. The concentration of Mg2+ is: = 0.50 M Mg2+ The number of grams of Mg(OH)2 not dissolved is:
0.015 mol MgSO 4 [Mg2+] = 0.0300 L soln
1 mol Mg 2 + 1 mol MgSO4
58.32 g Mg(OH)2 g Mg(OH)2 = (0.045 mol Mg(OH)2) = 2.6 g Mg(OH)2 1 mol Mg(OH)2 5.111
First, calculate the number of moles HCl based on the titration according to the following equation:
102
Chapter 5
NaOH(aq) + HCl(aq) NaCl(aq) + H2O
0.105 mol NaOH 1 mol HCl –3 mol HCl = (23.25 mL NaOH) = 2.44 × 10 mol HCl 1000 mL NaOH 1 mol NaOH Next, determine the concentration of the HCl solution: 2.44 × 10–3 mol ÷ 0.01975 L = 0.124 M HCl 5.112
(a)
The balanced equation for the titration is: NaOH(aq) + HC2H3O2(aq) NaC2H3O2(aq) + H2O
0.368 mol NaOH 1 mol HC2 H3O2 mol HC2H3O2 = (28.28 mL NaOH) 1000 mL NaOH 1 mol NaOH = 1.04 × 10–2 mol HC2H3O2 1.04 × 10–2 mol/0.0125 L = 0.833 M HC2H2O2 (b)
First convert the density of vinegar to a value appropriate for one liter of solution: 1.01 g/mL × 1000 mL/L = 1010 g/L
We know that one liter of this vinegar contains 0.833 mol of acetic acid so we can determine the mass of acetic acid that is present in one liter of this vinegar:
60.05 g HC2 H3O2 g HC 2 H3O 2 = ( 0.833 mol HC 2 H3O 2 ) 1 mol HC 2 H3O2 = 50.0 g HC2 H3O2 The % by weight of acetic acid in vinegar solution is then given by the following: (50.0 g HC2H3O2/L ÷ 1010 g/L) × 100 = 4.95 % acetic acid This is the mass of acetic acid in one L of solution divided by the total mass of one L of solution, multiplied by 100%. 5.113
Since lactic acid is monoprotic, it reacts with sodium hydroxide on a one to one mole basis: 0.155 mol NaOH 1 mol HC3 H5 O3 (a) mol HC3H5O3 = (17.25 mL NaOH) 1000 mL NaOH 1 mol NaOH = 2.67 × 10–3 mol HC3H5O3
5.114
90.08 g HC3H5O3 (b) g HC3H5O3 = 2.67 x 10-3 mol x = 0.240 g 1 mol HC3H5O3 Note that ascorbic acid is diprotic. 0.0200 mol NaOH 1 mol H 2 C6 H 6 O6 g H2C6H6O6 = (21.46 mL NaOH) 1000 mL NaOH 2 mol NaOH 3.78 × 10–2 g H2C6H6O6 % H2C6H6O6 =
3.78 × 10 −2 g × 100% = 24.2% 0.1565 g
103
176.13 g H 2 C6 H 6 O6 = 1 mol H 2 C6 H 6 O6
Chapter 5
5.115
1 L Na 2SO 4 0.122 mol Na 2SO 4 1 mol PbSO 4 (a) mol Pb = (29.22 mL) 1 L Na 2SO4 1000 mL Na 2SO 4 1 mol Na 2SO 4
1 mol Pb 2+ 1 mol PbSO 4
= 3.56 x 10-3 mol Pb
1 L Na 2SO 4 0.122 mol Na 2SO 4 1 mol PbSO 4 (b) g Pb = (29.22 mL) 1 L Na 2SO4 1000 mL Na 2SO 4 1 mol Na 2SO 4
1 mol Pb 2+ 1 mol PbSO 4
207.2 g Pb2+ × = 0.7386 g Pb 1 mol Pb 2+ The percentage of Pb in the sample can be calculated as 0.7386 g Pb ×100% = 48.40% Pb in the sample. 1.526 g sample
5.116
1 mol BaSO 4 1 mol Ba 137.33 g Ba g Ba = (1.204 g BaSO 4 ) = 0.7085 g Ba 233.39 g BaSO 4 1 mol BaSO 4 1 mol Ba %Ba in the original ore = (0.7085 g Ba/1.856 g ore) × 100% = 38.17%
Additional Exercises 5.117
The equation for the reaction indicates that the two materials react in equimolar amounts, i.e. the stoichiometry is 1 to 1: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) (a)
Because this reaction is 1:1, we can see by inspection that the AgNO3 is the limiting reagent. We know this because the concentration of the AgNO3 is lower than the NaCl. Since we start with equal volumes, there are fewer moles of the AgNO3.
0.375 mol AgNO3 1 mol AgCl mol AgCl = (25.0 mL AgNO3 soln) 1000 mL AgNO3 soln 1 mol AgNO3 = 9.38 × 10–3 mol AgCl (b)
Assuming that AgCl is essentially insoluble, the concentration of silver ion can be said to be zero since all of the AgNO3 reacted. The number of moles of chloride ion would be reduced by the precipitation of 9.38 × 10–3 mol AgCl, such that the final number of moles of chloride ion would be: 0.0250 L × 0.460 mol/L – 9.38 × 10–3 mol = 2.12 × 10–3 mol Cl– The final concentration of Cl– is, therefore: 2.12 × 10–3 mol ÷ 0.0500 L = 0.0424 M Cl– All of the original number of moles of NO3– and of Na+ would still be present in solution, and their concentrations would be:
104
Chapter 5
For NO3–:
0.375 mol AgNO3 1 mol NO3− (25.0 mL AgNO3 soln) 1000 mL AgNO3 soln 1 mol AgNO3 − M NO3 = 1 L soln (50.0 mL soln) 1000 mL soln
= 0.188 M NO − 3
For Na+:
M Na + =
5.118
(a)
+ 0.460 mol NaCl 1 mol Na (25.0 mL NaCl soln) 1000 mL NaCl soln 1 mol NaCl
1 L soln (50.0 mL soln) 1000 mL soln
= 0.230 M Na +
3Ca(NO3)2(aq) + 2Na3PO4(aq) Ca3(PO4)2(s) + 6NaNO3(aq) First, determine the initial number of moles of Ca2+ ion that are present:
0.140 mol Ca(NO3 )2 1 mol Ca 2+ mol Ca2+ = (38.0 mL Ca(NO3)2) 1000 mL Ca(NO3 )2 soln 1 mol Ca(NO3 )2
= 5.32 × 10–3 mol Ca2+ Next, determine the initial number of moles of phosphate ion that are present:
0.185 mol Na 3 PO 4 1 mol PO43− mol PO43– = (35.0 mL Na3PO4) 1000 mL Na 3 PO 4 soln 1 mol Na 3 PO4
= 6.48 × 10–3 mol PO43– Now determine the number of moles of calcium ion that are required to react with this much phosphate ion, and compare the result to the amount of calcium ion that is available:
3 mol Ca 2+ = 9.72 × 10–3 mol Ca2+ mol Ca2+ = (6.48 × 10–3 mol PO43–) 2 mol PO 3− 4 Since there is not this much Ca2+ available according to the above calculation, then we can conclude that Ca2+ must be the limiting reagent, and that subsequent calculations should be based on the number of moles of it that are present: 1 mol Ca 3 (PO 4 )2 310.2 g Ca 3 (PO 4 )2 g Ca3(PO4)2 = (5.32 × 10–3 mol Ca2+) 3 mol Ca 2+ 1 mol Ca 3 (PO 4 )2 = 0.550 g Ca3(PO4)2 (b)
If we assume that the Ca3(PO4)2 is completely insoluble, then its concentration may be said to be essentially zero. The concentrations of the other ions are determined as follows: For nitrate:
M NO3− =
0.140 mol Ca(NO3 )2 2 mol NO3− 3 ) 2 soln 1 mol Ca(NO3 ) 2
( 38.0 mL Ca(NO3 )2 soln ) 1000 mL Ca(NO ( (38.0 + 35.0) mL soln) )
1 L soln 1000 mL soln
= 0.146 M NO3−
105
Chapter 5
For Na+:
M Na + =
0.185 mol Na 3 PO 4 3 mol Na + 3 PO 4 soln 1 mol Na 3 PO 4
( 35.0 mL Na 3PO4 soln ) 1000 mL Na
( (38.0 + 35.0) mL soln) )
1 L soln 1000 mL soln
= 0.266 M Na+ For phosphate, we determine the number of moles that react with calcium:
2 mol PO 3− 4 = 3.55 × 10–3 mol PO 3– mol PO43– = (5.32 × 10–3 mol Ca2+) 4 3 mol Ca 2+ and subtract from the original number of moles that were present: mol PO43– = 6.48 × 10–3 mol PO43– – 3.55 × 10–3 mol PO43– = 2.93 × 10–3 mol PO43– This allows a calculation of the final phosphate concentration: M PO 43− =
5.119
5.120
5.121
(a) (b) (c) (d)
2.93 × 10−3 mol PO 43− = 0.0401 M PO43– 1 L soln ( (38.0 + 35.0) mL soln) ) 1000 mL soln
strong electrolyte nonelectrolyte strong electrolyte nonelectrolyte
(e) (f) (g) (h)
weak electrolyte nonelectrolyte strong electrolyte weak electrolyte
(a)
molecular: ionic: net ionic:
CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + H2O + CO2(g) CaCO3(s) + 2H+(aq) + 2NO3–(aq) Ca2+(aq) + 2NO3–(aq) + H2O + CO2(g) CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g)
(b)
molecular: ionic: net ionic:
CaCO3(s) + H2SO4(aq) CaSO4(s) + H2O + CO2(g) CaCO3(s) + 2H+(aq) + SO42–(aq) Ca2+(aq) + SO42–(aq) + H2O + CO2(g) CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g)
(c)
molecular: ionic net ionic:
FeS(s) + 2HBr(aq) FeBr2(aq) + H2S(g) FeS(s) + 2H+(aq) + 2Br–(aq) Fe2+(aq) + 2Br–(aq) + H2S(g) FeS(s) + 2H+(aq) Fe2+(aq) + H2S(g)
(d)
molecular: ionic: net ionic:
2KOH(aq) + SnCl2(aq) 2KCl(aq) + Sn(OH)2(s) 2K+(aq) + 2OH–(aq) + Sn2+(aq) + 2Cl–(aq) 2K+(aq) + 2Cl–(aq) + Sn(OH)2(s) 2OH–(aq) + Sn2+(aq) Sn(OH)2(s)
0.0300 mol KOH 1 mol HC9 H 7 O 4 180.16 g HC9 H 7 O 4 g HC9H7O4 = 29.40 mL KOH 1000 mL KOH 1 mol KOH 1 mol HC9 H 7 O 4 = 0.1589 g HC9H7O4
106
Chapter 5
Percentage by weight =
5.122
0.1589 g HC9 H 7 O4 = 63.56% aspirin in the sample 0.250 g sample
(a)
3Ba2+(aq) + 2Al3+(aq) + 6OH–(aq) + 3SO42–(aq) 3BaSO4(s) + 2Al(OH)3(s)
(b)
Because we know the amounts of both starting materials this is a limiting reactant problem. So start by assuming that the barium hydroxide is the limiting reactant.
0.270 mol Ba(OH)2 g BaSO4 = (45.0 mL Ba(OH) 2 ) 1000 mL Ba(OH)2
1 mol Ba 2 + 1 mol Ba(OH)2
1 mol BaSO 4 233.39 g BaSO 4 = 2.84 g BaSO 4 1 mol Ba 2+ 1 mol BaSO4 Now assume Al2(SO4)3 is the limiting reactant.
0.330 mol Al2 (SO 4 )3 3 mol SO 42 − g BaSO4 = (28.0 mL Al2 (SO 4 )3 ) 1000 mL Al2 (SO 4 )3 1 mol Al2 (SO4 )3 1 mol BaSO 4 1 mol SO 2− 4
233.39 g BaSO 4 = 6.47 g BaSO 4 1 mol BaSO 4
Therefore the barium hydroxide is the limiting reactant. Now we can calculate the mass of aluminum hydroxide that is produced.
0.270 mol Ba(OH) 2 g Al(OH)3 = (45.0 mL Ba(OH) 2 ) 1000 mL Ba(OH)2
2 mol OH − 1 mol Ba(OH)2
1 mol Al(OH)3 78.00 g Al(OH)3 = 0.632 g Al(OH)3 3 mol OH − 1 mol Al(OH)3 The total mass of the precipitate is 2.84 g + 0.632 g = 3.47 g (c)
All of the barium ion and hydroxide ion are reacted so the concentration of each is 0. We started with the following: 0.330 mol Al2 (SO 4 )3 2 mol Al3+ mol Al3+ = (28.0mL Al2 (SO 4 )3 ) 1000 mL Al2 (SO 4 )3 1 mol Al2 (SO 4 )3
= 1.85 × 10−2 moles Al3+ 0.330 mol Al2 (SO 4 )3 3 mol SO 42 − mol SO4 2 − = (28.0 mL Al2 (SO4 )3 ) 1000 mL Al2 (SO 4 )3 1 mol Al2 (SO 4 )3 = 2.77 × 10−2 moles SO 4 2− In precipitating the Al(OH)3 above, we used 8.1 × 10–3 mol Al leaving (1.85 × 10–2 – 8.1 × 10–3) = 1.0 × 10–2 mol Al3+ in solution, i.e., the resulting concentration of Al3+ is 1.0 × 10–2 mol / (0.0450 + 0.0280) L = 0.137 M Al3+.
107
Chapter 5
Similarly for SO42–, the concentration of SO42– remaining
2.77 × 10−2 mol − 1.22 × 10−2 mol = 0.212 M SO 4 2− (0.0450 + 0.0280) L where 1.22 × 10–2 mol SO42– represents the amount precipitated as BaSO4.
5.123
Since the number of moles in the final solution must be equal to the number of moles contributed by both solutions, the equation MfVf = MiVi may be used, and the volumes of the final solution must equal the volumes of the two solution combined. Vf = V1 + V2 Vf = 30.0 mL + V2
0.25 mol 0.10 mol 0.45 mol 1000 mL ( Vf ) = 1000 mL ( V2 ) + 1000 mL ( 30 mL ) 0.25 mol 0.10 mol 0.45 mol 1000 mL ( 30 mL + V2 ) = 1000 mL ( V2 ) + 1000 mL ( 30 mL ) 0.25 mol 0.25 mol 0.10 mol 0.45 mol 1000 mL ( 30 mL ) + 1000 mL ( V2 ) = 1000 mL ( V2 ) + 1000 mL ( 30 mL ) 0.45 mol 0.25 mol 0.25 mol 0.10 mol 1000 mL ( 30 mL ) - 1000 mL ( 30 mL ) = 1000 mL ( V2 ) − 1000 mL ( V2 ) multiply through by 1000 mL (0.45 mol)(30 mL) – (0.25 mol)(30 mL) = (0.25 mol)(V2) – (0.10 mol)(V2) 6.0 mol mL = (0.15 mol)V2 (6.0 mol mL) ÷ (0.15 mol) = V2 V2 = 40 mL 5.124 +
−
Na ( aq ) + OH ( aq ) +
+
−
+ (CH 3 ) 2 NH 2 (aq ) + Cl ( aq )
−
→ Na ( aq ) + Cl ( aq ) + (CH 3 ) 2 NH(aq ) + H 2 O(l )
net ionic equation: −
OH ( aq )
+
+ (CH 3 ) 2 NH 2 (aq ) → (CH 3 ) 2 NH(aq ) + H 2 O(l )
12.4 g (CH 3 ) 2 NH 2 Cl x
1 mol (CH 3 ) 2 NH 2 Cl 81.545 g
1 mol NaOH
x
1 mol (CH 3 ) 2 NH 2 Cl
6.08 g NaOH required
108
x
39.997 g mol NaOH
=
Chapter 6
Practice Exercises 6.1
6.2
6.3
2Na(s) + O2(g) Na2O2(s) Oxygen is reduced since it gains electrons. Sodium is oxidized since it loses electrons. 2Al(s) + 3Cl2(g) 2AlCl3(aq) Aluminum is oxidized and is, therefore, the reducing agent. Chlorine is reduced and is, therefore, the oxidizing agent. Fe2O3(s) + 2Al(s) 2Fe(s) + Al2O3(s) Fe2O3 is reduced and is, therefore, the oxiding agent. Al is oxidized and is, therefore, the reducing agent.
6.4
ClO2–: O –2 Cl +3
6.5
(a) Ni +2; Cl –1 (b) Mg +2; Ti +4; O –2 (c) K +1; Cr +6; O –2 (d) H +1; P +5, O –2 (e) V +3; C 0; H +1; O –2 (f) N -3; H +1
6.6
There is a total charge of +8, divided over three atoms, so the average charge is +8/3.
6.7
(a) Mo +3; Cl -1 (b) Mo +4; S -2 (c) Mo +6; O -2, Cl -1 (d) Mo +6; P -3
6.8
First the oxidation numbers of all atoms must be found. N2O5 + 3NaHCO3 2NaNO3 + 2CO2 + H2O Reactants: N = +5 O = –2
Products: N = +5 O = –2 Na = +1
Na = +1 C = +4 H = +1 O = –2 C = +4 O = –2 H = +1 O = –2 None of the oxidation numbers change, therefore it is not an oxidation reaction. KClO3 + 3HNO2 KCl + 3HNO3 Reactants: K = +1
Products: K = +1
109
Chapter 6
Cl = +5 O = –2
Cl = –1 O = -2
H = +1 N = +3 O = –2
H = +1 N = +5 O = –2
The oxidation numbers for K and Na do not change. However, the oxidation numbers for the chlorine atom decreases. The oxidation numbers for nitrogen increase.
6.9
Therefore, KClO3 is reduced and HNO2 is oxidized. This means KClO3 is the oxidizing agent and HNO2 is the reducing agent. This reaction is the redox reaction. In the other reaction, the oxidation numbers of the atoms do not change. First the oxidation numbers of all atoms must be found. Cl2 + 2NaClO2 2ClO2 + 2NaCl Reactants: Cl = 0
Products: Cl = +4 O = –2
Na = +1 Cl = +3 O = –2
Na = +1 Cl = –1
The oxidation numbers for O and Na do not change. However, the oxidation numbers for all chlorine atoms change. There is no simple way to tell which chlorines are reduced and which are oxidized in this reaction. One analysis would have the Cl in Cl2 end up as the Cl in NaCl, while the Cl in NaClO2 ends up as the Cl in ClO2. In this case Cl2 is reduced and is the oxidizing agent, while NaClO2 is oxidized and is the reducing agent. 6.10
6.11
If H2O2 acts as an oxidizing agent, it gets reduced itself in the process. Examining the oxidation numbers: H 2O 2 H = +1, O = –1 H 2O H = +1, O = –2 O2 O=0 If H2O2 is reduced it must form water, since the oxidation number of oxygen drops from –1 to –2 in the formation of water (a reduction).
The product is therefore water. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) The oxidation numbers for the atoms must be determined. Reactants Products C = -4 C = +4 H = +1 H = +1 O=0 O = -2 110
Chapter 6
The oxidation number for C increases and the oxidation number for O decreases. Therefore, CH4 undergoes oxidation and O2 undergoes reduction. This means that CH4 is the reducing agent and O2 is the oxidizing agent. 6.12
Al(s) + Cu2+(aq) Al3+(aq) + Cu(s) First, we break the reaction above into half-reactions: Al(s) Al3+(aq) Cu2+(aq) Cu(s) Each half-reaction is already balanced with respect to atoms, so next we add electrons to balance the charges on both sides of the equations: Al(s) Al3+(aq) + 3e– 2e– + Cu2+(aq) Cu(s) Next, we multiply both equations so that the electrons gained equals the electrons lost, 2(Al(s) Al3+(aq) + 3e–) 3(2e– + Cu2+(aq) Cu(s)) which gives us: 2Al(s) 2Al3+(aq) + 6e– 6e– + 3Cu2+(aq) 3Cu(s) Now, by adding the half-reactions back together, we have our balanced equation: 2Al(s) + 3Cu2+(aq) 2Al3+(aq) + 3Cu(s)
6.13
TcO4− + Sn2+ Tc4+ + Sn4+ First, we break the reaction above into half-reactions: TcO4– Tc4+ Sn2+ Sn4+ Each half-reaction is already balanced with respect to atoms other than O and H, so next we balance the O atoms by using water: TcO4– Tc4+ + 4H2O Sn2+ Sn4+ + Now we balance H by using H : 8H+ + TcO4– Tc4+ + 4H2O Sn2+ Sn4+ Next, we add electrons to balance the charges on both sides of the equations: 3e– + 8H+ + TcO4– Tc4+ + 4H2O Sn2+ Sn4+ + 2e– We multiply the equations so that the electrons gained equals the electrons lost, 2(3e– + 8H+ + TcO4– Tc4+ + 4H2O) 3(Sn2+ Sn4+ + 2e–) which gives us: 6e– + 16H+ + 2TcO4– 2Tc4+ + 8H2O 3Sn2+ 3Sn4+ + 6e– Now, by adding the half-reactions back together, we have our balanced equation: 3Sn2+ + 16H+ + 2TcO4– 2Tc4+ + 8H2O + 3Sn4+
111
Chapter 6
6.14
6.15
(Cu Cu2+ + 2e–) × 4 2NO3– + 10H+ + 8e– N2O + 5H2O 4Cu + 2NO3– + 10H+ 4Cu2+ + N2O +5H2O (C5H5N + 12H2O → NO2 + 5CO2 + 29H+ + 29 e-) x 4 + (O2 + 4H + 4 e → 2H2O) x 29 4NO2 + 20CO2 + 10H2O 4C5H5N + 29O2 → Since we have water on both sides we need to clean up the reaction so water is only on one side of the reaction. Subtracting 48H2O from both sides gives us: 4C5H5N + 29O2 → 4NO2 + 20CO2 + 10H2O
6.16
2H2O + SO2 SO42– + 4H+ + 2e– 4OH– + 2H2O + SO2 SO42– + 4H+ + 2e– + 4OH– 4OH– + 2H2O + SO2 SO42– + 2e– + 4H2O 4OH– + SO2 SO42– + 2e– + 2H2O
6.17
(MnO4– + 4H+ + 3e– MnO2 + 2H2O) × 2 (C2O42– + 2H2O 2CO32– + 4H+ + 2e–) × 3 2MnO4– + 3C2O42– + 2H2O 2MnO2 + 6CO32– + 4H+ Adding 4OH– to both sides of the above equation we get: 2MnO4– + 3C2O42– + 2H2O + 4OH– 2MnO2 + 6CO32– + 4H2O which simplifies to give: 2MnO4– + 3C2O42– + 4OH– 2MnO2 + 6CO32– + 2H2O First balance in acidic solution using H2O to balance oxygen atoms and H+ to balance hydrogen atoms. (Br2 + 2e→ 2Br- ) x5 Br2 + 6H2O → 2BrO3 + 12 H+ + 10e6Br2 + 6H2O → 10Br- + 2BrO3- + 12H+
6.18
Then, to balance the reaction in base use the above balanced reaction in acid and add 12OH- to each side. 6Br2 + 6H2O + 12OH- → 10Br- + 2BrO3- + 12H+ + 12OH6Br2 + 6H2O + 12OH- → 10Br - + 2BrO3- + 12H2O 6Br2 + 12OH → 10Br + 2BrO3- + 6H2O 6.19
Zn + H2O → ZnO + 2H+ + 2e2MnO2 + 2H+ + 2e→ Mn2O3 + H2O Zn + 2MnO2 → ZnO + Mn2O3 This reaction is balanced and does not contain either H+ or OH- so it is correctly balanced for a basic solution and for an acidic solution.
6.20
Zn + H+ Zn2+ + H2 Divide the reaction into two half reactions and balance the number of atoms Zn Zn2+ 2H+ H2 Balance the charges with electrons Zn Zn2+ + 2e– 2H+ + 2e– H2
6.21
(a)
molecular: Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) ionic: Mg(s) + 2H+(aq) + 2Cl–(aq) Mg2+(aq) + 2Cl–(aq) + H2(g) net ionic: Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) 112
Chapter 6
molecular: 2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) ionic: 2Al(s) + 6H+(aq) + 6Cl–(aq) 2Al3+(aq) + 6Cl–(aq) + 3H2(g) net ionic: 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g) → H2S + 4H2O H2SO4 + 8H+ + 8e+ (2HI → I2 + 2H + 2e ) x 4 H2SO4 + 8HI → H2S + 4I2 + 4H2O (b)
6.22
6.23
Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq)
6.24
(a) 2Al(s) + 3Cu2+(aq) 2Al3+(aq) + 3Cu(s) (b) Ag(s) + Mg2+(aq) No reaction
6.25 6.26 6.27 6.28 6.29
2C20H42(s) + 21O2(g) 40C(s) + 42H2O(g) 2C5H8(g) + 9O2(g) 10CO2(g) + 8H2O(g) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) 2Sr(s) + O2(g) 2SrO(s) 4Fe(s) + 3O2(g) 2Fe2O3(s)
Review Questions 6.1
(a) (b)
Oxidation is the loss of one or more electrons. Reduction is the gain of one or more electrons. The oxidation number decreases in a reduction and increases in an oxidation.
6.2
The number of electrons involved in both the reduction and the oxidation must be the same; only those electrons that come from the reductant and go to the oxidant are involved. No electrons from external or uninvolved sources are allowed to enter the process, and there cannot be any electrons left unreacted at the end of the process. An oxidizing agent is the species that is reduced or gains electrons in an oxidation reduction reaction. A reducing agent is the species that is oxidized or loses electrons in an oxidation reduction reaction.
6.3
Oxidation state and oxidation number are synonyms. Thus, the As is in the +3 oxidation state.
6.4
The oxidation state of nitrogen is +4 in both reactants and products. The oxygen atoms are -2 in the reactants and products. Thus, the reaction is not a redox reaction. The oxidation state of Cr is +6 in both reactants and products. No other element changes oxidation state. Therefore this is not a redox reaction.
6.5
This change in oxidation number represents a reduction of nitrogen by 5 units, and it requires that nitrogen gain 5 electrons.
6.6
The equation is not balanced since the charge is different on either side of the arrow. It is easily balanced by inspection to give: 2Ag+ + Fe 2Ag + Fe2+
113
Chapter 6
6.7
(a) (b)
+9 charge on the left, +1 charge on the right; add 8 electrons to the left side. 0 charge on the left, +6 charge on the right; add 6 electrons to the right side.
6.8
(a) (b)
is a reduction is an oxidation
6.9
It is a reaction in which one element replaces another in a compound.
6.10
A "nonoxidizing acid" is one in which the H+ ion is the strongest oxidizing agent. That is, the anion of the acid is not itself a better oxidizing agent than H+. Examples are HCl and H2SO4. The oxidizing agent in a nonoxidizing acid is the H+.
6.11
NO3–(aq)
6.12
The metal must be below hydrogen in the activity series in order for it to react with HCl.
6.13
The best reducing agents are those that are most easily oxidized and are found at the bottom of the activity series. The best oxidizing agents are those that are most easily reduced and are found at the top of the activity series. (See Table 6.3 for clarification.)
6.14
This would be any metal higher (less reactive) than hydrogen, i.e. gold, mercury, silver, and copper.
6.15
The most active types of metals will react with water for example: 2Cs(s) + 2H2O 2CsOH(aq) + H2(g) 2Rb(s) + 2H2O 2RbOH(aq) + H2(g) Ba(s) + 2H2O Ba(OH)2(aq) + H2(g) Up to Pb(s) + 2H2O Pb(OH)2(aq) + H2(g)
6.16
Based on the activity series, the manganese is oxidized and therefore is the reducing agent.
6.17
Combustion is the rapid reaction of a substance with oxygen, which is accompanied by the evolution of light and heat.
6.18
Historically, the reaction of a substance with oxygen was termed oxidation. Now we realize that reaction with oxygen most typically means that oxygen acquires electrons from the substance with which it reacts. The oxidation of a substance is, therefore, taken to represent the loss of electrons by a substance, whether the substance has reacted with oxygen or with another oxidizing agent.
6.19
a) CO2(g) and H2O(g) b) CO(g), CO2(g) and H2O(g) c) C(s) and H2O(g) The products will be CO2, H2O, and SO2: 2C2H6S(l)+ 9O2(g) 4CO2(g) + 2SO2(g) + 6H2O(g)
6.20
6.21
The other product is water: 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)
114
Chapter 6
6.22
The reaction of sulfur with air produced sulfur dioxide, SO2 gas. The limiting reagent is sulfur. Air contains approximately 20 % O2 and there is an excess of air present in combustion.
Review Problems 6.23 The sum of the oxidation numbers should be equal to the total charge: (a) Se: –2 (b) S: +6, O: –2 (c) S: 0 (d) As: +3; H: –1 6.24
The sum of the oxidation numbers should be equal to the total charge: (a) ClO4–: Cl +7, O –2 (b) Cl–: Cl –1 (c) SF6: S +6, F –1 (d) Au(NO3)3: Au +3, N +5,O –2
6.25
The sum of the oxidation numbers should be equal to the total charge: (a) K: +1 (c) K: +1 O: –2 O: –2 Br: +1 Br: +5 (b) K: +1 (d) K: +1 O: –2 O: –2 Br: +3 Br: +7 The sum of the oxidation numbers should be equal to the total charge.
6.26
6.27
6.28
(a)
Cl: –1 Mn: +2
(c)
O: –2 Mn: +6
(b)
O: –2 Mn: +7
(d)
O: –2 Mn: +4
The sum of the oxidation numbers should be zero: (a) S: +6 (c) Ca: +2 O: –1 Pb: +2 O: –2 (b) Cl: –1 (d) S: +2 Zr: +4 Cl: –1 (a) Sr: +2 (c) O: +2 O: –2 F: –1 I: +5 (b) Cr: +3 (d) H: +1 S: –2 F: –1 O: 0
6.29
Ti +3; N -3
6.30
Zr is in the +4 oxidation state and O is -2.
115
Chapter 6
6.31
O3; oxidation number of O is 0
6.32
The free radical OH is a molecular species. Thus, oxidations states of H and O must add up to zero. Rule 4 in the hierarchy for assigning oxidation numbers is that H is +1. Rule 5 assigns an oxidation number of -2 to O. Since the oxidations numbers do not add to aero we have a conflict. That means rule 4 takes precedence and O must then have an oxidation number of -1 in OH.
6.33
Cl2(aq) + H2O H+(aq) + Cl–(aq) + HOCl(aq) In the forward direction: The oxidation number of the chlorine atoms decreases from 0 to –1. Therefore Cl2 is reduced. However, in HOCl, chlorine has an oxidation number of +1, so Cl2 also oxidized! (One atom is reduced, the other is oxidized.) In the reverse direction: The Cl– ion begins with an oxidation number of –1 and ends with an oxidation number of 0. Therefore the Cl– ion is oxidized: This means Cl– is the reducing agent. Since the oxidation number of H+ does not change, HOCl must be the oxidizing agent.
6.34
N is both reduced and oxidized. N is reduced in the conversion of NO2 to NO, a two electron reduction −
step, and N is oxidized on the conversion of NO2 to NO 3 , a one electron oxidation step. Thus, NO2 is both the oxidizing agent and the reducing agent. 6.35
(a) (b) (c) (d)
6.36
(a) (b) (c) (d)
6.37
substance reduced (and oxidizing agent): HNO3 substance oxidized (and reducing agent): H3AsO3 substance reduced (and oxidizing agent): HOCl substance oxidized (and reducing agent): NaI substance reduced (and oxidizing agent): KMnO4 substance oxidized (and reducing agent): H2C2O4 substance reduced (and oxidizing agent): H2SO4 substance oxidized (and reducing agent): Al substance reduced (and oxidizing agent): H2SO4 substance oxidized (and reducing agent): Cu substance reduced (and oxidizing agent): HNO3 substance oxidized (and reducing agent): SO2 substance reduced (and oxidizing agent): H2SO4 substance oxidized (and reducing agent): Zn substance reduced (and oxidizing agent): HNO3 substance oxidized (and reducing agent): I2
(a)
2S2O32– S4O62– + 2e– OCl– + 2H+ + 2e– Cl– + H2O OCl– + 2S2O32– + 2H+ S4O62– + Cl– + H2O
(b)
(NO3– + 2H+ + e– NO2 + H2O) × 2 Cu Cu2+ + 2e– 2NO3– + Cu + 4H+ 2NO2 + Cu2+ + 2H2O 116
Chapter 6
(c)
IO3– + 6H+ + 6e– I– + 3H2O (H2O + H3AsO3 H3AsO4 + 2H+ + 2e–) × 3 IO3– + 3H3AsO3 + 6H+ + 3H2O I– + 3H3AsO4 + 3H2O + 6H+ which simplifies to give: 3H3AsO3 + IO3– I– + 3H3AsO4
(d)
SO42– + 4H+ + 2e– SO2 + 2H2O Zn Zn2+ + 2e– Zn + SO42– + 4H+ Zn2+ + SO2 + 2H2O
(e)
NO3– + 10H+ + 8e– NH4+ + 3H2O (Zn Zn2+ + 2e–) × 4 NO3– + 4Zn + 10H+ 4Zn2+ + NH4+ + 3H2O
(f)
2Cr3+ + 7H2O Cr2O72– + 14H+ + 6e– (BiO3– + 6H+ + 2e– Bi3+ + 3H2O) × 3 2Cr3+ + 3BiO3– + 18H+ + 7H2O Cr2O72– + 14H+ + 3Bi3+ + 9H2O which simplifies to give: 2Cr3+ + 3BiO3– + 4H+ Cr2O72– + 3Bi3+ + 2H2O
(g)
I2 + 6H2O 2IO3– + 12H+ + 10e– (OCl– + 2H+ + 2e– Cl– + H2O) × 5 I2 + 5OCl– + H2O 2IO3– + 5Cl– + 2H+
(h)
(Mn2+ + 4H2O MnO4– + 8H+ + 5e–) × 2 (BiO3– + 6H+ + 2e– Bi3+ + 3H2O) × 5 2Mn2+ + 5BiO3– + 30H+ + 8H2O 2MnO4– + 5Bi3+ + 16H+ + 15H2O which simplifies to: 2Mn2+ + 5BiO3– + 14H+ 2MnO4– + 5Bi3+ + 7H2O (H3AsO3 + H2O H3AsO4 + 2H+ + 2e–) × 3 Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O 3H3AsO3 + Cr2O72– + 3H2O + 14H+ 3H3AsO4 + 2Cr3+ + 6H+ + 7H2O which simplifies to give: 3H3AsO3 + Cr2O72– + 8H+ 3H3AsO4 + 2Cr3+ + 4H2O
(i)
6.38
(a)
(Sn + 2H2O SnO2 + 4H+ + 4e–) × 3 (NO3– + 4H+ + 3e– NO + 2H2O) × 4 3Sn + 4NO3– + 16H+ + 6H2O 3SnO2 + 12H+ + 4NO + 8H2O which simplifies to: 3Sn + 4NO3– + 4H+ 3SnO2 + 4NO + 2H2O
(b)
PbO2 + 2Cl– + 4H+ + 2e– PbCl2 + 2H2O 2Cl– Cl2 + 2e– PbO2 + 4Cl– + 4H+ PbCl2 + Cl2 + 2H2O
(c)
Ag Ag+ + e– NO3– + 2H+ + e– NO2 + H2O Ag + 2H+ + NO3– Ag+ + NO2 + H2O 117
Chapter 6
6.39
(d)
(Fe3+ + e– Fe2+) × 4 2NH3OH+ N2O + H2O + 6H+ + 4e– 4Fe3+ + 2NH3OH+ 4Fe2+ + N2O + 6H+ + H2O
(e)
2I– I2 + 2e– (HNO2 + H+ + e– NO + H2O) × 2 2I– + 2HNO2 + 2H+ I2 + 2NO + 2H2O
(f)
H2C2O4 2CO2 + 2H+ + 2e– (HNO2 + H+ + e– NO + H2O) × 2 H2C2O4 + 2HNO2 2CO2 + 2NO + 2H2O
(g)
(HNO2 + H2O NO3– + 3H+ + 2e–) × 5 (MnO4– + 8H+ + 5e– Mn2+ + 4H2O) × 2 5HNO2 + 2MnO4– + 16H+ + 5H2O 5NO3– + 2Mn2+ + 15H+ + 8H2O which simplifies to give: 5HNO2 + 2MnO4– + H+ 5NO3– + 2Mn2+ + 3H2O
(h)
(H3PO2 + 2H2O H3PO4 + 4H+ + 4e–) × 3 (Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O) × 2 3H3PO2 + 2Cr2O72– + 28H+ + 6H2O 3H3PO4 + 4Cr3+ + 14H2O + 12H+ which simplifies to: 3H3PO2 + 2Cr2O72– + 16H+ 3H3PO4 + 4Cr3+ + 8H2O
(i)
XeF2 + 2e– Xe + 2F– 2Cl– Cl2 + 2e– XeF2 + 2Cl– Xe + Cl2 + 2F–
For redox reactions in basic solution, we proceed to balance the half reactions as if they were in acid solution, and then add enough OH– to each side of the resulting equation in order to neutralize (titrate) all of the H+. This gives a corresponding amount of water (H+ + OH– → H2O) on one side of the equation, and an excess of OH– on the other side of the equation, as befits a reaction in basic solution. (a) (CrO42– + 4H+ + 3e– CrO2– + 2H2O) × 2 (S2– S + 2e–) × 3 2CrO42– + 3S2– + 8H+ 2CrO2– + 3S + 4H2O Adding 8OH– to both sides of the above equation we obtain: 2CrO42– + 3S2– + 8H2O 2CrO2– + 8OH– + 3S + 4H2O which simplifies to: 2CrO42– + 3S2– + 4H2O 2CrO2– + 3S + 8OH– (b) (C2O42– 2CO2 + 2e–) × 3 (MnO4– + 4H+ + 3e– MnO2 + 2H2O) × 2 3C2O42– + 2MnO4– + 8H+ 6CO2 + 2MnO2 + 4H2O Adding 8OH– to both sides of the above equation we get: 3C2O42– + 2MnO4– + 8H2O 6CO2 + 2MnO2 + 4H2O + 8OH– which simplifies to give: 3C2O42– + 2MnO4– + 4H2O 6CO2 + 2MnO2 + 8OH–
118
Chapter 6
(c)
(d)
(e)
6.40
(ClO3– + 6H+ + 6e– Cl– + 3H2O) × 4 (N2H4 + 2H2O 2NO + 8H+ + 8e–) × 3 4ClO3– + 3N2H4 + 24H+ + 6H2O 4Cl– + 6NO + 12H2O + 24H+ which needs no OH–, because it simplifies directly to: 4ClO3– + 3N2H4 4Cl– + 6NO + 6H2O NiO2 + 2H+ + 2e– Ni(OH)2 2Mn(OH)2 Mn2O3 + H2O + 2H+ + 2e– NiO2 + 2Mn(OH)2 Ni(OH)2 + Mn2O3 + H2O (SO32– + H2O SO42– + 2H+ + 2e–) × 3 (MnO4– + 4H+ + 3e– MnO2 + 2H2O) × 2 3SO32– + 3H2O + 8H+ + 2MnO4– 3SO42– + 6H+ + 2MnO2 + 4H2O Adding 8OH– to both sides of the equation we obtain: 3SO32– + 11H2O + 2MnO4– 3SO42– + 10H2O + 2MnO2 + 2OH– which simplifies to: 3SO32– + 2MnO4– + H2O 3SO42– + 2MnO2 + 2OH–
(a)
(CrO2– + 2H2O CrO42– + 4H+ + 3e–) × 2 (S2O82– + 2e– 2SO42–) × 3 3S2O82– + 2CrO2– + 4H2O 2CrO42– + 6SO42– + 8H+ Adding 8OH– to both sides of this equation: 3S2O82– + 2CrO2– + 4H2O + 8OH– 2CrO42– + 6SO42– + 8H2O which simplifies to give: 3S2O82– + 2CrO2– + 8OH– 2CrO42– + 6SO42– + 4H2O
(b)
(SO32– + H2O SO42– + 2H+ + 2e–) × 3 (CrO42– + 4H+ + 3e– CrO2– + 2H2O) × 2 3SO32– + 2CrO42– + 8H+ + 3H2O 3SO42– + 2CrO2– + 6H+ + 4H2O Adding 8OH– to both sides of the equation we get: 3SO32– + 2CrO42– + 11H2O 3SO42– + 2CrO2– + 2OH– + 10H2O which simplifies to: 3SO32– + 2CrO42– + H2O 3SO42– + 2CrO2– + 2OH–
(c)
(O2 + 2H+ + 2e– H2O2) × 2 N2H4 N2 + 4H+ + 4e– 2O2 + 4H+ + N2H4 2H2O2 + N2 + 4H+ which simplifies to: 2O2 + N2H4 2H2O2 + N2
(d)
(Fe(OH)2 + OH– Fe(OH)3 + e–) × 4 O2 + 2H2O + 4e– 4OH– 4Fe(OH)2 + O2 + 4OH– + 2H2O 4Fe(OH)3 + 4OH– which simplifies to: 4Fe(OH)2 + O2 + 2H2O 4Fe(OH)3
(e)
(Au + 4CN– Au(CN)4– + 3e–) × 4 (O2 + 2H2O + 4e– 4OH–) × 3 4Au + 16CN– + 3O2 + 6H2O 4Au(CN)4– + 12OH–
119
Chapter 6
6.41
6.42 6.43
NO3- + 10H+ + 8e→ NH4+ + 3H2O 2+ (Mg → Mg + 2e ) x 4 10H+(aq) + NO3-(aq) + 4Mg(s) → NH4+(aq) + 4Mg2+(aq) + 3H2O(aq) 2I– + Cl2 I2 + 2Cl– – + – (OCl + 2H + 2e Cl– + H2O) × 4 S2O32– + 5H2O 2SO42– + 10H+ + 8e– 4OCl– + S2O32– + 5H2O + 8H+ 4Cl– + 2SO42– + 10H+ + 4H2O which simplifies to: 4OCl– + S2O32– + H2O 4Cl– + 2SO42– + 2H+
6.44
(H2C2O4 2CO2 + 2H+ + 2e–) × 3 K2Cr2O7 + 14H+ + 6e– 2K+ + 2Cr3+ + 7H2O 3H2C2O4 + K2Cr2O7 + 14H+ 6CO2 + 2K+ + 2Cr3+ + 6H+ + 7H2O which simplifies to: 3H2C2O4 + K2Cr2O7 + 8H+ 6CO2 + 2K+ + 2Cr3+ + 7H2O
6.45
O3 + 6H+ + 6e– 3H2O Br– + 3H2O BrO3– + 6H+ + 6e– O3 + Br– + 3H2O + 6H+ BrO3– + 3H2O + 6H+ which simplifies to: O3 + Br– BrO3–
6.46
(Cl2 + 2e– 2Cl–) × 4 S2O32– + 5H2O → 2SO42– + 10H+ + 8e– 4Cl2 + S2O32– + 5H2O 8Cl– + 2SO42– + 10H+
6.47
(a)
M: Mn(s) + 2HCl(aq) MnCl2(aq) + H2(g) I: Mn(s) + 2H+(aq)+ 2Cl–(aq) Mn2+(aq) + 2Cl–(aq) + H2(g) NI: Mn(s) + 2H+(aq) Mn2+(aq) + H2(g)
(b)
M: Cd(s) + 2HCl(aq) CdCl2(aq) + H2(g) I: Cd(s) + 2H+(aq) + 2Cl–(aq) Cd2+(aq) +2Cl–(aq) + H2(g) NI: Cd(s) + 2H+(aq) Cd2+(aq) + H2(g)
(c)
M: Sn(s) + 2HCl(aq) SnCl2(aq) + H2(g) I: Sn(s) + 2H+(aq) + 2Cl–(aq) Sn2+(aq) + 2Cl–(aq) + H2(g) NI: Sn(s) + 2H+(aq) Sn2+(aq) + H2(g)
(a)
M: Ni(s) + H2SO4(aq) NiSO4(aq) + H2(g) I: Ni(s) + 2H+(aq)+ SO42–(aq) Ni2+(aq) + SO42–(aq) + H2(g) NI: Ni(s) + 2H+(aq) Ni2+(aq)+ H2(g)
(b)
M: 2Cr(s) + 3H2SO4(aq) Cr2(SO4)3(aq) + 3H2(g) I: 2Cr(s) + 6H+(aq) + 3SO42–(aq) 2Cr3+(aq) + 3SO42–(aq) + 3H2(g) NI: 2Cr(s) + 6H+(aq) 2Cr3+(aq) + 3H2(g)
(c)
M: 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g) I: 2Al(s) + 6H+(aq) + 3SO42–(aq) 2Al3+(aq) + 3SO42–(aq) + 3H2(g) NI: 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g)
6.48
120
Chapter 6
6.49 6.50
a) 3Ag(s) + 4HNO3(aq) 3AgNO3(aq) + 2H2O + NO(g) (b) Ag(s) + 2HNO3(aq) AgNO3(aq) + H2O + NO2(aq) Cu(s) → Cu2+(aq) + 2e– H2SO4(aq) + 2H+(aq) + 2e– SO2(g) + 2H2O Cu(s) + H2SO4(aq)+ 2H+(aq) Cu2+(aq) + SO2(g) + 2H2O
6.51
In each case, the reaction should proceed to give the less reactive of the two metals, together with the ion of the more reactive of the two metals. The reactivity is taken from the reactivity series table 6.2. (a) N.R. (b) Ba(s) + Mn2+(aq) Mn(s) + Ba2+(aq) (c) Sr(s) + Fe2+(aq) Fe(s) + Sr2+(aq) (d) Mn(s) + Cu2+(aq) Cu(s) + Mn2+(aq)
6.52
In each case, the reaction should proceed to give the less reactive of the two metals, together with the ion of the more reactive of the two metals. The reactivity is taken from the reactivity series table 6.2. (a) Mn(s) + Pb2+(aq) Mn2+(aq) + Pb(s) (b) N.R. (c) 2Al(s) + 3Sn2+(aq) 2Al3+(aq) + 3Sn(s) (d) NR
6.53
Increasing ease of oxidation: Y, U, Ga
6.54
Increasing ease of oxidation: Sc, V, Tl
6.55
The equation given shows that Cd is more active than Ru. Coupled with the information in Review Problem 6.53, we also see that Cd is more active than Tl. This means that in a mixture of Cd and Tl+, Cd will be oxidized and Tl+ will be reduced: Cd(s) + 2TlCl(aq) CdCl2(aq) + 2Tl(s) (The Tl(s) and the Cd(NO3)2(aq) will not react.)
6.56
The observation shows that Mg is more active than Ni, however, the information is not sufficient to determine which is easier to oxidize, Mg or Mo. Therefore it cannot be determined which reaction will occur spontaneously.
6.57
5Mg(s) +5Zn2+(aq) + 10Cl-(aq)
6.58
2Mg(s ) + 4NO 3− (aq ) + 8H + ( aq ) → 2Mg 2 + (s ) + 4NO3− + 4H 2 O(l ) + 4NO 2 (g )
6.59
10Li(s) + 10H2O(l)
6.60
10K(s ) + 10H 2 O(l ) → 10K + ( aq ) + 10OH − (aq ) + 5H 2 ( g )
6.61
(a) (b) (c)
→
→
5Mg2+(aq) + 10Cl-(aq) + 5Zin(s)
5H2(g) + 10OH-(aq) + 10Li-(aq)
2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(g) C3H8 (g) + 5O2(g) 3CO2(g) + 4H2O(g) C21H44(s) + 32O2(g) 21CO2(g) + 22H2O(g)
121
Chapter 6
6.62
(a) (b) (c)
2C12H26(l) + 37O2(g) 24CO2(g) + 26H2O(g) C18H36(l) + 27O2(g) 18CO2(g) + 18H2O(g) C7H8(l) + 9O2(g) 7CO2(g) + 4H2O(g)
6.63
(a)
2C8H18(l) + 17O2(g) 16CO(g) + 18H2O(g) 2C3H8(g) + 7O2(g) 6CO(g) + 8H2O(g) 2C21H44(s) + 43O2(g) 42CO(g) + 44H2O(g)
(b)
2C8H18(l) + 9O2(g) 16C(s) + 18H2O(g) C3H8(g) + 2O2(g) 3C(s) + 4H2O(g) C21H44(s) + 11O2(g) 21C(s) + 22H2O(g)
(a)
2C12H26(l) + 25O2(g) 24CO(g) + 26H2O(g) C18H36(l) + 18O2(g) 18CO(g) + 18H2O(g) 2C7H8(l) + 11O2(g) 14CO(g) + 8H2O(g)
(b)
2C12H26(l) + 13O2(g) 24C(s) + 26H2O(g) C18H36(l) + 9O2(g) 18C(s) + 18H2O(g) C7H8(l) + 2O2(g) 7C(s) + 4H2O(g)
6.64
6.65
CH3CH2OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)
6.66
C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)
6.67
2(CH3)2S(g) + 9O2(g) 4CO2(g) + 6H2O(g) + 2SO2(g)
6.68
C4H4S(l) + 6O2(g) 4CO2(g) + 2H2O(g) + SO2(g)
6.69
(a) (b) (c) (d)
2Zn(s) + O2(g) 2ZnO(s) 4Al(s) + 3O2(g) 2Al2O3(s) 2Mg(s) + O2(g) 2MgO(s) 4Fe(s) + 3O2(g) 2Fe2O3(s)
6.70
(a) (b) (c) (d)
2Be(s) + O2(g) 2BeO(s) 4Li(s) + O2(g) 2Li2O(s) 2Ba(s) + O2(g) 2BaO(s) 4Bi(s) + 3O2(g) 2Bi2O3(s)
6.71
a)
IO3– + 6H+ + 6e– I– + 3H2O [SO32– + H2O SO42– + 2H+ + 2e–] × 3 IO3– + 3SO32– + 6H+ + 3H2O I– + 3SO42– + 3H2O + 6H+ Which simplifies to: IO3– + 3SO32– I– + 3SO42–
(b)
1 mol NaIO3 3 mol Na 2SO3 126.0 g Na 2SO3 g Na2SO3 = (6.25 g NaIO3) 197.9 g NaIO3 1 mol NaIO3 1 mol Na 2SO3 = 11.9 g Na2SO3
122
Chapter 6
6.72
(a)
(b)
[Mn2+(aq) + 4H2O MnO4–(aq) + 8H+(aq) + 5e–] × 2 [BiO3–(aq) + 6H+(aq) + 2e– Bi3+(aq) + 3H2O(l)] × 5 2Mn2+(aq) + 5BiO3–(aq) + 8H2O + 30H+(aq) 2MnO4–(aq) + 5Bi3+(aq) + 15H2O(l) + 16H+(aq) which simplifies to give: 2Mn2+(aq) + 5BiO3–(aq) + 14H+(aq) 2MnO4–(aq) + 5Bi3+(aq) + 7H2O(l)
1 g 1 mol MnSO 4 mg NaBiO3 = (22.5 mg MnSO4) 1000 mg 151.0 g MnSO 4
1 mol Mn 2+ 1 mol MnSO 4
5 mol BiO − 1 mol NaBiO 280.0 g NaBiO 1000 mg NaBiO 3 3 3 3 × 2 mol Mn 2+ 1 mol BiO − 1 mol NaBiO3 1 g NaBiO3 3 = 104 mg NaBiO3 6.73
Cu + 2Ag+ Cu2+ + 2Ag
1 mol Ag 1 mol Cu 63.546 g Cu g Cu = (12.0 g Ag) = 3.53 g Cu 107.868 g Ag 2 mol Ag 1 mol Cu 6.74
Al(s) + 3AgNO3(aq) 3Ag(s) + Al(NO3)3(aq)
1 mol AgNO3 1 mol Al 26.98 g Al g Al = (25.0 g AgNO3) = 1.32 g Al 169.9 g AgNO3 3 mol AgNO3 1 mol Al 6.75
(a)
(b)
[MnO4– + 8H+ + 5e– Mn2+ + 4H2O] × 2 [Sn2+ Sn4+ + 2e–] × 5 2MnO4– +5Sn2+ + 16H+ 2Mn2+ + 5Sn4+ + 8H2O
0.250 mol SnCl2 mL KMnO4 = (35.0 mL SnCl2) 1000 mL SnCl2 1 mol KMnO 4 × 1 mol MnO − 4
6.76
(a)
(b)
(a)
2 mol MnO 4 − 5 mol Sn 2+
1000 mL KMnO 4 = 15.2 mL KMnO 4 0.230 mol KMnO 4
[HSO3– + H2O SO42– + 3H+ + 2e–] × 3 ClO3– + 6H+ + 6e– Cl– + 3H2O 3HSO3– + ClO3– + 3H2O + 6H+ 3SO42– + 9H+ + Cl– + 3H2O Which simplifies to: 3HSO3– + ClO3– 3SO42– + 3H+ + Cl–
0.450 mol NaHSO3 1 mol HSO3− mL NaClO3 = (45.0 mL NaHSO3) 1000 mL NaHSO3 1 mol NaHSO3 1 mol ClO − 3 × 3 mol HSO − 3
6.77
1 mol Sn 2+ 1 mol SnCl2
1 mol NaClO 1000 mL NaClO 3 3 = 45.0 mL NaClO3 1 mol ClO − 0.150 mol NaClO3 3
1 mol NaIO3 1 mol IO3 – 3 mol I3 – mol of I3– = 0.0421 g NaIO3 – 197.89 g NaIO3 1 mol NaIO3 1 mol IO3 = 6.38 × 10–4 mol I3– 123
Chapter 6
(b)
6.38 × 10 –4 mol I – 1000 mL 3 Molarity of I3– = = 6.38 × 10–3 M I3– 1 L 100 mL 1LI − 6.38 × 10−3 mol I − 3 3 g SO2 = 2.47 mL I3– − − 1 L I3 1000 mL I3 1 mol SO 2 1 mol I − 3
(c)
64.07 g SO 2 = 1.01 × 10–3 g SO 2 1 mol SO 2 The density of the wine was 0.96 g/mL and the SO2 concentration was 1.01 × 10–3 g SO2/in 50 mL 1.01× 10−3 g SO 2 = 2.02 × 10–5 g SO2/mL 50 mL In 1 mL of solution there are 0.96 g of wine and 2.02 × 10–5 g SO2 Therefore the percentage of SO2 in the wine is concentration SO2 =
2.02 ×10−5 g SO 2 × 100% = 2.10 × 10–3 % 0.96 g wine
6.78
2.02 ×10−5 g SO 2 × 106 ppm = 21 ppm 0.96 g wine
(d)
ppm SO2 =
(a)
1 mol KIO3 1 mol IO3 – 3 mol I3 – Molarity of I3– solution = 0.462 g KIO3 – 214.00 g KIO3 1 mol KIO3 1 mol IO3
1 – × = 0.0259 M I3 0.2500 L (b)
− 1 L 0.0259 mol I3 g (NH4)2S2O3 = 27.99 mL I3– solution 1L 1000 mL
2 mol S O 2 − 2 3 1 mol I − 3
1 mol (NH ) S O 148.24 g (NH ) S O 4 2 2 3 4 2 2 3 = 0.2149 g (NH4)2S2O3 × 1 mol S O 2– 1 mol (NH 4 )2S2 O3 2 3
6.79
(c)
0.2149 g (NH 4 ) 2 S2 O3 % by mass = × 100% = 98.6% (NH4)2S2O3 in sample 0.2180g sample
(a)
0.02100 mol S O 2 − 2 3 mol Cu2+ = (30.06 mL S2O32–) 1000 mL S O 2 − 2 3
1 mol I − 3 2 mol S O 2− 2 3
2 mol Cu 2+ 1 mol I − 3
= 6.313 × 10 −4 mol Cu 2+ g Cu = (6.313 × 10–4 mol Cu) × (63.546 g Cu/mol Cu) = 4.011 × 10–2 g Cu % Cu = (4.011 × 10–2 g Cu/0.4875 g sample) × 100 = 8.229%
(b)
1 mol CuCO3 123.56 g CuCO3 g CuCO3 = (6.313 × 10–4 mol Cu) = 0.07801 g CuCO3 1 mol Cu 1 mol CuCO3 0.07801 g CuCO3 % CuCO3 = × 100% = 16.00% 0.4875 g sample 124
Chapter 6
6.80
(a)
2+
0.0281 mol MnO
2–
g Fe = 41.89 mL MnO4 1000 mL MnO 4
2− 4 2−
5 mol Fe2 + 1 mol MnO 2 − 4
55.845 g Fe 2+ 1 mol Fe2+
= 0.329 g Fe2+ 0.329 g % Fe = × 100% = 21.3% Fe 1.543 g (b)
6.81
(a)
1 mol Fe 1 mol Fe3O 4 231.55 g Fe3O4 g Fe3O4 = (0.329 g Fe) 55.845 g Fe 3 mol Fe 1 mol Fe3O4 = 0.455 g Fe3O4 0.455 g % Fe = × 100% = 29.5% 1.543 g 0.02000 mol KMnO 4 g H2O2 = (29.10 mL KMnO4) 1000 mL KMnO 4 5 mol H O 2 2 × 2 mol MnO − 4
(b)
6.82
1 mol MnO 4 − 1 mol KMnO 4
34.02 g H O 2 2 = 0.04949 g H O 2 2 1 mol H 2 O2
(0.04949 g H2O2/1.650 g sample) × 100% = 3.000% H2O2
0.01000 mol KMnO4 g NaNO2 = (15.35 mL KMnO4) 1000 mL KMnO 4
1 mol MnO 4 2− 1 mol KMnO 4
5 mol HNO 1 mol NaNO 68.995 g NaNO 2 2 2 × 2 mol MnO 2− 1 mol HNO2 1 mol NaNO 2 4 –2 = 2.648 × 10 g NaNO2 % NaNO2 = (2.648 × 10–2 g NaNO2 / 1.250 g sample) × 100% = 2.118% 6.83
(a)
2CrO42– + 3SO32– + H2O 2CrO2– + 3SO42– + 2OH–
(b)
1 mol Na 2SO3 mol CrO42– = (3.18 g Na2SO3) 126.04 g Na 2SO3 1 mol SO 2 − 2 mol CrO 2− 3 4 × 1 mol Na 2SO3 3 mol SO 2 − 3
= 1.68 × 10–2 mol CrO42–
Since there is one mole of Cr in each mole of CrO42–, then the above number of moles of CrO42– is also equal to the number of moles of Cr that were present: 0.0168 mol Cr × 52.00 g/mol = 0.875 g Cr in the original alloy. (c)
(0.875 g Cr/3.450 g sample) × 100% = 25.4% Cr
125
Chapter 6
6.84
(a)
(b)
(Sn2+ Sn4+ + 2e–) × 3 Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O 3Sn2+ + Cr2O72– + 14H+ 3Sn4+ + 2Cr3+ + 7H2O
1 mol Na 2 Cr2 O7 g Sn = (0.368 g Na2Cr2O7) 262.0 g Na 2 Cr2 O7 3 mol Sn 2+ × 2− 1 mol Cr2 O7
6.85
1 mol Cr2 O7 2− 1 mol Na 2 Cr2 O7
1 mol Sn 118.7 g Sn = 0.500 g Sn 1 mol Sn 2+ 1 mol Sn
(c)
(0.500 g Sn / 1.50 g solder) × 100% = 33.3%
(a)
0.1000 mol KMnO 4 mol C2O42– = (31.06 mL KMnO4) 1000 mL KMnO 4
5 mol C2 O 4 2− 2 mol KMnO 4
= 7.765 × 10–3 mol C2O42–
6.86
(b)
The stoichiometry for calcium is as follows: 1 mol C2O42– = 1 mol Ca2+ = 1 mol CaCl2 Thus the number of grams of CaCl2 is given simply by: 7.765 × 10–3 mol CaCl2 × 110.98 g/mol = 0.8618 g CaCl2
(c)
(0.8618 g/3.876 g) × 100 = 22.23% CaCl2
(a)
0.3000 mol Na 2S2 O3 1 mol I3− mol I3– = (28.75 mL Na2S2O3) 1000 mL Na 2S2 O3 2 mol Na 2S2 O3 = 4.313 × 10–3 mol I3–
(b)
(c)
2 mol NO − 2 mol NO2– = (4.313 × 10–3 mol I3–) 1 mol I − 3
= 8.626 × 10–3 mol NO2–
1 mol NaNO 68.995 g NaNO 2 2 = 0.5952 g NaNO g NaNO2 = (8.626 × 10–3 mol NO2–) 2 1 mol NO − 1 mol NaNO 2 2 % NaNO2 = (0.5952 g NaNO2 / 0.958 g sample) × 100% = 62.1% NaNO2
Additional Exercises 6.87
H2S(aq) + Cl2(g) → S(s) + 2H+(aq) + 2Cl-(aq)
6.88
Bromine is both oxidized and reduced. The reduction of Br2 leads to Br- with an oxidation state of -1 and The oxidation of Br2 leads to OBr- where bromine has an oxidation state of +1. The net ionic equation is: Br2(l) + H2O(l) → Br-(aq) + OBr-(aq) +2H+(aq)
6.89
This is an example of an esterification reaction, the reaction of an organic acid with an organic base, an alcohol.
126
Chapter 6
CH3(CH2)6COOH (l) + CH3OH(l) → CH3(CH2)6COOCH3 (l) + H2O(l) 2CH3(CH2)6COOCH3 (l) + 25O2(g)
0.877
g CH 3 ( CH 2 )6 COOCH 3
x
1000 mL
mL
3319
g
x 2 gal x
44.01 g mol CO 2
x
3.785 L
L
gal
1 mol CH 3 ( CH 2 )6 COOCH 3
gal x
→ 18CO2(g) + 18H2O(l)
x
158.24 g
g
= 3319
gal
18 mol CO 2 2 mol CH 3 ( CH 2 )6 COOCH 3
4
= 16.6 × 10 g CO 2 produced
6.90
Total charge = –4 = (charge of phosphorus atoms) + (charge of oxygen atoms) Charge of oxygens = 7(–2) = –14 Charge of sulfur atoms = –4 –(–14) = +10 +10 spread out over 2 phosphorus atoms gives a charge of: +10/2 per P atom, or: Oxidation number of P = +5
6.91
Element oxidized: Element reduced: Oxidizing agent: Reducing agent:
Cl Cl NaOCl NaClO2
6.92
(a)
(b)
6.93
The first reaction demonstrates that Al is more readily oxidized than Cu. The second reaction demonstrates that Al is more readily oxidized than Fe. Reaction 3 demonstrates that Fe is more readily oxidized than Pb. Reaction 4 demonstrates that Fe is more readily oxidized than Cu. The fifth reaction demonstrates that Al is more readily oxidized than Pb. The last reaction demonstrates that Pb is more readily oxidized than Cu.
6.94
–2
0
(c)
+4
(d)
+4
Altogether, the above facts constitute the following trend of increasing ease of oxidation: Cu < Pb < Fe < Al No, the first, fourth and fifth reactions were not necessary.
6.95
Any metal that is lower than hydrogen in the activity series shown in Table 6.3 of the text will react with H+: (b) manganese and (d) aluminum.
6.96
We choose the metal that is lower (more reactive) in the activity series shown in Table 6.3: (a) aluminum (b) zinc (c) magnesium
6.97
(a) (b) (c) (d) (e)
Ca + Zn2+ Ca2+ + Zn Zn + Cu2+ Zn2+ + Cu 2K + Mn2+ 2K+ + Mn Al + Ca2+ N.R Fe + Pb2+ Fe2+ + Pb 127
Chapter 6
6.98
C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l)
6.99
(a)
2NBr3 + 6e– N2 + 6Br– 2NBr3 + 6H2O N2 + 6HOBr + 6H+ + 6e– 4NBr3 + 6H2O 2N2 + 6HOBr + 6Br– + 6H+ Add 6OH– to both sides of the above equation: 4NBr3 + 6H2O + 6OH– 2N2 + 6HOBr + 6Br– + 6H2O which simplifies to give: 4NBr3 + 6OH– 2N2 + 6HOBr + 6Br– or 2NBr3 + 3OH– N2 + 3HOBr + 3Br–
(b)
(Cl2 + 2e– 2Cl–) × 5 Cl2 + 6H2O 2ClO3– + 12H+ + 10e– 6Cl2 + 6H2O 2ClO3– + 10Cl– + 12H+ Adding 12OH– to both sides gives: 6Cl2 + 6H2O + 12OH– 2ClO3– + 10Cl– + 12H2O which simplifies to: 6Cl2 + 12OH– 2ClO3– + 10Cl– + 6H2O or 3Cl2 + 6OH– ClO3– + 5Cl– + 3H2O
(c)
H2SeO3 + 4H+ + 4e– Se + 3H2O (H2S S + 2H+ + 2e–) × 2 2H2S + H2SeO3 + 4H+ 2S + 4H+ + Se + 3H2O which simplifies to give: 2H2S + H2SeO3 2S + Se + 3H2O
(d)
MnO2 + 4H+ + 2e– Mn2+ + 2H2O 2SO32– S2O62– + 2e– 2SO32– + MnO2 + 4H+ Mn2+ + S2O62– + 2H2O
(e)
XeO3 + 6H+ + 6e– Xe + 3H2O (2I– I2 + 2e–) × 3 XeO3 + 6I– + 6H+ 3I2 + Xe + 3H2O
(f)
(CN)2 + 2e– 2CN– (CN)2 + 2H2O 2OCN– + 4H+ + 2e– 2(CN)2 + 2H2O 2CN– + 2OCN– + 4H+ Adding 4OH– to both sides gives: 2(CN)2 + 4OH– + 2H2O 2CN– + 2OCN– + 4H2O which simplifies to: (CN)2 + 2OH– CN–+ OCN– + H2O
6.100
First, balance the equation: 4H+ + 2e– + PbO2 Pb2+ + 2H2O 2Cl– Cl2 +2e– 4H+ + PbO2 +2Cl– Pb2+ + 2H2O + Cl2
128
Chapter 6
Then calculate the number of grams of PbO2.
1 mol Cl2 1 mol PbO 2 239.2 g PbO 2 g PbO2 = 13.5 g Cl2 = 45.5 g PbO2 70.91 g Cl2 1 mol Cl2 1 mol PbO 2 6.101
The oxidation state of cerium in the reactant ion is +4. The number of moles of this ion in the reactant solution is: 0.0150 M × 0.02500 L = 3.75 × 10–4 mol Ce4+ The number of moles of electrons that come from the Fe2+ reducing agent is: 0.0320 M × 0.02344 L = 7.50 × 10–4 mol e– The ratio of moles of electrons to moles of Ce4+ reactant is therefore 2:1, and we conclude that the product is Ce2+.
6.102
Cu + 2Ag+ 2Ag + Cu2+ The number of moles of Ag+ available for the reaction is 0.125 M × 0.255 L = 0.0319 mol Ag+ Since the stoichiometry is 2/1, the number of moles of Cu2+ ion that are consumed is 0.0319 ÷ 2 = 0.0159 mol. The mass of copper consumed is 0.0159 mol × 63.546 g/mol = 1.01 g. The amount of unreacted copper is thus: 12.340 g – 1.01 g = 11.33 g Cu. The mass of Ag that is formed is: 0.0319 mol × 108 g/mol = 3.45 g Ag. The final mass of the bar is: 11.33 g + 3.45 g = 14.78 g.
6.103
The reaction that occurs is 2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq). If we assume that there is excess copper available, we need to determine the number of moles of Ag that will be produced. The number of moles of Ag+ available for the reaction is 0.250 M × 0.0500 L = 0.0125 mol Ag+ We can determine the amount of copper consumed from the balanced equation. Since the stoichiometry is 2/1, the number of moles of Cu2+ ion that are consumed is 0.0125 ÷ 2 = 0.00625 mol. Convert this nmber of moles to a number of grams: 0.00625 mol × 63.546 g/mol = 0.397 g. The amount of unreacted copper is thus: 32.00 g – 0.397 g = 31.60 g Cu. The mass of Ag that is formed is: 0.0125 mol × 107.9 g/mol = 1.35 g Ag. The final mass of the bar will include the unreacted copper and the silver that is formed: 31.60 g + 1.35 g = 32.95 g.
6.104
The tarnishing reaction is: 8Ag(s) + O2(g) + 4HS-(aq) + 2H2O(l)
→ 4Ag2S(s) + 2H2(g) + 4OH-(aq)
The polishing reaction is: 3Ag2S(s) + 2 Al(s) → 6Ag(s) + Al2S3(s) 6.105
2Ti4+(aq) + Zn(s) → 2Ti3+(aq) + Zn2+(aq) 8SO2(g) + 16 H2(g)
→ S8(s) + 16H2O(l)
The activity series only lists metals in order of reactivity, not ions, non-metals, or molecules. 129
Chapter 6
6.106
First we need a balanced equation: Cl2 + 2e– 2Cl– S2O32– + 5H2O 2SO42– + 10H+ + 8e– 4Cl2 + S2O32– +5H2O 8Cl– + 2SO42– + 10H+
1 mol Cl2 1 mol Na 2S2 O3 158.132 g Na 2S2 O3 g Na2S2O3 = (5.00 g Cl2) = 2.79 g Na2S2O3 70.906 g Cl2 4 mol Cl2 1 mol Na 2S2 O3 6.107
(a)
(b)
(Sn2+ Sn4+ + 2e–) × 5 (MnO4– + 8H+ + 5e– Mn2+ + 4H2O) × 2 5Sn2+ + 2MnO4– + 16H+ 5Sn4+ + 2Mn2+ + 8H2O
0.0500 mol KMnO 4 g Sn = (17.46 mL KMnO4 soln) 1000 mL KMnO 4 1 mol Sn × 1 mol Sn 2+
1 mol MnO 4 − 1 mol KMnO 4
5 mol Sn 2+ 2 mol MnO − 4
118.71 g Sn = 0.259 g Sn 1 mol Sn
0.259 g Sn × 100% = 37.8% Sn 0.6850 g sample
(c)
% Sn =
(d)
0.0500 mol KMnO 4 g SnO2 = (17.46 mL KMnO4 soln) 1000 mL KMnO 4 1 mol SnO 2 × 1 mol Sn 2+ % SnO2 =
1 mol MnO 4 − 1 mol KMnO 4
5 mol Sn 2+ 2 mol MnO − 4
150.71 g SnO2 = 0.329 g SnO2 1 mol SnO 2
0.329 g SnO2 × 100% = 48.0% SnO2 0.6850 g sample
Multi-Concept Problems
6.108
0.0500 mol S O 2– 2 3 mg KIO3 = 22.61 mL S2O3 1000 mL –
1 mol I – 3 2 mol S O 2– 2 3
1 mol IO – 3 3 mol I – 3
1 mol KIO 214.00 g KIO 1000 mg KIO 3 3 3 × = 40.32 mg KIO3 1 mol IO – 1 mol KIO3 1 g KIO 3 3 6.109
It is first necessary to write a balanced equation for the reaction of MnO4– with Sn2+. Sn2+ Sn4+ + 2e– MnO4– + 8H+ + 5e– Mn2+ + 4H2O 5Sn2+ + 2MnO4– + 16H+ 5Sn4+ + 2Mn2+ + 8H2O Then, we calculate the original moles of Sn2+ in the 50.0 mL of 0.0300 M SnCl2 solution.
0.0300 mol Sn 2+ mol Sn2+ = (50.0 mL Sn2+) = 0.0015 mol Sn2+ 1000 mL The moles of Sn2+ that were titrated are calculated by multiplying (remembering to include stoichiometry) molarity (0.0100 M) by volume of titrant (0.02728 L).
130
Chapter 6
0.0100 mol MnO − 5 mol Sn 2+ 4 mol Sn2+ = (0.02728 L MnO4–) = 6.82 × 10–4 mol Sn2+ 2 mol MnO − 1 L 4 This number of moles of tin ion remaining is subtracted from the total that was available in the 50.0 mL portion that was titrated, mol Sn2+ remaining = 0.0015 mol Sn2+ – 6.82 × 10–4 mol Sn2+ = 8.18 × 10–4 mol Sn2+ and the answer is converted to the number of moles of MnO4– that had reacted with this number of moles of Sn2+. 2 mol MnO − 4 = 3.27 × 10–4 mol MnO – mol MnO4– = (8.18 × 10–4 mol Sn2+) 4 5 mol Sn 2 + Multiply this number by 10 to get the moles of MnO4– that had not reacted with the SO2 in the original 500 mL of 0.0200 M KMnO4. 3.27 × 10–4 mol MnO4– × 10 = 3.27 × 10–3 mol MnO4– By difference, calculate the moles of SO2 that had reacted, which is equal to the number of moles of S in the original sample. 0.0200 mol MnO − 4 mol MnO4– added to SO2 = (500 mL MnO4) 1000 mL – = 0.0100 mol MnO4 Balanced reaction of SO2 and MnO4– SO2 + 2H2O SO42– + 4H+ + 2e– MnO4– + 8H+ + 5e– Mn2+ + 4H2O 5SO2 + 2MnO4– + 2H2O 2Mn2+ + 5SO42– + 4H+
5 mol SO 2 mol SO2 = (0.0100 mol MnO4– – 3.27 × 10–3 mol MnO4–) 2 mol MnO − 4 The mass of S is calculated by dividing moles by atomic mass,
= 0.0168 mol SO2
1 mol S 32.067 g S g SO2 = (0.0168 mol SO2) = 0.540 g S 1 mol SO 2 1 mol S and the percentage of S in the original sample is the mass of S divided by the total sample mass, times 100. 0.540 g S %S= × 100% = 51.7% S 1.045 g sample 6.110
The balanced equation is the place to start. 6Fe2+(aq) + Cr2O72–(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O() Because amounts of both reagents are specified, we must work a limiting reactant problem to find out which of the two reactants is completely consumed. From the number of moles of this reactant that disappear, we can calculate the number of moles of H+ that react. This amount is subtracted from the initial number of moles of hydrogen ion, and the amount of titrant is calculated by dividing moles by molarity of NaOH solution. If Fe2+ is the limiting reactant:
131
Chapter 6
mL NaOH = 0.400 mol H + − 400 mL Fe 2+
(
14 mol H + 2+ 2+ 6 mol Fe
mol Fe ) 0.060 1000 mL Fe
2+
1 mol NaOH 1 mol H +
1000 mL NaOH × = 34,400 mL NaOH 0.0100 mol NaOH If Cr2O72– is the limiting reactant: mL NaOH =
0.400 mol H + − 300 mL Cr2 O7 2−
(
2− 2 7 2− 2 7
mol Cr O ) 0.0200 1000 mL Cr O
14 mol H + 1 mol Cr O 2− 2 7
1 mol NaOH 1 mol H +
1000 mL NaOH × = 31,600 mL NaOH 0.0100 mol NaOH Na2Cr2O7 is the limiting reagent, therefore 31,600 mL of NaOH is needed. 6.111
The balanced equation for the reaction is: 5H2C2O4 + 2MnO4– + 6H+ 10CO2 + 2Mn2+ + 8H2O
M KMO4 =
6.112
1 mol K 2 C 2 O 4 2 mol KMnO 4 (0.1244 g K 2 C 2 O4 ) 166.2 g K 2 C 2 O4 5 mol K 2 C2 O 4 = 0.02149 M KMnO 4 1L (13.93 mL KMnO 4 ) 1000 mL
The balanced equation for the oxidation-reduction reaction is: 3H2C2O4 + Cr2O72– + 8H+ 6CO2 + 2Cr3+ + 7H2O
0.200 moles K 2 Cr2 O7 3 moles H 2 C 2 O4 mol H2C2O4 = (6.25 mL K2Cr2O7) 1000 mL K 2 Cr2 O7 1 mole K 2 Cr2 O7 = 3.75 × 10–3 mol H2C2O4 So, if we titrate the same oxalic acid solution using NaOH we will need:
2 moles NaOH ml NaOH = (3.75 × 10–3 mol H2C2O4) 1 mole H 2 C2 O 4 6.113
1000 mL NaOH = 16.7 mL NaOH 0.450 moles NaOH
The balanced redox reaction is: 5H2C2O4 + 2MnO4- + 6H+
→ 2Mn2+ + 10CO2 + 8H2O
0.02000 moles KMnO 4 5moles H 2 C2 O 4 mol H2C2O4 = (18.30 mL KMnO4) 1000 mL KMnO 4 2mole KMnO4 = 9.15 × 10–4 mol H2C2O4
132
Chapter 6
The sample of oxalic acid to be titrated with NaOH is 25.00 mL so we need to determine the number of moles of oxalic acid in this sample size.
9.15 × 10 –4 mol H 2 C 2 O 4 15.00 mL
x 25.00 mL = 1.525 x 10 −3 mol H 2 C 2 O 4
The balanced neutralization reaction is: 2OH- + H2C2O4
→ C2O42- + 2H2O
The molarity of the NaOH is given by:
1.525 x 10 −3 mol H 2 C 2 O 4 x
6.114
2 mol NaOH 1 mol H 2 C 2 O 4
x
1 19.69 mL NaOH
x
1000 mL L
=
0.1549 M NaOH To begin, determine balanced equations for the reaction of SO32– with CrO42– and for S2O32– with CrO42–. 3SO32– + 2CrO42– + H2O 3SO42– + 2CrO2– + 2OH– 3S2O32– + 8CrO42– + H2O 6SO42– + 8CrO2– + 2OH– We know that the amount of CrO42– that reacted is
0.0500 moles CrO 2 − 4 mol CrO42– = (80 mL CrO42–) 1000 mL CrO 2 − 4 2– We also know the amount of SO4 produced 1 mol BaSO 4 mol SO42– = (0.9336 g BaO4–) 233.39 g BaSO 4
= 4.00 × 10–3 mol CrO42–
1 mol SO 42 − 1 mol BaSO 4
= 4.00 × 10–3 mol SO42
Let x = moles SO32– and y = moles S2O32– in the 100 mL sample, from the balanced equations and the known quantities we can write:
1 mol SO 2− 4 4.00 × 10–3 mol SO42– = x 1 mol SO 2− 3
2 mol SO 2 − 4 + y 1 mol S O 2 − 2 3
2 mol CrO 2 − 4 4.00 × 10–3 mol CrO42– = x 3 mol SO 2 − 3
8 mol CrO 2 − 4 + y 3 mol S O 2 − 2 3
We can solve for x and y from these and calculate the initial concentration of SO32– and S2O32–. 1 mol SO 2 − 3 2 mol SO 4 2− 1 mol SO32− –3 2– x = 4.00 × 10 mol SO4 2− – y 2− 1 mol SO 4 1 mol S2 O32− 1 mol SO 4
1 mol SO 2 − 3 x = 4.00 × 10–3 1
2 mol SO 2− 3 – y 1 mol S O 2 − 2 3
2 mol SO 2− 3 x = 4.00 × 10–3 mol SO32– – y 1 mol S2 O32 − 133
Chapter 6
Substitute x
2 mol CrO 2 − 4 4.00 × 10–3 mol CrO42– = x 3 mol SO 2 − 3 –3 2– 4.00 × 10 mol CrO4 =
8 mol CrO 2 − 4 + y 3 mol S O 2 − 2 3
2 mol SO 2− -3 2− 3 4.00 × 10 mol SO3 - y 2− 1 mol S O 2 3
2 mol CrO 2 − 4 3 mol SO 2 − 3
8 mol CrO 2− 4 + y 3 mol S O 2 − 2 3
4.00 × 10–3 mol CrO42– =
2 mol CrO 2 − 4 4.00 × 10−3 mol SO32 − 3 mol SO 2 − 3 8 mol CrO 2 − 4 + y 3 mol S O 2− 2 3 4.00 × 10–3 mol CrO42– =
−
2 mol SO 2− 3 y 1 mol S O 2− 2 3
2 mol CrO 2 − 4 3 mol SO 2 − 3
1 mol CrO 2 − 4 2.667 × 10−3 mol CrO 42 − − 1.333 y 1 mol S O 2− 2 3 1 mol CrO 2− 4 1.333 × 10–3 mol CrO42– = 1.333 y 1 mol S O 2 − 2 3 –3 2– y = 1.000 × 10 mol S2O3
8 mol CrO 2− 4 + y 3 mol S O 2 − 2 3
Substitute y back into the x equation:
x = 4.00 × 10–3 mol SO32– –
2 mol SO 2 3 y 1 mol S O 2 2 3
x = 4.00 × 10–3 mol SO32– – 1.000 × 10-3 mol S2 O32 –3 2– x = 4.00 × 10 mol SO3 – 2.00 × 10–3 mol SO32– x = 2.00 × 10–3 mol SO32– Therefore the concentration of SO32– is: 2.00 ×10−3 mol SO32− = 0.0200 M SO32– 0.100 L solution And the concentration of S2O32– is 1.00 ×10−3 mol S2 O32− = 0.0100 M S2O32– 0.100 L solution
134
2 mol SO 2 3 1 mol S O 2 2 3
Chapter 7
Practice Exercises 7.1
∆Twater = 30.0 °C – 20.0 °C = 10.0 °C
4.184 J q gained by water = (10.0 °C)(250 g H2O) = 10,460 J g °C q lost by ball bearing = –q gained by water = –10,460 J C = q/∆T C = –10,460 J/(30.0 °C–220 °C) = 55.1 J/°C 7.2
The amount of heat transferred into the water is: J = (255 g H2O)(4.184 J g-1 °C-1)(30.0 °C – 25.0 °C) = 5335 J 1 kJ kJ =(5335 J) = 5.34 kJ 1000 J 1 cal cal = (5335 J) = 1275 cal 4.184 J 1 kcal kcal = (1275 cal) = 1.28 kcal 1000 cal
7.3
549 J = (7.54 g H2O)(0.712 J g-1 °C-1)(Tf °C – 25.0 °C) Tf = 127.30C
7.4
1 mol CH3OH mol CH3OH = (2.85 g CH3OH) = 0.0890 mol CH3OH 32.04 g CH3OH −715 kJ kJ heat released = (0.0890 mol CH3OH) = –63.6 kJ 1 mol CH3OH 63.6 kJ Heat capacity = = 12.4 kJ °C–1 29.19 C − 24.05 C
7.5
8.930 kJ Heat absorbed by calorimeter = (25.51 °C – 20.00 °C) = 49.2 kJ 1 °C
1 mol C mol C = (1.50 g C) = 0.125 mol C 12.01 g C ∆E = energy/mol = 49.2 kJ/0.125 mol C = 394 kJ/mol C 7.6
Since the mole ratio of NaOH to HCl is 1:1 the number of moles of NaOH equals the number of moles of HCl, therefore the amount of heat needed to neutralize HCl equals the amount of heat needed to react NaOH, or –58 kJ mol–1 NaOH
7.7
q = specific heat × mass × temperature change = 4.184 J/g °C × (175 g + 4.90 g) × (14.9 °C – 10.0 °C) = 3.7 × 103 J = 3.7 kJ of heat released by the process.
135
Chapter 7
This should then be converted to a value representing kJ per mole of reactant, remembering that the sign of ∆H is to be negative, since the process releases heat energy to surroundings. The number of moles of sulfuric acid is: 1 mol H 2SO4 –2 mol H2SO4 = (4.90 g H2SO4) = 5.00 × 10 mol H2SO4 98.06 g H SO 2 4 and the enthalpy change in kJ/mole is given by: 3.7 kJ ÷ 0.0500 moles = 74 kJ/mole 7.8
q = specific heat × mass × temperature change Assume a density of 1.00 g/mL for water = 4.184 J/g °C × (20.0 g + 75.0 g) × (25.7 °C – 10.4 °C) = 6.08 × 103 J = 6.08 kJ of heat absorbed by the process.
This should then be converted to a value representing kJ per mole of reactant, remembering that the sign of ∆H is to be positive, since the process absorbs heat energy from the surroundings. The number of moles of NH4NO3 acid is: 1 mol NH 4 NO3 –1 mol NH4NO3 = 20.0 g H2SO4) = 2.50 × 10 mol NH4NO3 80.04 g NH NO 3 4 and the enthalpy change in kJ/mole is given by: 6.08 kJ ÷ 0.25 moles = 24.3 kJ/mole 7.9 7.10
1 4
CH4(g) +
1 2
O 2( g )
1 4
CO2(g) +
1 2
H2O(l)
∆H = -890.5 ÷ 4 = –222.6 kJ
We can proceed by multiplying both the equation and the thermochemical value of Example 7.6 by 2.5:
2H2(g) + O2(g) 2 H2O(l) 5H2(g) + 2.5O2(g) 5H2O(l)
∆H = –571.8 kJ ∆H = –571.8 kJ x 2.5 = –1429.5 kJ
7.11 2 Cu(s) + O 2(g) -169 kJ -310 kJ Cu 2O(s) + 1/2 O 2(g) -141 kJ 2 CuO(s)
136
Chapter 7
7.12 NO(g) + 1/2O2(g) -56.6 kJ +90.4 kJ NO2(g) +33.8 kJ 1/2N2 (g) + O2(g)
7.13
H 2( g ) +
1 2
O 2( g ) H 2O
∆H = –285.9 kJ
Reverse the reaction: H2O H2(g) +
1 2
O 2( g )
∆H = +285.9 kJ
Multiply the reaction by 3: 3H2O(l) 3H2(g) + 7.14
3 2
O 2( g )
∆H = +857.7 kJ
For this problem: divide the second reaction by two: 1 {2N O(g) + 3O (g) 4NO (g)} ∆H = 1 ( –28.0 kJ) 2 2 2 2 2 N2O(g) +
3 2
O2(g) 2NO2(g)} ∆H = –14.0 kJ
Reverse the first reaction 2NO2(g) 2NO(g) + O2(g)
∆H = +113.2 kJ
Add the reactions together: N2O(g) + 32 O2(g) 2NO2(g)
∆H = –14.0 kJ
2NO2(g) 2NO(g) + O2(g) ∆H = +113.2 kJ _________________________________________ N2O(g) + 12 O2(g) 2NO(g) ∆H = +99.2 kJ 7.15
This problem requires that we add the reverse of the second equation (remembering to change the sign of the associated ∆H value) to the first equation: C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l), 2CO2(g) + 3H2O(l) C2H5OH(l) + 3O2(g),
∆H° = –1411.1 kJ ∆H° = +1367.1 kJ
which gives the following net equation and value for ∆H°: C2H4(g) + H2O(l) C2H5OH(l)
∆H° = –44.0 kJ
7.16
1 moles C3 H6 O 1790.4 kJ/mol kJ = (12.5 g C3H6O) = 385 kJ 58.077 g C3H6 O 1 moles C8 H18
7.17
5450.5 kJ/mol 6 kJ = (480 mol C8H18) = 2.62 × 10 kJ 1 moles C8 H18
7.18
1 2
N2(g) + 2H2(g) +
1 2
Cl2(g) NH4Cl(s)
∆Hf = –315.4 kJ
137
Chapter 7
1 2
3 2
O2(g) NaHCO3(s),
7.19
Na(s) +
7.20
∆H° = sum ∆Hf [products] – sum ∆Hf [reactants]
H2(g) + C(s) +
∆Hf = –947.7 kJ/mol
∆H° = { ∆Hf [CaSO4(s)] + 2 ∆Hf [HCl(g)]} – { ∆Hf [CaCl2(s)] + ∆Hf [H2SO4(l)]} ∆H° = {[1 mol × (–1432.7 kJ/mol)] + [2 mol × (–92.30 kJ/mol)]} – {[1 mol × (–795.0 kJ/mol)] + [1 mol × (–811.32 kJ/mol)]} ∆H° = -10.98 kJ 7.21
S(s) +
3 2
O2(g) SO3(g) ∆Hf = –395.2 kJ/mol
S(s) + O2(g) SO2(g) ∆Hf = –296.9 kJ/mol Reverse the first reaction and add the two reactions together to get SO3(g) SO2(g) +
1 2
O2(g) ∆Hf = +98.3 kJ
∆H° = sum ∆Hf [products] – sum ∆Hf [reactants] ∆H° = { ∆Hf [SO2(g)] +
1 2
∆Hf [O2(g)]} – ∆Hf [SO3(s)]
∆H° = {[1 mol × (–296.9 kJ/mol)] + [ 12 mol × 0 kJ/mol]} – [1 mol × (–395.2 kJ/mol)] ∆H° = +98.3 kJ The answers for the enthalpy of reaction are the same using either method. 7.22
a)
∆H° = sum ∆Hf [products] – sum ∆Hf [reactants] = 2 ∆Hf [NO2(g)] – {2 ∆Hf [NO(g)] + ∆Hf [O2(g)]} = 2 mol × 33.8 kJ/mol – [2 mol × 90.37 kJ/mol + 1 mol × 0 kJ/mol] = –113.1 kJ
b)
∆H° = { ∆Hf [H2O(l)] + ∆Hf [NaCl(s)]} – { ∆Hf [NaOH(s)] + ∆Hf [HCl(g)]} = [(–285.9 kJ/mol) + (–411.0 kJ/mol)] – [(–426.8 kJ/mol) + (–92.30 kJ/mol)] = –177.8 kJ
Review Questions 7.1
(a) (b)
Energy is something that matter possesses by virtue of an ability to do work. Work is the energy expended in moving an opposing force through some particular distance.
7.2
(a) (b)
Kinetic energy is the energy of motion. Potential energy is stored energy.
7.3 7.4
1 )mv2 where m is the mass of the object and v is its velocity. 2 The Law of Conservation of Energy states that the energy of the universe is constant: it can be neither created nor destroyed but only transferred and transformed. The energy of the child on a swing is all potential energy when she is at the top of the arc. As she descends, the energy is converted to kinetic energy. At the bottom at the arc, all the energy is kinetic energy. The potential energy increases as she rises and is completely potential energy at the top of the arc. Kinetic energy = (
138
Chapter 7
7.5
1 1 )m(30 mph)2 K.E.2 = ( )m(60 mph)2 2 2 1 1 2 K.E.1 = ( )m (900 mph ) K.E.2 = ( )m (3600 mph2) 2 2 K.E.2 3600 = =4 K.E.1 900 There is a four-fold increase in the kinetic energy due to the doubling of the speed of the car. K.E. = (
1 )mv2 2
K.E.1 = (
1 )m1v2 2
K.E.1 = (
m2 = 2m1
K.E.2 = (
1 )2m1v2 2
K.E.2 2m1 = =2 K.E.1 m1 There is a two-fold increase in the kinetic energy due to the doubling of the mass of the truck 7.6
Chemical energy is the potential energy in substances, which changes into other forms of energy when substances undergo chemical reactions.
7.7
(a) (b) (c) (d)
7.8
Heat is a form of energy that is transferred between objects, and is the molecular kinetic energy possessed by molecules as a result of the temperature of the sample. Heat is related to the total kinetic energy of the molecules and temperature is related to the average kinetic energy.
7.9
Thermal equilibrium is when two objects in contact with each other are at the same temperature. The molecules in the hot object are moving with more kinetic energy than the colder object. As heat is transferred from the hot object to the cold one, the atoms in the hot object slow down and the atoms in the cold speed up until they have the same average kinetic energy.
7.10
The SI unit of energy is a joule (kg m2 s-2). 1 E = ( )(75 kg)(45 m s-2)2 = 76,000 J 2 1 cal cal = (76,000 J) = 18,000 cal 4.184 J
7.11
The heat produced by combustion of gasoline does no useful work. It is expended into the surroundings.
7.12
The internal energy is the sum of the molecular kinetic energy and the potential energy. The change in internal energy is defined as the difference in internal energy between the energy of the products and the energy of the reactants.
7.13
(a) (b) (c) (b)
7.14
a, b, c
increases increases increases decreases
point c point d It will increase The height of the curve at point A will decrease. The maximum of the curve will be lower and shifted to the right.
139
Chapter 7
7.15
The quart of boiling water has more heat and will cause a more severe burn because the quart has more water, so that the quart has more kinetic energy.
7.16
The first diagram represents an isolated system. Isolated systems cannot transfer mass or energy across its boundary. The second diagram represents a closed system. Closed systems cannot transfer mass but can transfer energy across its boundary. The third diagram represents an open system. Open systems transfer mass and energy across boundaries.
7.17
The state of a system in chemistry is usually specified by its current conditions such as its chemical composition, its pressure, its temperature and its volume. A state function is a quantity whose value depends only on the initial and final states of the system and not on the path taken by the system to get from the initial to the final state.
7.18
The system is that part of the universe under study and separated from the surroundings by a real or an imaginary boundary. The surroundings are that part of the universe other than the system being studied and separated from the system by a real or an imaginary boundary. An isolated system does not allow matter or energy to be transferred between the system and the surroundings. The closed system can absorb or release energy but not mass across the boundary between the system and its surrounding.
7.19
(a) (b) (c)
7.20
The energy depends directly on the specific heat, so the material with the large specific heat requires the higher energy input for the 5 °C rise in temperature.
7.21
Heat capacity is an extensive property and is proportional to the mass of the sample. Specific heat is an intensive property.
7.22
A negative value for heat means that heat is released from the system, it is exothermic.
7.23
If object A has twice the specific heat and twice the mass of object B, and the same amount of heat is applied to both objects, the temperature change of A will be one-fourth the temperature change in B. qB = mBsB∆t q = amount of heat applied sB = specific heat of B 2sB = specific heat of A mB = mass of B 2mB = mass of B q 1 q 1 q = ∆tB = ∆ tA = = ∆t B 2m B × 2s B 4 m Bs B 4 ms
7.24
(a) (b)
7.25
Exothermic; chemical energy decreases; q is negative.
7.26
Endothermic; chemical energy increases; q is positive.
7.27
The potential energy of gasoline and oxygen is higher than the potential energy of carbon dioxide and water vapor since energy is released as heat in the reaction.
specific heat molar heat capacity heat capacity
Kinetic energy must increase. The temperature of the system must increase because kinetic energy increased.
140
Chapter 7
7.28
∆E = q + w The change in internal energy is the sum of the heat absorbed by the system and the work done on it by the surroundings.
7.29
H = E + pV
7.30
∆H > 0 for an endothermic change.
7.31
Since energy is conserved, the enthalpy of the surroundings must decrease by 100 kJ.
7.32
The products are CO2, H2O and heat. The heat measured is ∆E since the volume did not change.
7.33
w = –p∆V
7.34
∆H = ∆E + P ∆V For this reaction, ∆V is approximately zero since the moles of gas for the products equals the moles of gas for the reactants, so ∆H = ∆E.
7.35
Since the heats of reaction will in general depend on temperature and pressure, we need some standard set of values for temperature and pressure so that comparisons of various heats of reaction are made under identical conditions. The standard temperature is 25 °C, slightly above room temperature, and the standard pressure is 1 atmosphere.
7.36
A thermochemical equation contains the value for the associated ∆H.
7.37
Fractional coefficients are permitted because thermochemical properties are extensive (they depend on the amount of material present). If 1/2 the amount of materials reacts, 1/2 the amount of heat will be generated or required. The coefficients of thermochemical equations are moles of substance.
7.38
∆H is a state function.
7.39
The reaction must (1) produce one mole of a compound at 25 °C and 1 atm, and (2) the reactants must be elements in their standard states.
Review Problems 7.40
∆E = q + w = 32 J – 56 J = –24 J
7.41
∆E = q + w = 31 J + 63 J = 94 J
7.42
Here, ∆E must = 0 in order for there to be no change in energy for the cycle. ∆E = q + w 0 = q + (–120 J) q = +120 J
7.43
7.44 7.45
∆E = q + w −585 J = q + (−495 J) q = −90 J If the engine absorbs 275 J of heat, the maximum amount of work it can do is 275 J.
∆E = q + w 255 = 375 + w w = −120 J 141
Chapter 7
7.46
7.47
∆T = 15.0 °C – 25.0 °C = –10.0 °C 18.02 g H 2O –1 –1 J = (2.25 mol H2O) × × 4.184 J g °C × –10.0 °C = –1696 J 1 mol H O 2 1 cal cal = (–1696 J) × = –405 cal 4.184 J q = mCp∆T Assume that the density of water is 1.00 g mL-1 at these temperatures so that the mass of water is the same as its volume.
q = 215 g x 4.184 J g-1 0C-1 x (99.0 0C – 25.0 0C) = 6.66 x 104 J 7.48
J heat released = (95.0 g Fe)(0.4498 J g-1 °C-1)(35.0 °C – 85.0 °C) = –2137 J J heat released by the iron = –J heat absorbed by water = 2137 J 2137 J = 51.1 g H O g H2O = 2 −1 −1 C 35.0 C − 25.0 C 4.184 J g
(
7.49
)(
)
The heat gained by the water must equal the heat lost by the copper. To get the correct signs for the heat transfer, remember heat lost is negative and heat gained is positive. mCu x 0.387 J g-1 0C-1 x (32.3 0C – 67.0 0C) = 100.0 g x 4.184 J g-1 0C-1 x (32.3 0C – 27.0 0C) mCu = 165 g of copper
7.50
(a) (b) (c) (d)
J = 4.184 J g–1 °C–1 × 100 g × 4.0 °C = 1.67 × 103 J 1.67 × 103 J 1.67 × 103 J/(100 – 28.0)°C = 23.2 J °C–1 23.2 J °C–1 ÷ 5.00 g = 4.64 J g–1 °C–1
7.51
(a) (b) (c)
J = 4.184 J g–1 °C–1 × 200.0 g × 1.50 °C = 1.255 × 103 J 1.255 × 103 J 1.26 × 103 J = 0.387 J g-1 °C–1 × (120 °C – 26.50 °C) × X g X = 34.7 g
7.52
0.4498 J 55.847 g Fe J = = 25.12 1 mol Fe mol C mol C g C
7.53
0.586 cal 4.184 J 46.08 g C2 H5 OH = = 113 J mol C mol C g C cal 1 mol C2 H5 OH
7.54
4.18 J g–1 °C–1 × (6.37 × 103 g) × (58.65 – 60.25) °C = –4.26 × 104 J = –42.6 kJ
7.55
4.18 J g–1 °C–1 × (4.95 × 103 g) × (27.31 – 25.24) °C = 4.28 × 104 J = 42.8 kJ
7.56
HNO3(aq) + KOH(aq) KNO3(aq) + H2O(l) Keep in mind that the total mass (assume the densities to be 1.00 g/mL for the solutions) must be considered in this calculation, and that both liquids, once mixed, undergo the same temperature increase: heat = (4.18 J/g °C) × (55.0 g + 55.0 g) × (31.8 °C – 23.5 °C) = 3.8 × 103 J of heat energy released
J
J
142
Chapter 7
Next determine the number of moles of reactant involved in the reaction: 0.0550 L × 1.3 mol/L = 0.072 mol of acid and of base. 1 kJ 3.8 × 103 J kJ 1000 J = − 53 kJ Thus the enthalpy change is: = mol mol ( 0.072 mol )
(
7.57
)
HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(l) Keep in mind that the total mass (assume the densities to be 1.00 g/mL for the solutions) must be considered in this calculation, and that both liquids, once mixed, undergo the same temperature increase: heat = (4.18 J/g °C) × (75.0 g + 45.0 g) × (23.4 °C – 22.4 °C) = 5.02 × 102 J of heat energy released Next determine the number of moles of reactant involved in the reaction: 0.0750 L x 1.07 mol/L = 0.080 mol of acid 0.0450 L x 1.78 mol/L = 0.080 mol of base. 1 kJ 5.02 × 102 J kJ 1000 J = − 6.3 kJ Thus the enthalpy change is: = mol mol ( 0.080 mol )
(
)
7.58
(a) (b) (c)
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) J = (97.1 kJ/°C)(27.282 °C – 25.000 °C) = 222 kJ = 2.22 × 105 J ∆H° = – 222 kJ/mol
7.59
(a) (b)
C7H8(l) + 9O2(g) 7CO2(g) + 4H2O(l) J = (45.06 kJ/°C)(26.413 °C – 25.000 °C) = 63.67 kJ = 6.367 × 104 J 6.367 × 104 J 92.14 g C7 H8 J = (1 mol C7H8) = 3.911 × 106 J 1 mol C7 H8 1.500 g
(c)
7.60
(a)
(b)
7.61
(a) (b)
Multiply the given equation by 2. 6CO(g) + 3O2(g) 6CO2(g), ∆H° = –1698 kJ
2 849 kJ 2.00 mol CO2
5 2 566 kJ 3mol CO 2
Divide the given equation by 2. 2NH3(g) + 7/2 O2(g) 2NO2(g) + 3H2O(g) ∆H° = –1132 kJ/2 = –566 kJ Divide the given equation by 4 NH3(g) + 7/4 O2(g) NO2(g) + 3/2 H2O(g) ∆H° = –1132 kJ/4 = –283 kJ
7.62
−1 mol mg 1203 kJ (7.89 g Mg) = –195 kJ of heat are evolved 24.31g mg 2 mol mg
7.63
1 mol CH3OH −1199 kJ kJ = (59.0 g CH3OH) = –1104 kJ 32.04 g CH3OH 2 mol CH3OH
7.64
1 mol CH 4 16.04 g CH 4 −690 kJ = 13.8 g CH4 −802 kJ 1 mol CH 4
143
Chapter 7
7.65
32.04 g 2 mol CH3OH g CH3OH = -750 kJ x = 40.1 g − 1199 kJ 1 mol CH3OH
7.66
7.67 NO(g) + 1/2O2(g) -56.6 kJ +90.4 kJ NO2(g) +33.8 kJ 1/2N2 (g) + O2(g)
The enthalpy change for the reaction NO(g) + 7.68
1 2
O2(g) NO2(g) is –56.6 kJ as seen in the figure above.
Since NO2 does not appear in the desired overall reaction, the two steps are to be manipulated in such a manner so as to remove it by cancellation. Add the second equation to the inverse of the first, remembering to change the sign of the first equation, since it is to be reversed: 2NO2(g) N2O4(g), 2NO(g) + O2(g) 2NO2(g),
∆H° = –57.93 kJ ∆H° = –113.14 kJ
Adding, we have: 2NO(g) + O2(g) N2O4(g), 7.69
7.70
∆H° = –171.07 kJ
Reverse the first equation, multiply the result by two, and add it to the second equation: 2KCl(s) + 2H2O(l ) 2HCl(g) + 2KOH(s), H2SO4(l) + 2KOH(s) K2SO4(s) + 2H2O(l),
∆H° = 407.2 kJ ∆H° = –342.4 kJ
Adding gives us: 2KCl(s) + H2SO4(l) 2HCl(g) + K2SO4(s),
∆H° = 64.8 kJ
If we label the four known thermochemical equations consecutively, 1, 2, 3, and 4, then the sum is made in the following way: Divide equation #3 by two, and reverse all of the other equations (#1, #2, and #4), while also dividing each by two: 144
Chapter 7
1 2
Na2O(s) + HCl(g)
NaNO2(s) 1 2 1 2
NO(g) + H2O(l) +
1 2
1 2
H2O(l) + NaCl(s),
Na2O(s) +
1 2
NO2(g) +
1 2
∆H° = –253.66 kJ
NO(g),
∆H° = +213.57 kJ
1 NO (g) 1 N O(g) + 1 O (g), 2 2 2 2 2 2 1 O (g) + 1 N O(g) HNO (l), 2 2 2 2 2
∆H° = –21.34 kJ ∆H° = –17.18 kJ
Adding gives: HCl(g) + NaNO2(s) HNO2(l) + NaCl(s), 7.71
7.72
Reverse the second and the third thermochemical equations and add them to the first: CaO(s) + 2HCl(aq) CaCl2(aq) + H2O(l), Ca(OH)2(s) CaO(s) + H2O(l), Ca(OH)2(aq) Ca(OH)2(s),
∆H° = –186 kJ ∆H° = 65.1 kJ ∆H° = 12.6 kJ
Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) + 2H2O(l),
∆H° = –108 kJ
1 2
Multiply all of the equations by 1 2 1 2 1 2 1 2
7.73
∆H° = –78.61 kJ
and add them together.
1 Cl (g) 1 CaOCl (s) 2 2 2 2 H2O(l) + 12 CaOCl2(s) + NaBr(s) NaCl(s) Ca(OH)2(s) 12 CaO(s) + 12 H2O(l) Cl2(g) + NaBr(s) NaCl(s) + 12 Br2(l)
CaO(s) +
The equation we want is Cu(s) +
1 2
∆ H° = +
1 2
Ca(OH)2(s) +
1 2
1 2
(–110.9kJ)
Br2(l) ∆H°= ∆H°=
∆H°= 1 2 1 2
1 2
(–60.2 kJ)
(+65.1 kJ) (–106 kJ) = –53 kJ
O2(g) → CuO(s). If we multiply all reactions by
1 2
and reverse the
second reaction we get: Cu(s) + 1 2 1 2
S(s)
SO2(g)
1 2
1 2
Cu2S(s)
S(s) +
1 2
1 2
∆ H° =
O 2( g )
Cu2S(s) + O2(g) CuO(s) +
Cu(s) + 7.74
1 2
∆ H° = 1 2
SO2(g)
O2(g) CuO(s)
∆ H° =
(–79.5 kJ) (+297 kJ) (–527.5 kJ)
∆H° = –155 kJ
We need to eliminate the NO2 from the two equations. To do this, multiply the first reaction by 3 and the second reaction by two and add them together. 12NH3(g) + 21O2(g) 12NO2(g) + 18H2O(g) 12NO2(g) + 16NH3(g) 14N2(g) + 24H2O(g) 28NH3(g) + 21O2(g) 14N2(g) + 42H2O(g) Now divide this equation by 7 to get 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)
7.75
1 2 1 2 1 2
∆H° = 3(–1132 kJ) ∆H° = 2(–2740 kJ) ∆H° = –8876 kJ ∆H° = 1/7(–8876 kJ) = –1268 kJ
Multiply the second equation by two and add them together: 3Mg(s) + 2NH3(g) Mg3N2(s) + 3H2(g) N2(g) + 3H2(g) 2NH3(g) 3Mg(s) + N2(g) Mg3N2(s)
∆H° = –371 kJ ∆H° = 2(–46 kJ) ∆H° = –463 kJ
145
Chapter 7
7.76
The heat of formation is defined as the enthalpy change when one mole of a compound is produced from its elements in their standard states. Only (c) satisfies this requirement. For choice (a,) reactant CO(NH2)2 is not an element For choice (b), O and H are not the standard states for oxygen and nitrogen gas. The state for C is not given as graphite or as a solid. For choice (d), the reaction is not balanced
7.77
The heat of formation is defined as the enthalpy change when one mole of a compound is produced from its elements in their standard states. Only (b) satisfies this requirement. For choices (a) and (c) the reactants are not elements in their standard state. (a) has molecules as reactants and (c) has atoms. Choice (d) might look okay but it does not fit the definition that one mole of product is formed.
7.78
7.79
7.80
∆Hf = –333.19 kJ/mol
(a)
C(s) + ½ O2(g) + N2(g) + 2H2(g) → CO (NH2)2(s)
(b)
2 C(s) + 3H2(g) → C2H6(g)
(c)
2K(s) +
(d)
Na(s) + ½ H2(g) + C(s) +
(a)
Mg(s) + Cl2(g) + 2H2(g) + O2(g) MgCl2·2H2O(s)
(b)
N2(g) + 4H2(g) + 2Cr(s) +
(c)
P(s,white) +
(a)
∆Hf = ∆Hf [O2(g)] + 2 ∆Hf [H2O(l)] – 2 ∆Hf ∆Hf [H2O2(l)]
1 8
∆Hf = –84.667 kJ/mol ∆Hf = –1433.7 kJ/mol
S8(s) + 2O2(g) → K2SO4(s)
1 2
O2(g) +
3 2
3 2
7 2
O 2( g ) →
NaHCO3(s)
O2(g) (NH4)2Cr2O7(s)
∆Hf = – 947.7 kJ/mol ∆Hf = –1280 kJ ∆Hf = –1807 kJ ∆Hf = –558.5 kJ
Cl2(g) POCl3(g)
∆Hf = 0 kJ/mol + [2 mol × (–285.9 kJ/mol)] – [2 × (–187.6 kJ/mol)] = –196.6 kJ
7.81
7.82
(b)
∆Hf = ∆Hf [H2O(l)] + ∆Hf [NaCl(s)] – ∆Hf [HCl(g)] – ∆Hf [NaOH(s)] = [1 mol × (–285.9 kJ/mol)] + [1 mol × (–411.0 kJ/mol)] – [1 mol × (–92.30 kJ/mol)] – [1 mol × (–426.8 kJ/mol)] = –177.8 kJ
(a)
∆Hf = ∆Hf [HCl(g)] + ∆Hf [CH3Cl(g)] – ∆Hf [CH4(g)] – ∆Hf [Cl2(g)] = [1 mol × (–92.30 kJ/mol)] + [1 mol × (–82.0 kJ/mol)] – [1 mol × (–74.848 kJ/mol)] – [1 mol × (0.0 kJ/mol)] = –99.5 kJ
(b)
∆Hf = ∆Hf [H2O(l)] + ∆Hf [CO(NH2)2(s)] – 2 ∆Hf [NH3(g)] – ∆Hf [CO2(g)] = [1 mol × (–285.9 kJ/mol)] + [1 mol × (–333.19 kJ/mol)] – [2 mol × (–46.19 kJ/mol) – [1 mol × (–393.5 kJ/mol)] = –133.2 kJ
C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l) 146
∆Hocombustion = –5.65 × 103 kJ/mol
Chapter 7
∆Hocombustion = Σ∆Hf°(products) – Σ∆Hf°(reactants) =[12 mol CO2 × ∆Hf (CO2(g)) + 11 mol H2O × ∆Hf (H2O(l))] – [1 mol C12H22O11 × ∆Hf (C12H22O11(s)) + 12 mol O2 × ∆Hf (O2(g))] Rearranging and realizing the ∆Hf° O2(g) = 0 we get ∆Hf°(C12H22O11(s)) = 12∆Hf°(CO2(g)) + 11∆Hf°(H2O(l)) – ∆Hocombustion = 12(–393kJ) + 11(–285.9 kJ) – (–5.65 × 103 kJ) = –2.21 × 103 kJ 7.83
∆Hf°(C2H2(g)) = =
1 2
1 2
{4∆Hf°(CO2(g)) + 2∆Hf°(H2O(l)) – ∆Hocombustion }
{4(–393.5 kJ) + 2(–285.9 kJ) – (–2599.3 kJ) = +226.8 kJ
Additional Exercises 7.84
Water has the highest specific heat and therefore will take the greatest amount of heat to raise its temperature as given amount. Lead has the smallest specific heat, though gold is on 0.001 J g-1 0C-1 higher, so it will increase the most in temperature for the given amount of applied heat.
7.85
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) The heat of neutralization is released to three "independent" components of the system, all of which undergo the same temperature increase: ∆T = 20.610 °C – 16.784 °C = 3.826 °C. Also, the total heat capacity of the system is the sum of the three heat capacities: heat capacityHCl + heat capacityNaOH + heat capacitycalorimeter = (4.031 J g–1 °C–1 × 610.29 g) + (4.046 J g–1 °C–1 × 615.31 g) + 77.99 J °C–1 = 5028 J °C–1 The heat flow to the system is thus: heat = 5028 J °C–1 × 3.826 °C = 1.924 × 104 J = 19.24 kJ and the heat of neutralization is the negative of this value, since the neutralization process is exothermic: ∆H = –19.24 kJ ÷ 0.33183 mol = –57.98 kJ/mol
7.86
KE =
1 2
2
mv =
1
x 4.003
2
g mol
x
1mol
x
23
6.02 x 10 atoms
1kg
x (1.32 x 10
3
1000 g
s
KE = 5.79 x 10 −21 J 7.87
Heat lost = (2000 g)(0.803 J/g °C)(95 °C – x °C) = (1606 J/°C)(95 °C – x °C) =152,600 J – (1606 J)(x °C) 1000 mL 1 g 4.184 J Heat gained = (2.00 L) x - 22 C 1L 1 mL g C = (8368 J/°C)(x °C – 22 °C) = (8368 J)(x °C) – 184,100 J Where x = final temperature we can solve for x since heat lost = heat gained 152,600 J – (1606 J)(x °C) = (8368 J)(x °C) – 184,100 J 336,700 J = (9974 J)(x °C) 33.8 °C
(
147
m
)
)
2
Chapter 7
7.88
7.89
Multiply the first reaction by 1/2: 1 Fe O (s) + 3 CO(g) Fe(s) + 2 3 2 2
3 2
∆ H° =
1 2
(–28 kJ)
Reverse the second reaction AND multiply by 1/6: 1 Fe O (s) + 1 CO (g) 1 Fe O (s) + 1 CO(g) 3 4 2 2 3 3 6 2 6
∆ H° =
1 6
(+59 kJ)
Reverse the third reaction AND multiply by 1/3: FeO(s) + 13 CO2(g) 13 Fe3O4(s) + 13 CO(g)
∆ H° =
1 3
(–38 kJ)
Now add the equations together: FeO(s) + CO(g) Fe(s) + CO2(g)
∆H° = –16.8 kJ
CO2(g)
∆H°reaction = [ ∆Hf (CO2(g)) + ∆Hf (Fe(s))] – [ ∆Hf (FeO(s)) + ∆Hf (CO(g))] Rearranging, and remembering that ∆Hf Fe(s) = 0
∆Hf (FeO(s)) = ∆Hf (CO2(g)) – ∆Hf (CO(g)) – ∆H°reaction ∆Hf = –393.5 kJ – (–110.5 kJ) – (–16.8 kJ) = – 266.2 kJ 7.90
∆H° = (8 mol × ∆Hf° [CO2(g)]) + (10 mol × ∆Hf° [H2O(l)]) + (2 mol × ∆Hf° [N2(g)]) – (4 mol × ∆Hf°[C2H5NO2(s)]) – (9 mol × ∆Hf°[O2(g)]) The value of ∆H° for the given combustion reaction is: ∆H° = 4 mol × (–973.49 kJ/mol) = –3894.0 kJ The standard enthalpy of formation for each substance is taken from Table 7.2: –3894.0 kJ = (8 mol × (–393.5 kJ/mol)) + (10 mol × (–285.9 kJ/mol)) – (4 mol × ∆Hf°[C2H5NO2]) Solving for the desired enthalpy of formation, we have:
∆Hf [C2H5NO2] = –528.3 kJ/mol 7.91
The equation may be written as:
1 2
H 2( g ) +
1 2
Br2(l) HBr(g); ∆Hf° = –36 kJ
To obtain ∆H, combine the equations in the following manner: Br2(aq) + 2KCl(aq) Cl2(g) + 2KBr(aq) H2(g) + Cl2(g) 2HCl(g) ∆H° = –184 kJ 2HCl(aq) + 2KOH(aq) 2KCl(aq) + 2H2O(l) 2KBr(aq) + 2H2O(l) 2HBr(aq) + 2KOH(aq) ∆H° = –154 kJ 2HCl(g) 2HCl(aq) 2HBr(aq) 2HBr(g) ∆H° = 160 kJ Br2(l) Br2(aq) ∆H° = –4.2 kJ
∆H° = 96.2 kJ ∆H° = –115 kJ ∆H° = 115 kJ
Add all of the above to get; H2(g) + Br2(l) 2HBr(g);
∆H° = –86 kJ
Now divide this equation by two to give the thermochemical equation for the formation of 1 mol of HBr(g): 1 2
H 2( g ) +
1 2
Br2(l) HBr(g);
∆H° = –43 kJ
Comparing this value to the ∆Hf° value listed in Appendix C.2 and at the outset of this problem, we see that this experimental data indicates a value that is close to the reported value.
148
Chapter 7
7.92
The equation we want is: 2C(s) + H2(g) C2H2(g) Take this one step at a time. Start with the fourth equation: CaC2(s) + 2H2O(l) Ca(OH)2(s) + C2H2(g) ∆H° = –126 kJ Add the reverse of the first equation and get rid of calcium hydroxide: Ca(OH)2(s) CaO(s) + H2O(l) ∆H° = +65.3 kJ Add the second equation to eliminate the CaO(s): ∆H° = +462.3 kJ CaO(s) + 5/2C(s) CaC2(s) + 1/2CO2(g) This also eliminates the CaC2(s) we had from the first equation. To eliminate the H2O, reverse the last equation AND multiply by 1/2: H2(g) + 1/2O2(g) H2O(l) ∆H° = –286 kJ Now get rid of the CO by reversing the fifth equation AND multiplying by 1/2: CO(g) C(s) + 1/2O2(g) ∆H° = +110 kJ Add the equations together to get: 2C(s) + H2(g) C2H2(g) ∆H° = 225.6 kJ
7.93
C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l) ∆Hcombustion = Σ∆Hf°(products) – Σ∆Hf°(reactants) = [12 ∆Hf°(CO2(g))+ 11 ∆Hf°(H2O(l))] – [∆Hf°(C12H22O11(s)) + 12∆Hf°(O2(g))] = [12(–393.5 kJ) + 11(–285.9 kJ)] – [(–2230 kJ) + 12(0 kJ)] = –5.64 × 103 kJ/mol This is the amount of heat liberated for 1 mol of sucrose. Thus, for 56.8 g we have,
1 mol −5640 kJ kJ = (56.8 g) = –936 kJ 342.3 g 1 mol 7.94
1 2 1 2 1 2
HCHO2(l) +
1 2
H2O(l)
1 2
CH3OH(l) +
1 2
O 2( g )
∆H° = +206kJ
CO(g) + H2(g) 12 CH3OH(l)
∆H° = –64 kJ
1 2
∆H° = –17 kJ
HCHO2(l)
CO(g) +
1 2
H2O(l)
Add these together: HCHO2(l) + H2(g) CH3OH(l) +
1 2
O 2( g )
7.95
Simply add the two together to get: O3(g) + O(g) 2O2(g) ∆H° = –394 kJ
7.96
K.E. =
1 2
mv2 =
1 2
∆H° = +125 kJ
(14.0 tons) ×
2000 lb 453.6 g 1 kg 45.0 mi 1 hr 5280 ft 30.48 cm 1 m 1 ton 1 lb 1000 g hr 3600 s 1 mi 1 ft 100 cm = 2.57 × 106 J = 2.57 × 103 kJ 3.785 L 1000 mL 1 g g water = (5.00 gal) = 18900 g water 1 gal 1 L 1 mL Increasing the temperature of water: 2.57 × 106 J = (4.184 J g–1 °C–1)(18900 g)(∆T) ∆T = 32 °C 149
2
Chapter 7
7.97
The desired reaction is given below: CO2(g) + 2H2O(l)
→ CH4(g) + 2O2(g)
∆Hreact = ∆Hf°(CH4) - ∆Hf°(CO2) + 2∆Hf°(H2O(l)) = -74.848 kJ mol-1 – (-393.5 kJ mol-1 + (-285.9 kJ mol-1) = +890.4 kJ Since ∆Hreact equals the amount of work needed for this reaction, w = +890.4 kJ. It is difficult to carry out this reaction as it is a non-spontaneous reaction requiring an input of energy. However, when energy is input to this mixture it is almost impossible to control what product will be formed. 7.98
(a)
(b) Water: q < 0 (c)
Zinc: q > 0
qlost + qgained = 0 Also, we must assume the density of water at these temperatures is 1.00 g mL-1 50 g x 4.184 J g-1 0C-1 x (Tf – 32.4 0C) + 5.25 g x 0.39 J g-1 0C-1 x (Tf – 0.50 0C) = 0 Tf = 32.1 0C
7.99
Na2CO3(s) + 2HNO3(aq) →
2Na2+(aq) + 2NO3−(aq) + H2O + CO2(g)
∆Hreact = 2[∆Hf°(Na+) + ∆Hf°(NO3−)] + ∆Hf°(CO2) + ∆Hf°(H2O(l) – [∆Hf°(Na2CO3) +2Hf°(HNO3(aq)] ∆Hreact = 2 ×[ (-240.12 kJ mol−1) + (−205.0 kJ mol−1)] + (−393.5 kJ mol−1) + (−285.9 kJ mol−1) – [(−1131 kJ mol-1) + 2 ×(−205.0 kJ mol−1)] ∆Hreact = −28.64 kJ NaHCO3(s) + HNO3(aq) →
Na2+(aq) + NO3−(aq) + H2O + CO2(g)
∆Hreact = ∆Hf°(Na+) + ∆Hf°(NO3-) + ∆Hf°(CO2) + ∆Hf°(H2O(l) – [∆Hf°(NaHCO3) +Hf°(HNO3(aq)] ∆Hreact = (−240.12 kJ mol−1) + (−205.0 kJ mol−1) + (−393.5 kJ mol−1) + (−285.9 kJ mol−1) – [(−947.7 kJ mol−1) + (−205.0 kJ mol−1)] ∆Hreact = +28.18 kJ Na2CO3 gives off the greatest amount of heat. The reaction of HNO3 with NaHCO3 is endothermic.
150
Chapter 7
7.100
We must assume the density of the solution is 1.00 g mL-1 and the specific heat of the solution is the same as water, 4.184 J g-1 0C-1. The heat generated by the reaction of the salts with water is given by:
q = 100.0 g x 4.184 J g-1 0C-1 x 5.33 0C = 2230 J Since the temperature of the water rose when the salts dissolved this value should be negative; -2230 kJ Since the reaction involves the formation of a solution from two solids we need to determine the enthalpy of solution of the two salts. Also, it would be useful to have the enthalpy in units of kJ g-1. NH4Cl(s) → NH4+(aq) + Cl- (aq) ∆Hsoln = -132.5 kJ mol-1 + (-167.2 kJ mol-1) – (-315.4 kJ mol-1) = +15.7 kJ mol-1 +15.7 kJ mol-1 x
1 mol
= 0.294 kJ g-1
53.49g
CaCl2(s) → Ca2+(aq) + 2Cl-(aq) ∆Hsoln = -542.83 kJ mol-1 + 2 x (-167.2 kJ mol-1) – (-795.0 kJ mol-1) = -82.23 kJ mol-1 -82.23 kJ mol-1 x
1 mol 110.99g
= -0.741 kJ g-1
Now we need to set up an equation for the heat generated when the two solids dissolve. We will let x be the mass of NH4Cl. Then, the mass of CaCl2 would be 4.56 – x.
x (0.294 kJ g-1) + (4.56 – x)(-0.741 kJ g-1) = -2.230 kJ x = 1.11 g of NH4Cl 7.101
4.56 g – 1.11 g = 3.45 g CaCl2
The reaction for the combustion of ethanol is: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
∆Hcomb = 2∆Hf°(CO2) + 3∆Hf°(H2O(l) – ∆Hf°C2H5OH) = 2 x (-393.5 kJ mol-1) + 3 x (-285.8 kJ mol-1) – (-277.63 kJ mol-1) = -1366.77 kJ mol-1 of ethanol It would be helpful to have this value in kJ per g of ethanol -1366.77 kJ mol-1 C2H5OH x
1 gal ethanol x
3.785 L gal
7.102
1 mol
= -29.67 kJ g-1
46.07 g
x 1000
mL
x
0.787 g
L
x (29.67
mL
kJ g
Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g)
∆Hreact = ∆Hf°(Ca(OH)2) – 2∆Hf°(H2O) = -986.59 kJ mol-1 – 2(-285.9 kJ mol-1) = -415.1 kJ mol-1 Ca(OH)2
151
) = 8.84. x 104 kJ released
Chapter 7
-415.1 kJ mol-1 Ca(OH)2 x
1 mol 74.09 g
= -5.60 kJ g-1 Ca(OH)2
K(s) + H2O(l) → K+(aq) + OH-(aq) + 1/2H2(g)
∆Hreact = ∆Hf°(K+) + ∆Hf°(OH-) - ∆Hf°(H2O(l) = -252.4 kJ mol-1 + (-230.0 kJ mol-1) – (-285.9 kJ mol-1) = -195.5 kJ mol-1 KOH -195.5 kJ mol-1 KOH x
1 mol
= -3.50 kJ g-1 KOH 56.106 g The reaction of Ca with water gives off more heat per mole and per gram. Now consider the oxidation of the two metals. Ca → Ca2+ + 2e-
K → K+ + e-
Calcium is a two electron oxidation and potassium is a one electron oxidation. 1 mol Ca(OH) 2
-415.1 kJ mol-1 Ca(OH)2 x
-195.5 kJ mol-1 x
2 mol e
1 mol KOH 1 mol e
−
−
= -207.55 kJ mol-1 e-
= -195.5 kJ mol-1 e-
Ca reacting with water gives off more heat per mole of electrons. 7.103
The number of moles of HCl in a 2.5 L bottle can be determined by the following: 12.0 mol HCl L-1 x 2.5 L = 30 mol HCl The number moles of ammonia in the solution can be determined by the following: 13.4 mol NH3(aq) L-1 x 2.5 L = 33.5 mol NH3(aq) The neutralization reaction involves one mole each of the acid and base. Note that we have an excess of NH3, 33.5 mol NH3 versus 30 mol HCl, and thus HCl is the limiting reagent. 50.49 kJ mol-1 HCl x 30 mol HCl = 1515 kJ of heat would be evolved in this reaction.
152
Chapter 8
Practice Exercises
8.1
1× 10−9 m λ = (588 nm) = 5.88 × 10–7 m 1 nm ν=
8.2
8.3
8.4
8.5
3.00 × 108 m / s c = = 5.10 × 1014 s–1 = 5.10 × 1014 Hz − 7 λ 5.88 × 10 m
1×10−6 m λ = (10.9 µm) = 1.09 × 10–5 m 1 µm ν=
2.998 × 108 m / s c = = 2.75 × 1013 s–1 = 2.75 × 1013 Hz − 5 λ 1.09 × 10 m
λ=
c 2.998 × 108 m s −1 = ν 92.3 × 106 s −1
= 3.25 m
1 1 1 = 109,678 cm–1 × (0.0625 – 0.02778) = 109,678 cm −1 × − 2 2 λ 6 4 1 3 –1 = 3.808 × 10 cm λ λ = 2.63 × 10–4 cm = 2.63 µm 1 1 1 = 109,678 cm–1 × (0.2500 – 0.1111) = 109,678 cm −1 × − 2 2 λ 3 2 1 4 –1 = 1.5233 × 10 cm λ λ = 6.565 × 10–5 cm = 656.5 nm, which is red.
8.6
(a) n = 4, l = 2 (b) n = 5, l = 3 (c) n = 7, l = 0
8.7
When n = 2, = 0, 1. Thus we have s, and p subshells. When n = 5, = 0, 1, 2, 3, 4. Thus we have s, p, d, f, and g subshells. The number of subshells spans the values: 0,1,2,3,---, n – 1 Thus, Shell 1: 1 subshell Shell 2: 2 subshells Shell 3: 3 subshells Shell 4: 4 subshells Shell 5: 5 subshells Shell 6: 6 subshells Mg: 1s22s22p63s2 Ge: 1s22s22p63s23p63d104s24p2 Cd: 1s22s22p63s23p63d104s24p64d105s2 Gd: 1s22s22p63s23p63d104s24p64d104f75s25p65d16s2
8.8
(a) (b) (c) (d)
8.9
The electron configuration of an element follows the periodic table. The electrons are filled in the order of the periodic table and the energy levels are determined by the row the element is in and the subshell is
153
Chapter 8
given by the column, the first two columns are the s-block, the last six columns are the p-block, the d-block has ten columns, and the f-block has 14 columns. 8.10
8.11
(b)
O: 1s22s22p4 S: 1s22s22p63s23p4 Se: 1s22s22p63s23p63d104s24p4 P: 1s22s22p63s23p3 N: 1s22s22p3 Sb: 1s22s22p63s23p63d104s24p64d105s25p3 The elements have the same number of electrons in the valence shell, and the only differences between the valence shells are the energy levels.
(a)
Na:
(b)
S
(c)
Fe
(a)
1s
8.12
2s
2p
(a)
Mg
(b)
0 unpaired electrons Ge
(c)
2 unpaired electrons Cd
1s
1s
2s
2s
2p
2p
3s
3p
4s
3s
3p
3d
4s
4p
3s
3p
3d
4s
4p
5s
(d)
3d
4d
0 unpaired electrons Gd 1s
2s
5s
4d
2p
3s
3p
5p
3d 6s
4p 5d
4f
8 unpaired electrons 8.13
Yes, Ti, Cr, Fe, Ni, and the elements in their groups have even numbers of electrons and are paramagnetic. Additionally, oxygen has eight electrons, but it is paramagnetic since it has two unpaired electrons in the 2p orbitals.
8.14
(a)
P: [Ne]3s23p3 [Ne] 3s
3p
(3 unpaired electrons)
154
Chapter 8
(b)
Sn: [Kr]4d105s25p2 [Kr] 4d
5s
5p
(2 unpaired electrons)
8.15
Based on the definition of valence, there are no examples where more than 8 electrons would occupy the valence shell. For representative elements the valence shell is defined as the occupied shell with the highest value of n. In the ground state atom, only s and p electrons fit this definition. The transition elements have outer electron configurations: (n-1)dn nsm so the valence shell is the ns subshell.
8.16
(a) Se: 4s24p4
(b) Sn: 5s25p2
(c) I: 5s25p5
8.17
(a) Sn
(b) Ga
(c) Cr
(d) S2–
8.18
(a) P
(b) Fe3+
(c) Fe
(d) Cl–
8.19
(a) Be
(b) C
8.20
(a) C2+
(b) Mg2+
Review Questions 8.1
Light is a form of energy that results from small oscillations in the electrical and magnetic properties of particles.
8.2
In general, frequency describes the number of times an event occurs in a finite time period. The frequency of light is the number of times a wave crest passes a specific point in space in a given time interval. The symbol for frequency is the Greek letter “nu”, ν, and has the units of inverse seconds, s–1.
8.3
Wavelength is the distance between consecutive maxima of a wave. The symbol is the Greek letter “lambda”, λ.
8.4
See Figure 8.2.
8.5
The amplitude affects the brightness of light. The color of light is affected by the wavelength or frequency. The energy of the light is affected by the frequency or wavelength.
8.6
gamma rays < X rays < ultraviolet < visible < infrared < microwaves < TV waves
8.7
By the visible spectrum, we mean that narrow portion of the electromagnetic spectrum to which our eyes are sensitive. These are the wavelengths from about 400 nm to 750 nm.
8.8
violet < blue < green < yellow < orange < red
8.9
λν = c, where c is a constant equal to the speed of light, λ is the wavelength and ν is the frequency.
8.10
E = hν, where E is the energy, h is Planck’s constant and ν is the frequency.
8.11
A photon is one unit of electromagnetic radiation whose energy is the product of h and ν.
8.12
Since ν = c/λ, we can substitute into equation 8.2 to get E = hc/λ.
8.13
(a) (b) (c)
infrared visible light X-rays 155
Chapter 8
8.14
(d) ultraviolet light The quantum is the lowest possible packet of energy, that of a single photon.
8.15
An atomic spectrum consists of a series of discrete (selected, definite and reproducible) frequencies (and therefore of discrete energies) that are emitted by atoms that have been excited. The particular values for the emission frequencies are characteristic of the element at hand. In contrast, a continuous spectrum, such as that emitted by the sun or another hot, glowing object, contains all frequencies and, therefore, photons of all energies.
8.16
An electron in an atom can have only certain specific values for energy. Aside from these discrete energies, other energies are not allowed. When an excited atom loses energy, not just any arbitrary amount can be lost, only specific amounts of energy can be lost. In other words, the energy of an electron is quantized.
8.17
Bohr proposed a model similar in design to a solar system. The nucleus is at the center and the electrons orbit the nucleus in specific orbits that are a constant fixed distance from the nucleus.
8.18
When an electron falls from an orbit of higher energy (larger radius) to an orbit of lower energy (smaller radius), the energy that is released appears as a photon with the appropriate frequency. The energy of the photon is the same as the difference in energy between the two orbits.
8.19
The lowest energy state of an atom is termed the ground state.
8.20
Bohr's model was a success because it accounted for the spectrum of the hydrogen atom, but it failed to account for spectra of more complex atoms.
8.21
Very small particles have properties that are reminiscent of both a particle and a wave. Massive particles also have this but the wavelike properties are too small to be observed.
8.22
Diffraction is a phenomenon caused by the constructive or the destructive interference of two or more waves. The fact that electrons and other subatomic particles exhibit diffraction supports the theory that matter is correctly considered to have wave nature.
8.23
To determine whether a beam was behaving as a wave or as a stream of particles, a diffraction experiment would have to be done.
8.24
Wave/particle duality is that light and matter have both wave-like properties and particle-like properties.
8.25
In a traveling wave, the positions of the peaks and nodes change with time. In a standing wave, the peaks and nodes remain in the same positions.
8.26
The collapsing atom paradox comes from classical mechanics and asks, "Why doesn't the electron fall into the nucleus?"
8.27
Quantum mechanics resolves the collapsing atom paradox by allowing only certain energy levels for the electron. The electron located in the nucleus does not give one of the allowed energies.
8.28
Wave mechanics and quantum mechanics.
8.29
This is the orbital of the electron.
8.30
First, we are interested in the energies of orbitals, because it is the energies of the various orbitals that determine which orbitals are occupied by the electrons of the atom. Secondly, we are interested in the shapes and orientations of the various orbitals, because this is important in determining how atoms form bonds in chemical compounds
156
Chapter 8
8.31
n = 1, 2, 3, 4, 5, …
8.32
(a)
8.33
Every shell contains the possibility that = 0.
8.34
(a)
8.35
Yes, if the value for for this electron is 3 or larger.
8.36
The only impossible set of legitimate quantum numbers would be an electron having the exact same values for the four quantum numbers. Recall the Pauli Exclusion Principle.
8.37
The electron behaves like a magnet, because the revolving charge (spin) of the electron creates a magnetic field. Recall that the electron does not actually revolve but behaves in a fashion reminiscent of an electromagnet.
8.38
Atoms with unpaired electrons are termed paramagnetic.
8.39
No two electrons in the same atom can have exactly the same set of values for all of the four quantum numbers. This limits the allowed number of electrons per orbital to two, since with other quantum numbers being necessarily the same, two electrons in the same orbital must at least have different values of ms.
8.40
ms = +1/2 or –1/2
8.41
The distribution of electrons among the orbitals of an atom.
8.42
The orbitals within a given shell are arranged in the following order of increasing energy: s < p < d < f.
8.43
The energies of the subshells are quantized.
8.44
The orbitals of a given subshell have the same energy.
8.45
Li Be B C (a) (b)
8.46
n=1
1
(b)
(b)
n=3
3
(c)
5
(d)
7
1s22s1 N 1s22s22p3 2 2 1s 2s O 1s22s22p4 2 2 1 1s 2s 2p F 1s22s22p5 2 2 2 1s 2s 2p Ne 1s22s22p6 1 5 5 1 Cr [Ar]4s 3d or [Ar]3d 4s Cu [Ar]4s13d10 or [Ar]3d104s1
8.47
[Kr]5s14d10
8.48
Elements in a given group generally have the same electron configuration except that the value for n is different, and corresponds to the row in which the element is found. O: [He]2s22p4 S: [Ne]3s23p4 Se: [Ar]4s23d104p4 Te: [Kr]5s24d105p4 Po: [Xe]6s24f145d106p4
8.49
The valence shell is the occupied shell having the largest value of n. The valence electrons are those electrons in the valence shell.
157
Chapter 8
8.50
As explained by the Heisenberg Uncertainly Principle, the position and momentum of an electron cannot be known with 100% certainty, therefore it is impossible to know where an electron is, but we can determine the probability of finding an electron in a given space. An electron is visualized as being within a cloud around the nucleus. This electron cloud defines a volume in space where the probability of finding an electron is high.
8.51
(a)
8.52
As n becomes larger, the orbital becomes larger.
8.53
The three p orbitals of a given p subshell are oriented at right angles (90°) to one another.
8.54
A nodal plane is a plane in which there is zero probability of finding an electron.
8.55
A radial node is a spherical shaped node on which the electron density is zero.
8.56
A p orbital has 1 nodal plane; a d orbital has 2 nodal planes.
8.57
See Figure 8.27.
8.58
The effective nuclear charge is the net nuclear charge that an electron actually experiences. It is different from the formal nuclear charge because of the varying imperfect ways in which one electron is shielded from the nuclear charge by the other electrons that are present. The effective nuclear charge remains nearly constant from top to bottom in any one group of the periodic table, and it increases from left to right in any one row of the periodic table.
8.59
The larger atoms are found in the lower left corner of the periodic table; the smaller atoms are found in the upper right corner of the periodic table.
8.60
The size changes within a transition series are more gradual because, whereas the "outer" electrons are in an s subshell, the electrons that are added from one element to another enter an inner (n – 1)d subshell.
8.61
Ionization energy is the energy that is needed in order to remove an electron from a gaseous atom or ion. These are positive values because the force of attraction between an electron and the nucleus of either an atom or a positive ion (cation) must be overcome in order to remove the electron.
8.62
(a) (b)
8.63
Ionization energy increases from left to right in a row of the periodic table because the effective nuclear charge increases from left to right. The latter trend occurs because of the consequences of the increasingly imperfect shielding of electrons by other electrons within the same level. The ionization energy decreases down a periodic table group because the electrons reside farther from the nucleus with each successive quantum level that is occupied. The farther the electrons are from the nucleus, the less tightly they are held by the nucleus.
8.64
Removing a second electron involves pulling it away from a greater positive charge because of the positive charge created by the removal of the first electron. Hence, more energy must be spent to ionize the second electron than the first.
8.65
The fifth electron must be removed from a different, lower n shell. Also, an electron must be removed from an ion having a 4+ rather than only a 3+ charge.
8.66
The last valence electron of aluminum begins the occupation of the 3p set of orbitals. This electron is therefore well shielded from the nuclear charge by the 3s electrons.
See Figures 8.22a and 8.23
(b)
See Figures 8.24a, 8.25 and 8.26.
O(g) O+(g) + e– O2+(g) O3+(g) + e–
158
Chapter 8
8.67
The last valence electron of sulfur is placed in an orbital with another electron, and the two electrons are required to be spin-paired by the Pauli exclusion principle. This destabilizes the last electron, making it easier to ionize than is the case for the last electron of phosphorus.
8.68
Electron affinity is the enthalpy change associated with the addition of an electron to a gaseous atom: X(g) + e– X–(g)
8.69
S(g) + e– S–(g), first electron affinity S–(g) + e– S2–(g), second electron affinity The first electron affinity is exothermic, while the second electron affinity should be endothermic because work must be done to force the electron into the negative S– ion.
8.70
The value for fluorine is low because of the especially small size of the atom, which causes addition of an electron to be relatively unfavorable. The comparison between chlorine and bromine follows the normal trend on descent of a group, the electron affinity decreasing with ionization energy.
8.71
The second electron affinity is always unfavorable (endothermic) because it requires that a second electron be forced onto an ion that is already negative.
8.72
The electron affinity becomes more negative (exothermic) as effective nuclear charge increases. Fluorine therefore has the more exothermic electron affinity because it has the larger effective nuclear charge.
Review Problems The number used for the speed of light, c, depends on the number of significant figures. For one to three significant figures, the value for c is 3.00 × 108 m/s, for four significant figures, the value for c is 2.998 × 108 m/s. 8.73
ν=
c 3.00 × 108 m/s = λ 563 × 10 −9 m
8.74
ν=
c 3.00 × 108 m/s = 1.01 × 1015 s–1 = 1.01 × 1015 Hz = λ 295 × 10 −9 m
8.75
ν=
c 3.00 × 108 m/s = 4.38 × 1013 s–1 = 4.38 × 1013 Hz = −6 λ 6.85 × 10 m
8.76
ν=
c 3.00 × 108 m/s = 6.25 × 1014 s–1 = 6.25 × 1014 Hz = λ 0.48 × 10 −6 m
8.77
295 nm = 295 × 10–9 m c 3.00 × 108 m/s = 1.02 × 1015 s–1 = 1.02 × 1015 Hz ν= = λ 295 × 10 −9 m
8.78
8.79
= 5.33 × 1014 s–1 = 5.33 × 1014 Hz
c 2.99792458 × 108 m/s = 4.73670145 × 1014 s–1 = 4.73670145 × 1014 Hz = −9 λ 632.99139822 × 10 m The speed of light is only given to 9 significant figures in the text. ν=
101.1 MHz = 101.1 × 106 Hz = 101.1 × 106 s–1 c 2.998 × 108 m/s = 2.965 m λ= = ν 101.1 × 106 s −1
159
Chapter 8
8.80
5.09 × 1014 Hz = 5.09 × 1014 s–1 c 3.00 × 108 m/s 1 nm λ= = = 5.89 × 10 −7 m × = 589 nm 14 − 1 ν 5.09 × 10 s 1 × 10 −9 m
(
)
c 3.00 × 108 m/s = 5.0 × 106 m = 5.0 × 103 km = −1 ν 60 s
8.81
λ=
8.82
1.50 × 1018 Hz = 1.50 × 1018 s–1 c 3.00 × 108 m/s = 2.00 × 10–10 m = 2.00 × 10–1 nm = 2.00 × 102 pm λ= = 18 -1 ν 1.50 × 10 s
8.83
E = hν = 6.63 × 10–34 J s × (4.0 × 1014 s–1) = 2.7 × 10–19 J 2.7 × 10−19 J 6.02 × 1023 photons J = = 1.6 × 105 J mol–1 mol 1 photon 1 mol
8.84
E = hν = hc/λ, and 436 nm = 436 × 10–9 m E=
8.85
8.86
8.89
8.90
(
)
m
s
)
= 4.56 × 10–19 J
violet (see Figure 8.7)
(b)
ν = c/λ = 2.998 × 108 m s–1 / 410.3 × 10–9 m = 7.307 × 1014 s–1
(c)
E = hν = (6.626 × 10–34 J s) × (7.307 × 1014 s–1) = 4.842 × 10–19 J
(a)
yellow (See Figure 8.7) 3.00 × 10 8 m c s = 5.09 × 1014 s 1 ν = = λ 589 × 10 9 m E = hν = (6.63 × 10–34 Js) × (5.09 × 1014 s–1) = 3.37 × 10–19 J
(c)
8.88
)(
(a)
(b)
8.87
(
6.63 × 10−34 J s 3.00 × 108 hc = λ 436 × 10−9 m
1 1 1 3 –1 −1 = 109,678 cm −1 × − = 109,678 cm × ( 0.1111 − 0.02778 ) = 9.140 × 10 cm 2 2 λ 6 3 λ = 1.094 × 10–4 cm = 1094 nm We would not expect to see the light since it is not in the visible region. 1 1 1 4 –1 −1 = 109,678 cm −1 × − = 109,678 cm × ( 0.250 − 0.0400 ) = 2.303 × 10 cm 2 λ 52 2 λ = 434.2 nm, this will be purple and will just be visible 1 1 1 = 5.758 × 103 cm–1 = 109,678 cm −1 × − 2 2 λ 10 4 λ = 1.737 × 10–6 m, this is in the infrared region 1 1 1 –1 −1 = 109,678 cm −1 × − = 109,678 cm × (1 − 0.0625 ) = 102,823 cm 2 λ 42 1 λ = 9.73 × 10–6 cm = 97.3 nm, which is in the ultraviolet region.
160
Chapter 8
E=
(
)(
)
6.626 x 10 −34 J s 3.00 x 108 m s −1 hc = = 2.04 x 10 −18 J λ 97.3 x 10 −9 m
8.91
(a)
s
(b)
d
8.92
(a)
4
(b)
2
8.93
(a)
n = 4, = 1
(b)
n = 6, = 0
8.94
(a)
n=4, =1
(b)
n = 6, = 3
8.95
0, 1, 2, 3, 4
8.96
n=8
8.97
(a)
8.98
–5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5
8.99
When m = –4 the minimum value of is 4 and the minimum value of n is 5.
8.100
If l = 5 we should have 2l + 1 orbitals which gives eleven orbitals. There are eleven values for m: –5, –4, – 3, –2, –1, 0, 1, 2, 3, 4, and 5. n m ms 3 1 –1 +1/2 3 1 –1 –1/2 3 1 0 +1/2 3 1 0 –1/2 3 1 +1 +1/2 3 1 +1 –1/2
8.101
8.102
n 3 3 3 3 3 3 3 3 3 3
m = 2, 1, 0, –1 or –2
2 2 2 2 2 2 2 2 2 2
m –2 –2 –1 –1 0 0 1 1 2 2
(b)
ms +1/2 –1/2 +1/2 –1/2 +1/2 –1/2 +1/2 –1/2 +1/2 –1/2
8.103
15 electrons have = 1, 10 electrons have = 2
8.104
12 electrons have = 0, 12 electrons have m = 1
8.105
(a) (b) (c) (d)
P Ca V Sb
m = 4, 3, 2, 1, 0, –1, –2, –3, or –4
1s22s22p63s23p3 1s22s22p63s23p64s2 1s22s22p63s23p63d 34s2 1s22s22p63s23p63d 104s24p64d 105s25p3 161
Chapter 8
1s22s22p63s23p64s23d 104p3 1s22s22p63s23p5 1s22s22p63s23p64s23d8 1s22s22p63s23p2
8.106
(a) (b) (c) (d)
As Cl Ni Si
8.107
(a) (b) (c) (d) (e)
Ga : [Ar]3d104s24p1, ∴ one unpaired electron, paramagnetic Ba : [Xe]6s2, ∴ zero unpaired electron, not paramagnetic Cr : [Ar]4s13d5, ∴ six unpaired electrons, paramagnetic Si : [Ne]3s23p2, ∴ two unpaired electrons, paramagnetic Ne : [He]2s22p6, ∴ zero unpaired electrons, not paramagnetic
8.108
(a) (b) (c) (d)
Ba is [Xe]6s2, zero unpaired electrons: diamagnetic Se is [Ar]4s23d104p4, two unpaired electrons: paramagnetic Zn is [Ar]4s23d10, zero unpaired electrons: diamagnetic Si is [Ne]3s23p2, two unpaired electrons: paramagnetic
8.109
(a) (b) (c)
Cu : [Ar]4s13d10, one unpaired electron Se : [Ar]3d104s24p4, two unpaired electrons Ca : [Ar]4s2, zero unpaired electron
8.110
(a) (b) (c)
Cs is [Xe]6s1, 1 unpaired electron S is [Ne]3s23p4, 2 unpaired electrons Ni is [Ar]4s23d8, 2 unpaired electrons
8.111
(a) (b) (c) (d) (e)
Fe Rb Sn Cl Sb
[Ar]4s23d6 [Kr]5s1 [Kr] 4d105s25p2 [Ne] 3s23p5 [Kr] 4d105s25p3
8.112
(a) (b) (c) (d) (e)
Al Se Ba Sb Gd
[Ne]3s23p1 [Ar]4s23d104p4 [Xe]6s2 [Kr]5s24d105p3 [Xe]6s24f75d1
(a)
Na
(b)
Sc
8.113
1s
1s
8.114
a)
As:
(b)
Ni:
1s
or [Ar]3d104s24p4 or [Kr]4d105s25p3 or [Xe]4f75d16s2
2s
2p
3s
2s
2p
3s
3p
4s
3d
2s
2p
3s
3p
4s
3d
162
4p
Chapter 8
8.115
(a)
Cr [Ar]
(b)
I [Kr]
(c)
In [Kr]
(d)
Cs [Xe]
(a)
Al
4s
3d
4p
5s
4d
5p
5s
4d
5p
6s
8.116
[Ne]
(b)
3s
3p
4s
3d
4p
5s
4d
5p
Se [Ar]
(c)
Ba
(d)
Sb [Kr]
8.117
The value corresponds to the row in which the element resides: (a) 4 (b) 6 (c) 4 (d) 3
8.118
The value corresponds to the row in which the element resides: (a) 3 (b) 4 (c) 6 (d) 5
8.119
(a) K
4s1
(b) Si 3s23p2
(c) Sr 5s2
(d) Te 5s25p4
8.120
(a) Mg 3s2
(b) Br 4s24p5
(c) Ga 4s24p1
(d) Pb 6s26p2
8.121
(a)
K
(b)
Si
(c)
Sr
(d)
Te
4s
3s
3p
5s
5s
8.122
(a)
5p
Mg 3s
3p
163
Chapter 8
(b)
Br
(c)
Ga
(d)
Pb
4s
4s
6s
8.123
(a)
4p
4p
6p
8.124
(a)
There are 10 core electrons in Mg so the valence electron would see 12 – 10 or +2 as the effective nuclear charge. There are 10 core electrons in P so the valence electrons would see 15 – 10 or +5 as the effective nuclear charge. There are 2 core electrons in O so the valence electrons would see 8 – 2 or +6 as the effective nuclear charge. 2 (b) 4 (c) 7
8.125
(a)
Na
(b)
Pt
8.126
(a)
Al
(b)
Tl
8.127
Te
8.128
Since these atoms and ions all have the same number of electrons, the size should be inversely related to the positive charge: Mg2+ < Na+ < Ne < F– < O2– < N3–
8.129
Cations are generally smaller than the corresponding atom, and anions are generally larger than the corresponding atom: (a) Ca (b) S2– (c) Fe2+
8.130
Cations are generally smaller than the corresponding atom, and anions are generally larger than the corresponding atom: (a) S2– (b) Al (c) Au+
8.131
(a)
Si
(b)
Ca
(c)
Se
8.132
(a)
Li
(b)
F
(c)
F
8.133
(a)
Br
(b)
As
8.134
(a)
S
(b)
N
8.135
The element with the largest difference between the second and third ionization potential would be the element with two valence electrons. The third ionization would remove an electron from the core, which is much higher in energy than removing electrons from valence levels. Mg has the valence structure:
8.136
Si
(b) (c)
164
Chapter 8
Additional Exercises
8.137
(speed = wavelength × frequency) v = λ × ν v λ= ν 330 m s (a) longest wavelength = = 16. m 20 s −1 330 m s = 0.016 m shortest wavelength = 20,000 s −1 (b)
8.138
1500 m s = 75 m 20 s −1 1500 m s shortest wavelength = = 0.075 m 20,000 s −1
longest wavelength =
We proceed by calculating the energy of a single photon: E=
(
)(
6.626 × 10 −34 J s 3.00 × 108 m s hc = λ 3.00 × 10 −3 m
(
)
) = 6.63 × 10
−23
J
It requires 4.184 J to increase the temperature of 1.00 g of water by 1 Celsius degree. So, photons = 4.184 J ÷ 6.63 × 10–23 J/photon = 6.32 × 1022 photons
8.139
1 1 1 = 109,678 cm −1 × − n 2 n 2 λ 1 2
1 1 nm 1 m = 410.3 nm 1× 10 −9 m 100 cm
1 1 1 −1 × − = 109,678 cm −5 2 cm x2 2 4.103 × 10 2.437 × 104 cm −1 −1
109678 cm 1 = − 0.222 2 4 x x = 6
=
1 1 − = 0.222 4 x2
1
8.140
This corresponds to the special case in the Rydberg equation for which n1 = 1 and n2 = ∞. 1 1 1 –1 = 109,678 cm −1 × − = 109,678 cm −1 × (1 − 0 ) = 109,678 cm 2 2 λ 1 ∞ λ = 9.12 × 10–6 cm = 91.2 nm.
8.141
A transition from high energy to low energy may result in light emission. The transition from 5p4d and 4d2s are the only possibilities.
8.142
(a)
Start by calculating λ 1 1 1 5 –1 = 109,678 cm −1 × − = 109,678 cm −1 × (1.000 − 0.04000 ) = 1.053 × 10 cm 2 2 λ 5 1 165
Chapter 8
(b)
(c)
λ = 9.498 × 10–6 cm = 94.98 nm, which is in the ultraviolet region of the visible spectrum. 1 1 1 4 –1 −1 = 109,678 cm −1 × − = 109,678 cm × ( 0.2500 − 0.0625 ) = 2.056 × 10 cm 2 λ 42 2 λ = 4.863 × 10–5 cm = 486.3 nm, which is in the visible region of the spectrum. 1 1 1 3 –1 = 109,678 cm −1 × − = 109,678 cm −1 × ( 0.0625 − 0.0278 ) = 3.808 × 10 cm 2 2 λ 6 4 λ = 2.626 × 10–4 cm = 2626 nm, which is in the infrared region of the spectrum.
8.143
(a) (b) (c) (d)
A d shell does not exist for n = 2. Using the shorthand notation of [Ar], we imply that 3s shell is already filled. There is nothing wrong. [Na] is not a core group of electrons.
8.144
(a)
This diagram violates the aufbau principle. Specifically, the s-orbital should be filled before filling the higher energy p-orbitals. This diagram violates the aufbau principle. Specifically, the lower energy s-orbital should be filled completely before filling the p-orbitals. This diagram violates the aufbau principle. Specifically, the lower energy s-orbital should be filled completely before filling the p-orbitals. This diagram violates the Pauli Exclusion Principle since 2 electrons have the same set of quantum numbers. This electron distribution is impossible.
(b) (c) (d)
8.145
13
8.146
The 4s electron is lost; n = 4, = 0, m = 0, ms = +1/2
8.147
(a)
This corresponds to the special case in the Rydberg equation for which n1 = 1 and n2 = ∞. For a single atom, we have: 1 1 1 = 109,678 cm −1 × − = 109,678 cm −1 × (1 − 0 ) = 109,678 cm −1 2 2 λ ∞ 1 λ = 9.12 × 10–6 cm = 91.2 nm. Converting to energy, we have: E = hc/λ
(
)(
6.626 × 10 −34 J s 3.00 × 108 m s hc E= = λ 91.2 × 10 −9 m
(
(b)
)
)
= 2.18 × 10–18 J
Conversion to kJ/mol gives us: 2.18 × 10−18 kJ/mol = photon
J 1 kJ 6.02 × 1023 photons = 1.31 × 103 kJ/mol 1000 J mole
166
Chapter 8
8.148 CI 3s
3p
+ 1e-
CI 3s
3p
Thus the chloride ion, Cl–, has the electron configuration of [Ar]. Since the chloride ion has a closed shell electron configuration, the addition of a second electron is extremely difficult. 8.149
8.150
We simply reverse the electron affinities of the corresponding ions. (a) F–(g) F(g) + e–, ∆H° = 328 kJ/mol (b) O–(g) O(g) + e–, ∆H° = 141 kJ/mol (c) O2–(g) O–(g) + e–, ∆H° = –844 kJ/mol The last of these is exothermic, meaning that loss of an electron from the oxide ion is favorable from the standpoint of enthalpy. In problem 8.149 parts b and c we can determine that O(g) + 2e– O2–(g) ∆H = +703 kJ From table 8.2 we can see that O(g) O+(g) + e– ∆H = +1314 kJ/mol. It takes more energy to ionize oxygen than to create O2–(g)
Multi-Concept Problems 8.151
E= v =
2E m
1 mv2 2
first determine E for a single particle
1 mole –18 E = (2080 × 103 J/mol) = 3.454 × 10 J/particle 6.022 × 1023 particles v=
8.152
(a)
2(3.454 × 10 −18 J) 9.109 × 10−31 kg
= 2.754 × 106 m/sec
We must first calculate the energy in joules of a mole of photons.
(
)(
)
6.63 × 10 −34 J s 3.00 × 108 m/s hc = 3.32 × 10–19 J/photon = −9 λ 600 × 10 m (3.32 × 10–19 J/photon)(6.02 × 1023 photons/mol) = 2.00 × 105 J/mol
E=
Next, we calculate the heat transfer problem as in Chapter 7: Heat = (specific heat)(mass)(change in temperature) 2.00 × 105 J = 4.18 J g–1 °C–1 × X g × 5.0 °C X = 9.57 × 103 g
167
Chapter 8
(b)
8.153
To solve this problem use E = hc/λ, where λ is our unknown quantity. λ=
8.154
E = 6.63 × 10–19 J/photon E = 3.99 × 105 J/mol X = 1.91 × 104 g
(
)(
6.63 × 10 −34 J s 3.00 × 108 m s hc = E 328 × 103 J mol
(
)
)
( 6.02 × 10
23 photons mol
) = 3.65 × 10
–7
m = 365 nm
From Table 8.2 we find that the first ionization energy for sodium is 496 kJ mol-1. We need to determine of a wavelength of 23.7 nm would have enough energy to remove the 3s1 electron. If this wavelength corresponds to an energy greater than 496 kJ mol-1, then the excess energy would be observed as the electron’s kinetic energy. E = hν = hc/λ E = [6.63 x 10-34 J s x 3.00 x 108 m s-1/23.7 x 10-9 m] x 6.02 x 1023 atoms mol-1 = 5.05 x 106 J or 5050 kJ The maximum kinetic energy of the ionized electron would be the difference between the energy of the photon and the ionization energy.
KE = 5050 kJ – 496 kJ = 4554 kJ 8.155
The ionization energies of B are given in Table 8.2. All of the following values are in kJ mol-1. 1st IE = 800
2nd IE = 2426
3rd IE = 3659
4th IE = 25,020
5th IE = 32,820
To determine the kinetic energy of the ionized electron we need to first determine the energy of the ionizing photon. The kinetic energy of each ejected electron will be the difference between this energy and the energy required to ionize the electron. Ephoton = [6.63 x 10-34 J s x 3.00 x 108 m s-1/23.7 x 10-9 m] x 6.02 x 1023 atoms mol-1 = 5.05 x 106 J Or 5050 kJ IE
KE
1st 2nd 3rd
5050 – 800 kJ = 4250 kJ 5050 – 2426 kJ = 2624 kJ 5050 – 3659 kJ = 1391 kJ
The photon’s energy is not great enough to remove electrons beyond these. The 4th ionization energy is 25,020 kJ or about 5 times the energy of the ionizing photon.
168
Chapter 9
Practice Exercises 9.1
There is one electron missing, and it should go into the 5s orbital, and the 5p orbital should be empty. 1s12s22p63s23p63d104s24p64d105s2
9.2
Cr: [Ar]3d54s1 (a) Cr2+: [Ar]3d4 (b) Cr3+: [Ar]3d3 (c) Cr6+: [Ar]
9.3
The 4s electron and one 3d electron are lost. The 4s electron and two 3d electrons are lost. The 4s electron and all of the 3d electrons are lost.
S2–: [Ne] 3s23p6 Cl–: [Ne] 3s23p6 The electron configurations are identical.
9.4 I
Ca
Ca2+
I
+
2
9.5 O
Mg2+
Mg
+
9.6
aldedyde
amine
acid
ketone
alcohol
169
O
2-
I
Chapter 9
9.7
(a) CH3NHCH3 will produce a basic solution (b) HCOOH will produce an acidic solution (c)
9.8
µ=q×r
1.602 × 10−19 C q = 0.167 e– = 2.675 × 10–20 C − 1e r = 154.6 pm = 154.6 × 10–12 m q = (2.675 × 10–20 C) × (154.6 × 10–12 m) = 4.136 × 10–30 C m in debey units: 1D q = 4.136 × 10–30 C m = 1.24 D −30 Cm 3.34 × 10 9.9
q=
µ r
3.34 ×10−30 C m µ = 9.00 D = 3.006 × 10–29 C m 1 D –12 r = 236 pm = 236 × 10 m 3.006 ×10−29 C m q= = 1.27 × 10–19 C 236 × 10−12 m In electron charges 1 e− q = 1.27 × 10–19 C = 0.795 e– 1.602 × 10−19 C On the sodium the charge is +0.795 e– and on the chlorine, the charge is –0.795 e–. This would be 79.5% positive charge on the Na and 79.5% negative charge on the Cl. 9.10
The bond is polar and the Cl carries the negative charge.
9.11
(a)
Br
(b)
Cl
(c)
9.12 O H
O
P
O
H
O
Number of valence electrons: O 6 each and 4 O atoms total 24 electrons P 5 electrons H 1 each and 2 H atoms total 2 electrons Negative charge 1 electron Total 32 valence electrons
170
Cl
Chapter 9
9.13
SO2
O
S
O
NO3– O O
N
O
HClO3
H3AsO4 H O H
O
As
O
H
O
9.14
9.15
SO2 has 18 valence electrons SeO42– has 32 valence electrons NO+ has 10 valence electrons F
O
+
H
F
H
N
H
H O
S
_
O
O O
N
O
F F
Cl
O
F
H
O
Cl O
171
O
Chapter 9
9.16
The negative sign should be on the oxygen, so two of the oxygen atoms should have a single bond and three lone pairs and the sulfur should have one double bond, two single bonds, and a lone pair.
9.17 (a)
(b)
_
-2 N
9.18
N +1
O +1
S
C
N
0
0
-1
In each of these problems, we try to minimize the formal charges in order to determine the preferred Lewis structure. This frequently means violating the octet rule by expanding the octet. Of course, this can only be done for atoms beyond the second period as the atoms in the first and second periods will never expand the octet. (a) SO2 O
(b)
HClO3
(c)
H3PO4
H
S
O
O
O
H
P
O
H
O
9.19
There is no difference between the coordinate covalent bond and the other covalent bonds.
172
Chapter 9
9.20 H
H+
+
O
H
O
H
coordinate covalent bond 9.21 There are four resonance structures. 3O O O
P
O
O
3-
P
O
O
P
3-
O
O
O
O
O
O
3-
O
P
O
O
9.22
O
H
H
O
O
C
O
O
C
O
9.23
_
O O
Br
O
_
O O
Br
O
_
O O
Br
O
Review Questions 9.1
A stable compound forms from a collection of atoms when bonding results in a net lowering of the energy. The process of bonding the atoms together must release energy to make the bonded compound more stable than the original collection.
9.2
The ionic bond is the attraction between positive and negative ions in an ionic compound. It is largely an electrostatic attraction, and it gives rise to the lattice energy of the ionic compound. The lattice energy is the net gain in stability when the gaseous ions are brought together to form the crystalline ionic compound.
9.3
Ionic bonds tend to form upon combining an element having a high EA with an element having a low IE.
9.4
The lattice energy is the energy necessary to separate a mole of an ionic solid into its constituent ions in the gas phase. It is also the energy that is released on forming the ionic solid from the gaseous ions. As discussed in the answer to review questions 9.1 and 9.2, it is the lattice energy that is primarily responsible for the stability of ionic compounds.
9.5
Magnesium ([Ne]3s2) can achieve the electron configuration of the nearest noble gas (Ne) by losing only two electrons: Mg2+ 1s22s22p6. Magnesium will not form the Mg3+ ion because an extremely high amount of energy would be required to break into the 2s22p6 core to remove an electron.
173
Chapter 9
9.6
When chlorine gains one electron to form Cl–, it has filled an orbital and achieved a noble gas configuration. To make the Cl2– ion, an electron would have to be placed in the next higher shell. The amount of energy required for this to occur is extremely high and makes the creation of a Cl2– ion energetically unfavorable.
9.7
Many of the transition metals in Period 4 have an 4s2 outer-shell electron configuration. Since these characteristically are the first electrons to be lost when a transition metal atom is ionized, it is common that a 2+ ion should be formed.
9.8
The largest difference in ionization energies would be between the third and the fourth successive ionization energies because of the extremely high amount of energy that would be required to break into the 2s22p6 core to remove an electron. (a) Al2O3 (b) BeO (c) NaCl
9.9
9.10
The valence electrons are primarily responsible for chemical bonds.
9.11
(a) (b) (c) (d)
9.12
Ionic bonding does not occur between two nonmetal elements because more energy must be provided in the form of IE and EA than can be recovered from the lattice energy.
9.13
As two hydrogen atoms approach each other in forming the H2 molecule, the electron density of the two atoms shifts to the region between the two nuclei.
9.14
The energy drops to some optimum or minimum value when the nuclei have become separated by the distance called the bond distance. The electron spins become paired.
9.15
Bond formation is always exothermic.
9.16
The bond distance in a covalent bond is determined by a balance (compromise) between the separate attractions of the nuclei for the electron density that is found between them, and the repulsions between the like-charged nuclei and those between the like-charged electrons. These attractions and repulsions oppose one another, and a bond distance is achieved that maximizes the attraction while minimizing the repulsions.
9.17
The octet rule is the expectation that atoms tend to lose or gain electrons until they achieve a noble gas-type electron configuration, namely eight electrons in the valence, or outer-most, shell. It is the stability of the closed-shell electron configuration of a noble gas that accounts for this.
9.18
(a)
9.19
The valence shell of a period 2 element can hold only eight electrons since only two types of subshells, s and p, are available in this period. The valence shell of elements in row three can hold as many as eighteen electrons. This results because there are three types of subshells, s, p and d, that can hold electrons. The s subshells hold 2 electrons, p subshells hold 6 electrons and d subshells hold 10 electrons. The total is, therefore, 18 electrons.
9.20
(a) (b) (c)
correct incorrect correct incorrect
one
(b)
four
(c)
two
(d)
three
(e)
one
single bond: a covalent bond formed by the sharing of one pair of electrons. double bond: a covalent bond formed by the sharing of two pairs of electrons. triple bond: a covalent bond formed by the sharing of three pairs of electrons.
174
Chapter 9
9.21
H
C
N
9.22
Since the outer shell (or valence shell) of hydrogen can hold only two electrons, hydrogen is not said to obey the octet rule. It does, however, readily satisfy its requirement for a closed shell electron configuration through the formation of one covalent bond.
9.23
Methane
Ethane
Propane
9.24
The is an example of an alkane, carbon-hydrogen compounds with single bonds. The general formula for alkanes is CnH2n+2 so we should have 14 hydrogen atoms. The formula is C6H14, and is named n-hexane. 175
Chapter 9
9.25
There are three isomers of pentane.
9.26
A carbonyl group is shown below.
This group is found in organic acids such as benzoic acid (shown below), ketones, aldehydes, amides, and esters.
9.27
acid ketone alcohol aldehyde amine hydrocarbon
9.28
CH3CH2COOH(aq) + H2O
→
CH3CH2COO-(aq) + H3O+(aq)
The acid is propanoic acid, a weak acid
176
Chapter 9
The anion formed is the propanoate ions:
CH3NH2CH2CH3+(aq) + OH-(aq)
9.29
CH3NHCH2CH3(aq) + H2O
9.30
CH3CH2COOH(aq) + CH3NHCH2CH3(aq)
9.31
Ethylene
9.32
A polar covalent bond is one in which the electrons of the bond are not shared equally by the atoms of the bond, and this causes one end of the linkage to carry a partial negative charge while the other end carries a corresponding partial positive charge. In other words, there is a dipole in a polar bond.
9.33
µ=q×r A dipole moment is the product of the amount of charge on one end of a polar bond (which behaves as a dipole) and the distance between the two partial charges that compose the dipole. It is also normally taken to be the product of the charge in the dipole of a polar bond and the internuclear distance in the polar bond.
→
→
CH3CH2COO-(aq) + CH3NH2CH2CH3+(aq) acetylene
1 debye = 3.34 × 10–30 coulomb × meter 9.34
Electronegativity is the attraction that an atom has for the electrons in chemical bonds to that atom. Pauling based his scale of electronegativity on the greater bond energy polar bonds have than would be expected if the opposite ends of the bonds were electrically neutral.
9.35
Fluorine has the largest electronegativity, whereas oxygen has the second largest electronegativity.
9.36
(b) and (d) are the only ones with an electronegativity difference greater than 1.7.
9.37
Elements having low electronegativities are metals. Elements with low ionization potentials and low electron affinities tend to have low electronegativities.
9.38
This has to do with the ease of oxidation. Aluminum is easier to oxidize than iron.
177
Chapter 9
9.39
The most reactive metals are in the left-most groups of the periodic table, namely the metals of Groups IA and IIA. The least reactive metals are found in the second and third rows of the transition elements.
9.40
The lower the electronegativity, the more reactive is the metal.
9.41
calcium > iron > silver > iridium
9.42
(a) (b) (c) (d) (e) (f)
NR 2NaI(aq) + Cl2(g) 2NaCl(aq) + I2(s) 2KCl(aq) + F2(aq) 2KF(aq) + Cl2(g) CaBr2(aq) + Cl2(g) CaCl2(aq) + Br2(l) 2AlBr3(aq) + 3F2(g) 2AlF3(aq) + 3Br2(l) NR
9.43
(a) (d)
F2 S8
9.44
(a) (b) (c)
four electrons or two bonding pairs six electrons, all in bonding pairs two electrons for each hydrogen
9.45
We expect a minimum of ten electrons, in bonding pairs between As and each of five Cl atoms.
9.46
N is in the second period, it can only have an octet of electrons, so it cannot form five bonds. Whereas, As is in the fourth row, it can have an expanded octet and form five bonds; therefore, it can form AsCl3 and AsCl5.
9.47
Bond length is the distance between the nuclei of two atoms that are linked by a covalent bond. Bond energy is the energy that is needed to break a chemical bond; conversely, the energy that is released when a chemical bond is formed.
9.48
The number of electron pairs that comprise a covalent linkage (bond) between two atoms is called the bond order. As bond order increases, so does the strength of the bond. Therefore, as bond order increases, bond energy increases and bond length decreases, in keeping with increased bond strength.
9.49
The H — Cl bond energy is defined to be the energy required to break the bond to give atoms of H and Cl, not the ions H+ and Cl–. In other words, when the bond is broken between the H and the Cl atoms, one of the two electrons of the bond must go to each of the atoms. When ions are obtained, however, both electrons of the bond are given to the chlorine atom and the lattice energy would be the more appropriate term to use.
9.50
The formal charge on an atom is its apparent charge in a particular Lewis diagram. For any particular atom in a Lewis diagram, the formal charge is calculated by subtracting the number of bonds to the atom, and the number of unshared electrons on the atom, from the number of valence electrons that the unbonded atom normally has in the ground state.
9.51
It is generally held that the Lewis structure having the smallest formal charges is most favored. This is based on the notion that either an accumulation of too much charge, or too much charge separation, is destabilizing.
9.52
The formal charges on the atoms in HCl are 0. The actual charges are H: +0.17e– and Cl: –0.17e–. The formal charges arise from the bookkeeping in Lewis structures, and are not the same as actual charges.
9.53
A coordinate covalent bond is one in which both electrons of the bond are contributed by (or donated by) only one of the atoms that are linked by the bond.
(b) (e)
P4 Cl2
(c) (f)
178
Br2 S8
Chapter 9
9.54
Once formed, a coordinate covalent bond is no different than any other covalent bond.
9.55 Cl Cl
Cl
B
O
Cl
H
Cl
H
B
O
Cl
H
H
9.56
Lewis structures are often inadequate, without the concept of resonance, to describe the distribution of electrons in certain molecules and ions. Sometimes more than one Lewis structure is needed to describe a molecule. These structures differ only in the placement of electrons. Resonance more fully describes the distribution of electrons in a molecule.
9.57
A resonance hybrid is the true structure of a molecule or polyatomic ion, whereas the various resonance structures that are used to depict the hybrid do not individually have any reality. The hybrid is a mix, or average, of the various resonance structures that compose it.
9.58 H
H H
C
H
C C H
H
C
C
C C
H
C
C
H
H
C C
H
H
H
The resonance structures of benzene are more stable than the ring containing three carbon-carbon double bonds because six bonds with a bond order of 1.5 is more stable than three single bond and three double bonds together. 9.59 H
H
C C
H
H
C
H
HH H C
C H
C
C C
H
H
H
C
H
H HH C
C
C
C C
H
H
H
C
H
H
HH H C
C
C
C C
H
H
H
H
C
H
H HH C
C
C
C C
H
H
H
C
H C
C
C
H
H
Review Problems 9.60
Na(g) Na+(g) + e– Cl(g) + e– Cl–(g) Na(g) + Cl(g) Na+(g) + Cl–(g)
IE = 496 kJ/mol EA = –348 kJ/mol ∆H = 148 kJ/mol
Na(g) Na+(g) + e– Na+(g) Na2+(g) + e– 2Cl(g) + 2e– 2Cl–(g) Na(g) + 2Cl(g) Na2+(g) + 2Cl–(g)
IE = 496 kJ/mol IE = 4563 kJ/mol EA = 2(–348 kJ/mol) ∆H = 4.36 ×103 kJ/mol
In order for NaCl2 to be more stable than NaCl, the lattice energy should be almost 30 times larger. 4.36 ×103 kJ/148 kJ = 29.5
179
Chapter 9
9.61 Mg2+(g) + O 2-(g) +844 kJ Mg2+(g)
+ O(g) -141 kJ
Mg2+(g) + O- (g)
+1450 kJ
Mg+(g) + O(g)
-3642 kJ
+737 kJ Mg(g) + O(g) +150 kJ Mg(s) + 1/2O2(g) -602 kJ
MgO(s)
The lattice energy for NaCl is 787 kJ. The large difference in lattice energy is a result of charge (divalent versus univalent) and size of ions. 9.62
Calcium loses two electrons: Ca Ca2+ + 2e– [Ar]4s2 Chlorine gains an electron: Cl + e– Cl– [Ne]3s23p5 [Ar] To keep the overall change of the formula unit neutral, two Cl– ions combine with one Ca2+ ion to form CaCl2: Ca2+ + 2Cl– CaCl2
9.63
Nitrogen (1s22s22p3) gains three electrons to achieve the electron configuration of the next noble gas, neon: N3– 1s22s22p6 Lithium (1s22s1) loses an electron to achieve the electron configuration of the closest noble gas helium: Li1+ 1s2 To keep the overall change of the formula unit neutral, three Li+ ions combine with one N3– ion to form Li3N: 3Li+ + N3– Li3N
9.64
Sn2+: [Kr] 4d105S2 Sn4+: [Kr] 4d10
9.65
Bi3+: Bi5+:
[Xe]6s24f145d10 [Xe]4f145d10
9.66
Zn2+:
[Ar]3d10 no unpaired electrons 180
Chapter 9
9.67
Co3+:
9.68
(a)
[Ar]3d6 4 unpaired electrons (b)
(c)
(d) (e)
9.69
(a)
(b) Ge
K
(d)
As
(e) Br
9.70
(c)
Se
(a)
(b)
(c)
9.71
(a)
Mg
Mg 2+ +
S
S
2-
(b) Cl
Mg
Cl
N
Mg
Mg2+
+
2 Cl
(c) Mg
9.72
N
Mg
µ=q×r
3.34 ×10−30 C ⋅ m 1m µ = 0.16 D = q × (115 pm) 12 10 pm 1D 5.34 × 10–31 C·m = q × (115 × 10–12 m) q = 4.64 × 10–21 C 1 e– q = 4.64 × 10–21 C = 0.029 e– 1.60 × 10−19 C 181
3 Mg2+ + 2
N
3-
Chapter 9
The charge on the oxygen is –0.029 e– and the charge on the nitrogen is +0.029 e–. The nitrogen atom is positive. 9.73
µ=q×r
3.34 ×10−30 C ⋅ m 1m µ = 1.42 D = q × (176 pm) 12 10 pm 1D 4.74 × 10–30 C·m = q × (176 × 10–12 m) q = 2.69 × 10–20 C 1 e– q = 2.69 × 10–20 C = 0.17 e– 1.60 ×10-19 C The charge on the fluorine is –0.17 e– and the charge on the bromine is +0.17 e–. The bromine atom is positive. 9.74
µ=q×r
3.34 ×10−30 C ⋅ m 1m µ = 1.83 D = q × (91.7 pm) 12 10 pm 1D 6.11 × 10–30 C·m = q × (91.7 × 10–12 m) q = 6.66 × 10–20 C 1 e– q = 6.69 × 10–20 C = 0.42 e– 1.60 ×10-19 C The charge on the fluorine is –0.42 e– and the charge on the hydrogen is +0.42 e–. The fluorine atom is negative. 9.75
µ=q×r
3.34 ×10−30 C ⋅ m 1m µ = 7.88 D = q × (0.255 nm) 1 D 109 nm 2.63 × 10–29 C·m = q × (0.255 × 10–9 m) q = 1.03 × 10–19 C 1 e– q = 1.03 × 10–19 C = 0.64 e– 1.60 ×10-19 C The charge on the fluorine is –0.64 e– and the charge on the cesium is +0.64 e–. The cesium atom is positive.
9.76
9.77
1 mole J /molecule = (155 × 103 J/mol) 23 6.022 × 10 molecules = 2.57 × 10–19 J/molecule We can use Hess’s Law, Tables 8.2 and 8.3 to answer this question. (1)
H(g) → H+(g) + e-
IE = 1312 kJ mol-1
(2)
Cl(g) + e- →
EA = -348 kJ mol-1
(3)
HCl(g) → H(g) + Cl(g)
Cl-(g)
BE = 431 kJ mol-1
If we reverse all three chemical equations and add them together the results are: H+(g) + Cl-(g)
→ HCl(g) 182
Chapter 9
The energy released from this reaction is: ∆H = -1312 kJ mol-1 + 348 kJ mol-1 – 431 kJ mol-1 = -1395 kJ mol-1 9.78
E = hν =
hc λ
λ =
hc E
1 mol –19 E = (348 × 103 J/mol) = 5.78 × 10 J/molecule 23 6.022 × 10 molecules
( 6.626 × 10 λ=
−34
)(
J sec 3.00 × 108 m s −1
(5.78 × 10
−19
J
)
) = 3.44 × 10
–7
m = 344 nm
Ultraviolet region 9.79
hc E 1 mol –19 E = (242.6× 103 J/mol) = 4.029 × 10 J/molecule 23 6.022 × 10 molecules
E = hν =
hc λ
( 6.626 × 10 λ= 9.80
λ =
−34
(
)(
J sec 3.00 × 108 m s −1
4.029 × 10
−19
J
)
) = 4.934 × 10
(a)
(b)
(c) H
9.81
+
Br
H – Br
(a)
(b)
183
–7
m = 493.4 nm
Chapter 9
(c)
(d)
9.82
(a) (b) (c)
9.83
(a)
(b) (c)
9.84
We predict the formula H2Te because tellurium, being in Group VIA, needs only two additional electrons (one each from two hydrogen atoms) in order to complete its octet. Phosphorus, being in Group VA, needs three electrons from hydrogen atoms in order to complete its octet, and we predict the formula PH3. Carbon is in Group IVB, and it needs four electrons (and hence four hydrogen atoms) to complete its octet: CH4. Each chlorine atom needs one further electron in order to achieve an octet, and the phosphorus atom requires three electrons from an appropriate number of chlorine atoms. We conclude that a phosphorus atom is bonded to three chlorine atoms and that each chlorine atom is bonded only once to the phosphorus atom: PCl3. Since carbon needs four additional electrons, the formula must be CF4. In this arrangement, each fluorine atom acquires the one additional electron that is needed to reach its octet. Each halogen atom needs only one additional electron from the other: ICl.
Here we choose the atom with the smaller electronegativity: (a) Cl (b) S (c) P (d)
S
9.85
(a)
N
9.86
Here we choose the linkage that has the greatest difference in electronegativities between the atoms of the bond: P—O.
9.87
The least polar bond of the four is P—I because it is the bond that has the smallest difference in electronegativities between the linked atoms.
9.88
(a)
I
(b)
I
(c)
F
(d)
(b)
+
Cl
_ O
Cl
As
Cl
O
Cl
Cl
(c)
(d) F
H
O
N
O
184
Xe
F
Chapter 9
9.89
(a)
(b) F
F
F
F Te
Cl
F
F F
F
F
(c)
(d) _ F F
F P
F
F
F F
F
9.90
F Xe
(a)
F
(b) Cl Cl
Si
Cl
F
Cl
F
(c)
P
F
(d) H H
9.91
P
Cl
H
(a)
S
Cl
(b) O H
O
I
O O
H
(c)
O
C
O
H
(d)
+
Cl
_
O H
O
C
Cl
P Cl
O
185
Cl
Chapter 9
9.92
(a)
(b)
9.93
(a)
(b)
9.94
(a)
(b) H H
As
H
H
(c)
O
Cl
O
(d) H O H
O H
9.95
Se
O
O
O
As
H
O
H
(a)
O
(b) 1
H
Cl Cl
H
O
H
Cl P
Cl
(c)
Cl
(d) 3–
O O
P O
186
O
Chapter 9
9.96
(a)
(b) O
O H
9.97
C
Cl
H
(a)
S
Cl
(b) Cl Cl
Ge
Cl
2-
O
Cl
O
(c)
C
O
(d) 3-
O O
P
O 2-
O
9.98
O
(a)
O
(b) 1-
O 0
0
H
-1
Cl
O
O
1-
0
+1
O
S 2+
O
(c) 1+ 0
1O
9.99
S
O
(a)
(b) 0 O
0
1+ F
0
Cl
N 1+
O
1-
F
12+
F
S
F 0
0 O
N
0
(c) O
F
1-
187
O 0
1-
Chapter 9
9.100
9.101 1-
O 0
Cl
S
O 0
Cl
0
Cl
0
S
Cl
0
0
1+
9.102
The formal charges on all of the atoms of the left structure are zero, therefore, the potential energy of this molecule is lower and it is more stable. Group 2 atoms normally have only two bonds.
9.103
The one on the right is better. The one on the left gives one chlorine atom a formal charge of +1, which is not a likely situation for such a highly electronegative element.
9.104 H
O
H
H
+
H
O
H
+
H
9.105
9.106
The average bond order is 4/3 2-
O O
C
O
2-
O O
C
188
O
2-
O O
C
O
Chapter 9
9.107 O
O N
O
O
N
O N
O
O
O
O
N
N O
O
N
O
O N
O
N
O
O
The average bond order is 3/2.
9.108
bond order = 1 1/3
bond order = 1½
Both have resonance structures. SO42- has a greater bond order, therefore the S–O bonds will be shorter than the S–O bond in SO32−
9.109
The Lewis structures of the four must be compared. Carbon monoxide has a triple bond, :O≡C:; the formate ion (see text) has an average CO bond order of 1.5; CO2 has double bonds; and CO32– has a bond order of 1.33 (from review problem 9.106) The order of increasing CO bond length is: CO < CO2 < HCO2– < CO32–
9.110 O
C
O
O
C
O
These are not preferred structures, because in each Lewis diagram, one oxygen bears a formal charge of +1 whereas the other bears a formal charge of –1. The structure with the formal charges of zero has a lower potential energy and is more stable.
9.111
0 0
0
O
Cl
1-
O
O 1O 0
O
0
O
189
O 0
Cl 0
Chapter 9
_
O O
O
Cl
O
_ Cl
O
O
Cl
Cl
Cl
O
Cl
_
O
O
O
O
O
O
O
_
O
O
_
O
O O
_
O
O
_
O O
Cl
O
O
The average bond order in ClO3– is 1.67 and the average bond order in ClO4– is 1.75. Since ClO4– has the larger bond order, it will have the shorter bond length.
Additional Exercises
9.112 Ca 2+(g) + 2Cl(g) -348 kJ x 2 +1146 kJ
Ca 2+(g) + 2Cl-(g)
Ca +(g) + 2Cl(g) +589.5 kJ Ca(g) + 2Cl(g)
-2269
+192 kJ Ca(s) + 2Cl(g) +242.6 kJ Ca(s) + Cl2 (g)
-795 kJ
CaCl2(s)
190
Chapter 9
9.113
Electron affinity for Br = –331.4 kJ/mol 9.114 Cl Cl
Sn
Cl
Cl
9.115
(a)
9.116
Two electrons 5s2
9.117
(a)
Carbon nearly always makes four bonds and this structure has only 3. Additionally, the formal charge on the carbon atom is –1. Also, there are too many electrons. The structure should have 24 but there are 26 shown.
(b)
The formal charge on the oxygen atom bonded to the hydrogen is +1 which is unlikely since it is the most electronegative element in the molecule.
(c)
Carbon never makes more than four bonds.
9.118
AlCl3
(b)
SiO2
(c)
SeF2
(d)
SeCl2
The two structures are not resonance structures because the atoms do not have the same connectivity. The first structure with the carbon in the middle is a resonance structure of COCl2. The other structure is not because the atoms are in a different order, and have a different connectivity. The latter structure is not reasonable one because carbon has a –2 formal charge and oxygen has a +2 formal charge.
191
Chapter 9
9.119
Given:
Better (lower energy) structure:
0 H
1+ N
0
N
N
1+
H
1+ N 0
2-
N
N -1
9.120
9.121
Of the four resonance structures, the one in the upper left is the best possible structure. Each of the terminal nitrogen atoms has a –1 formal charge and the central atom has a formal charge of +1. The structure in the upper right is also a good structure except that the formal charge on the rightmost nitrogen atom is –2 and the central nitrogen is +1 while the leftmost nitrogen has a formal charge of zero. The bottom two resonance structures are both poor because each violates the octet rule. The left bottom structure has too many electrons around the central nitrogen atom while the right bottom structure has too few electrons on the central atom.
9.122
The average bond orders are: NO2 (1.50), NO2– (1.50), NO3– (1.33). Therefore the NO bond lengths should vary in the following way: NO3– > NO2≅NO2−
9.123 Cl
9.124
S
S
Cl
Because this is an acid it is probable that the hydrogen atom is bonded to the most electronegative atom. For this molecule, that is O. So, the structures HOCN and HONC are the most likely.
9.125
3.34 × 10 8.59 D 1D –19 q = 1.32 × 10 C
−30
C⋅m
µ=q×r 1m = q × 217 pm 1012 pm
The charge of an electron is 1.60 × 10–19 C. The amount of charge is
1.32 × 10−19 C 1.60 × 10−19 C
Therefore, the amount of charge on the K is +0.826 e– and the amount of charge on the F is –0.826 e–. 9.126
The formula for the hydrocarbon is C3H8 and the proper name for this compound is propane.
192
Chapter 9
Multi-Concept Problems 9.127
9.128
9.129 H H N
H
C
H
H
C
C
C H
H
C
C
H
C
H
H
H
H
H
H
Moles of acid
12.5 g x C3 H5COOH x
1 mol 86.09 g
= 0.145 mol
Moles of Base
17.4 g C4 H9 NH 2 x
1 mol 73.14 g
= 0.239 mol
Since the reaction involves one mole of acid reacting with one mole of base, the base is in excess. Grams of unreacted base = (0.239 mol – 0.145 mol) x 73.14 g mol-1 = 6.78 g 9.130
Let q be the amount of energy released in the formation of 1 mol of H2 molecules from H atoms: 435 kJ/mol, the single bond energy for hydrogen. q = specific heat × mass × ∆T ∴ mass = q ÷ (specific heat × ∆T)
193
O
C H
H H
H
C
C H
H
C
H H
C
H
H
N
C
O
H H
O
H H
H C
H
H
O
H
H
H
Chapter 9
g H 2O =
9.131
(435 × 103 J)
( 4.184
J g C
) (100
C − 25 C
)
= 1.4 × 103 g
E = energy required to atomize H2 (bond energy) + IEH + EAH E = 435 kJ/mol + 1312 kJ/mol + (–73 kJ/mol) 1 mol 1000 J –18 E = 1674 kJ/mol = 2.78 × 10 J/molecule 23 1 kJ 6.022 × 10 molecule E = hν
E=
hc λ
2.78 × 10–18 J/molecule =
(6.626 × 10−34 Js)(3.00 × 108 m/s) λ
7.15 × 10–8 m = 71.5 nm (These are very high energy photons, in the X-ray region of the electromagnetic spectrum.) 9.132
First we need to determine the empirical formula of the compound using the combustion analysis data. Since the data is in mg it is just as convenient to work in mmoles, 1/1000 of a mole, rather than convert masses to grams. Molar masses remains numerically the same, that is mg mmol-1 rather than g mol-1. mmol C = 37.54 mg CO 2 x mmol H = 7.684 mg H 2 O x
1 mmol CO 2 44.01 mg 1 mmol H 2 O 18.01 mg
x x
1 mmol C mmol CO 2 2 mmol H mmol H 2 O
= 0.853 mmol = 0.853 mmol
To determine the moles of oxygen on the compound we need to subtract the mass of C and H from the sample mass of 38.40 mg. mg of C = 0.812 mmol x
12.01 mg C
= 9.75 mg mmol 1.008 mg H mg of H = 0.853 mmol H x = 0.860 mg mmol
mg of O = 38.40 mg –(9.75 mg C + 0.860 mg H) = 27.82 mg mmol of O = 27.79 mg O x empirical formula:
1 mmol O 16.00 g
= 1.74 mmol
C0.853H0.853O1.74
Divide by the smallest mmol value to obtain whole numbers. CHO2.14 This is most likely CHO2 taking into consideration experimental error in the combustion data.
The molar mass is 90 and the empirical mass is 45 so there are two empirical units per molecule. The molecular formula is C2H2O4 The titration data can be used to determine how many acid protons are in the compound.
194
Chapter 9
Moles of base used = 14.28 mL x
Mole of acid = 40.2 mg x
mmoles of acid x
0.0625 mmol
1 mmol 90 mg
mL
= 0.8925 mmol
= 0.447 mmol
number of acid protons
= mmol of base mmol of acid number of acid protons 0.8925 mmol = =2 mmol of acid 0.447 mmol
Thus, we know that both protons in the compound are acid protons.
The Lewis structure for the compound, oxalic acid, is shown below.
195
Chapter 10
Practice Exercises 10.1
SeF6 should have an octahedral shape (Figure 10.4) because it has six electron pairs around the central atom.
10.2
SbCl5 should have a trigonal bipyramidal shape (Figure 10.4) because, like PCl5, it has five electron pairs around the central atom.
10.3
HArF should have a linear shape (Figure 10.7) because although it has five electron pairs around the central Ar atom, only two are being used for bonding.
10.4
IBr2− should have a linear shape (Figure 10.7) because although it has five electron pairs around the central I atom, only two are being used for bonding.
10.5
RnF4 should have a square planar shape (Figure 10.10) because although it has six electron pairs around the central Xe atom, only four are being used for bonding.
10.6
In SO32–, there are three bond pairs and one lone pair of electrons at the sulfur atom, and as shown in Figure 10.5, this ion has a trigonal pyramidal shape. In PbCl4, there are four bonding domains (4 single bonds) around the lead atom, and as shown in Figure10.4, this molecule has a tetrahedral shape. In XeO4, there are four bond pairs of electrons around the Xe atom, and as shown in Figure 10.5, this molecule is tetrahedral. In OF2, there are two bond pairs and two lone pairs of electrons around the oxygen atom, and as shown in Figure 10.5, this molecule is bent.
10.7
SF4 is distorted tetrahedral and has one lone pair of electrons on the sulfur, therefore it is polar.
10.8
(a) TeF6 is octahedral, and it is not polar. (b) SeO2 is bent, and it is polar. (c) BrCl is polar because there is a difference in electronegativity between Br and Cl. (d) AsH3, like NH3, is pyramidal, and it is polar. (e) CF2Cl2 is polar, because there is a difference in electronegativity between F and Cl.
10.9
The H–Cl bond is formed by the overlap of the half–filled 1s atomic orbital of a H atom with the half–filled 3p valence orbital of a Cl atom: Cl atom in HCl (x = H electron): x 3s
3p
The overlap that gives rise to the H–Cl bond is that of a 1s orbital of H with a 3p orbital of Cl:
10.10
The half–filled 1s atomic orbital of each H atom overlaps with a half–filled 3p atomic orbital of the P atom, to give three P–H bonds. This should give a bond angle of 90°. P atom in PH3 (x = H electron):
196
Chapter 10
x 3s
x
x
3p
The orbital overlap that forms the P–H bond combines a 1s orbital of hydrogen with a 3p orbital of phosphorus (note: only half of each p orbital is shown):
z y x
10.11
BF3 uses sp2 hybridized orbitals since it has only three bonding electron pairs and no lone pairs of electrons. The sp2 hybrid orbitals on the B, x = F electron x
x
x
2p
sp 2
10.12
BeF2 uses sp hybridized orbitals since it has only two bonding electron pairs and no lone pairs of electrons. The sp hybrid orbitals on the Be; x = F electron x
x
2p
sp
10.13
sp3 x sp
x
x
x
3
10.14
Since there are five bonding pairs of electrons on the central phosphorous atom, we choose sp3d hybridization for the P atom. Each of phosphorous’s five sp3d hybrid orbitals overlaps with a 3p atomic orbital of a chlorine atom to form a total of five P–Cl single bonds. Four of the 3d atomic orbitals of P remain unhybridized.
10.15
VSEPR theory predicts that AsCl5 will be trigonal bipyramidal. Since there are five bonding pairs of electrons on the central arsenic atom, we choose sp3d hybridization for the As atom as a trigonal bipyramid. Each of arsenic's five sp3d hybrid orbitals overlaps with a 3p atomic orbital of a chlorine atom to form a total of five As–Cl single bonds. Four of the 4d atomic orbitals of As remain unhybridized.
10.16
sp3d2
10.17
(a) (b)
PCl3 has four pair of electrons around the central atom, P, so the hybridization is sp3 BrCl3 has five pair of electrons around the central atom, Br, so the hybridization is sp3d
197
Chapter 10
10.18
NH3 is sp3 hybridized. Three of the electron pairs are use for bonding with the three hydrogens. The fourth pair of electrons is a lone pair of electrons. This pair of electrons is used for the formation of the bond between the nitrogen of NH3 and the hydrogen ion, H+.
10.19
Since there are six bonding pairs of electrons on the central phosphorous atom, we choose sp3d2 hybridization for the P atom. Each of phosphorus’s six sp3d2 hybrid orbitals overlaps with a 3p atomic orbital of a chlorine atom to form a total of six P–Cl single bonds. Three of the 3d atomic orbitals of P remain unhybridized. P atom in PCl6– (x = Cl electron): x
x
x
x
x
x
3d
sp3 d 2
The ion is octahedral because six atoms and no lone pairs surround the central atom.
10.20
Atom 1has three electron domains: sp2 Atom 2 has four electron domains: sp3 2 Atom 3 has three electron domains: sp There are 10 σ bonds and 2 π bonds in the molecule.
10.21
Atom 1has two electron domains: sp Atom 2 has three electron domains: sp2 3 Atom 3 has four electron domains: sp There are 9 σ bonds and 3 π bonds in the molecule.
10.22
CN– has 10 valence electons and the MO diagram is similar to that of C and N. The bond order of the ion is 3 and this does agree with the Lewis structure. σ∗2pz π∗ 2px
π∗ 2spy
σ2pz π2px
π2py
σ∗2s σ2s
10.23
NO has 11 valence electrons, and the MO diagram is similar to that shown in Table 10.1 for O2, except that one fewer electron is employed at the highest energy level
198
Chapter 10
σ∗2pz π∗ 2px
π∗ 2spy
π2px
π2py
σ2pz σ∗2s σ2s
(8 bonding e ) − (3 antibonding e ) = 5 Bond Order = −
−
2 The bond order is calculated to be 5/2.
2
Review Questions 10.1
10.2
(a) (b) (c) (a) (b)
See Figure 10.4. The angles are 120°. See Figures 10.4 and 10.5 The bond angles are 109.5°. See Figures 10.4 and 10.10. The bond angles are 90°. See Figure 10.4. The bond angle is 180°. See Figures 10.4 and 10.7. The bond angles are 120° between the equatorial bonds and 180° between the two axial bonds.
10.3
VSEPR Theory is based on the principle that adjacent electron pairs repel one another, and that these destabilizing repulsions are reduced to a minimum when electron pairs stay as far apart as possible.
10.4
An electron domain is a region in space where electrons can be found.
10.5
In HCHO, there are three bonding domains around the C; one bonding domain around each H, and one bonding domain around O and two nonbonding domains around O.
10.6
(a)
90° bond angles
(b)
120° bond angle between equatorial bonds and 180° bond angle between axial bonds
(c)
90° bond angles between equatorial bonds and 90° bond angles between equatorial bonds and axial bond
(a) (b) (c) (d)
Planar triangular, otherwise known as trigonal planar Octahedral Tetrahedral Trigonal bipyramidal
10.7
199
Chapter 10
10.8
Polar molecules attract one another, and that influences the physical and chemical properties of substances.
10.9
A bond's dipole moment is depicted with an arrow having a + sign on one end, where the "barb" of the arrow is taken to represent the location of the opposing negative charge of the dipole:
10.10
A molecule will be polar if the bonds are polar and these polar bonds are arranged in such a geometry that they do not cancel one another.
10.11
A molecule having polar bonds will be nonpolar only if the bond dipoles are arranged so as to cancel one another's effect.
10.12 O
O S
The individual bond dipoles do not cancel one another. 10.13
Both VB and MO theory have wave mechanics as their theoretical basis. In each theory, bonds are considered to arise from the overlap of orbitals. Valence bond theory uses the overlap of atomic orbitals while molecular orbital theory uses the molecular orbitals to describe the bonding.
10.14
Lewis structures do not explain how atoms share electrons, nor do they explain why molecules adopt particular shapes. Also, electrons are known to be delocalized, a fact which only MO theory addresses from the beginning. Also, odd-electron systems can be more effectively discussed with MO theory.
10.15
VB theory views atoms coming together with their orbitals already containing specific electrons. The bonds that are formed according to VB theory do so by the overlap of orbitals on neighboring atoms, and this is accompanied by the pairing (sharing) of the electrons that are contained in the orbitals. MO theory, on the other hand, considers a molecule to be a collection of positive nuclei surrounded by a set of molecular orbitals, which, by definition, belong to the molecule as a whole, rather than to any specific atom. The electrons of the molecule are distributed among the molecular orbitals according to the same rules that govern the filling of atomic orbitals.
10.16
Orbital overlap occurs when orbitals from different atoms share the same space. This overlap provides a more stable region for the electrons, which find themselves under the stabilizing influence of the positive charge of two nuclei.
10.17
The greater the orbital overlap the greater the bond energy.
10.18
This is the same as the HF molecule, shown in Figure 10.16. A half–filled valence p orbital of Br overlaps with the half–filled 1s orbital of the hydrogen atom.
10.19
Hybrid orbitals provide better overlap than do atomic orbitals, and this results in stronger bonds.
10.20
These are shown in Figures 10.21.
200
Chapter 10
10.21
sp3d
sp3d2
10.22
Elements in period 2 do not have a d subshell in the valence level.
10.23
Lewis structures are something like shorthand representations of VB descriptions of molecules.
10.24
The VSEPR model gives the shape of the electron domains around the central atom; this information is used to predict the hybridization of the atom.
10.25
90°
10.26
sp3 C atom sp3 N atom
sp3 O atom
There are zero, one and two lone pairs, respectively, on these atoms when sp3 hybridization is utilized to form bonds. 10.27
(a) H
c H
H
H
(b)
N H
H
H
(c)
O H H
10.28
This angle would have had to be 90°, the angle between one atomic p orbital and another.
201
Chapter 10
10.29
See Figure 10.25. Valence shell for boron
2s
See Figure 10.33 Valence shell for carbon
2p
2s
sp2 hybridized boron atom
sp2
2p
sp2 hybridized carbon atom
sp2
2p
2p
10.30
sp3d – trigonal bipyramid
10.31
The ammonium ion has a tetrahedral geometry, with bond angles of 109.5°. + H N H
sp3d2 – octahedral
H H
10.32
The geometry of the boron BCl3 is trigonal planar; after the addition of the water molecule, it becomes tetrahedral. The hybridization changes from sp2 to sp3. The geometry of the oxygen in the water molecule changes from bent to trigonal pyramidal after coordination to the boron atom. Its hybridization does not change, but remains at sp3.
10.33
σ bond – The electron density is concentrated along an imaginary straight line joining the nuclei of the bonded atoms.
10.34
π bond – The electron density lies above and below an imaginary straight line joining the bonded nuclei. The characteristic side–to–side overlap of p atomic orbitals that characterizes π bonds is destroyed upon rotation about the bond axis. This is not the case for a σ bond, because regardless of rotation, a σ bond is still effective at overlap.
10.35
See Figure 10.33.
10.36
See Figure 10.35.
10.37
Two resonance hybrids should be drawn.
sp2 hybrid orbitals are used for each carbon atom. Consequently, each bond angle is 120°. See Figure 9.42. 10.38
If an electron is forced to occupy the higher energy MO, the molecule loses stability and the bond is made weaker than if an electron, or a pair of electrons, occupies the lower energy (bonding) MO.
202
Chapter 10
10.39
The overlap will create a bonding MO.
10.40
There are two regions of overall between the two 3d orbitals so this forms a π
10.41
In the hypothetical molecule He2, both the bonding and the antibonding MO are doubly occupied, and the net bond order is zero, as shown in Figure 10.39. H2 has a bond order of 1
10.42
As shown in Table 9.1, the highest energy electrons in dioxygen occupy the doubly degenerate π– antibonding level:
↑
↑
π*2p x
π*2p y
orbital.
Since their spins are unpaired, the molecule is paramagnetic. 10.43
As shown in Table 10.1, the bond order of Li2 is 1.0. The bond order of Be2 would be zero. Yes, Be2+ could exist since the bond order would be 1/2.
10.44
(a)
10.45
As bond order increases, bond energy (and strength) increases.
10.46
See Figure 10.40.
10.47
A delocalized MO is one that extends over more than two nuclei. Benzene is planar and each carbon atom has a half-filled p orbital perpendicular to the plane of the molecule. These orbitals overlap to form a pi bonding system throughout the molecule.
10.48
Delocalization increases stability.
10.49
Delocalization energy is a term used in MO theory to mean essentially the same thing as the term resonance energy, which derives from VB theory. They both represent the additional stability associated with a spreading out of electron density.
10.50
For conductors, metals, the valence band and the conduction band are the same so electrons can easily move throughout the metal.
2.5
(b)
1.5
(c)
1.5
For insulators, non-metals, the valence band is filled and the gap between the valence band and conduction band is very large so electrons cannot move into the conduction band. For semiconductors, the valence band is filled but the separation between the valence band and the conduction band is small so electrons can easily be promoted into the conduction. 10.51
The band containing the valence electrons of the element is called the valence band. The conduction band is the empty or partial filled band existing throughout the element. 203
Chapter 10
10.52
Increasing the temperature of a semiconductor excites more electrons from the valence band into the conduction band.
10.53
The 2s band is completely filled and much lower in energy than the conduction band. It is not possible to excite electrons from the 2s band into the conduction band. Calcium’s valence band is formed from the 4s electrons and therefore, it is that band that conducts electrons.
10.54
p- type transistors are those formed by doping silicon with elements containing three valence electrons, the 3A family of elements. The doping creates a site with a positive “hole.” It is the movement of the “hole” throughout the semiconductor that results in current flow. An n-type transistor if formed by doping silicon with elements containing five valence electrons, the 5A family of elements. Doping results in an extra electron at the doping site. This electron can move into the conduction band and cause current flow.
10.55
Any element in the 3A family such as boron or gallium would form a p-type germanium transistor. Any element in the 5A family such as phosphorus or arsenic would form an n-type germanium transistor.
10.56
A solar cell operates on the principle that light falling on the cell disturbs the equilibrium of the p-n junction causing a flow of current. LEDs operate on the opposite principle of a solar cell. That is, a current flows through the p-n junction causing an emission of light.
10.57
Of the nonmetals, only the noble gases exist in nature as isolated atoms.
10.58
Period 2 elements form strong π bonds because the atom size is small which enables atoms to approach each other very closely. Period 3 elements (and all other elements) are too large to accommodate the close proximity required to form strong π bonds. Because the period 3 and higher elements are large and form only weak π bonds, it is much more effective to form only σ bonds than to form a σ bond and a π bond.
10.59
Because the halogens need only a single electron to complete their valence shell, these elements do not require the formation of π bonds. The formation of a single σ bond creates an extremely stable diatomic molecule.
10.60
An allotrope is a different structure or physical form of an element. An isotope is a form of an atom that differs from other atoms of the same element in the number of neutrons in the nucleus. An example is H and D, two of the isotopes of hydrogen.
10.61
The allotropes of oxygen are dioxygen, O2, and ozone, O3.
204
Chapter 10
10.62
The two unpaired electrons are located in the π antibonding molecular orbitals. σ2px*
π2py*, π 2pz* π2py, π 2pz
σ2px
σ2s*
σ2s
The overall bond order is (8Be– – 4ABe–)/2 = 2. 10.63 O
O
O
The molecule is nonlinear and uses sp2 hybrid orbitals. There is a lone pair of electrons on the central oxygen, consequently, ozone is indeed a polar molecule. Additionally, the formal charges suggest that O3 is polar. 10.64
The presence of ozone in the upper atmosphere shields the earth from harmful ultraviolet radiation.
10.65
Diamond is a network covalent solid in which each carbon atom is the corner of a tetrahedron. Each carbon atom is bonded to four other carbon atoms using sp3 hybrid orbitals.
10.66
Graphene is a monolayer of graphite and is formed by the overlap of neighboring 2p orbitals to form a πtype bonding throughout the layer. Carbon forms sheets of sp2 – sp2 sigma (σ) bonds.
10.67
The weak π bonds that exist between the sheets of sp2 bonded carbon are easily broken. The lubricating properties of graphite are a result of the weak attraction between parallel sheets.
10.68
C60, or buckminsterfullerene, resembles a soccer ball. It consists of hexagonal and pentagonal arrangements of carbon atoms. This arrangement naturally assumes a spherical geometry. The molecule is named in honor of R. Buckminster Fuller, an architect. Other roughly spherical arrangements of carbon atoms are also possible. While not perfectly spherical, they resemble buckminsterfullerene and are given the general name of fullerenes.
10.69
Carbon nanotubes have the same molecular structure as graphite, but instead of forming sheets, the layer of carbon curves and forms a tube.
205
Chapter 10
10.70
S8
10.71
White phosphorous, P4, is a tetrahedron. See Figure 10.53
. 10.72
The bond angle in P4 is 60°. Using only p orbitals, a bond angle of 90° is expected. Because the observed bond angle is much less, the molecule is extremely reactive.
10.73
Red phosphorus consists of tetrahedra joined as in Figure 10.54. While it too is used in explosives, it is much less reactive than white phosphorus.
10.74
Black phosphorus, like graphite, exists as parallel sheets of atoms.
10.75
Silicon has a molecular structure identical to diamond, i.e., a network covalent solid arranged in a tetrahedral manner. Silicon, unlike carbon, does not form multiple bonds. Consequently, there is no graphite-like allotrope of silicon.
Review Problems 10.76
(a) (b) (c) (d) (e)
Bent (central N atom has two single bonds and two lone pairs) Planar triangular (central C atom has three bonding domains—a double bond and two single bonds) T-shaped (central I atom has three single bonds and two lone pairs) Linear (central Br atom has two single bonds and three lone pairs) Planar triangular (central Ga atom has three single bonds and no lone pairs)
10.77
(a) (c) (e)
trigonal pyramidal tetrahedral linear
10.78
(a) (b) (c) (d) (e)
Bent (central F atom has four electron domains; two are lone pair) Trigonal bipyramidal (central As atom has five electron domains) Trigonal pyramidal (central As atom has four electron domains; one is a lone pair) Trigonal pyramidal (central Sb atom has four electron domains; one is a lone lair) Bent (central Se atom has four electron domains; two are lone pair)
10.79
(a) (c) (e)
distorted tetrahedral nonlinear tetrahedral
10.80
(a) (b) (c) (d) (e)
Tetrahedral (central I atom has four electron domains) Square planar (central I atom has six electron domains; two are lone pair) Octahedral (central Te atom has six electron domains) Tetrahedral (central Si atom has four electron domains) Linear (central I atom has five electron domains; three are lone pair)
10.81
(a) (c)
linear T–shaped
(b) (d)
(b) (d)
(b) (d)
tetrahedral nonlinear (bent)
octahedral tetrahedral
square planar trigonal pyramidal 206
Chapter 10
(e)
planar triangular
10.82
PF4−
The central P atom has five electron domains, one is a lone pair
10.83
AsH3
10.84
180°
10.85
all angles 120°
10.86
(a) (c) (e)
109.5° 120° 109.5°
(b) (d)
109.5° 180°
10.87
(a) (c) (e)
90° 180° 109.5°
(b) (d)
109.5° 109.5°
10.88
The ones that are polar are (a), (b), and (c). The last two have symmetrical structures, and although individual bonds in these substances are polar bonds, the geometry of the bonds serves to cause the individual dipole moments of the various bonds to cancel one another.
10.89
I3− contains no polar bonds. BeCl2 is linear so the individual dipoles cancel each other. C2H6 has atoms
There are two electron domains around each carbon atom so the molecule is linear
symmetrically distributed and is non-polar. SO2 and SbCl52− both have a lone pair of electrons and have a bent shape and square pyramid shape respectively, and are therefore polar molecules. 10.90
All are polar. (a), (b), (c) and (e) have asymmetrical structures, and (d) only has one bond, which is polar.
10.91
OH− is linear and polar. Each of the other structures have one or more lone pairs on the central atom and are therefore polar.
10.92
In SF6, although the individual bonds in this substance are polar bonds, the geometry of the bonds is symmetrical which serves to cause the individual dipole moments of the various bonds to cancel one another. In SF5Br, one of the six bonds has a different polarity so the individual dipole moments of the various bonds do not cancel one another.
10.93
In SiCl4, although the individual bonds in this substance are polar bonds, the geometry of the bonds is symmetrical which serves to cause the individual dipole moments of the various bonds to cancel one another. In SiHCl3, one of the four bonds has a different polarity so the individual dipole moments of the various bonds do not cancel one another.
10.94
The 1s atomic orbitals of the hydrogen atoms overlap with the mutually perpendicular p atomic orbitals of the selenium atom. Se atom in H2Se (x = H electrons): x 4s
10.95
x
4p
This is shown in Figure 10.18 for F2 . Since Cl2 is in the same family the bonding will be similar, using the n = 3 shell rather than the n = 2 shell. We can diagram it showing the orbitals of one of the chlorine atoms: Each Cl atom (x = an electron from the other Cl atom):
207
Chapter 10
10.96
Atomic B:
2s 2p Hybridized B: (x = a Cl electron)
10.97
(a)
Si atom in SiCl4 (x = a Cl electron) x sp
(b)
x
x
x
3
P atom in PCl5 (x = a Cl electron) x
x
x
x
x 3d
3
sp d
10.98
(a)
There are three bonds to the central Cl atom, plus one lone pair of electrons. The geometry of the electron pairs is tetrahedral so the Cl atom is to be sp3 hybridized:
_ O O
Br O
(b)
There are two atoms bonded to the central selenium atom, and one lone pair on the central selenium. The geometry of the electron pairs is that of a planar triangle, and the hybridization of the Se atom is sp2: O
Se
O
One other resonance structure should also be drawn for SeO2. (c)
There are two bonds to the central S atom, as well as two lone pairs. The S atom is to be sp3 hybridized, and the geometry of the electron pairs is tetrahedral. Cl
10.99
(a)
S
Cl
Five Cl atoms and one pair of electrons surround the central Sb atom in an octahedral geometry, and the hybridization of Sb is sp3d2.
208
Chapter 10
_ Cl
Cl Sb
Cl
Cl Cl
(b)
Four Br atoms are bonded to the Se atom, plus the Se atom has a lone pairs of electrons. This requires the Se atom to be sp3d hybridized, and the geometry is a see-saw. Br Br
Se
Br
Br
(c)
The central B atom is bonded to three F atoms. The molecule has trigonal planar geometry. This requires sp3 hybridization of B. F B
F
F
10.100 (a)
There are three bonds to As and one lone pair at As, requiring As to be sp3 hybridized. The Lewis diagram Cl
As
Cl
Cl
The hybrid orbital diagram for As: (x = a Cl electron) x
x
x
sp 3
(b)
There are three atoms bonded to the central Cl atom, and it also has two lone pairs of electrons. The hybridization of Cl is thus sp3d. The Lewis diagram F
Cl
F
F The hybrid orbital diagram for Cl: (x = a F electron)
x
x
3d
sp 3d
209
Chapter 10
10.101 (a)
Sb has three bonds to Cl atoms and two lone pairs and is therefore sp3d hybridized. The Lewis diagram
The hybrid orbital diagram for Sb: (x = a Cl electron)
(b)
Se has two bonds to Cl atoms and two lone pairs of electrons, requiring it to be sp3 hybridized. The Lewis diagram Se Cl
Cl
The hybrid orbital diagram for Se: (x = a Cl electron) x
x
sp
10.102 We can consider that this ion is formed by reaction of SbF5 with F–. The antimony atom accepts a pair of electrons from fluoride: Sb in SbF6–: (xx = an electron pair from the donor F–) xx sp 3d 2
10.103 This is an octahedral ion with sp3d2 hybridized tin: Sn atom in SnCl62– (xx = an electron pair from the donor Cl–) x
x
x
x xx xx
3d
3 2
sp d
10.104 (a)
N in the C=N system:
2p 2
sp
210
Chapter 10
2p sp2 sp2 sp2
(b)
sigma bond
pi bond
C
C
N
N
(c) 120 o
H 120 o H
10.105 (a)
C
N
120o
120 o H
sp– hybridized N atom:
2p
sp
(b)
The σ bonds: C
N
The π bonds:
C
N
(c)
For HCN, the only difference from (b) is the formation of another σ bond when an H 1s orbital overlaps the C sp hybrid orbital.
(d)
The HCN bond angle should be 180°.
10.106 Each carbon atom is sp2 hybridized, and each C–Cl bond is formed by the overlap of an sp2 hybrid of carbon with a p atomic orbital of a chlorine atom. The C=C double bond consists first of a C–C σ bond formed by "head on" overlap of sp2 hybrids from each C atom. Secondly, the C=C double bond consists of a side–to–side overlap of unhybridized p orbitals of each C atom, to give one π bond. The molecule is planar, and the expected bond angles are all 120°.
211
Chapter 10 10.107 The central carbon uses sp2 orbitals to bond to the hydrogen s orbitals and the oxygen sp2 orbitals. 10.108 1.
10.109 1. 2. 3. 4.
sp3
2.
sp2
3.
sp
4.
sp2
one σ bond one σ bond and two π bonds one σ bond This requires sp2 hybridization of B.
10.110 Here we pick the one with the higher bond order. (a)
O 2+
(b)
O2
(c)
N2
10.111 NO has 11 valence electrons, and the MO diagram is similar to that shown in Table 10.1 for O2, except that one fewer electron is employed at the highest energy level The bond order for NO is calculated to be 5/2:
(8 bonding e ) − (3 antibonding e ) = 5 Bond Order = −
−
2 2 If we remove one electron to form NO+, the bond order becomes 3 (there are only two antibonding electrons). The larger bond order indicates a shorter bond length.
10.112 (a)
N 2+
(b)
NO
(c)
O 2–
10.113 All of the molecules or ions are paramagnetic except N2. 10.114 σ∗s 1s
2p 2p σs
2s 2s Hydrogen Oxygen There are two electrons in bonding MOs and three electrons in nonbonding MOs. The net bond order is 1.
212
Chapter 10
10.115 σ∗p
B
N
π∗ p
2p
2p σp πp σ∗s
2s
2s σs
The molecule would be diamagnetic and the net bond order in the molecule would be 2.0.
Additional Exercises 10.116 planar triangular 10.117 Five valence electrons Group VA X
Cl Cl Three valence electrons Group IIIA Cl Cl
X Cl
Cl
Seven valence electrons Group VIIA Cl X
Cl
Cl It is unlikely that X would be in Group VI because it would have an unpaired electron since X would have six valence electrons and three would be used for bonding with the Cl, and that would leave three remaining electrons.
10.118 Since the angles are nearer to 90° than to 109°, we must use atomic rather than hybrid orbitals on antimony. 10.119
(a) (b) (c) (d) (e)
planar triangular tetrahedral trigonal bipyramidal octahedral T-shaped square planar Trigonal pyramidal trigonal bipyramidal linear planar 213
Chapter 10
10.120 Structure (a) 10.121 The normal C–C–C angle for an sp3 hybridized carbon atom is 109.5°. The 60° bond angle in cyclopropane is much less than this optimum bond angle. This means that the bonding within the ring cannot be accomplished through the desirable "head on" overlap of hybrid orbitals from each C atom. As a result, the overlap of the hybrid orbitals in cyclopropane is less effective than that in the more normal, noncyclic propane molecule, and this makes the C–C bonds in cyclopropane comparatively weaker than those in the noncyclic molecule. We can also say that there is a severe "ring strain" in the molecule. 10.122 (a) (b) (c)
PF3 is a pyramidal molecule and uses sp3 hybrid orbitals. The expected bond angle is 109.5°. Using the unhybridized p orbitals, we would anticipate a bond angle of 90°. The observed bond angle is almost exactly the average of the bond angles listed in parts (a) and (b) above. So, neither hybrid orbitals nor unhybridized atomic orbitals explain the observed bond angle.
10.123 The arrangement around an sp2 hybridized atom may have 120° bond angles (which well accommodates a six–membered ring, the geometrical arrangement around an atom that is sp hybridized must be linear, as in C – C ≡ C – C systems. 10.124 The arrangement of the atoms is trigonal bipyramidal. Recall that the bond angle between equatorial atoms is 120°. The bond angle from the equatorial position to the axial position is 90°. Due to the smaller bond angles, the atoms in the axial positions create more repulsions. The structure with the least amount of total repulsion is preferred; the statement implies that the more electronegative atoms create less repulsion; therefore the more electronegative atom should be placed in the axial position. Since fluorine is more electronegative than chlorine, the F atoms will be in the axial positions and the Cl atoms will be in the equatorial positions. The molecule is non–polar. F Cl
P
Cl Cl
F
10.125 The lone pair of electrons repels the Te–F bonding pairs, causing the F–Te–F angles to be smaller than in the ideal trigonal bipyramid. F This angle less than 180 o F Te
This angle less than 120o F
F
In BrF5, the lone pair is located perpendicular to the plane made up of four of the F atoms , giving a square pyramid geometry to the molecule. The angle between the four F in the plane and the f above the plane will be less than 900 due to lone pair repulsion. The F-Br-F angles in the plan will remain 900.
214
Chapter 10
10.126 The double bonds are predicted to be between S and O atoms. Hence, the Cl–S–Cl angle diminishes under the influence of the S=O double bonds. O S Cl
strong repulsion O Cl
weak repulsion
10.127 This is a π bond, since overlap is side to side rather than the end to end. Also, consider that no bond rotation is possible here without breaking the bond since overlap occurs both above and below the bond axis. The diagrams below show bonding and antibonding combinations, respectively.
10.128 Only the px orbital can form a π bond with dxz, if the internuclear axis is the z axis. Multi-Concept Problems 10.129 (a) (b)
The O–O–N bond angle is about 109.5° and the O–N–O bond angle is around 120°. sp2
215
Chapter 10
(c)
The nitrate ion benefits from three stable resonance structures (or electron delocalization) which stabilize it. The peroxynitrite ion does not.
10.130 The polarity in the N–H bond, as measured by the difference in electronegativities between N (3.1) and H (2.1) adds to the polarity created by the sp3 hybridized lone pair on the N atom. However, the higher electronegativity of fluorine (4.1) causes the N–F bond polarity to oppose the polarity associated with the hybridized lone pair on the N atom. Thus in the first case, a net dipole exists in the molecule, whereas in the second case, the polarity of the N–F bonds cancel the polarity of the hybrid lone pair. 10.131 (a)
The C–C single bonds are formed from head–to–head overlap of C atom sp2 hybrids. This leaves one unhybridized atomic p orbital on each carbon atom, and each such atomic orbital is oriented perpendicular to the plane of the molecule.
(b)
Sideways or π type overlap is expected between the first and the second carbon atoms, as well as between the third and the fourth carbon atoms. However, since all of these atomic p orbitals are properly aligned, there can be continuous π type overlap between all four carbon atoms.
(d)
The situation described in part (b) is delocalized. We expect completely delocalized π type bonding among the carbon atoms. Therefore, the center C-C bond has some π character and is thus shorter than a normal C-C single bond.
10.132 The compound is of the form PnBrm. From the mass of AgBr formed we can determine the mass of Br in the original compound. The mass of Br will also allow the determination of the number of moles of Br in the phosphorus compound. Then, using the original mass of the phosphorus compound, we can determine the mass and moles of P. Using this information , we can determine the empirical formula of the compound.
0.508 g AgBr x
1mol AgBr 187.77 g
Grams of Br = 2.705 x 10
−3
x
1 mol Br mol AgBr
mol Br x
= 2.705 x 10−3 mol Br
79.904 g mol
= 0.216
Grams of P = 0.244 g Pm Brn - 0.216 g Br = 0.028 g Moles of P = 0.028 g P x
1 mol P 30.97 g
= 8.98 x 10−4
Empirical formula: P0.000898 Br0.002705
or PBr3
Since P has five valence electrons, forming PBr3 results in three P-Br bonds and a lone-pair of electrons on P. The geometry of the compound is trigonal pyramid and the hybridization is sp3. Since the P-Br bonds are polar the geometry of PBr3 results in a polar molecule.
216
Chapter 11
Practice Exercises 11.1
14.7 psi psi = 730 mm Hg = 14.1 psi 760 mm Hg 29.921 in. Hg in. Hg = 730 mm Hg = 28.7 in. Hg 760 mm Hg
11.2
1 bar 1 atm 101,325 Pascal Pascals = 888 mbar = 88,800 Pascal 1 atm 1000 mbar 1.013 bar 1 bar 1 atm 760 torr torr = 888 mbar = 666 torr 1000 mbar 1.013 bar 1 atm
11.3
10 mm Hg mm Hg = 25 cm Hg = 250 mm Hg 1 cm Hg 760 mm Hg mm Hg = 770 torr = 770 mm Hg 760 torr The maximum pressure = 770 mm Hg + [(250 mm Hg) × 2] = 1270 mm Hg The minimum pressure = 770 mm Hg – [(250 mm Hg) × 2] = 270 mmHg
11.4
The pressure of the gas in the manometer is the pressure of the atmosphere less the pressure of the mercury, 11.7 cm Hg. Using the pressure in the atmosphere from the previous example: 10 mm Hg mm Hg = 11.7 cm Hg = 117 mm Hg 1 cm Hg 770 mm Hg – 117 mm Hg = 653 mm Hg
11.5
11.6
P1V1 P2 V2 = T1 T2 V2 = 3V1 and T2 = 2T1 P1V1 P2 3V1 = T1 2T1 P2 = 2/3 P1 The pressure must change by 2/3. Since volume is to decrease, pressure must increase, and we multiply the starting pressure by a volume ratio that is larger than one. Also, since P1V1 = P2V2, we can solve for P2:
PV P2 = 1 1 = V2 11.7
( 740 torr )(880 mL ) ( 870 mL )
= 750 torr
In general the combined gas law equation is:
PVT P2 = 1 1 2 = T1V2
P1V1 PV = 2 2 , and in particular, for this problem, we have: T1 T2
( 745 torr ) ( 950 m3 ) ( 333.2 K )
(1150 m ) ( 298.2 K ) 3
217
= 688 torr
Chapter 11
11.8
11.9
11.10
When gases are held at the same temperature and pressure, and dispensed in this fashion during chemical reactions, then they react in a ratio of volumes that is equal to the ratio of the coefficients (moles) in the balanced chemical equation for the given reaction. We can, therefore, directly use the stoichiometry of the balanced chemical equation to determine the combining ratio of the gas volumes: 2 volume O2 L O2 = (4.50 L CH4) = 9.00 L O2 1 volume CH 4
2 volume O2 L O2 = (6.75 L CH4) = 13.50 L O2 1 volume CH 4 13.50 L O2 ×100% 20.9% O2 = x L air 100% L air = (13.50 L O2) = 64.6 L air 20.9% 1 mL O2 2 mL V(NO) 1 volume O 2 VO 2 = x 184 mL NO = 92 mL O2 2 volume NO
V2 = 11.11
( 723 torr )( 92.0 mL )( 295 K ) ( 755 torr )( 318 K )
= 81.7 mL O2
First determine the number of moles of CO2 in the tank: PV n= RT 1 atm P = 2000 psig = 140 atm 14.696 psig ( 30.48 cm )3 1 mL 1L 3 V = 6.0 ft 3 3 1000 mL = 170 L 1 cm (1 ft )
L ⋅ atm mol ⋅ K T = 22 °C + 273 = 295 K R = 0.0821
mol CO2 in the tank =
(140 atm )(170 L ) L ⋅ atm 295 K ) ( 0.0821 mol ⋅ K )(
= 980 mol CO2
Then find the total number of grams of CO2 in the tank, MW CO2 = 44.01 g/mol
44.01 g CO 2 g CO2 in the tank = 980 mol CO2 = 43,000 g CO2 1 mol CO 2 Amount of solid CO2 = 43,000 g CO2 × 0.35 = 15,000 g solid CO2 11.12
n=
PV = RT
( 57.8 atm )(12.0 L ) L atm 298 K ) ( 0.0821 mol K )(
= 28.3 moles gas
28.3 mol Ar (39.95 g Ar/mol) = 1,130 g Ar
218
Chapter 11
11.13
Find the number of moles of argon (1.0000 atm )( 0.54423 L ) PV n= = = 0.024281 mol argon L atm 273.15 K RT 0.082057 mol ) K (
(
)
39.948 g arg on Mass of the argon = (0.024281 mol argon) = 0.96998 g argon 1 mol arg on The mass of the flask = 735.6898 g – 0.96998 g air = 734.7198 g Mass of the organic compound = 736.13106 g – 734.7198 g = 1.4113 g The number of moles of the organic compound equals the number of moles of air: 1.4113 g organic compound MW = = 58.12 g/mol 0.024281 mole organic compound The unknown could be butane, MW = 58.1 g/mol 11.14
Since PV = nRT, then n = PV/RT 1 atm 685 torr ) ( ( 0.300 L ) PV 760 torr n = = = 0.0110 moles gas RT 0.0821 L atm ( 300.2 K )
(
mol K
)
1.45 g = 132 g mol−1 0.0110 mol The gas must be xenon. molar mass =
11.15
28.8 g air 1 mol air Density of air = 1 mol air = 1.29 g/L 1 mol air 22.4 L 222.0 g Rn 1 mol Rn Density of radon at STP = 1 mol Rn = 9.91 g/L 1 mol Rn 22.4 L Since radon is almost eight times denser than air, the sensor should be in the lowest point in the house: the basement.
11.16
d = m/V Taking 1.00 mol SO2: m = 64.1 g
nRT V= = P
density = 11.17
(1.00 mol ) ( 0.0821
L atm mol K
) ( 268.2 K )
1 atm 96.5 kPa 101.325 kPa
= 23.1 L = 23,100 mL
64.1 g = 2.77 g/L 23.1 L
In general PV = nRT, where n = mass × formula mass. Thus
PV =
mass RT formula mass
We can rearrange this equation to get;
219
Chapter 11
formula mass
(mass/V)RT dRT = P P
(5.60 g L ) ( 0.0821 −1
L atm mol K
) ( 296.2 K )
= 138 g mol–1 1 atm ( 750 torr ) 760 torr The empirical mass is 69 g mol–1. The ratio of the molecular mass to the empirical mass is formula mass =
138 g mol-1
=2 69 g mol-1 Therefore, the molecular formula is 2 times the empirical formula, i.e., P2F4. 11.18
formula mass
(mass/V)RT dRT = P P
(5.55 g L ) ( 0.0821 formula mass = −1
L atm mol K
(1.25 atm )
) ( 313.2 K ) = 114 g mol
–1
Since the compound contains C and H it could be an alkane, CnH2n+2, an alkene, CVnH2n, or an alkyne, CnH2n-2. 9 C and 6 H 8 C and 18 H 7 C and 30 H 6 C and 42 H 5 C and 54 H 4 C and 66 H 3 C and 78 H 2 C and 90 H 1 C and 102 H The most probable compound is C8H18 also known as octane. 11.19
CS2(g) + 3O2(g) 2SO2(g) + CO2(g)
1 mol CS2 mol of CS2 = 11.0 g CS2 = 0.1445 mol CS2 76.131 g CS2 1 mol CO 2 mol CO2 = 0.1445 mol CS2 = 0.1445 mol CO2 1 mol CS2 2 mol SO 2 mol SO2 = 0.1445 mol CS2 = 0.2890 mol SO2 1 mol CS2 L CO2 =
L SO2 =
L⋅atm 301 K ( 0.1445 mol CO2 ) ( 0.0821 mol ) ⋅K ) (
1 atm 883 torr 760 torr L⋅atm 301 K ( 0.2890 mol SO2 ) ( 0.0821 mol ) ⋅K ) (
1 atm 883 torr 760 torr The total number of liters is = 9.21 L.
220
= 3.07 L CO2
= 6.14 L SO2
Chapter 11
11.20
CaCO3(s) CaO(s) + CO2(s) mol CO2 = PV = RT
( 738 torr )
(
1 atm ( 0.257 L ) 760 torr = 0.01027 mol CO2 L atm 296 K 0.0821 mol ( ) K
)
1 mol CaCO3 mol CaCO3 = 0.01027 mol CO2 = 0.01027 mol CaCO3 1 mol CO 2 100.09 g CaCO3 g CaCO3 = 0.01027 mol CaCO3 = 1.03 g CaCO3 1 mol CaCO3 11.21
1 mol Ar mol Ar = 11.0 g Ar = 0.275 mol Ar 39.95 g Ar PAr =
( 0.275 mol Ar ) ( 0.0821
L ⋅atm mol⋅K
) (301 K )
= 6.80 atm
1.00 L
1 mol N 2 mol N2 = 10.6 g N2 = 0.378 mol N2 28.02 g N 2 P N2 =
( 0.378 mol N 2 ) ( 0.0821
L ⋅atm mol⋅K
) ( 301 K )
= 9.34 atm
1.00 L
1 mol O 2 mol O2 = 14.3 O2 = 0.447 mol O2 32.00 g O 2 P O2 =
( 0.447 mol O 2 ) ( 0.0821
L ⋅atm mol⋅ K
) ( 301 K )
= 11.04 atm
1.00 L
Ptotal = PAr + PN2 + PO2 = 6.80 atm + 9.34 atm + 11.04 atm = 27.18 atm 11.22
We can determine the pressure due to the oxygen since Ptotal = PN2 + PO2. PO2 = Ptotal – PN2 = 237.0 atm – 115.0 atm = 122.0 atm. We can now use the ideal gas law to determine the number of moles of O2:
PV (122.0 atm)(17.00 L) = 84.8 mol O2 = L atm RT 0.0821 mol K (298 K) 32.0 g O 2 g O2 = (84.8 mol O2) = 2713 g O2 1 mol O 2 n=
11.23
The total pressure is the pressure of the methane and the pressure of the water. We can determine the pressure of the methane by subtracting the pressure of the water from the total pressure. The pressure of the water is determined by the temperature of the sample. At 28 °C, the partial pressure of water is 31.82 torr. PCH4 = Ttotal – Pwater = 775 torr – 28.3 torr = 747 torr The pressure in the flask is 743.18 torr.
221
Chapter 11
1 atm (747 torr) (2.50 L) 760 torr = 0.0994 mol CH4 L atm 0.0821 (301 K) mol K First we find the partial pressure of nitrogen, using the vapor pressure of water at 15 °C: PN2 = Ptotal – Pwater = 745 torr – 12.79 torr = 732 torr. PV mol CH4 = = RT
11.24
To calculate the volume of the nitrogen we can use the combined gas law P1V1 PV = 2 2 T1 T2 For this problem,
V2 = 11.25
P1V1T2 (732.2 torr)(0.317 L)(273 K) = = 289 mL P2 T1 (760 torr)(288 K)
Since the stoichiometric ratio of the SO2 and SO3 are the same, the pressure in the flask after the reaction when the only substance in the flask is SO3 will be the same as the pressure in flask when there is just SO2, 0.750 atm. After the reaction additional oxygen gas is added to the flask.
1 atm O 2 P(O2) added to flask after reaction = 0.750 atm SO2 x = 0.375 atm 2 atm SO 2 The total pressure of SO2 and O2 = 0.750 atm + 0.375 atm = 1.125 atm 11.26
Find the number of moles of both the H2 and NO then find the mol fractions. 1 mol H 2 mol H2 = (2.15 g H2) = 1.07 mol H2 2.016 g H 2
1 mol NO mol NO = (34.0 g NO) = 1.13 mol NO 30.01 g NO 1.07 mol H 2 χH2 = = 0.486 1.13 mol NO + 1.07 mol H 2 1.13 mol NO = 0.514 1.13 mol NO + 1.07 mol H 2 PH2 = (Ptotal)(χH2) = (2.05 atm)(0.486) = 0.996 atm χNO =
PNO = (Ptotal)(χNO) = (2.05 atm)(0.514) = 1.05 atm 11.27
11.28
The mole fraction is defined in Equation 11.5: PO 2 116 torr XO2 = = = 0.147 or 14.7% Ptotal 788 torr
effusion rate (Br − 81) = effusion rate (Br − 79)
M Br −79 = M Br −81
78.9 = 0.988 80.9
222
Chapter 11
11.29
Use Equation 11.7;
effusion rate (HX) = effusion rate (HCl)
M HCl M HX
effusion rate (HX) M HX = M HCl × effusion rate (HCl)
2
= 36.46 g mol−1 × (1.88)2 = 128.9 g mol−1
The unknown gas must be HI.
Review Questions 11.1
The reason it hurts more to be jabbed by a point of a pencil rather than the eraser, even though the force is the same, is because the area of the point is smaller than the area of the eraser, and therefore, the pressure is higher.
11.2
(a) (b) (c) (d) (e) (f)
1 atm 101.325 kPa and 101.325 kPa 1 atm 1 torr 1 mm Hg and 1 mm Hg 1 torr
1.013 bar 101,325 Pa and 101,325 Pa 1.013 bar 760 torr 1 atm and 1 atm 760 torr 760 torr 101,325 Pa and 101,325 Pa 760 torr 1 bar 0.9868 atm and 0.9868 atm 1 bar
11.3
Since the density of water is approximately 13 times smaller than that of mercury, a barometer constructed with water as the moveable liquid would have to be some 13 times longer than one constructed using mercury. Also, the vapor pressure of water is large enough that the closed end of the barometer may fill with sufficient water vapor so as to affect atmospheric pressure readings. In fact, the measurement of atmospheric pressure at normal temperatures would be about 18 torr too low, due to the presence of water vapor in the closed end of the barometer.
11.4
A closed–end manometer reads pressure without the need to correct for atmospheric pressure.
11.5
(a)
(b)
(c)
(d)
Temperature–Volume Law: The volume of a given mass of a gas is directly proportional to the Kelvin temperature, provided the pressure is held constant: V ∝ T or V1/T1 = V2/T2, at constant P. This is Charles' Law. Temperature–Pressure Law: The pressure of a gas is directly proportional to the Kelvin temperature, provided the volume is held constant: P ∝ T or P1/T1 = P2/T2, at constant V. This is Gay-Lussac’s Law. Pressure–Volume Law: The volume of a given mass of a gas is inversely proportional to the pressure, provided the temperature is held constant: V ∝ 1/P or P1V1 = P2V2, at constant T. This is Boyle's Law. Combined Gas Law: The pressure and volume of a gas are directly proportional to the Kelvin temperature, provided the number of particles is held constant. PV ∝ T or P1V1/T1 = P2V2/T2, at constant n.
223
Chapter 11
11.6
(a) (b) (c) (d)
number of moles and temperature number of moles and pressure number of moles and volume number of moles
11.7
An ideal gas obeys the gas laws over all pressures and temperatures. A real gas behaves most like an ideal gas at low pressures and high temperatures.
11.8
PV = nRT (a) Plot temperature in Kelvin versus Volume. (b) Plot temperature in Kelvin versus Pressure. (c) Plot Pressure versus the inverse Volume.
11.9
Ptotal = Pa + Pb + Pc + ···
11.10
Mole fraction is the ratio of the number of moles of one component of a mixture to the total number of moles of all components.
11.11
Middle drawing Left drawing: A: 0.500 atm Middle drawing: A: 0.600 atm Right drawing A: 0.667 atm
B: 0.500 atm B: 0.400 atm B: 0.333 atm
11.12
Diffusion is the spontaneous intermingling of one substance with another while effusion is the movement of a gas through a very tiny opening into a region of lower pressure. dB MB effusion rate (A) = = effusion rate (B) dA MA
11.13
A gas consists of hard, super small or volumeless particles in random motion, and the particles neither attract nor repel one another.
11.14
The temperature and pressure will decrease.
11.15
The increase in temperature requires an increase in kinetic energy. This can happen only if the gas velocities increase. Higher velocities cause the gas particles to strike the walls of the container with more force, and this in turn causes the container to expand if a constant pressure is to be maintained.
11.16
The increase in temperature causes an increase in the force with which the gas particles strike the container walls. If the container cannot expand, an increase in pressure must result.
11.17
The minimum temperature corresponds to zero kinetic energy, which is accomplished only when velocity is zero. In other words, the molecules have ceased all movement.
11.18
The answer (c) NH3 will have the largest ν rms since it has the lowest molecular mass.
11.19
(a) As the pressure of a gas increases, the rate of effusion should increase since the molecules will hit the walls of the container more frequently and with greater force. If the molecules hit more frequently, they are more likely to go through the small openings in the walls of the container. (b) As the temperature of the gas increases, the rate of effusion will increase since temperature is proportional to kinetic energy which is dependent on the velocity of the particles. The faster the particles move, the more likely they are to hit the walls and pass through the small openings.
224
Chapter 11
11.20
It is not true that the gas particles occupy no volume themselves, apart from the volume between the gas particles. Also, it is not true that the gas particles exert no force on one another. In other words, real molecules occupy space and attract or repel one another. Because of short-range interactions, it is also not true that particles travel always in straight paths.
11.21
A small value for the constant a suggests that the gas molecules have weak forces of attraction among themselves.
11.22
(b) has a larger value of the van der Waals constant b, since it is a larger molecule.
11.23
Under the same conditions of T and V, the pressure of a real gas is less than the pressure of an ideal gas because real gases do not have perfectly elastic collisions and may clump together and stick to the walls of the container, thus decreasing the number of collisions the gas makes. The volume of a real gas is greater than the volume of an ideal gas because the atoms and molecules take up space.
11.24
The helium atoms are moving faster than the argon atoms because they have less mass.
Review Problems 11.25
11.26
11.27
11.28
(a)
760 torr torr = (1.26 atm) = 958 torr 1 atm
(b)
1 atm atm = (740 torr) = 0.974 atm 760 torr
(c)
760 torr mm Hg = 738 torr = 738 mm Hg 760 mm Hg
(d)
760 torr torr = (1.45 × 103 Pa) = 10.9 torr 5 1.01325 × 10 Pa
(a)
760 torr torr = (0.835 atm) = 635 torr 1 atm
(b)
1 atm atm = (950 torr) = 1.3 atm 760 torr
(c)
760 torr torr = 75 mm Hg = 75 torr 760 mm Hg
(d)
1.013 bar bar = 1.36 kPa = 0.0136 bar 101.32 kPa
(a)
760 torr torr = (0.329 atm) = 250 torr 1 atm
(b)
760 torr torr = (0.460 atm) = 350 torr 1 atm
(a)
1 atm atm = (595 torr) = 0.783 atm 760 torr
(b)
1 atm atm = (160 torr) = 0.211 atm 760 torr
225
Chapter 11
(c)
11.29
1 atm –4 atm = (0.300 torr) = 3.95 × 10 atm 760 torr
765 torr – 720 torr = 45 torr
760 mm Hg 45 torr = 45 mm Hg 760 torr
1 cm cm Hg = (45 mm Hg) = 4.5 cm Hg 10 mm
gas
4.5 cm
11.30
820 torr – 750 torr = 70 torr
760 mm Hg 70 torr = 70 mm Hg 760 torr
1 cm cm Hg = (70 mm Hg) = 7.0 cm Hg 10 mm
gas
7.0 cm
11.31
760 torr 72 mm Hg = 72 torr 760 mm Hg
745 torr + 72 torr = 817 torr
226
Chapter 11
11.32
760 torr 76 mm Hg = 76 torr 760 mm Hg
749 torr – 76 torr = 673 torr
11.33
In a closed-end manometer the difference in height of the mercury levels in the two arms corresponds to the pressure of the gas. Therefore, the pressure of the gas is 125 mm Hg. 760 torr 125 mm Hg = 125 torr 760 mm Hg
11.34
The closed-end manometer data indicates that the pressure inside the flask is 236 mm Hg. The open-end manometer data indicate that Patm = 512 mm Hg + 236 mm Hg = 748 mm Hg.
11.35
Use Boyle’s Law to solve for the second volume:
V2 = 11.36
P1V1 = P2
( 255 mL )( 755 torr ) 365 torr
= 527 mL
P 1V 1 = P 2V 2 PV V2 = 1 1 since the pump has a fixed diameter, the length of the tube is proportional to its volume P2 P 11 = P 22
P 1 atm(75.0 cm) 2 = 1 1 = = 13.6 cm P2 5.50 atm 11.37
Use Charles’s Law to solve the second volume:
VT 3.25 L (369 K) V2 = 1 2 = = 3.77 L T1 318 K 11.38
Use Charles’s Law to solve for the second volume:
VT 2.75 L (258 K) V2 = 1 2 = = 2.39 L T1 297 K 11.39
Compare pressure change to temperature to solve for temperature change: T2 =
11.40
1076 K – 273 K = 803°C
PV PV In general the combined gas law equation is: 1 1 = 2 2 , and in particular, for this problem since the T1 T2 volume does not change, we have: P2 =
11.41
P2 T1 (1708 torr) (538 K) = 1076 K = P1 854 torr
P1T2 = T1
( 45 lb in ) (316 K) −2
285.2 K
= 50 lb in −2
PV PV In general the combined gas law equation is: 1 1 = 2 2 , and in particular, for this problem, we have: T1 T2
227
Chapter 11
P2 =
11.42
11.43
11.44
11.45
11.46
P1V1T2 (765 torr)(2.58 L)(345.2 K) = = 816 torr T1V2 (297.2 K)(2.81 L)
PV PV In general the combined gas law equation is: 1 1 = 2 2 , and in particular, for this problem, we have: T1 T2 PVT (0.985 atm)(643 mL)(336.2 K) P2 = 1 1 2 = = 1.05 atm T1V2 (289.2 K)(698 mL) PV PV In general the combined gas law equation is 1 1 = 2 2 , and in particular, for this problem, we have: T1 T2 PVT (745 torr)(2.68 L)(648.2 K) V2 = 1 1 2 = = 5.69 L T1P2 (297.2 K)(765 torr) PV PV In general the combined gas law equation is: 1 1 = 2 2 , and in particular, for this problem, we have: T1 T2 PVT (741 torr)(319 mL)(306.2 K) V2 = 1 1 2 = = 324 mL T1P2 (291.2 K)(765 torr) PV PV In general the combined gas law equation is: 1 1 = 2 2 , and in particular, for this problem, we have: T1 T2 PVT (373 torr)(9.45 L)(293.2 K) T2 = 2 2 1 = = 219.8 K = − 53 C P1V1 (761 torr)(6.18 L) PV PV In general the combined gas law equation is: 1 1 = 2 2 , and in particular, for this problem, we have: T1 T2 P2 V2 T1 (2.00 atm)(222 mL)(298.2 K) = = 193 K = − 80.2 C T2 = P1V1 (1.51 atm)(455 mL)
11.47
L atm 1000 mL 760 torr mL torr R = 0.0821 = 6.24 × 104 mol K 1 L 1 atm mol K
11.48
If PV = nRT, then R = PV/nT. Let P = 1 atm = 101,325 Pa, T = 273 K, and n = 1. Next, express the volume of the standard mole using the units m3, instead of L, remembering that 22.4 L = 22,400 cm3:
1m m3 = 22, 400 cm3 × 100 cm (101,325 Pa ) 0.0224 m3 R = (1 mole )( 273 K )
(
11.49
3
= 0.0224 m3
)
3 −1 −1 = 8.31 m Pa mol K
1 mol L atm 0.287 g 0.0821 (293.2 K) mol K nRT 32.0 g V = = = 0.228 L P 1 atm 748 torr 760 torr 228
Chapter 11
11.50
11.51
11.52
11.53
1 mol L atm 2.46 g 0.0821 (295.2 K) 28.0 g mol K nRT V = = = 2.14 L P 1 atm 756 torr 760 torr 1 mol L atm 12.8 g 0.0821 (300.2 K) mol K 760 torr nRT 32.0 g V = = = 3.28 atm = 2490 torr P (3.00 L) 1 atm
1 mol L atm ( 381 K ) 12.0 g 0.0821 mol K nRT 18.0 g P = = = 5.80 atm V ( 3.60 L )
n =
PV = RT
( 735 torr )
atm ( 0.0265 L ) 44.0 g 760 torr = 1.07 × 10−3 mol = 0.0471 g L atm 1 mol 0.0821 mol K ( 293.2 K )
(
11.54
PV n = = RT
(( 758 torr ) (
1 atm 760 torr
) ) ( 0.255 L )
L atm 0.0821 mol K ( 300.2 K )
= 1.03 x 10−2 mol
16.0 g grams of methane = 1.03 × 10−2 mol = 0.165 g 1 mol CH 4
(
11.55
(a) (b) (c) (d)
11.56
)
30.1 g C2 H 6 1 mol –1 density C2H6 = = 1.34 g L 1 mol C2 H 6 22.4 L 28.0 g N 2 1 mol –1 density N2 = = 1.25 g L 1 mol N 22.4 L 2 70.9 g Cl2 1 mol –1 density Cl2 = = 3.17 g L 1 mol Cl 22.4 L 2 39.9 g Ar 1 mol –1 density Ar = 22.4 L = 1.78 g L 1 mol Ar
(a)
20.2 g Ne 1 mol –1 density Ne = = 0.902 g L 1 mol Ne 22.4 L
(b)
32.0 g O 2 density O2 = 1 mol O 2
1 mol –1 = 1.43 g L 22.4 L
229
)
Chapter 11
11.57
(c)
16.0 g CH 4 density CH4 = 1 mol CH 4
1 mol –1 = 0.714 g L 22.4 L
(d)
88.0 g CF4 1 mol –1 density CF4 = = 3.93 g L 1 mol CF 22.4 L 4
In general PV = nRT, where n = mass ÷ formula mass. Thus mass PV = RT (formula mass) and we arrive at the formula for the density (mass divided by volume) of a gas:
d=
P × (formula mass) RT
( 7601 atmtorr ) (32.0 g/mol) L atm (301.2 K) ( 0.0821 mol K)
(742 torr) d=
d = 1.26 g/L for O 2 11.58
In general PV = nRT, where n = mass ÷ formula mass. Thus mass PV = RT (formula mass) and we arrive at the formula for the density (mass divided by volume) of a gas:
d=
P × (formula mass) RT
( 7601 atmtorr ) (39.95 g/mol) L atm (293.80 K) ( 0.0821 mol K)
(748.0 torr) d=
d = 1.63 g/L for Ar 11.59
n=
First determine the number of moles from the ideal gas law:
(
)
(
1 atm 1L (10.0 torr) 760 255 mL ) 1000 PV torr ( mL = L atm RT 0.0821 (298.2 K)
(
mol K
)
)
= 1.37 × 10−4 mol
Now calculate the molecular mass:
mass molecular mass = = # of moles
11.60 11.61
1g (12.1 mg ) ( 1000 mg )
1.37 × 10−4 mol
dRT mass RT molecular mass = = = P PV
= 88.2 g/mol
(1.13 g/L ) 0.0821 (755 torr)
L atm ( 295 K ) mol K
( 7601 atmtorr )
The balanced equation is 2C4H10 + 13O2 8CO2 + 10H2O 13 mL O 2 3 mL O2 = (225 mL C4 H10 ) = 1.46 × 10 mL O 2 2 mL C H 4 10
230
= 27.6 g/mol
Chapter 11
11.62
11.63
The balanced equation is 2C6H14 + 19O2 12CO2 + 14H2O 19 mL O2 3 mL O2 = (855 mL CO 2 ) = 1.35 × 10 mL O2 12 mL CO 2
1 mol C3 H 6 mol C3H6 = (22.5 g C3H6) = 0.535 mol C3H6 42.08 g C3 H 6 1 mol H 2 mol H2 = (0.535 mol C3H6) = 0.535 mol H2 1 mol C3 H 6
(
)
L atm (297.2 K) (0.535 mol H 2 ) 0.0821 mol nRT K V= = = 13.4 L H2 1 atm P (740 torr)
( 760 torr )
11.64
11.65
1 mole HNO3 mol HNO3 = (12.0 g HNO3) = 0.190 mol HNO3 63.01 g HNO3 3 moles NO 2 mol NO2 = (0.190 mol HNO3) = 0.286 mol NO2 2 moles HNO3 L atm 0.286 moles NO 2 ) 0.0821 ( ( 298 K ) nRT mol K V= = = 7.07 L or 7.07 × 103 mL P 1 atm ( 752 torr ) 760 torr CH4 + 2O2 CO2 + 2H2O
n CH 4 =
PV = RT
( 725 torr )
1 atm −3 22.4 × 10 L 760 torr = 8.45 × 10–4 mol CH4 L atm 0.0821 mol K ( 308.2 K )
11.66
n NH3
)
2 mol O 2 –3 mol O2 = (8.45 × 10–4 mol CH4) = 1.69 × 10 mol O2 1 mol CH 4 L atm 1.69 × 10−3 moles 0.0821 ( 300.2 K ) nRT mol K = = = 4.83 × 10−2 L = 36.3 mL O2 P 1 atm ( 654 torr ) 760 torr
(
VO 2
(
PV = = RT
)
( 825 torr )
1 atm −3 33.6 × 10 L 760 torr = 1.11 × 10–3 mol NH3 L atm 0.0821 mol K ( 400 K )
(
)
6 mol H 2 O –3 mol H2O = 1.11 x 10-3 mol NH3 = 1.67 × 10 mol H2O 4 mol NH 3 L atm 1.67 × 10−3 moles 0.0821 ( 591 K ) nRT mol K VH 2O = = = 8.36 × 10−2 L = 83.6 mL P 1 atm ( 735 torr ) 760 torr
(
)
231
Chapter 11
11.67
2CO + O2 2CO2
( 683 torr )
1 atm ( 0.300 L ) 760 torr moles CO = = 1.10 × 10−2 moles L atm 0.0821 mol K ( 298.2 K ) 1 atm ( 715 torr ) ( 0.155 L ) 760 torr moles O 2 = = 4.46 × 10−3 moles L atm 0.0821 mol K ( 398.2 K ) ∴ O2 is the limiting reactant 2 mol CO 2 −3 moles CO 2 = (4.46 × 10−3 moles O2 ) = 8.92 × 10 moles CO2 1 mol O 2
(8.92 × 10 V =
11.68
L atm mol 0.0821 ( 300.2 K ) mol K = 2.24 × 10−1 L ⇒ 224 mL 1 atm ( 745 torr ) 760 torr
−3
( 0.750 atm ) ( 0.300 L )
= 9.10 × 10−3 moles L atm 0.0821 mol K ( 301 K ) 0.780 atm 0.220 L) ( )( moles O 2 = = 6.47 × 10−3 moles L atm 0.0821 mol K ( 323 K ) Assume NH3 is the limiting reagent. 2 mol N 2 −3 moles N 2 = (9.10 × 10−3 moles NH3 ) = 4.55 × 10 moles N 2 4 moles NH 3 Assume O2 is the limiting reagent: 2 mol N 2 −3 moles N 2 = (6.47 × 10−3 moles O 2 ) = 4.31 × 10 mol N 2 ∴ O2 is limiting reactant 3 mol O2 L atm 1000 mL 4.31 × 10−3 mol 0.0821 ( 373 K ) 1 L mol K = 178 mL mL N 2 = 0.740 atm moles NH3 =
(
11.69
)
)
PTot = PN2 + PO2 + PHe PTot =
(20 cm)(10 mm/cm)(1 torr/mm) + 155 torr + (0.450 atm)(760 torr/atm)
PTot = 200 torr + 155 torr + 342 torr = 697 torr 11.70
PTot = PN2 + PO2 + PCO2 PCO2 = PTot – PN2 –PO2
10 mm Hg 1 torr PCO2 = 740 torr – (12.0 cm Hg ) 1 cm Hg 1 mm Hg
100 mm Hg 1 torr – 4.25 dm Hg = 195 torr 1 dm Hg 1 mm Hg 232
Chapter 11
11.71
Assume all gases behave ideally and recall that 1 mole of an ideal gas at 0 °C and 1 atm occupies a volume of 22.4 L. Therefore, the moles of gas equals the pressure of gas in atm: (RT/V = 1.000 atm mol-1 ) PN2 = 0.30 atm PO2 = 0.20 atm PHe = 0.40 atm PCO2 = 0.10 atm
760 torr PN2 = 0.30 atm = 228 torr 1 atm 760 torr PO2 = 0.20 atm = 152 torr 1 atm 760 torr PHe = 0.40 atm = 304 torr 1 atm 760 torr PCO2 = 0.10 atm = 76 torr 1 atm 1 bar PN2 = 0.30 atm = 0.304 bar 0.9868 atm 1 bar PO2 = 0.20 atm = 0.203 bar 0.9868 atm 1 bar PHe = 0.40 atm = 0.405 bar 0.9868 atm 1 bar PCO2 = 0.10 atm = 0.101 bar 0.9868 atm 11.72
PCO2 = 845 torr – 322 torr = 523 torr
523 torr n CO 2 = (0.200 mol) = 0.124 moles 845 torr 11.73
Ptotal = (PCO + PH2O) PH2O = 19.8 torr at 22 °C, from Table 11.2. PCO = 754 – 19.8 torr = 734 torr P1V1/T1 = P2V2/T2 , and
V2 = 11.74
P1V1T2 (734.2 torr)(0.297 L)(298.2 K) = = 0.290 L ⇒ 290. mL T1P2 (295.2 K)(760 torr)
Ptotal = PH2 + PH2O PH2O = 23.76 torr at 25 °C, from Table 11.2. PH2 = Ptotal – PH2O = 742 – 23.76 = 718 torr The temperature stays constant so, P1V1 = P2V2, and PV (718 torr)(262 mL) V2 = 1 1 = = 248 mL P2 (760 torr) 233
Chapter 11
11.75
From Table 11.2, the vapor pressure of water at 20 °C is 17.54 torr. Thus only (742 – 17.54) = 724 torr is due to "dry" methane. In other words, the fraction of the wet methane sample that is pure methane is 724/742 = 0.976. The question can now be phrased: What volume of wet methane, when multiplied by 0.976, equals 244 mL? Volume "wet" methane × 0.976 = 244 mL Volume "wet" methane = 244 mL/0.976 = 250 mL In other words, one must collect 250 total mL of "wet methane" gas in order to have collected the equivalent of 244 mL of pure methane.
11.76
First convert the needed amount of oxygen at 760 torr to the volume that would correspond to the laboratory conditions of 746 torr: P1V1 = P2V2 or V2 = P1V1/P2 V2 = 275 mL × 760 torr/746 torr = 280 mL of dry oxygen gas The wet sample of oxygen gas will also be collected at atmospheric pressure in the lab of 746 torr. The vapor pressure of water at 15 °C is equal to 12.8 torr (from Table 11.2), and the wet sample will have the following partial pressure of oxygen, once it is collected: PO2 = Ptotal – PH2O = 746 – 12.8 = 733 torr of oxygen in the wet sample. Thus the wet sample of oxygen is composed of the following % oxygen: % oxygen in the wet sample = 733/746 × 100 = 98.3 % The question now becomes what amount of a wet sample of oxygen will contain the equivalent of 280 mL of pure oxygen, if the wet sample is only 98.3 % oxygen (and 1.7 % water). 0.983 × Vwet = 280 mL, hence Vwet = 285 mL. This means that 285 mL of a wet sample of oxygen must be collected in order to obtain as much oxygen as would be present in 280 mL of a pure sample of oxygen.
11.77
Effusion rates for gases are inversely proportional to the square root of the gas density, and the gas with the lower density ought to effuse more rapidly. Nitrogen in this problem has the higher effusion rate because it has the lower density:
rate(N 2 ) 1.96 g L−1 = = 1.25 rate(CO 2 ) 1.25 g L−1 11.78
Ethylene, C2H4, the lightest of these three, diffuses the most rapidly, and Cl2, the heaviest, will diffuse the slowest. Cl2 < SO2 < C2H4
11.79
The relative rates are inversely proportional to the square roots of their molecular masses:
rate( 235 UF6 ) rate( 238 UF6 )
=
molar mass ( 238 UF6 ) molar mass ( 235 UF6 )
=
352 g mol−1 349 g mol−1
= 1.0043
Meaning that the rate of effusion of the 235UF6 is only 1.0043 times faster than the 238UF6 isotope.
234
Chapter 11
11.80
Use equation 11.7
M C3 H 8
effusion rate x = effusion rate C3 H8
Mx
effusion rate C3 H8 M x = M C3 H 8 effusion rate x 1 = 44.1 g/mol 1.65
2
2
= 16.2 g/mol
Additional Exercises 11.81
We found that 1 atm = 33.9 ft of water. This is equivalent to 33.9 ft × 12 in./ft = 407 in. of water, which in this problem is equal to the height of a water column that is uniformly 1.00 in.2 in diameter. Next, we convert the given density of water from the units g/mL to the units lb/in.3: 3
lb 1.00 g 1 lb 1 mL 2.54 cm = = 0.0361 454 g 3 3 1 in. 1.00 mL in. in.3 1 cm The area of the total column of water is now calculated: 1.00 in.2 × 407 in. = 407 in.3, along with the mass of the total column of water: 407 in.3 × 0.0361 lb/in.3 = 14.7 lb. Finally, we can determine the pressure (force/unit area) that corresponds to one atm: 1 atm = 14.7 lb ÷ 1.00 in.2 = 14.7 lb/in.2 lb
11.82
6.0 in × 3.2 in 2 Total footprint = (4 tires) = 76.8 in tire 3500 lb = 45.6 lb/in 2 Total pressure = 2 76.8 in Gauge pressure = 45.6 lb/in 2 - 14.7 lb/in 2 = 30.9 lb/in 2
11.83
2000 lb 5 Total weight = (45.6 tons + 8.3 tons) = 1.08 × 10 lbs 1 ton Total pressure = 85 psi + 14.7 psi = 99.7 psi/tire 1.08 × 105 lbs
number of tires =
= 10.8 tires (99.7 lbs in −2 /tire)(100 in 2 ) The minimum number of wheels is 12 since tires are mounted in multiples of 2. 11.84
Assume a 1 sq in. cylinder of water
12 in. V = (12,468 ft) 1 in.2 1 ft
(
)
3
2.54 cm 1 mL 6 = (149616 in.3 ) = 2.4518 × 10 mL 3 1 in. 1 cm
1 lb 1.025 g 3 Mass = (2.4518 × 106 mL) = (2.51306 × 106 g) = 5.54026 × 10 lb 1 mL 453.6 g 1 atm Pressure = (5.54026 × 103 lb in −2 ) = 376.89 atm −2 14.7 lb in 11.85
From the data we know that the pressure in flask 1 is greater than atmospheric pressure, and greater than the pressure in flask 2. The pressure in flask 1 can be determined from the manometer data. The pressure in flask 1 is:
235
Chapter 11
760 mm Hg 1 cm P = (0.827 atm) 10 mm + 11.40 cm = 74.25 cm Hg 1 atm The pressure in flask 2 is lower than flask 1 0.826 g mL−1 P = 74.25 cm Hg – (15.85 cm oil) = 73.29 cm Hg = 732.9 torr 13.6 g mL−1 11.86
To calculate the pressure at 100 ft assume a cylinder of water 100 ft long and 1 in2. 3
2.54 cm 1 mL 1.025 g 1 lb 12 in mass = (100 ft) (1 in )2 = 44.4 lb 1 ft 1 in 1 cm3 1 mL 453.6 g 1 atm = 3.02 atm P = (44.4 lb in −2 ) −2 14.7 lb in Since the pressure decreases by a factor of 3, the volume must increase by a factor of 3. Divers exhale to decrease the amount of gas in their lungs, so it does not expand to a volume larger than the divers lungs. 11.87
First calculate the initial volume (V1) and the final volume (V2) of the cylinder, using the given geometrical data, noting that the radius is half the diameter (10.7/2 = 5.35 cm): V1 = π × (5.35 cm)2 × 13.4 cm = 1.20 × 103 cm3 V2 = π × (5.35 cm)2 × (13.4 cm – 12.7 cm) = 62.9 cm3
PV PV In general the combined gas law equation is: 1 1 = 2 2 , and in particular, for this problem, we have: T1 T2 PVT (34.0 atm)(62.9 cm3 )(364 K) T2 = 2 2 1 = = 649 K = 376 C 3 3 P1V1 (1.00 atm)(1.20 × 10 cm ) 11.88
First convert the temperature data to the Kelvin scale: 273 + 5/9(58.0 – 32.0) = 287 K and 273 + 5/9(102 – 32) = 312 K. Next, calculate the final pressure at the gauge, taking into account the temperature change only:
PT (64.7 lb in.−2 )(312 K) P2 = 1 2 = = 70.3 lb in.−2 T1 (287 K) This represents the actual pressure inside the tire. The pressure gauge measures only the difference between the pressure inside the tire and the pressure outside the tire (atmospheric pressure). Hence the gauge reading is equal to the internal pressure of the tire less atmospheric pressure: (70.3 – 14.7) lb/in2 = 55.6 lb/in2 11.89
The temperatures must first be converted to Kelvin:
5 5 × ( F − 32) = × ( − 50 − 32) = − 46 C or 227 K 9 9 5 5 C= × ( F − 32) = × (120 − 32) = 49 C or 322 K 9 9 Next, the pressure calculation is done using the following equation: PT (32 lb in.−2 )(322 K) P2 = 1 2 = = 45 lb in.−2 T1 (227 K)
C=
236
Chapter 11
11.90
Using the ideal gas law, determine the number of moles of H2 and O2 gas initially present: For hydrogen:
n=
(1250 torr ) PV = RT 0.0821
(
( 7601 atmtorr ) ( 0.400 L ) = 2.52 × 10−2 mol H 2 L atm 318 K ) mol K ) (
for oxygen:
n=
( 740 torr ) PV = RT 0.0821
(
( 7601 atmtorr ) ( 0.300 L ) = 1.19 × 10−2 mol O 2 L atm 298 K ) mol K ) (
This problem is an example of a limiting reactant problem in that we know the amounts of H2 and O2 initially present. Since 1 mol of O2 reacts completely with 2 mol of H2, we can see, by inspection, that there is excess H2 present. Using the amounts calculated above, we can make 2.38 × 10–2 mol of H2O and have an excess of 1.4 × 10–3 mol of H2. Thus, the total amount of gas present after complete reaction is 2.52 × 10–2 mol. Using this value for n, we can calculate the final pressure in the reaction vessel:
(
)(
L atm 2.52 × 10−2 mol 0.0821 mol nRT K P= = V ( 0.500 L )
11.91
) ( 395 K ) = 1.63 atm = 1.24 × 103 torr
We first need to determine the pressure inside the apparatus. Since the water level is 8.5 cm higher inside than outside, the pressure inside the container is lower than the pressure outside. To determine the inside pressure, we first need to convert 8.5 cm of water to an equivalent dimension for mercury. This is done using the density of mercury: PHg = 85 mm/13.6 = 6.25 mm (where the density of mercury, 13.6 g/mL, has been used.) Pinside = Poutside – PHg = 746 torr – 6 torr = 740 torr. In order to determine the PH2, we need to subtract the vapor pressure of water at 24 °C. This value may be found in Table 11.2 and Appendix C.5 and is equal to 22.4 torr. The PH2 = Pinside – PH2O = 740 torr – 22.4 torr = 717 torr. Now, we can use the ideal gas law in order to determine the number of moles of H2 present;
(
)
(
)
1 atm 1L 16.35 mL ) 1000 ( 717 torr ) 760 PV torr ( mL n= = = 6.33 × 10−4 mol H 2 L atm RT 0.0821 ( 297 K )
(
mol K
)
The balanced equation described in this problem is: Zn(s) +2HCl(aq) ZnCl2(aq) + H2(g) By inspection we can see that 1 mole of Zn(s) reacts to form 1mole of H2(g) and we must have reacted 6.33 × 10–4 mol Zn in this reaction.
65.39 g Zn −2 g Zn = 6.33 × 10−4 mol Zn = 4.14 × 10 g Zn 1 mol Zn
(
11.92
)
2H2 + O2 2H2O
1 mol H 2 mol H2 = (14.8 g H2) = 7.33 mol H2 2.02 g H 2 1 mol O 2 mol O2 = (92.3 g O2) = 2.88 mol O2 32.0 g O 2 ∴O2 is the limiting reactant 2 mol H 2 O mol H2O = (2.88 mol O2) = 5.76 mol H2 1 mol O 2
237
Chapter 11
2 mol H 2 mol H2 needed = (2.88 mol O2 = 5.76 mol H2 1 mol O2 Remaining mol H2 = 7.33 mol H2 – 5.76 mol H2 = 1.57 mol H2 L atm ( 5.76 mol ) 0.0821 ( 435 K ) mol K PH 2O = = 15.8 atm 13.0 L L atm (1.57 mol ) 0.0821 ( 435 K ) mol K PH 2 = = 4.31 atm 13.0 L PTot = 15.8 atm + 4.31 atm = 20.1 atm 11.93
Note: 0.08747 mg/mL = 0.08747 g/L
dRT molecular mass = = P
( 0.08747 g/L ) 0.0821
L atm ( 290.2 K ) mol K = 2.08 g/mol 1 atm ( 760 torr) 760 torr
This gas must be H2. 11.94
(a)
The equation can be rearranged to give:
0.04489 ×
V(P − PH 2O ) 273 + t °C
= %N × W
This means that the left side of the above equation should be obtainable simply from the ideal gas law, applied to the nitrogen case. If PV = nRT, then for nitrogen: PV = (mass nitrogen)/(28.01 g/mol) × RT, and the mass of nitrogen that is collected is given by: (mass nitrogen) = PV(28.01)/RT, where R = 82.1 mL atm/K mol × 760 torr/atm = 6.24 × 104 mL torr/K mol. Using this value for R in the above equation, we have the following result for the mass of nitrogen, remembering that the pressure of nitrogen is less than the total pressure, by an amount equal to the vapor pressure of water:
(mass nitrogen) =
28.01 × V × (Ptotal − PH 2O )
( 6.24 × 10
4 mL torr mol K
)( 273 + C)
Finally, it is only necessary to realize that the value
28.01 6.24 × 104
× 100 = 0.04489
is exactly the value given in the problem. (b)
% N = 0.04489 ×
(18.90 mL)(746 torr − 22.1 torr) = 8.639 % ( 0.2394 g )( 273.15 + 23.80 )
Multi-Concept Problems 11.95
(a)
Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) Calculate the number of moles of hydrogen:
238
Chapter 11
n=
(
)
1 atm 12.0 L ) ( 765 torr ) 760 PV torr ( = = L atm 293.2 K RT 0.0821 mol ( ) K
(
)
0.502 mol H 2
and the number of moles of zinc:
1 mol Zn mol Zn = ( 0.502 mol H 2 ) = 0.502 mol Zn 1 mol H 2 The number of grams of zinc needed is, therefore: 65.39 g Zn g Zn = (0.502 mol Zn) = 32.8 g Zn 1 mol Zn
(b)
2 mol HCl mol HCl = (0.502 mol Zn) = 1.004 mol HCl 1 mol Zn 1000 mL HCl mL HCl = (1.004 mol HCl) = 126 mL HCl 8.00 mol HCl
11.96
First determine the % by mass S and O in the sample: % S = 1.448 g/3.620 g × 100 = 40.00 % S % O = 2.172 g/3.620 g × 100 = 60.00 % O Before we determine the molecular formula of the compound we need to find the empirical formula. Determine the number of moles of S and O in a sample of the material weighing 100 g exactly, in order to make the conversion from % by mass to grams straightforward: In 100 g of the material, there are 40.00 g S and 60.00 g O: 40.00 g S ÷ 32.07 g/mol = 1.247 mol S 60.00 g O ÷ 16.00 g/mol = 3.750 mol O Dividing each of these mole amounts by the smaller of the two gives the relative mole amounts of S and O in the material: for S, 1.247 mol ÷ 1.247 mol = 1.000 relative moles, for O, 3.750 mol ÷ 1.247 mol= 3.007 relative moles, and the empirical formula is, therefore, SO3. We determine the formula mass of the material by use of the ideal gas law:
n=
(
)
1 atm 1.120 L ) ( 750 torr ) 760 PV torr ( = = L atm 298.2 K RT 0.0821 mol ( ) K
(
)
0.0451 mol
The formula mass is given by the mass in grams (given in the problem) divided by the moles determined here: formula mass = 3.620 g ÷ 0.0451 mol = 80.2 g mol–1. Since this is equal to the formula mass of the empirical unit determined in step (b) above, namely SO3, then the molecular formula is also SO3. 11.97
(a)
Ptotal = 746.0 torr = PH2O + PN2 PN2 = 746.0 torr – 22.1 torr = 723.9 torr Now, use the ideal gas equation to determine the moles of N2 that have been collected:
239
Chapter 11
n=
(
)
(
)
1 atm 1L 18.90 mL ) 1000 ( 723.9 torr ) 760 PV torr ( mL = = L atm RT 0.0821 mol K ( 296.95 K )
(
)
7.384 × 10−4 mol N 2
Then the mass of nitrogen that has been collected is determined: 7.384 × 10–4 mol N2 × 28.0 g/mol = 2.068 × 10–2 g N2. Next, the % by mass nitrogen in the material is calculated: % N = (0.02068 g)/(0.2394 g) × 100 = 8.638 % N (b)
mass of C in the sample:
1 mole CO2 1 mol C g C = (17.57 × 10−3 g CO2 ) 44.01 g CO 2 1 mol CO 2
12.01 g C 1 mol C
= 4.795 × 10−3 g C mass of H in the sample:
1 mole H 2 O 2 mol Η 1.008 g H g H = (4.319 × 10−3 g H 2 O) 18.02 g H 2 O 1 mol H 2 O 1 mol H
= 4.832 × 10−4 g H mass of N in the sample:
8.638 g N −4 g N = (6.478 × 10−3 g sample) = 5.596 × 10 g N 100 g sample mass of O in the sample = total mass – (mass C + H + N) mg O = 6.478 mg sample − ( 4.795 mg C + 0.4832 mg H + 0.5596 mg N )
= 0.640 mg O Next we convert each of these mass amounts into the corresponding mole values: for C, 4.795 × 10–3 g ÷ 12.01 g/mol = 3.993 × 10–4 mol C for H, 4.832 × 10–4 g ÷ 1.008 g/mol = 4.794 × 10–4 mol H for N, 5.596 × 10–4 g ÷ 14.01 g/mol = 3.994 × 10–5 mol N for O, 6.40 × 10–4 g ÷ 16.00 g/mol = 4.00 × 10–5 mol O Last, we convert these mole amounts into relative mole amounts by dividing each by the smallest of the four: for C, 3.993 × 10–4 mol/ 3.994 × 10–5 mol = 9.998 for H, 4.794 × 10–4 mol/ 3.994 × 10–5 mol = 12.00 for N, 3.994 × 10–5 mol/ 3.994 × 10–5 mol = 1.000 for O, 4.00 × 10–5 mol/ 3.994 × 10–5 mol = 1.00 The empirical formula is therefore C10H12NO % C = (4.795 x 10-3 g C/6.478 x 10-3 g sample) x 100 = 74.02 % % H = (4.832 x 10-4g H/6.478 x 10-3 g sample) x 100 = 7.46 % The formula mass of the empirical unit is 162. Since this is half the value of the known molecular mass, the molecular formula must be twice the empirical formula, C20H24N2O2. 11.98
(a)
We begin by converting the dimensions of the room into cm: 42 ft × 30.48 cm/ft = 1.3 × 103 cm, 24 ft × 30.48 cm/ft = 7.3× 102 cm, 8.6 ft × 30.48 cm/ft = 2.6 × 102 cm. Next, the volume of the room is determined: V = (1.3 × 103 cm)(7.3 × 102 cm)(2.6 × 102 cm) = 2.4 × 108 cm3. Since there are 1000 cm3 in a liter, volume is: V = 2.4 × 105 L
240
Chapter 11
The calculation of the amount of H2S goes as follows:
0.15 L H 2S L H 2S = 2.4 × 105 L space = 3.6 × 10−5 L H 2S 1 × 109 L space
(
(b)
)
Convert volume (in liters) to moles at STP:
1 mol H 2S −6 mol H 2S = 3.6 × 10−5 L H 2S = 1.6 × 10 mol H 2S 22.4 L H S 2
(
)
Since the stoichiometry is 1:1, we require the same number of moles of Na2S:
1000 mL Na 2S mL Na 2S = 1.6 × 10−6 mol Na 2S 0.100 mol Na 2S
(
)
= 1.6 × 10−2 mL Na 2S 11.99
Cl2 + SO32– + H2O 2Cl– + SO42– + 2H+
0.200 moles Na 2SO3 1 mole SO32− 1 mole Cl2 2− 1000 mL Na 2SO3 1 mole Na 2SO3 1 mole SO3
( 50.0 mL Na 2SO3 )
moles Cl2 =
= 1.00 × 10−2 moles Cl2
(1.00 × 10 VCl2 =
L atm moles 0.0821 ( 298 K ) mol K = 0.253 L = 253 mL 1 atm ( 734 torr ) 760 torr
)
−2
11.100 Ptotal = 740 torr = PH2 + Pwater The vapor pressure of water at 25 °C is available in Table 11.2: 23.76 torr. Hence: PH2 = (740 – 24) torr = 716 torr Next, we calculate the number of moles of hydrogen gas that this represents:
(
)
1 atm 0.335 L ) ( 716 torr ) 760 PV torr ( n= = = L atm 298.2 K RT 0.0821 mol ) K (
(
)
0.0129 mol H 2
The balanced chemical equation is: Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) and the quantities of the reagents that are needed are:
1 mol Zn g Zn = (0.0129 mol H 2 ) 1 mol H 2
65.39 g Zn = 0.844 g Zn 1 mol Zn
2 mol HCl 1000 mL HCl mL HCl = (0.0129 mol H 2 ) = 4.30 mL HCl 1 mol H 2 6.00 mol HCl 11.101 This is a limiting reactant problem. First we need to calculate the moles of dry CO2 that can be produced from the given quantities of CaCO3 and HCl:
241
Chapter 11
1 mol CaCO3 mol CaCO3 = (12.3 g CaCO3 ) = 0.123 mol CaCO3 100.09 g CaCO3 0.250 mol HCl mol HCl = (185 mL HCl) = 0.0463 mol HCl 1000 mL HCl Thus, HCl is limiting and we use this to determine the moles of CO2 that can be produced: 1 mol CO 2 mol CO 2 = (0.0463 mol HCl) = 0.0231 mol CO 2 2 mol HCl The pressure of the dry CO2 is 745 torr. Finally, the volume of this “dry” CO2 is calculated using the ideal gas equation:
(
) ( 293.2 K ) = 0.567 L CO or 567 mL 2 ( 760 torr )
L atm ( 0.0231 mol ) 0.0821 mol nRT K V= = 1 atm P ( 745 torr )
11.102 formula mass =
dRT (mass)RT = P PV
(
) (
L atm (298.2 K) (6.3 × 10−3 g) 0.0821 mol K formula mass = 1 atm 1L (11 torr) 760 torr (385 mL) 1000 mL
(
)
)
–1
formula mass = 28 g mol
The formula weights of the boron hydrides are: BH3, 13.8 B2H6, 27.7 B4H10, 53.3 And we conclude that the sample must have been B2H6.
242
Chapter 12
Practice Exercises 12.1
(a) (b)
CH3CH2CH2CH2CH3 < CH3CH2OH < KBr CH3CH2OCH2CH3 < CH3CH2NH2 < HOCH2CH2CH2CH2OH
12.2
Propylamine would have a substantially higher boiling point because of its ability to form hydrogen bonds (there are N–H bonds in propylamine, but not in trimethylamine.)
12.3
The piston should be pushed in. This will decrease the volume and increase the pressure, and when equilibrium is re–established, there will be fewer molecules in the gas phase.
12.4
The number of molecules in the vapor will decrease, and the number of molecules in the liquid will increase, but the sum of the molecules in the vapor and the liquid remains the same.
12.5
Decreasing the temperature will decrease the vapor pressure and therefore, decrease the number of molecules in the vapor state. The volume will remain constant as long as the external pressure on the piston is also decrease by the same amount as the product of nT.
12.6
The boiling point is most likely (a) less than 10 °C above 100 °C.
12.7
We use the curve for water, and find that at 330 torr, the boiling point is approximately 75 °C.
12.8
You will need to calculate the joules of heat lost for the following: (a) benzene vapor at 105.00C cools to it boiling point, 80.10C (b) benzene vapor, at 80.10C condenses (c) benzene liquid cools from 80.10c to 25.00C q = 55 x 1.92 J g-1 0C-1 x (105.0 – 80.1 0C) + 55 g x (1 mol/78.11) x 30,770 J/mol + 55 g x 1.8 J g-1 0C-1 x (80.1 0c – 25.0 0C) = 3.0 x 104 J or 3.0 x 101 kJ
12.9
The heat released when 10 g of water vapor condenses is: q = 10 g x (1 mol/18.01 g) x 43,900 J mol-1 + 10 g x 4,184 J g-1 0C-1 x (1000C – 37 0C) = 2.7 x 104 J or 2.7 x 101 kJ The heat content of 10 g of water at 100 0C cooling to 370C is q = 10 g x 4.184 J g-1 0C-1 x (1000C – 37 0C) = 2.6 x 103 J or 2.6 kJ
12.10
ln
P1 ∆Hvap 1 1 = − P2 R T 2 T1
45.37mm 30,100kJ mol−1 1 1 = − = −2.458 −1 −1 P2 8.314 J mol K 335.4 K 273.2 K ln 45.37 − ln P 2 = −2.458 ln
− ln P 2 = −2.458 − 3.815 = −6.273
P 2 = 530 mm Hg
243
Chapter 12
12.11
ln
P1 ∆Hvap 1 1 = − P2 R T 2 T1
ln
0.0992 40,500 J mol −1 1 1 = − −1 −1 1.00 8.314 J mol K T 2 300.5 K
ln
0.0992 40,500 J mol −1 1 1 = − −1 −1 1.00 8.314 J mol K T 2 300.5 K
1 −2.311 = 4871.3 − 3.338 x 10−3 T2
T 2 = 350.5 K 12.12
Adding heat will shift the equilibrium to the right, producing more vapor. This increase in the amount of vapor causes a corresponding increase in the pressure, such that the vapor pressure generally increases with increasing temperature.
12.13
Boiling Endothermic Melting Endothermic Condensing Exothermic Subliming Endothermic Freezing Exothermic No, each physical change is always exothermic, or always endothermic as shown.
12.14
The line from the triple point to the critical point is the vapor pressure curve, see Figure 12.23.
12.15
Refer to the phase diagram for water, Figure 12.29. We "move" along a horizontal line marked for a pressure of 2.15 torr. At –20 °C, the sample is a solid. If we bring the temperature from –20 °C to 50 °C, keeping the pressure constant at 2.15 torr, the sample becomes a gas. The process is thus solid gas, i.e. sublimation.
12.16
As diagramed in Figure 12.29, this falls in the liquid region.
12.17
Since chromium crystallizes in a body centered cubic structure there are Cr atoms located at the corners of the cube and in the center of the cube. At the corner 1/8 of a Cr atom occupies the cube. There are eight corners of 8 x (1/8 Cr) = 1 Cr atom. In addition, there is a Cr atom in the center of the cube. Thus, there are a total of two Cr atoms per unit cell for a bcc structure.
12.18
For cesium: 8 corners × 1/8 Cs+ per corner = 1 Cs+ For chloride: 1 Cl− in center, Total: 1 Cl− Thus, the ratio is 1 to 1. This matches the observed composition of CsCl.
12.19
The compound is an organic molecule and the solid is held together by dipole–dipole attractions and London forces. It is also a soft solid with a low melting point, so it is a molecular crystal.
12.20
Because this is a high melting, hard material, it must be a covalent or network solid. Covalent bonds link the various atoms of the crystal.
12.21
Since the melt does not conduct electricity, it is not an ionic substance. The softness and the low melting point suggest that this is a molecular solid, and indeed the formula is most properly written S8. 244
Chapter 12
Review Questions 12.1
Intermolecular forces in liquids and solids are more important than in gases because the molecules and atoms of liquid and solid samples are so much closer together than they are in a gas.
12.2
The transfer of a gas from one container to another may be accompanied by either a change in shape or volume, or both. Only the shape of a liquid may change when its container is altered; the volume of a liquid does not change when the liquid is transferred to a new container. A solid changes neither its shape nor its volume when it is transferred into a new container.
12.3
The intermolecular attractive forces are strongest for a solid and weakest for a gas with the liquid state in between.
12.4
The potential energy for intermolecular forces is a function of 1/dn where d is the separation between two chemical species and n is an integer. For dipole-dipole interactions n = 3. The inverse function falls off in value very rapidly as d increases.
12.5
It is the intramolecular forces, the bonds, that are responsible for the chemical properties, not the intermolecular forces. For the physical properties, the intermolecular forces are: dipole-dipole attractions, hydrogen bonds, London forces, and ion–dipole attractions.
12.6
London forces are diagrammed in Figure 12.3. These weak forces of attraction are caused by instantaneous dipoles that attract induced dipoles in neighboring molecules. London forces increase in strength with increasing molecular size, as illustrated in Figure 12.4 and as shown by the data of Table 12.1 and Table 12.2. London forces increase in strength as the number of atoms in a molecule increases, as illustrated in Figure 12.5. London forces decrease the more compact the molecule is compared to a more chainlike molecule with the same number and types of atom as illustrated in Figure 12.6.
12.7
Polarizability is a measure of the ease with which the electron cloud is distorted. If an electron cloud is easily polarizable, that is large and easily deformed, then instantaneous dipoles and induced dipoles form without much difficulty and stronger London forces are experienced by that molecule.
12.8
Dipole–dipole interactions arise from the attraction of the permanent dipole moment of one molecule with that of an adjacent molecule, the positive end of one dipole being drawn to the negative end of the other dipole. This is diagrammed in Figure 12.2 of the text.
12.9
These are fluorine, oxygen and nitrogen, which have small atomic size and high electronegativities.
12.10
Since these are both nonpolar molecular substances, the only type of intermolecular force that we need to consider is London forces. The larger molecule has the greater London force of attraction and hence the higher boiling point: C8H18.
12.11
Whereas the ether has no O–H linkage, ethanol does. Therefore, ethanol can have hydrogen bonding between molecules and ether cannot. Ethanol thus has stronger intermolecular forces, and its boiling point is consequently higher.
12.12
Covalent bonds are normally about 100 times stronger than normal dipole–dipole attractions; hydrogen bonds are about 5–10 times stronger than dipole–dipole attractions.
12.13
Ion-dipole interactions increase with increase in ion charge so Al3+ would have a stronger interaction with water than would Na+.
245
Chapter 12 12.14
(a) (b)
O2- would have stronger ion–induced dipole because the O2– radius is smaller than S2–, so the charge density will be higher. Al3+ will have the stronger ion–induced dipole because Al3+ has a higher charge and smaller size than Mg2+, so it will have a higher charge density.
12.15
Physical properties that depend on tightness of packing: compressibility and diffusion. Physical properties that depend on the strengths of intermolecular interactions: retention of volume and shape, surface tension, wetting of a surface by a liquid, viscosity, evaporation, and sublimation.
12.16
The particles of a gas are free to move randomly, and thus to diffuse readily. Diffusion in liquids is comparatively slower because of the more numerous collisions that a molecule in a liquid sample must undergo in traveling from place to place. The particles of a solid are not free to move from place to place in a solid sample.
12.17
The rate of diffusion should increase because the molecules move faster at the higher temperature.
12.18
Surface tension is a measure of the work necessary to increase the surface area of a mass of liquid. Molecules at the surface of a liquid have no other molecules above them, and they consequently are attracted only to those molecules that are next to them – namely, those in the interior of the liquid. This is illustrated in Figure 12.14.
12.19
Water should have the greater surface tension because it has the stronger intermolecular forces, i.e., hydrogen bonding.
12.20
Wetting – spreading a liquid across a surface to form a thin film. Surfactant – a substance that lowers surface tension in a liquid and thereby promotes wetting.
12.21
There is no intermolecular force common to both polyethylene and water that can allow for wetting. The surface tension of water, which is high, is not disrupted by any effective interaction between water and polyethylene.
12.22
Glycerol ought to wet the surface of glass quite nicely, because the dipolar bonds at the surface of glass can interact strongly with the polar O–H groups of glycerol.
12.23
Viscosity increases as the magnitude of intermolecular forces increases. Acetone has a larger dipole, 2.91 D than water, 1.85 D; however, water has hydrogen bonding interactions which acetone does not have. Therefore, water should be more viscose than acetone. Ethylene glycol has a larger dipole than water, 2.28 D, and like water it also has hydrogen bonding. In addition, ethylene glycol is a larger molecule than water so it possesses greater London forces between molecules. Therefore, the order, based on intermolecular force arguments is correct.
12.24
Since it is the high energy molecules in a sample that are the first to evaporate, the remaining molecules have a lower average kinetic energy. A reduction in kinetic energy corresponds to a decrease in temperature.
12.25
Raising the temperature of the sample increases the fraction of molecules in the sample that have enough kinetic energy to escape by evaporation.
12.26
An increase in surface area causes an increase in the rate of evaporation because, when the surface area is increased, there are more molecules in position at the surface of the liquid sample, where they are capable of evaporation. The stronger the intermolecular forces, the less readily a substance can evaporate.
12.27
The snow dissipates by sublimation. Even at low temperatures (by human standard) many compounds have measurable vapor pressures.
246
Chapter 12 12.28
(a) (b) (c) (d) (e) (f)
sublimation deposition vaporization condensation melting freezing
12.29
After the molecule is in the vapor phase for a while its kinetic energy will be less because it collided with other molecules in the vapor phase and transferred some of its energy to the other molecules. This molecule is not likely to bounce out of the surface of the liquid unless it regains some kinetic energy from the liquid.
12.30
This happens because of the loss in kinetic energy that the colliding molecule experiences when it hits the surface molecules. The colliding molecule has less kinetic energy after striking the surface, and its ability to escape subsequently from the liquid is momentarily diminished by the presence of intermolecular forces of attraction generated upon mixing.
12.31
A dynamic equilibrium is established if the liquid evaporates into a sealed container. It is termed a dynamic equilibrium because opposing processes (evaporation and condensation) continue to take place, once the condition of equilibrium has been achieved. At equilibrium, the rate of condensation is equal to the rate of evaporation, and there is consequently no net change in the number of molecules in the vapor or in the liquid.
12.32
A dynamic equilibrium is achieved when a solid is held at its melting temperature. At this point, particles are melting and freezing at an equivalent rate. This is the melting or freezing point of a substance.
12.33
Yes. This is the sublimation process.
12.34
Equilibrium vapor pressure is the pressure exerted by a vapor that is in equilibrium with its liquid. It is a dynamic equilibrium because events have not ceased. Liquid continues to evaporate once the state of equilibrium has been reached, but the rate of evaporation is equal to the rate of condensation. These two opposing processes occur at equal rates, such that there is no further change in the amount of either the liquid or the gas.
12.35
Changing the volume only upsets the equilibrium for a moment, provided the volume is not increased to a point that all the liquid evaporates at which no equilibrium would exist. After sufficient time has elapsed, the rates of evaporation and condensation again become equal to one another, and the same condition of equilibrium is achieved. The vapor pressure (or the ease of evaporation) only depends on the strength of intermolecular forces in the liquid sample.
12.36
The equilibrium vapor pressure is governed only by the strength of the attractive forces within the liquid and by the temperature.
12.37
Raising the temperature increases the vapor pressure by imparting enough kinetic energy (for evaporation) to more of the liquid molecules.
12.38
At the temperature of the cool glass, the equilibrium vapor pressure of the water is lower than the partial pressure of water in the air. The air in contact with the cool glass is induced to relinquish some of its water, and condensation occurs.
12.39
In humid air, the rate of condensation on the skin is more nearly equal to the rate of evaporation from the skin, and the net rate of evaporation of perspiration from the skin is low. The cooling effect of the evaporation of perspiration is low, and our bodies are cooled only slowly under such conditions. In dry air, however, perspiration evaporates more rapidly, and the cooling effect is high.
247
Chapter 12 12.40
The boiling point is the temperature at which the liquid boils and the vapor pressure of the liquid is equal to the prevailing atmospheric pressure. The normal boiling point is the boiling point of a liquid when the atmospheric pressure is 760 mm Hg.
12.41
This happens because boiling is a process that is a function of pressure. Since the vapor pressure varies with temperature, the boiling point must also change as the pressure changes.
12.42
At about 80 °C
12.43
Ethanol vapor is present inside the bubbles of boiling ethanol.
12.44
Even at higher temperatures, the contents of the radiator do not boil, because the pressure in the system increases with temperature, since the system is closed. The boiling point of the liquid is higher because the pressure is higher.
12.45
Inside the lighter, the liquid butane is in equilibrium with its vapor, which exerts a pressure somewhat above normal atmospheric pressure. This keeps the butane as a liquid.
12.46
Since H2Se is larger than H2S, its London forces are stronger than in H2S. Because water is capable of hydrogen bonding, whereas H2S is not, its boiling point is higher than that of H2S.
12.47
The hydrogen bond network in HF is less extensive than in water, because it is a monohydride not a dihydride, that is HF can only donate one hydrogen for a hydrogen bond while water can donate two hydrogens.
12.48
(a) (b) (c) (d) (e) (f) (g) (h)
12.49
The heat of vaporization of a molecular substance is generally larger than the heat of fusion, because, in vaporization, the molecules undergo much larger changes in their distance of separation (and require the disruption of much stronger intermolecular forces) than is true of melting. The heat of sublimation is typically larger than the heat of vaporization of a liquid because sublimation involves a greater change in intermolecular separation, a larger disruption of intermolecular forces of attraction, and hence a larger change in potential energy.
12.50
The heat of condensation is exothermic, and it is equal in magnitude, but opposite in sign to the heat of vaporization (which is endothermic).
12.51
The moisture which powers the hurricane condenses when the hurricane travels over cold water. As the amount of vapor in the storm decreases, its energy decreases.
12.52
The substance with the larger molar heat of vaporization has the stronger intermolecular forces. This is ethanol, which has hydrogen bonding, whereas ethyl acetate does not.
12.53
Steam releases a considerable amount of energy in the form of condensation energy as opposed to simply cooling liquid water.
12.54
CH4 < CF4 < HCl < HF
12.55
According to the Clausius-Clapeyron equation, vapor pressure increases as temperature increases.
1, 3, and 5 2 and 4 2 4 The heat of vaporization is larger. This is the temperature of line 4. This is the temperature of line 2. Line 3 would descend lower in temperature than line 4, before rising to the temperature of line 4.
248
Chapter 12
12.56
The gas constant used in the Clausius-Clapeyron equation is in units of Kelvin for temperature so you must use Kelvin not Celsius temperatures.
12.57
When a system at equilibrium is modified so as to upset the equilibrium, the system will respond in a manner which enables the equilibrium to be reestablished. See also Section 12.9.
12.58
By "position of equilibrium" we mean the relative amounts of the various reactants and products that exist in the equilibrium mixture.
12.59
This is an endothermic system, and adding heat to the system will shift the position of the equilibrium to the right, producing a new equilibrium mixture having more liquid and less solid. Some of the solid melts when heat is added to the system.
12.60
Changing the temperature disrupts the sublimation, deposition equilibrium that was established. By lowering the temperature, the kinetic energy of the molecules will decrease and more deposition will occur, giving rise to a lower vapor pressure. This will continue until a new equilibrium is established in which the rate of sublimation and deposition are again equal.
12.61
An increase in pressure should favor the system with the lower volume, i.e. the solid. Therefore, if the substance is at its melting point at a pressure of one atmosphere, and then if the pressure were to be increased, more solid would form at the expense of liquid – that is, more of the substance would freeze. If melting were to be accomplished at the higher pressure, it would require a temperature that is higher than the normal melting temperature. The phase diagram would be similar to the one for carbon dioxide, see Fig. 12.31.
12.62
Critical temperature – the temperature above which the substance can not exist as a liquid, regardless of the applied pressure. It is, therefore, the temperature above which a gas cannot be made to liquefy, regardless of the amount of pressure that is applied. Critical pressure – the vapor pressure of a liquid at the liquid's critical temperature. A critical temperature and critical pressure together constitute a substance's critical point.
12.63
A supercritical fluid is a substance at a temperature above its critical temperature. Supercritical CO2 is used to decaffeinate coffee because it replaces organic solvents such as methylene chloride and ethyl acetate which cannot be completely removed from the coffee bean. Supercritical CO2 can remove as much as 97% of the caffeine.
12.64
Solid, liquid and gas are all in equilibrium at the triple point.
12.65
Carbon dioxide does not have a normal boiling point because its triple point lies above one atmosphere. Thus, the liquid–vapor equilibrium that is taken to represent the boiling point does not exist at the pressure (1 atm) conventionally used to designate the "normal" boiling point.
12.66
The critical temperature of hydrogen is below room temperature because, at room temperature, it cannot be liquefied by the application of pressure. The critical temperature of butane is above room temperature, because butane can be liquefied by the application of pressure.
12.67
Crystalline solids have an ordered internal structure while amorphous solids do not have the long-range repetitive order of crystalline solids.
12.68
A lattice is a set of points that have the same repeat distances and are arranged along lines oriented at the same angles. A unit cell is the smallest repeating unit of the lattice that can be used to define the lattice.
12.69
The entire crystal lattice of the solid can be generated by repeated use of the unit cell only. 249
Chapter 12
12.70
a)
b)
12.71
No, it is not a unit cell of the substance. If the unit cell is move in one direction, to the right for example, then the smaller atom would be replace by a larger atom. This is not repeating unit.
12.72
(a) Cl- ions
(b) Na+ ions
Each unit cell is a face centered cubic cell. Unit cell (a) has chloride ions in the corners and face and unit cell (b) has sodium ions in the corners and face. Also, see Figures 12.35 and 12.40. 12.73
These structures are both of the face–centered cubic variety. They differ only in the length of an edge for a unit cell, that is, the length of the cube edge is different in the two metals. Silver might be expected to have a face–centered unit cell, also.
12.74
Zinc sulfide has a face centered cubic lattice. Calcium fluoride also has a face centered cubic lattice.
12.75
Although there are only fourteen different kinds of lattice geometries that can fill space, there are essentially an infinite variety of cell dimensions that can be adopted by substances.
12.76
nλ = 2d sinθ n = an integer (1, 2, 3, . . .) λ = wavelength of the X–rays d = the interplane spacing in the crystal θ = the angle of incidence and the angle of reflectance of X–rays to the various crystal planes.
12.77
These are not 1:1 ionic substances. The cubic unit cell of NaCl contains the same number of sodium and chloride ions. 250
Chapter 12
12.78
The lattice positions are occupied by metal cations, which are then surrounded by the core electrons of the metal. A sea of valence electrons encompasses the entire metallic solid.
12.79
(a) (b) (c)
12.80
Covalent crystals are also termed network solids because they are constructed of atoms that are covalently bonded to one another, giving a giant interlocking network.
12.81
Amorphous means, literally, without form. It is taken here to represent a solid that does not have the regular, repeating geometrical form normally associated with a crystal lattice.
12.82
An amorphous solid is a noncrystalline solid. It is a solid that lacks the long–range order that characterizes a crystalline substance. When cooled, a liquid that will form an amorphous solid gradually becomes viscous and slowly hardens to give a glass, or a supercooled liquid. When crystalline solids are broken, the angles are regular and the faces are flat. When amorphous solids are broken, the faces are smooth and flat.
dipole–dipole, London forces or hydrogen bonds electrostatic forces covalent bonds
Review Problems 12.83
Dimethyl ether has the faster rate of vaporization, since it does not have hydrogen bonds, as does ethanol.
12.84
Dimethyl ether should have a higher vapor pressure since it has weaker intermolecular forces. Ethanol has a higher boiling point since it has stronger intermolecular forces of attraction. London forces are possible in them all. Where another intermolecular force can operate, it is generally stronger than London forces, and this other type of interaction overshadows the importance of the London force. The substances in the list that can have dipole–dipole attractions are those with permanent dipole moments: (a), (b), and (c) PCl5, (d), is a non–polar molecular substance. (CH3)2NH, (a), has hydrogen bonding.
12.85
12.86
(a) (b) (c) (d)
London forces, dipole-dipole London forces, dipole-dipole London forces London forces
12.87
Chloroform would be expected to display larger dipole-dipole attractions because it has a larger dipole moment than bromoform. (Chlorine has a higher electronegativity which results in each C–Cl bond having a larger dipole than each C–Br bond.) On the other hand, bromoform would be expected to show stronger London forces due to having larger electron clouds which are more polarizable than those of chlorine. Since bromoform in fact has a higher boiling point that chloroform, we must conclude that it experiences stronger intermolecular attractions than chloroform, which can only be due to London forces. Therefore, London forces are more important in determining the boiling points of these two compounds.
12.88
NO2 is capable of forming a liquid at atmospheric pressure while CO2 does not, this suggests that NO2 has stronger intermolecular attractions than CO2. Since CO2 and NO2 cannot form hydrogen bonds, the next strongest intermolecular interactions are dipole-dipole interactions; therefore, NO2 probably has dipoledipole interactions. This would require NO2 to have a dipole which is only possible if it is bent, while CO2 is linear.
12.89
London forces are higher in chains than in branched isomers. Therefore, octane has higher London forces between molecules than 2,2,3,3-tetramethylbutane so octane would be more viscose.
251
Chapter 12 12.90
Wetting results from intermolecular interactions allowing a chemical to spread itself out over a surface. A wetting agent reduces the surface tension of a material. Since grease is a non-polar substance diethyl ether should be a better wetting agent than propylene glycol for a greasy glass surface. Diethyl ether is less polar than propylene glycol (1,2-propanediol). Diethyl ether will spread out over a greasy surface through London force interactions between diethyl ether and grease. In addition to the difference in polarity, 1,2propanediol would be able to wet a surface if it could hydrogen bond to the surface but grease does not allow for hydrogen bonding.
12.91
Ethanol, because it has H-bonding.
12.92
The London forces are stronger in CS2 because the larger S atoms are more easily polarized than O atoms. Consequently, CS2 has a higher boiling point than CO2.
12.93
diethyl ether < acetone < benzene < water < acetic acid
12.94
diethyl ether < ethanol < water < ethylene glycol
12.95
Compound
Intermolecular Forces Broken
(a) CH3CH2OH
London, Dipole-Dipole, Hydrogen Bonding
(b) H3CCN
London, Dipole-Dipole
(c) NaCl
London, Ionic
12.96
(a) (b) (c)
London forces dipole-dipole and London forces London forces
12.97
1 mol H 2 O 43.9 kJ kJ = (165 g H2O) = 402 kJ 18.015 g H 2 O 1 mol H 2 O
12.98
1 mol C3 H 6 O −30.3 kJ kJ = (6.25 g C3H6O) = –3.26 kJ 58.1 g C3 H 6 O 1 mol C3 H 6 O
12.99
We can approach this problem by first asking either of two equivalent questions about the system: how much heat energy (q) is needed in order to melt the entire sample of solid water (105 g), or how much energy is lost when the liquid water (45.0 g) is cooled to the freezing point? Regardless, there is only one final temperature for the combined (150.0 g) sample, and we need to know if this temperature is at the melting point (0 °C, at which temperature some solid water remains in equilibrium with a certain amount of liquid water) or above the melting point (at which temperature all of the solid water will have melted). Heat flow supposing that all of the solid water is melted: q = 6.01 kJ/mole × 105 g × 1 mol/18.0 g = 35.1 kJ Heat flow on cooling the liquid water to the freezing point: q = 45.0 g × 4.18 J/g °C × 85 °C = 1.60 × 104 J = 16.0 kJ The lesser of these two values is the correct one, and we conclude that 16.0 kJ of heat energy will be transferred from the liquid to the solid, and that the final temperature of the mixture will be 0 °C. The system will be an equilibrium mixture weighing 150 g and having some solid and some liquid in equilibrium with one another. The amount of solid that must melt in order to decrease the temperature of
252
Chapter 12
45.0 g of water from 85 °C to 0 °C is: 16.0 kJ ÷ 6.01 kJ/mol = 2.66 mol of solid water. 2.66 mol × 18.0 g/mol = 47.9 g of water must melt. (a) (b)
The final temperature will be 0 °C. 47.9 g of solid water must melt.
12.100 The amount of heat gained by melting all the benzene is: 1 mol benzene 9.92 kJ kJ = (10.0 g benzene) = 1.27 kJ 78.11 g benzene 1 mol Assuming all of this heat is removed from the water and using: Heat = mass × specific heat × ∆T ∆T =
heat = mass × specific heat
1.27 × 103 J
(10.0 g ) ( 4.184 J/gC )
= 30.4 °C
The melting point of benzene is 5.5 oC. Therefore, the water will cool to 5.5 oC, at which point some solid benzene and the liquid water will be in equilibrium. Thus, not all of the benzene can be melted. 12.101
ln
P1 ∆Hvap 1 1 = − P2 R T 2 T1
1atm 59, 200J mol−1 1 1 = − −1 −1 8.314 J mol K 295 K 629.9 K P2 ln1 − ln P 2 = 12.833
ln
− ln P2 = 12.833
P 2 = 2.67 x 10−6 atm
12.102
ln
P1 ∆Hvap 1 1 = − P2 R T 2 T1
1atm 43,900J mol −1 1 1 = − −1 −1 8.314 J mol K 345 K 373 K P2 ln1 − ln P 2 = 1.1489
ln
− ln P2 = 1.1489 P 2 = 0.317 atm 12.103
ln
P1 ∆Hvap 1 1 = − P2 R T 2 T1
425mm ∆Hvap 1 1 = − −1 −1 1.0mm 8.314 J mol K 273.2-74.3 K 273.2 + 18.7 K 6.052 = (1.927 x 10−4 J −1 mol)∆Hvap ln
253
Chapter 12
∆Hvap = 31, 400 J mol−1 12.104
ln
P1 ∆Hvap 1 1 = − P2 R T 2 T1
ln
755mm ∆Hvap 1 1 = − −1 −1 7.23mm 8.314 J mol K 273.2+95.1 K 273.2 + 197.1 K
4.648 = (7.083x 10−5 J −1 mol)∆Hvap ∆Hvap = 65,600 J mol−1 12.105
P (atm)
1.0
0.30
-15.0 -10.0 T (° C)
90
12.106 Sublimation is possible only below a pressure of 0.30 atm, as marked on the phase diagram. The density of the solid is higher than that of the liquid. Notice that the line separating the solid from the liquid slopes to the right, in contrast to the diagram for water, Figure 12.29 of the text. 12.107 (a) solid
(b) gas
(c) liquid
(d) solid, liquid, and gas
12.108 The solid–liquid line slants toward the right. 12.109 For zinc:
4 surrounding center = 4 Zn2+ For sulfide: 8 corners × 1/8 S2− per corner = 1 S2– 6 faces × 1/2 S2− per face = 3 S2– Total = 4 S2–
12.110 A cube has six faces and eight corners. Each of the six face atoms is shared by two adjacent unit cells: 6 × 1/2 = 3 atoms. The eight corner atoms are each shared by eight unit cells: 8 × 1/8 = 1 atom. The total number atoms to be assigned to any one cell is thus 3 + 1 = 4. (6 × 1/2 copper atoms) + (8 × 1/8 copper atoms) = 4 copper atoms 12.111 From figure 12.38, we can see that the length of the diagonal of the cell = 4r, where r = radius of the atom. According to the Pythagorean theorem, a2 + b 2 = c2 254
Chapter 12 for a right triangle. Since a = b here, we may re–write this as 2l2 = c2, where l = length of the edge of the unit cell. As mentioned above, the diagonal of the unit cell = 4r, so we may say that 2l2 = (4r)2 l2 = (4r)2/2 l2 = 16r2/2 l2 = 8r2 l=
8r 2
Finally, substituting the value provided for r in the problem, l =
8 (1.24 Å )
2
= 3.51 Å. Using the
conversion factor 1pm = 100 Å, this is 351 pm. 12.112 The following diagram is appropriate:
The face diagonal is 4 times the radius of the atom. The Pythagorean theorem is: diagonal2 = edge2 + edge2. Hence we have: [4(144 pm)]2 = 2 × edge2. Solving for the edge length we get 407 pm. 12.113 Each edge is composed of 2 × radius of the cation plus 2 × radius of the anion. The edge is therefore 2 × 133 + 2 × 195 = 656 pm. 12.114 2 × rNa + dCl = 564.0 pm 2 × 95 pm + dCl = 564.0 pm dCl = 374.0 pm 12.115 Using the Bragg equation (eqn. 12.5), n λ = 2d sinθ (a) n(229pm) = 2(1,000)sinθ 0.1145n = sinθ θ = 6.57° (b) n(229pm) = 2(250)sinθ 0.458n = sinθ θ = 27.3° 12.116 nλ = 2d sin θ nλ d= 2sin θ At 20.0°
1 × 141 pm 2 × sin 20.0° d = 206 pm d=
At 27.4° d=
1 × 141 pm 2 × sin 27.4°
255
Chapter 12
d = 153 pm At 35.8° 1 × 141 pm 2 × sin 35.8° d = 121 pm d=
12.117 According to the Pythagorean theorem, a2 + b 2 = c2 for a right triangle. First, we need to find the length of a diagonal on a face of the unit cell. Since a = b here, we may re-write this as 2l2 = c2, where l = length of the edge of the unit cell and c = the diagonal length. Using the given 412.3 pm as the length of the edge, c = 583.1 pm. The diagonal length inside the cell from corner to opposite corner may now be found by the same theorem: a2 + b 2 = c2 (412.3)2 + (583.1)2 = c2 c = 714.1 pm This diagonal length inside the cell from corner to opposite corner is due to 1 Cs+ ion and 1 Cl− ion (see Figure 12.41). Therefore: 2rCs+ + 2rCl− = 714.1pm 2rCs+ + 2(181pm) = 714.1pm 2rCs+ = 352 pm rCs+ = 176 pm 12.118 (2 × rRb) + (2 × rCl) = 658 pm (2 × rRb) + (2 × 181 pm) = 658 pm rRb = 148.0 pm 12.119 This must be a molecular solid, because if it were ionic it would be high–melting, and the melt would conduct. 12.120 This is a covalent network solid. 12.121 This is a metallic solid. 12.122 This is a molecular solid. 12.123 (a) (b) (c)
molecular ionic ionic
(d) (e) (f)
metallic covalent molecular
(g)
ionic
12.124 (a) (b) (c)
molecular molecular metallic
(d) (e) (f)
ionic covalent ionic
(g)
molecular
12.125 Intermolecular (between different particles): The principal attractive forces are ion–ion forces and ion– dipole forces. These are of overwhelming strength compared to London forces, which do technically exist. Intramolecular (within certain particles): The phosphate ion (PO43−) is a polyatomic ion whose atoms are held together by covalent bonds. Although a covalent bond is not an “attraction” in itself, the attractive forces which make up these bonds are valence electron’s attractions to the nuclei of neighboring atoms in the polyatomic ion.
256
Chapter 12
12.126 The pressure of the water is: PH 2O = ( 0.82)(17.54 torr ) = 14.38 torr = 14.38 mm Hg The number of moles of water is calculated from the ideal gas equation: PV = nRT PV n= RT 1 atm (14.38 torr ) 760 torr (10.0 L) n= = 7.87 × 10–3 mol H2O 0.0821 L atm mol K ( 293 K )
(
)
The number of grams of water is calculated from the molecular mass of water: 18.02 g H 2 O g H2O = 7.87 × 10–3 mol H2O = 0.142 g H2O 1 mol H 2 O 12.127 Yes, because of the possibility of a weak hydrogen bond between the carbonyl oxygen of acetone and an OH group of water. 12.128 Using Hess’s Law, sublimation may be considered equivalent to melting followed by vaporization. ∆Hsublimation = ∆Hfusion + ∆Hvaporization = 10.8 kJ/mol + 24.3 kJ/mol = 35.1 kJ 12.129 The oxidation state of the chromium in CrO3 is Cr6+ and for Cr2O3, it is Cr3+. Since Cr6+ has a higher charge, it pulls electron density from the anion, O2–, towards itself to a greater extent than the Cr3+. Therefore, there will be more covalent character between the Cr6+ and O2– as compared to Cr3+ and O2–. This will cause the melting point of CrO3 to be lower than Cr2O3. 12.130 As the cooling takes place, the average kinetic energy of the gas molecules decreases, and the attractive forces that can operate among the gas molecules become able to bind the various molecules together in a process that leads to condensation. The air thus loses much of its moisture on the ascending side of the mountain. On descending the other side of the mountain, the air is compressed, and the temperature rises according to Charles' Law. The relative humidity of the air is now very low for two reasons: much of the moisture was released on the other side of the mountain range, and now that the temperature is higher, there is far less than the maximum allowable water content. The coast of California receives the rain since the air releases its moisture to the west of the mountains and the valleys east of the mountains do not receive any rain. 12.131 We start by determining the volume of one unit cell: 407.86 pm = 407.86 × 10–12 m = 407.86 × 10–10 cm Vol = (407.86 × 10–10 cm)3 = 6.7847 × 10–23 cm3 Next, we calculate the volume per atom, remembering that each unit cell contains a total of 4 gold atoms: 6.7847 × 10–23 cm3 / 4 atoms = 1.6962 × 10–23 cm3 This value is multiplied by the density: 1.6962 × 10−23 cm3 19.31 g = 3.275 × 10−22 g/atom cm3 atom Finally, it is necessary to divide this value into the atomic mass of gold: 196.97 g/mol = 6.014 × 1023 atoms/mol 3.275 × 10−22 g/atom
12.132 From Figure 12.38, we can see that the length of the diagonal of the cell = 4r, where r = radius of the atom. According to the Pythagorean theorem, a2 + b 2 = c2 for a right triangle. Since a = b here, we may re–write this as 2l2 = c2, 257
Chapter 12 where l = length of the edge of the unit cell. As mentioned above, the diagonal of the unit cell = 4r, so we may say that 2l2 = (4r)2 2l2 = 16r2 (0.125)l2 = r2
( 0.125 ) l2
=r
Finally, substituting the value provided for l in the problem, r = 12.133 (a) (b) (c)
( 0.125)( 407.86 )2
= 144.20 pm.
face-centered cubic simple cubic triclinic
12.134 Radius = 0.5diameter = 0.50 pm = r Simple Cubic: Volume of cube: (2r)3 = (2 × 0.50 pm)3 = 1 pm3 Volume of atoms: 1 atom in unit cell 4 4 Volume of atoms = πr3 = π(0.5 pm)3 = 0.524 pm3 3 3 Volume of empty space = Volume of cube – Volume of atoms = 1 pm3 – 0.524 pm3 = 0.476 pm3 Body-Centered Cubic: Volume of cube: (edge)3 (4r)2 = 3(edge)2 edge = a (4 × 0.50 pm)2 = 3(a)2 a = 1.15 pm Volume of cube = a3 = (1.15 pm)3 = 1.52 pm3 Volume of atoms: 2 atoms in unit cell 4 4 Volume of atoms = 2( πr3) = 2( π(0.5 pm)3) = 1.05 pm3 3 3 Volume of empty space = Volume of cube – Volume of atoms = 1.52 pm3 – 1.05 pm3 = 0.47 pm3 Face-Centered Cubic: Volume of cube: (edge)3 (4r)2 = 2(edge)2 edge = a (4 × 0.50 pm)2 = 2(a)2 a = 1.41 pm Volume of cube = a3 = (1.41 pm)3 = 2.83 pm3 Volume of atoms: 4 atoms in unit cell 4 4 Volume of atoms = 4( πr3) = 4( π(0.5 pm)3) = 2.09 pm3 3 3 Volume of empty space = Volume of cube – Volume of atoms = 2.83 pm3 – 2.09 pm3 = 0.74 pm3 The efficiency of packing is determined by dividing the volume occupied by the atoms divided by the volume occupied by the cube: Simple Cubic: 0.524 pm3 ÷ 1 pm3 = 0.524 Body-Centered Cubic 1.05 pm3 ÷ 1.52 pm3 = 0.691 Face-Centered Cubic 2.09 pm3 ÷ 2.83 pm3 = 0.739 12.135 a = edge of a unit cell First determine the mass of 1 atom of silver in grams:
258
Chapter 12
107.87 g Ag 1 mol Ag mass of 1 atom of silver = = 1.791 × 10–22 g /atom Ag 23 1 mol Ag 6.022 × 10 atoms Ag Then determine the volume of each unit cell and the number of atoms in each unit cell. (a) Simple Cubic Lattice: 2r = a volume of the unit cell = a3 = (2r)3 = (2 × 144 pm)3 = 2.39 × 107 pm3 3 1 cm 7 3 = 2.39 × 10–23 cm3/unit cell 2.39 × 10 pm 10 10 pm number of atoms in the unit cell = 1 1 atom/unit cell 1.791×10−22 g 1 atom Ag 1 unit cell = 7.49 g cm–3 Density = mass/volume = 1 atom Ag 1 unit cell 2.39 × 10−23 cm3 (b)
(c)
Body-Centered Cubic Lattice: (4r)2 = 3a2 (4 × 144 pm)2 = 3a2 a = 333 pm Volume of the unit cell = a3 = (333 pm)3 = (333 pm)3 = 3.69 × 107 pm3 3 1 cm = 3.69 × 10–23 cm3/unit cell 3.69 × 107 pm3 10 10 pm number of atoms in the unit cell = 2 2 atom/unit cell 1.791×10−22 g 2 atom Ag 1 unit cell Density = mass/volume = = 9.70 g cm–3 1 atom Ag 1 unit cell 3.69 × 10−23 cm3 Face-Centered Cubic Lattice: (4r)2 = 2a2 (4 × 144 pm)2 = 2a2 a = 407 pm Volume of the unit cell = a3 = (407 pm)3 = (407 pm)3 = 6.76 × 107 pm3 3 1 cm 7 3 –23 3 6.76 × 10 pm 10 = 6.76× 10 cm /unit cell 10 pm
number of atoms in the unit cell = 4 4 atom/unit cell 1.791×10−22 g 4 atom Ag 1 unit cell Density = mass/volume = 1 atom Ag 1 unit cell 6.76 × 10−23 cm3 Silver has a face-centered cubic lattic. 12.136 Using the Bragg equation (eqn. 12.5), nλ = 2d sin θ 154= 2dsin(12.8°) 77 = d(0.222) d = 347 pm The mass of the unit cell would be: (4 potassium ions × 39.10) + (4 chloride ions × 35.45) = 298.20 amu and its volume would be: (347 pm)3 = 4.20 × 107 pm3 259
–3 = 10.6 g cm
Chapter 12
Therefore, its density would be: d = m/V = 7.10 × 10–6 amu/pm3 12.137 Clouds form when the humid air of a warm front encounters the cool, relatively dry air of a cold front because the moisture in the air condenses.
Multi-Concept Problems 12.138 270 Calories is equal to 270,000 cal. The density of ice at − 25 oC is 0.9202 g cm−3 (CRC Handbook of Chemistry and Physics, 91st ed., 2010-11) Mass of water = 0.9202 g cm−3 x 75 mL ice = 69.0 g The energy available to heat water is: 270,000 cal x 4.184 J/cal = 1.13 x 106 J Consider the processes that will occur when this heat is absorbed by the water. (a) (b) (c) (d) (e)
Solid at -25 oC warms to 0 oC Solid melts Liquid warms to 100 oC Liquid vaporizes at 100 oC Vapor heats to a temperature higher than 100 oC
q = m C p∆ T q = ∆Hfus q = m C p∆ T q = ∆Hvap q = m C p∆ T
Start with steps (a) through (c) to determine the amount of heat used and then determine how much heat is left.
q = m C p∆ T q = 69.0 g x 2.25
J o
x 25 o C = 3881 J
g C Step (b) Melt the solid at 0 oC
q = ∆Hfus
q = 6.02
kJ 1000 J 1 mol x x 69.0 g x = 23,063 J mol kJ 18.01 g
Step (c) Heat the liquid from 0 oC to 100 oC.
q = 69.0 g x 4.18
J g oC
x 100 o C = 28,842 J
How much heat have we used so far? The sum of steps (a), (b), and (c) equals 55,786 J. We have lots of heat available to vaporize the water. Step (d) Vaporize the water at 100 oC
q = ∆Hvap
q = 43.9
kJ 1000 J 1 mol x x 69.0 g x = 168,189 J mol kJ 18.01 g
Now we have used 223,975 J. Round this answer to 2.24 x 105. We still have heat left over. 260
Chapter 12
1.13 x 106 J – 0.224 x 106 J = 9.06 x 105 J is available to heat the vapor. Our final task is to determine the temperature of the vapor when this heat is absorbed.
q = m C p∆ T 9.06 x 105 J = 69.0 g x 2.08
J g oC
x ∆T
∆T = 6312 oC Therefore, the final temperature of the vapor would be (6312 + 100) oC or 6412 oC At this temperature a significant portion of the water would undergo bond breaking and H2O molecules would not exist. Rather, we would have OH and some atomic species in the vapor phase. 12.139 Freeze-drying is based on the fact that water can readily sublime when the temperature and pressure are reduced below the triple point of water. Assume that the density of water is 1.00 g/mL for this problem.
q = ∆Hsub
q = 49.9
1 mol H 2 O kJ x 150 g H 2 O x = 416 kJ mol 18.01 g
Water changes from a solid directly to a gas during the freeze-dry process; that is, water sublimes. To determine how much water vapor exists under these conditions we need to use the Universal gas law. PV=nRT = gRT/(MW)
Thus, g = PV(MW)/RT
L atm x (273.2 - 80)K / 0.082 760 torr mol H 2 O K mol g = 7.48 x 10-6 g of water in the vapor phase under these conditions. g = 0.001 torr x
1 atm
x 5 L x 18.01
g
If it takes one hour to remove 1 mL then it would require 150 hours to remove 150 mL.
261
Chapter 13
Practice Exercises 13.1
CH2S = kHPH2S
0.11 mol H 2S 34.08g H 2S –1 CH2S = 1 mol H S = 3.7 g L L 2 3.7 g L–1 H2S = kH (1.0 atm HsS) kH = 3.7 g L–1 Hydrogen sulfide is more soluble in water than nitrogen and oxygen. Hydrogen sulfide reacts with the water to form hydronium ions and HS–. PH2S = 1.0 atm
13.2
With one atmosphere of air, the concentrations of the gases in the water depends on the partial pressures of the gases. 0.00430 g O 2 159 mm Hg Oxygen: CO2 = = 0.899 mg O2/ 100 mL 100 mL H 2 O 760 mm Hg g O2 in 125 g of water = 0.899 mg O2/ 100 mL x 125 mL = 1.12 g
0.00190 g N 2 Nitrogen CN2 = 100 mL H 2 O
593 mm Hg = 1.48 mg N2/ 100 mL 760 mm Hg
g N2 in 125 mL of water = 1.48 mg N2/ 100 mL x 125 mL = 1.85 g 13.3
A 10% w/w solution of sucrose will need 10 grams of sucrose for each 100 g of solution. For a solution with 45.0 g of sucrose: 45.0 g sucrose 10% solution = x g solution x = 450 g solution g water = 450 g solution – 45.0 g sucrose g water = 405 g water 1 cm3 mL water = (405 g water) = 405.7 mL water 0.9982 g
13.4
The total mass of the solution is to be 25.0 g. If the solution is to be 1.00 % (w/w) NaBr, then the mass of NaBr will be: 25.0 g × 1.00 g NaBr/100 g solution = 0.250 g NaBr. We therefore need 0.250 g of NaBr and (25.0 – 0.250) = 24.75 g H2O. The volume of water that is needed is: 24.75 g/0.988 g/mL = 25.0 mL H2O.
13.5
An HCl solution that is 37 % (w/w) has 37 grams of HCl for every 1.0 × 102 grams of solution. 1.0 × 102 g solution g solution = (7.5 g HCl) = 2.0 × 101 g solution 37 g HCl We need to know the number of moles of Na2SO4 and the number of kg of water. 44.00 g Na2SO4 ÷ 142.0 g/mol = 0.3099 mol Na2SO4 250 g H2O × 1 kg/1000 g = 0.250 kg H2O
13.6
The molality is thus given by: m = 0.3099 mol/0.25 kg = 1.239 mol Na2SO4/kg H2O = 1.239 m Molarity is moles solute per liters of solution. The moles of solute is the same for molarity and molality but the volume of solution would be larger than the kilograms of solvent so M < m.
262
Chapter 13
13.7
0.050 mol CH3OH 32.0 g CH3OH g CH3OH for 0.050 m = 0.200 kg kg H 2 O 1 mol CH3OH = 0.320 g 0.100 mol CH3OH 32.0 g CH3OH g CH3OH for 0.100 m = 0.200 kg kg H 2 O 1 mol CH3OH = 0.640 g 0.150 mol CH3OH 32.0 g CH3OH g CH3OH for 0.150 m = 0.200 kg kg H 2 O 1 mol CH3OH = 0.960 g 0.200 mol CH3OH 32.0 g CH3OH g CH3OH for 0.200 m = 0.200 kg kg H 2 O 1 mol CH3OH = 1.28 g 0.250 mol CH3OH 32.0 g CH3OH g CH3OH for 0.250 m = (0.200 kg H2O) kg H 2 O 1 mol CH3OH = 1.60 g CH3OH
13.8
First we need to find the number of grams of Fe(NO3)3 for each kg of solvent.
241.86 g Fe ( NO3 ) 3 = 206.3 g Fe(NO ) 0.853 m Fe(NO3)3 = 0.853 mol Fe(NO3)3 3 3 1 mol Fe ( NO3 ) 3 Then we need to find how the ratio of the moles of Fe(NO3)3 to the mass of the solution: 0.853 mol Fe ( NO3 )3 ratio = = 7.072 × 10–4 mol Fe(NO3)3 / g solution 1000 g H 2 O + 206.3 g Fe ( NO3 )3 (a)
1 g solvent = 28.3 g solution g solution = (0.0200 mol Fe(NO3)3) 7.07 ×10−4 mol Fe ( NO3 ) 3
(b)
1mol Fe ( NO3 ) 3 g sol'n = (0.0500 mol Fe3+) 1 mol Fe3+
1 g solvent = 70.7 g sol'n 7.07 ×10−4 mol Fe ( NO3 ) 3
1mol Fe ( NO3 ) 1 g solvent 3 = 1.41 g sol'n g sol'n = (0.00300 mol NO3–) 3 mol NO − 7.07 ×10−4 mol Fe ( NO ) 3 3 3 If a solution is 52% NaOH, then it has 52 g of NaOH for each 100 g of solution. The mass of water is 48 g of water for 52 g of NaOH. To calculate the molality of the solution, we need to find the moles of NaOH for each kilogram of water. 52 g NaOH 1 mol NaOH 1000 g H 2 O m= = 27 mol NaOH/kg H2O = 27 m NaOH 48 g H 2 O 40.0 g NaOH 1 kg H 2 O (c)
13.9
13.10
If a solution is 37.0% (w/w) HCl, then 37.0% of the mass of any sample of such a solution is HCl and (100.0 – 37.0) = 63.0% of the mass is water. In order to determine the molality of the solution, we can conveniently choose 100.0 g of the solution as a starting point. Then 37.0 g of this solution are HCl and 63.0 g are H2O. For molality, we need to know the number of moles of HCl and the mass in kg of the solvent: 37.0 g HCl ÷ 36.46 g/mol = 1.01 mol HCl 63.0 g H2O × 1 kg/1000 g = 0.0630 kg H2O 263
Chapter 13
molality = mol HCl/kg H2O = 1.01 mol/0.0630 kg = 16.1 m 13.11
40.0 % HBr means 40 g of HBr per 100 g of solution. Since molarity is defined as moles of solute per liter of solution we need to determine the volume of solution. The density of the solution allows us to determine the volume of the solution
40 g HBr 1mol HBr 1.38 g 1000mL = 6.82 M x x x 100 g solution 80.91g mL L 13.12
First determine the number of moles of Al(NO3)3 dissolved in the liter of water. 1 mol Al ( NO3 ) 3 = 0.00469 mol Al(NO3)3 mol Al(NO3)3 = (1.00 g Al(NO3)3) 212.996 g Al ( NO3 ) 3 Next find the mass of the water: 1000 mL H 2 O 0.9982 g H 2 O g H2O = (1.00 L H2O) = 998.2 g H2O 1 L H 2 O 1 mL H 2 O To find the molarity of the solution, first we need to find the mass of the solution, and then the volume of the solution: g solution = 998.2 g H2O + 1.00 g Al(NO3)3 = 999.2 g solution 1 mL solution 1 L solution L solution = (999.2 solution) = 1.0003 L 0.9989 g solution 1000 mL solution M of solution =
0.00469 mol Al ( NO3 )3
= 0.00469 M Al(NO3)3 1.0003 L solution The molality of the solution can also be determined 0.00469 mol Al ( NO3 )3 m of solution = = 0.00470 m Al(NO3)3 0.9982 kg H 2 O
13.13
First determine the number of moles of each component of the solution: For C16H22O4, 20.0 g/278 g/mol = 0.0719 mol For C5H12, 50.0 g/72.2 g/mol = 0.692 mol The mole fraction of solvent is: 0.692 mol/(0.692 mol + 0.0719 mol) = 0.906 Using Raoult's Law, we next find the vapor pressure to expect for the solution, which arises only from the solvent (since the solute is known to be nonvolatile): Psolvent = χsolvent × P°solvent = 0.906 × 541 torr = 4.90 x 102 torr
13.14
Pacetone = χacetone × P°acetone
1 mol acetone mol acetone = (156 g acetone) = 2.690 mol acetone 58.0 mol acetone Do not round your answers until the end. The moles of stearic acid is small compared to the moles of acetone and rounding error may give you too high of a mass of stearic acid. 2.690 mol acetone 155 torr = x 162 torr 2.690 mol acetone + x mol stearic acid 2.690 mol acetone 0.957 = 2.690 mol acetone + x mol stearic acid 2.574 mol acetone + 0.957x mol stearic acid = 2.690 mol acetone 0.957x mol stearic acid = 0.116 x mol stearic acid = 0.121 mol stearic acid 264
Chapter 13
Finally solve to find the number of grams of stearic acid 284.5 g stearic acid g stearic acid = (0.121 mol stearic acid) = 34.5 g stearic acid 1 mol stearic acid 13.15
Pcyclohexane = χcyclohexane × P°cyclohexane = 0.750 × 66.9 torr = 50.2 torr Ptoluene = χtoluene × P°toluene = 0.250 × 21.1 torr = 5.28 torr Ptotal = Pcyclohexane + Ptoluene = 33.5 torr + 10.6 torr = 55.4 torr
13.16
First we need to find the moles of the cyclohexane and the moles of toluene. 1 mol cyclohexane mol cylcohexane: = (122 g cyclohexane) = 1.450 mol cyclohexane 84.15 g cyclohexane
1 mol toluene mol toluene = (122 g toluene) = 1.324 mol toluene 92.14 g toluene Now, find the χcyclohexane and the χtoluene 1.450 mol cyclohexane χcyclohexane = = 0.523 1.450 mol cyclohexane + 1.324 mol toluene χtoluene =1 – χcyclohexane = 1 – 0.523 = 0.477 Pcyclohexane = χcyclohexane × P°cyclohexane = 0.523 × 66.9 torr = 35.0 torr Ptoluene = χtoluene × P°toluene = 0.477 × 21.1 torr = 10.1 torr Ptotal = Pcyclohexane + Ptoluene = 35.0 torr + 10.1 torr = 45.1 torr 13.17
First convert 0F to 0C. 235 0F = 112.78 0C and 240 0F = 115.56 0C The corresponding boiling point elevations are then 12.78 0C and 15.56 0C respectively. The molality of the two solutions is given as: m =12.78 0C/ 0.51 0C m-1 = 25.06 m m = 15.56 0C/0/51 0C m-1 = 30.51 m The solution that boils at 235 0F has the following mass percent of sugar:
25.06mol C12H22O11 1000 g H2O
x
342 g
=
mol C12H22O11
8.57 g C12H22O11 g H2O
Total grams of the solution = 8.57 g C12H22O11 + 1.00 g H2O = 9.57 g Percent sugar =
8.57 g x100 = 89.5% 9.57 g
For the solution that boils at 240 0F
30.51mol C12H22O11 1000 g H2O Percent sugar =
x
10.43 g 11.43 g
342 g mol C12H22O11
=
10.43 g C12H22O11 g H2O
x100 = 91.2%
The mass percent range for the solutions is 89.5 % to 91.2 %
13.18
∆Tb = Kb × m = 0.51 °C m–1× x m = 2.36 °C x m = 4.627 m To find the number of grams of glucose, first we need to find the number of moles of glucose. 265
Chapter 13
mol glucose = (m solution)(kg solvent) mol glucose = (4.627 m)(0.255 kg H2O) = 1.18 mol glucose 180.9 g glucose g glucose = (1.18 mol glucose) = 213 g glucose 1 mol glucose 13.19
It is first necessary to obtain the values of the freezing point of pure benzene and the value of Kf for benzene from Table 13.4 of the text. We proceed to determine the number of moles of solute that are present and that have caused this depression in the freezing point: ∆T = Kfm ∴ m = ∆T/Kf = (5.45 °C – 4.13 °C)/(5.07 °C kg mol–1) = 0.260 m Next, use this molality to determine the number of moles of solute that must be present: 0.260 mol solute/kg solvent × 0.0850 kg solvent = 0.0221 mol solute Last, determine the formula mass of the solute: 3.46 g/0.0221 mol = 157 g mol-1
13.20
To find the molar mass of the substance, first, we need to find the molality of the solution from the freezing point depression, and then using the 5.0% (wt/wt) amount, determine the moles of the solute.
∆T 80.2 o C − 77.3 o C = = 0.420 m Kf 6.9 o C m −1 Assume there is 100 g of solution: 5 g unknown substance 5.0% (wt/wt) = 5 g unknown substance + 95 g naphthalene We have 95 g of naphthalene, or 0.095 kg naphthalene and 5 g of the unknown. Using the equation for molality, we can determine the number of moles of the unknown mol unknown = (m solution)(kg solvent) = (0.420 m)(0.095 kg naphthalene) = 0.0399 mol unknown 5.0 g unknown molar mass = = 125 g mol–1 0.0399 mol unknown
m=
13.21
Use the equation Π = MRT: 1 mol protein ( 5 g protein ) 235, 000 g protein = 2.13 × 10–4 M solution M= 0.1000 L solution R = 0.0821 L atm/K mol T = 4.0 + 273.2 = 277.2 K Π = (2.13 × 10–4 M solution)(0.0821 L atm/K mol)(277.2 K) = 4.84 × 10–3 atm 760 mm Hg mm Hg = 4.84 × 10–3 atm = 3.68 mm Hg 1 atm
13.6 mm H 2 O mm H2O = 3.68 mm Hg = 50.0 mm H2O 1 mm Hg 13.22
We can use the equation Π = MRT: Π = (0.0115 M)(0.0821 L atm/K mol)(310 K) Π = 0.293 atm Π = 0.293 atm x 760 torr atm-1 = 222 torr To determine the boiling and freezing temperatures of the solution we can assume that the molality is equal to the molarity. At low concentrations the two values are nearly identical. Tf = 0 0C –m Kf = -0.0115 m x 1.86 0C m-1 = -0.021 0C Tbp = 100 0C + m Kbp = 100 + 0.0115 m x 0.51 0C m-1 = 100.006 0C Note that significant figures rules were not used for the boiling point answer. 266
Chapter 13
13.23
Π = MRT
10 mm H 2 O 1.00 g mL−1 1 atm Π = 6.45 cm water = 6.24 × 10–3 atm −1 760 mm Hg 1 cm H O 2 13.6 g mL R = 0.0821 L atm / mol K T = 277 K
(
)
6.24 × 10−3 atm Π M= = = 2.74 × 10–4 mol L–1 RT ( 0.0821 L atm/mol K )( 277 K ) mol protein = (2.74 × 10–3 mol L–1)(0.1000 L) = 2.74 × 10–5 mol 0.1372 g protein molar mass = = 5.00 x 103 g mol–1 2.74 × 10−5 mol protein 13.24
We can use the equation Π = MRT, remembering to convert pressure to atm: 1 atm atm = (25.0 torr) = 0.0329 atm 760 torr Π = 0.0329 atm = M × (0.0821 L atm/K mol)(298 K) M = 1.34 × 10–3 mol L–1 mol = 1.34 × 10–3 mol L–1 × 0.100 L = 1.34 × 10–4 mol formula mass =
72.4 × 10−3 g 1.34 × 10
−4
= 5.38 × 102 g mol–1
mol
13.25
For the solution as if the solute were 100% dissociated: ∆T = (1.86 °C m–1)(2 × 0.237 m) = 0.882 °C and the freezing point should be –0.882 °C. For the solution as if the solute were 0% dissociated: ∆T = (1.86 °C m–1)(1 × 0.237 m) = 0.441 °C and the freezing point should be –0.441 °C.
13.26
Use the freezing point depression equation: ∆T = Kfm Remember that there are two moles of ions for each mole of MgSO4. Kf water = 1.86 °C m–1 (a) For 0.1 m MgSO4 m = 0.2 m ∆T = (1.86 °C m–1)(0.2 m) = 0.372 °C thus, Tf = − 0.372 oC (b)
m = 0.02 m For 0.01 m MgSO4 –1 ∆T = (1.86 °C m )(0.02 m) = 0.0372 °C thus, Tf = − 0.0372 °C
(c)
m = 0.002 m For 0.001 m MgSO4 ∆T = (1.86 °C m–1)(0.002 m) = 0.00372 °C thus, Tf = − 0.00372 °C
The first freezing point depression could be measured using a laboratory thermometer that can measure 1 °C increments. Review Questions 13.1
This event, diagrammed in Figure 13.1, is due to the tendency for all systems to proceed spontaneously towards a state with a higher degree of randomness (disorder).
13.2
First, the tendency towards randomness drives the solution process, and second, the new forces of attraction between solute and solvent molecules drive the process. Thus the relative degree of solute–solute, solvent– solvent and solute–solvent interactions will determine if a solute is soluble in a solvent.
267
Chapter 13
13.3
Since water and methanol both have OH groups, there can be hydrogen bonding between a water molecule and a methanol molecule. This allows any proportion of methyl alcohol in water to be nearly as stable as either separate water samples or separate methyl alcohol samples.
13.4
Water molecules are tightly linked to one another by hydrogen bonding. In hexane, however, which is a nonpolar organic substance, we have only weak London forces of attraction. This means that hexane as a solute in water offers no advantage in attraction to individual water molecules, and the solvent is therefore not disrupted to allow the solute to dissolve.
13.5
The dipole moments of water molecules can be oriented so as to stabilize both the dissolved cation and the dissolved anion.
13.6
There is no solvating force provided by carbon tetrachloride that can overcome and offset the very strong ion–ion forces of the solid KCl sample.
13.7
Since the enthalpy of solution is positive, the process is endothermic. The system thus requires heat for the dissolving process, and the heat flow should cause the temperature to decrease as the solute dissolves.
13.8
The lattice energy is numerically larger since that step is endothermic, that is it requires energy to separate the particles.
13.9
The Al3+ ion, having the greater positive charge, should have the larger hydration energy.
13.10
When a gas dissolves in a liquid, there is no endothermic step analogous to the lattice energy of a solid. The only enthalpy change is the one associated with hydration, and this is always negative.
13.11
There is a greater attraction between water and acetone molecules in the resulting solution than there is among acetone molecules in the starting pure solute or water molecules in the starting pure solvent.
13.12
The disruption of ethyl alcohol and the disruption of hexane together cost more energy than is gained on formation of the solution. This is because the two liquids are not alike; ethyl alcohol is a polar substance with hydrogen bonding, whereas hexane is a nonpolar liquid having only London forces.
13.13
If the solution becomes cool, it is an endothermic process and the ∆Hsoln is positive. This is because the solution is absorbing heat from the surroundings. The solubility is likely to increase with increasing temperature because heat is required for the reaction to proceed, so increasing the temperature increases the amount of heat available.
13.14
The fact that the ∆Hsoln value for the formation of a mixture of A and B is zero, implies that the relative strengths of A–A, B–B, and A–B intermolecular attractions are similar.
13.15
We can estimate from Figure 13.10 that the solubility of NH4NO3 in 100 g of H2O is 550 g at 75 °C and 165 g at 15 °C. The amount of solid that will crystallize is the difference between these two solubilities, namely 550 – 165= 385 g in 100 g of solvent. However, we have 125 g of solvent. 385 g NH 4 NO3 100 g solvent
x 125 g solvent = 481 g NH 4 NO3 will separate out of the solution.
13.16
Oxygen solubility increases as the temperature decreases. The larger fish will need more oxygen and will be found in the colder areas of lake bottoms.
13.17
Henry's Law is the statement, applied to the dissolving of a gas in a solvent, that at a given temperature, the concentration (Cg) of the gas in a solution is directly proportional to the partial pressure (Pg) of the gas on the solution, where k in the following equation is the constant of proportionality: Cg = k × Pg. As
268
Chapter 13
discussed in the text, an alternate statement expresses the relationship of concentration at one pressure P1 to the concentration that would exist at some new pressure P2: C1/P1 = C2/P2 13.18
The atmospheric pressure on a mountain is less than the atmospheric pressure at sea level. From Henry's Law, as the partial pressure of oxygen decreases, the concentration of the oxygen also decreases. Therefore, there is less oxygen to sustain life in mountain streams.
13.19
Ammonia is more soluble in water than nitrogen because ammonia is able to hydrogen bond with solvent molecules, whereas nitrogen cannot. Nitrogen is a nonpolar molecular substance, whereas ammonia is a polar substance capable of hydrogen bonding. Also, ammonia reacts with water to form nonvolatile ions: NH3(g) + H2O(l) NH4+(aq) + OH–(aq) Hydrogen chloride would have a high solubility in water because it ionizes in water: HCl(g) + H2O(l) H3O+(aq) + Cl–(aq)
13.20
When the cap is removed from a bottle of carbonated beverage, the liquid fizzes because CO2 is being released from the liquid. When the cap is on, the CO2 fills the space above the liquid until equilibrium is established between the gas and the liquid. After the cap is removed, the equilibrium is disrupted and more of the gas leaves the solution. This is the fizzing.
13.21
(a) (b) (c) (d)
mole fraction = moles component/total number of moles mole percent = mole fraction × 100% molality = moles solute/kg solvent percent by mass = (mass component/total mass) × 100%
The maximum value for mole fraction is 1, and the maximum value for mole percent and percent by mass is 100 %. 13.22
Molality is independent of temperature. Molarity decreases with increasing temperature because the volume of the solvent increases with increasing temperature.
13.23
The molarity will be greater than 1.0. Since the density of the solution is greater than one, the mass of the solution in kg will be greater than its volume in liters.
13.24
A colligative property of a solution is one that depends only on the molal concentration of the solute particles, and not on the identity of the solute.
13.25
A solution is ideal if the sum of the partial pressures of the components of the solution equals the observed vapor pressure of the solution, i.e., if the solution obeys Raoult's Law. Also, it should be true that the heat of solution is nearly zero.
13.26
A positive deviation indicates that the vapor pressure of the real solution is greater than expected if the solution behaved ideally. Positive deviations result when mixtures with weaker intermolecular forces of attraction between the two substances as compared to the intermolecular forces of the pure substances are formed.
13.27
When a solute is dissolved in a solvent, the vapor pressure is lowered. As a result, the boiling point is increased to a temperature where the vapor pressure is high enough to once again allow boiling to occur. This affect also reduces the triple point and the entire solid-liquid equilibrium curve on a phase diagram shifts to lower temperatures. The net result is a lowering of the freezing point.
13.28
These are semipermeable because only certain substances are able to pass through the membrane. A nonpermeable material would allow nothing to pass through.
269
Chapter 13
13.29
An osmotic membrane allows only solvent to pass, whereas a dialyzing membrane allows solvated ions of a certain minimum size to pass as well as solvent molecules. A dialyzing membrane prevents the passage of only certain solute particles, usually those of large size, such as colloid particles.
13.30
The side of the membrane less concentrated in solute will be more concentrated in solvent. Therefore, the escaping tendency of the solvent will be greater than on the side of the membrane less concentrated in solute. The solvent will shift through the membrane from the side less concentrated in solute to the side more concentrated in solute.
13.31
The solution that loses solvent into the other solution is the one with the lower molarity.
13.32
In each case, the osmotic pressure Π is given by the equation: Π = M × R × T. Since we do not know either the density of the solution or the volume of the solution, we cannot convert values for % by mass into molarities. However, we do know that glucose, having the smaller molecular mass, has the higher molarity, and we conclude that it will have the larger osmotic pressure.
13.33
By the "association of solute particles" we mean that some particles are attracted to others, or that solvent does not perfectly insulate solute particles from attachment to one another. This is another way of saying that there is less than 100% dissociation or dissolution of solute in such a solution. Colligative properties depend on the number of particles in solution. If association of solute occurs, this decreases the number of independent particles in solution. Therefore, the freezing point will not be lowered as much as expected, and the boiling point and osmotic pressure will not increase as much as expected.
13.34
If a cell is placed in a solution, the concentration of salts in the solution will affect the cell. If the solution is hypertonic, the concentration of salts is higher than the concentration of salts in the cell. If the solution is hypotonic, then the concentration of salts is lower than the concentration of salts in the cell.
13.35
Ionic compounds dissociate in solution. The dissociation results in an increase in the number of particles in the solution, i.e., one NaCl “molecule” will dissociate creating two ions; Na+ and Cl–. Colligative properties depend upon the concentration of particles so any compound that dissociates into multiple particles will have pronounced effects on colligative properties.
13.36
The van't Hoff factor is the ratio of the value for a colligative property as actually measured to that value of the colligative property that is expected in the complete absence of any solute dissociation. A van't Hoff factor of one is expected for all nondissociating molecular solutes. A van’t Hoff factor greater than one indicates a dissociation of the solute. A van’t Hoff factor less than one indicates association of the solute. If the van’t Hoff factor is 0.5, then this indicates the formation of dimers.
13.37
The solute that dissolves to produce the greater number of ions, Na2CO3, gives the solution with the larger boiling point elevation and, thus, the higher boiling point.
13.38
(a) (b) (c) (d) (e) (f)
suspension solution or a suspension. It depends on the type of apple juice solution solution suspension. The oils from the lemon are not soluble on water. suspension
The laser can be used to test light scattering by the liquid. 13.39
Sodium stearate is a soap and will stabilize a water in oil emulsion. The hydrophilic head (carboxylate group) of the molecule attaches to water and the hydrophobic tail (non-polar section) of the molecule attaches to the oil molecules.
270
Chapter 13
13.40
13.41
A colloidal particle of BaSO4 in solution has a charged surface due to the adsorption of excess positive or negative ions on its surface as it grows. This ionic atmosphere has a defined volume. Consider a colloidal particle that has an excess of Ba2+ ions adsorbed on its surface. The colloidal particle will be repelled by other positively charged colloidal particles when they come within each others ionic boundary. The addition of an electrolyte, such as an acid or salt, to this solution decreases the volume of the ionic boundary. The negative ions of the electrolyte will attach to a positively charged colloidal particle reducing its overall charge. As a result, colloidal particles can come closer together and thus form larger aggregates. Also, heating the solution can be used to increase the kinetic energy of the particles allowing them to collide with enough energy to overcome repulsion and thus increasing particle size. A micelle is a cluster of surfactant molecules dispersed in a colloidal suspension. Micelles form due to the structure of surfactant molecules. They have a hydrophilic head and a hydrophobic tail. Surfactant molecules aggregate in a manner that will minimize contact between the hydrophobic end with water molecules. This results in a spherical structure.
Review Problems 13.42
This is to be very much like that shown in Figure 13.5: The lattice energy is defined as: K+(g) + Cl–(g) KCl(s) (a) (b)
13.43
13.44
KCl(s) K+(g) + Cl–(g), K+(g) + Cl–(g) K+(aq) + Cl–(aq), KCl(s) K+(aq) + Cl–(aq),
∆H° = –715 kJ mol–1
∆H° = +715 kJ mol–1 ∆H° = –686 kJ mol–1 ∆H° = +29 kJ mol–1
∆Hsoln = ∆Hlattice energy + ∆Hhydration ∆Hlattice energy = ∆Hsoln – ∆Hhydration ∆Hlattice energy = –56 kJ mol–1 – (–894 kJ mol–1) = 838 kJ mol–1
C1 C 2 = P1 P2
(
)
0.025 g L−1 × (1.4 atm ) C1 × P2 C2 = = = 0.035 g/L. P1 (1.0 atm ) 13.45
We can compare the solubility that is actually observed with the predicted solubility based on Henry's Law. If the actual and the predicted solubilities are the same, we conclude that the gas obeys Henry's Law. We proceed as in Review Problem 12.40: C1 C 2 = P1 P2
(
)
0.018 g L−1 × ( 620 torr ) C1 × P2 = = 0.015 g/L. C2 = P1 ( 740 torr ) The calculated value of C2 is the same as the observed value, and we conclude that over this pressure range, nitrogen does obey Henry's Law.
13.46
C1 C 2 = P1 P2
(
)
0.015 g L−1 × ( 2.0 atm ) C1 × P2 C2 = = = 0.030 g L–1 P1 (1.0 atm ) 13.47
Cgas = kH × Pgas 271
Chapter 13
CO2 = 0.0039 g O2/100 mL solution kH = 13.48
PO2 = 1.0 atm
3.9 ×10−5 g mL–1 = 3.9 × 10– 5 g mL–1 atm–1 1.0 atm
One liter of solution has a mass of: 1000 mL solution 1.07 g solution g solution = 1 L solution = 1,070 g 1 L solution 1 mL solution According to the given molarity, it contains 3.000 mol NaCl. This has a mass of: 58.45 g NaCl g NaCl = 3.000 mol NaCl = 175.4 g NaCl 1 mol NaCl Thus, the mass of water in 1 L solution must be: 1,070 g – 175.4 g = 895 g water
3.000 mol NaCl m= = 3.35 m 0.895 kg solvent 13.49
Since the density of the solution is 1.00 g mL–1, the molarity and molality are the same 0.158 mol CH3O 2 H 1 L soln 1 mL soln molality of the acetic acid solution = 1 L soln 1000 mL soln 1.00 g soln
1000 g soln × = 0.158 m 1 kg soln 13.50
13.51
27.5 g glucose ÷ 180.16 g/mol = 0.153 mol glucose molality = 0.153 mol glucose/1.00 kg solvent = 0.153 m mole fraction = moles glucose/total moles moles glucose = 0.153 1 mole H 2 O moles H2O = (1.00 × 103 g H2O) = 55.5 mol H2O 18.01 g H 2 O 0.153 χglucose = = 2.75 × 10–3 55.5 + 0.153 27.5 g glucose mass % = × 100% = 2.68% 1000 g H 2 O + 27.5 g glucose
1 mol NaCl mol of NaCl = 11.5 g NaCl = 0.197 mol NaCl 58.44 g NaCl molality = 0.197 mol NaCl/1.00 kg H2O = 0.197 molal mass % = 11.5/1011.5 g × 100% = 1.14% ( mol NaCl ) × 100% mole % = ( total mol ) moles NaCl = 0.197 moles ( 0.197 mol NaCl ) mole % = × 100% = 0.354% ( 55.5 mol H 2 O + 0.197 mol NaCl )
272
Chapter 13
Since the density of water is 1.00 g/mL, the volume of 1 kg is 1 L. Thus, the molarity is: 0.197 mol/1.00 L = 0.197 M. A solvent must have a density close to 1 g/mL for this to happen. Also, the volume of the solvent must not change appreciably on addition of the solute. 13.52
We need to know the mole amounts of both components of the mixture. It is convenient to work from an amount of solution that contains 1.35 mol of ethyl alcohol and, therefore, 1.00 kg of solvent. Convert the number of moles into mass amounts as follows: For CH3CH2OH: g ethanol 1.35 mol ethanol 46.08 g ethanol 1 kg water = = 62.2 g ethanol/1000 g water g solution 1 kg water 1 mol ethanol 1000 g water Mass % ethanol = (mass ethanol/(total solution mass) × 100% Mass % ethanol = (62.2 g ethanol/(1,000 g water + 62.2 g ethanol) × 100% = 5.86%
13.53
If we have 100.0 g of the solution, then 19.5 g is NaCl and the remainder, 80.5 g, is water. We need to know the number of moles of NaCl and the number of kg of water: 1 mol NaCl mol NaCl = (19.5 g NaCl) = 0.334 mol NaCl; 58.44 g NaCl
1 kg H 2 O –2 kg H2O = (80.5 g) = 8.05 × 10 kg H2O. 1000 g H O 2 0.334 mol NaCl Molality = = 4.14 m NaCl. 8.05 ×10−2 kg H 2 O 13.54
If we assume 100 g of solution we have 6.85 g NH3 and 93.15 g H2O. 1 mole NH3 mol NH3 = (6.85 g NH3) = 0.402 mol NH3 17.03 g NH3
1 kg kg H2O = (93.15 g H2O) = 0.09315 kg H2O 1000 g 0.402 moles NH3 m= = 4.32 m 0.09315 kg H 2 O 1 mole H 2 O mol H2O = (93.15 g H2O) = 5.169 mol H2O 18.0 g H 2 O 0.402 mol NH3 mole percent = × 100% = 7.22% ( 0.402 mol NH3 + 5.169 mol H 2 O ) 13.55
Assume 1 mole total.
60.10 g g C3H8O = 0.250 mol = 15.02 g C3H8O 1 mol mol H2O = 0.750 mol 18.0 g H 2 O g H2O = (0.750 mol H2O) = 13.5 g H2O 1 mol H 2 O C3H8O mass % =
15.02 g × 100% = 52.7% 13.5 g + 15.02 g
273
Chapter 13
Molality =
13.56
0.250 mol = 18.5 m 1 kg 13.5 g ( ) 1000 g
If we choose, for convenience, an amount of solution that contains 1 kg of solvent, then it also contains 0.363 moles of NaNO3. The number of moles of solvent is: 1 mole H 2 O mol H2O = (1000 g) = 55.6 mol H2O 18.02 g H 2 O Now, convert the number of moles to a number of grams: for NaNO3, 0.363 mol × 85.0 g/mol = 30.9 g; for H2O, 1000 g was assumed and the percent (w/w) values are: % NaNO3 = 30.9 g/1030.9 g × 100% = 3.00% % H2O = 1000 g/1030.9 g × 100 = 97.0% To determine the molar concentration of NaNO3 assume 1 kg of solvent which would then contain 0.363 mole of NaNO3 or 30.9 g NaNO3. The total mass of the solution would be 1000 g + 30.9 g = 1031 g of solution. Now, the ratio of moles of solute to grams of solution is 0.363 mol NaNO3/1031 g solution. From this calculate the molarity of the solution 0.363 mol NaNO3 1.0185 g soln 1000 mL soln M of solution = = 0.359 M NaNO3 1031 g solution 1 mL soln 1 L soln χNaNO3 =
13.57
(a)
(b)
(c)
0.363 mol NaNO3 = 6.49 × 10–3 55.6 mol H 2 O + 0.363 mol NaNO3
If the sample is 1.89 mol% H2SO4, then an amount of the solution that contains 1.89 mol of H2SO4 also contains (100 – 1.89) = 98.11 mol water. We can calculate the molality if we know the number of moles of H2SO4 and the number of kg of solvent. The latter is determined as follows: kg H2O = 98.11 mol H2O × 18.02 g/mol × 1 kg/1000 g = 1.768 kg H2O. 1.89 mol H 2SO 4 Molality = = 1.07 m H2SO4. 1.768 kg H 2 O The mass of H2SO4 in the above sample is: 1.89 mol × 98.1 g/mol = 185 g H2SO4. The total mass of the solution is then equal to [185 g + (1.768 × 103 g)] = 1.953 × 103 g, and the % (w/w) values are: for H2SO4, 185 g H 2SO 4 Mass % H2SO4 = × 100 = 9.47%; 1953 g total mass for H2O, 1768 g H 2 O Mass % water = × 100 = 90.53%. 1953 g total mass If we have on hand 100 mL (0.100 L) of this solution, it will have a mass that can be determined using its known density: 1.0645 g solution mass solution = (100.0 mL) = 106.4 g of solution. 1 mL solution Since this solution has 9.49% (w/w) H2SO4, the mass of H2SO4 in 0.100 L of the solution is: Mass H2SO4 = 106.4 g × 0.0949 = 10.1 g H2SO4. The number of moles of H2SO4 is thus: 1 mol H 2SO 4 mol H2SO4 = (10.1 g H2SO4) = 0.103 mol H2SO4. 98.1 g H 2SO 4 The molarity is the number of moles of H2SO4 divided by the volume of solution: 0.103 mol H 2SO4 Molarity = = 1.03 M H2SO4. 0.100 L solution
274
Chapter 13
13.58
Psolution = P°solvent × χsolvent We need to determine χsolvent:
1 mol mol glucose = (55.0 g) = 0.305 mol glucose 180.2 g 1 mol H 2 O mol H2O = (125 g H2O) = 6.94 mol H2O 18.02 g H 2 O The total number of moles is thus: 6.94 mol + 0.305 mol = 7.25 mol and the mole fraction of the solvent is: 6.94 mol solvent χsolvent = = 0.958. Therefore, 7.25 mol solution Psolution = 23.8 torr × 0.958 = 22.8 torr. 13.59
In 100 g of the mixture we have the following mole amounts: 65.0 g H2O ÷ 18.02 g/mol = 3.61 mol H2O 35 g C2H6O2 ÷ 62.07 g/mol = 0.564 mol ethylene glycol 3.61 mol H 2 O χH2O = = 0.865 3.61 mol H 2 O + 0.564 mol ethylene glycol Psolution = P°solvent × χsolvent = 17.5 torr × 0.865 = 15.1 torr
13.60
Pbenzene = χbenzene × P°benzene Ptoluene = χtoluene × P°toluene PTot = Pbenzene + PToluene 1 mol mol benzene = (35.0 g) = 0.448 mol benzene 78.11 g
1 mol mol toluene = (65.0 g) = 0.705 mol toluene 92.14 g 0.448 χbenzene = = 0.389 0.448 + 0.705 0.705 χtoluene = = 0.611 0.448 + 0.705 Pbenzene = (0.389)(93.4 torr)= 36.3 torr Ptoluene = (0.611)(26.9 torr)= 16.4 torr PTotal = 36.3 torr + 16.4 torr = 52.7 torr 13.61
Assume 50 g of each substance: 1 mol pentane mol pentane = (50 g pentane) = 0.693 mol 72.15 g pentane
1 mol heptane mol heptane = (50 g heptane) = 0.499 mol heptane 100.21 g heptane 0.693 χpentane = = 0.581 0.693 + 0.499 0.499 χheptane = = 0.419 0.693 + 0.499 Ppentane = χpentane × P°pentane = 0.581 × 422 torr = 245 torr 275
Chapter 13
Pheptane = χheptane × P°heptane = 0.419 × 36 torr = 15.1 torr PTotal = Ppentane + Pheptane = (245 + 15.1) torr = 260 torr
13.62
The following relationships are to be established: PTotal = 84 torr = (P°benzene × χbenzene) + (P°toluene × χtoluene). The relationship between the two mole fractions is: χbenzene = 1 – χtoluene, since the sum of the two mole fractions is one. Substituting this expression for χbenzene into the first equation gives: 84 torr = [P°benzene × (1 – χToluene)] + [P°Toluene × χToluene], 84 torr = [180 torr × (1 – χToluene)] + [60 torr × χToluene]. Solving for χToluene we get: 120 × χToluene = 96, χToluene = 0.80 and χbenzene = 0.20. The mole % values are to be 80 mol% toluene and 20 mol% benzene.
13.63
χCH3OH= P/P° = 145 torr/164 torr = 0.884
1 mol CH3OH mol CH3OH = (115 g) = 3.59 mol CH3OH 32.0 g CH3OH 3.59 mol CH 3OH 0.884 = 3.59 mol CH3OH + x mol C3 H5 ( OH )3 3.59 mol CH3OH + x mol C3H5(OH)3 = x=
( 3.59 mol ) 0.884
3.59 mol CH3OH 0.884
– 3.59 mol = 0.471 mol
92.1 g g C3H5(OH)3 = (0.471 mol) = 43.4 g C3H5(OH)3 1 mol 13.64
(a)
(b) (c)
511 torr = 0.971 526 torr P χsolute = 1 – χsolvent = 0.029 1 mol x = 2.99 × 10–2 moles We know 0.971 = 1 mol + x mol 8.3 g molar mass = = 278 g/mol 2.99 ×10−2 mol
χsolvent=
P
=
13.65
Psolvent = χsolvent × P°solvent 336.0 torr = χsolvent × 400.0 torr χsolvent = 0.8400 χsolute = 1 – 0.8400 = 0.1600 The number of moles of solvent is: 33.25 g ÷ 109.0 g/mol = 0.3050 mol and the following expression for mole fraction of solvent can be solved to determine the number of moles of solute We know 0.8400 = 0.3050 mol/(0.3050 mol + x), x = 5.81 × 10–2 moles Molar mass = 18.26 g/5.81 × 10–2 mol = 314.3 g/mol
13.66
∆Tf = Kfm m = ∆Tf/Kf = 3.00 °C/1.86 °C kg/mol = 1.61 mol/kg 1 kg kg = (175 g) = 0.175 kg 1000 g
276
Chapter 13
1.61 mol mol = (0.175 kg) = 0.282 mol 1 kg 342.3 g g = (0.282 mol) = 96.4 g 1 mol 13.67
∆Tb = Kbm
5 oC Tc = (270 °F – 32 °F) = 132 °C 9 oC (132 °C – 100 °C) = (0.51 °C kg/mol) × m m = 63 m 63 mol sucrose = 0.53 χsucrose = 1 mol H 2 O 1000 g water + 63 mol sucrose 18.01 g H 2 O 13.68
∆Tb = Kbm ∆Tb = 81.7 °C – 80.2 °C = 1.5 °C m= ∆Tb/Kb = 1.5 °C/2.53 °C kg/mol = 0.593 mol/kg mol solute = (0.593 mol/kg)(0.100 kg) = 0.0593 mol molar mass = (14 g)/0.0593 mol = 240 g/mol
13.69
∆T = (5.45 – 3.45) = 2.00 °C = Kf × m = 5.07 °C kg mol–1 × m m = 0.394 mol solute/kg solvent 0.394 mol/kg benzene × 0.200 kg benzene = 0.0788 mol solute and the molecular mass is: 12.00 g/0.0788 mol = 152 g/mol
13.70
∆Tf = Kfm m = ∆Tf/Kf = 0.337 °C/5.07 °C kg/mol = 0.0665 mol/kg 0.0665 mol mol = (0.5 kg) = 0.0332 mol 1 kg
4.26 g = 128.3 g/mol 0.0332 mol The empirical formula has a mass of 64.1 g/mol. So the molecular formula is C8H4N2. molar mass =
13.71
(a)
(b)
For convenience we choose to work with 100 g of the compound, and then to convert the mass amounts of each element found in this compound into mole amounts: for C, 42.86 g ÷ 12.01 g/mol = 3.569 mol C for H, 2.40 g ÷ 1.01 g/mol = 2.38 mol H for N, 16.67 g ÷ 14.01 g/mol = 1.190 mol N for O, 38.07 g ÷ 16.00 g/mol = 2.379 mol O The relative mole amounts that represent the empirical formula are determined by dividing the above mole amounts each by the smallest mole amount: for C: 3.569 mol ÷ 1.190 mol = 2.999 for H: 2.37 mol ÷ 1.190 mol = 1.99 for N: 1.190 mol ÷ 1.190 mol = 1.000 for O: 2.379 mol ÷ 1.190 mol = 1.999 and the empirical formula is C3H2NO2. ∆Tb = 1.84 °C = Kb × m = 2.53 °C kg mol–1 × m m = 0.727 mol solute/kg benzene. 277
Chapter 13
The number of moles of solute is: 0.727 mol/kg benzene × 0.045 kg benzene = 0.0327 mol, and the formula mass is: 5.5g/0.0327 mol = 168 g/mol. Since the mass of the empirical unit is 84, the molecular formula must be twice the empirical formula, namely C6H4N2O4. 13.72
(a)
If the equation is correct, the units on both sides of the equation should be g/mol. The units on the right side of this equation are:
(g) × (L atm mol−1 K −1 ) × (K) = g/mol L × atm which is correct. (b)
Π = MRT = (n/V)RT, n = ΠV/RT This means that we can calculate the number of moles of solute in one L of solution, as follows:
n=
( 0.021 torr ) (1 atm 760 torr ) (1.0 L ) ( 0.0821 L atm mol−1 K −1 ) ( 298 K )
= 1.1 × 10−6 mol
The molecular mass is the mass in 1 L divided by the number of moles in 1 L: 2.0 g/1.1 × 10–6 mol = 1.8 × 106 g/mol 13.73
Π = MRT = (n/V)RT, n = ΠV/RT 1 atm ( 3.74 torr ) (1 L ) 760 torr n= = 2.00 × 10−4 mol Latm 0.0821 mol K ( 300 K ) 0.400 g molar mass = = 2.00 × 103 g/mol −4 2.00 × 10 mol
13.74
The equation for the vapor pressure is: Psolution = P°H2O × χH2O Where P°H2O is 17.5 torr. To calculate the vapor pressure we need to find the mole fraction of water first. χH2O = moles H2O/(moles H2O + moles NaCl) Calculate the moles of NaCl in 21.5 g 1 mol NaCl mol NaCl = ( 21.5 g NaCl ) = 0.368 moles NaCl 58.44 g NaCl When NaCl dissolves in water, Na+ and Cl– are formed. So, for every mole of NaCl that dissolves, two moles of ions are formed. For this solution, the number of moles of ions is 0.736. The number of moles of solvent (water) is: 1 mol H 2 O mol H 2 O = (100 g H 2 O ) = 5.55 moles H 2 O 18.02 g H 2 O Calculate the mole fraction as ( moles H 2 O ) 5.55 mol χ H2O = = = 0.883 (moles H 2 O + moles NaCl) (5.55 mol + 0.736 mol) The vapor pressure is then Psolution = P°H2O × χH2O = 17.5 torr × 0.883 = 15.5 torr
13.75
χH2O =
P P
o
=
38.7 torr = 0.917 42.2 torr
278
Chapter 13
1 g 1 mol mol H 2 O = (175 mL) = 9.71 mol 1 mL 18.02 g 9.71 mol 9.71 mol + x mol X = 0.883 mole Since the van't Hoff factor for AlCl3 is 4, we need: 0.883 mol/4 = 0.221 mol AlCl3 133.3 g g AlCl3 = (0.221 mol) = 29.4 g AlCl3 1 mol 0.917 =
13.76
Assume 100 mL of solution, that is, 2.0 g NaCl and 0.100 L of solution: Π = MRT 1 mol NaCl (2.0 g NaCl) 58.45 g NaCl = 0.34 M M = 0.100 L For every NaCl there are two ions produces so M = 0.68 M 760 torr 4 ∏ = (0.68 M)(0.0821 L atm/mol K)(298 K) = 1.3 × 10 torr 1 atm
13.77
Π = MRT For each ion, multiply the concentration by 24.47 L atm/mol Ion Cl– Na+ Mg2+ SO42– Ca2+ K+ HCO3–
Molality (mol/L) 0.566 0.486 0.055 0.029 0.011 0.011 0.002
Π (atm) 13.9 11.9 1.35 0.710 0.269 0.269 0.0489
Adding these together we get Π = 28.4 atm. Thus, a pressure greater than 28.4 atm is needed to desalinate seawater by reverse osmosis. 13.78
CaCl2 Ca2+ + 2Cl–; van’t Hoff factor, i = 3 ∆Tf = i × Kf × m = (3)(1.86 °C m–1)(0.24 m) = 1.3 °C The freezing point is –1.3 °C.
13.79
If we assume that mercury(I) nitrate has the formula HgNO3, we predict a freezing point of –0.37 °C, ∆Tf = i × Kf × m = 2 × 1.86 °C/m × 0.10, m = 0.37 °C. However, the observed freezing point depression is lower than this. So, assume that the correct formula of the compound is Hg2(NO3)2 where the mercury ion is dimeric and divalent, i.e., Hg22+. Assuming that the person who prepared the solution thought it was HgNO3 when they calculated the molality, the concentration of the solution based on the correct formula for the compound would be recalculated correctly as 0.050 m since the true formula mass is twice that assumed, and ∆Tf = i × Kf × m = 3 × 1.86 °C/m × 0.050 m = 0.28 °C. Therefore, the dissociation produces three ions and the equation is: Hg2(NO3)2 Hg22+ + 2NO3–
279
Chapter 13
13.80
Any electrolyte such as NiSO4, that dissociated to give 2 ions, if fully dissociated should have a van't Hoff factor of 2.
13.81
MgSO4 Mg2+ + SO42–, the van’t Hoff factor is expected to be two.
13.82
∆Tf = i × Kf × m i = ∆Tf/Kf × m = 0.415°C/(1.86 °C m–1)(0.118 m) = 1.89
13.83
To solve this problem, we need to assume the density of the solution is 1 g/mL. From problem 13.82 we know that most of the LiCl has dissociated. As a result, the affect of the dissociated ions will increase the osmotic pressure. Π = MRT If we consider the dissociation Π = iMRT Π = (1.89)(0.118 mol L–1)(0.0821 L atm/mol K)(283 K)(760 torr/1atm) = 3.94 × 103 torr
Additional Exercises 13.84
The partial pressure of N2 in air is: PN2 = 1.00 atm(78 mol%) = 0.78 atm Therefore, according to Henry’s Law, the amount of N2 dissolved per liter of blood at 1.00 atm is: (1 L)(0.015 g/L)(0.78/1.00) = 0.012 g N2 0.012 g N2 (1 mol N2/28.0 g N2) = 0.00043 mol N2 The amount of N2 dissolved per liter of blood at 4.00 atm would be four times that, or: 0.0017 mol N2 The amount of nitrogen released per liter of blood upon quickly surfacing is the difference between the two, or (0.0017 mol – 0.00043 mol) = 0.0013 mol N2. The volume of that gas at 1.00 atm and 37 °C would be given by the ideal gas law: PV = nRT V = nRT/P V = (0.0013 mol N2)(0.0821 L·atm/mol·K)[(273+37) K]/1 atm V = 0.033 L = 33 mL N2 per liter of blood
13.85
Let A = CCl4 and B = unknown PTot = PA + PB PTot = χAPA° + χBPB° We also know that χA + χB = 1 So, PTot = (1 – χB)PA° + χBPB° PTot – PA° = χB(PB° – PA°)
P − PAo 137 torr − 143 torr = χB = Tot = 0.103 Po − P o 85 torr − 143 torr A B χA = 0.897 χCCl4 =
mol CCl4 , rearranging, we get mol CCl4 + mol unknown
280
Chapter 13
moles unknown = mol CCl4 1 − 1 χCCl 4 1 mol CCl4 mol CCl4 = (400 g CCl4) = 2.60 mol CCl4 153.8 g CCl4 mol unknown = 2.60 mol CCl4 1 − 1 = 0.299 mol unknown 0.897 molar mass =
43.3 g = 145 g / mol 0.299 moles
13.86
Π = MRT = (0.0100 mol/L)(0.0821 L atm K–1 mol–1)(298 K) = 0.245 atm 0.245 atm × 760 torr/atm = 186 torr.
13.87
(a)
(b) (c)
Since –40.0°F is equal to –40.0°C, the following expression applies: ∆T = Kfm, so 40 °C = (1.86 °C kg mol–1) × m, m = 40/1.86 mol/kg = 21.5 molal Therefore, 21.5 moles must be added to 1 kg of water. 62.1 g 1.00 mL 3 mL = (21.5 moles) = 1.2 × 10 mL 1 mol 1.11 g There are 946 mL in one quart. Thus, for 1 qt of water we are to have 946 mL, and the required number of quarts of ethylene glycol is:
1.2 × 103 mL C H O 1 g H O 946 mL H O 1 qt C H O 2 6 2 2 6 2 2 2 1 mL H 2 O 1 qt H 2 O 946 mL C2 H 6 O 2 1000 g H 2 O = 1.2 qt C2H6O2 The proper ratio of ethylene glycol to water is 1.2 qt to 1 qt. qt C2 H 6 O 2 1 qt H 2 O
13.88
13.89
(a)
The height difference is proportional to the osmotic pressure, therefore ∏ may be calculated by converting the height difference to the height of a mercury column in mm, which is equal to the pressure in torr (1 mm Hg = 1 torr): hHg = hsolution × (dsolution/dHg) = (12.6 mm) × (1.00 g/mL/13.6 g/mL) = 0.926 mm Hg P = 0.926 torr
(b)
∏ = MRT
(c)
1atm (0.926 torr) 760 torr = 4.98 × 10–5 M M = ∏/RT = L atm 0.0821 mol K (298 K) Since this is a dilute solution and the solute does not dissociate, we can assume that the molarity and molality are equivalent. So, ∆Tf = kfm = (1.86 °C m–1)(4.98 × 10–5 m) = 9.26 × 10–5 °C Freezing point will be –9.26 × 10–5 °C
(d)
The magnitude of the temperature change is too small to measure.
(a)
Since the molarity of the solution is 4.613 mol/L, then one L of this solution contains: 281
Chapter 13
(b) 13.90
(a)
(b)
(c) 13.91
4.613 mol × 46.07 g/mol = 212.5 g C2H5OH. The mass of the total 1 L of solution is: 1000 mL × 0.9677 g/mL = 967.7 g. The mass of water is thus 967.7 g – 212.5 g = 755.2 g H2O, and the molality is: 4.613 mol C2H5OH/0.7552 kg H2O = 6.108 m. % (w/w) C2H5OH = (212.5 g/967.7 g) × 100% = 21.96% ∆Tb = kbm = (0.51 °C m–1)(1.25 m) = 0.64 °C ∆Tb = 100 °C + 0.64°C = 100.64 °C ∆Tb = ikbm = (4)(0.51 °C m–1)(1.25 m) = 2.55 °C ∆Tb = 100 °C + 2.55°C = 102.55 °C i = 0.229 °C/0.51 °C = 0.449
Colligative properties are based on the number of particles in solution. In each of the two cases we are told to assume complete dissociation. Thus, one formula unit of AlCl3 contributes 4 ions and one formula unit of Na2SO4contributes 3 ions to the solution. The total molality, mT, of the solution is the sum of the molalities of the two salts. mT = m1(AlCl3) + m2(Na2SO4) The molality of AlCl3 is 4/7 of mT and the molality of Na2SO4 is 3/7 of mT.
∆Tf = i × Kf × mT o
2.65 o C = 7 x 1.86
C x mT molal
mT = 0.2035 m Since the total molality is small we can assume that molarity and molality are approximately the same value. M(AlCl3) =
4 x 0.2035 m = 0.116 M 7
M(Na2SO4) =
3 x 0.2035 m = 0.0872 M 7
Multi-Concept Problems 13.92
From the boiling point elevation we can determine the total molality of the solution. ∆Tf = i × Kf × mT 0
4.6 o C = 5 x 0.51
C x mT molal
mT = 1.804 m The molality of the KCl solution is:
282
Chapter 13
2 x 1.804 m = 0.7216 m 5 The molality of the Fe(NO3)2 is:
3 x 1.804 m = 1.0824 m 5 From the titration we can determine the molarity of the iron(II) nitrate. 6Fe2+ + Cr2O72- + 14H+ →
6Fe3+ + 2Cr3+ + 7H2O
2-
36.3 mL x
0.220 mmol Cr2 O 7 mL
x
6 mmol Fe 2+ mmol Cr2 O72-
= 47.916 mmol Fe 2+
Or 47.916 x 10-3 mol Fe2+ We can determine the volume of solvent from the calculated molality of the iron(II) solution and the solution’s density. First, we need to determine the mass of solvent containing 4.7916 x 10-2 mol of iron salt.
1 kg solvent x 0.04792 mol = 4.427 x 10−2 kg solvent 1.0824 mol Fe(NO3 ) 2 Or 44.27 g of solvent
Mass of Fe(NO3 ) 2 = 0.04792 mol x
179.857 g = 8.62 g mol Fe(NO3 ) 2
Mass of solution = 8.62 g Fe(NO3)2 + 44.27 g H2O = 52.89 g Volume of solution = 52.89 g x
M{Fe(NO3)2} =
1 mL = 51.25 mL 1.032 g
47.916 mmol Fe(NO3 ) 2 = 0.93 M 51.25 mL
The molarity of the KCl solution would be: M{KCl} =
13.93
0.7216 mol KCl 1 x 4.427 x 10−2 kg solvent x = 0.62 M kg solvent 5.125 x 10−2 L
Osmotic pressure is given by Π = iMRT Since this is a mixture, the molarity, M, is the sum of the individual solutes in solution. BaSO4 will precipitate out of solution. Therefore, we need to determine how much Ba2+ and SO42− will be removed from the solution. Before reaction:
283
Chapter 13 mmol Ba2+ = 25.00 mL x 0.200 M = 5 mmol mmol NO3− = 2 x 5 mmol = 10 mmol mmol SO42− = 14.00 mL x 0.250 M = 3.5 mmol mmol K+ = 2 x 3.5 mmol = 7 mmol SO42− is the limiting reagent. After reaction we have: 1.5 mmol of Ba2+ 10 mmol of NO3− 7 mmol of K+
iM =
1.5 mmol + 10 mmol + 7 mmol = 0.474 M 39.00 mL
Π = 0.474 M x 0.0821 L atm K-1 mol-1 x 298 K = 11.6 atm 11.6 atm x 760 torr atm-1 = 8.82 x 103 torr 13.94
There are two isomers of C2H6O, ethanol, and dimethyl ether. Shown below are the abbreviated structural forms of the two isomers.
(a) Both are expected to be liquids at 25 oC. Ethanol possesses a dipole moment, 1.68 Debye, it can hydrogen bond with itself, and has significant London forces. Dimethyl ether has a dipole, 1.30 Debye, and possesses significant London forces. (b) Both should be soluble in water since they are both polar and can interact with polar water molecules. Also, ethanol will hydrogen bond with water. Neither should be significantly soluble in non-polar pentane. (c) Ethanol will form hydrogen bonds due the presence of the OH group and unshared electrons on the oxygen atom. Dimethyl ether cannot hydrogen bond as it does not have a hydrogen attached to the oxygen atom. (d) Both are non-electrolytes since they do not significantly ionize in water. The hydrogen on the OH group is not an ionizable hydrogen. 13.95
The reaction is given below. 3Sn2+ + Cr2O72- + 14H+ →
0.155 L x
3Sn4+ + 2Cr3+ + 7H2O
0.650 mol Sn 2+ = 0.1008mol Sn 2 + L
0.1008 mol Sn 2 + x
1 mol Cr2 O72 − 3 mol Sn 2 +
= 0.223
mol Cr2 O72 − xV L
284
Chapter 13
V = 0.151 L or 151 mL required To determine the osmotic pressure we first need to determine the molarity of the resulting solution.
M(Sn 2 + ) =
0.1008 mol Sn 2 + = 0.329 M (0.155 L + 0.151 L)
M(Cr2 O72− ) =
1 mol Cr2 O72 − 0.1008 mol Sn 2 + x = 0.110 M (0.155 L + 0.151 L) 3 mol Sn 2 +
Using these molarities and the number of particles created for each solution when the reaction is complete, we can determine the osmotic pressure. Be sure to include the spectator ions when determining the value of “i.” Π = iMRT 2− mol Cr2 O7 mol Sn 2 + −1 −1 Π = 4 x 0.110 + 3 x 0.329 x 0.0821 L atm K mol x 298 K L L
Π = 34.9 atm
Π = 34.9 atm x 760 torr atm-1 = 2.65 x 104 torr
In this solution to the problem a value of i = 4 was used for the dichromate since 2 moles of Cr3+ are produced by the reaction. Alternately, you could determine the molar concentration of the Cr3+ and then use the molar concentration of the dichromate to determine the molar concentration of the K+ ions. 13.96
(a)
(b)
The formula masses are Na2Cr2O7·2H2O: 298 g/mol, C3H8O: 60.1 g/mol, and C3H6O: 58.1 g/mol. 1 mol C3 H8 O 1 mol Na 2 Cr2 O7 • 2H 2 O g Na 2 Cr2 O7 • 2H 2 O = (21.4 g C3 H8 O) 3 mol C3 H8 O 60.1 g C3 H8 O
298 g Na 2 Cr2 O7 • 2H 2 O × = 35.4 g Na 2 Cr2 O7 • 2H 2 O 1 mol Na 2 Cr2 O7 • 2H 2 O The theoretical yield is: 1 mol C3 H8 O 3 mol C3 H 6 O 58.1 g C3 H 6 O g C3H6O = (21.4 g C3H8O) = 20.7 g C3H8O 60.1 g C3 H8 O 3 mol C3 H8 O 1 mol C3 H 6 O The percent yield is therefore: 12.4/20.7 × 100% = 59.9%
(c)
First, we determine the number of grams of C, H, and O that are found in the products, and then the % by mass of C, H, and O that were present in the sample that was analyzed by combustion, i.e. the by–product:
12.011 g C –3 g C = (22.368 × 10–3 g CO2) = 6.1046 × 10 g C 44.010 g CO 2 and the % C is: (6.1046 × 10–3 g/8.654 × 10–3 g) × 100% = 70.54% C 2.0159 g H –3 g H = (10.655 × 10–3 g H2O) = 1.1923 × 10 g H 18.015 g H O 2 and the % H is: (1.1923 × 10–3 g H/8.654 × 10–3 g) × 100% = 13.78% H For O, the mass is the total mass minus that of C and H in the sample that was analyzed: 285
Chapter 13
8.654 × 10–3 g total – (6.1046 × 10–3 g C + 1.1923 × 10–3 g H) = 1.357 × 10–3 g O and the % O is: (1.357 × 10–3 g)/(8.654 × 10–3 g) × 100% = 15.68% O. Alternatively, we could have determined the amount of oxygen by using the mass % values, realizing that the sum of the mass percent values should be 100. Next, we convert these mass amounts for C, H, and O into mole amounts by dividing the amount of each element by the atomic mass of each element: For C, 6.1046 × 10–3 g C ÷ 12.011 g/mol = 0.50825 × 10–3 mol C For H, 1.1923 × 10–3 g H ÷ 1.0079 g/mol = 1.1829 × 10–3 mol H For O, 1.357 × 10–3 g O ÷ 16.00 g/mol = 0.08481 × 10–3 mol O Lastly, these are converted to relative mole amounts by dividing each of the above mole amounts by the smallest of the three (We can ignore the 10–3 term since it is common to all three components): For C, 0.50825 mol/0.08481 mol = 5.993 For H, 1.1829 mol/0.08481 mol = 13.95 For O, 0.08481 mol/0.08481 mol = 1.000 and the empirical formula is given by this ratio of relative mole amounts, namely C6H14O. (d)
∆Tf = Kfm, (5.45 °C – 4.87 °C) = (5.07 °C/m) × m, m = 0.11 molal, and there are 0.11 moles of solute dissolved in each kg of solvent. Thus, the number of moles of solute that have been used here is: 0.11 mol/kg × 0.1150 kg = 1.3 × 10–2 mol solute. The formula mass is thus: 1.338 g/0.013 mol = 102 g/mol. Since the empirical formula has this same mass, we conclude that the molecular formula is the same as the empirical formula, i.e. C6H14O.
286
Chapter 14
Practice Exercises 14.1
Using the coefficients of the reaction, we find that the ratio of iodide production to sulfite disappearance in 1:3, and the ratio of sulfate production to sulfite disappearance is 3:3. ⎛ 1 mol I − ⎞ Rate of production of I– = (2.4 × 10–4 mol L–1 s–1) ⎜ ⎟ = 8.0 × 10–5 mol L–1 s–1 ⎜ 3 mol SO 2 − ⎟ 3 ⎝ ⎠ ⎛ 3 mol SO 2 − ⎞ 4 Rate of production of SO42– = (2.4 × 10–4 mol L–1 s–1) ⎜ ⎟ = 2.4 × 10–4 mol L–1 s–1 ⎜ 3 mol SO 2 − ⎟ 3 ⎝ ⎠
14.2
From the coefficients in the balanced equation we see that, for every two moles of SO2 that is produced, 2 moles of H2S are consumed, three moles of O2 are consumed, and two moles of H2O are produced. ⎛ 3 mol O 2 ⎞ ⎛ 0.30 mol ⎞ −1 −1 Rate of disappearance of O2 = ⎜ ⎟⎜ ⎟ = 0.45 mol L s 2 mol SO L s ⎝ ⎠ 2 ⎠ ⎝
14.3
14.4
⎛ 2 mol H 2S ⎞ ⎛ 0.30 mol ⎞ −1 −1 Rate of disappearance of H2S = ⎜ ⎟⎜ ⎟ = 0.30 mol L s 2 mol SO L s ⎝ ⎠ 2 ⎠ ⎝ The rate of the reaction at 2.00 minutes (120 s) is equal to the slope of the tangent to the curve at 120 s. After drawing the tangent, the slope can be estimated as follows: −1 −1 ⎫ ⎧⎪ [ HI ] ⎫ ⎧ − HI final [ ]initial ⎪ ⎪ 0 mol L − 0.075 mol L ⎪ = –2.1 × 10–4 mol L–1 s–1 ratewith respect to HI = ⎨ ⎬=⎨ ⎬ t final − t initail 360 s − 0 s ⎪⎩ ⎪⎭ ⎪⎩ ⎭⎪
The rate of the reaction after 250 seconds have elapsed is equal to the slope of the tangent to the curve at 250 seconds. First draw the tangent, and then estimate its slope as follows, where A is taken to represent one point on the tangent, and B is taken to represent another point on the tangent: ⎛ A (mol/L) − B (mol/L) ⎞ change in concentration rate = ⎜ ⎟ = A (s) − B (s) change in time ⎝ ⎠ A value near 1 × 10–4 mol L–1 s–1 is correct.
14.5
14.6
(a)
Using the rate law: Rate = k[NO]2[H2] Substitute in the concentration and the rate and solve for the rate constant: 7.86 × 10–3 mol L–1 s–1 = k(2 × 10–6 mol L–1)2(2 × 10–6 mol L–1) k = 9.8 × 1014 L2 mol–2 s–1 (b) The units can be derived from the equation: mol L–1 s–1 = k(mol L–1)2(mol L–1) mol L–1 s–1 = k (mol3 L–3) mol L−1 s −1 = L2 mol–2 s–1 k= 3 −3 mol L (a) First use the given data in the rate law: Rate = k[HI]2 2.5 × 10–4 mol L–1 s–1 = k[5.58 × 10–2 mol/L]2 k = 8.0 × 10–2 L mol–1 s–1 (b) L mol–1 s–1
14.7
The order of the reaction with respect to a given substance is the exponent to which that substance is raised in the rate law: order of the reaction with respect to [BrO3–] = 1 order of the reaction with respect to [SO32–] = 1 287
Chapter 14 overall order of the reaction = 1 + 1 = 2 14.8
The rate law is second order with respect to Cl2 and first order with respect to NO. Therefore the exponent for the Cl2 is two and the exponent for NO is one: Rate = k[Cl2]2[NO]
14.9
Since the rate law is Rate = k[Br2] the order with respect to Br2 is 1 and the order with respect to HCO2H is zero. The overall order of the reaction is 1.
14.10
In each case, k = rate/[A][B]2, and the units of k are L2 mol–2 s–1. Each calculation is performed as follows, using the second data set as the example: 0.40 mol L−1 s −1 k= = 2.0 × 102 L2 mol−2 s −1 2 0.20 mol L−1 0.10 mol L−1
(
)(
)
Each of the other data sets also gives the same value: k = 2.0 × 102 L2 mol–2 s–1. 14.11
The rate law is: rate = k[A][B]2 (a) If the concentration of B is tripled, then the rate will increase nine–fold, rate = k[A][3B]2 rate = 9k[A][B]2 (b) If the concentration of A is tripled, then the rate will increase three–fold, rate = k[3A][B]2 rate = 3k[A][B]2 (c) If the concentration of A is tripled, and the concentration of B is halved, then the rate will decrease to three fourths, or seventy five percent 1 rate = k[3A][ B]2 2 3 rate = k[A][B]2 4
rate = k [ NO] [ H 2 ] n
14.12 (a)
m
To find the rate law, take two reactions in which the concentration of one of the reactants is held constant, compare the two reactions and solve for the exponent: For NO, use the first two reactions:
k [ NO]1 [ H 2 ]1 n
k [ NO]2
n
m
[ H2 ]2
m
=
rate1 rate2
k ⎡ 0.40 × 10−4 mol L−1 ⎤ ⎡ 0.30 × 10−4 mol L−1 ⎤ ⎣ ⎦ ⎣ ⎦
n
m
n
m
k ⎡ 0.80 × 10−4 mol L−1 ⎤ ⎡ 0.30 × 10−4 mol L−1 ⎤ ⎣ ⎦ ⎣ ⎦ n
⎡ 0.40 × 10−4 mol L−1 ⎤ −8 −1 −1 ⎣ ⎦ = 1.0 × 10 mol L s n 4.0 × 10−8 mol L−1 s −1 ⎡ 0.80 × 10−4 mol L−1 ⎤ ⎣ ⎦ n
1 ⎛1⎞ n=2 ⎜2⎟ = 4 ⎝ ⎠ For H2, use the second two reactions:
288
=
1.0 × 10−8 mol L−1 s −1 4.0 × 10−8 mol L−1 s −1
Chapter 14
k [ NO]2
[ H2 ]2m n m k [ NO]3 [ H 2 ]3 n
=
rate2 rate3
k ⎡ 0.80 × 10−4 mol L−1 ⎤ ⎡ 0.30 × 10−4 mol L−1 ⎤ ⎣ ⎦ ⎣ ⎦
n
m
n
m
k ⎡ 0.80 × 10−4 mol L−1 ⎤ ⎡ 0.60 × 10−4 mol L−1 ⎤ ⎣ ⎦ ⎣ ⎦ ⎡ 0.30 × 10−4 mol L−1 ⎤ ⎣ ⎦
m
⎡ 0.60 × 10−4 mol L−1 ⎤ ⎣ ⎦
m
⎛1⎞ ⎜2⎟ ⎝ ⎠
m
=
1 2
=
=
4.0 × 10−8 mol L−1 s −1 8.0 × 10−8 mol L−1 s −1
4.0 × 10−8 mol L−1 s −1 8.0 × 10−8 mol L−1 s −1
m=1
rate = k [ NO] [ H 2 ] 2
(b)
1
To find the value for the rate constant, choose one of the reactions and use the values for the concentrations and rate and solve for the rate constant:
rate = k [ NO] [ H 2 ] 2
1 2
(c)
1
1.0 ×10−8 mol L−1 s−1 = k ⎡0.40 ×10−4 mol L−1 ⎤ ⎡0.30 × 10−4 mol L−1 ⎤ ⎣ ⎦ ⎣ ⎦ k = 2.1 × 105 L2 mol–2 s–1 The units for the rate constant can be determined from the rate law and cancelling the units: 2
1
mol L−1 s−1 = k ⎡ mol L−1 ⎤ ⎡ mol L−1 ⎤ ⎣ ⎦ ⎣ ⎦ ⎡ mol L−1 s −1 ⎤ ⎣ ⎦ = L2 mol–2 s–1 k= 2 1 ⎡ mol L−1 ⎤ ⎡ mol L−1 ⎤ ⎣ ⎦ ⎣ ⎦ 14.13
(a)
The rate law will likely take the form rate = k[A]n, where n is the order of the reaction with respect to A. rate2 k [ A ]2 k(0.36) n 2.22x10−4 = = = rate1 k [ A ] n k(0.10) n 6.17x10−5 1 n
3.6n = 3.6
Therefore, n = 1
rate3 k [ A ]3 k(0.58)n 3.58x10−4 = = = rate1 k [ A ] n k(0.10)n 6.17x10−5 1 n
5.8n = 5.8
Therefore, n = 1
rate3 k [ A ]3 k(0.58)n 3.58x10−4 = = = rate1 k [ A ] n k(0.36)n 2.22x10−4 2 n
1.6n = 1.6
Therefore, n = 1
289
Chapter 14
(b)
rate = k[sucrose] (6.17 × 10–4 mol L–1 s–1) = k(0.10 mol L–1) k = 6.17 × 10–4 s–1 The rate constant is 6.17 × 10–4 s–1 The other two data sets give the same value for k. This reaction is actually a pseudo-first order reaction. We cannot determine whether H+ or water are part of the rate expression. The experiment did not vary the acid concentration so we do not know how the rate might change with respect to acid and since it is an aqueous reaction we cannot determine the effect changing water’s concentration would have on the rate as the concentration of water is so large that its concetration would not change significantly during the reaction.
14.14
(a)
The rate law will likely take the form rate = k[A]n[B]n', where n and n' are the order of the reaction with respect to A and B, respectively. rate1 k [ A ]2 [ B]2 k(0.40)n (0.30)m 1.00x10−4 x = = = rate2 k [ A ] n [ B] m k(0.60)n (0.30) m 2.25x10−4 3 3 n
m
0.6667n = 0.44444 n=2 On comparing the second and third lines of data, neither the concentration of A nor the concentration of B are held constant, but we know that the reaction is second–order with respect to A and we can solve the two equations for the order with respect to B: rate2 k [ A ]2 [ B]2 = rate3 k [ A ] 2 [ B] m 3 3 2
m
2.25 × 10−4 mol L−1 s −1 1.60 × 10−3 mol L−1 s −1 2.25 × 10−4 1.60 ×10−3
=
( 0.30 M )m = 2 m k ( 0.80 M ) ( 0.60 M ) 2
( 0.36 )( 0.30 M )m ( 0.64 )( 0.60 M )m
⎛1⎞ 0.141 = 0.563 ⎜ ⎟ ⎝2⎠ ⎛1⎞ 0.25 = ⎜ ⎟ ⎝2⎠
k ( 0.60 M )
m
m
m
1 ⎛1⎞ = 4 ⎜⎝ 2 ⎟⎠ m=2 Rate = k[A]2[B]2
(b)
For the rate constant, use one of the experiments and insert the values and solve for k Rate = k[A]2[B]2 1.00 × 10–4 mol L–1 s–1 = k[0.40 mol L–1]2[0.30 mol L–1]2 k = 6.9 × 10–3 L3 mol–3 s–1
(c)
The units for the rate constant are 290
Chapter 14 mol L−1 s −1
( mol L ) ( mol L ) 2
−1
(d) 14.15
−1
2
= L3 mol–3 s–1
The overall order for this reaction is 2 + 2 = 4
Since this is a first order reaction, then we can use the integrated rate law for a first order reaction: [ A ]0 ln = kt [ A ]t If only 5% of the active ingredient can decompose in two years, then 95% must remain, therefore, [A]0 = 100, [A]t = 95, and t = 2 yr [ A ]0 ln = kt [ A ]t 100 = k ( 2 yr ) 95 k = 2.56 × 10–2 yr–1 ln
14.16
(a)
In order to find the concentration at specific time for a first order reaction, we substitute into equation 14.5, first converting the time to seconds: t = 2 hr × 3600 s/hr = 7200 s ln
[ A]0 [ A ]t
= kt
⎡ [A ] ⎤ 0 ⎥ = antiln [ kt ] antiln ⎢ ln ⎢⎣ [ A ]t ⎥⎦ [ A]0 = antiln [ kt ] [ A ]t
[ A]0 [ A ]t
(
)
= antiln ⎡ 6.17 × 10−4 s −1 ( 7200 s ) ⎤ ⎣⎢ ⎦⎥
0.40 M = 84.98 [ A ]t
[ A ]t
=
0.40 M 84.98
= 4.71 x 10 − 3 M
(b) Again, we use equation 14.5, this time solving for time: [ A]0 ln = kt [ A ]t
[ A ]0 1 1 0.40 M = = 466 s × ln × ln − − 4 1 k 0.30 M [ A]t 6.17 × 10 s 4.66 × 102 s × 1 min/60 s = 7.8 min t=
291
Chapter 14 14.17
For a first–order reaction: 0.693 0.693 t1/2 = = = 1.12 × 103 s k 6.17 × 10−4 s −1
t1/2 = 1.12 × 103 s ×
1 min 60 s
= 18.7 min If we refer to the chart given in the text in example 14.8, we see that two half lives will have passed if there is to be only one quarter of the original amount of material remaining. This corresponds to: 18.7 min per half–life × 2 half lives = 37.4 min 14.18
From practice exercise 14.15, the rate constant is: k = 2.56 × 10–2 yr–1, and use the half–life of a first order reaction equation: ln 2 ln 2 = 27.1 yr t 12 = = k 2.56 × 10−2 yr −1
14.19
Recall that for a first order process 0.693 k= t1/ 2 So k = ln
0.693 = 4.86 × 10–2/days. Also, 14.26 days
[A] 0 = kt [A] t
Assume was start with 100 g of 32 P ln [A]t = ln[A]0 − kt = ln100 − 4.86x10−2 days x (60 days)
ln[A]0 = 1.689 [A]0 = 5.41 g or 5.41 % Alternative method
⎡1⎤ ⎢⎣ 2 ⎥⎦
t / t1/2
= [ 0.5]
60 days /14.26 days
= 0.0541g
Thus, 0.0541 grams would be left if you started with 1.00 grams, or 5.41 % was left. 14.20
Recall that for a first order process 0.693 k= t1/ 2 So k = ln
0.693 = 1.21 × 10–4/yr. Also, 5730 yr
[A] 0 = kt [A] t
t =
[A] 0 1 1 ln = k [A] t 1.21 × 10
4
/yr
ln
10 = 1.90 × 104 yrs 1
292
Chapter 14
14.21
Use the value of the rate constant for C–14 from the previous practice exercise: 1.21 × 10–4 y–1. To find the upper and lower limits of dates before present, set the concentration of the C–14 to 5% and 95%, respectively, and then solve for the time: For samples with less than 5% of the C–14 remaining: [A] 0 ln = kt [A] t [A] 0 1 1 100 = 2.48 × 104 yr ln ln = − 4 1 k [A] t 5 1.21 × 10 yr Therefore the upper limit of dates is 24,800 years before present. t =
For samples with more than 95% of the C–14 remaining: [A] 0 ln = kt [A] t [A] 0 1 1 100 = 4.24 × 102 yr ln ln = −4 −1 k [A] t 95 1.21 × 10 yr Therefore the lower limit of dates is 424 years before present. t =
14.22
This is a second–order reaction, and we use equation 14.10: 1 1 = kt − [ NOCl]t [ NOCl]0 1 − 0.010 M] [
(
)
1 = 0.020 L mol−1 s −1 × t 0.040 M [ ]
t = 3.8 × 103 s t = 3.8 × 103 s × 1 min/60 s = 63 min 14.23
This is the same reaction as in the previous practice exercise, so it is a second–order reaction, and we use equation 14.10 and k = 0.020 L mol–1 s–1: 1 1 − = kt [ NOCl]t [ NOCl]0 1 − [0.00035 M ]
(
)
1 = 0.020 L mol−1 s −1 × t [ x M]
We need to find the time in seconds: From 10:35 to 3:15, 4 hours and 40 minutes has elapsed, or 280 minutes ⎛ 60 s ⎞ 4 s = (280 min) ⎜ ⎟ = 1.68 × 10 s 1 min ⎝ ⎠ 1 1 = 0.020 L mol−1 s −1 × 1.68 × 104 s − [0.00035 M ] [ x M ]
(
−
(
) (
) (
1 = 3.36 × 102 L mol−1 − 2.86 × 103 L mol−1 x M [ ]
)
)
4.0 × 10–4 M 14.24
Rate = k[NO2]2 To find the rate constant solve the rate law with the given data: 4.42 × 10–7 mol L–1s–1 = k(6.54 × 10–4 mol L–1)2
293
Chapter 14
k=
4.42 × 10 −7 mol L−1 s −1
( 6.54 ×10
−4
−1
mol L
)
2
= 1.03 L mol–1 s–1
The half life of the system is found using the half–life equation for a second–order reaction 1 t1/ 2 = k × ( initial concentration of reactant ) t1/ 2 =
(1.03 L mol
1
− 1 −1
s
)( 6.54 ×10
−4
−1
mol L
)
= 1.48 × 103 s
14.25
The reaction is first–order. A second–order reaction should have a half–life that depends on the initial concentration according to equation 14.11.
14.26
Use the equation 14.15: −Ea ⎡ 1 k 1⎤ ln 2 = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ For the 5% decomposition over 2 years, or 104.4 weeks 1 1 − =k t [ B]t [ B]0 1 1 1 = k (104.4 weeks ) − [95 M ] [100 M ] 1
k1 = 5.04 × 10–6 M–1 weeks–1 For the 5% decomposition over 1 week 1 1 − =k t B B [ ]t [ ]0 2 1 1 = k (1 week ) − 95 M 100 [ ] [ M] 2
k2 = 5.26 × 10–4 M–1 weeks–1 Now use the values for k1, k2, T1 = 298 K, and Ea = 154 × 103 J mol–1 −Ea ⎡ 1 k 1⎤ ln 2 = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦
⎡ 5.26 ×10−4 M −1 weeks−1 ⎤ −154 ×103 J mol−1 ⎡ 1 1 ⎤ − ln ⎢ ⎥ = ⎥ −6 −1 −1 −1 −1 ⎢ T 298 K ⎦ ⎢⎣ 5.04 ×10 M weeks ⎥⎦ 8.314 J mol K ⎣ 2 ⎡ 1 ⎤ ln(1.04 × 102) = (–1.85 × 104 K) ⎢ − 3.36 × 10−3 K −1 ⎥ ⎣ T2 ⎦
(
(
)
)
⎡ 1 ⎤ –2.51 × 10–4 K–1 = ⎢ − 3.36 × 10−3 K −1 ⎥ ⎣ T2 ⎦ 1 = 3.11 × 10–3 K–1 T2 T2 = 322 K or 49 °C
Alternatively,
294
Chapter 14
ln
104.4k −154 x 103 J mol−1 = k 8.314 J mol−1 K −1
⎡1 1 ⎤ x⎢ − ⎥ ⎣ T2 298K ⎦
T2 = 322 K 14.27
(a) Use equation 14.15: ln
−Ea ⎡ 1 k2 1⎤ − = ⎢ ⎥ k1 R ⎣ T2 T1 ⎦
⎡ 23 L mol−1 s −1 ⎤ −E a 1 ⎤ ⎡ 1 − ln ⎢ ⎥ = −1 −1 −1 −1 ⎢ 678 K 628 K ⎥⎦ ⎣ ⎢⎣ 3.2 L mol s ⎥⎦ 8.314 J mol K
Solving for Ea gives 1.4 × 105 J/mol = 1.4 × 102 kJ/mol (b) We again use equation 14.12, substituting the values: k1 = 3.2 L mol–1 s–1 at T1 = 628 K k2 = ? at T2 = 583 K ln
−Ea ⎡ 1 k2 1⎤ − = ⎢ ⎥ k1 R ⎣ T2 T1 ⎦
k2 ⎡ ⎤ −1.4 × 105 J mol−1 ⎡ 1 1 ⎤ − ln ⎢ = ⎥ −1 −1 −1 −1 ⎢ 583 K 628 K ⎥⎦ ⎣ 8.314 J mol K ⎣ 3.2 L mol s ⎦ k2 ⎡ ⎤ ln ⎢ =2.0697 −1 −1 ⎥ ⎣ 3.2 L mol s ⎦
k2 = (e-2.0697) x 3.2 Solving for k2 gives 0.40 L mol–1 s–1.
14.28
−Ea ⎡ 1 k2 1⎤ − = ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ ⎡ 3.37x104 M −1 s −1 ⎤ 1 ⎤ −9.31 × 104 J mol−1 ⎡ 1 ln ⎢ − ⎥ = ⎢ ⎥ 3 −1 −1 −1 −1 600 K ⎦ 8.314 J mol K ⎣ T2 ⎣⎢ 3.37x10 M s ⎦⎥ ln
2.056 x 10-4 = −
1 1 + T2 600
T2 = 684 K 14.29
(a), (b), and (e) may be elementary processes. Equations (c), (d), and (f) are not elementary processes because they have more than two molecules colliding at one time, and this is very unlikely.
14.30
If the reaction occurs in a single step, one molecule of each reactant must be involved, according to the balanced equation. Therefore, the rate law is expected to be: Rate = k[NO][O3].
14.31
The slow step (second step) of the mechanism determines the rate law:
295
Chapter 14 Rate = k[NO2Cl]1[Cl]1 However, Cl is an intermediate and cannot be part of the rate law expression. We need to solve for the concentration of Cl by using the first step of the mechanism. Assuming that the first step is an equilibrium step, the rates of the forward and reverse reactions are equal: Rate = kforward[NO2Cl] = kreverse[Cl][NO2] Solving for [Cl] we get k [ NO Cl] [Cl] = kf NO2 r [ 2] Substituting into the rate law expression for the second step yields: k [ NO 2 Cl]
2
Rate =
[ NO2 ]
, where all the constants have been combined into one new constant.
Review Questions
14.1
Student responses will vary. (a) combustion of gasoline (b) cooking an egg in boiling water (c) curing of cement
14.2
A collision between only two molecules is much more probable than the simultaneous collision of three molecules. We therefore conclude that a reaction involving a two–body collision is faster (i.e. will occur more frequently) than one requiring a three–body collision.
14.3
The instantaneous rate of reaction is the rate of the reaction at a particular moment. The average rate of reaction is the average rate for the reaction over the time of the whole reaction. This includes the very rapid rates when the concentration of reactants is high and the very slow rates when the concentration of reactants is low.
14.4
A tangent line to the curve of concentration as a function of time at time zero is drawn. The slope of this line is determined which is the initial instantaneous rate of reaction.
14.5
A homogeneous reaction is one in which all reactants and products are in the same phase. An example would be: 2H2(g) + O2(g) J 2H2O(g) A heterogeneous reaction is one in which all reactants and products are not in the same phase. An example would be: 25O2(g) + 2C8H18(l) J 16CO2(g) + 18H2O(g)
14.6
Chemical reactions that are carried out in solution take place smoothly because the reactants can mingle effectively at the molecular level.
14.7
Heterogeneous reactions are most affected by the extent of surface contact between the reactant phases.
14.8
In a heterogeneous reaction, the smaller the particle size, the faster the rate. This is because decreasing the particle size increases the surface area of the material, thereby increasing contact with another reactant phase.
14.9
This illustrates the effect of concentration.
14.10
Reaction rate generally increases with increasing temperature.
14.11
In cool weather, the rates of metabolic reactions of cold–blooded insects decrease, because of the effect of temperature on rate.
296
Chapter 14 14.12
The low temperature causes the rate of metabolism to be very low.
14.13
Reaction rate has the units mol L–1 s–1, or molar per second (M s–1). Reaction rates are reported as positive values regardless of whether the species increases or decreases during the recaction.
14.14
The units are, in each case, whatever is required to give the units of rate (mol L–1 s–1) to the overall rate law: (b) L mol–1 s–1 (c) L2 mol–2 s–1 (a) s–1
14.15
A zero–order reaction has no dependence on concentration while the first–order reaction is linearly dependent on concentration and the second-order reaction is inversely dependent on concentration.
14.16
No, the coefficients of a balanced chemical reaction do not predict with any certainty the exponents in a rate law. These exponents must be determined through experiment.
14.17
This is the first case in Table 14.4 of the text, that is a zero-order reaction.
14.18
This is the fourth case Table 14.4 of the text, that is a first-order reaction.
14.19
This is the seventh case in Table 14.4, and the rate increases by a factor of 22 = 4.
14.20
This is the eleventh case in Table 14.4, and the order of the reaction with respect to the reactant is 3.
14.21
Since the substrate concentration does not influence rate, its concentration does not appear in the rate law, and the order of the reaction with respect to substrate is zero.
14.22
The half–life of a first–order reaction is unaffected by the initial concentration.
14.23
The half–life of a second–order reaction is inversely proportional to the initial concentration, as expressed in equation 14.11.
14.24
The half-life of a zeroth-order reaction is directly proportional to initial concentration.
14.25
First order [ A]0 ln = kt [ A]t Second order 1 1 − = kt [ A]t [ A]0
t1/2 =
[ A]t
1/2
1 1 [ A]0 2
=
1 [ A]0 2
−
ln
1 = kt [ A]0 1/ 2
[ A]0 1 [ A]0 2
1 1 [ A]0 2
−
t1/ 2 =
= kt1/2
1 2 [ A]0 2
= kt1/ 2 =
ln 2 0.693 = k k
2 2 1 − = [ A]0 2 [ A]0 [ A]0
1 k [ A]0
Zeroth order
[ A]t = −kt + [ A]0 14.26
1 [ A]0 = −kt1/ 2 + [ A]0 2
kt1/ 2 = [ A]0 −
1 [ A]0 2
t1/ 2 =
For the zero–order reaction, substitute half the value of [A]0 for the value of [A]t in the equation [A]0 – [A]t = kt [A]0 – 1/2[A]0 = kt1/2 1/2[A]0 = kt1/2
297
[ A]0 2k
Chapter 14
t 12 =
14.27
[A]0 2k
To answer this question refer to the integrated equations for each order. Graph (b) represents a kinetics plot for a first order reaction. Graph (c) represents a kinetics plot for a second order reaction. Graph (a) represents a kinetics plot for a zeroth order reaction.
14.28
According to collision theory, the rate is proportional to the number of collisions per second among the reactants.
14.29
The effectiveness of collisions is influenced by the orientation of the reactants and by the activation energy.
14.30
This happens because a larger fraction of the reactant molecules possess the minimum energy necessary to surpass Ea.
Potential Energy
14.31
Ea (forward)
Ea (reverse)
Products ∆Hreaction
Reactants Reaction Coordinate
14.32
Potential Energy
Transition State
Reactants ∆H Products Reaction Coordinate
14.33
Breaking a strong bond requires a large input of energy, hence a large Ea.
298
Chapter 14 14.34
An elementary process is an actual collision event that occurs during the reaction. It is one of the key events that moves the reaction along in the stepwise process that leads to the overall reaction that is observed. It is thus one step in a potentially multi–step mechanism.
14.35
The rate–determining step in a mechanism is the slowest step.
14.36
The rate law for a reaction is based on the rate–determining step.
14.37
Adding all of the steps gives: 2NO + 2H2 J N2 + 2H2O The intermediates are N2O2 and N2O.
14.38
For such a mechanism, the rate law should be: rate = k[NO2][CO]
14.39
Since this is not the same as the observed rate law, this is not a reasonable mechanism to propose. Add all of the steps together: CO + O2 + NO J CO2 + NO2
14.40
On adding together all of the steps in each separate mechanism, we get, in each case: OCl– + I– J Cl– + OI–
14.41
The predicted rate law is based on the rate–determining step: rate = k[NO2]2
14.42
A catalyst changes the mechanism of a reaction, and provides a reaction path having a smaller activation energy.
14.43
A homogeneous catalyst is one that is present in the same phase as the reactants. It is used in one step of a cycle, but regenerated in a subsequent step, so that in a net sense, it is not consumed.
14.44
Adsorption is a clinging to a surface. Absorption involves a penetration below the surface, as in the action of a sponge. Heterogeneous catalysis involves adsorption.
14.45
The catalytic converter promotes oxidation of unburned hydrocarbons, as well as the decomposition of nitrogen oxide pollutants. Lead poisons the catalyst and renders it ineffective.
Review Problems
14.46
Since they are in a 1-to-1 mol ratio, the rate of formation of SO2 is equal and opposite to the rate of consumption of SO2Cl2. This is equal to the slope of the curve at any point on the graph (see below). At 200 min, we obtain a value of about 1 × 10–4 M/min. At 600 minutes, this has decreased to about 7 × 10–5 M/min.
299
Chapter 14
14.47
The slope of the tangent to the curve at each time is the negative of the rate at each time: Rate60 = 8.5 × 10–4 mol L–1 s–1 Rate120 = 4.0 × 10–4 mol L–1 s–1
14.48
This is determined by the coefficients of the balanced chemical equation. For every mole of N2 that reacts, 3 mol of H2 will react. Thus the rate of disappearance of hydrogen is three times the rate of disappearance of nitrogen. Similarly, the rate of disappearance of N2 is half the rate of appearance of NH3, or NH3 appears twice as fast as N2 disappears.
14.49
From the coefficients in the balanced equation we see that, for every mole of B that reacts, 2 mol of A are consumed, and three mol of C are produced. This means that A will be consumed twice as fast as B, and C will be produced three times faster than B is consumed. rate of disappearance of A = 2(–0.25) = –0.50 mol L–1 s–1 rate of appearance of C = 3(0.25) = 0.75 mol L–1 s–1
14.50
(a) (b)
rate for O2 = –1.35 mol L–1 s–1 × 19/2 = –12.8 mol L–1 s–1 By convention, this is reported as a positive number: 12.8 mol L–1 s–1 rate for CO2 = +1.35 mol L–1 s–1 × 12/2 = 8.10 mol L–1 s–1 300
Chapter 14 (c)
rate for H2O = +1.35 mol L–1 s–1 × 14/2 = 9.45 mol L–1 s–1
14.51
We rewrite the balanced chemical equation to make the problem easier to answer: N2O5 J 2NO2 + 1/2O2. Thus, the rates of formation of NO2 and O2 will be, respectively, twice and one half the rate of disappearance of N2O5. rate of formation of NO2 = 2(4.2 × 10–6) = 8.4 × 10–6 mol L–1 s–1 rate of formation of O2 = 1/2(4.2 × 10–6) = 2.1 × 10–6 mol L–1 s–1
14.52
(a)
−
(b)
Rate =
(a)
−
(b)
Rate =
14.53
14.54
d [CH3Cl] 1 d [Cl2] d [CCl4] 1 d [HCl] =− = = dt 3 dt dt 3 dt
1 (0.029) M s −1 = 9.7 x 10-3 M s-1 3
1 d [PH3] d [P4] 1 d [H 2 ] = = 4 dt dt 6 dt 1 (0.34) M s −1 = 0.085 M s-1 4
The rate can be found by simply inserting the given concentration values: rate = (5.0 × 105 L5 mol–5 s–1)[H2SeO3][I–]3[H+]2 rate = (5.0 × 105 L5 mol–5 s–1)(3.5 × 10–2 mol/L)(3.0 × 10–3 mol/L)3(1.5 × 10–3 mol/L)2 rate = 1.1 × 10–9 mol L–1 s–1
14.55
rate = (1.3 × 1011 L mol–1 s–1)(1.0 × 10–7 mol/L)(1.0 × 10–7 mol/L) rate = 1.3 × 10–3 mol L–1 s–1
14.56
rate = (7.1 × 109 L2 mol–2 s–1)(2.5 × 10–3 mol/L)2(6.2 × 10–2 mol/L) rate = 2.8 × 103 mol L–1 s–1
14.57
rate = (1.0 × 10–5 s–1)(2.0 × 10–3 mol/L) = 2.0 × 10–8 mol L–1 s–1
14.58
For a zero order reaction the rate is independent of concentration so R = k Therefore, the rate of the reaction is 6.4 x 102 M s-1.
14.59
Rate = k[Rn] = 0.0125 s-1 x 1.0 x 10-9 mol L-1 = 1.25 x 10-11 M s-1
14.60
On comparing the data of the first and second experiments, we find that, the concentration of N is unchanged, and the concentration of M has been doubled, causing a doubling of the rate. This corresponds to the fourth case in Table 14.4, and we conclude that the order of the reaction with respect to M is 1. In the second and third experiments, we have a different result. When the concentration of M is held constant, the concentration of N is tripled, causing an increase in the rate by a factor of nine. This constitutes the eighth case in Table 14.4, and we conclude that the order of the reaction with respect to N is 2. This means that the overall rate expression is: rate = k[M][N]2 and we can solve for the value of k by substituting the appropriate data: 5.0 × 10–3 mol L–1 s–1 = k × [0.020 mol/L][0.010 mol/L]2 k = 2.5 × 103 L2 mol–2 s–1
14.61
First, compare the first and second experiments: there has been an increase in concentration by a factor of 2. This has caused an increase in rate of:
301
Chapter 14 5.90 × 10−5 mol L−1 s −1 2.95 × 10 −5 mol L−1 s −1
= 2.00 = 21
This is the fourth case in Table 14.4, and the order of the reaction is found to be 1. Similarly, on comparing the first and third experiments, an increase in concentration by a factor of 3 has caused an increase in rate by a factor of: 8.85 × 10−5 mol L−1 s −1 = 3.00 = 31 2.95 × 10−5 mol L−1 s −1 This is the fifth case in Table 14.4, and we again conclude that the order is 1. rate = k[C3H6] We can use any of the three sets of data to solve for k. Using the first data set gives: 2.95 × 10–5 mol L–1 s–1 = k(0.050 mol L–1) k = 5.9 × 10–4 s–1 14.62
The reaction is first–order in OCl–, because an increase in concentration by a factor of 4.75, while holding the concentration of I– constant (compare the first and second experiments of the table), has caused an increase in rate by a factor of 4.751 = 4.75. The order of reaction with respect to I– is also 1, as is demonstrated by a comparison of the first and third experiments. rate = k[OCl–][I–] Using the last data set: 1.05 × 105 mol L–1 s–1 = k[1.6 × 10–3 mol/L][9.6 × 10–3 mol/L] k = 6.8 × 109 L mol–1 s–1
14.63
Compare the data of the first and second experiments, in which the concentration of NO is held constant and the concentration of O2 is increased by a factor of 4. Since this caused a rate increase by a factor of 28.4/7.10 = 41, we conclude that the order of the reaction with respect to O2 is one (case number six in Table 14.4). In the second and third experiments, an increase in the concentration of NO by a factor of 3 (while holding the concentration of O2 constant) caused a rate increase by a factor of 255.6/28.4 = 9. This is the eighth case in Table 14.4, and the order is seen to be two. rate = k[O2][NO]2 We can use any of the three sets of data to solve for k. Using the first data set gives: 7.10 mol L–1 s–1 = k[1.0 × 10–3 mol/L][1.0 × 10–3 mol/L]2 k = 7.10 × 109 L2 mol–2 s–1
14.64
Compare the first and second experiments. On increasing the ICl concentration by a factor of 6.5, the rate is found to increase by a factor of 6.5 = 6.51, and the order of the reaction with respect to ICl is 1 (see Tables 14.3 and 14.4). In the first and third experiments, the concentration of ICl is constant, whereas the concentration of H2 in the first experiment is 1.35 times that in the third. This causes a change in the rate by a factor of 1.36, and the rate law is found to be: rate = k[ICl][H2]. Using the data of the first experiment: 1.5 × 10–3 mol L–1 s–1 = k[0.12 mol L–1][0.12 mol L–1] k = 1.04 × 10–1 L mol–1 s–1
14.65
In the first, fourth, and fifth experiments, the concentration of OH– has been made to increase while the [(CH3)3CBr] is left unchanged. In each of these experiments, there is no change in rate. This means that the rate is independent of [OH–], and the order of reaction with respect to OH– is zero. The concentration of (CH3)3CBr increases by a factor of 2.42 from the first to the second experiment, and by a factor of 3.84 from the first to the third experiment, as the OH– concentration is held constant. There is a corresponding 2.4–fold (i.e. 2.41) increase in rate from the first to the second experiment, and there is a 3.8–fold (i.e. 3.81) 302
Chapter 14 increase in rate from the first to the third experiment. In both cases, we conclude that the order with respect to (CH3)3CBr is one. rate = k[(CH3)3CBr] Using the third set of data gives: 3.8 × 10–3 mol L–1 s–1 = k[7.3 × 10–1 mol/L] k = 5.2 × 10–3 s–1 14.66
A graph of ln [SO2Cl2]t versus t will yield a straight line if the data obeys a first–order rate law.
These data do yield a straight line when ln [SO2Cl2]t is plotted against the time, t. The slope of this line equals –k. Plotting the data provided and using linear regression to fit the data to a straight line yields a value of 1.32 × 10–3 min–1 for k. 14.67
Since it is the plot of 1/conc. that gives a straight line, the order of the reaction with respect to CH3CHO is two. The rate constant is given by the slope directly: k = 0.0771 M–1 s–1
14.68
(a)
The time involved must be converted to a value in seconds: 2 hr × 3600 s/hr = 7.2 × 103 s, and then we make use of equation 14.5, where x is taken to represent the desired SO2Cl2 concentration: 0.0065 M ln = (2.2 × 10−5 s −1 )(7.2 × 103 s) x x = 5.5 × 10–3 M
(b)
The time is converted to a value having the units seconds 24 hr × 3600 s/hr = 8.64 × 104 s, and then we use equation 14.5, where x is taken to represent the desired SO2Cl2 concentration: 0.0065 M ln = (2.2 × 10−5 s−1 )(8.64 × 104 s) x x = 9.7 × 10–4 M 303
Chapter 14
14.69
1 1 − = kt [ A]t [ A]0 1 1 = kt + [ A]t [ A]0 (a) 1
= 0.0771 M −1 s −1 x 30 min x
1
= 142.1 M −1
1
= 0.0771 M −1 s −1 x 180 min x
1
= 836.0 M −1
[ A]30 min
60 s 1 + min 0.300 M
[ A]30 min [ A]30min = 7.0 x 10−3 M (b)
[ A]180min
[ A]180min [ A]180min = 14.70
60 s 1 + min 0.300 M
1.2 x 10−3 M
Use equation 14.5 and assume the initial concentration is 100 and the final concentration is, therefore, 25. The units are immaterial in this case since they will cancel out in the calculation: [ A ]0 ln = kt [ A ]t 100 = k(85.0 min) 25 Solving for k we get 1.63 × 10–2 min–1 ln
14.71
1 1 − = kt [ A]t [ A]0 1 1 − = k x18.7 min 0.075 M 1 M
14.72
k = 0.66 M −1 min −1 Any consistent set of units for expressing concentration may be used in equation 14.5, where we let A represent the drug that is involved: [ A ]0 ln = kt [ A ]t
ln
25.0 mg kg
= k(120 min) 15.0 mg kg Solving for k we get 4.26 × 10–3 min–1
304
Chapter 14
14.73
ln
ln
[ A]0 [ A]t
= kt
1.1 x 10−6 g L−1 0.30 x 10−6 g L−1
= k x 180 s
k = 7.2 x 10−3 s −1 14.74
We use the equation: 1 1 = kt − HI HI [ ]t [ ]0
(
)
1 1 = 1.6 × 10−3 L mol−1 s −1 × t − ⎡9.5 × 10−4 M ⎤ ⎡ 6.7 × 10−2 M ⎤ ⎣ ⎦ ⎣ ⎦ Solving for t gives: t = 6.5 × 105 s or t = (6.5 × 105 s) × 1 min/60 s = 1.1 × 104 min
14.75
1 1 − = kt A A [ ]t [ ]0
1 1 − = 0.556 L mol−1 s −1 x t 0.050 M 0.50 M t = 32 s
14.76
Use the equation, taking time in minutes; 3 hr = 180 min. x ln = 4.26 × 10−3 min −1 (180 min ) 5.0 mg kg
(
)
x = 11 mg/kg 14.77
Use equation 14.10, where the time is: (2.5 × 103 min) × 60 s/min = 1.5 × 105 s 1 1 − = kt [ B]t [ B]0 1 −4
−
⎡3.7 × 10 M ⎤ ⎣ ⎦ 1 = 2.5 × 103 L mol−1 HI [ ]0
[ HI]0 14.78
(
)(
1 = 1.6 × 10−3 L mol−1 s −1 1.5 × 105 s [ HI]0
= 4.1 × 10−4 M
⎛ 60 min ⎞ ⎛ 1 half life ⎞ half lives = (2.0 hrs) ⎜ = 6.0 half lives ⎝ 1 hr ⎟⎠ ⎜⎝ 20 min ⎟⎠ Six half lives correspond to the following fraction of original material remaining:
Fraction remaining 1/2 1/4 1/8 1/16 1/32 1/64
Number of half lives 1 2 3 4 5 6 305
)
Chapter 14
Alternatively, since 6 half-lives have occurred 0.56 = 0.0156
or 1/64
14.79
Since 1/2 of the Sr–90 decays every half–life, it will take 5 half–lives, or 5 × 28 yrs = 140 yrs, for the Sr–90 to decay to 1/32 of its present amount.
14.80
It requires approximately 500 min (as determined from the graph) for the concentration of SO2Cl2 to decrease from 0.100 M to 0.050 M, i.e., to decrease to half its initial concentration. Likewise, in another 500 minutes, the concentration decreases by half again, i.e. from 0.050 M to 0.025 M. This means that the half–life of the reaction is independent of the initial concentration, and we conclude that the reaction is first–order in SO2Cl2.
14.81
From the graph, we see that the first half–life is achieved in about 65 seconds, because this is the amount of time it requires for the concentration to decrease from its initial value (0.200 M) to half its initial value (0.100 M). The second half–life period, i.e., the time required for the concentration to decrease from 0.100 M to 0.050 M, is longer, namely about 125 seconds (this is determined by estimating the time when the concentration is 0.050 M and subtracting the time when the concentration is 0.100 M). Since the half–life does depend on concentration; we conclude that the reaction is not first–order. In fact, the data are consistent with second–order kinetics, because the value of the half–life decreases in proportion to the inverse of the initial concentration. That is, as the initial concentration for any half–life period becomes smaller, the half–life becomes larger. t
14.82
⎡ 1 ⎤ t1/2 ⎢⎣ 2 ⎥⎦ = 0.05 t
⎡ 1 ⎤10 hrs = 0.05 ⎢⎣ 2 ⎥⎦ t log 0.5 = log 0.05 10 hrs
t = 43.2 hrs 14.83
We can use the half-life to determine the value of k. 4.88 x 103 s = 0.693/k
[ A]0 ln [ A]t 14.84
= kt
ln
k = 1.42 x 10-4 s-1
0.024 M = 1.42 x 10−4 x t 0.0040 M
t = 1.3 x 104 s
In order to solve this problem, it must be assumed that all of the argon–40 that is found in the rock must have come from the potassium–40, i.e., that the rock contains no other source of argon–40. If the above assumption is valid, then any argon–40 that is found in the rock represents an equivalent amount of potassium–40, since the stoichiometry is 1:1. Since equal amounts of potassium–40 and argon–40 have been found, this indicates that the amount of potassium–40 that remains is exactly half the amount that was
306
Chapter 14 present originally. In other words, the potassium–40 has undergone one half–life of decay by the time of the analysis. The rock is thus seen to be 1.3 × 109 years old. 14.85
Using equation 14.9 we may determine how long it has been since the tree died. ⎛r ⎞ ln ⎜ o ⎟ = 1.2 × 10 −4 t where ro = 1.2 x 10−12 ⎝ rt ⎠
(
)
⎛ 1.2 × 10−12 ⎞ ln ⎜ ⎟ = 1.21 × 10−4 t ⎜ 4.8 × 10−14 ⎟ ⎝ ⎠ Taking the natural log we determine:
(
)
⎛ ⎞ ⎛ 1.2 × 10−12 1 t=⎜ ⎟ ln ⎜ ⎝ 1.21 × 10−4 ⎠ ⎜⎝ 4.8 × 10−14
14.86
⎞ ⎟ = 2.7 × 104 yr ⎟ ⎠
Use equation 14.8 and 14.9. r ln 0 = kt rt ⎛ 1.2 × 10−12 ⎞ ln ⎜ ⎟ = 1.21 × 10−4 y −1 9.0 × 103 y ⎜ ⎟ r t ⎝ ⎠ − 12 ⎛ 1.2 × 10 ⎞ ⎜ ⎟ = exp ⎡⎢ 1.21 × 10−4 y−1 9.0 × 103 y ⎤⎥ ⎜ ⎟ ⎣ ⎦ r t ⎝ ⎠ ⎛ 1.2 × 10−12 ⎞ ⎜ ⎟ = rt = 4.0 × 10−13 ⎜ ⎟ 2.97 ⎝ ⎠
(
)(
(
)
)(
)
14.87
In a fashion similar to that outlined in the answer to Review Problem 14.84, we conclude that, in order for the rock to be one half–life old (1.3 × 109 yr), there must be equal amounts of the two isotopes, one having been formed by decay of the other. The answer is, thus, 1.16 × 10–7 mol of potassium–40.
14.88
The graph is prepared exactly as in example 14.12 of the text. The slope is found using linear regression, to be: –9.5 × 103 K. Thus –9.5 × 103 K = –Ea/R Ea = –(–9.5 × 103 K)(8.314 J K–1 mol–1) = 7.9 × 104 J/mol = 79 kJ/mol Using the equation, we proceed as follows: −Ea ⎡ 1 k 1⎤ ln 2 = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦
⎡ 1.94 × 10−3 L mol−1 s −1 ⎤ −E a 1 ⎤ ⎡ 1 ln ⎢ − ⎥ = −4 −1 −1 −1 −1 ⎢ 673 K 593 K ⎥⎦ ⎣ ⎣⎢ 2.88 × 10 L mol s ⎦⎥ 8.314 J mol K 2.00 × 10−4 K −1
× Ea 8.314 J mol−1 K −1 Ea = 7.93 × 104 J/mol = 79.3 kJ/mol 1.907 =
14.89
The graph is prepared exactly as in example 14.15 of the text. The slope is found using linear regression, to be: –1.67 × 104 K. Thus –1.67 × 104 K = –Ea/R 307
Chapter 14 Ea = –(–1.67 × 104 K)(8.314 J K–1 mol–1) = 1.39 × 105 J/mol = 139 kJ/mol Using equation 13.12 we have: −Ea ⎡ 1 k2 1⎤ − = ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ ⎡ 1.08 × 10−1 L mol−1 s −1 ⎤ −E a 1 ⎤ ⎡ 1 ln ⎢ − ⎥ = −2 −1 −1 −1 −1 ⎢ 503 K 478 K ⎥⎦ ⎣ ⎣⎢ 1.91 × 10 L mol s ⎦⎥ 8.314 J mol K
ln
1.732 =
1.04 × 10−4 K −1
× Ea
8.314 J mol−1 K −1
Ea = 1.38 × 105 J/mol = 138 kJ/mol 14.90
Using the equation we have: −Ea ⎡ 1 k2 1⎤ = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ ⎡ 1.0 × 10−3 L mol−1 s −1 ⎤ −Ea 1 ⎤ ⎡ 1 ln ⎢ − ⎥ = −5 −1 −1 −1 −1 ⎢ 403 K 373 K ⎥⎦ ⎣ ⎢⎣ 9.3 × 10 L mol s ⎥⎦ 8.314 J mol K ln
2.37 =
2.00 × 10−4 K −1 8.314 J mol−1 K −1
× Ea
Ea = 9.89 × 104 J/mol = 99 kJ/mol ⎛ −E ⎞ Equation states k = A exp ⎜ a ⎟ ⎝ RT ⎠ k A= ⎛ −E ⎞ exp ⎜ a ⎟ ⎝ RT ⎠
=
9.3 × 10−5 L mol−1 s −1 ⎛ −9.89 × 104 J ⎞ mol ⎟ exp ⎜ ⎜ 8.314 J mol K ( 373 K ) ⎟ ⎝ ⎠
(
)
9
= 6.6 × 10 L mol
−1
s −1
⎛ ⎞ −9.89 × 104 J / mol k = 6.6 × 109 L mol−1 s −1 exp ⎜ ⎟ ⎜ 8.314 J mol−1 K −1 x 423 K ⎟ ⎝ ⎠ k = 4.0 × 10−3 L mol-1 s-1
308
Chapter 14 14.91
(a)
Substituting into the equation: −Ea ⎡ 1 k 1⎤ − ln 2 = ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ ⎡ 1.1 × 10−5 L mol−1 ln ⎢ −6 −1 ⎢⎣ 1.3 × 10 L mol 6.34 × 10−5 K −1
× Ea 8.314 J mol−1 K −1 Ea = 2.8 × 105 J/mol = 2.8 × 102 kJ/mol ⎛ −E ⎞ Equation states k = A exp ⎜ a ⎟ ⎝ RT ⎠ 2.1 =
(b)
−Ea s−1 ⎤ 1 ⎤ ⎡ 1 − ⎥ = −1 −1 −1 ⎢ 703 K 673 K ⎥⎦ ⎣ s ⎥⎦ 8.314 J mol K
A=
=
k ⎛ −E ⎞ exp ⎜ a ⎟ ⎝ RT ⎠
1.1 × 10−5 L mol−1 s−1 ⎛ −2.8 × 105 J ⎞ mol ⎟ exp ⎜ ⎜ 8.314 J mol K ( 703 K ) ⎟ ⎝ ⎠
(
)
= 7.0 × 1015 L mol −1 s −1
(c)
The equation can be used directly now that the value of A is known: ⎛ −E ⎞ k = A exp ⎜ a ⎟ ⎝ RT ⎠ ⎛ −2.8 × 105 J mol−1 ⎞ ⎟ = 7.0 × 1015 L mol−1 s −1 ⎜ ⎜ 8.314 J mol K ( 598 K ) ⎟ ⎝ ⎠
(
)
(
)
= 2.4 × 10 −9 L mol −1 s − 1
14.92
Use equation 14.13: (a)
⎛ −E ⎞ k = A exp ⎜ a ⎟ ⎝ RT ⎠ ⎛ −103 × 103 J mol−1 ⎞ ⎟ = 4.3 × 1013 s −1 exp ⎜ ⎜ 8.314 J mol K ( 310 K ) ⎟ ⎝ ⎠
(
)
(
)
= 1.9 × 10 −4 s −1
(b)
⎛ −E ⎞ k = A exp ⎜ a ⎟ ⎝ RT ⎠ ⎛ −103 × 103 J mol−1 ⎞ ⎟ = 4.3 × 1013 s −1 exp ⎜ ⎜ 8.314 J mol K ( 390 K ) ⎟ ⎝ ⎠
(
)
(
)
= 6.9 × 10 −1 s −1
309
Chapter 14
14.93
Substitute into the equation: −Ea ⎡ 1 k 1⎤ ln 2 = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ k2 ⎡ ⎤ −108 × 103 J mol−1 ⎡ 1 1 ⎤ − ln ⎢ = −5 − 1 ⎥ −1 −1 ⎢ 328 K 308 K ⎥⎦ ⎣ 8.314 J mol K ⎣ 6.2 × 10 s ⎦ ln(k2/6.2 × 10–5 s–1) = 2.57 k2 = 6.2 × 10–5 s–1 × exp(2.57) = 8.1 × 10–4 s–1
14.94
If this reaction occurs in one step then the rate law is the product of the concentrations raised to the appropriate powers. Rate = k [AB][C]
14.95
See Review Question 14.41 for information about this mechanism. The know rate law is Rate = k[NO2]2. The proposed mechanism is: NO2 + NO2
→
NO3 + NO (slow)
NO3 + CO
→
CO2 + NO2
(fast)
From the slow, rate determining step, we can write a proposed rate law. Rate = k[NO2]2 14.96
An intermediate is produced in one step and used up in another. The intermediate in this reaction is N2O2. The overall reaction is the sum of the two steps 2NO(g) + O2(g)
'
2NO2(g)
The rate law is determined from the slow step. Rate = k[N2O2][O2] An intermediate cannot be part of the rate law expression. We can use the first step, which is an equilibrium step to solve for the concentration of N2O2. Rate =
k f [NO]2 = kr [N 2 O2 ]
Solve for [N2O2] to get
[N2O2] =
kf kr
[NO]2
Rate = k[NO]2 [O 2 ] where all of the constants have been combined into a new constant 14.97
An intermediate is formed in one step of a mechanism and then used in another step. Intermediates in this mechanism are Cl and CCl3 Sum the three steps to obtain the overall, balanced equation for the reaction. 310
Chapter 14
Cl2(g) + CHCl3(g)
→
HCl(g) + CCl4(g)
The rate law is obtained from the slow step. Rate = k[Cl][CHCl3] However, Cl is an intermediate and does not occur in the overall reaction. To obtain a rate equation in terms of reactants or products we need to replace [Cl]. 2 Cl] [ K eq = [Cl2 ]
From the first step,
[Cl] = {K eq [Cl2 ]}
1/ 2
{
1/ 2 } [CHCl3 ] = kK eq1/ 2 [Cl2 ]1/ 2 [CHCl3 ] = k ' [Cl2 ]1/ 2 [CHCl3 ]
Rate = k K eq [ Cl 2 ]
Additional Exercises
14.98
From the tangent lines we can obtain the instantaneous rate at 1000 s and 9000 s. For the decomposition of cyclobutane: At 1000 s the slope is − 9.6 x 10-7 M s-1 which is the rate of decomposition. At 9000 s the slope is − 3.8 x 10-7 M s-1 which is the rate of decomposition For the formation of ethylene:
311
Chapter 14 At 1000 s the slope is 1.9 x 10-6 M s-1 which is the rate of formation of. At 9000 s the slope is 7.5 x 10-7 M s-1 The rate of formation of ethylene is twice the rate of decomposition of cyclobutane. 14.99
k = 0.693/12.5 y = 5.54 × 10–2 y–1 ln
(
)
1 = 5.54 × 10−2 y −1 × t 0.1
t = 41.6 y
14.100
Reaction is endothermic 14.101 From Section 14.4, the fraction remaining after n half-lives is 1/2n. Therefore, we have: 0.810 = 1/2n Inverting both sides of the equation gives: 1.23 = 2n We can solve this for now using trial and error: We know that 21 = 2, and that 20 = 1. Therefore n must be between 0 and 1. We start by trying n = 0.50: 20.50 = 1.41 This is too high, so we might try 0.30: 20.30 = 1.23 This is just what we were looking for, therefore n = 0.30. Since one half life of C-14 is 5730 years, one might estimate the age of the mummy as (0.30)(5730) = about 1,700 years old. 14.102 The chemical product is BaCl2.Recall that for a first order process k = So k = 0.693/30 yr = 2.30 × 10–2/yr. Then, [A] 0 ln = kt [A] t [A] t = [A] 0 exp(- kt) [A] 0 = exp[− (2.30 × 10 2 /yr)(120 yr)] [A] t
312
0.693 t1 2
Chapter 14
[A] 0 = 6.33 × 10 2 [A] t so 6.3% of the original sample remains.
14.103 (a) The reaction rate is first order so as the concentration of reactants doubles the rate will double. (b) The reaction rate is second order so as the concentration of reactants doubles the rate will increase by a factor of four. (c) The reaction rate is zeroth order so the rate is independent of concentration. There is no change in the rate. (d) The reaction rate is second order so as the concentration of reactants double the rate increases by a factor of four. (e) The reaction rate is third order so as the concentration of reactants double the rate increases by a factor of eight. 14.104 NO2 + O3 J NO3 + O2 NO3 + NO2 J N2O5
(slow) (fast)
rate = k1[A]2 rate = k–1[A2]1 rate = k2[A2]1[E]1 2A + E J B + C The rates for the forward and reverse directions of step one are set equal to each other in order to arrive at an expression for the intermediate [A2] in terms of the reactant [A]: k1[A]2 = k–1[A2] k [A2] = 1 [A]2 k −1 This is substituted into the rate law for question (c) above, giving a rate expression that is written k using only observable reactants: rate = k2 1 [A]2[E]1 k −1
14.105 (a) (b) (c) (d) (e)
14.106 First, determine a value for Ea using the equation: −Ea ⎡ 1 k2 1⎤ = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ ⎡ 2.25 × 10−5 min −1 ⎤ −Ea 1 ⎤ ⎡ 1 − ln ⎢ ⎥ = −6 −1 −1 −1 ⎢ 348 K 338 K ⎥⎦ ⎣ ⎣⎢ 5.84 × 10 min ⎦⎥ 8.314 J mol K
ln
1.35 =
8.50 × 10−5 K −1 8.314 J mol−1 K −1
× Ea
Ea = 1.32 × 105 J/mol = 132 kJ/mol Next, use this value of Ea and the data at 75 °C to calculate a rate constant at 85 °C: ln
−Ea ⎡ 1 k2 1⎤ = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦
313
Chapter 14 k2 ⎡ ⎤ −1.32 × 105 J mol−1 ⎡ 1 1 ⎤ ln ⎢ = ⎢ 358 K − 348 K ⎥ −5 −1 ⎥ −1 −1 ⎣ ⎦ 8.314 J mol K ⎣ 2.25 × 10 min ⎦ k2 ⎡ ⎤ ln ⎢ = 1.27 −5 −1 ⎥ ⎣ 2.25 × 10 min ⎦
k2 = (2.25 × 10–5 min–1) × exp(1.27) = 8.02 × 10–5 min–1. Finally, use the first–order rate expression to determine time: ln
0.0025 M = (8.02 × 10−5 min −1 ) × t 0.0015 M
Solving for t we get 6.37 × 103 min. 14.107 Taking the log of both sides of this equation, we have: log(rate) = log k + n × log[A] Thus, if we plot log(rate) vs. log[A], the slope should be equal to the value for n, and the intercept is equal to log k. A plot of the data is used to determine slopes of the tangents to the curve at time = 150, 300, 450, and 600 min. The negatives of these slopes are equal to the value of the rate constants at these particular times during the reaction. (Notice that the rate decreases with time.) The four rates are found to be approximately: 1.1 × 10–4 mol L–1 min–1, at t = 150 min 8.9 × 10–5 mol L–1 min–1, at t = 300 min 7.2 × 10–5 mol L–1 min–1, at t = 450 min 6.0 × 10–5 mol L–1 min–1, at t = 600 min Suitable values for [SO2Cl2] at each of these times can be read from the graph of concentration vs. time. Once each of these concentration values has been converted to log[SO2Cl2], the plot of log rate vs. log [SO2Cl2] can be constructed:
The slope of this straight line has the value 1.0, which is the order of the reaction with respect to [SO2Cl2].
314
Chapter 14 14.108 −Ea ⎡ 1 k2 1⎤ = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ −Ea 1 ⎤ ⎡2⎤ ⎡ 1 ln ⎢ ⎥ = − ⎢ 1 1 − − 298 K ⎥⎦ ⎣ 1 ⎦ 8.314 J mol K ⎣ 308 K –5 (1.31 × 10 mol/J)(Ea) = 0.693 Ea = 5.29 × 104 J/mol = 52.9 kJ/mol ln
14.109 The rate law for a first order reaction is given by, Rate = k[A]. If the rate constant is doubled then the rate of the reaction will double. 14.110 The catalyzed reaction path is represented by the blue line. A catalyzed reaction path is lower in energy than an un-catalyzed reaction path. Since there are two activation energies shown for the catalyzed reaction there are two steps in the mechanism. The slowest step would be the first step in the mechanism as that step has the highest activation energy of the two steps in the mechanism. 14.111 To solve this problem, plot the data provided as 1/T vs 1/t where T is the absolute temperature and 1/t is proportional to the rate constant. t (min) 10 9 8 7 6
T (K) 291 293 294 295 297
1/T 0.003436 0.003412 0.003401 0.003389 0.003367
ln(1/t) –2.302585 –2.197224 –2.079441 –1.945910 –1.791759
13.6
288
0.003472
–2.608
315
Chapter 14 The slope of the graph is equal to –Ea/R, therefore: –7,704 = –Ea/R 7,704R = Ea 7,704 K(8.314 J/mol·K) = Ea Ea = 64,050 J/mol Ea = 64 kJ/mol From the straight–line equation, we can determine the time needed to develop the film at 15 °C is 14 min. 14.112 (a)
The number of chirps in eight seconds is simply the temperature minus 4. So, in order, there will be 6, 11, 16 and 21 chirps at these temperatures.
The data are plotted below. (b) T 1/T ln (chirps) # of chirps 6 283 0.003534 1.791759 11 288 0.003472 2.397895 16 293 0.003413 2.772589 21 298 0.003356 3.044522
The slope of the line is equal to the Activation energy divided by R so Ea = 5.81 × 104 J/mol. (c)
At 35 °C the cricket would make 32 chirps.
14.113 Taking note of the inverse relationship between the reaction rate constant, k, and the cooking time, t, we set up equation 14.15 in the following manner: −Ea ⎡ 1 k 1⎤ ln 2 = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦ ⎡ 1t ⎤ −Ea ⎡ 1 1⎤ ln ⎢ 2 ⎥ = − ⎢ ⎥ 1 R T T ⎢ t1 ⎥ 1⎦ ⎣ 2 ⎣ ⎦ We are provided with some subtle but key information about the physical conditions. For instance, the 3– minute traditional egg provides a cooking time of 3 minutes at the normal boiling point of water, 100 °C or 373 K. We are also given the atmospheric pressure (355 torr) on Mt. McKinley where the cooking is to be carried out at a lower temperature. At 355 torr, H2O will boil when its vapor pressure equals 355 torr. The temperature corresponding to this pressure is 80 °C or 323 K (See Appendix). Thus, with the given value of the activation energy, i.e., Ea = 418 kJ/mol, we can proceed with the calculation to obtain t2:
316
Chapter 14
⎡ 1t ⎤ −418 × 103 J/mol ⎡ 1 1 ⎤ 2 ⎥ = ln ⎢ − ⎢ 1 8.314 J/mol K 373 K ⎥⎦ ⎢⎣ 3 min ⎥⎦ ⎣ 353 K ⎡ 3 min ⎤ ln ⎢ ⎥ = − 7.64 ⎣ t2 ⎦ 3 min = exp( − 7.64) t2 t 2 = 3 min/exp( − 7.64) = 6.2 × 103 min = 104 hrs Thus, to get the same degree of protein denaturization, it would take roughly 4 days to cook the egg at an atmospheric pressure of 355 torr as opposed to cooking the egg at normal atmospheric pressure.
14.114 (a) (b) (c)
The first step, in which a free radical is produced, is the initiation step. Both the second and third steps are propagating steps since HBr, the desired product, and an additional free radical are produced. The final step in which two bromine free radicals recombine to give a bromine molecule is the termination step.
The presence of the additional reaction step serves to decrease the concentration of HBr. 14.115 Consider a 1.00 g sample of radioactive material. After 10 half lives you would have (1/2)10 grams or 9.76 x 10-4 grams of material remaining. Therefore, the percent of material remaining would be, ⎛ 9.76 x 10 −4 g ⎞ −2 ⎜⎜ ⎟⎟ x 100 = 9.76 x 10 % 1.00 g ⎝ ⎠
Multi-Concept Problems
14.116 The units on the rate constant tells us that this is a second order reaction. Rate = k [C2H5OH]2 Since all chemicals are in the gaseous state the total pressure, at any given time, is given by, PT = P(C2H5OH) + P(C2H4) + P(H2). We can relate pressure to molar concentration using PV=nRT P = (n/V)RT = [X]RT or [X] = P/(RT) At time t the ethanol concentration we be reduced by –x due to conversion to ethylene and hydrogen and each of the products will increase in concentration by +x. Now set up a table showing initial molar concentrations and concentrations at time t.
Initial
[C2H5OH]
[C2H4]
[H2]
0.020
0
0
+x
+x
Time t 0.020 – x
Using our equation for total pressure, PT, we can write, 1.4 atm = (0.0200 – x)RT + xRT + xRT 1.4/(RT) = 0.0200 + x 317
Chapter 14 x =1.4/(0.0821 L atm K-1 mol-1 x (327 + 273)K - 0.0200M x=0.00842M
Thus, during the time necessary to achieve a pressure of 1.4 atm the reaction has converted 0.00842 moles per liter to products. [C2H5OH]0 = 0.0200M
[C2H5OH]t =0.0158M
1 1 − = kt [ A]t [ A]0
1 1 − = 4.00 x 10−5 L mol−1 s −1 t [0.0158]t [0.0200]0
t = 3.32 x 105 s or 92 hrs
14.117 We can determine the temperature that the reaction mixture would achieve if all of the heat is supplied at once by using the enthalpy of reaction and the heat capacity of the system. q = Ccal∆T
40 kJ mol-1 x 1000 J kJ-1 x 10,000 mol = 7.5 x 106 J oC-1 ∆T ∆T = 53.3 oC Thus, the final temperature of the mixture would be 78 oC or 351 K ln
ln
ln
−Ea ⎡ 1 k2 1⎤ = − ⎢ ⎥ k1 R ⎣ T2 T1 ⎦
1000 J 1 ⎤ ⎡ 1 kJ ⎢ 351 K − 298 K ⎥ −1 −1 ⎣ ⎦ 8.314 J K mol
−50 kJ mol−1 x
k2
=
k2
= 3.047
1x10−5 s−1
1x10−5 s−1
k2 = 2 x 10-4 s-1 If the temperature of the reaction mixture was 182 oC and all of the heat released at once the final temperature would be: 182 oC + 53 oC = 235 oC so the reaction would indeed reach 199 oC.
318
Chapter 15
Practice Exercises 15.1
2N2O3 + O2 4NO2
[ H 2 O]2 = K c [ H 2 ]2 [O2 ]
[CO2 ][ H 2O]2 [CH 4 ][ O2 ]2
(b)
= Kc
15.2
(a)
15.3
Since the starting equation has been reversed and divided by two, we must invert the equilibrium constant, and then take the square root: Kc = 1.2 × 10–13
15.4
If we divide both equations by 2 and reverse the second we get: Kc = 5.7 × 1045 Kc = 3.3 × 10–41
CO(g) + 1/2O2(g) CO2(g) H2O(g) H2(g) + 1/2O2(g)
Note that when we divide the equation by two, we need to take the square root of the rate constant. When we reverse the reaction, we need to take the inverse. Adding these equations we get the desired equation so we need simply multiply the values for Kc in order to obtain the new value: Kc = 1.9 × 105 2
15.5
( PN O ) KP = 2 ( PN ) ( PO ) 2
2
15.6
KP =
( PHI )2
( PH )( PI ) 2
15.7
2
2
Use the equation K p = K c ( RT )
K p = K c ( RT ) 15.8
∆n g
∆n g
. In this reaction, ∆ng = 3 – 2 = 1, so
(
= 7.3 × 1034
) (( 0.0821
L atm mol K
1
) ( 298 K ) )
= 1.8 × 1036
We would expect KP to be smaller than Kc since ∆ng is negative. Use the equation:
K p = K c ( RT ) Kc =
∆n g
Kp
( RT )∆n
g
In this case, ∆ng = (1 – 3) = –2, and we have: Kp 3.8 × 10−2 Kc = = = 57 −2 L atm ( RT )∆n g 0.0821 mol K ( 473 K )
((
15.9
Kc =
)
1 NH3 ( g ) HCl ( g ) 319
)
Chapter 15
15.10
1
(a)
Kc =
(b)
K c = Na + (aq) OH − (aq) [ H 2 (g) ]
(c)
K c = Ag + CrO 42 −
(d)
Ca +2 (aq) HCO − (aq) 3 Kc = CO (aq) [ 2 ]
[Cl2 (g)] 2
15.11
2
To solve this problem, the first step is to find Q, the mass action expression, and then compare it to the value for Kc. If Q is larger than Kc, then there are more products than the equilibrium concentration and the reaction will move to reactants. If Q is smaller than Kc, then there are more reactants than the equilibrium concentrations and the reaction will move to products. For this reaction, the concentrations of H2, Br2 and HBr are all equal. If we set them to x, then we see that they cancel to give Q = 1: Q=
[ HBr ]2 = [ x ]2 = 1 [ H 2 ][ Br2 ] [ x ][ x ]
Q is larger than Kc, therefore the reaction will move to reactants. 15.12
Reaction (a) will proceed the least amount towards completion followed by reaction (c) and finally, reaction (b) will proceed farthest to completion since it has the largest value for Kc.
15.13
Only the gases will be affected by the volume of the container. But the stoichiometric ratio of the reactants to products is 5:5, so as the volume is changes, the number of moles of gas will not change, therefore the reaction will not change, and there will be no change in the amount of H3PO4.
15.14
(a) (b) (c)
(d)
The equilibrium will shift to the right, decreasing the concentration of Cl2 at equilibrium, and consuming some of the added PCl3. The value of Kp will be unchanged. The equilibrium will shift to the left, consuming some of the added PCl5 and increasing the amount of Cl2 at equilibrium. The value of Kp will be unchanged. For any exothermic equilibrium, an increase in temperature causes the equilibrium to shift to the left, in order to remove energy in response to the stress. This equilibrium is shifted to the left, making more Cl2 and more PCl3 at the new equilibrium. The value of Kp is given by the following: PPCl5 Kp = PPCl3 × PCl 2 In this system, an increase in temperature (which causes an increase in the equilibrium concentrations of both PCl3 and Cl2 and a decrease in the equilibrium concentration of PCl5) causes an increase in the denominator of the above expression as well as a decrease in the numerator of the above expression. Both of these changes serve to decrease the value of Kp. Decreasing the container volume for a gaseous system will produce an increase in partial pressures for all gaseous reactants and products. In order to lower the increase in partial pressures, the equilibrium will shift so as to favor the product side having the smaller number of gaseous molecules, in this case to the right. This shift will decrease the amount of Cl2 and PCl3 at equilibrium, and it will increase the amount of PCl5 at equilibrium. While the change in volume will change the position of the equilibrium, it does not change the value for Kp.
320
Chapter 15
15.15
2CO(g) + O2(g) 2CO2(g) Using the stoichiometry of the reaction we can see that for every mol of O2 that is used, twice as much CO will react and twice as much CO2 will be produced. Consequently, if the [O2] decreases by 0.030 mol/L, the [CO] decreases by 0.060 mol/L and [CO2] increases by 0.060 mol/L.
15.16
Kc =
15.17
(a)
(b)
(c)
(d)
15.18
Kc =
[CO2 ][ H 2 ] [CO][ H 2 O]
=
(0.150)(0.200) = 4.06 (0.180)(0.0411)
The initial concentrations were: [PCl3] = 0.200 mol/1.00 L = 0.200 M [Cl2] = 0.100 mol/1.00 L = 0.100 M [PCl5] = 0.00 mol/1.00 L = 0.000 M The change in concentration of PCl3 was (0.200 – 0.120) M = 0.080 mol/L. The other materials must have undergone changes in concentration that are dictated by the coefficients of the balanced chemical equation, namely: PCl3 + Cl2 PCl5 so both PCl3 and Cl2 have decreased by 0.080 M and PCl5 has increased by 0.080 M. As stated in the problem, the equilibrium concentration of PCl3 is 0.120 M. The equilibrium concentration of PCl5 is 0.080 M since initially there was no PCl5. The equilibrium concentration of Cl2 equals the initial concentration minus the amount that reacted, 0.100 M – 0.080 M = 0.020 M. PCl5 (0.080) Kc = = = 33 (0.120)(0.020) PCl3 [ PCl2 ]
[ NO2 ]2 [ N 2O4 ] –3
4.61 × 10 =
[ NO2 ]2
0.0466 2 [NO2] = 1.04 × 10–2 M 15.19
CH3CO 2 C2 H5 [ H 2 O ] (0.910)(0.00850) Kc = = = 4.10 C2 H5 OH (0.210) C 2 H5 OH CH3CO 2 H [C2H5OH] = 8.98 × 10–3 M
15.20
Initially we have [H2] = [I2] = 0.200 M.
I C E
[H2] 0.200 –x 0.200–x
[I2] 0.200 –x 0.200–x
[HI] – +2x +2x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ HI]2 [ H 2 ][ I2 ]
=
(2x)2 = 49.5 ( 0.200 − x )( 0.200 − x )
321
Chapter 15
Take the square root of both sides of this equation to get;
2x = 7.04 . This equation is easily ( 0.200 − x )
solved giving x = 0.156. The substances then have the following concentrations at equilibrium: [H2] = [I2] = 0.200 – 0.156 = 0.044 M, [HI] = 2(0.156) = 0.312 M. 15.21
Initially we have [H2] = 0.200 M, [I2] = 0.100 M.
I C E
[H2] 0.200 –x 0.200–x
[I2] 0.100 –x 0.100–x
[HI] – +2x +2x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ HI]2 [ H 2 ][ I2 ]
=
(2x)2 = 49.5 ( 0.200 − x )( 0.100 − x )
4x2 = 49.5(0.0200 – 0.300x + x2) 45.5x2 –14.9x + 0.990 = 0 Solve the quadratic equation:
−b ± b2 − 4ac − ( −14.9 ) ± x= = 2a
(14.9 )2 − 4 ( 45.5 )( 0.990 ) 2 ( 45.5 )
= 0.0.0934
The substances then have the following concentrations at equilibrium: [H2] = 0.200 – 0.0934 = 0.107 M [I2] = 0.100 – 0.0934 = 0.0066 M [HI] = 2(0.0.0934) = 0.1868 M. 15.22
2NH3(g) N2(g) + 3H2(g) I C E
[NH3] 0.041 –2x 0.041–x
[N2] – +x x
[H2] – +3x +3x
Substitute the values for the equilibrium concentrations into the mass action expression: Kc =
[ N 2 ][ H 2 ]3 2
3 x )( 3x ) ( = ( 0.041 − x )2
= 2.3 × 10–9
NH3 We will assume that x is small compared to the concentration of NH3, so the equation will simplify to:
Kc =
27x 4
( 0.041)
2
= 1.4 × 10–13
4
x = 1.4 × 10–13 x = 6.2 × 10–4 [N2] = 6.2 × 10–4 [H2] = 1.9 × 10–3 15.23
N2(g) + O2(g) 2NO(g)
I C E
[N2] 0.033 –x 0.033–x
[O2] 0.00810 –x 0.00810–x 322
[NO] – +2x +2x
Chapter 15
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ NO]2 [ N 2 ][ O2 ]
=
(2x)2 = 4.8 × 10−31 ( 0.033 − x )( 0.00810 − x )
If we assume that x << 0.033 and x << 0.00810, we can simplify this equation. (Because the value of Kc is so low, this assumption should be valid.) The equation simplifies as:
Kc =
(2x)2 = 4.8 × 10−31 ( 0.033)( 0.00810 )
This equation is easily solved to give x = 5.7 × 10–18 M. The equilibrium concentration of NO is 2x according to the ICE table so, [NO] = 1.1 × 10–17 M. Review Questions 15.1
See Figure 15.1.
15.2
The reactants are the substances to the left of the arrow in a chemical equation, and in terms of equilibrium, the concentration of the reactants are in the denominator in the mass action expression. The products are the substances to the right of the arrow in a chemical equation, and in terms of equilibrium, the concentration of the reactants are in the numerator in the mass action expression.
15.3
An equilibrium law is the statement that the reaction quotient, i.e., the value of Kc must be equal to the numerical value of the mass action expression once equilibrium is attained. The reaction quotient is the numerical value of the mass action expression.
15.4
The reaction quotient becomes equal to the value of Kc once equilibrium has been attained.
15.5
If the value of Q is less than Kc, then the ratio of the numerator, or products, to the dominator, or reactants, is less than the ratio for the equilibrium values. In order to restore equilibrium the reaction must move to increase the ratio, or increase the amount of products. Therefore, the reaction will proceed in the forward direction.
15.6
By convention, the products are always written into the numerator and the reactants are written into the denominator of the mass action expression.
15.7
This equilibrium constant is small, and we do not expect the equilibrium to favor products.
15.8
The increasing tendency to go to completion is: (a) < (c) < (b), based on the relative magnitudes of Kc.
15.9
K p = K c × ( RT )
∆n g
where ∆ng is equal to the change in the number of moles of gaseous material on going from reactants to products. The value to be used for the gas constant, R, is 0.0821 L atm K–1 mol–1. 15.10
For the ith component of a gas mixture, the partial pressure is given by the ideal gas law:
Pi =
n i RT = MiRT, since ni/Vi is equal to molarity, Mi. Vi
The proportionality constant between Pi and the molarity of the ith gas, Mi, is the product RT.
323
Chapter 15
15.11
In a homogeneous equilibrium, all of the reactants and products are in the same phase. In heterogeneous equilibria, at least two different phases are found among the reactants and products.
15.12
This is possible because their concentrations are constants that are incorporated into the numerical values of equilibrium constants. The concentrations may be considered constant because as the solid or liquid loses (or gains) mass, it loses (or gains) a volume proportional to the mass.
15.13
The value for ∆ng, is the difference between the number of moles of reactant gas and the number of moles of product gas: ∆ng, = (moles of gaseous products) – (moles of gaseous reactants) ∆ng, = 2 mol – 0 mol = 2
15.14
When a system at equilibrium is disturbed so that the equilibrium is upset, the system changes in a way that opposes the disturbance and returns the system to a state of equilibrium.
15.15
If a reactant is added to the reaction, then the value of the reaction quotient will decrease since the concentrations of the reactants is in the denominator of the reaction quotient. In order to re–establish equilibrium, the reaction will have to shift to the products.
15.16
Since the volume is decreased, initially, the pressure will be increased. This reaction will then adjust to relieve the excess pressure. This is done by moving to the side with fewer moles of gaseous substance. In the reaction: N2O4(g) 2NO2(g), this is on the left side of the equation.
15.17
(a) (b) (c)
15.18
This is an exothermic reaction. By increasing the temperature, shifts the equilibrium to the left, more reactants are formed, therefore the denominator becomes larger, and the KP decreases. This is an endothermic reaction. By increasing the temperature, the equilibrium is shifted to the right and more products are formed, and the numerator increases and KP increases. This is the same as part (a): an exothermic reaction and the increase in temperature shifts the reaction to the left and increases the amount of the reactants and KP decreases.
A catalyst affects the rate of the reaction, but does not affect the position of the equilibrium. The catalyst affects both the forward and the reverse reaction to the same extent, so the ratios of the forward and the reverse reactions for the catalyzed and uncatalyzed reactions do not change.
Review Problems 2
15.19
(a)
Kc =
(b)
Kc =
(c)
Kc =
POCl3 2
PCl3 [ O 2 ]
[SO2 ]2 [ O2 ] 2
SO3
(d)
Kc =
[ NO2 ]2 [ H 2 O]8 [ N 2 H 4 ][ H 2 O2 ]6
(e)
Kc =
[SO2 ][ HCl]2 [SOCl2 ][ H 2 O]
[ NO]2 [ H 2 O]2 [ N 2 H 4 ][ O2 ]2
324
Chapter 15
3
15.20
(a)
NCl3 [ HCl] Kc = [ Cl2 ]3 NH3
(d)
[ PCl2 Br ][ PClBr2 ]
(b)
Kc =
(c)
Kc =
(a)
( PPOCl ) Kp = 2 ( PPCl ) ( PO )
PBr3 PCl3
(e)
Kc =
(d)
( PNO ) ( PH O ) Kp = 6 ( PN H )( PH O )
2
2
(e)
( PNO )2 ( PH O )
(a)
( PN H )( PO ) 4
2
2
2 PSO ) ( PHCl ) ( Kp = ( PSOCl )( PH O ) 2
2
2
2
2
3 PNCl ) ( PHCl ) ( Kp = 3 ( PCl ) ( PNH ) 3
2
(b)
4
2
2
15.22
2
2
3
8
2
2
2
( PSO ) ( PO ) Kp = 2 ( PSO ) Kp =
[ Br2 ][ F2 ]5
2
3
2
(c)
BrF5
[ HNO2 ]2 [ NO][ NO2 ][ H 2 O]
3
(b)
[ HOCl]2 [ H 2 O][Cl2 O] 2
2
15.21
Kc =
(d)
Kp =
2
2
( PPCl Br )( PPClBr ) Kc = ( PPCl )( PPBr ) 2
3
( PH O )( PCl O ) 2
3
2
( PHOCl )2
(e)
3
( PBrF ) Kc = 5 ( PBr )( PF ) 5
2
2
2
(c)
( PHNO ) Kc = ( PNO ) ( PNO )( PH O ) 2
2
2
(a)
Ag(NH ) + 3 2 Kc = Ag + NH 2 3
15.24
(a)
H O + ClO − 3 Kc = [ HClO]
15.25
The first equation has been reversed in making the second equation. We therefore take the inverse of the value of the first equilibrium constant in order to determine a value for the second equilibrium constant: K = 1 × 1085
15.23
325
Cd(SCN) 2 − 4
(b)
Kc =
(b)
2− HCO3 SO 4 Kc = CO 2 [ HSO ] 4 3
Cd 2+ SCN −
4
Chapter 15
15.26
If we reverse the second reaction and double it, we can then add the first and second reactions together to obtain the desired reaction. When we double the second reaction we square the rate constant. When we reverse it, we invert the rate constant. Consequently, we get the following: 2CH4(g) C2H6(g) + H2(g)
Kc = 9.5 × 10–13
2CH3OH(g) + 2H2(g) 2CH4(g) + 2H2O(g)
Kc = 1.3 × 1041
Adding these equations we get: Kc = 1.2 × 1029
2CH3OH(g) + H2(g) C2H6(g) + 2H2O(g)
Where the Kc for the final reaction is the product of the Kc for the first reaction and the Kc for the “modified” second reaction.
15.27
Kc =
(a)
[ HCl]2 [ H 2 ][Cl2 ]
(b)
Kc =
[ HCl] [ H 2 ] [Cl2 ] 1
2
1
2
Kc for reaction (b) is the square root of Kc for reaction (a). 15.28
Kc =
[ H 2 ][Cl2 ] [ HCl]2
This is equal to 1/Kc for Review Problem 15.27(a). 15.29
M = P/RT
( 755 torr )
1 atm 760 torr M= = 0.0381 M L atm 318 K 0.0821 mol ) K (
(
)
15.30
P = M × RT = (0.0250 mol/L)(0.0821 L atm/K mol)(418 K) P = 0.858 atm
15.31
b, since ∆ng = 0
15.32
a, since ∆ng = 0
15.33
Kp = Kc × (RT)∆ng Kp = 11[(0.0821 L atm K–1 mol–1)(498 K)]–2 = (5.98 × 10–4) × Kc Kp = 6.3 × 10−3
15.34
Kp = Kc × (RT)∆ng Kp = 2.9 × 104[(0.0821 L atm K–1 mol–1)(673 K)]1 Kp = 1.6 × 106
15.35
Kp = Kc × (RT)∆ng 2.7 × 10−2= Kc[(0.0821 L atm K–1 mol–1)(773 K)]1 Kc = 4.2 × 10−4 326
Chapter 15
15.36
Kp = Kc × (RT)∆ng 4.7 × 1057 = Kc × [(0.0821 L atm K–1 mol–1)(573 K)]–1 Kc = 2.2 × 1059
15.37
Kp = Kc × (RT)∆ng 5.4 × 10−5 = Kc[(0.0821 L atm K–1 mol–1)(1046 K)]–2 Kc = 0.40
15.38
Kp = Kc × (RT)∆ng
15.39
L atm 668 K K p = 8.4 × 10−4 0.0821 mol ) K ( Kp = 4.6 × 10−2 In each case we get approximately 55.5 M: (a) 1 g 1 mol H 2 O mol H 2 O = ( 25.0 mL H 2 O ) = 1.39 mol H 2 O 1 mL 18.02 g H 2 O
(
1
)
1.39 mol H 2 O 1000 mL M= = 55.6 M 25.0 mL H 2 O 1 L (b)
1 g 1 mol H 2 O mol H 2 O = (150.0 mL H 2 O ) = 8.324 mol H 2 O 1 mL 18.02 g H 2 O 8.324 mol H 2 O 1000 mL M= = 55.49 M 150.0 mL H 2 O 1 L (c)
1000 mL 1 g 1 mol H 2 O mol H 2 O = (1.00 L H 2 O ) = 55.5 mol H 2 O 1 L 1 mL 18.02 g H 2 O 55.5 mol H 2 O M= = 55.5 M 1.00 L H 2 O 15.40
The density is 2.164 g/mL.
2.164 g 1 mol NaCl mol NaCl = 12.0 cm3 NaCl = 0.4444 mol NaCl 1 cm3 58.44 g NaCl
(
)
0.4444 mol NaCl M= = 37.03 M 0.0120 L NaCl 1 mol NaCl mol NaCl = ( 25.0 g NaCl ) = 0.4278 mol NaCl 58.44 g NaCl 1 cm3 NaCl 1 L NaCl L NaCl = 25.0 g NaCl = 1.155 × 10−2 L NaCl 2.164 g NaCl 1000 cm3 NaCl 0.4278 mol NaCl M= = 37.04 M 1.155 × 10−2 L NaCl
327
Chapter 15
15.41
15.42
(a)
Kc =
[CO]2 [O2 ]
(d)
(b)
K c = [ H 2 O ][SO 2 ]
(e)
[ H 2 O][ CO2 ] [ HF]2 5 K c = [ H 2O]
(c)
Kc =
[CH 4 ][ CO2 ] [ H 2 O]2
(a)
Kc =
[CO2 ] [SO2 ]
(d)
K c = [ H 2O]
(b)
Cl− Kc = − Br
(c)
K c = Cu 2+ OH −
Kc =
Kc =
(e)
[ N 2 ][ H 2 O]3 2
NH3
2
15.43 [HCl] 0.150 –2x 0.150–2x
I C E
[HI] – +2x +2x
[Cl2] – +x +x
Note: Since the I2(s) has a constant concentration, it may be neglected.
Kc = Kc =
[HI]2 [Cl2 ] [HCl]2
= 1.6 × 10−34
( 2x )2 ( x ) (0.150 − 2x)2
= 1.6 × 10−34
Because the value of Kc is so small, we make the simplifying assumption that (0.100 – 2x) ≈ 0.100, and the above equation becomes:
Kc = Kc =
[HI]2 [Cl2 ]
= 1.6 × 10−34
2
[HCl]
( 2x )2 ( x ) (0.150)
2
= 1.6 × 10−34
4x3 = 3.6 × 10–36; ∴ x = 9.65 × 10–13, and the above assumption is seen to have been valid. [HI] = 2x = 1.93 × 10–12 M [Cl2] = x = 9.65 × 10–13 M [HCl] = (0.100 – 2x) ≈ 0.150 M
328
Chapter 15
15.44
AgCl(s) + Br–(aq) AgBr(s) + Cl–(aq) [Br–] 0.12 –x 0.12–x
I C E
[Cl–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression gives: Cl− x = Kc = = 360 0.12 − x Br − Solving for the value of x: x = [Cl–] = 0.117 M and [Br–] = 0.12 – 0.117 = 0.0032 M. If we concern ourselves with significant figures, the [Br–] ≈ 0.00 M.
15.45
(a) (b) (c) (d) (e)
The system shifts to the right to consume some of the added methane. The system shifts to the left to consume some of the added hydrogen. The system shifts to the right to make some more carbon disulfide. The system shifts to the left to decrease the amount of gaseous moles. The system shifts to the right to absorb some of the added heat.
15.46
(a) (b) (c) (d) (e)
increase; we are adding a reactant decrease; we are removing a reactant increase; decreasing the volume favors the side with fewer gas molecules no change; a catalyst increases the rate but does not affect the concentration decrease; this is an exothermic reaction so heat may be considered a product
15.47
(a) (d)
right right
15.48
(a) (c)
increase increase
15.49
The mass action expression for this equilibrium is: [PCl5 ] Kc = = 0.18 [PCl3 ][PCl2 ]
(b) (e)
left no effect
(b) (d)
(c) (f)
left left
increase decrease
(a) The value for the reaction quotient for this system is: ( 0.00500 ) Q= = 4.96 ( 0.0420 )( 0.0240 ) This is not the value of the equilibrium constant, and we conclude that the system is not at equilibrium. (b)
15.50
Since the value of the reaction quotient for this system is larger than that of the equilibrium constant, the system must shift to the left to reach equilibrium.
The mass action expression for the system is: [ NO]SO3 = 85.0 Kc = [SO2 ][ NO2 ] 329
Chapter 15
and the reaction quotient for the system is:
[ NO]SO3 [SO2 ][ NO2 ]
Q=
15.51
=
( 0.0250 )( 0.0400 ) ( 0.00250 )( 0.00350 )
= 114
(a)
The system is not at equilibrium because Q ≠ Kc.
(b)
Since the value of the reaction quotient is larger than the value of the equilibrium constant, the system must shift to the left in order to reach equilibrium.
CH3OH CH3OH Kc = = = 0.500 2 [CO][ H 2 ] ( 0.175)( 0.225)2 [CH3OH] = 4.43 × 10–3 M.
15.52
The mass action expression is:
KP =
2 PNH
3
PN 2 PH3
= 0.018 2
Solving for [H2] gives:
PH 2
=3
2 PNH
3
PN 2 K P
=
3
( 20.5)2 (1.28) (0.018)
= 26.3 atm
15.53
CH3OH ( 0.00261) = 0.398 Kc = = [CO][ H 2 ]2 ( 0.105)( 0.250 )2
15.54
KP =
PC2 H5OH PC 2 H 4 PH 2O
=
( 6.99 ) ( 0.575)(1.30 )
= 9.35
15.55 I C E
[HBr] 0.075 –2x 0.075–2x
[H2] – +x +x
[Br2] – +x +x
The problem tell us that [Br2] = 0.0955 M = x at equilibrium. Using the ICE table as a guide we see that the equilibrium concentrations are; [H2] = [Br2] = 0.0955 M and [HBr] = 0.500–2(0.0955) = 0.309 M.
Kc =
[ H 2 ][ Br2 ] [ HBr ]2
=
( 0.0955)( 0.0955) ( 0.075 )2
= 0.0955
330
Chapter 15
15.56
Kc =
CH2O(g) H2(g) + CO(g)
[CH2O] 0.100 –x 0.100–x
I C E
[ H 2 ][CO] [CH 2 O]
[H2] – +x +x
[CO] – +x +x
The problem states that the equilibrium concentration of CH2O is 0.066 M. Therefore, x = 0.034 M = [H2] = [CO] at equilibrium. Substituting these values into the mass action expression we can determine Kc:
Kc =
15.57
[ H 2 ][CO] [CH 2 O]
(0.034)(0.034) = 1.8 × 10−2 0.066
=
According to the problem, the concentration of NO2 increases in the course of this reaction. This means our ICE table will look like the following: [NO2] 0.0560 +x 0.0560 + x
I C E
[NO] 0.294 +x 0.294 + x
[N2O] 0.184 –x 0.184 – x
[O2] 0.377 –x 0.377 – x
The problem tell us that [NO2] = 0.118 M = 0.0560 + x at equilibrium. Solving we get; x = 0.062 M. Using the ICE table as a guide we see that the equilibrium concentrations are; [NO] = 0.356 M, [N2O] = 0.122 M and [O2] = 0.315 M.
15.58
Kc =
[ N 2 O][ O2 ] [ NO2 ][ NO]
Kc =
[ NO2 ]4 [ N 2 O]2 [ O2 ]3
=
( 0.122 )( 0.315) ( 0.118)( 0.356 )
I C E
= 0.915
[N2O] 0.020 –2x 0.020 – 2x
[O2] 0.0560 –3x 0.0560 – 3x
At equilibrium [NO2] = 4x = 0.020 M x = 0.0050M So, [N2O] = 0.020 – 2(0.0050) = 0.010 M [O2] = 0.0560 – 3(0.0050) = 0.041M
Kc =
[0.020]4 [0.010]2 [ 0.041]3
= 23
331
[NO2] – +4x +4x
Chapter 15
15.59
2BrCl Br2 + Cl2
I C E
[Br2] – +x +x
[BrCl] 0.075 –2x 0.075–2x
[Cl2] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression gives: [ Br2 ][Cl2 ] = ( x )( x ) = 0.145 Kc = ( 0.075 − 2x )2 [ BrCl]2 Take the square root of both sides to get
Kc =
x = 0.381 0.075 − 2x
Solving for x gives: x = 0.016 M = [Br2] = [Cl2]
15.60
2BrCl Br2 + Cl2 In this problem, the reaction will shift from right to left as written above. This is due to the presence of products and the absence of reactants.
I C E
[BrCl] – +2x +2x
[Br2] 0.0225 –x 0.0225–x
[Cl2] 0.0225 –x 0.0225–x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ Br2 ][Cl2 ] [ BrCl]2
=
( 0.0225 − x )( 0.0225 − x ) ( 2x )2
= 0.145
Take the square root of both sides to get
Kc =
0.0225 − x = 0.381 2x
Solving for x gives: x = 0.0128 M. The individual concentrations are then: [Br2] = [Cl2] = 0.0225 – 0.0128 = 0.00973 M, [BrCl] = 2(0.0128) = 0.0256 M. 15.61
The initial concentrations are each 0.280 mol/2.00 L = 0.140 M.
I C E
[SO3] 0.140 –x 0.140–x
[NO] 0.140 –x 0.140–x
[NO2] – +x +x
[SO2] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression gives: 332
Chapter 15
Kc =
[ NO2 ][SO2 ] SO3 [ NO ]
=
( x )( x ) ( 0.140 − x )( 0.140 − x )
= 0.500
Taking the square root of both sides of this equation gives: 0.707 = x/(0.140 – x) Solving for x we have: 1.707(x) = 0.09898, x = 0.0580 mol/L = [NO2] = [SO2], [NO] = [SO3] = 0.140 – x = 0.0820 mol/L 15.62
The initial concentrations are each 0.140 M. [SO3] – +x +x
I C E
[NO] – +x +x
[NO2] 0.140 –x 0.140–x
[SO2] 0.140 –x 0.140–x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ NO2 ][SO2 ] SO3 [ NO ]
=
( 0.140 − x )( 0.140 − x ) ( x )( x )
= 0.500
Taking the square root of both sides of this equation gives: 0.707 = (0.140 – x)/x. Solving for x we have: 1.707(x) = 0.140 ∴ x = 0.0820 mol/L = [NO] = [SO3], [NO2] = [SO2] = 0.140 – x = 0.0580 mol/L Although these two systems reach different equilibrium positions, the equilibrium concentrations of the four substances in each experiment give the same value for the equilibrium constant when substituted into the mass action expression. An alternative method for solving problems where the reaction proceeds from right to left, such as this problem, is to reverse the reaction written and use this “new” reaction as the basis for the problem. In doing this, the mass action expression and the equilibrium constant must change to reflect the “new” reaction. As outlined in Section 15.2, “when the direction of an equation is reversed, the new equilibrium constant is the reciprocal of the original” and the “new” mass action expression is also the reciprocal of the original. Try solving this problem using this method. Your answer should, of course, remain the same as stated above.
15.63
The initial concentrations are all 1.00 mol/100 L = 0.0100 M. Since the initial concentrations are all the same, the reaction quotient is equal to 1.0, and we conclude that the system must shift to the left to reach equilibrium since Q > Kc. [CO] 0.0150 +x 0.0150+x
I C E
[H2O] 0.0150 +x 0.0150+x
[CO2] 0.0150 –x 0.0150–x
[H2] 0.0150 –x 0.0150–x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[CO2 ][ H 2 ] [CO][ H 2 O]
=
( 0.0150 − x )( 0.0150 − x ) ( 0.0150 + x )( 0.0150 + x )
= 0.400
333
Chapter 15
We take the square root of both sides of the above equation:
( 0.0150 − x ) ( 0.0150 + x )
= 0.632
and (0.632)(0.0150 + x) = 0.0150 – x (1.632)x = 5.52 × 10–3, or x = 3.38 × 10–3 mol/L The equilibrium concentrations are then: [H2] = [CO2] = (0.0150 – 3.38 × 10–3) = 0.0116 M, [CO] = [H2O] = (0.0150 + 3.38 × 10–3) = 0.0184 M. 15.64
If we substitute the initial concentrations of 0.0400 M for each component of this mixture into the reaction quotient expression that results from the balanced equation in the text, we see that the reaction quotient equals one. Since Q > Kc, the reaction will proceed from right to left as written, i.e., [Br2] and [Cl2] will decrease and [BrCl] will increase.
I C E
[BrCl] 0.0450 + 2x 0.0450 + 2x
[Br2] 0.0450 –x 0.0450 – x
[Cl2] 0.0450 –x 0.0450 – x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ Br2 ][Cl2 ] [ BrCl]2
=
( 0.0450 − x )( 0.0450 − x ) ( 0.0450 + 2x )2
= 0.145
Take the square root of both sides of this equation to get;
( 0.0450 − x ) ( 0.0450 + 2x )
= 0.381 .
This equation is easily solved giving x = 0.0158 M. The substances then have the following concentrations at equilibrium: [Cl2] = [Br2] = 0.0450 – 0.0158 = 0.0292 M, [BrCl] = 0.0450 + 2(0.0158) = 0.0766 M.
15.65 I C E
Kc =
[ H 2 ][Cl2 ] [ HCl]2
=
[HCl] 0.0450 –2x 0.0450–2x
( x )( x ) ( 0.0450 − 2x )
2
[H2] – +x +x
[Cl2] – +x +x
= 3.2 × 10−34
Because Kc is so exceedingly small, we can make the simplifying assumption that x is also small enough to make (0.0450 – 2x) ≈ 0.0450. Thus we have: 3.2 × 10–34 = (x)2/(0.0450)2 Taking the square root of both sides, and solving for the value of x gives: x = 8.0 × 10–19 M = [H2] = [Cl2] [HCl] = (0.0450 – x) ≈ 0.0450 mol/L 334
Chapter 15
15.66
The initial concentrations are : 0.250 mol/4.00 L = 0.0625 M N2O 0.350 mol/4.00 L = 0.0875 M NO2
I C E
Kc =
[ NO]3 [ N 2 O][ NO2 ]
[N2O] 0.0625 –x 0.0625–x
[NO2] 0.0875 –x 0.0875–x
[NO] – +3x +3x
= 1.4 × 10−10
(3x)3 (0.0625)(0.0875) 3 x = 2.84 × 10–14 x = 3.05 × 10–5 Kc =
Where we have assumed that x << 0.0625. Solving we get x = 3.05 × 10−5. Therefore, the molar concentration of NO would be 3x or 3(3.05 × 10−5) M = 9.15 × 10−5 M.
15.67
Kc =
[ CO]2 [O2 ] [ CO2 ]2
= 6.4 × 10−7
I C E
[CO2] 2.25 × 10–2 –2x 2.25 × 10–2 – 2x
[ 2x ]2 [ x ]
Kc =
2.25 × 10−2 – 2x Assume x << 2.25 × 10–2. 4x 3 (2.25 × 10−2 ) 2
2
[CO] – +2x +2x
[O2] – +x +x
= 6.4 × 10−7
= 6.4 × 10−7
x = 4.3 × 10−4
[CO] = 2x = 8.6 × 10−4 M
15.68
Kc =
[ H 2 ]2 [O2 ] [ H 2 O]2
= 6.0 × 10−28
I C E
[H2O] 0.025 mol/5.00L –2x 5.0 × 10–3 – 2x
[H2] – +2x +2x
335
[O2] – +x +x
Chapter 15
[ 2x ]2 [ x ]
Kc =
5.0 × 10 – 2x Assume 2x << 5.0 × 10–3. −3
4x 3 −3 2
2
= 6.0 × 10−28
= 6.0 × 10−28
x = 1.6 × 10−11
(5.0 × 10 )
[H 2 ] = 1.6 × 10−11 M, [O 2 ] = 1.6 × 10−11 M
15.69
We first approach the problem in the normal fashion with an initial concentration of PCl5 = 0.013 M.
I C E
[PCl3] – +x +x
[Cl2] – +x +x
[PCl5] 0.0175 –x 0.0175–x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
PCl5 ( 0.0175 − x ) = = 0.18 PCl3 [ Cl2 ] ( x )( x )
rearranging; 0.18x 2 + x − 0.0175 = 0 We next attempt to use the quadratic equation to solve for the value of x, setting a = 0.18; b = 1; c = –0.013. However, we find that unless we carry one more significant figure than is allowed, the quadratic formula for this problem gives us a concentration of zero for PCl5. A better solution is obtained by "allowing" the initial equilibrium to shift completely to the left, giving us a new initial situation from which to work:
I C E
[PCl3] 0.0175 –x 0.0175–x
[Cl2] 0.0175 –x 0.0175–x
[PCl5] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
PCl5 (+x ) = = 0.18 PCl3 [ Cl2 ] ( 0.0175 − x )( 0.0175 − x )
Now, we may assume that x << 0.0175. The equation is simplified and we solve for x x = [PCl5] = 5.5 × 10–5 M. 15.70
All concentrations are a factor of ten less than the stated value due to the container size. Since Q < Kc for this reaction, the reaction proceeds as written, i.e., from left to right.
I C E
[SO2] 0.0100 –x 0.0100–x
[NO2] 0.00600 –x 0.00600–x 336
[NO] 0.00800 +x 0.00800+x
[SO3] 0.0120 +x 0.0120+x
Chapter 15
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ NO]SO3 [SO2 ][ NO2 ]
=
( 0.00800 + x )( 0.0120 + x ) ( 0.0100 − x )( 0.00600 − x )
= 85.0
0.00510 – 1.36(x) + 85.0(x)2 = 9.60 × 10–5 + 0.0200(x) + (x)2 84.0(x)2 – 1.38(x) + 0.00500 = 0 The quadratic equation is used, with the following values: a = 84.0; b = –1.38; c = 0.00500. Upon solving for x, we find that only the negative root is sensible: x = 0.00542 M. [SO2] = 0.0100 – 0.00542 = 0.0046 M [NO2] = 0.00600 – 0.00542 = 0.00058 M [NO] = 0.00800 + 0.00542 = 0.0134 M [SO3] = 0.0120 + 0.00542 = 0.0174 M 15.71 [SO3] 0.0750 –x 0.0750
I C E
[NO] 0.0875 –x 0.0875
[NO2] – +x +x
[SO2] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ NO2 ][SO2 ] SO3 [ NO ]
=
( x )( x ) ( 0.0750 − x )( 0.0875 − x )
= 0.500
Since the equilibrium constant is not much larger than either of the values 0.0750 or 0.0875, we cannot neglect the size of x in the above expression. A simplifying assumption is not therefore possible, and we must solve for the value of x using the quadratic equation. Multiplying out the above denominator, collecting like terms, and putting the result into the standard quadratic form gives: 0.500x2 + (8.13 × 10–2)x – (3.28 × 10–3) = 0
−8.13 × 10−2 ± x=
(8.13 × 10 ) −2
2
−
(
4(0.500) −3.28 × 10−3
2(0.500)
) = 0.0335 M
using the (+) root. So, [NO2] = [SO2] = 0.0335 M 15.72
If we substitute the initial concentrations for each component of this mixture into the reaction quotient expression that results from the balanced equation in the text, we see that the reaction quotient equals 0.375. Since Q > Kc for this situation, the reaction will proceed from right to left, i.e., [Br2] and [Cl2] will decrease and [BrCl] will increase.
I C E
[BrCl] 0.0400 +2x 0.0400+2x
[Br2] 0.0300 –x 0.0300–x
[Cl2] 0.0200 –x 0.0200–x
Substituting the above values for equilibrium concentrations into the mass action expression gives: 337
Chapter 15
Kc =
[ Br2 ][Cl2 ] [ BrCl]2
=
( 0.0300 − x )( 0.0200 − x ) ( 0.0400 + 2x )2
= 0.145
Rearranging we get the following quadratic expression: 0 = 0.42x2 – 0.0732x + 0.000368. Solving and using the negative root gives x = 0.0052 M. The individual concentrations are: [Cl2] = 0.0200 – 0.0052 = 0.0148 M, [Br2] = 0.0300 – 0.0052 = 0.0248 M, [BrCl] = 0.0400 + 2(0.0052) = 0.0504 M.
15.73
Kc =
[ CO][ H 2 O] [ HCHO2 ]
= 4.3 × 105
Since Kc is large, start by assuming all of the HCHO2 decomposes to give CO and H2O
I C E
Kc =
[ 0.200][ 0.200] [x]
[HCHO2] – +x +x
[CO] 0.200 –x 0.200 –x
[H2O] 0.200 –x 0.200 –x
= 4.3 × 105
x = 9.3 × 10–8 so, at equilibrium [CO] = [H2O] = 0.200 – x = 0.200 M 15.74
Because of the very large value of Kc, we start by realizing that nearly all of the 0.100 moles of H2 will react with 0.100 moles of Br2 to form 0.200 moles of HBr. This brings us to a new "initial" condition that is more realistically close to the true equilibrium condition, i.e., [HBr] = (0.200 moles/10.0 L) = 0.0200 M, [Br2] = (0.100 moles/10.0 L) = 0.0100 M, [H2] = 0 M. Next, we proceed in the normal fashion, allowing 2x mol/L of HBr to disappear, and making x mol/L each of H2 and Br2 in order to reach equilibrium:
I C E
[H2] – +x +x
[Br2] 0.0100 +x 0.0100+x
[HBr] 0.0200 –2x 0.0200–2x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ HBr ]2 [ H 2 ][ Br2 ]
=
( 0.0200 − 2x )2 ( + x )( 0.0100 + x )
= 2.0 × 109
We next make the assumptions that (0.0200 – 2x) ≈ 0.0200, and (0.0100 + x) ≈ 0.0100, giving: 2.0 × 109 = (0.0200)2/(0.0100)(x) x = 2.0 × 10–11 M = [H2]. The small size of x demonstrates that the assumptions made above were justified. [HBr] = 0.0200 – 2x = 0.0200 M [Br2] = 0.0100 + x = 0.0100 M
338
Chapter 15
Additional Exercises 15.75
(a)
The mass action expression is: 2
( PNO ) Kp = ( PN O ) 2
2
= 0.140 atm
4
Solving the above expression for the partial pressure of NO2, we get:
( 0.225 atm )( 0.140 atm ) = 0.177 atm
PNO2 = PN 2O 4 × K p = (b)
Ptotal = PNO2 + PN2O4 = 0.177 + 0.225 = 0.402 atm −8
15.76
3
[ NO]3 = ( 4.7 ×10 ) = 1.88 ×10−20 Kc = [ N 2 O][ NO2 ] ( 0.023)( 0.24 ) ∆n K p = K c ( RT ) = (1.88 × 10–20)(0.0821 × 298)1 = 4.6 × 10–19 g
5
15.77
Kc =
H5 IO6 7
IO − H + 3
7
= 1 × 10–85
H5IO6 is a weak acid and must be part of the equilibrium expression. As ions, both H+ and IO3– are part of the expression. H2O is ignored since it is a pure liquid and I2 is ignored since it is a solid at 25°C. 15.78
The initial concentrations are: [NO2] = (0.250 mol/4.00 L) = 0.0625 M [NO] = (0.275 mol/4.00 L) = 0.0688 M [N2O] = (0.175 mol/4.00 L) = 0.0438 M [O2] = (0.225 mol/4.00 L) = 0.0563 M Substituting these values into the mass action expression we determine: [ N 2 O][ O2 ] = ( 0.0438)( 0.0563) = 0.573 Q= [ NO2 ][ NO] ( 0.0625)( 0.0688) Since Q
I C E
[NO] 0.0688 –x 0.0688 – x
[N2O] 0.0438 +x 0.0438 + x
[O2] 0.0563 +x 0.0563 + x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ N 2 O][ O2 ] [ NO2 ][ NO]
=
( 0.0438 + x )( 0.0563 + x ) ( 0.0625 − x )( 0.0688 − x )
= 0.914
To solve we need to use the quadratic equation. Expanding the above calculation we get:
339
Chapter 15
15.79
0.086x2 + 0.220x – 0.00146 = 0 Solving we get x = 0.00581. So, [NO2] = 0.0625 – x = 0.0567 M [NO] = 0.0688 – x = 0.0630 M [N2O] = 0.0438 + x = 0.0380 M [O2] = 0.0563 + x = 0.0505 M First, calculate a value for Kc using a rearranged form of equation 15.9 and setting the value for ∆n to –1:
Kc =
Kp
( RT )
∆n g
=
1.5 × 1018 0.0821 L atm mol K
(
) ( 300 K )
−1
= 3.7 × 1019
The value for the equilibrium constant is very large, indicating that the equilibrium lies far to the right. We therefore anticipate that the initial conditions are unrealistic. The system is "allowed" to come to a more realistic new initial set of concentrations, by reaction of all of the starting amount of NO, to give a stoichiometric amount of N2O and NO2. Only then can we solve for the equilibrium concentrations in the usual manner:
I C E
[NO] – +3x +3x
[N2O] 0.015 –x 0.015 – x
[NO2] 0.015 –x 0.015 – x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ N 2 O][ NO2 ] [ NO]3
=
(0.015 − x)(0.015 − x) (3x)
3
= 3.7 × 1019
The simplifying assumption can be made that the value for x is much smaller than the number 0.015. Upon solving for x we get: 27x3 = 6.1 × 10–24 x = 6.1 × 10–9 M –8 [NO] = 3x = 1.8 × 10 M, [N2O] = [NO2] = 0.015 – x = 0.015 M. 15.80
(a) (b) (c) (d)
Equilibrium is unaffected by the addition of a solid. Equilibrium is unaffected by the removal of a solid (provided it is not completely removed.) The equilibrium will shift to the left. The equilibrium will shift to the right.
15.81
The initial concentrations are: [NO2] = (0.250 mol/5.00 L) = 0.0500 M, [NO] = (0.250 mol/5.00 L) = 0.0500 M, [N2O] = (0.250 mol/5.00 L) = 0.0500 M, [O2] = (0.250 mol/5.00 L) = 0.0500 M. Substituting these values into the mass action expression we determine:
Q=
[ N 2 O][O2 ] [ NO2 ][ NO]
=
( 0.0500 )( 0.0500 ) ( 0.0500 )( 0.0500 )
= 1.00
Since Q > Kc, this reaction will proceed from right to left as written. The ICE table becomes:
I C E
[NO2] 0.0500 +x 0.0500 + x
[NO] 0.0500 +x 0.0500 + x
340
[N2O] 0.0500 –x 0.0500 – x
[O2] 0.0500 –x 0.0500 – x
Chapter 15
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[ N 2 O][ O2 ] [ NO2 ][ NO]
=
( 0.0500 − x )( 0.0500 − x ) ( 0.0500 + x )( 0.0500 + x )
Taking the square root of both sides we get K c =
= 0.914
( 0.0500 − x ) ( 0.0500 + x )
= 0.956
Solving for x we get: x = 1.1 × 10–3. Thus, at equilibrium we have: [N2O] = [O2] = 0.0489 M, [NO2] = [NO] = 0.0511 M. If 0.050 moles of NO2 are added (note: 0.050 mol/5.00 L = 0.01 M NO2 added), the ICE table becomes: [NO2] 0.0511 + 0.01 –x 0.0611 – x
I C E
[NO] 0.0511 –x 0.0511 – x
[N2O] 0.0489 +x 0.0489 + x
[O2] 0.0489 +x 0.0489 + x
Substituting these values into the mass action expression gives:
Kc =
[ N 2 O][ O2 ] [ NO2 ][ NO]
=
( 0.0489 + x )( 0.0489 + x ) ( 0.0611 − x )( 0.0511 − x )
= 0.914
Solving the quadratic expression, 0.086x2 + 0.200 x – 4.6 × 10–4 = 0, we get x = 0.00581. Thus, at equilibrium, we now have: [N2O] = 0.0547 M [O2] = 0.0547 M [NO2] = 0.0553 M [NO] = 0.0453 M 15.82
Kc =
[ N 2 O][O2 ] [ NO2 ][ NO]
= 0.914
Let z = the initial concentration.
I C E
Kc =
[x] [x] [ z -x ] [ z - x ]
[NO2] z –x z–x
[NO] z –x z –x
[N2O] – +x +x
= 0.914
Take the square roots of both sides to get x = 0.956 x = 0.0600 from the data in the problem so z − x 0.0600 = 0.956 solving for z we get z = 0.12 M z − 0.0600 0.12 mol moles NO = moles NO 2 = ( 5.00 L ) = 0.60 moles L
341
[O2] – +x +x
Chapter 15
15.83
(a)
F2(g) + PF3(g) PF5(g) PF5(g) F2(g) + PF3(g)
(b) (c)
Because the volume is smaller, LeChateliers Principle states that the equilibrium will favor the products. Consequently, equilibrium will be shifted to the right in a smaller volume. As can be seen here, a higher proportion of the reactants have reacted to form more PF5 as compared to the case in (a). 15.84
We start by determining the initial concentration of NO2: [NO2] = 0.25 mol/4.00 L = 0.0625 M. Next, this value is used in an equilibrium law type calculation using the equation: 3NO2 N2O5 + NO
I C E
[NO2] 0.0625 –3x 0.0625–3x
[N2O5] – +x +x
[NO] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
N 2 O5 [ NO ] x2 Kc = = = 1.0 × 10−11 3 3 ( 0.0625 − 3x ) [ NO2 ] Assume 3x << 0.0625
x2 3
= 1.0 × 10−11
( 0.0625)
342
Chapter 15 x = 4.9 × 10–8 [N2O5] = 4.9 × 10–8 M mol N2O5 = (4.9 × 10–8 M)(4.00 L) = 2.0 × 10–7 mol N2O5
15.85
The concentration of the water needs to be calculated: 1 atm 17.5 torr P 760 torr M= = = 9.57 × 10–4 M −1 −1 RT 0.0821 L atm mol K ( 293 K )
(
)
[NO] 2.59 × 10–3 –x 2.59 × 10–3 – x
I C E
[NO2] 2.59 × 10–3 –x 2.59 × 10–3 – x
[H2O] 9.57 × 10–4 –x 9.57 × 10–4 – x
4.0 × 10–4 = 2x x = 2.0 × 10–4 M [NO] = [NO2] = 2.59 × 10–3 – 2.0 × 10–4 = 2.39 × 10–3 M [H2O] = 9.57 × 10–4 – 2.0 × 10–4 = 7.57 × 10–4 M −4
Kc =
2
( 4.0 ×10 ) [ HNO2 ]2 = [ NO][ NO2 ][ H 2 O] ( 2.39 ×10−3 )( 2.39 ×10−3 )( 7.57 ×10−4 ) 2
15.86
Kc =
NH3
[ H 2 ]3 [ N 2 ] ( 0.030 )2 0.135 = [ H 2 ]3 ( 0.50 ) [ H 2 ]3 =0.0133
[H2] = 0.237 M Partial Pressures: PNH3 = (0.030 M)(0.08214 L atm mol–1 K–1)(373 K) = 0.92 atm
PH 2 = (0.237 M)(0.08214 L atm mol–1 K–1)(373 K) = 7.26 atm PN 2 = (0.50 M)(0.08214 L atm mol–1 K–1)(373 K) = 15.3 atm 2
Kp
( PNH ) = 3 ( PH ) ( PN ) 3
2
=
( 0.92 )2 ( 7.26 )3 (15.3)
= 1.44 × 10–4
2
KP using Equation 15.9:
K P = K C ( RT )
∆n g
∆ng = 2 – 4 = –2 mol KP = 0.135[(0.08214 L atm mol–1 K–1)(373)]–2 = 1.44 × 10–4
343
= 37
[HNO2] – +2x 4.0 × 10–4
Chapter 15
Multi-Concept Problems 15.87
First, calculate the number of moles of CO used in the experiment, using the ideal gas law:
n=
( 0.177 atm )( 2.00 L ) PV = = 0.0145 mol L atm 298 K RT 0.0821 mol ( ) K
(
)
Next, calculate the partial pressure of CO at 400 °C:
P=
(
L atm ( 0.0145 mol ) 0.0821 mol nRT K = V 2.00 L ( )
) ( 673 K ) = 0.401 atm
Next, we calculate the number of moles of water that are supplied to the reaction, and convert to partial pressure for water, using the ideal gas equation:
1 mol H 2 O mol H 2 O = ( 0.391 g H 2 O ) = 0.0217 mol H2O 18.02 g H 2 O
P=
(
L atm ( 0.0217 mol ) 0.0821 mol nRT K = V 2.00 L ( )
) ( 673 K ) = 0.600 atm
Finally, we solve for the equilibrium partial pressure in the usual manner:
I C E
PHCHO2
PCO
P H 2O
– +x +x
0.401 –x 0.401–x
0.600 –x 0.600–x
Substituting the above values for equilibrium partial pressures into the mass action expression gives:
Kp =
( PCO ) ( PH O ) 2
( PHCHO )
=
(0.401 − x)(0.600 − x) = 1.6 × 106 x
2
Because Kp is so large, we assume x << 0.401 and x << 0.600. We then solve for PHCHO2 = x = 1.5 × 10–7 atm.
15.88
Kc =
[CO]2 [O2 ] [CO2 ]2
= 6.4 x 10−7
I C E
[CO2] 0.0100 –2x 0.0100–2x
[CO]
[O2]
0 +2x +2x
0 +x +x
344
Chapter 15
Kc =
(2x)2 (x) (0.0100 − 2x)2
x = 2.437 x 10-4 M PCO =
nRT V
= 6.4 x 10−7
round answer to 2.44 x 10−4 −4
= [CO]RT = 2 x 2.44 x 10 M x 0.0821 L atm K
−1
mol
−1
x 2273 K
PCO = 9.11 x 10−2 atm Molecules of O2 = 2.44 x 10-4 mol L-1 x 1.00 L x 6.02 x 1023 molecules O2 mol-1 = 1.47 x 1020 Density is defined as the mass of substance per volume. During the reaction the volume of the container does not change and since we must conserve mass during a chemical reaction, the total number of grams of substances does not change. Therefore, the density of the mixture does not change during the reaction. We can use bond energies to determine the enthalpy of reaction. 4 C=O bonds are broken: 4 mol x 799 kJ/mol = 3196 kJ 1 O=O bond is formed: 1 mol x (− 494 kJ/mol) = − 494 kJ 2 C≡ O bonds are formed: 2 mol x (− 1072 kJ/mol) = − 2144 kJ The total enthalpy change for the reaction, as written, is + 558 kJ. This is the amount of heat that would have been required to form a mole of O2. At equilibrium we have formed 2.44 x 10−4 moles of O2. Therefore, the heat required is: (+558 kJ/mol O2) x 2.44 x 10−4 mol O2 = +0.136 kJ or +136 J 15.89
Kc =
[ NO][SO3 ] [SO2 ][ NO2 ]
= 85.0
The initial concentrations are: [SO2] = (0.0255 mol/10.00 L) = 0.00255 M, [NO2] = (0.0600 mol/10.00 L) = 0.00600 M, [NO] = (0.0800 mol/10.00 L) = 0.00800 M, [SO3] = (0.0446 mol/10.00 L) = 0.00446 M. Calculate Q to determine the direction of the reaction.
Q=
(0.00800)(0.00446) = 2.3 (0.00255)(0.00600)
Therefore, the products are favored and the reaction will proceed to the right.
I C E
[SO2] 0.00255 –x 0.00255 – x
[NO2] 0.00600 –x 0.00600 – x
(0.00800 + x)(0.00446 + x) = 85.0 (0.00255 − x)(0.00600 − x) x = 0.002324 345
[NO] 0.00800 +x 0.00800+ x
[SO3] 0.00446 +x 0.00446+x
Chapter 15
[SO2] = 0.000226 M
[NO2] = 0.00368 M
[NO] = 0.0103 M
[SO3] = 0.00678 M
Grams of each gas: SO2 NO2 NO SO3
0.000226 mol L-1 x 10.0 L x 64.06 g mol-1 = 0.146 g 0.00368 mol L-1 x 10.0 L x 46.00 g mol-1 = 1.69 g 0.0103 mol L-1 x 10.0 L x 30.01 g mol-1 = 3.09 g 0.00678 mol L-1 z 10.0 L x 80.06 g mol-1 = 5.46 g
Total pressure of the mixture at equilibrium: P = (n/V) RT = [Xi]RT where Xi is the molar concentration of the ith species. PT = P(SO2) + P(NO2) + P(NO) + P(SO3) PT = (0.000226 + 0.00368 + 0.0103 + 0.00678) mol L-1 x 0.0821 L atm K mol-1 x (460 + 273) K PT = 1.26 atm
15.90
Kc =
[ NO]3 [ N 2 O][ NO2 ]
= 1.4 x 10−10
The molar concentration of the 300 mL sample of gas is given by: 800 torr n P 760 torr atm −1 = = V RT 0.0821 L atm K −1 mol−1 x (25 + 273)K
n = 4.30 x 10−2 M V The concentration of the gas when placed in the 4.00 L container is given by: 4.30 x 10-2 M x (0.300L/4.00L) = 3.23 x 10-3 M
I C E
[N2O] 0 +x +x
[NO2] 0 +x +x
[NO] 3.23 x 10-3 -3x 3.23 x 10-3 -3x
(3.23 x 10−3 -3x)3 = 1.4 x 10−10 (x)(x) Be very careful about rounding your coefficients in the cubic equation. You should carry several decimal places. The equilibrium constant is very small so most of the NO is converted to N2O and NO2. Thus, if you are not careful you will end up with an answer for x that is larger than 3.23 x 10-3 M. x = 1.075 x 10-3 M At equilibrium:
346
Chapter 15 [N2O] = 1.075 x 10-3 M
[NO] = 5.00 x 10-6 M
PT = (2 x 1.075 x 10-3 + 5.00 x 10-6) M x 0.0821 L atm K-1 mol-1 x 473 K = 8.37 x 10-2 atm 15.91
HCHO2(g) CO(g) + H2O(g) Since there is no HCHO2(g) present initially, we know this reaction will move from right to left as written.
I C E
[HCHO2] – +x +x
[CO] 0.20 –x 0.20–x
[H2O] 0.30 –x 0.30–x
Substituting the above values for equilibrium concentrations into the mass action expression gives:
Kc =
[CO][ H 2 O] [ HCHO2 ]
=
( 0.20 − x )( 0.30 − x ) (x)
= 2.9 × 104
Since Kc is so large for the forward direction, we do not anticipate that the reverse reaction will proceed to any appreciable extent. Therefore, we may assume x << 0.20 and x << 0.30. This simplifies the problem immensely and we may solve for x. x = [HCHO2] = 2.1 × 10–6 M. At equilibrium we have: 0.2 moles of CO or 0.2 x 46 g/mol = 5.6 g 0.3 moesl of H2O or 0.3 x 18.0 g/mol = 5.4 g 2.1 x 10-6 moles of HCHO2 or 0 g The density is: (5.6 + 5.6)g/1 L = 11.0 g/L 15.92
(a) The density of the gas before equilibrium: Grams of N2O4 = 0.0245 mol N 2 O 4 x
Grams of NO2 = 0.0116 mol x
92.01 g = 2.25 g mol
46.00 g = 0.534 g mol
Density = (2.25 + 0.534)g/4.50 L = 0.620 g L-1 (b) Matter is not destroyed during a chemical reaction. Therefore, the density of the mixture cannot change since the total mass of the system does not change and the volume remains constant. Density = 0.620 g L-1
347
Chapter 16
Practice Exercises 16.1
Conjugate acid base pairs (a), (c), and (f) (b) The conjugate base of HI is I– (d) The conjugate base of HNO2 is NO2– and the conjugate base of NH4+ is NH3 (e) The conjugate acid of CO32– is HCO3– and the conjugate acid of CN– is HCN
16.2
In each case the conjugate base is obtained by removing a proton from the acid: (a) OH– (b) I– (c) NO2– (d) H2PO4– 2– 3– – (e) HPO4 (f) PO4 (g) HS (h) NH3
16.3
In each case the conjugate acid is obtained by adding a proton to the base: (a) H 2O 2 (b) HSO4– (c) HCO3– (d) HCN + (e) NH3 (f) NH4 (g) H3PO4 (h) H2PO4–
16.4
HCN and CN–
16.5
The Brønsted acids are H2PO4–(aq) and H2CO3(aq) The Brønsted bases are HCO3–(aq) and HPO42–(aq)
16.6
conjugate pair ┌────────────────────┐ PO43–(aq) + HC2H3O2(aq) HPO42–(aq) + C2H3O2–(aq) base acid acid base └─────────────────────┘ conjugate pair
16.7
(a) (b) (c) (d) (e) (f) (g) (h)
16.8
HPO42–(aq) + OH–(aq) PO43–(aq) + H2O; HPO42– acting as an acid HPO42–(aq) + H3O+(aq) H2PO4– + H2O; HPO42– acting as a base
16.9
HSO4–(aq) + HPO42–(aq) SO42−(aq) + H2PO4–(aq)
16.10
The substances on the right because they are the weaker acid and base.
16.11
(a) (b) (c) (d) (e)
HF < HBr < HI PH3 < H2S < HCl H2O < H2Se < H2Te AsH3 < H2Se < HBr PH3 < H2Se < HI
16.12
(a) (b) (c)
HBr is the stronger acid since binary acid strength increases from left to right within a period. H2Te is the stronger acid since binary acid strength increases from top to bottom within a group. H2S since acid strength increases from top to bottom within a group.
16.13
(a)
HClO3
H2PO4– HPO42– H 2S H3PO4 NH4+ H 2O HI HNO2
HCl and Cl–
amphoteric since it can both accept and donate a proton amphoteric since it can both accept and donate a proton amphoteric since it can both accept and donate a proton not amphoteric: it can only donate protons not amphoteric: it can only donate protons amphoteric since it can both accept and donate a proton not amphoteric: it can only donate protons not amphoteric: it can only donate protons
(b)
H2SO4 348
Chapter 16
16.14 16.15
(a) (a)
H3AsO4 HIO4
(b) (b)
H2TeO4 H2TeO4
16.16
(a)
H2SO4
(b)
H3AsO4
16.17
The acid strength decreases as follows:
(c)
H3AsO4
FCH2COOH > ClCH2COOH > BrCH2COOH 16.18
(a) (b)
(c)
16.19
(a) (b) (c)
NH3 is the Lewis base since it has an unshared pair of electrons. H+ is the Lewis acid since it can accept a pair of electrons Na2O is the Lewis base (O2- would also be a correct answer since sodium salts are soluble, assuming the reactions are in water) SeO3 is the Lewis acid Ag+ is the Lewis acid NH3 is the Lewis base Fluoride ions have a filled octet of electrons and are likely to behave as Lewis bases, i.e., electron pair donors. BeCl2 is a likely Lewis acid since it has an incomplete shell. The Be atom has only two valence electrons and it can easily accept a pair of electrons. It could reasonably be considered a potential Lewis base since it contains three oxygens, each with lone pairs and partial negative charges. However, it is more effective as a Lewis acid, since the central sulfur bears a significant positive charge.
Review Questions 16.1
A Brønsted acid is a proton donor, whereas a Brønsted base is a proton acceptor.
16.2
In a conjugate acid–base pair, the acid has one more hydrogen ion than does the base.
16.3
H2SO4 is not the conjugate acid of SO42– because H2SO4 has two more hydrogen ions than does SO42–. HSO4– is the conjugate acid of SO42–.
16.4
An amphoteric substance can act either as an acid or as a base. HCl(aq) + H2O H3O+(aq) + Cl–(aq), in which water serves as a base NH3(g) + H2O OH–(aq) + NH4+(aq), in which water serves as an acid
16.5
A compound that can act either as a proton donor or a proton acceptor.
16.6
Acid strength increases from left to right in the same period and from top to bottom in the same family.
16.7
(a) (b)
16.8
16.9
The relative strength of the binary acids increases from top to bottom in a group of the periodic table, so we expect HAt to be a stronger acid than HI. The relative strength of oxoacids increases from bottom to top in a group of the periodic table, so we expect HAtO4 to be a weaker acid that HBrO3.
In nitric acid, there are more oxygen atoms bound to the nitrogen atom than in nitrous acid. As the number of oxygen atoms increases, the “pull” on the electrons in the OH bond increases withdrawing electrons away from the hydrogen atom. This makes it easier to lose a hydrogen ion. The relative strengths of the binary acids increases from top to bottom in a group of the periodic table, so we expect H2S to be a stronger acid than H2O. This is because the H–S bond strength is lower than the H– O bond strength.
349
Chapter 16
16.10
CH3CH2O– is a stronger Brønsted base than CH3CH2S–. The S–H bond is weaker than the O–H bond, therefore, CH3CH2SH is a stronger acid than CH3CH2OH this in turn makes the conjugate base of the stronger acid a weaker base.
16.11
There are more oxygen atoms not attached to protons in HClO4 than in H2SeO4. In addition, the chlorine atom is more electronegative than selenium which makes the OH bond more polarized in HClO4.
16.12
conjugate pair ┌───────────────────────┐ HOCl(aq) + H2O H3O+(aq) + OCl–(aq) acid base acid base └─────┘ conjugate pair OCl– is a stronger base than water and H3O+ is a stronger acid than HOCl
16.13
The equilibrium would lie to the left, more so than to the right, so as to lie in the direction which favors formation of the weaker acid and base.
16.14
C2H3O2– is a stronger base than NO2–.
16.15
This equilibrium should lie to the left, because if HNO3 is a strong acid, then, by definition, NO3– must be a weak base.
16.16
Both HClO4 and HNO3 are 100% ionized in water and, therefore, appear to be the same strength. To distinguish between these two, we would need a solvent that is a weaker proton acceptor than H2O.
16.17
Water, formic acid, and acetic acid have similar but different dissociation constants. Hence, it is possible to distinguish between them. In NH3 solution, formic acid and acetic acid are fully dissociated and appear the same strength.
16.18
The molecule on the right will be the stronger acid because the very electronegative F atoms will stabilize the resulting ion.
16.19
The hydrogen on the tri-chloro compound is more acidic because the high electronegativity of Cl will stabilize the ion. So Cl2HCOO− is the stronger conjugate base.
16.20
Lewis acid: Electron pair acceptor Lewis base: Electron pair donor
16.21
The H+ ion accepts a pair of electrons from the oxygen atom of the water molecule. This makes the H+ ion the Lewis acid and the H2O molecule is the Lewis base.
16.22
H
H
H
F
C
N
B
H
H
F
F
H
350
H
H
F
C
N
B
H
H
F
F
Chapter 16
16.23
The Lewis base is the water molecule and the Lewis acid is the CO2. 16.24
The O2– ion has a complete valence shell. It can donate electron pairs functioning as a Lewis base. It cannot accept additional electrons and, therefore, cannot serve as a Lewis acid.
16.25
SbF5 has an empty orbital, it can accept a pair of electrons to form a Lewis acid-base adduct.
16.26
In the reaction of calcium with oxygen, there is a complete transfer of the electron pair from the calcium to the oxygen, then the electrostatic attractions bind the two ions together. In a Lewis acid–base reaction, a covalent bond is formed between the two atoms and the electrons are shared.
16.27
The element would be classified as a metal if it had a characteristic metallic luster, was a good conductor of heat and electricity, was ductile and malleable, formed ionic compounds with nonmetals and formed an oxide that was a basic anhydride. The element would be a nonmetal if it formed covalent bonds with other nonmetals.
16.28
Metal oxides are typically basic so this must be a non-metal oxide.
16.29 H
H O
H O
H
B
H
O
H
O
O
B
H
O
H
H
H O
O
B
H
O
H
O
O
O
B
H O
O H
16.30 16.31
O
O
H H H
H
H
This is due to the equilibrium: Cr(H2O)63+ + H2O Cr(H2O)5OH2+ + H3O+ The charge density on the Fe3+ is higher, so it is more acidic.
351
+ H+
Chapter 16
16.32 16.33
Because they have such a small charge density, the metal ions have little impact and essentially zero interaction with H2O in an acid–base sense. (a) H2SO4 (b) H2CO3 (c) H3PO4
16.34
(a) (b) (c)
CrO3 CrO Cr2O3
16.35
(a) (b)
Al2O3 + 6H+ 2Al3+ + 3H2O Al2O3 + 2OH– 2AlO2– + H2O
Review Problems (b) (e)
H2CO3 H2Cr2O7
(c)
CH3NH3+
CO32– HSO3–
(c)
SCN–
16.36
(a) (d)
HCl NH3
16.37
(a) (d)
HPO42– (b) IO2– (e)
16.38
(a)
conjugate pair ┌────────────────────┐ HNO3 + N2H4 N2H5+ + NO3– acid base acid base └───────┘ conjugate pair
(b)
conjugate pair ┌────────────┐ NH3 + N2H5+ NH4+ + N2H4 base acid acid base └────────────┘ conjugate pair
(c)
conjugate pair ┌──────────────────────┐ H2PO4– + CO32– HCO3– + HPO42– acid base acid base └────────┘ conjugate pair
(d)
conjugate pair ┌──────────────────────┐ HIO3 + HC2O4– H2C2O4 + IO3– acid base acid base └────────┘ conjugate pair
352
Chapter 16
16.39
(a)
conjugate pair ┌────────────────────┐ HSO4– + SO32– HSO3– + SO42– acid base acid base └───────┘ conjugate pair
(b)
conjugate pair ┌───────────────┐ H2O + S2– HS– + OH– acid base acid base └─────┘ conjugate pair
(c)
conjugate pair ┌─────────────────┐ H3O+ + CN– HCN + H2O acid base acid base └─────┘ conjugate pair
(d)
conjugate pair ┌─────────────────┐ + – H2Se + H2O H3O + HSe acid base acid base └─────┘ conjugate pair
16.40
(a) (b) (c)
HBr, HBr bond is weaker HF, more electronegative F polarizes and weakens the bond HBr, larger Br forms a weaker bond with H
16.41
(a) (b) (c)
H2Se, larger central atom forms a weaker bond with H HI, more electronegative atom PH3, larger central atom
16.42
(a) (b)
HClO3, because it has more oxygen atoms H3PO4, because it has more lone oxygen atoms
16.43
(a) (b)
HIO4, because it has more oxygen atoms H3AsO4, because it has more oxygen atoms
16.44
(a) (b) (c)
HClO3, because Cl is more electronegative HClO3, because the charge is more evenly distributed HBrO4, because the negative charge is more evenly distributed
16.45
(a) (b) (c)
H3PO4, since P is more electronegative HNO3, because N is more electronegative (HNO3 is a strong acid) HClO4, because Cl is more electronegative
353
Chapter 16
16.46
Cr(H2O)63+ (aq) + H2O(l)
16.47
Mg(OH)2 is an ionic species consisting of Mg2+ions and OH- ions. When the compound dissociates in water it produces OH- ions so the compound is a basic species.
→
Cr(H2O)5OH2+ (aq) + H3O+ (aq)
Si(OH)4 is a molecular compound with the OH groups attached to the silicon atom. This bonding weakens the OH bond so that a hydrogen ion will dissociate whe the compound is dissolved in water. Thus, Si(OH)4 is an acidic species. 16.48 H H
H
C
N
H
H
+
H+
H
Lewis Acid
C
N
H
H
H
Lewis base
16.49
Lewis acid:
BF3
F–
Lewis Base: F
F
+
F
B
F
F
F
B F
16.50
354
F
Chapter 16
16.51
Cl
Cl
Al
Al
Cl
Cl
Cl Al
Cl
Cl
Cl
16.52
SO2 Lewis acid
H2O Lewis base
16.53 2O
S
O
Cl
O
S
O
2O
Lewis acid SO2
Lewis base O2−
Lewis base
Lewis acid
16.54
355
O
Cl Al
Cl
Cl
Chapter 16
16.55 2O S
+
O
Lewis bases:
2-
O C
O O
C
O
O
+
O
O2– on CO32–
Lewis acids:
S
O
O
SO2 and CO2
Additional Exercises 16.56
Acid: CH3NH3+ Base: CH3NH–
16.57
Molecular equation NaNH2 + NH4Cl →
NaCl +2NH3
Ionic equation Na+ + NH2- + NH4+ + Cl- → Net ionic equation NH2- + NH4+ →
Na+ + Cl- + 2NH3
2NH3
NH4+ is an acid and NH2- is a base 16.58
SOCl2 + Na2SO3 →
2SO2 + 2NaCl
SOCl2 is an acid and Na2SO3 is a base. From a Lewis definition SO2+ is an electron acceptor and SO32- is an electron donor. SO2+ accepts an electron pair from one of the oxygen atoms on SO32- and the SO32- unit loses the oxygen atom. 16.59
Ethanamide, also known as acetamide, is a stronger acid and therefore a weaker base than NH3. The H3CCO unit withdraws electron density from the NH2 group thus making it easier for the hydrogen atoms on the NH2 unit to dissociate. Ka for ethanamide is 2.34 x 10-1 while Ka for NH3 is 5.56 x 10-10
16.60
NH3 is a stronger base than NCl3. This is due to the difference in the electron withdrawing strength of Cl compared with H. Cl is more electronegative than H and will withdraw more electron density from the nitrogen than hydrogen. Thus, the nitrogen atom in NCl3 has a greater positive charge than it does in NH3. The greater positive charge means that is more difficult to add H+ to NCl3 than it is to add H+ to NH3.
16.61
Ethanedioic acid (pKa1 = 1.23, pKa2 = 4.19) is stronger than butanedioic acid (pKa1 = 2.82, pKa2 = 5.70). For ethanedioic acid the two COOH groups are connected to each other. In the case of butanedioic acid the COOH groups are separated by CH2 groups. The closer the two COOH groups are to one another the more influence they have on weakening the O-H bond due to their electron withdrawing effects. In essence, you dilute the electronic affect of neighboring electron withdrawing groups as you increase their separation from the group they are influencing.
16.62
HSO4– + H2O → H2SO4 + OH− HSO4– + H2O → SO42− + H3O+
356
Chapter 16
16.63
(a) (b)
16.64
In H2O2, the oxygen atom helps to stabilize the HO2– ion making it easier for H2O2 to lose a proton than can H2O. acidic
Acid-base properties are based on the structure of the compound and on bond strengths. In the case of KOH you have an ionic compound consisting of K+ ions and OH- ions. Therefore, when added to water KOH will dissociate into these ions. Thus, KOH is a source of OH- ions and is a base. In the case of Al(H2O)3(OH)3 the OH groups are covalently bonded to the Al atom. Depending on the conditions, the OH- group can react with an H+ or an OH- react with the Al(H2O)3(OH)3. In acid conditions an OH- on Al(H2O)3(OH)3 will react with the H+. Complete neutralization will result in Al(H2O)63+. In base OH- is added to the Al(H2O)3(OH)3 molecule resulting in [Al(H2O)2(OH)4]- being formed. In the case of HClO3, the Cl and O atoms are very electronegative, withdrawing electrons and thus weakening the O-H bond. This increases the positive charge on the H making it a very acidic hydrogen.
16.65
The equilibrium lies to the right since reactions favor the weaker acid and base.
16.66
acids: bases:
16.67
NH3OH+ is the stronger acid
16.68
HClO3 will ionize more in the more weakly acidic solvent, H2O. H2F+ is a very strong acid and the second equilibrium will be to the left.
(CH3)2NH2+, NH3OH+ NH2OH, (CH3)2NH
Multi-Concept Problems 16.69
The structure with the bulkier H2O groups at 1800 apart is more stable. 16.70
The predominate species in solution can be determined from the appropriate equilibrium constants.
357
Chapter 16
Arsenic acid and arsenous acid are both weak acids therefore, the predominate species will be the molecular acid forms. The two salts are soluble and the equilibrium constants for their respective anion reactions with water are very small so the predominate species will be the anions. The predominate species in solution are: H3AsO4, H3AsO3, AsO43-, and AsO33-
O OH
As
OH OH
HO
As
OH
OH
Arsenic acid
Arsenous acid
3-
3-
O O
As
O
O
Arsenate ion
O
As
O
O
Arsenite ion
H3AsO + H3AsO3 6H+ + AsO43- + AsO33–
358
Chapter 17
Practice Exercises 17.1
[H3O+] = 12 M 1.0 x 10-14 = [H3O+][OH-] = 12 [OH-] [OH–] = 8.3 × 10–16 M
17.2
1.00 x 10-14 = [H3O+][OH-] [H3O+] = 1.00 x 10-14/7.8 x 10-6 = 1.3 x 10-9 The solution is basic
17.3
pOH = 14 – pH = 14 – 4.25 = 9.75 [H3O+] = 10–4.25 = 5.6 × 10–5 M [OH–] = 10–9.75 = 1.8 × 10–10 M
17.4
pH = –log[H3O+] = –log[3.2 × 10–5] = 4.49 pOH = 14.00 – pH = 14.00 – 4.49 = 9.51
17.5
In general, we have the following relationships between pH, [H+], and [OH–]: [H3O+] = 10–pH [H3O+][OH−] = 1.00 × 10–14 (a)
(b)
(c)
(d)
(e)
[H3O+] = 10–2.90 = 1.3 × 10–3 M [OH–] = 1.00 × 10–14/1.3 × 10–3 M = 7.7 × 10–12 M The solution is acidic. [H3O+] = 10–3.85 = 1.4 × 10–4 M [OH–] = 1.00 × 10–14/1.4 × 10–4 M = 7.1 × 10–11 M The solution is acidic. [H3O+] = 10–10.81 = 1.549 × 10–11 M = 1.6 x 10-11M [OH–] = 1.00 × 10–14/1.549 × 10–11 M = 6.456 × 10–4 M = 6.5x 10-4M The solution is basic. [H3O+] = 10–4.11 = 7.8 × 10–5 M [OH–] = 1.00 × 10–14/7.8 × 10–5 M = 1.3 × 10–10 M The solution is acidic. [H3O+] = 10–11.61 = 2.4 × 10–12 M [OH–] = 1.00 × 10–14/2.4 × 10–12 M = 4.1 × 10–3 M The solution is basic.
17.6
[H3O+] = 0.0050 M pH = –log[H+] = –log[0.0050] = 2.30 pOH = 14.0 – pH = 14.00 – 2.30 = 11.70
17.7
First determine the number of moles of KOH and then the molarity of the solution 1 mol KOH mol KOH = (1.20 g KOH) = 0.0214 mol KOH 56.11 g KOH
0.0214 mol KOH = 0.0855 M KOH 0.250 L solution pOH = –log 0.0855 M = 1.07 pH = 14 – 1.068 = 12.93 [H3O+] = 1.2 × 10–13 M molarity =
359
Chapter 17
17.8
[H3O+] = 10–5.5 = 3.2 × 10–6 M
17.9
(a)
HC2H3O2 + H2O H3O+ + C2H3O2– Ka =
(b)
(CH3)3NH+ + H2O H3O+ + (CH3)3N
Ka = (c)
(a)
[H3O + ][H 2 PO 4 − ] H 3 PO 4
[H3O + ][CHO2 − ] [ HCHO2 ]
(CH3)2NH2+ + H2O H3O+ + (CH3)2NH
Ka = (c)
( CH ) NH + 3 3
HCHO2 + H2O H3O+ + CHO2– Ka =
(b)
[H3O + ][ ( CH3 )3 N]
H3PO4 + H2O H3O+ + H2PO4– Ka =
17.10
[H3O+ ][C2 H3O2 − ] HC2 H3O2
[H3O+ ][(CH3 ) 2 NH ] (CH3 ) 2 NH 2 +
H2PO4– + H2O H3O+ + HPO42– Ka =
[H3O + ][HPO4 2− ] H 2 PO4 −
17.11
The smaller the value of pKa, the stronger the acid. The acids stronger than acetic acid and weaker than formic acid from Table 17.2 are barbituric acid and hydrazoic acid..
17.12
The acid with the smaller pKa (HA) is the strongest acid. Since pKa = – log Ka, Ka = 10–pKa For HA: Ka = 10–3.16 = 6.9 × 10–4 For HB: Ka = 10–4.14 = 7.2 × 10–5
17.13
(a)
(CH3)3N + H2O (CH3)3NH+ + OH– Kb =
(b)
SO32– + H2O HSO3– + OH– Kb =
(c)
[(CH3 )3 NH + ][OH − ] (CH3 )3 N [HSO3− ][OH − ] SO32 −
NH2OH + H2O NH3OH+ + OH– Kb =
[NH3OH + ][OH − ] [ NH 2 OH ]
360
Chapter 17
17.14
(a)
HS– + H2O H2S + OH– Kb =
(b)
H2PO4– + H2O H3PO4 + OH– Kb =
(c)
[H 2 PO 4 − ][OH − ] HPO 42 −
HCO3– + H2O H2CO3 + OH– Kb =
(e)
[H3 PO4 ][OH − ] H 2 PO 4−
HPO42– + H2O H2PO4– + OH– Kb =
(d)
[H 2S][OH − ] HS−
[H 2 CO3 ][OH − ] HCO3−
HSO3– + H2O H2SO3 + OH– Kb =
[H 2SO3 ][OH − ] HSO3−
17.15
For conjugate acid base pairs, Ka × Kb = Kw: Kb = Kw ÷ Ka = 1.0 × 10–14 ÷ 2.3 × 10–11 = 4.3 × 10–4
17.16
For conjugate acid base pairs, Ka × Kb = Kw: Ka = Kw ÷ Kb = 1.0 × 10–14 ÷ 5.5 × 10–11 = 1.8 × 10–4
17.17
HSal + H2O H3O+ + Sal–
H O + Sal− 3 Ka = [ HSal] [H3O+] = 10–pH = 10–1.83 = 0.0148 M
I C E
[HSal] 0.200 –x 0.200–x
[H3O+] – +x +x
[Sal–] – +x +x
We know the [H+] is 0.0148 M x = 0.0148 [HSal] = 0.200 – 0.0148 = 0.185 M [Sal–] = 0.0148 M H O + Sal− 3 = ( 0.0148 )( 0.0148 ) = 1.2 × 10–3 Ka = ( 0.185 ) [ HSal] pKa = –log (Ka) = –log (1.18 × 10–3) = 2.93
361
Chapter 17
17.18
HBu + H2O H3O+ + Bu– Ka =
[H3O+ ][Bu − ] [ HBu ]
I C E
[H3O+] – +x +x
[HBu] 0.01000 –x 0.01000–x
[Bu–] – +x +x
We know that the acid is 4.0% ionized so x = 0.01000 M × 0.040 = 0.00040 M. Therefore, our equilibrium concentrations are [H+] = [Bu–] = 0.00040 M, and [HBu] = 0.01000 M – 0.00040 M = 0.00960 M. Substituting these values into the mass action expression gives: (0.00040)(0.00040) = 1.7 × 10–5 Ka = 0.00960 pKa = –log(Ka) = –log(1.7 × 10–5) = 4.78 17.19
We will use the symbol Mor and HMor+ for the base and its conjugate acid respectively: Mor + H2O HMor+ + OH–
Kb =
[HMor + ][OH − ] [ Mor ]
I C E
[HMor+] – +x +x
[Mor] 0.010 –x 0.010–x
[OH–] – +x +x
At equilibrium, the pOH = 3.90. The [OH–] = 10–pOH = 10–3.90 = 1.26 × 10–4 M = x. Substituting these values into the mass action expression gives:
Kb =
(1.26 × 10−4 )(1.26 × 10−4 ) −4
= 1.6 × 10–6
0.010 − 1.26 × 10 pKb = –log(Kb) = –log(1.61 × 10–6) = 5.79 17.20
HC6H5CO2 + H2O H3O+ + C6H5CO2-
C H CO − H O+ 6 5 2 3 = 6.5 × 10–5 Ka = HC6 H5 CO2
I C E
HC6H5CO2 0.023 –x 0.023–x
[H3O+] – +x +x
C6H5CO2– +x +x
Assume that x << 0.050 and substitute the equilibrium values into the mass action expression to get:
C H CO − H O+ 6 5 2 3 = 6.5 × 10–5 Ka = HC H CO 6 5 2 362
Chapter 17
Ka =
( x )( x ) ( 0.023)
= 6.5 × 10–5
x = 1.2 × 10–3 [H3O+] = x = 1.2 × 10–3M pH = –log[H3O+] = –log(1.2 × 10–3) = 2.91 17.21
C5NH4COOH + H2O H3O+ + C5NH4COO– Ka =
[H 3O + ][C5 NH 4 COO − ] = 1.4 × 10−5 C NH COOH 4 5
I C E
[H3O+] – +x +x
[C5NH4COOH] 0.050 –x 0.050 – x
[C5NH4COO–] – +x +x
Assume that x << 0.050 and substitute the equilibrium values into the mass action expression to get: [x][x] Ka = = 1.4 × 10−5 [ 0.050] Solving for x we determine that x = 8.4 × 10–4 M = [H+]. pH = –log[H+] = –log(8.4 × 10–4) = 3.08 17.22
C6H5NH2 + H2O C6H5NH3+ + OH– Kb =
[C6 H 5 NH 3+ ][OH − ] = 3.9 × 10−10 C6 H 5 NH 2
I C E
[C6H5NH2] 0.025 –x 0.025 – x
[C6H5NH3+] – +x +x
[OH–] – +x +x
Assume that x << 0.025 and substitute the equilibrium values into the mass action expression to get: ( x )( x ) = 3.9 × 10–10 Kb = ( 0.025 ) x = 3.12 × 10–6 [OH–] = 3.12 × 10–6 pOH = –log[OH–] = –log(3.12 × 10–6) = 5.51 17.23
C5H5N + H2O C5H5NH+ + OH– Kb =
[C5 H 5 NH + ][OH − ] = 1.7 × 10 −9 C5 H 5 N
I C E
[C5H5N] 0.010 –x 0.010 – x
[C5H5NH+] – +x +x
[OH–] – +x +x
Assume that x << 0.010 and substitute the equilibrium values into the mass action expression to get:
363
Chapter 17
Kb =
[x][x] = 1.7 × 10−9 [ 0.010]
Solving for x we determine that x = 4.1 × 10–6 M = [OH–]. pOH = –log[OH–] = –log(4.1 × 10–6) = 5.38 pH = 14.00 – pOH = 8.62 17.24
We will use the notation phenol and phenolate for the acid and its conjugate base, respectively: phenol H+ + phenolate
Ka =
I C E
[H + ][phenolate] = 1.3 × 10−10 [ phenol] [phenol] 0.15 –x 0.15 – x
[H+] – +x +x
[phenolate] – +x +x
If we assume that x << 0.15, a good assumption based upon the size of Ka, we can substitute the equilibrium values in to the mass action expression to get: (x)(x) Ka = = 1.3 × 10−10 0.15 Solving gives x = 4.42 × 10–6 M = [H+] pH = –log[H+] = –log(4.42 × 10–6) = 5.35 17.25
Examine the ions in solution one at a time: The cation for acidity and the anion for basicity. (a) NaNO2 Na+ is an ion of a Group I metal and is not acidic NO2– is the conjugate base of HNO2, a weak acid, therefore it is a weak base. The solution should be basic. (b) KCl K+ is an ion of a Group I metal and is not acidic Cl– is the conjugate base of HCl, a strong acid, therefore it is not a strong base. The solution should be neutral. (c) NH4Br NH4+ is the conjugate acid of NH3, a weak base, therefore it is a weak acid Br– is the conjugate base of HBr, a strong acid, therefore it is not a strong base. The solution should be acidic.
17.26
Examine the ions in solution one at a time: The cation for acidity and the anion for basicity. (a) NaNO3 Na+ is an ion of a Group I metal and is not acidic NO3– is the conjugate base of HNO3, a strong acid, therefore it is not a strong base. The solution should be neutral. (b) KOCl K+ is an ion of a Group I metal and is not acidic OCl– is the conjugate base of HOCl, a weak acid, therefore it is a weak base. The solution should be basic. (c) NH4NO3 NH4+ is the conjugate acid of NH3, a weak base, therefore it is a weak acid NO3– is the conjugate base of HNO3, a strong acid, therefore it is not a strong base. The solution should be acidic.
364
Chapter 17
17.27
The chloride ion is neutral since it is the conjugate base of a strong acid, HCl. The ion CH3NH3+ forms from the reaction of CH3NH2 with water and will act as a weak acid. CH3NH3+ + H2O H3O+ + CH3NH2
K w 1.0 ×10−14 = = 2.3 × 10–11 −4 Kb 4.4 ×10
Ka =
Ka =
CH3 NH 2 H3O +
CH NH + 3 3 The concentration of the salt, CH3NH3Cl, in water is 1 mol CH3 NH3Cl 1 M CH3NH3C = (25.0 g CH3NH3Cl) = 0.740 M CH3NH3Cl 67.53 g CH NH Cl 0.500 L 3 3
[CH3NH3+] [H3O+] I 0.740 – C –x +x E 0.740 – x +x Assume that x is small relative to 0.740 M CH3NH3Cl. ( x )( x ) = 2.3 × 10–11 Ka = ( 0.740 )
[CH3NH2] – +x +x
x = 4.13 × 10–6 = [H+] pH = –log[H+] = –log(4.13 × 10–6) = 5.39 17.28
The sodium ion is neutral since it is the salt of the strong base, NaOH. The nitrite ion is basic since it is the salt of nitrous acid, HNO2, a weak acid. The equilibrium we are interested in for this problem is: NO2– + H2O HNO2 + OH–.
Kb =
[HNO2 ][OH − ] − NO2
In order to determine the value for Kb recall that Ka Kb = Kw. We can look for the value of Ka for HNO2 Kb = 1.0 × 10–14 ÷ 4.6 × 10–4 = 2.17 × 10–11.
I C E
[NO2–] 0.10 –x 0.10 – x
[HNO2] – +x +x
[OH–] – +x +x
Assume that x << 0.10 and substitute the equilibrium values into the mass action expression to get: [x][x] Kb = = 2.17 × 10−11 0.10 [ ] Solving we determine that x = 1.47 × 10–6 M = [OH–]. pOH = –log[OH–] = –log(1.2 × 10–6) = 5.83 pH = 14.00 – pOH = 8.17 17.29
Upon mixing, the NH3 will react with HBr to form NH4Br. The initial concentration of the NH4Br will be: 0.20 mol NH 3 mol NH3 = (500 mL solution) = 0.10 mol NH3 1000 mL solution 0.20 mol HBr mol HBr = (500 mL solution) = 0.10 mol HBr 1000 mL solution
365
Chapter 17
The NH3 and HBr are in a 1:1 ratio, therefore the number of moles of NH4Br is 0.10. The volume of the solution is 500 mL + 500 mL = 1000 mL = 1.0 L The concentration of NH4Br is: 0.10 mol NH 4 Br M NH4Br = = 0.10 M NH4Br 1.00 L solution As previously determined, a solution of NH4Br will be acidic since NH4+ is the salt of a weak base and Br– is the salt of a strong acid. As in the previous Practice Exercise, we need to determine the value for the dissociation constant using the relationship Ka × Kb = Kw and the value of Kb for NH3 as listed in Table 17.3. Ka = 1.0 × 10–14 ÷ 1.8 × 10–5 = 5.6 × 10–10. The equilibrium reaction is: NH4+ NH3 + H+
I C E
[NH4+] 0.10 –x 0.10 – x
[NH3] – +x +x
[H+] – +x +x
Assume that x << 0.10 and substitute the equilibrium values into the mass action expression to get: [x][x] Ka = = 5.6 × 10 −10 [ 0.10] Solving we determine that x = 7.5 × 10–6 M = [H+]. pH = –log[H+] = –log(7.5 × 10–6) = 5.13 17.30
NH3 is a weak base and its conjugate, the ammonium ion, is a weak acid. The cyanide ion is the conjugate base of HCN, a weak acid. In order to determine if the solution is acidic or basic, we need to determine the relative strength of the two components. Use the relationship Ka × Kb = Kw in order to determine the dissociation constants for the cyanide ion and the ammonium ion. Ka(NH4+) = Kw ÷ Kb(NH3) = 1.0 × 10–14 ÷ 1.8 × 10–5 = 5.6 × 10–10 Kb(CN–) = Kw ÷ Ka(HCN) = 1.0 × 10–14 ÷ 4.9 × 10–10 = 2.0 × 10–5 Since the Kb(CN–) is larger than the Ka(NH4+) the NH4CN solution will be basic.
17.31
Using the same logic as the previous question: NH4+ is the salt of a weak base, and is acidic NO2– is the salt of a weak acid, and is basic The relative strengths of the two components must be compared. Ka(NH4+) = Kw ÷ Kb(NH3) = 1.0 × 10–14 ÷ 1.8 × 10–5 = 5.6 × 10–10 Kb(NO2–) = Kw ÷ Ka(HNO2) = 1.0 × 10–14 ÷ 4.6 × 10–4 = 2.2 × 10–11 The Ka (NH4+) is larger than the Kb(NO2–), thus the solution will be acidic.
17.32
In the acetate buffer, there are HC2H3O2 and C2H3O2– present. Upon addition of a strong acid, the concentration of HC2H3O2 will increase: H+ + C2H3O2– HC2H3O2 When a strong base is added, it reacts with the acid to form more of the acetate ion; therefore the concentration of the acetic acid will decrease: HC2H3O2 + OH– C2H3O2– + H2O
17.33
(a) (b)
H+ + NH3 NH4+ OH– + NH4+ H2O + NH3
366
Chapter 17
17.34
The equation is: C2H3O2– + H2O HC2H3O2 + OH– Start by determining Kb for acetate ion using Kw = KaKb Kb = Kw/Ka = 1.0 × 10–14/1.8 × 10–5 = 5.56 × 10–10 HC 2 H 3O 2 OH − Kb = C H O − 2 3 2
I C E
Kb =
[C2H3O2–] 0.11 –x 0.11 – x
[HC2H3O2] 0.090 +x 0.090 +x
[OH–] – +x +x
[ x ][0.090 + x ] [0.11 - x ]
assume x << 0.090 and solve for x x = [OH ] = 6.79 × 10
10
pOH = 9.17 pH = 14.00 – 9.17 = 4.83, the difference is due to rounding errors 17.35
Find the concentrations of the acetic acid and the acetate ion: 1 mol HC2 H3O 2 1 M HC2H3O2 = (100.0 g HC2H3O2) = 1.665 M 60.053 g HC H O 1 L solution 2 3 2
1 mol HC 2 H3O 2 1 M NaC2H3O2 = (100.0 g HC2H3O2) = 1.219 M 82.035 g HC 2 H3O 2 1 L solution + – HC2H3O2 C2H3O2 + H C H O − H + 2 3 2 = 1.8 × 10–5 Ka = HC2 H3O 2
I C E
Ka =
[HC2H3O2] 1.665 –x 1.665 – x
(1.219 + x )( x ) (1.665 − x )
[C2H3O2–] 1.219 +x 1.219 +x
[H+] – +x +x
= 1.8 × 10–5
Assume that x is small and solve for x x = 2.5 × 10–5 pH = –log[H+] = –log(2.5 × 10–5) = 4.61 17.36
Use propanoic acid. It has a pKa of 4.89 which is within the range of pH = pKa ± 1 You may assume no volume change when adding the sodium salt of the acid to the solution. [salt ] pH = pKa + log [acid ] [acid] = 0.200 M
367
Chapter 17
5.25 = 4.89 + log
[salt ] [0.200]
[salt] = 0.458 M
0.458 mol NaC3 H5 O 2 96.06 g NaC3 H5 O 2 g salt = ( 0.5000 L ) = 22.0 g NaC3H5O2 1L 1 mol NaC3 H5 O2 Propanoic acid buffer, 22.1 g NaC3H5O2 Using the same calculations, the following buffers can also be used: Acetic acid buffer, 26.2 g NaC2H3O2 Hydrazoic acid buffer, 29.0 NaN3, Butanoic acid buffer, 29.6 g NaC4H7O2, 17.37
Yes, formic acid and sodium formate would make a good buffer solution since pKa = 3.74 and the desired pH is within one pH unit of this value. Desired [H+] = 10-3.90 = 1.259 x 10-4 M Using Equation 17.10; H + = K a ×
( mol HCHO2 )initial
( mol CHO ) −
2
initial
Rearranging and substituting the known values we get; H+ 1.259 × 10−4 = 0.70 mol = = Ka mol CHO 2 −initial 1.8 × 10−4
mol HCHO 2initial
0.1 mol HCHO = 0.70 x ( mol CHO2-) mol CHO2- = 0.1 mol HCHO/0.70 = 0.1428 mol CHO2(0.1428 mol NaCHO2)(68.007 g/1 mol) = 9.7 g NaCHO2. 17.38
The NaOH added to a buffer solution will react with the HC2H3O2. NaOH + HC2H3O2 NaC2H3O2 + H2O First, calculate the amount of acid and base after the addition of the NaOH: mol NaC2H3O2 = 1.00 mol NaC2H3O2 + 0.15 mol NaC2H3O2 = 1.15 mol NaC2H3O2 mol HC2H3O2 = 1.00 mol HC2H3O2 – 0.15 mol NaC2H3O2 = 0.85 mol HC2H3O2 1.15 mol NaC 2 H 3O 2 M NaC2H3O2 = 1 L solution 0.85 mol HC2 H 3O 2 M HC2H3O2 = 1 L solution salt ] [ (1.15) = 4.87 pH = pKa + log = 4.74 + log ( 0.85) [acid ] The pH of the initial solution was 4.74 since the concentration of the acid equals the concentration of the salt. The pH change: 4.74 – 4.87 = 0.13 pH units
17.39
Calculate the moles of the salt and the acid, then calculate the pH of the solution. 1 mol NH3 mol NH3 = (50.0 g NH3) = 2.94 mol NH3 17.03 g NH3 368
Chapter 17
1 mol NH 4 Cl mol NH4Cl = (50.0 g NH4Cl) = 0.935 mol NH4Cl 53.49 g NH 4 Cl NH4+ NH3 + H+ K 1.0 ×10−14 = 5.6 × 10–10 Ka = w = −5 Kb 1.8 ×10 pKa = –log(Ka) = –log(5.6 × 10–10) = 9.26 [salt ] = 9.26 + log ( 2.94 ) = 9.76 pH = pKa + log ( 0.935) [acid ] If 5.00 g of HCl is add, the solution will become more acidic: NH3 + H+ NH4+ 1 mol HCl mol HCl = (5.00 g HCl) = 0.137 mol HCl 36.46 g HCl The number of moles of NH4+ will increase by this amount and the number of moles of NH3 will decrease by this amount. mol NH3 = 2.94 mol NH3 – 0.137 mol = 2.80 mol NH3 mol NH4+ = 0.935 mol + 0.137 mol = 1.07 mol NH4+ [salt ] = 9.26 + log ( 2.80 ) = 9.67 pH = pKa + log (1.07 ) [acid ]
17.40
+
H3PO4 + H2O H3O + H2PO4
–
H O+ H PO − 3 2 4 Ka = H3 PO4
H O + HPO 2− 3 4 H2PO4– + H2O H3O+ + HPO42– Ka = H PO − 2 4 HPO4
17.41
2–
+
+ H2O H3O + PO4
3–
H O + PO 3− 3 4 Ka = HPO 2 − 4
[H+] is determined by the first protic equilibrium: H2C6H6O6 H+ + HC6H6O6– The mass action expression is: Ka1 = 8.0 × 10–5 = x2/0.10 x = [H+] = 2.8 × 10–3 M pH = –log(2.8 × 10–3) = 2.55 The concentration of the anion, [HC6H6O6–], is given almost entirely by the second ionization equilibrium: HC6H6O6– H+ + C6H6O62– for which the mass action expression is:
K a2
H + C H O 2− 6 6 6 = = 1.6 × 10−12 HC H O − 6 6 6
We have used the value for Ka2 from Table 17.4. Using the value of x from the first step above gives:
369
Chapter 17
( 2.8 × 10 ) C H O = ( 2.8 × 10 ) −3
1.6 × 10
−12
6
6 6
2−
−3
[HC6H6O6–] = 1.6 × 10–12 17.42
CO32–(aq) + H2O HCO3–(aq) + OH–(aq) Kb =
Kw 1.0 × 10−14 = = 2.1 × 10–4 −11 Ka2 4.7 × 10
HCO − OH − 3 Kb = 2.1 × 10–4 = 2 − CO 3 [CO32–] 0.10 –x 0.10–x
I C E
Kb = 2.1 × 10–4 =
[HCO3–] – +x +x
[OH–] – +x +x
( x )( x ) ( 0.10 − x )
x = [OH–] = 4.5 × 10–3 M pOH = –log(4.5 × 10–3) = 2.35 pH = 14.00 – 2.35 = 11.66 It is basic, but it is too basic to be a substitute for NaHCO3. 17.43
The equilibrium we are interested in for this problem is: SO32–(aq) + H2O HSO3–(aq) + OH–(aq) Kb = Kw/Ka2 = 1.0 × 10–14 / 6.4 × 10–8 = 1.6 × 10–7
K b = 1.6 × 10
−7
HSO − OH − 3 = SO 2− 3
[SO32–] [HSO3–] [OH–] I 0.20 – – C –x +x +x E 0.20–x +x +x Substituting these values into the mass action expression gives: (x)(x) K b = 1.6 × 10 −7 = 0.20 − x Assume that x << 0.20 and solving gives x = 1.8 × 10–4. x = [OH–] = 1.8 × 10–4 M pOH = –log(1.8 × 10–4) = 3.75 pH = 14.00 – pOH = 14.00 – 3.75 = 10.25 17.44
For weak polyprotic acids, the concentration of the polyvalent ions is equal to the volume of Kan where n is the valency. By analogy, the concentration of H2SO3 in 0.010 M Na2SO3 will be equal to Kb2 for SO32–.
17.45
(a) (b)
H2O, K+, HC2H3O2, H+, C2H3O2–, and OH– [H2O] > [K+] > [C2H3O2–] > [OH–] > [HC2H3O2] > [H+] H2O, HC2H3O2, H+, C2H3O2–, and OH– 370
Chapter 17
(c) (d)
17.46
[H2O] > [HC2H3O2] > [H+] = [C2H3O2–] > [OH–] H2O, K+, HC2H3O2, H+, C2H3O2–, and OH– [H2O] > [K+] > [OH–] > [C2H3O2–]> [HC2H3O2] > [H+] H2O, K+, HC2H3O2, H+, C2H3O2–, and OH– [H2O] > [HC2H3O2] > [K+] > [C2H3O2–]> [OH–] > [H+]
HCHO2 + H2O H3O+ + CHO2–
Ka =
[H3O + ][CHO 2 ] = 1.8 × 10−4 [HCHO 2 ]
(a) I C E
[HCHO2] 0.100 –x 0.100–x
[H3O+] – +x +x
[CHO2–] – +x +x
Assume x << 0.100. Solving we get x = [H3O+] = 4.2 × 10–3. The pH is 2.37. (b) (c)
[HCHO2] = [CHO2–] so [H3O+] = Ka = 1.8 × 10–4 and the pH = 3.74. 0.100 mol −3 mol base added = (15.0 mL) = 1.50 × 10 1000 mL 0.10 mol −3 mol acid initially present = (20.0 mL) = 2.00 × 10 1000 mL excess acid = 2.00 × 10 −3 − 1.50 × 10−3 = 0.50 × 10 −3 moles acid
[acid] =
0.50 × 10−3 moles = 1.43 × 10−2 M 1L (35 mL) 1000 mL
1.50 × 10 −3 moles = 4.29 × 10−2 M 1L (35 mL) 1000 mL Substituting into the equilibrium expression and solving we get [H3O+] = 6.00 × 10–5 and the pH = 4.22. [base] =
(d)
17.47
We now have a solution of formate ion with a concentration of 0.0500 M. We need Kb for formate ion: Kb = Kw/Ka = 5.6 × 10–11. If we set up the equilibrium problem and solve we get: [OH–] = 1.7 × 10–6. The pOH = 5.78 and the pH = 8.22. 0.15 mol −3 mol base added = (30.0 mL) = 4.50 × 10 1000 mL 0.20 mol −2 mol acid initially present = (50.0 mL) = 1.00 × 10 1000 mL excess acid = (1.00 × 10–3) – (4.50 × 10–3) = 5.50 × 10–3 mol acid
371
Chapter 17
5.50 × 10 −3 moles = 6.88 × 10−2 M 1L (80 mL) 1000 mL
[acid] =
4.50 × 10−3 moles = 5.63 × 10−2 M 1L (80 mL) 1000 mL
[base] =
HCHO2 + H2O H3O+ + CHO2–
Ka =
[H3O + ][CHO 2 ] [HCHO 2 ] I C E
[HCHO2] 6.88 × 10–2 –x 6.88 × 10–2–x
[H3O+] – +x +x
[CHO2–] 5.63 × 10–2 +x 5.63 × 10–2+x
Assume x << 5.63 × 10–2
Ka =
[x][5.63 × 10−2 ] −2
= 1.8 x 10−4
[6.88 × 10 ]
x = 2.20 × 10−4 = [H3O + ] pH = 3.66
Review Questions 17.1
H2O + H2O H3O+ + OH– Kw = [H3O+][OH–]
17.2
(a)
neutral solution acidic solution basic solution
[H3O+] = [OH–] = 1.0 × 10–7 M [H3O+] > [OH–] [H3O+] < [OH–]
(b)
neutral solution acidic solution basic solution
pH = 7.00 pH < 7.00 pH > 7.00
pOH = 7.00 pOH > 7.00 pOH < 7.00
17.3
pH + pOH = 14.00
17.4
The product of H+ and OH– is a constant, 1 × 10–14. By adding an acid to water the acid provides the H+ ions, this forces the water to decrease the amount of OH– in solution, thus suppressing the ionization of the water. The same effect occurs with the OH–, but it is the base that is forcing the water to not ionize.
17.5
The leveling effect of water is a result of the strength of water as an acid or a base. If it is the weaker acid or base in solution, then the other substance will ionize completely.
17.6
HI, HBr, HClO4, HCl, H2SO4(first dissociation), HNO3, HClO3
17.7
The reaction of the acid with a solvent, usually water, that causes the dissociation of a proton.
17.8
Strong bases are those that ionize 100 % producing an excess of OH- in solution.
372
Chapter 17
17.9
The key to a safe antacid is solubility. If the base releases OH- ions readily in water then damage to the esophagus will occur. A low soluble base such as Mg(OH)2 will dissolve in the stomach, not in the esophagus, and thus be safe for consumption.
17.10
HA + H2O H3O+ + A–
Ka = 17.11
17.12
17.13
17.14
17.15
[H3O+ ][A- ] [HA]
(a)
HNO2 H+ + NO2–
(b)
H3PO4 H+ + H2PO4–
(c)
HAsO42– H+ + AsO43–
(d)
(CH3)3NH+ H+ + (CH3)3N
(a)
H + NO − 2 Ka = [ HNO2 ]
(b)
H + H PO − 2 4 Ka = H3 PO 4
(c)
H + AsO 3− 4 Ka = 2− HAsO4
(d)
H + (CH ) N 3 3 Ka = (CH ) NH + 3 3
B + H2O HB+ + OH– Kb =
[HB+ ][OH ] [B]
(a)
(CH3)3N + H2O (CH3)3NH+ + OH–
(b)
AsO43– + H2O HAsO42– + OH–
(c)
NO2– + H2O HNO2 + OH–
(d)
(CH3)2N2H2 + H2O (CH3)2N2H3+ + OH–
(a)
(CH ) NH + OH − 3 3 Kb = (CH3 )3 N
(c)
Ka =
[ HNO2 ] OH − − NO 2
HAsO 4 2− OH −
(b)
Kb =
(d)
(CH ) N H + OH − 3 2 2 3 Kb = (CH ) N H 3 2 2 2
AsO 43−
17.16
In general, there is an inverse relationship between the strength of an acid and its conjugate base; the stronger the acid, the weaker its conjugate base. Since HCN has the larger value for pKa, we know that it is a weaker acid than HF. Accordingly, CN– is a stronger base than F–.
17.17
(a)
(b) CH3 H 3C
H2 C
N
H
N
H
373
H
Chapter 17
(c) H H
O
N
H
H
17.18
(a)
(b) CH3 H 3C
NH 2
N CH3
(c)
H
H
H
N
N
H
17.19
Percentage ionization is the number of moles ionized per liter of solution divided by the moles available per liter of solution times 100. moles ionized per liter Percentage ionization = × 100% moles available per liter
17.20
If [HA]initial < 400 × Ka or if the % ionization ≥ 5%, the initial concentration of the acid is not equal to the equilibrium concentration. The same argument holds true for bases.
17.21
Simplification works when [HA]initial ≥ 400 × Ka (a) 400 × (1.8 × 10–5) = 0.0072 0.0072 M < 0.020 M Use initial concentration (b)
(c)
(d)
400 × (4.4 × 10–4) = 0.176 0.176 M > 0.10 M
May not use initial concentration
400 × (1.7 × 10–6) = 6.8 × 10–4 6.8 × 10–4 M < 0.002 M
Use initial concentration
400 × (1.8 × 10–4) = 7.2 × 10–2 7.2 × 10–2 M > 0.050 M
May not use initial concentration
17.22
The acetylsalicylate ion is the anion of the salt of a weak acid and is therefore a weak base. The solution of the sodium salt will be basic since the sodium ion is neutral.
17.23
The oxalate ion hydrolyzes in water, to give a basic solution, according to the equation: C2O42– + H2O HC2O4– + OH–
17.24
Recall that the salt of a weak acid is basic, the cation of a salt of a weak base is acidic and the salt of a strong acid or base is neutral. (a)
acidic: NH4Br: NH4+ is the salt of NH3, a weak base.
374
Chapter 17 KF: F– is the anion of a salt of HF, a weak acid. KCN: CN– is the anion of a salt of HCN, a weak acid. KC2H3O2: C2H3O2– is the anion of a salt of HC2H3O2, a weak acid.
(b)
basic:
(c)
neutral: NaI: This is a salt of a strong acid, HI, and a strong base, NaOH. CsNO3: Metal ions with small charges are neutral and NO3– is the anion of a salt of HNO3, a strong acid. KBr: This is the salt of a strong acid, HBr, and a strong base, KOH.
17.25
The litmus paper will turn red indicating the presence of an acid. In section 16.4 it was stated that small, highly charged cations are acidic in water because the water molecules that surround the metal ion easily lose H+ ions. Al3+ was a specific example.
17.26
Kb for hydrazine is larger than Ka for acetic acid since the solution is acidic. The hydrazine reacts with water to form a weak acid. Since the solution is acidic, the more of the hydrazine has dissociated than the acetate ion has reacted, this occurs if Kb is larger than Ka.
17.27
Mg(OH)2 + 2NH4+ Mg2+ + 2NH3 + 2H2O
17.28
(a)
H2CO3 + OH– H2O + HCO3– H+ + HCO3– H2CO3
(b)
H2PO4– + OH– H2O + HPO42– H+ + HPO42– H2PO4–
(c)
NH4+ + OH– H2O + NH3 H+ + NH3 NH4+
(d)
C6H5OH + OH– H2O + C6H5O– C6H5O– + H+ C6H5OH
17.29
HPO42– + H+ H2PO4 – HPO42– + OH– H2O + PO43–
17.30
H2SO3 HSO3– + H+
Ka 1
17.31
HSO − H + 3 = H 2SO3
HSO3– SO32– + H+
Ka 2
SO 2− H + 3 = HSO − 3
H3C6H5O7 + H2O H3O+ + H2C6H5O7–
H O+ H C H O 3 2 6 5 7 K a1 = H3C6 H5 O7 H2C6H5O7– + H2O H3O+ + HC6H5O72–
375
Chapter 17
Ka2
H O+ HC H O 2– 3 6 5 7 = H C H O – 2 6 5 7
HC6H5O72– + H2O H3O+ + C6H5O73–
Ka3
H O+ C H O 3 3 6 5 7 = HC H O 2 6 5 7
17.32
Nearly all of the H+ in solution comes from the first ionization. The concentration of the conjugate base of a weak polyprotic acid once deprotonated is approximately equal to the value of the equilibrium constant. These approximations are usually valid because the value of the first ionization constant is much larger than the second ionization constant. If the two ionization constants are not very different, the approximation stated above will fail.
17.33
(a)
SO32– + H2O HSO3– + OH– HSO3– + H2O H2SO3 + OH–
(b)
PO43– + H2O HPO42– + OH– HPO42– + H2O H2PO4– + OH– H2PO4– + H2O H3PO4 + OH–
(c)
C4H4O62– + H2O HC4H4O6– + OH– HC4H4O6– + H2O H2C4H4O6 + OH–
17.34
Simplifying assumptions are essentially the same for polyprotic acids (see 17.32 above), and all of the hydroxide ion comes from the first hydrolysis. The assumptions are valid because the first dissociation constant is so much larger than the second.
17.35
The equivalence point is when the moles of acid exactly equal the moles of base. The end point is a measured value that approximates the equivalence point.
17.36
At the equivalence point, the titrated formic acid solution will be basic.
17.37
At the equivalence point, the titrated hydrazine solution will be acidic.
17.38
An acid-base indicator is a weak acid that changes color when it is converted from its acidic form to its basic form. Because indicators are weak acids, they will react with the titrant. Consequently, small amounts of indicator are used so the results will not be impaired excessively by the indicator.
17.39
At the equivalence point, all of the acetic acid will have been converted to acetate ions. Since this would be a solution of a weak base having a pH > 7, it would be argued that the use of methyl orange would be inappropriate since the color change occurs in the range pH = 3.2 to pH = 4.4. A better indicator for this titration would be thymol blue, or phenolphthalein which change in the pH range of 8-10.
Review Problems 17.40
D2O D+ + OD–, Kw = [D+] × [OD–] = 8.9 × 10–16 Since [D+] = [OD–], we can rewrite the above expression to give: 376
Chapter 17 8.9 × 10–16 = ([D+])2, [D+] = 3.0 × 10–8 M = [OD–] pD = –log[D+] = –log(3.0 × 10–8) = 7.52 pOD = –log[OD–] = –log(3.0 × 10–8) = 7.52 pKw = pD + pOD = 15.04 17.41
It should be stated from the outset that water at this temperature is neutral by definition, since [H+] = [OH–]. In other words, the self–ionization of water still occurs on a one–to–one mole basis: H2O H+ + OH–. Kw = 2.5 × 10–14 = [H+] × [OH–] + Since [H ] = [OH–], we can rewrite the above relationship: 2.5 × 10–14 = ([H+])2, ∴ [H+] = [OH–] = 1.6 × 10–7 M pH = –log[H+] = –log(1.6 × 10–7) = 6.80 pOH = –log[OH–] = –log(1.5 × 10–7) = 6.80 pKw = pH + pOH = 6.80 + 6.80 = 13.60 Alternatively, for the last calculation we can write: pKw = –log(Kw) = –log(2.5 × 10–14) = 13.60 Water is neutral at this temperature because the concentration of the hydrogen ion is the same as the concentration of the hydroxide ion.
17.42
At 25 °C, Kw = 1.0 × 10–14 = [H+] × [OH–]. Let x = [H+], for each of the following: (a)
(b)
(c)
(d)
17.43
x(0.0085) = 1.0 × 10–14 [H+] = (1.0 × 10–14) ÷ (0.0085) = 1.2 × 10−12 pH = –log[H+] = –log(1.2 × 10−12) = 11.93 pOH = 14 – pH = 14 – 11.93 = 2.07 x(5.3 × 10–5) = 1.0 × 10–14 [H+] = (1.0 × 10–14) ÷ (6.4 × 10–5) = 1.9 × 10–10 M pH = –log[H+] = –log(1.9 × 10–10) = 9.72 pOH = 14 – pH = 14 – 9.72 = 4.28 x(2.6 × 10–8) = 1.0 × 10–14 [H+] = (1.0 × 10–14) ÷ (2.6 × 10–8) = 3.8 × 10–7 M pH = –log[H+] = –log(3.8 × 10–7) = 6.41 pOH = 14 – pH = 14 – 6.41 = 7.59 x(7.8 × 10–12) = 1.0 × 10–14 [H+] = (1.0 × 10–14) ÷ (8.2 × 10–12) = 1.3 × 10–3 M pH = –log[H+] = –log(1.3 × 10–3) = 2.89 pOH = 14 – pH = 14 – 2.89 = 11.11
At 25 °C, Kw = 1.0 × 10–14 = [H+] × [OH–]. Let x = [OH–], for each of the following: (a)
(b)
(c)
(3.5 × 10–7)x = 1.0 × 10–14 [OH–] = (1.0 × 10–14) ÷ (3.5 × 10–7) = 2.9 × 10–8 M pH = –log[H+] = –log(3.5 × 10–7) = 6.46 pOH = 14 – pH = 14 – 6.46 = 7.54 (0.0017)x = 1.0 × 10–14 [OH–] = (1.0 × 10–14) ÷ (0.0017) = 5.9 × 10–12 M pH = –log[H+] = –log(0.0017) = 2.77 pOH = 14 – pH = 14 – 2.77 = 11.23 (2.5 × 10–11)x = 1.0 × 10–14 [OH–] = (1.0 × 10–14) ÷ (2.5 × 10–11) = 4.0 × 10–4 M pH = –log[H+] = –log(2.5 × 10–11) = 10.60 pOH = 14 – pH = 14 – 10.60 = 3.40
377
Chapter 17
(d)
(7.9 × 10–2)x = 1.0 × 10–14 [OH–] = (1.0 × 10–14) ÷ (7.9 × 10–2) = 1.3 × 10–13 M pH = –log[H+] = –log(7.9 × 10–2) = 1.10 pOH = 14 – pH = 14 – 1.10 = 12.90
17.44
pH = –log[H+] = –log(1.9 × 10–5) = 4.72
17.45
pH = –log[H+] = –log(1.4 × 10–5) = 4.85
17.46
[H+] = 10–pH and [OH–] = 10–pOH At 25 °C, pH + pOH = 14.00
17.47
17.48
(a)
[H+] = 10–pH = 10–7.96 = 1.1 × 10–8 M pOH = 14.00 – pH = 14.00 – 7.96 = 6.04 [OH–] = 10–pOH = 10–6.04 = 9.1 × 10–7
(b)
[H+] = 10–pH = 10–2.35 = 4.5 × 10–3 M pOH = 14.00 – pH = 14.00 – 2.35 = 11.65 [OH–] = 10–pOH = 10–11.65 = 2.2 × 10–12 M
(c)
[H+] = 10–pH = 10–10.85 = 1.4 × 10–11 M pOH = 14.00 – pH = 14.00 – 10.85 = 3.15 [OH–] = 10–pOH = 10–3.15 = 7.1 × 10–4 M
(d)
[H+] = 10–pH = 10–13.75 = 1.8 × 10–14 M pOH = 14.00 – pH = 14.00 – 13.75 = 0.25 [OH–] = 10–pOH = 10–0.25 = 5.6 × 10–1 M
(e)
[H+] = 10–pH = 10–6.25 = 5.6 × 10–7 M pOH = 14.00 – pH = 14.00 – 6.25 = 7.75 [OH–] = 10–pOH = 10–7.75 = 1.8 × 10–8 M
(a)
[OH–] = 10–pOH = 10–12.75 = 1.8 × 10–13 M pH = 14.00 – pOH = 14.00 – 12.75 = 1.25 [H+] = 10–pH = 10–1.25 = 5.6 × 10–2 M
(b)
[OH–] = 10–pOH = 10–6.56 = 2.8 × 10–7 M pH = 14.00 – pOH = 14.00 – 6.56 = 7.44 [H+] = 10–pH = 10–7.44 = 3.6 × 10–8 M
(c)
[OH–] = 10–pOH = 10–11.45 = 3.5 × 10–12 M pH = 14.00 – pOH = 14.00 – 11.45 = 2.55 [H+] = 10–pH = 10–2.55 = 2.8 × 10–3 M
(d)
[OH–] = 10–pOH = 10–4.72 = 1.9 × 10–5 M pH = 14.00 – pOH = 14.00 – 4.72 = 9.28 [H+] = 10–pH = 10–9.28 = 5.2 × 10–10 M
(e)
[OH–] = 10–pOH = 10–3.05 = 8.9 × 10–4 M pH = 14.00 – pOH = 14.00 – 3.05 = 10.95 [H+] = 10–pH = 10–10.95 = 1.1 × 10–11 M
[H+] = 10–pH = 10–5.7 = 2 × 10–6 M pOH = 14.00 – pH = 14.00 – 5.7 = 8.3 [OH–] = 10–pOH = 10–8.3 = 5 × 10–9 M
378
Chapter 17
17.49
[H+] = 10–pH = 10–3.16 = 6.92 × 10–4 M If the concentration of H+ is doubled: [H+] = 1.38 × 10–3 M pH = –log[H+] = –log(1.38 × 10–3) = 2.86
17.50
HNO3 is a strong acid so [H+] = [HNO3] = 0.0035 M pH = –log[H+] = –log(0.0035) = 2.46 pOH = 14.00 – pH = 14.00 – 2.46 = 11.54 [OH–] = 10–pOH = 10–11.54 = 2.9 × 10–12 M
17.51
HClO4 is a strong acid so [H+] = [HClO4] = 0.045 M pH = –log[H+] = –log(0.045) = 1.35 pOH = 14.00 – pH = 14.00 – 1.35 = 12.65 [OH–] = 10–pOH = 10–12.65 = 2.24 × 10–13 M If the concentration of H+ is doubled: [H+] = 0.045 M × 2 = 0.090 M pH = –log[H+] = –log(0.090) = 1.05 The pH drops by 0.3 pH units.
17.52
M OH − =
moles OH − 5.0 g NaOH 1 mole NaOH = 1.00 L solution 40.0 g NaOH L solution
1 mole OH − 1 mole NaOH
= 0.125 M OH − pOH = –log[OH–] = –log(0.125) = 0.903 pH = 14.00 – pOH = 14.00 – 0.903 = 13.10 [H+] = 10–pH = 10–13.10 = 8.0 × 10–14 M
M OH − = 17.53
moles OH − L solution
0.837 g Ba(OH) 2 1 mole Ba(OH)2 = 0.100 L solution 171.3 g Ba(OH)2
2 mole OH − 1 mole Ba(OH)2
= 0.0977 M OH − pOH = –log[OH–] = –log(9.77 × 10–2) = 1.01 pH = 14.00 – pOH = 14.00 – 1.01 = 12.99 [H+] = 10–pH = 10–12.99 = 1.02 × 10–13 M 17.54
pOH = 14.00 – pH = 14.00 – 11.40 = 2.60 [OH–] = 10–pOH = 10–2.60 = 2.5 × 10–3 M 2.5 × 10−3 mol OH − 1 mol Ca(OH)2 [ Ca(OH)2 ] = 1 L solution 2 mol OH − = 1.3 × 10 −3 M Ca(OH)2
pH = 10.40 pOH = 14 – 10.40 = 3.60 [OH–] = 10–pOH = 10–3.60 = 2.5 × 10–4 M
379
Chapter 17
2.5 × 10 −4 mol OH − 1 L solution
[ Ca(OH)2 ] =
1 mol Ca(OH) 2 2 mol OH −
= 1.3 × 10 −4 M Ca(OH)2 17.55
[H+] = 10–pH = 10–2.25 = 5.6 × 10–3 M 5.6 × 10−3 mol H + g HCl = ( 0.250 L ) 1 L solution
1 mol HCl 36.46 g HCl 1 mol H + 1 mol HCl
= 5.1 × 10−2 g HCl pH = 2.25 × 2 = 4.50 [H+] = 10–pH = 10–4.50 = 3.2 × 10–5 M 3.2 × 10−5 mol H + g HCl = ( 0.250 L ) 1 L solution
1 mol HCl 36.46 g HCl 1 mol H + 1 mol HCl
= 2.9 × 10−4 g HCl 17.56
Since NaOH is a strong base, [NaOH] = [OH–] = 0.0020 M OH–. (Next, we make the simplifying assumption that the amount of hydroxide ion formed from the dissociation of water is so small that we can neglect it in calculating pOH for the solution.) [H+] = 1 × 10–14 ÷ 0.0020 = 5.0 × 10–12 M H+ The only source of H+ is the autoionization of water. Therefore, the molarity of OH– from the ionization water is 5.0 × 10–12 M.
17.57
The H+ concentration from the ionization of water also gives the same OH– concentration. Since the only source of OH– is the water, the concentration of OH– is 3.4 × 10–11 M. The total H+ concentration can be determined from: 1 × 10–14 = [H+] × [OH–] = [H+] × (3.4 × 10–11) [H+] = 2.9 × 10–4 M H+
17.58
First calculate the molarity of OH- produced from the Ba(OH)2.
1mol Ba(OH)2 2mol OH − 1 −8 = 3.08 x 10 M 171.3 g mol Ba(OH)2 1L
2.64 x10−6 g Ba(OH)2
pOH = -log(3.08 x 10-8) = 7.51 pH = 14 – 7.51 = 6.49 Notice the pH calculated from the Ba(OH)2 is less than 7. We would expect the solution to be basic but since the amount of base added to water is so small, the salt is not the major contributor to the pH of the solution. Thus, we cannot neglect the ionization of water in this case, since water’s ionization is a significant contributor to the H+ and OH- concentrations in the solution. The problem is more complicated than when acid or base concentrations are larger than 10-7M. Let’s analyze all of the reactions: Ba(OH)2
→
2H2O H3O+ + OH-
Ba2+ + 2OH-
complete ionization of a strong base
Kw = [H3O+][OH-] = 1.00 x 10-14
380
Chapter 17
We know that the concentration of cations must equal the concentration of anions for the solution to be neutral. Therefore, [H3O+] + 2[Ba2+] = [OH-] Charge balance equation. Remember, the Ba2+ ion’s total charge is twice its concentration. We also know that [OH-] = Kw/[H3O+]. Thus, [H3O+] + 2[Ba2+] = Kw/[H3O+] [H3O+]2 + 2[Ba2+][H3O+] – Kw = 0 [Ba2+] = (1/2)[OH-] = (1/2) 3.08 x 10-8 M = 1.54 x 10-8 M [H3O+]2 + 3.08 x 10-8[H3O+] - 1.00 x 10-14 = 0 Solve using the quadratic equation. [H3O+] = 8.58 x 10-8 pH = 7.07 A basic solution, though just barely!
17.59
The total [H+] is from the HCl and from the dissociation of H2O. Since HCl is a strong acid, it will contribute 3.0 × 10–7 mol of H+ per liter of solution. We need to use the equilibrium expression to determine the amount of H+ contributed by the water. Kw = [H+][OH–] = (3.0 × 10–7 + x)(x) = 1.0 × 10–14 2 –7 –14 x + 3.0 × 10 x – 1.0 × 10 = 0
−3.0 × 10−7 ± x=
(3.0 ×10 ) −7
2
(
− 4 (1) −1.0 × 10−14
)
2 (1)
x = 3.03 × 10–8 Solving a quadratic equation we see that x = 3.03 × 10–8 = [OH–] = [H+] from water dissociation. So, [H+]total = 3.0 × 10–7 + 3.03 × 10–8 = 3.30 × 10–7 and pH = 6.48. 17.60
[H+] = 10–pH = 10–2.37 = 4.27 × 10–3 M
% ionization = 17.61
4.27 × 10−3 × 100% = 0.43% 1.0
NH3 + H2O NH4+ + OH–
[H3O+] = 10–11.32 = 4.79 x 10–12 [OH–] = 1.0 x 10-14/4.79 x 10-12 = 2.09 x 10-3 = [NH4+] Percentage ionization =
moles ionized per liter × 100% moles available per liter
2.09 x 10−3 Percentage ionization = x100 = 0.84 % 0.250
381
Chapter 17
17.62
At 25 °C, Ka × Kb = Kw Kb = Kw/Ka = 1.0 × 10–14 ÷ 1.8 × 10–5 = 5.6 × 10–10
17.63
At 25 °C, Ka × Kb = Kw Ka = Kw/Kb = 1.0 × 10–14 ÷ 1.0 × 10–10 = 1.0 × 10–4
17.64
(a) At 25 °C, Ka × Kb = Kw Kb = Kw/Ka = 1.0 × 10–14 ÷ 1.4 × 10–4 = 7.1 × 10–11 (b) Kb for acetic acid is 1.0 x 10-14/1.8 x 10-5 = 5.6 x 10-10. Therefore, since Kb for the lactate ion is smaller than that for the acetate ion, the lactate ion is a weaker base.
17.65
17.66
(a)
The conjugate base is IO3– pKb = 14.00 – pKa = 14.00 – 0.77 = 13.23 Kb = 10–pKb = 10–13.23 = 5.9 × 10–14
(b)
IO3– is a weaker base than an acetate anion, because its Kb value is smaller than that of an acetate anion.
[H+] = 10–2.87 = 1.35 × 10–3 M moles ionized per liter Percentage ionization = × 100% moles available per liter
pH = 2.87
Percentage ionization =
1.35 × 10−3 M × 100% = 0.54% 0.25 M
HA H+ + A–
I C E
Ka =
17.67
[HA] 0.25 – 1.35 × 10–3 0.25
[H+] – 1.35 × 10–3 1.35 × 10–3
[A–] – 1.35 × 10–3 1.35 × 10–3
H + A – 1.35 × 10−3 1.35 × 10−3 = = 7.3 × 10–6 HA 0.25 [ ] [ ]
moles ionized per liter × 100% moles available per liter Let x = moles of H+ in solution xM 0.025% = × 100% x = 8.75 × 10–6 M 0.035 M pH = –log[H+] = –log[8.75 × 10–6] = 5.06 Percentage ionization =
HA H+ + A–
I C E
[HA] 0.035 – 8.75 × 10–6 0.035
[H+] – 8.75 × 10–6 8.75 × 10–6
382
[A–] – 8.75 × 10–6 8.75 × 10–6
Chapter 17
H + A – 8.75 × 10−6 8.75 × 10−6 Ka = = = 2.2 × 10–9 [ HA ] [ 0.035] 17.68
HIO4 H+ + IO4–
H + IO − 4 Ka = [ HIO4 ]
I C E
[HIO4] 0.10 –x 0.10 – x
[H+] – +x +x
[IO4–] – +x +x
We know that at equilibrium [H+] = 0.038 M = x. The equilibrium concentrations of the other components of the mixture are: [HIO4] = 0.10 – x = 0.062 M and [IO4–] = x = 0.038 M. Substituting the above values for equilibrium concentrations into the mass action expression gives: (0.038)(0.038) = 2.3 × 10−2 0.062 pKa = –log(Ka) = –log(2.3 × 10–2) = 1.64 Ka =
17.69
HC2H2ClO2 H+ + C2H2ClO2–
H + C H ClO − 2 2 2 Ka = [ HC2 H 2 ClO2 ]
I C E
[HC2H2ClO2] 0.10 –x 0.10 – x
[H+] – +x +x
[C2H2ClO2–] – +x +x
pH = 1.96 [H+] = 10–1.96 = 0.011 M = x The equilibrium concentrations of the other components of the mixture are: [HC2H2ClO2] = 0.10 – x = 0.089 M and [C2H2ClO2–] = x = 0.011 M. Substituting the above values for equilibrium concentrations into the mass action expression gives: (0.011)(0.011) = 1.4 × 10−3 0.089 pKa = –log(Ka) = –log(1.4 × 10–3) = 2.87 Ka =
17.70
pOH = 14.00 – pH = 14.00 – 11.87 = 2.13 [OH–] = 10–pOH = 10–2.13 = 7.41 × 10–3 M CH3CH2NH2 + H2O CH3CH2NH3+ + OH–
383
Chapter 17
CH CH NH + OH − 3 2 3 Kb = CH3CH 2 NH 2
I C E
[CH3CH2NH3+] – +x +x
[CH3CH2NH2] 0.10 –x 0.10–x
[OH–] – +x +x
In the equilibrium analysis, the value of x is, therefore, equal to 7.41 × 10–3 M. Therefore, our equilibrium concentrations are [CH3CH2NH3+] = [OH–] = 7.41 × 10–3 M, and [CH3CH2NH2] = 0.10 M – 7.41 × 10–3 M = 0.0926 M. Substituting these values into the mass action expression gives:
( 7.41 × 10 )( 7.41 × 10 ) = 5.9 × 10 = −3
Kb
−3
−4
0.0926 pKb = –log(Kb) = –log(5.9 × 10–4) = 3.23 moles ionized per liter Percentage ionization = × 100% moles available per liter Percentage ionization =
17.71
7.4 × 10−3 × 100% = 7.4% 0.10
HONH2 + H2O HONH3+ + OH–
HONH + OH − 3 Kb = [ HONH 2 ] pOH = 14.00 – pH = 14.00 – 10.11 = 3.89 [OH–] = 10–3.88 = 1.3 × 10–4 M In the equilibrium analysis, the value of x is, therefore, equal to 1.3 × 10–4 M:
I C E
[HONH3+] – +x +x
[HONH2] 0.25 –x 0.25 – x
[OH–] – +x +x
Therefore, our equilibrium concentrations are [HONH3+] = [OH–] = 1.3 × 10–4 M, and [HONH2] = 0.15 M – 1.3 × 10–4 M = 0.15 M. Substituting these values into the mass action expression gives:
(1.3 × 10 )(1.3 × 10 ) = 1.1 × 10 = −4
Kb
−4
−7
0.15 pKb = –log(Kb) = –log(1.1 × 10–7) = 6.96 moles ionized per liter Percentage ionization = × 100% moles available per liter
384
Chapter 17
Percentage ionization =
17.72
1.3 × 10-4 × 100% = 0.087% 0.15
HC3H5O2 + H2O H3O+ + C3H5O2–
Ka =
[H 3O + ][C3 H 5 O 2 − ] = 1.4 × 10 −4 HC3 H 5 O2
[HC3H5O2] 0.125 I –x C 0.125 – x E Assume x << 0.125 [x][x] Ka = = 1.4 × 10−4 [ 0.125]
[H3O+] – +x +x
[C3H5O2–] – +x +x
x = 4.2 × 10−3 = [H3O+ ]
pH = 2.38 [HC3H5O2] = 0.121 M [H+] = 4.2 × 10–3 M [C3H5O2–] = 4.2 × 10–3 M 17.73
H2O2 H+ + HO2– Ka =
[H + ][HO 2 − ] = 2.4 × 10−12 [ H 2 O2 ]
I C E
[H2O2] 1.0 –x 1.0 – x
[H+] – +x +x
[HO2–] – +x +x
Substituting these values into the mass action expression gives: Ka =
(x)(x) = 2.4 × 10 −12 1.0 − x
Assuming x << 1.0 we determine that x = 1.6 × 10–6 M = [H+] [H+] = 1.6 × 10–6 M [HO2–] = 1.6 × 10–6 M [H2O2] = 1.0 M – (1.6 × 10–6 M) = 1.0 M pH = –log[H+] = 5.80 17.74
Kb = 10–pKb = 10–5.80 = 1.58 × 10–6 Cod + H2O HCod+ + OH–
HCod + OH − Kb = = 1.58 × 10−6 [ Cod ]
385
Chapter 17
I C E
[Cod] 0.025 –x 0.025 – x
[HCod+] – +x +x
[OH–] – +x +x
Substituting these values into the mass action expression gives: ( x )( x ) = 1.58 × 10−6 Kb = 0.025 − x If we assume that x << 0.025 we get; x2 = 3.95 × 10–8, x = 1.99 × 10–4 M = [OH–] pOH = –log[OH–] = –log(1.99 × 10–4) = 3.70 pH = 14.00 – pOH = 14.00 – 3.70 = 10.30 17.75
C6H5N + H2O HC6H5N+ + OH– Kb = 10–pKb = 10–8.77 = 1.70 × 10–9 HC H N + OH − 6 5 Kb = = 1.70 × 10−9 C6 H5 N
I C E
[C6H5N] 0.25 –x 0.25 – x
[HC6H5N+] – +x +x
[OH–] – +x +x
Assume x << 0.25
Kb =
( x )( x )
0.25 pOH = 4.69 17.76
= 4.25 × 10−10
x = 2.1 × 10−5 = [OH − ]
pH = 9.31
[H+] = 10–pH = 10–2.95 = 1.1 × 10–3 M HC2H3O2 H+ + C2H3O2– Ka =
[H + ][C 2 H 3O 2 − ] = 1.8 × 10−5 HC 2 H3O2
I C E
[HC2H3O2] Z –x Z–x
[H+] – +x +x
[C2H3O2–] – +x +x
Substituting these values into the mass action expression gives:
Ka =
(x)(x) = 1.8 × 10−5 Z−x 386
Chapter 17
Assuming x << Z and knowing that x = 1.1 × 10–3 M, we can solve for Z and find Z = 0.070. The initial concentration of HC2H3O2 is 0.070 M. 17.77
Let Z = initial concentration of NH3. NH3 + H2O NH4+ + OH–
NH + OH − 4 Kb = = 1.8 × 10−5 NH3 [NH3] Z –x Z–x
I C E
[NH4+] – +x +x
[OH–] – +x +x
Assume x << Z (x)(x) (x)(x) Kb = = 1.8 × 10−5 Z = Z 1.8 × 10−5 pOH = 14 – pH = 14 – 11.35 = 2.65 x = [OH–] = 10–pOH = 10–2.65 = 2.24 × 10–3 M Z = 0.28 M 0.28 mol mol NH3 = (500 mL) = 0.14 mol NH3 1000 mL
17.78
CN– + H2O HCN + OH–
Kb =
[ HCN ] OH
−
CN −
Kw 1.0 × 10−14 = = = 2.0 × 10−5 −10 Ka 4.9 × 10
[CN–] 0.0050 –x 0.0050 – x
I C E
[OH–] – +x +x
[HCN] – +x +x
Assume that x << 0.0050
Kb =
( x )( x ) 0.0050
= 2.0 × 10 5
x = 3.2 × 10 4 = [OH ]
Wait!!!! 3.2 × 10 4 is not << 0.0050 So, use the method of successive approximations.
( x )( x )
= 2.0 × 10−5 0.0050 − 0.00032 (x)(x) Kb = = 2.0 × 10−5 0.0050 − 0.00031
Kb =
x = 3.1× 10−4 x = 3.1 × 10−4
x = 3.1 × 10–4 = [OH–] pOH = 3.51 pH = 10.49 387
Chapter 17
17.79
HA + H2O H3O+ + A–
H O+ A − 3 = 1.4 × 10−3 Ka = [ HA ]
I C E
[HA] 0.020 –x 0.020 –x
[H3O+] – +x +x
[A–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.020 gives: ( x )( x ) = 1.36 × 10−3 Ka = x = 5.3 × 10–3 Wait!!! 5.3 × 10-3 is not << 0.020 0.020 Use the method of successive approximations (x)(x) Ka = = 1.4 × 10−3 x = 4.5 × 10−3 0.020 -0.0053 (x)(x) Ka = x = 4.6 × 10−3 = 1.4 × 10−3 0.020 − 0.0045 So, x = 4.6 × 10–3 = [H3O+] pH = 2.34 17.80
Ka = 10–pKa = 10–4.92 = 1.2 × 10–5 H–Paba H+ + Paba–
H + Paba − = 1.2 × 10−5 Ka = [ H − Paba ]
I C E
[H–Paba] 0.030 –x 0.030 – x
[H+] – +x +x
[Paba–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.030 gives:
Ka =
[ x ][ x ] [0.030]
= 1.2 × 10−5
x2 = 3.6 × 10–7, x = 6.0 × 10–4 M = [H+] pH = –log[H+] = –log(6.0 × 10–4) = 3.22 17.81
Ka = 10–pKa = 10–4.01 = 9.8 × 10–5 H–Bar H+ + Bar–
H + Bar − = 9.8 × 10−5 Ka = [ H − Bar ]
388
Chapter 17
I C E
[H–Bar] 0.020 –x 0.020 – x
[H+] – +x +x
[Bar–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.020 gives: ( x )( x ) Ka = = 9.8 × 10−5 x = 1.4 × 10−3 Wait!!!! 1.4 × 10−3 is not << 0.020 0.020 Use method of successive approximations
Ka =
(x)(x) = 9.8 × 10−5 0.020 -0.0014
x = 1.4 × 10−3
So, x = 1.4 × 10−3 = [H3O + ] pH = 2.85 17.82
NaCN will be basic in solution since CN– is a basic ion and Na+ is a neutral ion. CN– + H2O HCN + OH– For HCN, Ka = 4.9 × 10–10, we need Kb for CN–; Kb = Kw/Ka = (1.0 × 10–14) ÷ (4.9 × 10–10) = 2.0 × 10–5 [ HCN ] OH − Kb = = 2.0 × 10−5 − CN
I C E
[CN–] 0.25 –x 0.25 – x
[HCN] – +x +x
[OH–] – +x +x
Substituting these values into the mass action expression gives: ( x )( x ) = 2.0 × 10−5 Kb = 0.25 − x Assuming that x << 0.25 we can solve for x and determine; x = 2.2 × 10–3 M = [OH–] pOH = –log[OH–] = –log(2.2 × 10–3) = 2.65 pH = 14.00 – pOH = 14.00 – 2.65 = 11.35 Concentration of HCN is equal to that of hydroxide ion: 2.2 × 10–3 M 17.83
KNO2 will be basic in solution since NO2– is a basic ion and K+ is neutral. NO2– + H2O HNO2 + OH– For HNO2, Ka = 4.6 × 10–4, we need Kb for NO2–; Kb = Kw/Ka = (1.0 × 10–14) ÷ (4.6 × 10–4) = 2.2 × 10–11 [ HNO2 ] OH − Kb = = 2.2 × 10−11 NO 2−
389
Chapter 17
[NO2–] 0.35 –x 0.35–x
I C E
[HNO2] – +x +x
[OH–] – +x +x
Substituting these values into the mass action expression gives: ( x )( x ) = 2.2 × 10−11 Kb = 0.35 − x Assuming that x << 0.35 we can solve for x and determine the equilibrium concentrations; x = 2.8 × 10–6 M = [OH–] = [HNO2] pOH = –log[OH–] = –log(2.8 × 10–6) = 5.56 pH = 14.00 – pOH = 14.00 – 5.56 = 8.44 17.84
A solution of CH3NH3Cl will be acidic since the Cl– ion is neutral and the CH3NH3+ ion is acidic. CH3NH3+ H+ + CH3NH2 For CH3NH2, Kb = 4.4 × 10–4. We need Ka for CH3NH3+; Ka = Kw/Kb = (1.0 × 10–14) ÷ (4.4 × 10–4) = 2.3 × 10–11 H + CH NH 2 3 Ka = = 2.3 × 10−11 CH NH + 3 3
I C E
[CH3NH3+] 0.18 –x 0.18 – x
[H+] – +x +x
[CH3NH2] – +x +x
Substituting these values into the mass action expression gives: ( x )( x ) = 2.3 × 10−11 Ka = 0.18 − x Assuming that x << 0.18 we can solve for x and determine; x = 2.0 × 10–6 M = [H3O+] pH = –log[H3O+] = –log(2.0 × 10–6) = 5.69 17.85
Hydrazinium chloride is the acidic salt of hydrazine N2H4. N 2H 5+ + H 2O H 3O + + N 2H 4
H O+ [ N H ] 3 2 4 Ka = N H + 2 5
I C E
[N2H5+] 0.10 –x 0.10 – x
K b for hydrazine is 1.3 × 10−6 so K a =
[H3O+] – +x +x
[N2H4] – +x +x
Assume x << 0.10
390
Kw = 7.7 × 10−9 Kb
Chapter 17
( x )( x )
Ka =
0.10 pH = 4.56 17.86
= 7.7 × 10−9
x = 2.8 × 10−5 = [H3O+ ]
Let HNic symbolize the nicotinic acid Nic– + H2O HNic + OH–
Kb =
[ HNic] OH
−
Nic − [Nic–] 0.18 –x 0.18 – x
I C E
[HNic] – +x +x
[OH–] – +x +x
Assume x << 0.18 ( x )( x ) Kb = 0.18 We know x = [OH–] pH = 9.05 and pOH = 4.95 and [OH–] = 10–pOH = 1.1 × 10–5 = x
Kb = Ka =
17.87
(1.1 × 10−5 )2 = 7.0 × 10−10 0.18 1× 10 –14
7.0 × 10 –10
= 1.4 × 10–5
The reaction of the weak base in water is: B + H2O BH+ + OH– The salt will react as: BH+ H+ + B
H + [ B] Ka = BH +
I C E
[BH+] 0.15 –x 0.15 – x
[H+] – +x +x
[B] – +x +x
At equilibrium, the pH = 4.28 and [H+] = 5.2 × 10–5 M = x. Substituting these values into the mass action expression gives:
Ka
(5.2 × 10 )(5.2 × 10 ) = 1.8 × 10 = = 0.15 − x 0.15 − ( 5.2 × 10 ) ( x )( x )
−5
−5
−8
−5
Recall that Ka × Kb = Kw, so Kb = Kw ÷ Ka = 1.0 × 10–14 ÷ 1.8 × 10–8 = 5.6 × 10–7
391
Chapter 17
17.88
OCl– + H2O HOCl + OH–
Kb =
[ HOCl] OH
−
Kw 1.0 × 10−14 = = = 3.3 × 10 −7 −8 Ka 3.0 × 10
OCl 5.1 g NaOCl 1 mol NaOCl 1.0 g 1000 mL [OCl ] = 100 g solution 74.44 g NaOCl 1 mL 1 L = 0.68 M −
I C E
[OCl–] 0.68 –x 0.68 – x
[HOCl] – +x +x
[OH–] – +x +x
x = 4.7 × 10
4
Assume that x << 0.68
Kb =
( x )( x )
0.68 pOH = 3.33 pH = 10.67
17.89
= 3.3 × 10
7
= [OH ]
The conjugate acid BH+ is acidic due to the following equilibrium: BH+ H+ + B The base Y– has the following equilibrium: Y– + H2O HY + OH– If a solution of the salt BHY is basic, it can only mean that Y– is a stronger base than B: BH+ + Y– HY + B This can happen only if HY is a weaker acid than BH+. We conclude that pKa for HY is greater than 5.
17.90
HC2H3O2 H+ + C2H3O2– Ka =
[H + ][C 2 H3O 2 − ] = 1.8 × 10−5 HC 2 H3O2
I C E
[HC2H3O2] 0.14 –x 0.14 – x
[H+] – +x +x
[C2H3O2–] 0.27 +x 0.27 + x
Substituting these values into the mass action expression gives: (x)(0.27 + x) Ka = = 1.8 × 10 −5 0.14 − x
392
Chapter 17
Assume that x << 0.14 M and x << 0.27 M, then; 0.27 x × ≈ 1.8 × 10 −5 0.14
0.14 x ≈ × 1.8 × 10 −5 0.27 x ≈ 9.3 × 10–6 M = [H+] pH = –log [H+] = 5.03 17.91
C2H3O2– + H2O HC2H3O2 + OH– Kb =
[HC 2 H3O 2 ][OH − ] Kw 1.0 × 10−14 = = = 5.6 × 10 −10 −5 Ka C 2 H3O 2 − 1.8 × 10
I C E
[C2H3O2–] 0.27 –x 0.27 – x
[OH–] – +x +x
[HC2H3O2] 0.14 +x 0.14 + x
Substituting these values into the mass action expression gives: (0.14 + x)(x) Kb = = 5.6 × 10 −10 0.27 − x Assume that x << 0.14 M and x << 0.27 M, then; 0.14 x × ≈ 5.6 × 10 −10 0.27
0.27 x ≈ × 5.6 × 10−10 0.14 x ≈ 1.08 × 10–9 M = [OH–] pOH = –log [OH–] = 8.97 pH = 14.00 – pOH = 5.03 The answer is identical no matter which direction we choose to solve it. 17.92
The equilibrium we will consider in this problem is: NH3 + H2O NH4+ + OH–
NH + OH − 4 Kb = = 1.8 × 10 −5 NH3 =
( 0.50) OH 0.30
−
= 1.8 × 10−5
[OH–] = 1.1 × 10–5 M pOH = –log[OH–] = –log(1.1 × 10–5) = 4.97 pH = 14.00 – pOH = 14.00 – 4.97 = 9.03 17.93
Now we consider the equilibrium; NH4+ H+ + NH3 We know that KaKb = Kw, so Ka = Kw/Kb = 5.6 × 10–10. H + NH3 Ka = = 5.6 × 10−10 NH 4 + 393
Chapter 17
H + ( 0.30) = = 5.6 × 10−10 0.50 [H+] = 9.3 × 10–10 M and pH = 9.03 Notice that the two values calculated are the same. 17.94
Any H+ produced will react with NH3, reducing its concentration, to produce NH4+, thus increasing its concentration. H+ + NH3
→
NH4+
The amount of H+ produced in 35 seconds = (1.8 x 10-6 mol s-1)(35 s) = 6.3 x 10-5 mol The concentration of NH3 will decrease by: 6.3 x 10-5 mol/0.0025 L = 2.5 x 10-2 M and the NH4+ will increase by the same amount. 17.95
Any H+ consumed will react with C2H3O2− , reducing its concentration, to produce HC2H3O2 thus increasing its concentration. H+ + C2H3O2−
→
HC2H3O2
The amount of H+ consumed in 3.45 minutes = (1.8 x 10-7 mol min-1)(3.45 min) = 6.21 x 10-7 mol The concentration of C2H3O2− will decrease by: 6.21 x 10-7 mol/0.0025 L = 2.5 x 10-4 M and the HC2H3O will increase by the same amount. 17.96
The initial pH of the buffer is 5.03 as determined in Review Problem 17.90. The added acid, 0.025 mol, will react with the acetate ion present in the buffer solution. Assume the added acid reacts completely. For each mole of acid added, one mole of C2H3O2– is converted to HC2H3O2. Since 0.025 mol of acid is added; [HC2H3O2]final = (0.14 + 0.025) M = 0.165 M [C2H3O2–]final = (0.27 – 0.025) M = 0.245 M Now, substitute these values into the mass action expression to calculate the final [H+] in solution; H + (0.245) = 1.8 × 10−5 (0.165) [H+] = 1.2 × 10–5 mol L–1 and the pH = 4.92 The pH of the solution changes by 5.03 – 4.92 = –0.11 pH units upon addition of the acid.
17.97
The initial pH is 9.03 as calculated in Review Problem 17.92. For every mole of OH− added, one mol of NH4+ will be changed to one mol of NH3. Since we added 0.20 M x 0.025 L = 0.005 mol OH−: NH 4 + = (0.50 M x 0.500 L − 0.005)/0.525 L = 0.47 M final NH3 final = (0.30 M x 0.500 L + 0.005 M )/0.525 L = 0.295 M Using these new concentrations, we can calculate a new pH: NH + OH − (0.47) OH − 4 Kb = = = 1.8 × 10−5 0.295 NH3 [OH–] = 1.13 × 10–5 M, the pOH = 4.95 and the pH = 9.05
394
Chapter 17
As expected, when an base is added, the pH increases. In this problem, the pH increases by 0.02 pH units from 9.03 to 9.05. 17.98
The initial pH can be calculated using the Henderson-Hasselbalch equation.
[ HB + ] [ B] 0.20 pOH = (4.74) + log 0.25 pOH = pKb + log
pOH = 4.64 pH = 14.00 - 4.64 = 9.36 For every mole of H+ added, one mol of NH3 will be changed to one mol of NH4+. Since we added 6.3 x 10-5 mol H+:
6.3 x10 −5 mol NH 4 + = 0.20 M + = 0.2252 M final 0.0025 L NH3 final = 0.25 M −
6.3 x10 −5 mol = 0.2248 M 0.0025 L
Using these new concentrations, we can calculate a new pH: NH + OH − (0.2252) OH − 4 Kb = = = 1.8 × 10−5 0.2248 NH3 [OH–] = 1.8 × 10–5 M, the pOH = 4.74 and the pH = 9.26 As expected, when an acid is added, the pH decreases. In this problem, the pH decreases by 0.10 pH units from 9.36 to 9.26. 17.99
The initial pH can be calculated using the Henderson-Hasselbalch equation.
[ A− ] pH = pKa + log [ HA] 0.25 pH = (4.74) + log 0.15 pOH = 4.96
pH = 14.00 - 4.96 = 9.04
For every mole of H+ added, one mol of C2H3O2− will be changed to one mol of HC2H3O2. Since we added 6.21 x 10-7 mol H+: HC2 H 3O2 final = 0.15 M +
6.21x10−7 mol = 0.1502 M 0.0025 L
6.21x10−7 mol C H O − = 0.25 M − = 0.2498 M 2 3 2 final 0.0025 L Using these new concentrations, we can calculate a new pH:
395
Chapter 17
C H O − H+ + 2 3 2 = (0.2498) H = 1.8 × 10−5 Ka = 0.1502 HC2 H3O2 [H+] = 1.1 × 10–5 M, the pH = 4.96 and the pH = 9.04 We would expected when acid is removed the pH should increase. In this problem, the amount of acid consumed is so small that no effect on the pH can be determined.
17.100
pH = pK a + log
[anion ] [ acid ]
A− = pK a + log [ HA ]
4.00 = 4.74 + log( [NaC2H3O2]/[HC2H3O2] ) –0.74 = log[NaC2H3O2]/[HC2H3O2] [NaC2H3O2]/[HC2H3O2] = 0.18 [NaC2H3O2] = 0.18 × [HC2H3O2] = 0.18 × 0.12 = 0.0216 M Thus to the 1 L of acetic acid solution we add: 0.0216 mol NaC2H3O2 × 82.0 g/mol = 1.8 g NaC2H3O2.
17.101
pH = pK a + log
[anion ] [ acid ]
A− = pK a + log [ HA ]
3.50 = 3.74 + log([NaCHO2]/[HCHO2]) [NaCHO2]/[HCHO2] = 0.575 [NaCHO2] = 0.575 × [HCHO2] = 0.575 × 0.15 = 0.0863 M Thus, to the 1 L of formic acid solution we add: 0.0863 mol NaCHO2 × 68.0 g/mol = 5.9 g NaCHO2, assuming no increase in volume upon addition of the sodium formate. 17.102 The equilibrium is; HC2H3O2 H+ + C2H3O2– Ka =
[H + ][C 2 H3O 2 − ] = 1.8 × 10−5 HC 2 H3O2
H + (0.110) The initial pH is; = 1.8 × 10−5 , [H+] = 1.636 × 10–5 M and pH = 4.79. In this calculation 0.100 we are able to use either the molar concentration or the number of moles since the volume is constant in this portion of the problem. In order to calculate the change in pH, we need to determine the concentrations of HC2H3O2 and C2H3O2– after the complete reaction of the added acid. One mole of C2H3O2– will be consumed for every mole of acid added and one mole of HC2H3O2 will be produced. The number of moles of acid added is: + 0.100 mol HCl 1 mol H mol H + = (30.00 mL HCl) = 3.00 × 10−3 mol H + 1000 mL HCl 1 mol HCl Since the volume of the new concentrations of HC2H3O2 and C2H3O2– are: ( 0.100 mol + 0.00300 mol ) = 0.792 M HC2 H3O 2 final = 0.130 L ( 0.110 mol − 0.00300 mol ) C H O − = = 0.823 M 2 3 2 final 0.130 L Note: The new volume has been used in these calculations.
396
Chapter 17
[H + ][C2 H3O2 − ] [H + ](0.823) = = 1.8 × 10−5 HC H O (0.792) 2 3 2 [H+] = 1.73 × 10–5 M and pH = 4.76 Ka =
Notice that the change in pH is very small in spite of adding a strong acid. If the same amount of HCl were added to water, a completely different effect would be observed. Since HCl is a strong acid, the [H+] in a water solution will be the result of the strong acid dissociation. We do, of course, need to account for the dilution. Using the dilution equation, M1V1 = M2V2. 30 mL x
1L 1L x 0.100M = M x (125 + 30 mL) x 1000 mL 1000 mL
M = 0.0194 mol L–1 and the pH = 1.71. The change in pH in this case is 7.00 – 1.71 = 5.29 pH units. A significantly larger change!!
17.103 The initial pH is 9.03. We want to lower the pH to 8.98, a decrease of 0.050 pH units. First we will calculate the ratio of acid and its conjugate base: + [ cation ] = pK + log BH pOH = pK b + log a [ base] [ B] + 14.00–8.98 = 4.74 + log([NH4 ]/[NH3]) [NH4+]/[NH3] = 1.91 We need a final ratio of conjugate base to acid of 1.91. Initially there are 0.050 moles of ammonium ion and 0.030 moles of ammonia. We will add a strong acid using up some of the ammonia and forming additional ammonium. For every mole of ammonium that reacts, one mole of ammonia will be formed. We can solve the following equation: 0.050 moles ammonium + x moles H + 0.030 moles ammonia – x moles H +
= 1.91
Solving we determine x = 2.51 × 10–3 mole H+. This is the number of moles of HCl needed to change the pH by 0.05 pH units. 1 mol HCl 1000 mL mL = 2.51 × 10−3 mol H + = 17 mL . 1 mol H + 0.15 mol HCl Thus, we need to add only 17 mL to the 100 mL of buffer.
(
)
If we added 17 mL of 0.15 M HCl directly to 100 mL of pure water, we would find the pH = 1.66. Thus, the pH of pure water would change by 5.34 pH units as opposed to the 0.05 pH unit change in the buffer 17.104 H2C6H6O6 + H2O H3O+ + HC6H6O6–
Ka1 = 8.0 × 10 –5
HC6H6O6– + H2O H3O+ + C6H6O62–
I C E
[H2C6H6O6] 0.15 –x 0.15 – x
Ka2 = 1.6 × 10–12
[H3O+] – +x +x
[HC6H6O6–] – +x +x
Assume x << 0.15
397
Chapter 17
Ka =
[x][x] = 8.0 × 10−5 [0.15]
x = 3.5 × 10−3
[H 2 C6 H 6 O6 ] ≅ 0.15 M [H3O + ] = [HC6 H 6 O6− ] = 3.5 × 10−3 M The concentration of the ascorbate ion equals K a2 [C6 H 6 O6 2− ] = 1.6 × 10−12 M pH = 2.46 pOH = 11.54 [OH–] = 2.9 × 10–12 M 17.105 As we do with most diprotic weak acids, assume that all of the H+ results from the first dissociation and all of the divalent anion results from the second dissociation. And, since the second dissociation constant is much smaller than the first dissociation constant, the concentration of the divalent anion is equal to the second dissociation constant. H TeO − H + 5 6 = 2 × 10−8 H6TeO6 H5TeO6– + H+ Ka 1 = H 6 TeO6 –
H5TeO6 H4TeO6
I C E
Ka 1
2–
+ H
+
Ka 2
H TeO 2 − H + 4 6 = 1 × 10−11 = H TeO − 6 5
[H5TeO6–] – +x +x
[H6TeO6] 0.25 –x 0.25 – x
[H+] – +x +x
H TeO − H + 5 6 = (x) (x) = 2 × 10−8 = (0.25 - x) H 6 TeO6
Assume that x << 0.25 then; x2 = (0.25)(2 × 10–8) and x = 7.1 × 10–5 M [H4TeO62–] = Ka2 = 1 × 10–11 M [H+] = 7.1 × 10–5 M [H5TeO6–] = 7.1 × 10–5 M pH = – log[H+] = – log(7.1 × 10–5) = 4.2
–
+
17.106 H3PO4 + H2O H2PO4 + H3O K a1
–
H2PO4 + H2O HPO4
HPO4
2–
+ H2O PO4
3–
2–
+ H 3O
+ H3O
H PO − H O + 2 4 3 = 7.5 × 10−3 = H3 PO 4
+
K a2
+
HPO 2− H O + 4 3 = 6.2 × 10−8 = H PO − 2 4
K a3
The following assumptions are made:
398
PO 3− H O+ 4 3 = 4.2 × 10−13 = HPO 2− 4
Chapter 17
[H+]total ≈ [H+]first step [H2PO4–]total ≈ [H2PO4–]first step [HPO42–]total ≈ [HPO42–]second step The first dissociation: [H3PO4] 2.0 I –x C 2.0 – x E
K a1
[H2PO4–] – +x +x
[H3O+] – +x +x
H PO − H O + 2 4 3 = (x)(x) = = 7.5 × 10−3 (2.0 − x) H3 PO 4
Solve for x by successive approximations or by the quadratic equation: x = 0.12 M = [H3O+] = [H2PO4–] [H3PO4] = 2.0 – 0.12 = 1.9 M pH = –log(0.12) = 0.92 The second dissociation: [H2PO4–] [HPO42–] [H3O+] 0.12 – 0.12 I –x +x +x C 0.12 – x +x 0.12 + x E Assume that x << 0.12, therefore 0.12 – x ≈ 0.12 and 0.12 + x ≈ 0.12
(x)(0.12) = 6.2 × 10−8 (0.12) x = 6.2 × 10–8 = [HPO42–] K a2 =
The third dissociation: [HPO42–] [PO43–] [H3O+] –8 6.3 × 10 – 0.12 I –x +x +x C 6.3 × 10–8 – x +x 0.12 + x E Assume that x << 6.2 × 10–8, therefore 6.2 × 10–8 – x ≈ 6.2 × 10–8 and 0.12 + x ≈ 0.12 (x)(0.12) K a3 = = 4.2 × 10 −13 (6.2 × 10−8 ) Solving for x we get, x = 2.2 × 10–19 = [PO43–] 17.107 H3AsO4 + H2O H3O+ + H2AsO4–
Ka1 = 5.5 × 10–3
H2AsO4– + H2O H3O+ + HAsO42–
Ka2 = 1.7 × 10–7
HAsO42– + H2O H3O+ + AsO43– [HAsO42–] = 1.7 × 10–7
Ka3 = 5.1 × 10–12
The following assumptions are made: [H+]total ≈ [H+]first step [H2AsO4–]total ≈ [H2AsO4–]first step [HAsO42–]total ≈ [HAsO42–]second step
399
Chapter 17
Now solve using Ka1 = 5.5 × 10–3 for the other concentrations:
I C E
[H2AsO4–] – +x +x
[H3AsO4] 0.25 –x 0.25 – x
[H+] – +x +x
H AsO − H + 2 4 (x)(x) = K a1 = = 5.5 × 10−3 − H AsO 0.25 x ( ) 4 3 Solve for x by successive approximations or by the quadratic equation: x = 0.034 M = [H+] = [H2AsO4–] [H3AsO4] = 0.25– 0.034 = 0.216 M pH = –log(0.034) = 1.47 [H2AsO4–] = 0.034 M For the concentration of HAsO42– HAsO 2− H + 4 = 1.7 × 10–7 Ka2 = H AsO − 4 2
HAsO 2− ( 0.034 ) 4 Ka2 = = [HAsO42–] = 1.7 × 10–7 0.034 ( )
–
17.108 H3PO3 + H2O H2PO3 + H3O
+
Ka 1
H PO − H O + 2 3 3 = 5.0 × 10−2 = H3 PO3
HPO 2 − H O + 3 3 = 2.0 × 10−7 H2PO3 + H2O HPO3 + H3O Ka 2 = H PO − 2 3 To simplify the calculation, assume that the second dissociation does not contribute a significant amount of H+ to the final solution. Solving the equilibrium problem for the first dissociation gives: –
2–
I C E
Ka 1
[H3PO3] 1.0 –x 1.0 – x
+
[H2PO3–] – +x +x
[H3O+] – +x +x
H PO − H + 2 3 = (x) (x) = 5 × 10−2 = (1.0 − x) H3 PO3
Because K a 1 is so large, a quadratic equation must be solved. On doing so we learn that x = 0.20 M = [H3O+] = [H2PO3–]. pH = – log[H3O+] = – log(0.20) = 0.70 The [HPO32–] may be determined from the second ionization constant.
400
Chapter 17
I C E
Ka 2
[H2PO3–] 0.20 –x 0.20 – x
[HPO32–] – +x +x
[H3O+] 0.20 +x 0.20 + x
HPO 2− H O + 3 3 = (x) (0.20 + x) = 2.0 × 10−7 = − (0.20 − x) H PO 2 3
If we assume that x is small then 0.20 ± x ≈ 0.20. Then, x = [HPO32–] = 2.0 × 10–7 M.
+
17.109 H2C2O4 + H2O H3O + HC2O4
I C E
[H3O+] – +x +x
[H2C2O4] 0.20 –x 0.20 – x
[HC2O4–] – +x +x
( x )( x )
Ka = 6.0
H O + HC O − 3 2 4 = 6.0 × 10−2 Ka = [ H 2 C 2 O4 ]
–
( 0.20 − x ) ( x )( x ) × 10–2 = ( 0.20 − x )
–x2 – (6.0 × 10–2)x + (1.12 × 10–2) = 0
(
) ( −6.0 ×10 )
− −6.0 × 10−2 ± x=
−2
2
(
− 4 ( −1) 1.12 ×10−2
2 ( −1)
) = ( 6.0 ×10−2 ) ± (3.6 ×10−3 ) + ( 4.48 ×10−2 ) −2
( 6.0 ×10 ) ± ( 2.20 ×10 ) = −2
−1
−2 x = 8.0 × 10–2 = [H3O+], pH = 1.10
The hydrolysis equation is:
17.110
HSO − OH − 3 SO3 + H2O HSO3 + OH Kb = SO 2 − 3 In order to obtain Kb we will use the relationship Kw = Ka × Kb K 1.0 × 10−14 Kb = w = = 1.5 × 10−7 −8 Ka 6.6 × 10 2–
–
–
[SO32–] [HSO3–] [OH–] 0.24 – – I –x +x +x C 0.24 – x +x +x E Since Kb is so small, assume that x << 0.24 and we determine x = 1.9 × 10–4 M = [OH–] pOH = –log(1.9 × 10–4) = 3.72, pH = 14.00 – pOH = 10.28 401
Chapter 17
[SO32-] = 0.24 M
[HSO3-] = [OH-] = 1.9 x 10-4 M
[Na+] = 0.24 M [H3O+] = 1.00 x 10-14/1.9 x 10-4 = 5.3 x 10-11 The second ionization is very small compared to the first ionization so the [HSO3-], [OH-] and [H3O+] do not change significantly. For the concentration of H2SO3 HSO3– + H2O H2SO3 + OH–
K b2 =
− H 2SO3 OH
HSO − 3
Kw 1.0 × 10−14 = = 8.3 × 10−13 K a1 1.2 × 10−2
K b2 =
H 2SO3 1.9 × 10−4
K b2 =
1.9 ×10−4 [H2SO3] = 8.3 × 10–13
= 8.3 × 10–13
17.111 The hydrolysis equation is: CO3
2–
–
+ H2O HCO3 + OH
–
HCO − OH − 3 Kb = CO 2− 3
In order to obtain Kb we will use the relationship Kw = Ka × Kb Kb1 =
Kw 1.0 × 10−14 = = 1.8 × 10 −4 −11 Ka2 5.6 × 10
I C E
[CO32–] 0.33 –x 0.33 – x
1.8 × 10–4 =
[HCO3–] – +x +x
[OH–] – +x +x
( x )( x )
0.33 − x Since Kb is so small, assume that x << 0.33 and we determine x = 7.7 × 10–3 M = [OH–] pOH = –log(7.7 × 10–3) = 2.11, pH = 14.00 – pOH = 11.89 H 2 CO3 OH − – – HCO3 + H2O H2CO3 + OH K b2 = HCO − 3 For the concentration of H2CO3 Kw 1.0 × 10−14 K b2 = = = 2.3 × 10−8 −7 K a1 4.3 × 10
402
Chapter 17
K b2 =
H 2 CO3 7.7 × 10−3
7.7 × 10−3 [H2CO3] = 2.3 × 10–8
= 2.3 × 10–8
17.112 C6H5O73– + H2O HC6H5O72– + OH– HC H O 2− OH − 6 5 7 Kw 1.0 × 10−14 Kb = = = = 2.5 × 10 −8 Ka3 C6 H5 O7 3− 4.0 × 10−7
I C E
[C6H5O73–] 0.12 –x 0.12 – x
[HC6H5O72–] – +x +x
Assume x << 0.12 ( x )( x ) = 2.5 × 10−8 Kb = 0.12 pOH = 4.26 pH = 9.74
[OH–] – +x +x
x = 5.5 × 10−5 = [OH − ]
17.113 C2O42– + H2O HC2O4– + OH– HC O − OH − 2 4 K 1.0 × 10−14 Kb = = w = = 1.6 × 10−10 2− −5 Ka C2 O 4 5.4 × 10
I C E
[C2O42–] 0.22 –x 0.22 – x
[HC2O4–] – +x +x
Assume x << 0.22 ( x )( x ) = 1.6 × 10−10 Kb = 0.22 pOH = 5.23 pH = 8.77 3–
2–
17.114 PO4 + H2O HPO4 + OH 2–
x = 5.9 × 10−6 = [OH − ]
–
–
HPO4 + H2O H2PO4 + OH –
[OH–] – +x +x
H2PO4 + H2O H3PO4 + OH
–
–
K b1 =
Kw 1.0 x 10−14 = =2.4 × 10−2 Ka 3 4.2 x 10−13
Kb 2 =
Kw 1.0 x 10−14 = =1.6 × 10−7 −8 Ka 2 6.2 x 10
Kb 3 =
Kw 1.0 x 10−14 = = 1.3 × 10−12 K a1 7.5 x 10−3
By analogy with polyprotic acids, we know that [H2PO4–] = 1.6 × 10–7 403
Chapter 17 We need to solve the first equilibrium expression to determine [HPO42–].
K b1
HPO 2 − OH − 4 = 2.4 × 10−2 = − 3 PO 4
[PO43–] [HPO42–] [OH–] 0.50 – – I –x +x +x C 0.50 – x +x +x E Assume x << 0.50 ( x )( x ) = 2.4 × 10−2 Kb1 = 0.50 − x Use the quadratic equation to solve for x since Kb is so large x = 9.8 × 10–2 M = [OH–] = [HPO42–] pOH = 1.01 pH = 12.99 [PO43–] = 0.50 – 9.8 × 10–2 = 0.40 M Solve the third equilibrium expression to determine [H3PO4]: H3 PO4 OH − = 1.3 × 10−12 K b3 = H PO − 2 4 Substitute the calculated values of [H2PO4–] and [OH–] and solve for x:
(9.8 × 10 ) x = 1.3 × 10 (1.6 × 10 ) −2
K b3 =
−12
−7
x = H3 PO4 = 2.1 × 10−18 17.115 Because the pH is so high and is greater than pKa, the [CO32–] = 0.10 M. The added OH– reacts with most of the HCO3– to reform CO32–. 17.116 Since HCO2H and NaOH react in a 1:1 ratio: HCO2H + NaOH NaHCO2 + H2O we can use the equation Va × Ma = Vb × Mb to determine the volume of NaOH that is required to reach the equivalence point, i.e. the point at which the number of moles of NaOH is equal to the number of moles of HCO2H: VNaOH = 35.0 mL × 0.050/0.050 = 35.0 mL Thus the final volume at the equivalence point will be 35.0 + 35.0 = 70.0 mL. The concentration of NaHCO2 would then be: 0.035 mol/L × 0.050 L = 1.75 × 10–3 mol HCO2H = 1.75 × 10–3 mol NaHCO2 1.75 × 10–3 mol/0.070 L = 2.5 × 10–2 M NaHCO2 The hydrolysis of this salt at the equivalence point proceeds according to the following equilibrium: HCO2– + H2O HCO2H + OH–
Kb =
[ HCO2 H] OH HCO2 −
−
= 5.6 × 10−11
404
Chapter 17
I C E
[HCO2–] 0.025 –x 0.025 – x
[HCO2H] – +x +x
[OH–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming x << 0.050 gives: ( x )( x ) = 5.6 × 10−11 Kb = 0.025 x2 = 1.4 × 10–12 x = 1.2 × 10–6 M = [OH–] = [HCO2H] – pOH = –log[OH ] = –log(1.2 × 10–6) = 5.93 pH = 14.00 – pOH = 14.00 – 5.93 = 8.07 Cresol red would be a good indicator, since it has a color change near the pH at the equivalence point. 17.117 Since HBr and NH3 react in a 1:1 ratio: HBr + NH3 NH4+ + Br– we can use the equation Va × Ma = Vb × Mb to determine the volume of HBr that is required to reach the equivalence point, i.e. the point at which the number of moles of HBr is equal to the number of moles of NH3: VHBr = 25 mL × 0.12/0.12 = 25 mL Thus the final volume at the equivalence point will be 25 + 25 = 50 mL. The concentration of NH4+ would then be: 0.12 mol/L × 0.025 L = 3.0 × 10–3 mol NH3 = 3.0 × 10–3 mol NH4+ 3.0 × 10–3 mol/0.050 L = 6.0 × 10–2 M NH4+ At the equivalence point NH4+ dissociate according to the following equilibrium: NH4+ H+ + NH3 Ka =
+ NH3 H = 5.6 × 10−10 NH 4+
I C E
[NH4+] 0.060 –x 0.060 – x
[H+] – +x +x
[NH3] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming x << 0.050 gives:
Ka =
( x )( x ) 0.060
= 5.6 × 10−10
x = 5.8 × 10–6 M = [H+] pH = –log[H+] = –log(5.8 × 10–6) = 5.24 Bromocresol purple would be a good indicator, since it has a color change near the pH at the equivalence point.
405
Chapter 17
0.180 mol HC2 H3O2 –3 17.118 mol HC2H3O2 = (30.0 mL HC2H3O2) = 5.40 × 10 mol HC2H3O2 1000 mL HC H O 2 3 2 − 0.250 mol OH mol OH– = (45.0 mL OH–) = 1.13 × 10–2 mol OH– 1000 mL OH − excess OH– = (1.13 × 10–2) – (5.4 × 10–3) = 5.9 × 10–3 mol 5.9 × 10−3 moles = 7.87 × 10–2 M 1L (30.0 + 45.0 mL) 1000 mL pOH = 1.10 pH = 12.90 [OH − ] =
0.200 mol HC2 H3O 2 –3 17.119 mol HC2H3O2 = (25.0 mL HC2H3O2) = 5.00 × 10 mol HC2H3O2 1000 mL HC H O 2 3 2 − 0.400 mol OH mol OH– = (15.0 mL OH–) = 6.00 × 10–3 mol OH– 1000 mL OH − excess OH– = (6.00 × 10–3) – (5.00 × 10–3) = 1.00 × 10–3 mol 1.00 × 10−3 moles = 2.5 × 10–2 M 1L (25.0 + 15.0 mL) 1000 mL pOH = 1.60 pH = 12.40 [OH − ] =
17.120 (a)
HC2H3O2 + H2O H3O+ + C2H3O2–
I C E
[HC2H3O2] 0.1000 –x 0.1000 – x
[H3O+] – +x +x
H O + C H O − 3 2 3 2 = 1.8 × 10−5 Ka = HC 2 H3O 2 [C2H3O2–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.1000 gives: x = [H3O+] = 1.342 × 10–3 M. pH = –log [H3O+] = –log (1.342 × 10–3) = 2.87. (b)
When NaOH is added, it will react with the acetic acid present decreasing the amount in solution and producing additional acetate ion. Since this a one–to–one reaction, the number of moles of acetic acid will decrease by the same amount as the number of moles of NaOH added and the number of moles of acetate ion will increase by an identical amount. We must determine the number of moles of all ions present and calculate new concentrations accounting for dilution. 0.1000 moles HC 2 H3O 2 –3 mol HC2H3O2 = (0.07500 L solution) = 7.500 × 10 mol HC2H3O2 1 L solution − 0.1000 moles OH mol OH– = (0.02500 L solution) = 2.500 × 10–3 mol OH– 1 L solution 406
Chapter 17
HC 2 H3O 2 =
7.500 × 10−3 moles − 2.500 × 10−3 moles = 5.000 × 10–2 M 0.07500 L + 0.02500 L
−3 C H O − = 0 moles + 2.500 × 10 moles = 2.500 × 10–2 M 2 3 2 0.07500 L + 0.02500 L
C H O − 2 3 2 pH = pK a + log HC2 H3O2
( 2.500 × 10 ) = 4.44 (5.000 × 10 ) −2
= 4.7447 + log
−2
(c)
When half the acetic acid has been neutralized, there will be equal amounts of acetic acid and acetate ion present in the solution. At this point, pH = pKa = 4.74.
(d)
At the equivalence point, all of the acetic acid will have been converted to acetate ion. The concentration of the acetate ion will be half the original concentration of acetic acid since we have doubled the volume of the solution. We then need to solve the equilibrium problem that results when we have a solution that possesses a [C2H3O2–] = 0.05000 M. C2H3O2– + H2O HC2H3O2 + OH–
Kb =
HC2 H3O2 OH − C H O − 2 3 2
I C E
= 5.6 × 10−10
[C2H3O2–] 0.05000 –x 0.05000 – x
[HC2H3O2] – +x +x
[OH–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.05000 gives: x = [OH–] = 5.292 × 10–6 M. pOH = –log [OH–] = –log (5.292 × 10–6) = 5.2764. pH = 14.0000 – pOH = 14.0000 – 5.2764 = 8.72. 17.121
(a)
NH3 + H2O NH4+ + OH–
NH + OH − 4 = 1.8 × 10−5 Kb = NH3
I C E
[NH3] 0.1000 –x 0.1000 – x
[NH4+] – +x +x
[OH–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.1000 gives: x = [OH–] = 1.342 × 10–3 M. pOH = –log [OH–] = –log (1.342 × 10–3) = 2.8724. pH = 14.00 – pOH = 11.1276
407
Chapter 17
(b)
When HCl is added, it will react with the ammonia present decreasing the amount in solution and producing additional ammonium ion. Since this a one–to–one reaction, the number of moles of ammonia will decrease by the same amount as the number of moles of HCl added and the number of moles of ammonium ion will increase by an identical amount. We must determine the number of moles of all ions present and calculate new concentrations accounting for dilution. 0.1000 moles NH3 –3 mol NH3 = (0.05000 L solution) = 5.000 × 10 mol NH3 1 L solution 0.1000 moles H + mol H+ = (0.02000 L solution) = 2.000 × 10–3 mol H+ 1 L solution
NH3 =
5.000 × 10−3 moles − 2.000 × 10−3 moles = 4.286 × 10–2 M NH3 0.05000 L + 0.02000 L
−3 NH + = 0 moles + 2.000 × 10 moles = 2.857 × 10–2 M NH4+ 4 0.05000 L + 0.02000 L
NH + 2.857 × 10−2 4 pOH = pKb + log = 4.7447 + log NH3 4.286 × 10−2
( (
) = 4.5686 )
pH = 14.0000 – pOH = 9.4314 (c)
When half the ammonia has been neutralized, there will be equal amounts of ammonia and ammonium ion present in the solution. At this point, pH = pKb = 4.7447.
(d)
At the equivalence point, all of the ammonia will have been converted to ammonium ion. The concentration of the ammonium ion will be half the original concentration of ammonia since we have doubled the volume of the solution. We then need to solve the equilibrium problem that results when we have a solution that possesses a [NH4+] = 0.05000 M. NH4+ H+ + NH3
H + NH 3 Ka = = 5.6 × 10−10 NH + 4
I C E
[NH4+] 0.05000 –x 0.05000 – x
[H+] – +x +x
[NH3] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.05000 gives: x = [H+] = 5.292 × 10–6 M. pH = –log [H+] = –log (5.292 × 10–6) = 5.2764.
Additional Exercises 17.122 HNO2 + H2O H3O+ + NO2– Ka =
[H3O+ ][NO 2− ] = 7.1 × 10−4 HNO 2
408
Chapter 17
I C E
[HNO2] 1.0 1.0 – x 1.0 – x
[H3O+] – +x +x
[NO2–] – +x +x
Assume x << 1.0 [x][x] = 7.1 × 10−4 [1.0] x = 2.7 × 10–2 = [H3O+] Ka =
% ionization = ([H+]/[total acid]) × 100% = (2.7 × 10–2/1.0) × 100% = 2.7% At lower concentrations, the only change is the denominator in the final expression for Ka above. Calculating the answers, we have: 1.0 M = 2.7% ionization 0.10 M = 8.4% ionization 0.010 M = 27% ionization As a weak acid becomes more dilute, its % ionization goes up. 17.123 From the complete ionization of HCl: [H+] = 0.100 M We must solve an equilibrium problem in order to determine the contribution of H+ from the acetic acid. H + C H O − 2 3 2 + – HC2H3O2 H + C2H3O2 Ka = = 1.8 × 10−5 HC2 H3O 2
I C E
[HC2H3O2] 0.125 –x 0.125–x
[H+] 0.100 +x 0.100+x
[C2H3O2–] – +x +x
Substituting the above values for equilibrium concentrations into the mass action expression and assuming that x << 0.125 M and x << 0.100 M gives: x = [C2H3O2–] = 2.25 × 10–5 M. The total [H+] is therefore 0.100 + 2.25 × 10–5 = 0.100 M, and pH is 1.000. The weak acid contributes very little H+ to the final solution. 17.124 When these two solutions are mixed, the HCl will react with the ammonia producing NH4+. Since HCl is the limiting reactant, we will have an excess of ammonia. The resulting solution will be a buffer as it consists of ammonia, a weak base, and ammonium ion, its conjugate acid. We need to determine the amount of NH4+ that is produced and the amount of NH3 that remains: + 0.500 mol HCl 1 mol H mol H + = (225 mL HCl) = 0.1125 mol H + 1000 mL HCl 1 mol HCl mol NH3 = (mol NH3)initial – (mol H+)added 0.500 mol NH3 + mol NH3 = (375 mL) – 0.1125 mol H = 0.075 mol NH3 1000 mL NH 3 mol NH4+ = mol H+ = 0.1125 mol NH4+
409
Chapter 17
NH + 4 pOH = pK b + log = 4.74 + 0.18 = 4.92 NH3 pH = 14.00 – pOH = 9.08
17.125 (a)
(b)
We first need to determine the concentration of the components of the solution. mol HC2 H3O 2 12.5 g HC2 H3O 2 1 mol HC2 H3O 2 = = 0.269 M HC2 H3O2 = L soln 0.775 L 60.05 g HC2 H3O 2 mol NaC2 H3O 2 20.0 g NaC2 H3O 2 1 mol NaC2 H3O 2 = = 0.315 M NaC2 H3O2 = L soln 0.775 L 82.03 g NaC2 H3O 2 [ anion ] = 4.745 + log 0.315 = 4.813 pH = pK a + log 0.269 [ acid ] The added base will react with the acid present using it up and producing additional acetate ion. To solve this problem, determine the number of moles of base added and subtract this amount from the amount of acid present. Add the same amount to the amount of acetate ion present. Be sure to account for dilution when you determine the new concentrations. − 0.25 mol NaOH 1 mol OH mol OH − = (25.00 mL NaOH) = 0.0063 mol OH– 1000 mL NaOH 1 mol NaOH Consequently, the # mol of acetic acid will decrease by 0.0063 and the # of moles of acetate ion will increase by 0.0063. The new concentrations are: ( 0.208 − 0.0063) mol HC2 H3O 2 mol HC2 H3O 2 = HC2 H3O2 = = 0.252 M L soln 0.800 L − 0.244 + 0.0063) mol C H O − 2 3 2 C H O − = mol C2 H3O 2 = ( 2 3 2 L soln 0.800 L [ anion ] = 4.745 + log 0.313 = 4.839 pH = pK a + log 0.252 [ acid ]
(c)
This problem is identical to the previous except that we are adding acid to the original solution. Consequently, the acid concentration will increase and the acetate ion concentration will decrease. + 0.40 mol HCl 1 mol H mol H + = (25.0 mL HCl) = 0.010 mol H+ 1000 mL HCl 1 mol HCl ( 0.208 + 0.010 ) mol HC2 H3O 2 mol HC2 H3O 2 = = 0.273 M HC2 H3O2 = L soln 0.800 L − 0.244 − 0.010 ) mol C H O − 2 3 2 C H O − = mol C2 H3O 2 = ( 2 3 2 L soln 0.800 L anion [ ] = 4.745 + log 0.293 = 4.776 pH = pK a + log 0.273 [ acid ]
17.126 (a)
= 0.313 M
= 0.293 M
The equation that will be used is the normal Henderson–Hasselbach equation, namely: A− anion ] [ pH = pK a + log = pK a + log [ acid ] [ HA ] 410
Chapter 17 where A– = C2H3O2– and HA = HC2H3O2. We note further that the log term involves a ratio of concentrations, but that the volume remains constant in a process such as that to be analyzed here. Thus the log term may be replaced by a ratio of mole amounts, since volumes cancel:
( moles C H O ) −
pH = pK a + log
2
3 2
( moles HC2 H3O2 )
Thus we need only determine the number of moles of acid and conjugate base that remain in the buffer after the addition of a certain amount of H+ or OH–, in order to determine the pH of the buffer mixture after that addition. The buffer is changed in the following way by the addition of OH–: HC2H3O2 + OH– H2O + C2H3O2– In other words, if 0.0100 moles of OH– are added to the buffer, the amount of HC2H3O2 goes down by 0.0100 moles, whereas the amount of C2H3O2– goes up by 0.0100 moles. In addition, the pH of the solution will increase upon the addition of base. The buffer is changed in the following way by the addition of H+: C2H3O2– + H+ HC2H3O2 If 0.0100 mol of H+ are added, then the amount of C2H3O2– goes down by 0.0100 moles and the amount of HC2H3O2 goes up by 0.0100 moles. As before, the addition of acid will decrease the pH of the buffer system. For the general buffer mixture that contains x mol of C2H3O2– and y mol of HC2H3O2, we can apply the Henderson–Hasselbach equation, noting the maximum amount by which we want the pH to change (namely 0.10 units): For the case of added base:
5.12 + 0.10 = 4.74 + log
( x + 0.0100 ) moles C2 H3O2− ( x − 0.0100 ) moles HC2 H3O2
Simplifying and taking the antilog of both sides of the above equation gives: [x + 0.0100]/[y – 0.0100] = 3.0, which is now designated eq. 1. For the case of added acid:
5.12 − 0.10 = 4.74 + log
( x − 0.0100 ) moles C2 H3O2− ( x + 0.0100 ) moles HC2 H3O2
Simplifying and taking the antilog of both sides of the equation gives: [x – 0.0100]/[y + 0.0100] = 1.9, which is now designated eq. 2. The equations 1 and 2 as designated above are solved simultaneously, since they are two equations containing two unknowns: x = initial mol C2H3O2– = 0.15 mol y = initial mol HC2H3O2 = 6.3 × 10–2 mol These values are converted to grams as follows: 6.3 × 10–2 mol HC2H3O2 × 60.1 g/mol = 3.8 g HC2H3O2 0.15 mol NaC2H3O2 × 118 g/mol = 18 g NaC2H3O2•2H2O
411
Chapter 17
These are the minimum amounts of acid and conjugate base that would be required in order to prepare a buffer that would change pH by only 0.10 units on addition of either 0.0100 mol of OH– or 0.0100 mol of H+. (b)
6.3 × 10–2 mol/0.250 L = 0.25 M HC2H3O2 0.15 mol C2H3O2–/0.250 L = 0.60 M NaC2H3O2
(c)
0.0100 mol OH–/0.255 L = 3.92 × 10–2 M OH– pOH = –log[OH–] = –log(3.92 × 10–2) = 1.406 pH = 14.000 – pOH = 14.000 – 1.406 = 12.593
(d)
1.00 × 10–2 mol H+/0.255 L = 3.92 × 10–2 M H+ pH = –log[H+] = –log(3.92 x 10–2) = 1.406
17.127 Since the ammonium ion is the salt of a weak base, NH3, it is acidic. The fluoride ion is the salt of a weak acid, HF, so it is basic. In order to determine if the solution is acidic or basic, we need to determine the relative strength of the two components. Use the relationship Ka × Kb = Kw in order to determine the dissociation constants for the cyanide ion and the ammonium ion. Ka(NH4+) = Kw ÷ Kb(NH3) = 1.0 × 10–14 ÷ 1.8 × 10–5 = 5.6 × 10–10 Kb(F–) = Kw ÷ Ka(HF) = 1.0 × 10–14 ÷ 6.8 × 10–4 = 1.5 × 10–11 Since the Ka(NH4+) is larger than the Kb(F−) the NH4F solution will be acidic. 17.128 Using the equilibrium for the conjugate base of NH4+ (Kb = 1.8 × 10–5) and the conjugate acids of C2H3O2– (Ka = 1.8 × 10–5). Since the two equilibrium constants are equivalent, and the concentrations of the anion and cation are equivalent, therefore the concentrations of H+ and OH– are equivalent and the pH is 7. 17.129 In order to solve this problem we must first neutralize the HNO3 that is present in the solution. Since HNO3 is a monoprotic acid and NH3 is a monobasic base, they will react in a one to one stoichiometry. For every mole of NH3 added, one mole of HNO3 will be neutralized and one mole of NH4+ will be produced. There are initially 0.0125 moles HNO3 = (0.250 L)(0.050 mol L–1) present in the solution. Consequently, we must first add 0.0125 moles of NH3 to neutralize all of the HNO3. Now the question may be rephrased to ask, how much ammonia should be added to a solution that is 0.050 M in NH4+ so that the final pH is 9.26?
[NH3] Z Z–x Z–x
[NH4+] 0.050 0.050 + x 0.050 + x
( 0.050 + x )( x ) (Z − x)
= 1.8 × 10−5
I C E
Kb =
[OH–] – +x +x
The problem states that the equilibrium pH = 9.26 and, therefore, the pOH = 4.74 which implies that [OH–] = 1.8 × 10–5 M = x. Substituting this into the equilibrium expression and assuming that Z >> x we get
( 0.050 ) (1.8
× 10−5
)
= 1.8 × 10−5 Z and Z = 0.050 M. Since the volume of the solution is 250 mL, this concentration corresponds to 0.0125 moles NH3 = (0.250 L)(0.050 mol L–1).
412
Chapter 17
The total amount of NH3 that must be added to the original solution is thus 0.0125 moles + 0.0125 moles = 0.0250 mols. The first addition neutralizes the HNO3 and the second amount adjusts the pH. The volume of NH3 that must be added may be calculated using the ideal gas law;
(
) )
L atm (298 K) (0.0250 mol) 0.0821 mol nRT K V = = = 0.63 L = 630 mL 1 atm P (740 torr) 760 torr
(
17.130 Both HClO4 and HNO3 are 100% ionized in water and, therefore, appear to be the same strength. To distinguish between these two, we would need a solvent that is a weaker proton acceptor than H2O.
17.131 (a)
HSO4– + H2O H3O+ + SO42–
SO 2− H O + 4 3 Ka = [ HSO4 ]
= 1.2 × 10–2
(b)
I C E
[HSO4–] 0.010 –x 0.010 – x
[H3O+] – +x +x
[SO42–] – +x +x
[x][x] = 1.2 × 10−2 0.010 − x Solve the quadratic equation to get x = 6.5 × 10–3 = [H3O+] Ka =
[x][x] = 1.2 × 10−2 0.010
x = 1.1 × 10 −2 = [H3O+ ]
(c)
Ka =
(d)
The exact answer (obtained in part b) is 60% smaller than the approximate solution.
17.132 Ascorbic acid has the formula H2C6H6O6 and is a diprotic weak acid. We only need look at the first dissociation to determine the pH. HC H O − H + 6 6 6 = 8.0 × 10−5 – + H2C6H6O6 HC6H6O6 + H K a1 = H 2 C6 H 6 O6 6.0 g H 2 C6 H 6 O6 1 mole H 2 C6 H 6 O6 1000 mL H 2 C6 H 6 O6 = = 0.14 M 250 mL 176 g H 2 C6 H 6 O6 1 L
I C E
[H2C6H6O6] 0.14 –x 0.14 – x
[HC6H6O6–] – +x +x
[H+] – +x +x
HC H O − H + 6 6 6 = (x)(x) = 8.0 × 10−5 Ka 1 = H C H O (0.14 − x) 2 6 6 6
413
Chapter 17
Assuming that x << 0.14 and solving we find; x = [H+] = 3.3 × 10–3 M
pH = – log[H+] = – log (3.3 × 10–3) = 2.47
17.133 This exercise is an example of a titration using a strong acid and a strong base. Prior to reaching the equivalence point, there will be an excess of acid present. The amount of acid present may be determined by subtracting the number of moles of base added from the amount of acid initially present. We may then use the dilution equation, M1 × V1 = M2 × V2, to determine the new concentration of acid and then calculate the pH. After the equivalence point, there is an excess of base. The amount in excess is determined by subtracting the inital amount of acid present from the amount of base added. Again, the dilution equation may be used to determine the resulting concentration. Start by determining the amount of H+ initially present:
0.1000 moles H + moles H + = ( 25.00 mL solution ) 1000 mL solution = 2.500 × 10−3 moles H + (a)
Initially, there is only 0.1000 M HCl present: [H+] = 0.1000 M, pH = –log[H+] = 1.0000.
(b)
10.00 mL added base
0.1000 moles OH − moles OH − = (10.00 mL solution ) = 1.000 × 10–3 mol OH– 1000 mL solution −3 −3 2.500 × 10 moles − 1.000 × 10 moles + H = = 4.286 × 10–2 mol H+ 0.02500 L + 0.01000 L pH = –log[H+] = 1.3680
(c)
24.90 mL added base
0.1000 moles OH − moles OH − = ( 24.90 mL solution ) 1000 mL solution
= 2.490 × 10–3 mol OH–
−3 −3 H + = 2.500 × 10 moles − 2.490 × 10 moles = 2.004 × 10–4 mol H+ 0.02500 L + 0.02490 L pH = –log[H+] = 3.6981
(d)
24.99 mL added base
0.1000 moles OH − moles OH − = ( 24.99 mL solution ) 1000 mL solution
= 2.499 × 10–3 mol OH–
−3 −3 H + = 2.500 × 10 moles − 2.499 × 10 moles = 2.000 × 10–5 mol H+ 0.02500 L + 0.02499 L pH = –log[H+] = 4.6989
414
Chapter 17
(e)
25.00 mL added base This is the equivalence point and [H+] = 1.000 × 10–7 M pH = 7.000
(f)
25.01 mL added base We have now added excess base so we will subtract the number moles H+ initially present from the number moles OH– added. 0.1000 moles OH − moles OH − = ( 25.01 mL solution ) =2.501 × 10–3 mol OH– 1000 mL solution −3 −3 OH − = 2.501 × 10 moles − 2.500 × 10 moles = 2.000 × 10–5 M OH– 0.02501 L + 0.02500 L pOH = –log[OH–] = 4.6991 pH = 14.0000 – pOH = 9.3009
(g)
25.10 mL added base
0.1000 moles OH − moles OH − = ( 25.10 mL solution ) 1000 mL solution
=2.510 × 10–3 mol OH–
−3 −3 OH − = 2.510 × 10 moles − 2.500 × 10 moles = 1.996 × 10–4 M OH– 0.02510 L + 0.02500 L pOH = –log[OH–] = 3.6998 pH = 14.0000 – pOH = 10.3002
(h)
26.00 mL added base
0.1000 moles OH − moles OH − = ( 26.00 mL solution ) 1000 mL solution
=2.600 × 10–3 mol OH–
−3 −3 OH − = 2.600 × 10 moles − 2.500 × 10 moles = 1.961 × 10–3 M OH– 0.02600 L + 0.02500 L pOH = –log[OH–] = 2.708 pH = 14.0000 – pOH = 11.292
(i)
50.00 mL added base
0.1000 moles OH − moles OH − = ( 50.00 mL solution ) 1000 mL solution
= 5.000 × 10–3 mol OH–
−3 −3 OH − = 5.000 × 10 moles − 2.500 × 10 moles = 3.333 × 10–2 M OH– 0.05000 L + 0.02500 L pOH = –log[OH–] = 1.4771 pH = 14.0000 – pOH = 12.5229
415
Chapter 17
17.134 For the given mixture, pH does not equal pKa because the concentration of the acid does not equal the concentration of its salt. For each OH– ion added an additional F– will form and an HF molecule will dissociate: HF + H2O H3O+ +F– As OH– is added it reacts with the H3O+ causing additional HF to dissociate. For each H+ ion added an F– ion will react with it forming additional HF. Using the equation above, the added H+ reacts with F– producing HF. Initially, there are 5HF and 4F − ions. After addition of 3OH − ions we will have 2HF and 6F −. If 2H+ are added we will have 7HF and 2F −.
Multi-Concept Problems 17.135 A solution with pH = 4.00 would have: [H+] = 10–pH = 10–4.00
= 1.00 × 10–4 M = 1.00 × 10–4 mol [H+]/L solution = 0.000100 mol [H+]/L solution
We begin with: 1 L HCl soln 0.010 mol HCl mol HCl = 200 mL HCl soln 1,000 mL HCl soln 1 L HCl soln = 0.002 mol HCl = 0.002 mol H+ Our final molarity will be: M = mol [H+] /L solution. This can be written as: M = (0.002 – nNaOH)/(0.200 + x), where nNaOH = moles of NaOH added and, x = volume of 0.10 M NaOH added. But nNaOH is simply MNaOH·VNaOH, so M = (0.002 – 0.10x)/(0.200 + x) 0.000100 = (0.002 – 0.10x)/(0.200 + x) (0.200 + x)0.000100 = (0.002 – 0.10x)
416
Chapter 17
2.00 × 10−5 + 0.000100 x = 0.002 – 0.10x 0.1001 x = 0.00198 x = 0.0200, or 20.0 mL 20.0 mL of 0.10 M NaOH should be added to 200 mL of 0.010 M HCl 17.136 pOH = 14.00 – pH = 14.00 – 10.08 = 3.92 pOH = –log [OH–] [OH–] = 10–pOH = 10–3.92 = 1.2 × 10–4 M 1 Tbs has a volume of 14.8 mL 1.2 × 10−4 mol 1 mol Mg(OH) 2 58.32 g Mg(OH)2 grams Mg(OH) 2 = (29.6 mL) 2 mol OH − 1 mol Mg(OH)2 1000 mL = 1.0 × 10−4 g Mg(OH)2 17.137 First, we must find the concentration of the HCl solution. Since HCl is a strong acid, we know that [HCl] = [H+]. We can find [H+] from the pH: [H+] = 10–pH = 10–2.15 = 7.1 × 10–3 M H+ Now we can solve the problem using the given conversion factors: 7.1 × 10−3 mol H + 1 mol OH – 1000 mL OH – volume of 0.0100 M KOH = 325 mL 1000 mL solution 1 mol H + 0.0100 mol OH – = 231 mL 0.0100 M KOH 17.138 pOH = 14 – pH = 14 – 12.05 = 1.95 [OH–] = 10–pOH = 10–1.95 = 1.12 × 10–2 M OH– 1.1× 10−2 mol OH – 1 mol H + mol of H+ = 30.0 mL LiOH solution 1000 mL LiOH solution 1 mol OH – –4 + 3.4 × 10 mol H –2 M HNO3 = = 1.5 × 10 M HNO3 0.02275 L solution
= 3.4 × 10–4 mol H+
17.139 HCl and NaOH react in a one-to-one mole ratio. The first step is to determine the limiting reagent. Moles HCl = 38.0 mL x (1L/1000 mL) x 0.000200 M = 7.6 x 10-6 moles Moles NaOH = 40.0 mL x (1 L.1000 mL) x 0.000180 M = 7.2 x 10-6 moles The final solution will have an excess of H+. 7.6 x 10-6 mol H+ - 7.2 x 10-6 mol OH− = 4.0 x 10-7 mol H+ [H+] = 4.0 x 10-7 mol/(0.0380 + 0.040) L = 5.1 x 10-6 M pH = -log (5.1 x 10-6) = 5.29
17.140 The moles of HCl is determined from the ideal gas law:
(
)
1 atm PV ( 734 torr ) 760 torr ( 0.0100 L ) n= = = 3.947 × 10 –4 mol HCl L atm RT 0.0821 mol K 298 K
417
Chapter 17
The molarity of the solution is:
3.95 × 10 –4 mol HCl = 1.54 × 10–3 M HCl 0.257 L solution pH = –log[H+] = –log(1.54 × 10–3 M) = 2.813 M of solution =
17.141 From problem 17.140, we find there are 3.95 × 10–4 mol HCl present. We need to know how many mol NaOH are in the NaOH solution to react with the HCl. We can find [OH–] from the given pH: pOH = 14.00 – pH = 14.00 – 10.50 = 3.50 [OH–] = 10–pOH = 10–3.50 = 3.16 × 10–4 M –4 1 L NaOH soln 3.16 × 10 mol NaOH mol NaOH = 20.00 mL NaOH soln 1 L NaOH soln 1000 mL NaOH soln –6 = 6.32 × 10 mol NaOH
Since HCl and NaOH react in a 1–to–1 ratio, 6.32 × 10–6 mol of each will neutralize one another. This leaves (3.95 × 10–4 mol HCl – 6.32 × 10–6 mol HCl) = 3.89 × 10–4 mol HCl remaining. The concentration of H+ is now: [H+]
= mol H+/L solution = 3.89 × 10–4 mol/0.0200 L solution = 1.94 × 10–2 M
Therefore pH = –log [H+] = –log [1.94 × 10–2] = 1.71 17.142 Recall the ∆Tf = Kf m, where Kf is the molal freezing point depression constant for the solvent and m is the molality of the solution. We need to determine the molality of this solution. In order to do this, we need to determine the concentrations of the ions which result from the dissociation of the weak acid. Recall that each ion in the solution helps to lower the freezing point. HC2HO2Cl2 H+ + C2HO2Cl2– H + C HO Cl − 2 2 2 Ka = = 5.0 × 10−2 [ HC2 HO2 Cl2 ]
I C E
Ka =
[HC2HO2Cl2] 0.50 –x 0.50 – x
( x )( x ) ( 0.50 − x )
[H+] – +x +x
[C2HO2Cl2–] – +x +x
= 5.0 × 10−2
Because the value of Ka is relatively large, we must solve a quadratic equation in order to determine the H+ concentration. Doing so gives: x = 0.14 M = [H+] = [C2HO2Cl2–] [HC2HO2Cl2] = 0.50 – x = 0.36 M We now calculate the molality of the components of the solution. Because this is an aqueous solution with a density of 1.0 g/mL, the molality and the molarity have the same value: mH+ = 0.14, mC2HO2Cl2– = 0.14 and mHC2HO2Cl2 = 0.36.
418
Chapter 17
The total molality is 0.14 + 0.14 + 0.36 = 0.64 m. The freezing point depression constant for water equals –1.86 °C/m.
(
)
∆Tf = −1.86 o C / m ( 0.64 m ) = −1.19 o C . The freezing point of the solution is –1.19 °C. In solving this problem, we are assuming that the dichloroacetic acid that does not dissociate does not interact with other dichloroacetic acid molecules. In reality, there is some association between the acid molecules forming dimers through hydrogen bonding.
If all of the undissociated acid formed dimmers then the molality of the solution would be: m = 0.14 + 0.14 + 0.18 = 0.46 m.
Then ∆Tf = − 0.86 oC. All of the dichloroacetic acid molecules do not form dimmers. Therefore, the freezing point depression should be between − 0.86 oC and − 1.19 oC.
419
Chapter 18
Practice Exercises 18.1
Ba3(PO4)2(s) 3Ba2+(aq) + 2PO43+(aq) Ksp = [Ba2+]3[PO43+]2
18.2
(a)
18.3
TlI(s) Tl+(aq) + I–(aq) Ksp = [Tl+][I–] 1 mol TlI –5 mol TlI = 5.9 × 10–3 g = 1.78 × 10 mol TlI 331.3 g TlI
Ksp = [Ca2+][C2O42–]
(b)
Ksp = [Ag+]2[SO42–]
1.78 ×10−5 mol = 1.78 × 10–5 M 1L Ksp = (1.8 × 10–5)(1.8 × 10–5) = 3.2 × 10–10
[Tl+] = [I–] =
18.4
(
K sp = 2.15 × 10−3 18.5
Ksp = Pb 2+ F−
PbF2(s) Pb2+(aq) + 2F–(aq)
) ( 2 ( 2.15 × 10 )) −3
2
2
= 3.98 × 10–8
Ag3PO4(s) 3Ag+(aq) + PO42–(aq) Ksp = [Ag+]3[PO42–] The molar solubility of the salt is:
( 7.5x10
−3
gL−1 )
1mol Ag3PO4 = 1.79 x10 −5 M 418.57 g
Ksp (3 x 1.79 x 10-5)3(1.79 x 10-5) = 2.78 x 10-18 18.6
(a) AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.4 x 10-13 The molar solubility for AgBr, x, is equal to the molar concentration of either Ag+ or Br- in the solution since the stoichiometry is 1:1.
x2 = 5.4 x 10-13 x = 7.3 x 10-7 M (b) Ag2CO3(s) 2Ag+(aq) + CO32-(aq) The molar solubility of Ag2CO3, x, is equal to the molar concentration of CO32- in solution. Ksp = [Ag+]2[CO32-] = (2x)2(x) = 8.5 x 10-12
x = 1.3 x 10-4 M
420
Chapter 18
18.7
Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Let the molar solubility equal x. Ksp = [Ca2+]3[PO43-]2 = (3x)3(2x)2 = 2.0 x 10-20
x = 4.5 x 10-5 M 18.8
AgI(s) Ag+(aq) + I–(aq)
Ksp = [Ag+][I–] = 8.5 × 10–17
The initial concentration of I– is 2 × 0.20 M from the CaI2. [Ag+] [I–] I – 0.40 C +x +x E +x 0.40 + x Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = 8.5 × 10–17 = [Ag+][I–] = (x)(0.40 + x) We know that the value of Ksp is very small, and it suggests the simplifying assumption that (0.40 + x) ≈ 0.40: Hence, 8.5 × 10–17 = (0.40)x, and x = 2.125 × 10–16. The assumption that (0.40 + x) ≈ 0.40 is seen to be valid indeed. Thus the molar concentration of AgI in a 0.20 M CaI2 solution will be 2.1 × 10–16 M . In pure water, Ksp = 8.5 × 10–17 = [Ag+][I–] = (x)(x) x = [AgI(aq)] = 9.2 × 10–9 M (much more soluble) 18.9
Fe(OH)3(s) Fe3+(aq) + 3OH–(aq)
I C E
[Fe3+] – +x +x
Ksp = [Fe3+][OH–]3 = 2.8 × 10–39
[OH–] 0.050 + 3x 0.050 + 3x
Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = 2.8 × 10–39 = [Fe3+][OH–]3 = (x)[0.050 + 3x]3 We try to simplify by making the approximation that (0.050 + 3x) ≈ 0.050: 2.8 × 10–39 = (x)(0.050)3 or x = 2.24 × 10–35 Clearly the assumption that (0.050 + 3x) ≈ 0.050 is justified. Thus 2.24 × 10–35 M of Fe(OH)3 will dissolve in a 0.050 M sodium hydroxide solution. 18.10
The expression for Ksp is Ksp = [Ca2+][SO42–] = 2.4 × 10–5 and the ion product for this solution would be: [Ca2+][SO42–] = (2.5 × 10–3)(3.0 × 10–2) = 7.5 × 10–5 Since the ion product is larger than the value of Ksp, a precipitate will form. 421
Chapter 18
18.11
The solubility product constant is Ksp = [Ag+]2[CrO42–] = 1.2 × 10–12 and the ion product for this solution would be: [Ag+]2[CrO42–] = (4.8 × 10–5)2(3.4 × 10–4) = 7.8 × 10–13 Since the ion product is smaller than the value of Ksp, no precipitate will form.
18.12
We expect PbSO4(s) since nitrates are soluble. Because two solutions are to be mixed together, there will be a dilution of the concentrations of the various ions, and the diluted ion concentrations must be used. In general, on dilution, the following relationship is found for the concentrations of the initial solution (Mi) and the concentration of the final solution (Mf): MiVi = MfVf Thus the final or diluted concentrations are: Pb 2+ = 1.0 × 10−3 M 100.0 mL = 5.0 × 10−4 M 200.0 mL
(
)
SO 4 2− = 2.0 × 10 −3 M 100.0 mL = 1.0 × 10−3 M 200.0 mL
(
)
The value of the ion product for the final (diluted) solution is: [Pb2+][SO42–] = (5.0 × 10–4)(1.0 × 10–3) = 5.0 × 10–7 Since this is larger than the value of Ksp (2.53 × 10–8), a precipitate of PbSO4 is expected. 18.13
We expect a precipitate of PbCl2 since nitrates are soluble. We proceed as in Practice Exercise 12 MiVi = MfVf Pb 2+ = ( 0.10 M ) 50.0 mL = 0.071 M 70.0 mL Cl− = ( 0.040 M ) 20.0 mL = 0.011 M 70.0 mL The value of the ion product for such a solution would be: [Pb2+][Cl–]2 = (7.1 × 10–2)(1.1 × 10–2)2 = 8.6 × 10–6 Since the ion product is smaller than Ksp (1.7 × 10–5), a precipitate of PbCl2 is not expected.
18.14
Follow the exact procedure outlined in Example 18.8. Ksp = [Ca2+][SO42–]= 4.9 × 10–5 Ksp = [Ba2+][SO42–] = 1.1 × 10–10 CaSO4 is more soluble and will precipitate when: [SO42–] =
K sp
=
4.9 × 10 0.25
5
[Ca 2 + ] BaSO4 will precipitate when: K sp 1.1 × 10 = [SO42 − ] = 2+ 0.05 [Ba ]
= 2.0 × 10
4
10
= 2.2 × 10
9
BaSO4 will precipitate and CaSO4 will not precipitate if [SO42–] > 2.2 × 10–9 and [SO42–] < 2.0 × 10–4.
422
Chapter 18
18.15
Follow the exact procedure outlined in Example 18.8. Ksp = [Ca2+][OH–]2 = 5.0 × 10–6 Ksp = [Mg2+][OH–]2 = 5.6 × 10–12 Ca(OH)2 is more soluble and will precipitate when: 1/2
K sp [OH ] = [Ca 2 + ] –
5.0 × 10 = 0.20
6
1/2
= 5.0 × 10
3
Mg(OH)2 will precipitate when: 1/2
K sp [OH ] = [Ba 2+ ] −
5.6 × 10 = 0.1
12
1/2
= 7.4 × 10
6
Mg(OH)2 will precipitate and Ca(OH)2 will not precipitate if 7.4 x 10-6 < [OH–] < 5.0 x 10-3. 18.16
CoS will precipitate if the H+ concentration is too low. Solving for Q and then comparing Q to Kspa, we can determine whether or not CoS will precipitate.
K spa = Q=
[Co 2+ ][H 2S] [H + ]2
[Co 2 + ][H 2S] + 2
=
= 5 × 10−1 [0.005][0.10] −4 2
= 5 × 103
[H ] [3.16 × 10 ] Q > Kspa Since Q is greater than Kspa, then the reaction will move to reactants and CoS solid will form. 18.17
Consulting Table 18.2, we find that Fe2+ is much more soluble in acid than Hg2+. We want to make the H+ concentration large enough to prevent FeS from precipitating, but small enough that HgS does precipitate. First, we calculate the highest pH at which FeS will remain soluble, by using Kspa for FeS. (Recall that a saturated solution of H2S = 0.10 M.) K spa =
[Fe 2+ ][H 2S] + 2
=
[H ] + [H ] = 0.0013 M pH = –log [H+] = 2.9
[0.010][0.10] [H + ]2
= 6 × 102
Since Fe2+ is much more soluble in acid than Hg2+ we already know that this pH will precipitate HgS, but we can check it by using Kspa for HgS: K spa =
[Hg 2+ ][H 2S] + 2
[H ] [H+] = 2.2 × 1014 M
=
[0.010][0.10] [H + ]2
= 2 × 10−32
(This concentration is impossibly high, but it tells us that this much acid would be required to dissolve HgS at these concentrations.)
423
Chapter 18
18.18
BaC2O4(s) Ba2+(aq) + C2O42–(aq) [Ba2+] = 0.050 M 1.2 × 10–7 = (0.050)[C2O42–] [C2O42–] = 2.4 × 10–6 M
Ksp = 1.2 × 10–7 = [Ba2+][C2O42–]
H2C2O4 + H2O H3O+ + HC2O4– Ka1 = 6.0 × 10–2 HC2O4– + H2O H3O+ + C2O42–
Ka2 = 6.1 × 10–5
H2C2O4 + H2O 2H3O+ + C2O42– Ka = (6.1 × 10–5) × (6.0 × 10–2) = 3.7 × 10–6 [H2C2O4] = 0.10 [C2O42–] = 2.4 × 10–6 M 2
Ka = 3.7 × 10–6
2
H O+ C O 2− H O+ 2.4 × 10−6 3 3 2 4 = = ( 0.10 ) [ H 2 C2 O 4 ]
(
)
Since the amount of oxalate formed is so small, the concentration of oxalic acid is essentially unchanged. [H+] = 1.54 × 10–1 This is the minimum concentration of H+ that will prevent the formation of BaC2O4 precipitate. 18.19
Follow the exact procedure outlined in Example 18.10. Ksp = [Ca2+][CO32–]= 3.4 × 10–9 Ksp = [Ni2+][CO32–] = 1.4 × 10–7 NiCO3 is more soluble and will precipitate when: K sp 1.4 × 10 7 [CO32–] = = = 1.4 × 10 6 2+ 0.10 [Ni ] CaCO3 will precipitate when: K sp 3.4 × 10 9 [CO32− ] = = = 3.4 × 10 8 2+ 0.10 [Ca ] CaCO3 will precipitate and NiCO3 will not precipitate if [CO32–] > 3.4 × 10–8 and [CO32–] < 1.4 × 10–6. Now, using the equation in example 18.10 we get:
0.030 [H + ]2 = (2.4 × 10−17 ) [CO 2 − ] 3 NiCO3 will precipitate if: 0.030 −13 [H + ]2 = (2.4 × 10 −17 ) = 5.14 × 10 1.4 × 10 −6 [H+] = 7.17 × 10–7 pH = 6.14 CaCO3 will precipitate:
0.030 −11 [H + ]2 = (2.4 × 10−17 ) = 2.12 × 10 3.4 × 10−8 [H + ] = 4.6 × 10−6
pH = 5.34
So CaCO3 will precipitate and NiCO3 will not if the pH is maintained between pH = 5.34 and pH = 6.14
424
Chapter 18
18.20
The overall equilibrium is AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl–(aq)
Ag(NH ) + Cl− 3 2 Kc = 2 NH3 In order to obtain a value for Kc for this reaction, we need to use the expressions for Ksp of AgCl(s) and the Kform of Ag(NH3)2+:
K sp = Ag + Cl− = 1.8 × 10−10 Ag(NH ) + 3 2 K form = = 1.6 × 107 2 + Ag NH 3 Ag(NH ) + Cl− 3 2 = 2.9 × 10−3 K c = K sp × K form = 2 NH3 Now we may use an equilibrium table for the reaction in question:
I C E
[Ag(NH3)2+] – +x x
[NH3] 0.10 – 2x 0.10 – 2x
[Cl–] – +x x
Substituting these values into the mass action expression gives: K c = 2.9 × 10−3 =
(x)(x) (0.10 − 2x)2
(x) (0.10 − 2x) Solving for x we get, x = 4.9 × 10–3 M. The molar solubility of AgCl in 0.10 M NH3 is therefore 4.9 × 10–3 M. Take the square root of both sides to get 0.054 =
In order to determine the solubility in pure water, we simply look at Ksp AgCl(s) Ag+ (aq) + Cl–(aq)
Ksp = [Ag+][Cl–] = 1.8 × 10–10
At equilibrium; [Ag+] = [Cl–] = 1.3 × 10–5. Hence the molar solubility of AgCl in 0.10 M NH3 is about 380 times greater than in pure water. 18.21
We will use the information gathered for the last problem. Specifically, AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl–(aq)
Ag(NH ) + Cl− 3 2 = 2.9 × 10−3 Kc = 2 NH3
425
Chapter 18 If we completely dissolve 0.20 mol of AgCl, the equilibrium [Cl–] and [Ag(NH3)2+] will be 0.20 M in a one liter container. The question asks, therefore, what amount of NH3 must be initially present so that the equilibrium concentration of Cl– is 0.20 M? [NH3] Z – 2x Z – 2x
I C E
K c = 2.9 × 10−3 =
[Ag(NH3)2+] – +x x
[Cl–] – +x x
(x)(x)
(Z − 2x)2 Take the square root of both sides to get; 0.054 =
x 0.20 = Z − 2x Z − 0.40
We have substituted the known value of x. Solving for Z we get, Z = 4.1 M Consequently, we would need to add 4.1 moles of NH3 to a one liter container of 0.20 M AgCl in order to completely dissolve the AgCl.
Review Questions 18.1
The ion product is the quantity which results from the mass action expression and it is a product of the ion concentrations. The ion product constant, also called the solubility product constant, is equal to the product of the ion concentrations for a saturated solution of a sparingly soluble substance. 3
18.2
Ba 2+ PO 3− 4 Kc = Ba 3 ( PO 4 ) 2
2
The denominator of the mass action expression is a constant since the concentration of substance within a pure solid is constant. Consequently, we define a new constant which is the product of Kc and the concentration of the pure solid. 3
K sp = Ba 2+ PO43−
2
18.3
The addition of a common ion to a saturated solution lowers the solubility of a sparingly soluble ionic salt. According to Le Châtelier’s principle, addition of an ion to a saturated solution will shift the equilibrium so as to absorb as much of the added ion as possible. This results in the precipitation of the sparingly soluble salt. In the case of AgCl, Ksp = 1.8 × 10–10, addition of NaCl to a saturated solution containing Ag+ and Cl– will result in the precipitation of AgCl(s). The common ion in this case is Cl–. It is present in the solution due to the NaCl as well as the AgCl.
18.4
When the value of the ion product is greater than Ksp, a precipitate will form.
18.5
Ksp values do not give highly accurate compound solubilities for a number of reasons including: (a) Ksp values are determined at a specific temperature and unless you use an accurately controlled temperature bath, the measured solubility will not be precisely what was calculated. (b) Ksp values assume 100% dissociation into ions. This condition is a reasonable assumption for singly charged species but the assumption does not work well when working with charges greater than +1 or −1. Ion paring becomes a significant factor when ions charges are large and when concentrations are high.
426
Chapter 18
18.6
Salts such as NaCl are strong electrolytes. We assume they dissociate 100%, just like strong acids and bases. Therefore, just like we do not need a Ka or Kb value from strong acids and bases, there is no need for an equilibrium constant for highly soluble salts.
18.7
The oxide ion reacts with the water to produce hydroxide ion: O2– + H2O 2OH–
18.8
Na2S(s) 2Na+(aq) + HS–(aq) + OH–(aq)
18.9
(a)
CoS(s) Co2+(aq) + HS–(aq) + OH–(aq)
Ksp = [CO2+][HS–][OH–]
(b)
CoS(s) + 2H+(aq) Co2+(aq) + H2S(aq)
Co2+ [ H S] 2 K spa = 2 H+
18.10
As the pH of an oxalic acid solution is decreased, the hydrogen ion concentration is increased, this shifts the equilibrium to the undissociated acid, thus increasing the H2C2O4 concentration. The opposite is true as the pH of the solution is increased.
18.11
By combining the three dissociation expressions for phosphoric acid, we come up with a single expression that relates the hydrogen ion concentration and the phosphate ion concentration: H3PO4(aq) 3H+(aq) + PO43–(aq) K = K a1 × K a2 × K a3
18.12
Halides of Ag+ are insoluble while those of Co2+ and Cu2+ are soluble. Therefore, Ag+ can be separated from the mixture using 1 M HCl. Sulfides of Co2+ are insoluble in base but soluble in acid while sulfides of Cu2+ are insoluble in acid. Therefore, the addition of S2− in acidic solution will precipitate the Cu2+ leaving the Co2+ in solution. To precipitate the Co2+ increase the pH to a value > 9.
18.13
MgBr2(aq) + 2 AgCl(s) MgCl2(aq) + 2AgBr(s) AgCl and AgBr are both insoluble compounds. However, AgBr is less soluble than AgCl. When solid AgCl is added to an aqueous solution of MgBr2, some of the AgCl dissociates. The Ag+ ion reacts with the aqueous Br– to form insoluble AgBr. This disrupts the AgCl/Ag+ equilibrium and additional AgCl dissociates. With sufficient stirring, and perhaps a little heat, all of the AgCl will dissolve and AgBr will precipitate. The result is a solution of MgCl2 and solid AgBr.
18.14
NH3(aq) + H2O NH4+(aq) + OH–(aq) The addition of NH4Cl to the mixture of Mg(OH)2 and water causes the Mg(OH)2 to dissolve because the NH4+ reacts with the OH– and shifts the equilibrium towards the dissolution of the solid.
18.15
AgCl(s) Ag+(aq) + Cl–(aq) Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) Silver chloride is an insoluble solid. However, any Ag+ ions present react with added NH3 to form the Ag(NH3)2+ complex ion. According to Le Châtelier’s Principle, as NH3 is added to a solution containing Ag+ ions, the complex ion forms using up the Ag+ ions. This disrupts the equilibrium and forces AgCl to dissolve. 427
Chapter 18 Upon addition of HNO3, a strong acid, H+ reacts with NH3 to form NH4+. This disrupts the complex equilibria and causes the Ag(NH3)2+ to dissociate and form free Ag+ ions. Once again, the [Ag+] reaches a value sufficient in the presence of Cl– to precipitate AgCl in accordance with Le Châtelier’s Principle. 18.16
PbCl2(s) Pb2+(aq) + 2Cl–(aq) Pb2+(aq) + 3Cl–(aq) PbCl3–(aq)
PbCl − 3
K form =
Pb2+ Cl−
3
=
mol PbCl − 3 volume 3
− 2+ mol Pb volume mol Cl volume
( mol PbCl ) ( mol Pb )( mol Cl ) −
3
K form = ( volume )
×
3
2+
−
3
Notice that the above expression is the product of a ratio of mole amounts and a volume3 term. The constant Kform does not change on dilution, but the volume term is changed by dilution. This means that the ratio of moles term in the above expression must change on dilution, in order to hold the product constant. If the volume is doubled, the ratio of moles would have to become smaller by a factor of 8 (= 23) in order for the entire argument to have a constant value, i.e, in order for Kform to remain constant on dilution. This means that the concentrations of Pb2+ and Cl– must increase as the solution containing the complex ion is diluted. Eventually the ion product for PbCl2 will exceed the value of Ksp for PbCl2, resulting in its precipitation. 18.17
Any of the species that contains the conjugate base of a weak acid will be more soluble in acid. Thus, ZnS, Ca(OH)2, MgCO3 and PbF2 will increase in solubility in acid. ZnS(s) + 2H+(aq) Ca(OH)2(s) + 2H+(aq) MgCO3(s) + 2H+(aq) PbF2(s) + 2H+(aq)
→ → → →
Zn2+(aq) + H2S(aq) Ca2+(aq) + 2H2O Mg2+(aq) + CO2(g) + H2O Zn2+(aq) + 2HF
AgCl will increase in solubility if the base is NH3. AgCl is less soluble than Ag2O so it will not dissolve in other bases. AgOH is not favored in the reaction of silver salts with strong base. Rather, the reaction favors the formation of the oxide. AgCl(s) + 2NH3(aq)
→
Ag(NH3)2+(aq) + Cl-(aq)
None of the compound’s solubility is independent of pH.
Review Problems 18.18
(a)
BaF2(s) Ba2+ + 2F–
Ksp = [Ba2+][F–]2
(b)
MnCO3(s) Mn2+ + CO32–
Ksp = [Mn2+] [CO32–]
(c)
SrSO4(s) Sr2+ + SO42–
Ksp = [Sr2+][SO42–]
(d)
Ni(OH)2(s) Ni2+ + 2OH–
Ksp = [Ni2+][OH–]2
(e)
PbBr2(s) Pb2+ + 2Br–
Ksp = [Pb2+][Br–]2
(f)
Zn(CN)2(s) Zn2+ + 2CN–
Ksp = [Zn2+][CN–]2
428
Chapter 18
18.19
18.20
(a)
AgI(s) Ag+ + I–
Ksp = [Ag+][I–]
(b)
Ag3PO4(s) 3Ag+ + PO43–
Ksp = [Ag+]3[PO43–]
(c)
PbCrO4(s) Pb2+ + CrO42–
Ksp = [Pb2+][CrO42–]
(d)
Al(OH)3(s) Al3+ + 3OH–
Ksp = [Al3+][OH–]3
(e)
ZnCO3(s) Zn2+ + CO32–
Ksp = [Zn2+][CO32–]
(f)
Zn(OH)2(s) Zn2+ + 2OH–
Ksp = [Zn2+][OH–]2
PbCl2 Pb2+ + 2Cl– Ksp = [Pb2+][Cl–]2 At equilibrium [Pb2+] = 0.016 M and [Cl–] = 0.032 M so Ksp = (0.016)(0.032)2 = 1.6 × 10–5
18.21
[Zn2+] = [C2O4] = 7.9 × 10–3 M Ksp = [Zn2+][C2O4] = (7.9 × 10–3)2 = 6.2× 10–5
18.22
1 mole BaSO4 –5 mol BaSO4 = ( 0.00368 g BaSO 4 ) = 1.58 × 10 mol 233.3906 g BaSO 4 1.58 × 10 –5 mol = 1.05 × 10-5 m 1.50 L [Ba2+] = [SO42–] = 1.05 × 10–5 M
This is the molar solubility of BaSO4
Ksp = [Ba2+][SO42–] = (1.05 × 10–5)2 = 1.10 × 10–10 18.23
1 mol AgC2 H3O2 −3 mol AgC2 H3O 2 = (1.20 g AgC2 H3O 2 ) = 7.19 × 10 mol AgC2 H3O2 166.9 g AgC H O 2 3 2 7.19 × 10−3 mol AgC2 H3O 2 = 4.79 × 10–2 M 0.1500 L solution One mole of Ag+ and one mole of C2H3O2 will be produced for every mole of AgC2H3O2. Therefore, Ksp = [Ag+][C2H3O2–] = (4.79 × 10–2)2 = 2.29 × 10–3. [AgC2H3O2] =
18.24
BaF2 Ba2+(aq) + 2F–(aq) Ksp = [Ba2+][F–]-2 First find the concentration of the Ba2+ and F– that was in solution and then find the value for Ksp. Using the amount of BaF2 recovered; determine the number of moles of each ion, then find the concentration of each ion.
1 mol BaF2 –4 mole BaF2 = 0.165 g BaF2 = 9.41 × 10 mol BaF2 175.32 g BaF2 − 4 9.41 × 10 mol BaF 1 mol Ba 2+ 1000 mL 2 [Ba2+] = = 7.53 × 10–3 M Ba2+ 125 mL solution 1 mol BaF2 1 L 9.41 × 10−4 mol BaF 2 mol F− 1000 mL 2 [F–] = = 1.51 × 10–2 M F– 125 mL solution 1 mol BaF2 1 L Ksp = [Ba2+][F–] = (7.53 × 10–3)(1.51 × 10–2)2 = 1.7 × 10–6
429
Chapter 18
18.25
To solve this problem we need to realize that the concentration of the solution is equal to the number of moles of solid recovered divided by the volume of the solution, i.e., 0.325 g CaCrO 4 1 mole CaCrO 4 1000 mL [ CaCrO4 ] = 156.1 g CaCrO 1 L = 0.0267 M 78 mL 4 2+ 2– The equilibrium for this problem is CaCrO4(s) Ca + CrO4 Ksp = [Ca2+][CrO42–] = (0.0267)2 = 7.13 × 10–4
18.26
Ag3PO4(s) 3Ag+ + PO43– Ksp = [Ag+]3[PO43–] Ksp = [3(1.8 × 10–5)]3[1.8 × 10–5] = 2.8 × 10–18
18.27
Ba3(PO4)2(s) 3Ba2+ + 2PO43– Ksp = [Ba2+]3[PO43–]2 Ksp = [3(1.4 × 10–8)]3[2(1.4 × 10–8)]2 = 5.8 × 10–38
18.28
Compare the values of Ksp. The larger the value the greater the molar solubility. Ksp = 6.8 x 10-8 for MgCO3 Ksp = 1.5 x 10-10 for ZnCO3 Therefore, MgCO3 is more soluble in water.
18.29
Compare the values of Ksp. The larger the value the greater the molar solubility. Ksp = 1.1 x 10-10 for BaSO4 Ksp = 5.6 x 10-10 for SrCO3 Therefore, SrCO3 is more soluble in water.
18.30
PbBr2(s) Pb2+ + 2Br–
I C E
Ksp = [Pb2+][Br–]2
[Pb2+] – +x +x
[Br–] – + 2x + 2x
Ksp = (x)(2x)2 = 4x3 = 6.6 × 10–6, x = 18.31
Ag2CrO4(s) 2Ag+ + CrO42–
I C E
[Ag+] – + 2x + 2x
6.6 × 10−6 = 1.2 × 10−2 M 4
Ksp = [Ag+]2[CrO42–]
[CrO42–] – +x +x
Ksp = (2x)2(x) = 4x3 = 1.2 × 10–12, x = 18.32
3
3
1.2 × 10−12 = 6.7 × 10−5 M 4
Zn(CN)2(s) Zn2+(aq) + 2CN-(aq) For every mole of Zn2+ produced, 2 moles of CN- will be produced. Let x = [Zn2+] at equilibrium and [CN-] = 2x at equilibrium. Ksp = [Zn+2][CN–]2 = 3.0 × 10–16 = (x)(2x)2 = 4x3. Solving we find x = 4.2 × 10–6. Thus, the molar solubility of Zn(CN)2 is 4.2 × 10–6 moles/L. 430
Chapter 18
18.33
For every mole of Pb2+ produced, 2 moles of F– will be produced. Let x = [Pb2+] at equilibrium and [F–] = 2x at equilibrium. Ksp = [ Pb2+][F–]2 = 3.3 × 10–8 = (x)(2x)2 = 4x3. Solving we find x = 2.0 × 10–3 M. Thus, the molar solubility of PbF2 is 2.1 × 10–3 moles/L.
18.34
To solve this problem, determine the molar solubility for each compound. LiF: let x = [Li+] = [F–] Ksp = [Li+][F–] = x2 = 1.7 × 10–3 –2 x = 4.1 × 10 moles/L = molar solubility of LiF. BaF2: let x = [Ba2+]; [F–]2 = 2x Ksp = [Ba2+][F–]2 = (x)(2x)2 = 1.7 × 10–6 3 –6 –3 4x = 1.7 × 10 , and x = 7.5 × 10 M = molar solubility of BaF2. Because the molar solubility of LiF is greater than the molar solubility of BaF2, LiF is more soluble.
18.35
To solve this problem, first determine the molar solubility for each compound. AgCN: let x = [Ag+] = [CN–] Ksp = [Ag+][CN–] = x2 = 6.0 × 10–17 –8 x = 7.7 × 10 moles/L = molar solubility of AgCN. The number of grams of AgCN that will dissolve in 100 mL is;
7.7 × 10−9 moles 133.9 g g = (100 mL) = 1.0 × 10−7 g 1 mole 1000 mL Zn(CN)2: let x = [Zn2+], [CN–] = 2x; Ksp = [Zn2+][CN–]2 = (x)(2x)2 = 3 × 10–16 3 –16 –6 4x = 3 × 10 , and x = 4.2 × 10 M = molar solubility of Zn(CN)2. The number of grams of Zn(CN)2 that will dissolve in 100 mL is;
4.2 × 10−6 moles 117.4 g g = (100 mL) = 4.9 × 10−5 g 1 mole 1000 mL More Zn(CN)2 will dissolve in 100 mL of water so it has the larger solubility. 18.36
First determine the molar solubility of the MX salt. Let x = [M+] = [X–], Ksp = [M+][X–] = (x)(x) = 3.2 × 10–10 –5 x = 1.8 × 10 M. This is the equilibrium concentration of the two ions. For the MX3 salt, let x = equilibrium concentration of M3+, [X–] = 3x. Ksp = [M+][X–]3 = (x)(3x)3 = 27x4. The value of x in this expression is the value determined in the first part of this problem. So, Ksp = (27)(1.8 × 10–5)4 = 2.8 × 10–18
18.37
First determine the molar solubility of the M2X3 salt. M2X3 (s) 2M3+ (aq) + 3X2–(aq) Ksp = [M3+]2[X2–]3
I C E
[M3+] – + 2x + 2x
[X2–] – + 3x + 3x
Ksp = (2x)2(3x)3= 2.2 × 10–20 = 108x5
x = 4.6 × 10–5 M 431
Chapter 18 The molar solubility of this compound is 4.6 × 10–5 moles/L We want the molar solubility of the M2X compound to be twice the value just calculated or 9.2 × 10–5 moles/L. We need to solve the equilibrium expression: M2X (s) 2M+ (aq) + X2–(aq)
I C E
[M+] – + 2x + 2x
Ksp = [M+]2[X2–]
[X2–] – +x +x
Ksp = (2x)2(x) = 4x3 So, Ksp = 4(9.2 × 10–5)3 = 3.1 × 10–12
where x = 9.2 × 10–5 M
18.38
CaSO4(s) Ca2+(aq) + SO42–(aq) Ksp = [Ca2+][SO42–] = 4.9 × 10–5 2+ 2 2– let x = [Ca ] = [SO4 ] Ksp = x = 4.9 × 10–5 and x = 7.0 × 10–3 M. –3 The molar solubility of CaSO4 is 7.0 × 10 moles/L.
18.39
CaCO3(s) Ca2+ + CO32– Ksp = [Ca2+][CO32–] = 3.4 × 10–9 2+ 2– let x = [Ca ] = [CO3 ] x2 = 3.4 × 10–9; x = 5.8 × 10–5 M The solubility of CaCO3 is 5.8 × 10–5 M. So, 5.8 × 10–5 moles of CaCO3 dissolves in 1 L of H2O. If 100 mL of water are available, 5.8 × 10–6 moles of CaCO3 will dissolve. Converting this to g/100 mL we multiply by the molar mass of 100 g/mole and determine that 5.8 × 10–4 g will dissolve in 100 mL of water.
18.40
BaSO3(s) Ba2+ + SO32– Ksp = [Ba2+][SO32–] Ksp = (0.10)(8.0 × 10–6) = 8.0 × 10–7 In this problem, all of the Ba2+ comes from the BaCl2.
18.41
Ag2CrO4(s) 2Ag+ + CrO42– Ksp = [Ag+]2[CrO42–] Ksp = (0.10)(1.7 × 10–6)2 = 2.9 × 10–13 In this problem, all of the Ba2+ comes from the BaCl2.
18.42
(a)
CuCl(s) Cu+(aq) + Cl–(aq)
I C E
[Cu+] – +x +x
Ksp = x2 = 1.7 × 10–7 (b)
[Cl–] – +x +x ∴ x = molar solubility = 4.1 × 10–4 M
CuCl(s) Cu+(aq) + Cl–(aq)
I C E
[Cu+] – +x +x
Ksp = [Cu+][Cl–] = 1.7 × 10–7
Ksp = [Cu+][Cl–] = 1.7 × 10–7
[Cl–] 0.0175 +x 0.0175 + x
432
Chapter 18
Ksp = (x)(0.0175+x) = 1.7 × 10–7 Assume that x << 0.0175 ∴ x = molar solubility = 9.7 × 10–6 M (c)
CuCl(s) Cu+(aq) + Cl–(aq)
I C E
[Cu+] – +x +x
Ksp = [Cu+][Cl–] = 1.7 × 10–7
[Cl–] 0.175 +x 0.175 + x
Assume that x << 0.175 Ksp = (x)(0.175+x) = 1.7 × 10–7 ∴ x = molar solubility = 9.7 × 10–7 M (d) CuCl(s) Cu+(aq) + Cl–(aq) Ksp = [Cu+][Cl–] = 1.7 × 10–7 Note that the Cl– concentration equals (2)(0.150 M) since two moles of Cl– are produced for every mole of CaCl2.
I C E
[Cu+] – +x +x
[Cl–] 0.350 +x 0.350 + x
Ksp = (x)(0.350+x) = 1.7 × 10–7 Assume that x << 0.350 ∴ x = molar solubility = 4.9 × 10–7 M 18.43
AuCl3(s) Au3+ + 3Cl– (a)
Ksp = [Au3+][Cl–]3 = 3.2 × 10–25
let x = [Au3+]; then [Cl–] = 3x Ksp = (x)(3x)3 = 27x4 3.2 × 10 −25 = 3.3 × 10–7 M 27 The molar solubility of AuCl3 is 3.3 × 10–7 M in H2O.
x=
4
(b)
[Au3+] = x; [Cl–] = 0.012 + 3x Ksp = (x)(0.012 + 3x)3: Assume 3x << 0.012 Ksp = (x)(0.012)3 x = 1.9 × 10–19 M The molar solubility of AuCl3 is 1.9 × 10–19 M in 0.012 M HCl.
(c)
[Au3+] = x; [Cl–] = 0.024 + 3x Ksp = (x)(0.024 + 3x)3: Assume 3x << 0.024 Ksp = (x)(0.024)3 x = 2.3 × 10–20 M The molar solubility of AuCl3 is 2.3 × 10–20 M in 0.024 M MgCl2.
(d)
[Au3+] = 0.012 + x; [Cl–] = 3x Ksp = (0.012 + x)(3x)3: Assume x << 0.012 Ksp = (0.012)(3x)3
433
Chapter 18
3.2 × 10−25 = 1.0 × 10–9 M 27 The molar solubility of AuCl3 is 1.0 × 10–9 M in 0.012 M Au(NO3)3.
x=
3
18.44
Mg(OH)2 Mg2+ + 2OH– Ksp = [Mg2+][OH–]2 = 5.6 × 10–12 The concentration of OH– is determined from the pH: pOH = 14 – 12.25 = 1.75 [OH–] = 0.0356 M 2+ [Mg ] = x [OH–] = 0.0356 M 2 Ksp = x(0.0356) = 5.6 × 10–12 x = 4.4 × 10–9 M The molar solubility of Mg(OH)2 is 4.4 × 10–9 M in a solution with a pH of 12.25.
18.45
Al(OH)3(s) Al3+ + 3OH– Ksp = [Al3+][OH–]3 = 3 × 10–34 The concentration of OH– is determined from the pH: pOH = 14 – 9.75 = 4.25 [OH–] = 1.7 × 10–4 M 3+ [Al ] = x [OH–] = 1.7 × 10–4 M –4 3 Ksp = x(1.7 × 10 ) = 3 × 10–34 x = 6.1 × 10–23 The molar solubility of Al(OH)3 is 6.1 × 10–23 M in a solution with a pH of 9.75.
18.46
PbCl2(s) Pb2+ + 2Cl– [Cl–] = 0.10 M [Pb2+][Cl–]2 =[Pb2+][0.10]2 = 1.7 × 10–5 [Pb2+] = 1.7 × 10–3 M
18.47
PbBr2 will be less soluble in a solution of 0.10 M NaBr because the values of Q and Ksp for lead bromide are dependent upon the [Br–]2.
18.48
Ag2CrO4 (s) 2Ag+ (aq) + CrO42–(aq) (a) [Ag+] [CrO42–] 0.175 – I + 2x +x C 0.175 + 2x +x E
Ksp = [Pb2+][Cl–]2 = 1.7 × 10–5
Ksp = [Ag+]2[CrO42–] = 1.1 × 10–12
Ksp = (0.175+2x)2(x) Assume that x << 0.175 1.1 × 10–12 = (0.175)2(x) x = 3.6 × 10–11 The molar solubility is 3.6 × 10–11 moles/L (b)
I C E
[Ag+] – +2x +2x
[CrO42–] 0.175 +x 0.175 + x
Ksp = (2x)2(0.175+x) Assume that x << 0.175 1.1 × 10–12 = (2x)2(0.175) x = 1.3 × 10–6 The molar solubility is 1.3 × 10–6 moles/L.
434
Chapter 18
18.49
Mg(OH)2 (s) Mg2+(aq) + 2OH–(aq) (a) [Mg2+] [OH–] – 0.18 I +x 0.18 + 2x C +x 0.18 + 2x E
Ksp = [Mg2+][OH–]2 = 5.6 × 10–12
Ksp = (x)(0.18 + 2x)2 Assume 2x << 0.18 Ksp = (x)(0.18)2 = 5.6 × 10–12 x = 1.7 × 10–10 M The assumption is valid and the molar solubility of Mg(OH)2 in 0.18 M NaOH is 1.7 × 10–10 moles/L (b)
I C E
[Mg2+] 0.18 0.18 + x 0.18 + x
[OH–] – + 2x + 2x
Ksp = (0.18 + x)(2x)2 Assume x << 0.18 Ksp = (0.18)(2x)2 = 5.6 × 10–12 x = 6.6 × 10–6 M The assumption is valid and the molar solubility of Mg(OH)2 in 0.20 M MgSO4 is 2.6 × 10–6 moles/L 18.50
Fe(OH)2(s) Fe2+(aq) + 2 OH–(aq)
Ksp = Fe 2+ OH −
2
mol OH– = 2.20 g NaOH(1 mol/40.01 g NaOH) = 0.0550 mol NaOH [OH–] = mol OH–/L solution = 0.0550 mol/0.250 L = 0.22 M
I C E
[Fe2+] – +x x
[OH–] 0.22 + 2x 0.22 + 2x
We assume that x << 0.22, so that 0.22 + 2x ≈ 0.22, then we enter the equilibrium values of the above table into the Ksp expression: 2
Ksp = Fe 2+ OH − 4.9 × 10–17 = x(0.22)2 x = molar solubility = 1.0 × 10–15 M Next, we must determine how many moles of Fe(OH)2 are formed in the reaction. This is a limiting reactant problem. The number of moles of OH– is 0.0550 (see above). The number of moles of Fe2+ is (0.250 L)(0.10 mol/L) = 0.025 mol From the balanced equation at the top, we need two OH– for every one Fe2+. This would be 2(0.025 mol) = 0.050 mol OH–. Looking at the molar quantities above, we have more than enough OH– so, Fe2+ is our limiting reactant:
435
Chapter 18
0.025 mol Fe(OH)2 will form in 0.25 L solution. If dissolved, this would be a concentration of 0.025 mol/0.25 L = 0.10 M. But from above, the maximum molar solubility of is 1.0 × 10–15 M. This means that remainder of Fe(OH)2 in excess of this value precipitates: 0.10 – 1.0 × 10–15 ≈ 0.10 M. This works out to 0.25 L(0.10 mol/L) = 0.025 mol Fe(OH)2(89.8 g/mol) = 2.2 g solid Fe(OH)2 (essentially all of it). The remaining OH–, 0.005 mol, gives a concentration of OH– of
0.005 mol OH − = 0.02 M OH– 0.250 L 4.9 × 10–17 = [Fe2+][0.02]2 [Fe2+] = 1.2 × 10–13 M 18.51
Ni(OH)2(s) Ni2+(aq) + 2OH–(aq) mol OH– = 1.75 g NaOH(1 mol/40.01 g NaOH) = 0.0437 mol NaOH First, assume that all of the ions are in solution and no precipitate has formed. At that point the concentrations are:
0.0437 mol OH − = 0.175 M OH– 0.250 L [Ni2+] = 0.10 M Next, determine the limiting reactant: 0.10 mol Ni 2+ 2 mol OH − = 0.20 M OH– 1 L soln 1 mol Ni 2+ Only 0.175 M OH– is available, therefore OH– is the limiting reagent. Using the limiting reagent, assume that all of the OH– has precipitated [OH–] = 0.0 M 0.175 mol OH − 1 mol Ni 2+ [Ni2+] = 0.10 M Ni2+ – 2 mol OH − 1 L soln = 0.10 M Ni2+ – 0.088 M Ni2+ = 0.012 M Ni2+ From here, calculate the equilibrium concentrations: [Ni2+] [OH–] I 0.012 – C +x + 2x E 0.012 + x 2x We assume that x << 0.012, so that 0.012 + 2x ≈ 0.012, then we enter the equilibrium values of the above table into the Ksp expression: [OH–] =
2
Ksp = Ni 2+ OH − –16 5.5 × 10 = 0.012(x)2 x = 2.1 × 10–7 M The amount of Ni(OH)2 formed can be calculated from the amount of OH– added to the solution since essentially all of the OH– was precipitated: 1 mol Ni(OH)2 92.71 g Ni(OH)2 = 2.03 g Ni(OH)2 g Ni(OH)2 = 0.0437 mol OH– 2 mol OH − 1 mol Ni(OH)2 To determine the pH of the final solution, the [OH–] needs to be found: From the ICE table, the equilibrium concentration of OH– = 2x x = 2.1 × 10–7 M 2x = 4.2 × 10–7 M = [OH–] We need to add in the concentration of OH–,1 × 10–7 M, from the dissociation of water: [OH–] = 4.2 × 10–7 M + 1 × 10–7 = 5.2 × 10–7 M 436
Chapter 18 pOH = –log[OH–] = –log[5.2 × 10–7] = 6.28 pH = 14 – pOH = 14 – 6.28 = 7.72 18.52
Fe(OH)2(s) Fe2+(aq) + 2 OH–(aq)
2
Ksp = Fe 2+ OH − = 4.9 × 10–17
pH = 9.45 pOH = 14.00 – pH = 4.55 [OH–] = 10–4.55 = 2.82 × 10–5 M
I C E
[Fe2+] – +x x
[OH–] 2.82 × 10–5 + 2x (2.82 × 10–5) + 2x
Since Ksp for iron(II) hydroxide is so small, we can safely assume that 2x << 2.82 × 10–5, so that (2.82 × 10–5) + 2x ≈ 2.82 × 10–5, then we enter the equilibrium values of the above table into the Ksp expression: 2
Ksp = Fe 2+ OH − 4.9 × 10–17 = x(2.82 × 10–5)2 x = molar solubility = 6.2 × 10–8 M 18.53
(a)
Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = [Ca2+][OH–]2 = 5.0 × 10–5 let 2x = [OH–], [Ca2+] = x + 0.12 Ksp = (0.12 + x)(2x)2 = 5.0 × 10–5 By successive approximations, x = 0.0030 The molar solubility of Ca(OH)2 in 0.12 M CaCl2 is 0.0030 moles/L.
(b)
Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = [Ca2+][OH–]2 = 5.0 × 10–5 2+ – let x = [Ca ], [OH ] = 2x + 0.012 Ksp = (x)(2x + 0.012)2 = 5.0 × 10–5 By successive approximations, x = 4.1 x 10− 4 The molar solubility of Ca(OH)2 in 0.12 M NaOH is 4.1 x 10− 4 moles/L.
18.54
In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of Ksp. For PbCl2, Ksp = 1.7 × 10–5 (see Table 18.1). 2
Q = Pb2+ Cl− = (0.0105)(0.0084)2 = 7.41 × 10–7. Since Q < Ksp, no precipitate will form. 18.55
In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of Ksp. For AgC2H3O2, Ksp = 2.3 × 10–3. Q = Ag + C2 H3O2− = (0.0075)(0.250) = 1.9 × 10–3. Since Q < Ksp, no precipitate will form. (Note: The concentration of C2H3O2– is twice the concentration of Ca(C2H3O2)2 since one mole of Ca(C2H3O2)2 produces two moles of C2H3O2–).
18.56
To solve this problem, determine the value for Q and apply LeChâtelier’s Principle. (a)
Pb 2+ = (50.0 mL)(0.0100 moles/L)/(75.0 mL) = 6.67 × 10–3 − Br = (25.0 mL)(0.0200 moles/L)/(75.0 mL) = 6.67 × 10–3
437
Chapter 18
2
Q = Pb2+ Br − = (6.67 × 10–3)(6.67 × 10–3)2 = 2.96 × 10–7 For PbBr2, Ksp = 6.6 × 10–6 Since Q < Ksp, no precipitate will form. (b)
Pb 2+ = (50.0 mL)(0.0100 moles/L)/(75.0 mL) = 6.67 × 10–3 Br − = (25.0 mL)(0.200 moles/L)/(75.0 mL) = 6.67 × 10–2 2
Q = Pb2+ Br − = (6.67 × 10–3)(6.67 × 10–2)2 = 2.96 × 10–5 For PbBr2, Ksp = 6.6 × 10–6 Since Q > Ksp, a precipitate will form. 18.57
In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of Ksp. For AgC2H3O2, Ksp = 2.3 × 10–3. Ag + = (44.0 mL)(0.0500 M)/(89.0 mL) = 2.47 × 10–2 M C2 H3O2− = (45.0 mL)(0.0260 M)/(89.0 mL) = 1.31 × 10–2 M
Q = Ag + C2 H3O2− = (2.47 × 10–2)(1.31 × 10–2) = 3.25 × 10–4. Since Q < Ksp, no precipitate will form. 18.58
The precipitate that may form is PbBr2(s). To determine if a precipitate will form, a value for the reaction quotient, Q, must be calculated: Q = [Pb2+][Br–]2. In performing this calculation, the dilution of the ions must be considered: [Pb2+] = [Br–] = 0.0100 M. Q = [0.0100][0.0100]2 = 1.0 × 10–6 If Q > Ksp, a precipitate will form. Ksp for PbBr2(s) is 6.6 × 10–6. Therefore, a precipitate will not form. Hence, the concentrations calculated are the diluted concentrations. Since no precipitate forms, the concentrations are not equilibrium values.
18.59
Ksp = [Mn2+][OH–]2 = 1.6 × 10–13 Ksp = [Cd2+][OH–]2 = 7.2 × 10–15 Mn(OH)2 is more soluble, so we need to determine the hydroxide ion concentration when it begins to precipitate. Ksp = [Mn2+][OH–]2 = 1.6 × 10–13 = (0.10)(x)2 x = 1.3 × 10–6 M OH– – Use this value for the [OH ] to solve for the concentration of the [Cd2+] left in solution: Ksp = [Cd2+][OH–]2 = 7.2 × 10–15 = (x)[1.3 × 10–6]2 x = 4.5 × 10–3 M Cd2+
18.60
AgCl(s) Ag+ + Cl– AgI(s) Ag+ + I–
K sp = Ag + Cl− = 1.8 × 10–10 K sp = Ag + I − = 8.5 × 10–17
When AgNO3 is added to the solution, AgI will precipitate before any AgCl does due to the lower solubility of AgI. In order to answer the question, i.e., what is the [I–] when AgCl first precipitates, we need to find the minimum concentration of Ag+ that must be added to precipitate AgCl. Let x = [Ag+]; Ksp = (x)(0.050) = 1.8 × 10–10; x = 3.6 × 10–9 M When the AgCl starts to precipitate, the solution will have a [Ag+] of 3.6 × 10–9 M. Now we ask, what is the [I–] if [Ag+] = 3.6 × 10–9 M? So, K sp = Ag + I − = (3.6 × 10–9)(x) = 8.5 × 10–17; x = 2.3 × 10–8 M = [I–] 438
Chapter 18
18.61
This problem is similar to 18.60 except that the Ksp constants are closer in value. We first determine the minimum amount of SO42– that must be added to initiate the precipitation of CaSO4. CaSO4 will precipitate after SrSO4 due to its larger value for Ksp: Ksp(CaSO4) = 4.9 × 10–5 and Ksp(SrSO4) = 3.4 × 10–7 (see Table 18.1) (a) Let x = Ca 2+ ; K sp = Ca 2+ SO42− = (0.15)(x) = 4.9 × 10–5 x = SO42− = 3.3 × 10–4 M 2− When the SO4 = 3.3 × 10–4 the CaSO4 will start to precipitate. Now we ask, what is the
Sr 2+ if SO42− = 3.3 × 10–4 M? SrSO4 (s) Sr2+ (aq) + SO42–(aq)
I C E
[Sr2+] – +x +x
K sp = Sr 2+ SO42−
[SO42–] 3.3 × 10–4 3.3 × 10–4 + x 3.3 × 10–4 + x
K sp = Sr 2+ SO42− = (x)(1.6 × 10–4 + x) = 3.2 × 10–7 For this problem, we must solve the quadratic equation and we determine that x = 4.4 × 10–4 M. Thus, the [Sr2+] = 4.4 × 10–4 M when the CaSO4 starts to precipitate. (b)
Initially the solution had a concentration of 0.15 M. The solution now has a [Sr2+] = 4.4 × 10–4 M. So, the percentage of Sr2+ precipitated is;
0.15 − 4.4 × 10-4 × 100% = 99.7 % 0.15 18.62
The less soluble substance is PbS. We need to determine the minimum [H+] at which CoS will precipitate.
Co 2+ [ H 2S] (0.010)(0.1) K spa = = = 0.5 2 [H + ]2 H+
(from Table 18.2)
(0.010)(0.1) = 0.045 0.5 pH = –log[H+] = 1.35. At a pH lower than 1.35, PbS will precipitate and CoS will not. At larger values of pH, both PbS and CoS will precipitate. [H + ] =
18.63
The less soluble substance is SnS so we will determine the maximum amount of H+ that is permitted before MnS starts to precipitate. Mn 2+ [ H 2S] (0.010)(0.1) K spa = = = 3 × 107 (from Table 18.2) 2 + 2 [H ] H+
[H + ] =
(0.010)(0.1)
= 6 × 10−6 3 × 107 pH = –log[H+] = 5.2. At pH values equal to or less than 5.2, MnS will not precipitate. 439
Chapter 18
18.64
Cu(OH)2(s) Cu2+(aq) + 2 OH–(aq)
K sp = Cu 2+ OH −
2 2
4.8 × 10-20 = [ 0.10] OH − [OH–] = 6.9 × 10–10 M pOH = –log[OH–] = –log[6.9 × 10–10] = 9.2 pH = 14.00 –pOH = 4.8 4.8 is the pH below which all the Cu(OH)2 will be soluble. Mn(OH)2(s) Mn2+(aq) + 2 OH–(aq)
K sp = Mn 2+ OH −
2 2
1.6 × 10-13 = [ 0.10] OH − [OH−] = 1.3 × 10–6 M pOH = –log[OH−] = –log[1.3 × 10–6] = 5.9 pH = 14.00 – pOH = 8.1 8.1 is the pH below which all the Mn(OH)2 will be soluble. Therefore, from pH = 4.8–8.1 Mn(OH)2 will be soluble, but some Cu(OH)2 will precipitate out of solution. 18.65
The following reactions are possible: CaC2O4(s) Ca2+(aq) + C2O42–(aq)
Ksp = [Ca2+][C2O42–] = 2.3 × 10–9
MgC2O4(s) Mg2+(aq) + C2O4–2(aq)
Ksp = [Mg2+][C2O42–] = 4.8 × 10–6
K a1
H + HC2 O4 – = 6.5 × 10–2 = [ H 2 C2 O4 ]
HC2O4 (aq) H (aq) + C2O4 (aq)
Ka2
H + C2 O4 2– = 6.1 × 10–5 = HC2 O4 –
H2C2O4(aq) 2H+(aq) + C2O42–(aq)
H + C O 2– 2 4 KO = = 4.0 × 10–6 [ H 2 C 2 O4 ]
+
–
H2C2O4(aq) H (aq) + HC2O4 (aq)
–
+
2–
2
Assume that the concentration of H2C2O4 at the end of the process is 0.10 M First, calculate the concentration of C2O42–, at which the MgC2O4 will precipitate Ksp = [Mg2+][C2O42–] = 4.8 × 10–6 [Mg2+] = 0.10 M (0.10)(x) = 4.8 × 10–6 x = 4.8 × 10–5 M = [C2O42–] As long as the concentration of C2O42– is kept below 4.8 × 10–5 M, the Mg2+ will remain in solution In order for the concentration of oxalate to remain below = 4.8 × 10–5 M, the pH will be: 2
+
H + C2 O4 2– = 4.0 × 10–6 KO = H C O [ 2 2 4]
2–
H2C2O4(aq) 2H (aq) + C2O4 (aq)
440
Chapter 18
2
H + 4.8 × 10 –5 = 4.0 × 10–6 KO = 0.10 [ ] [H+] = 0.0913 M pH = 1.04 By keeping the pH < 1.04, the MgC2O4 will not precipitate. To precipitate the CaC2O4, a similar calculation needs to be done: Ksp = [Ca2+][C2O42–] = 2.3 × 10–9 [Ca2+] = 0.10 M (0.10)(x) = 2.3 × 10–9 x = 2.3 × 10–8 M = [C2O42–] As long as the concentration of C2O42– is kept above 2.3 × 10–8 M, the Ca2+ will precipitate In order for the concentration of oxalate to remain above = 2.3 × 10–8 M, the pH must be: 2
H2C2O4(aq) 2H+(aq) + C2O42–(aq)
Ka2
H + C O 2– 2 4 = = 4.0 × 10–6 [ H 2 C2 O4 ]
2
H + 2.3 × 10 –8 = 4.0 × 10–6 K= [0.10] [H+] = 4.17 M pH = 0.62
18.66
Ksp = [Ag+][CO32–]= 8.5 × 10–12 Ksp = [Ni2+][CO32–] = 1.4 × 10–7 Assume the solutions are equal molar with concentration of 0.10 M. NiCO3 is more soluble and will precipitate when: Ksp 1.4 × 10 7 = = 1.4 × 10 6 [CO32–] = 2+ 0.10 [Ni ] Ag2CO3 will precipitate when: Ksp 8.5 × 10 12 [CO32− ] = = = 8.5 × 10 [Ag + ]2 (0.10)2
10
Ag2CO3 will precipitate and NiCO3 will not precipitate if [CO32–] > 8.5 × 10–10 and [CO32–] < 1.4 × 10–6. Now, using the equation in example 18.10 we get: 0.030 [H + ]2 = (2.4 × 10−17 ) [CO 2 − ] 3 NiCO3 will precipitate if: 0.030 −13 [H + ]2 = (2.4 × 10−17 ) = 5.5 × 10 1.4 × 10−6 [H+] = 7.4 × 10–7 pH = 6.13 Ag2CO3 will precipitate:
441
Chapter 18
0.030 [H + ]2 = (2.4 × 10 −17 ) = 8.5 × 10−10 −10 8.5 × 10 [H + ] = 2.9 × 10−5
pH = 4.54
So Ag2CO3 will precipitate and NiCO3 will not if the pH is maintained between pH = 4.54 and pH = 6.13. 18.67
Ksp = [Zn2+][CO32–]= 1.5 × 10–10 Ksp = [Ni2+][CO32–] = 1.4 × 10–7 Assume the solutions are equal molar with concentration of 0.10 M. NiCO3 is more soluble and will precipitate when: Ksp 1.4 × 10 7 [CO32–] = = = 1.4 × 10 6 2+ 0.10 [Ni ] ZnCO3 will precipitate when: Ksp 1.5 × 10 10 [CO32− ] = = = 1.5 × 10 0.10 [Zn 2 + ]
9
ZnCO3 will precipitate and NiCO3 will not precipitate if [CO32–] > 1.5 × 10–9 and [CO32–] < 1.4 × 10–6. Now, using the equation in example 18.10 we get: 0.030 [H + ]2 = (2.4 × 10−17 ) [CO 2 − ] 3 NiCO3 will precipitate if: 0.030 −13 [H + ]2 = (2.4 × 10−17 ) = 5.1 × 10 1.4 × 10−6 [H+] = 7.2 × 10–7 pH = 6.14 ZnCO3 will precipitate: 0.030 [H + ]2 = (2.4 × 10−17 ) = 4.8 × 10−10 −9 1.5 × 10 [H + ] = 2.2 × 10−5
pH = 4.66
So ZnCO3 will precipitate and NiCO3 will not if the pH is maintained between pH = 4.66 and pH = 6.14.
18.68
(a)
2+
–
2–
K form =
Cu (aq) + 4Cl (aq) CuCl4 (aq)
+
–
–
CuCl4 2− Cu 2+ Cl−
4
AgI2 −
(b)
Ag (aq) + 2I (aq) AgI2 (aq)
K form =
(c)
Cr3+(aq) + 6NH3(aq) Cr(NH3)63+(aq)
Cr(NH3 )63+ K form = 3+ Cr NH3 6
442
Ag + I−
2
Chapter 18
18.69
(a)
(b)
18.70
18.71
2–
3–
K form =
Ag (aq) + 2S2O3 (aq) Ag(S2O3)2 (aq)
2+
2+
K form
Zn (aq) + 4NH3(aq) Zn(NH3)4 (aq)
4+
2–
2–
Ag(S O ) 3− 2 3 2 Ag + S O 2− 2 3
Zn(NH3 )4 2+ = 2+ Zn NH 4 3 SnS 2 − 3
(c)
Sn (aq) + 3S (aq) SnS3 (aq)
K form =
(a)
Co3+(aq) + 6NH3(aq) Co(NH3)63+(aq)
Co(NH3 )63+ K form = 3+ Co NH3 6
2+
–
2–
K form =
Sn 4+ S2−
3
HgI 42 −
(b)
Hg (aq) + 4I (aq) HgI4 (aq)
(c)
Fe2+(aq) + 6CN–(aq) Fe(CN)64–(aq)
K form =
(a)
Hg2+(aq) + 4NH3(aq) Hg(NH3)42+(aq)
Hg(NH3 )4 2 + K form = 2+ Hg NH3 4
(b)
(c)
18.72
+
(a)
(b)
4+
–
2–
K form =
Sn (aq) + 6F (aq) SnF6 (aq)
3+
–
3–
Fe (aq) + 6CN (aq) Fe(CN)6 (aq)
3+
3+
Co(NH3)6 (aq) Co (aq) + 6NH3(aq)
2-
2+
Hg 2+ I−
-
HgI4 (aq) Hg (aq) + 4I (aq)
443
K form =
4
Fe(CN)6 4− Fe 2+ CN −
6
SnF6 2− Sn 4+ F−
6
Fe(CN)63− Fe3+ CN −
K inst
Co3+ NH3 6 = + 3 Co(NH3 )6
K inst
Hg 2+ I− = Hg(I)4 2−
4
6
2
Chapter 18
(c)
18.73
(a)
(b)
(c)
2+
4–
–
K inst
Fe(CN)6 (aq) Fe (aq) + 6CN (aq)
2+
2+
Hg(NH3)4 (aq) Hg (aq) + 4NH3(aq)
2–
4+
–
SnF6 (aq) Sn (aq) + 6F (aq)
Fe(CN)63–(aq) Fe3+(aq) + 6CN–(aq)
Fe 2+ CN − = Fe(CN)6 4−
K inst
Hg 2+ NH3 4 = 2 + Hg(NH ) 3 4
K inst
Sn 4+ F− = SnF6 2−
K inst
Fe3+ CN − = Fe(CN)63−
6
6
6
18.74
Kc = Ksp × Kform = (1.7 × 10–5)(2.5 × 101) = 4.3 × 10–4
18.75
Kc = Ksp × Kform = (6.0 × 10–17)(5.3 × 1018) = 3.2 × 102
18.76
There are two events in this net process: one is the formation of a complex ion (an equilibrium which has an appropriate value for Kform), and the other is the dissolving of Fe(OH)3, which is governed by Ksp for the solid. Fe(OH)3(s) Fe3+(aq) + 3OH–(aq) 3+
–
3–
Fe (aq) + 6CN (aq) Fe(CN)6 (aq)
3
K sp = Fe3+ OH − = 2.8 × 10−39 K form =
Fe(CN)63− 3+
Fe
CN
−
6
= 1.0 × 1031
The net process is: Fe(OH)3(s) + 6CN–(aq) Fe(CN)63–(aq) + 3OH–(aq) The equilibrium constant for this process should be:
Fe(CN)63− OH − Kc = 6 CN −
3
The numerical value for the above Kc is equal to the product of Ksp for Fe(OH)3(s) and Kform for Fe(CN)63–, as can be seen by multiplying the mass action expressions for these two equilibria: Kc = Kform × Ksp = 2.8 × 10–8 Because Kform is so very large, we can assume that all of the dissolved iron ion is present in solution as the complex, thus: [Fe(CN)63–] = 0.12 mol/1.3 L = 0.092 M. Also the reaction stoichiometry shows that each
444
Chapter 18 iron ion that dissolves gives 3 OH– ions in solution, and we have: [OH–] = 0.092 × 3 = 0.28 M. We substitute these values into the Kc expression and rearrange to get:
[ CN −] =
6
Fe(CN) 3− OH − 6 Kc
3
(0.092)(0.28)3 =6 2.8 × 10−8 Thus we arrive at the concentration of cyanide ion that is required in order to satisfy the mass action requirements of the equilibrium: [CN–] = 6.45 mol L–1. Since this concentration of CN– must be present in 1.3 L, the number of moles of cyanide that are required is: 6.45 mol L–1 × 1.3 L = 8.39 mol CN–. Additionally, a certain amount of cyanide is needed to form the complex ion. The stoichiometry requires six times as much cyanide ion as iron ion. This is 0.12 moles × 6 = 0.72 mol. This brings the total required cyanide to (8.39 + 0.72) = 9.1 mol. 9.1 mol × 49.0 g/mol = 446 g NaCN are required. 18.77
AgBr(s) Ag+ + Br–
Ksp = 5.0 × 10–13
Ag+ + 2S2O32– Ag(S2O3)23–
Kf = 2.0 × 1013
AgBr(s) + 2S2O32– Ag(S2O3)23– + Br–
Kc = Ksp*Kf = 10
2–
I C E
[Br–] – +x +x
3–
[S2O3 ] 1.20 – 2x 1.20 – 2x
[Ag(S2O3)2 ] – +x +x
Note: Since the AgBr(s) has a constant concentration, it may be neglected.
Kc = Kc =
[Ag(S2 O3 ) 23− ][Br − ] [S2 O32− ]2
= 10
x2
= 10 (1.20 − 2x)2 To solve this equation, take the square root of both sides and then solve for x. x = 0.518 M = [Ag(S2O3)23–] Since 1 mole of AgBr produces 1 mole of Ag(S2O3)23–, we can determine the number of grams of AgBr that will dissolve in 125 mL. g AgBr = (0.125 L)(0.518 moles/L)(187.77 g/mole) = 12.2 g AgBr 18.78
The applicable equilibria are as follows: AgI(s) Ag+(aq) + I–(aq) +
–
–
Ag (aq) + 2I (aq) AgI2 (aq)
K sp = Ag + I − = 8.5 × 10-17 K form =
445
AgI − 2 + −
Ag I
2
= 1 × 1011
Chapter 18 When a solution of AgI2– is diluted, all of the concentrations of the species in Kform above decrease. However, the decrease of [I–] has more effect on equilibrium because its expression is squared. Hence, the denominator is decreased more than the numerator in the reaction quotient, Q. The system reacts according to Le Châtelier’s Principle, by moving to the left (toward reactants) to increase the value of [I–]. As the system moves to the left, more Ag+ is created, which has an effect on the first equilibrium above. Again, Le Châtelier’s Principle causes the reaction to move to the left to re-establish equilibrium, which produces AgI(s) precipitate. The two equations above may be combined and Kc found as follows: AgI − 2 Kc = = Ksp × Kform = 8.5 × 10–6 AgI(s) + I–(aq) AgI2–(aq) − I To answer the second question, we make a table and fill in what we know. We begin with 1.0 M I–. This is reduced by some amount (x) as it reacts with the silver ions, and [AgI2–] is increased by the same amount:
I C E
[I–] 1.0 –x 1.0 – x
[AgI2–] – +x +x
Now we insert the equilibrium values into the above equation: AgI − 2 Kc = = 8.5 × 10–6 − I [ x ] = 8.5 × 10–6 Kc = [1.0 − x ] x = 8.5 × 10–6 This value represents the change in concentration of I– which, from the balanced equation, equals the change in concentration of AgI(s). The given volume is 0.100 L, which allows us to find the amount of AgI reacting: 0.125 L(8.5× 10–6 mol/L) = 1.1 × 10–6 mol AgI 1.1 × 10–6 mol AgI(234.8 g/mol) = 2.5 × 10–4 g AgI 18.79
The applicable equilibria are as follows: AgI(s) Ag+(aq) + I–(aq)
Ag+(aq) + 2I–(aq) AgI2–(aq)
K sp = Ag + I− = 8.5 × 10−17 Ag ( I ) − 2 K = = 1 × 1011 form
2
Ag + I− The two equations above may be combined and Kc found as follows: Ag ( I ) − 2 – – Kc = = Ksp × Kform = 8.5 × 10–6 AgI(s) + I (aq) Ag(I)2 (aq) I− If all of the AgI dissolves, it will be in the form of Ag(I)2–, therefore the concentration of Ag(I)2– is:
446
Chapter 18
[Ag(I)2–] =
0.020 mol Ag ( I) 2
−
= 0.160 M Ag(I)2– 0.125 L solution 0.160 = 1.9 × 104 M [CN–] = 8.5 × 10−6 The amount of KI that must be added is: (1.9 × 104 M)(0.125 L) = 2350 mol KI g KI = (2350 mol KI)(166 g/mol) = 3.91 × 105 g KI
18.80
The applicable equilibria are as follows:
K sp = Ag + I− = 8.5 × 10−17 Ag ( CN ) − 2 – + – K form = = 5.3 × 1018 Ag (aq) + 2CN (aq) Ag(CN)2 (aq) 2 Ag + CN − The two equations above may be combined and Kc found as follows: Ag ( CN ) − I− 2 – – – Kc = = Ksp × Kform = 4.5 × 102 AgI(s) + 2CN (aq) Ag(CN)2 (aq) + I (aq) 2 CN − AgI(s) Ag+(aq) + I–(aq)
We begin with 0.010 M CN–. This is reduced by some amount (2x) as it reacts with the silver ions, and [AgI2–] is increased by x:
I C E
[CN–] 0.010 – 2x 0.010 – 2x
[Ag(CN)2–] – +x x
I– – +x x
Now we insert the equilibrium values into the above equation:
Kc =
Ag ( CN ) − I− 2
Kc =
2
CN − [ x ][ x ]
2
= 4.5 × 102
= 4.5 × 102
[ 0.010 − 2x ]
Take the square root of both sides and solve for x: x = 4.9 × 10–3 This value represents the change in concentration of I– which, from the balanced equation, equals the change in concentration of AgI(s). The molar solubility of AgI in 0.010 M KCN is 4.9 × 10–3 M. 18.81
In case (a), the formation constant is relatively small indicating that the complex is not very stable. At the same time, the extremely small value for Ksp indicates that the ML2 solid is very stable. Consequently, the solution will contain very little M2+. In case (b), the solubility of ML2 is even smaller than in case (a). However, the large value for the formation constant indicates that any M2+ ions in solution will react with any ligand present to form the complex ion. As a result, more of the ML2 solid will dissolve increasing the amount of M2+ in solution.
447
Chapter 18
18.82
Recall that Kinst = 1/ Kform. 2
K sp = Zn 2+ OH − = 3.0 × 10-17
Zn(OH)2 (s) Zn2+(aq) + 2OH–(aq) 2+
2+
K form
Zn (aq) + 4NH3 (aq) Zn(NH3)4 (aq)
Zn(NH ) 2+ 3 4 = =? Zn 2+ NH 4 3
Combined, this is: Zn(OH)2 (s) + 4NH3 (aq) Zn(NH3)42+(aq) + 2OH–(aq)
Zn(NH ) 2+ OH - 3 4 Kc = 4 NH3 [NH3] 1.0 – 4x 1.0 – 4x
I C E
Kc =
2
[Zn(NH3)42+] – +x x
[OH–] – + 2x 2x
[ x ][ 2x ]2 [1.0 - 4x ]4
The problem gives the molar solubility of Zn(OH)2 as 5.7 × 10–3 M. This means in one liter of 1.0 M NH3, x = 5.7 × 10–3 moles. Substituting this value in for x, we get Kc = 8.1 × 10–7. Kc = Ksp × Kform 8.1 × 10–7 = 3.0 × 10–17 × Kform Kform = 2.7 × 1010 Kinst = 1/Kform Kinst = 1/(2.7 × 109) = 3.7 × 10–11 18.83
2
Cu(OH)2 (s) Cu2+(aq) + 2OH–(aq) Cu2+(aq) + 4NH3 (aq) Cu(NH3)42+(aq) Combined, this is:
K sp = Cu 2+ OH − = 4.8 × 10-20 13 Kform = 1.1 x 10
Cu(OH)2 (s) + 4NH3 (aq) Cu(NH3)42+(aq) + 2OH–(aq)
Cu(NH ) 2 + OH - 3 4 Kc = 4 NH3
I C E
Kc =
[NH3] 2.2 – 4x 2.2 – 4x
[ x ][ 2x ]2 [ 2.2 − 4x ]4
2
= 5.3 × 10–7
[Cu(NH3)42+] – +x x
= 5.3 × 10–7
Solve for x using successive approximations. 448
[OH–] – + 2x 2x
Chapter 18 x = 1.5 × 10–2 [Cu(NH3)42+] = 1.5 × 10–2 2+ Since all of the Cu comes from the Cu(OH)2, the molar solubility of Cu(OH)2 is 1.5 × 10–2 M
Additional Exercises 18.84
Assume the more soluble compound is 1.00 M. Then the less soluble compound is 0.0001 M.
M1q + = 1.00 M
M 2q + = 0.0001 M
The counter ion has molar concentration = X q − K sp (1) K sp (2) 18.85
=
1.00 X q − 0.0001 X q −
=1 x 104
The student would not collect PbCl2 by adding HCl to the final solution. The Pb2+ would be preciptated in step (1) when H2S was added to the acidified solution. This precipitate was separated from the remaining solution. The first precipitate would consist of PbS and CdS while the second precipitate consists of NiS.
18.86
We must first calculate the solubility in terms of # mols/L, i.e., mol = 7.05 × 10−3 g 1 mol Mg(OH)2 = 1.21 × 10−4 M L L 58.32 g Mg(OH) 2 Next, use this to establish the individual ion concentrations based on the equilibrium: Mg(OH)2 Mg2+ + 2OH–
(
)
Mg 2+ = 1.21 × 10−4 M OH − = 2.42 × 10−4 M Finally, calculate Ksp using the standard expression: 2
K sp = Mg 2+ OH − = 1.21 × 10−4 2.42 × 10−4
18.87
+
(
)(
2+
K spa
FeS(s) + 2H (aq) Fe (aq) + H2S(aq)
I C E
K spa
[H+] 8 8 – 2x 8 – 2x
[Fe2+] – +x +x
Fe 2+ [ H S] 2 = (x)(x) = 600 = 2 (8 − 2x)2 H+
(from Table 18.2)
x = 24.5 (8 − 2x) Solving gives x = 3.92 M. FeS is very soluble in 8 M acid. 449
2
= 7.09 × 10−12
Fe2+ [ H S] 2 = 2 H +
[H2S] – +x +x
take the square root of both sides to get;
)
Chapter 18
18.88
In order to answer this question, we need the [OH–] at equilibrium. Ca(OH)2 is a sparingly soluble compound. According to Table 18.1, Ksp = 5.0 × 10–6. Ca(OH)2(s) Ca2+(aq) + 2OH–(aq)
K sp = Ca 2+ OH −
[Mg2+] [OH–] – – I +x + 2x C +x + 2x E Ksp = (x)(2x)2 = 4x3 = 5.0 × 10–6 x = 1.1 × 10–2 M, [OH–] = 2x = 2.2 × 10–2 M: The pH = 14.00 – pOH = 12.34.
2
pOH = –log[OH–] = 1.66
18.89
The reaction that will dissolve the Mg(OH)2 is: Mg(OH)2 + 2H+ Mg2+ + 2H2O The concentration of H+ in the solution, before any reaction has occurred between the acid and the Mg(OH)2, is (0.0225 L)(0.10 M HCl) = 2.25 × 10–3 mol HCl 2.25 × 10–3 mol/1.000 L = 2.25 × 10–3 M Since all of the H+ will react with the solid Mg(OH)2, the amount of Mg2+ in solution will be: 1 mol Mg 2+ 2.25 × 10–3 mol HCl = 1.1 × 10–3 mol Mg2+ 2 mol H + Find the equilibrium concentration of OH– with 1.1 × 10–3 M Mg2+ Ksp = [Mg2+][OH–]2 = 5.6 × 10–12 5.6 × 10–12 = (1.1 × 10–3)(x)2 x = 7.1 × 10–5 = [OH–] pOH = –log[OH–] = –log(7.1 × 10–5) = 4.15 pH = 14 – pOH = 14 – 4.15 = 9.85
18.90
In this problem, we have two simultaneous equilibria occurring: 2
K sp = Mn 2+ OH − = 1.6 × 10–13 1 Fe2+(aq) + 2OH–(aq) Fe(OH)2(s) = 1/(4.9 × 10–17) = 2.04 × 1016 Kc = 2 Fe 2+ OH - The second equilibrium represents the opposite equation from that of Ksp. Therefore, its value is 1/Ksp for Fe(OH)2. Mn(OH)2(s) Mn2+(aq) + 2OH–(aq)
Combined, and omitting spectator ions, this is: Mn(OH)2(s) + Fe2+(aq) Mn2+(aq) + Fe(OH)2(s)
Mn 2+ =K −13 Kc = 2.04 × 1016 = 3265 sp (Mn) ⋅ K c (Fe) = 1.6 × 10 Fe2 +
(
I C E
)(
[Fe2+] 0.120 –x 0.120 – x
[Mn2+] – +x x
450
)
Chapter 18
Mn 2+ Kc = Fe 2+
[ x]
3265 =
[ 0.120-x ] 391.8 – 3265x = x 391.8 = 3266x x = 0.1199 = 0.120 M Therefore, [Fe2+] = 0.120 – x = 0 M and [Mn2+] = 0.120 M Since Ksp for Fe(OH)2 and Mn(OH)2 are so small, we assume there is almost no free hydroxide ion present and therefore the pH would remain neutral, around 7. 18.91
(a)
The number of moles of the two reactants are: 0.12 M Ag+ × 0.065 L = 7.8 × 10–3 moles Ag+ 0.048 M Cl– × 0.070 L = 3.4 × 10–3 moles Cl– The precipitation of AgCl proceeds according to the following stoichiometry: Ag+ + Cl– AgCl(s). If we assume that the product is completely insoluble, then 3.4 × 10–3 moles of AgCl will be formed because Cl– is the limiting reagent (see above.) 143.3 g AgCl g AgCl = 3.4 × 10−3 mol AgCl = 0.48 g AgCl 1 mol AgCl
(
(b)
)
The silver ion concentration may be determined by calculating the amount of excess silver added to the solution: [Ag+] = (7.8 × 10–3 moles – 3.4 × 10–3 moles)/0.135 L = 3.3 × 10–2 M The concentrations of nitrate and sodium ions are easily calculated since they are spectators in this reaction: [NO3–] = (0.12 M)(65.0 mL)/(135.0 mL) = 7.8 × 10–2 M [Na+] = (0.048 M)(70.0 mL)/(135.0 mL) = 3.4 × 10–2 M In order to determine the chloride ion concentration, we need to solve the equilibrium expression. Specifically, we need to ask what is the chloride ion concentration in a saturated solution of AgCl that has a [Ag+] = 3.3 × 10–2 M. AgCl(s) Ag+ + Cl–
I C E
[Ag+] 0.033 +x 0.033 + x
Ksp = 1.8 × 10–10
[Cl–] – +x +x
K sp = Ag + Cl− = (0.033+x)(x) = 1.8 × 10–10 x = 5.0 × 10–9 M if we assume that x<<0.033
451
Chapter 18 Therefore, [Cl–] = 5.5 × 10–9 M. (c)
The percentage of the silver that has precipitated is: (3.4 × 10–3 moles)/(7.8 × 10–3 moles) × 100% = 43.6%
18.92
To solve this problem, recognize that for a solution having a density of 1.00 g mL–1, 1 ppm = 1 mg L–1. Therefore, the initial hard water solution has a concentration of 278 mg Ca2+ / 1 L solution. Converting to molar concentration: 278 mg Ca 2+ 1 g Ca 2+ 1 mol Ca 2+ mol Ca 2 + = 2+ L solution 1 L solution 1000 mg Ca 2+ 40.078 g Ca
= 6.94 × 10−3 M Ca 2 + The concentration of CO32– is:
mol CO32 − 1.00 g Na 2 CO3 = L solution 1 L solution
1 mol Na 2 CO3 105.99 g Na 2 CO3
1 mol CO32− 1 mol Na 2 CO3
= 9.43 × 10−3 M CO32 − Comparing the concentrations of Ca2+ and CO32–, we observe that Ca2+ is the limiting reactant. Because of the small value of Ksp, we can assume that CaCO3 will precipitate using all of the available Ca2+ and leaving 9.43 × 10–3 – 6.94 × 10–3 = 2.49 × 10–3 M CO32–. The question now becomes, how much Ca2+ will be present in a solution having a [CO32–] = 2.49 × 10–3 M? Use the solubility product constant for Ksp to answer this question.
K sp = 3.4 × 10−9 = Ca 2+ CO32− − 9 K sp 3.4 × 10 Ca 2+ = = = 1.4 × 10−6 M −3 CO 2− 2.49 × 10 3 Converting back to units of ppm (mg L–1) we get:
1.4 × 10−6 mol Ca 2+ ppm Ca 2+ = 1 L solution
40.08 g Ca 2+ 1 mol Ca 2+
1000 mg 1 g
= 5.5 × 10−2 ppm Ca 2+ 18.93
A saturated solution of La2(CO3)3 satisfies the following equilibrium expression: 2
K sp = 4.0 × 10−34 = La 3+ CO32−
3
If [La3+] = 0.010 M, then the carbonate concentration of a saturated solution is: CO 2− = 3
K sp 3
2
=
3
4.0 × 10−34 2
La 3+ ( 0.010 ) We do the same calculation for PbCO3: K sp = 7.4 × 10−14 = Pb2+ CO32−
452
= 1.6 × 10 −10
Chapter 18 If [Pb2+] = 0.010 M, then the carbonate concentration of a saturated solution is: K sp 7.4 × 10−14 CO 2− = = = 7.4 × 10−12 M 3 0.010 Pb2+ Therefore, at a carbonate ion concentration between 7.4 × 10–12 M and 1.6 × 10–10 M, PbCO3 will precipitate, but La2(CO3)3 will not precipitate. The upper limit for the carbonate ion concentration is therefore 1.6 × 10–10 M. The equilibrium we need to look at now is: H2CO3(aq) 2H+(aq) + CO32–(aq) The Ka for this reaction is the product of Ka1 and Ka2 for carbonic acid. From Table 17.4we see that Ka1 = 4.5 × 10–7 and Ka2 = 4.7 × 10–11. So the equilibrium expression and value for the reaction of interest is: 2
H + CO 2− 3 Ka = = K a 1 × K a 2 = 2.1 × 10−17 H 2 CO3 This equation is rearranged and the values above and the values given in the problem are substituted in order to determine the pH range over which PbCO3 will selectively precipitate:
H+ =
K a H 2 CO3 = CO32−
( 2.1 × 10 )(3.3 × 10 ) −17
−2
CO32 −
If we substitute CO32− = 7.4 × 10−12 M we determine H + = 3.1 × 10−4 M and the pH = 3.51. Substituting CO32− = 1.6 × 10−10 M , H + = 6.6 × 10−5 M and pH = 4.18. Consequently, if [H+] = 6.6 × 10–5 M (pH = 4.18), La2(CO3)3 will not precipitate but PbCO3 will precipitate. At pH = 3.51 and below, neither carbonate will precipitate. 18.94
(a)
Mg(OH)2(s) Mg2+ + 2OH– NH4+ + OH– NH3 + H2O Mg(OH)2(s) + 2NH4+(aq) Mg2+(aq) + 2H2O + 2NH3(aq)
(b)
We want all of the Mg(OH)2 to go into solution. The NH4+ reacts with any OH– produced in the dissociation of Mg(OH)2 thereby shifting the equilibrium to the right. Using the Ksp value for Mg(OH)2, we may find the hydroxide ion concentration under these conditions: Mg(OH)2(s) Mg2+ + 2OH–
K sp = Mg 2+ OH −
2
5.6 × 10−12 = [ 0.10 ] OH − OH − = 7.5 × 10 −6
2
Now we can use this value in the following, simultaneous equilibrium: NH4+(aq) + OH–(aq) H2O + NH3 (aq) Kc = 1/KbNH3 = 1/1.8 × 10–5 = 5.6 × 104 NH3 [ 0.20] Kc = = = 5.6 × 104 − + OH NH 7.5 × 10-6 NH + 4 4 453
Chapter 18
(We know that [NH3] = 0.20 M because in the equation below 2 moles of ammonia are formed for every one mole of magnesium ion: Mg(OH)2(s) + 2NH4+(aq) Mg2+(aq) + 2H2O + 2NH3 (aq)) Solving for [NH4+], we get 0.48 M. So the total [NH3] + [NH4+] = 0.20 + 0.48 = 0.68 M One must therefore add 0.68 mol NH4Cl to a liter of solution. (c)
18.95
The resulting solution will contain 0.20 mol of NH3. Solve the weak base equilibrium problem for NH3. The pH = 11.28. There are two reactions that have to be considered here: the dissociation of CaCO3 in water, CaCO3(s) Ca2+(aq) + CO32–(aq) Ksp = [Ca2+][CO32–] = 3.4 × 10–9 and the ionization of carbonate ion in water,
[OH − ][HCO3− ]
= 1.8 × 10−4 [CO32− ] Assuming that all of the CO32– reacts with the water, the net reaction is: CaCO3(s) + H2O(l) Ca2+(aq) + HCO3–(aq) + OH–(aq) Kc = Ksp × Kb = [Ca2+][HCO3–][OH–] = 6.1 × 10–13 CO32–(aq) + H2O HCO3–(aq) + OH–(aq)
Kb =
We can obtain [OH–] from the pH: pOH = 14 – pH = 14 – 8.50 = 5.50 [OH–] = 10–pOH = 10–5.50 = 3.2 × 10–6 M Assume that [Ca2+] = [HCO3–] = x Kc = 6.1 × 10–13 = (x)(3.2 × 10–6 M)(x) = (x)2(3.2 × 10–6 M) Solve for x: x = 4.4 × 10–4 M = [Ca2+] = [HCO3–] 2+ The [Ca ] is equal to the molar solubility. Thus, the molar solubility of CaCO3 is 4.4 × 10–4 M. 18.96
Fe(OH)3(s) Fe3+ + 3OH–
I C E
[Fe3+] – +x +x
Ksp = [Fe3+][OH–]3 = 2.8 × 10–39
[OH–] 1.0 × 10–7 + 3x 1.0 × 10–7 + 3x
Assume 3x << 1.0 × 10–7, Ksp = (x)(1.0 × 10–7)3, solving for x we get, x = 2.8 × 10–18 M. Thus, 2.8 × 10–18 mol of Fe(OH)3 dissolve in 1 L of water. 18.97
The reaction for this problem is the formation of Ag(NH3)2+: Ag(NH ) + 3 2 + 2+ Ag (aq) + 2NH3(aq) Ag(NH3) K form = = 1.6 × 107 2 Ag + NH 3 We can rearrange this equation and substitute the values from the text to determine the [Ag+]: Ag(NH ) + 2.8 × 10−3 3 2 Ag + = = = 1.8 × 10−10 M 2 2 7 1.6 × 10 ( 1) K form NH3
(
(
)
)
454
Chapter 18
18.98
Let x = mols of PbI2 that dissolve per liter; Let y = mols of PbBr2 that dissolve per liter. Then, at equilibrium, we have [Pb2+] = x + y, [I–] = 2x and [Br–] = 2y We know: PbBr2(s) Pb2+ + 2Br–
Ksp = 6.6 × 10–6 = [Pb2+][Br–]2
PbI2(s) Pb2+ + 2I–
Ksp = 9.8 × 10–9 = [Pb2+][I–]2
Substituting we get: 6.6 × 10–6 = (x + y)(2y)2 and 9.8 × 10–9 = (x + y)(2x)2 Solving for x and y we find: x = 4.85 × 10–4 y = 7.91 × 10–3 Thus, [Pb2+] = 8.40 × 10–3 M, [I–] = 9.70 × 10–4 M and [Br–] = 1.58 × 10–2 M Note: [Br–] > [I–] because PbBr2 is more soluble than PbI2. 18.99
Initially, both Ag+ and HC2H3O2 are at 1.0 M concentrations. These values will be used to determine if the AgC2H3O2 will precipitate. First, determine the concentration of the acetate ion from the equilibrium: HC2H3O2 H+ + C2H2O2–
H + C H O – 2 3 2 Ka = = 1.8 × 10–5 HC2 H3O 2
I C E
[HC2H3O2] 1.0 –x 1.0 – x
Assume x << 1.0 [x][x] Ka = = 1.8 × 10−5 [1.0]
[H3O+] – +x x
[C2H3O2–] – +x x
x = 4.2 × 10-3 = [C2 H3O 2 − ]
Next, using the concentration of the acetate ion, determine whether or not a precipitate will form. AgC2H3O2(s) Ag+ + C2H3O2– Ksp = [Ag+][C2H3O2–] Q = [Ag+][C2H3O2–] before equilibrium is established Q = (1.0 M)(4.2 × 103 M) = 4.2× 103 Q > Ksp therefore a precipitate will form. 18.100 [Ag+] = 0.200 M [H+] = 0.10 M First, the concentration of acetate ion needs to be determined at the point that the silver acetate precipitates: AgC2H3O2(s) Ag+ + C2H3O2– Ksp = [Ag+][C2H3O2–] = 2.3 × 10–3 Let x = [C2H3O2–] 2.3 × 10–3 = (0.200)(x) x = 1.2 × 10–2 M = [C2H3O2–]
455
Chapter 18 When NaC2H3O2 is added to the solution, the C2H3O2– will react with the H+ from the nitric acid to form HC2H3O2. This will give a concentration of 0.10 M HC2H3O2. This will then come to equilibrium Now, find the concentration of HC2H3O2 using the Ka for acetic acid: HC2H3O2 H+ + C2H3O2– [HC2H3O2](0.175 L) + [C2H3O2–](0.175 L) = mole NaC2H3O2 that needs to be added. [HC2H3O2] + [H+] = 0.10 M which is from the nitric acid [HC2H3O2] = 0.10 M – [H+] H + C H O − 2 3 2 Ka = HC2 H3O 2 + Let x = [H ] [ x ] 1.2 ×10 –2 -5 1.8 × 10 = [ 0.10 – x ] x = 1.5 × 10–4 M = [H+] [HC2H3O2] = 0.10 M – [H+] = 0.10 – 1.5 × 10–4 M = 9.99 × 10–4 M The amount of NaC2H3O2 to be added is: [HC2H3O2](0.175 L) + [C2H3O2–](0.175 L) = mole NaC2H3O2 [9.99 × 10–4 M](0.175 L) + [1.2 × 10–2 M](0.175 L) = 2.3 × 10–3 mole NaC2H3O2 The number of grams to be added is
82.03 g NaC2 H3O 2 g NaC2H3O2 = 2.3 × 10–3 mole NaC2H3O2 = 0.19 g NaC2H3O2 1 mol NaC H O 2 3 2 18.101 First, let’s examine the question to make clear what is happening. The solution contains 0.20 M Ag+ ions and 1.34 x 10− 3 M acetate ions. The ion product of these two (2.7 x 10− 4 ) is less than Ksp for silver acetate (2.3 × 10–3), so the silver acetate remains in solution. There are also H+ ions (H3O+) in the solution as a result of the following equilibrium (OAc– will symbolize acetate): H2O + HOAc H3O+(aq) + OAc–(aq) The amount of H3O+ may be found by using the Ka for acetic acid:
I C E
Ka
[H3O+] – +x X
[HOAc] 0.10 –x 0.10 – x
[OAc–] – +x x
H O + OAc− 3 = [ HOAc]
1.8 × 10-5 =
[ x ][ x ] [0.10
− x]
–3
x ≈ 1.3 × 10 So [H3O+] = 1.3 × 10–3 M. However, when F– is added to the solution (in the form of KF), the following equilibrium takes place: F–(aq) + H3O+(aq) H2O + HF
456
Chapter 18 This depletes H3O+ ions from the solution, which causes the first equilibrium above to move to the right, producing more acetate ions. When the acetate ion concentration hits some minimum value (determined by Ksp) silver acetate will precipitate. That value may be found as follows:
K sp = Ag + OAc− 2.3 × 10−3 = [ 0.20 ][ x ]
x = 0.012 mol/L So the problem becomes…How many grams of KF must be added such that the acetate concentration increases to 0.012 M? This is now a simultaneous equilibrium problem:
K a = 1.8 × 10−5
H2O + HOAc H3O+(aq) + OAc–(aq) F–(aq) + H3O+(aq) H2O + HF
Kc’ =
1 1 = = 1.5 × 103 K a 6.8 × 10−4
(Note the above equation is simply the reverse of that for the Ka of HF, so Kc’ = 1/Ka.) Combined, this becomes: F–(aq) + HOAc HF + OAc–(aq)
(
)(
)
K c = K a ⋅ K b = 1.8 × 10−5 1.5 × 103 = 2.7 × 10−2 Recall that we have already found the initial concentrations of OAc– and HOAc above. Using this information, and the fact that we want the final [OAc–] to be 1.2 × 10–2, we can begin to fill out the table below.
I C E
[F–] x – 0.011 ?
[HOAc] 0.099 – 0.011 0.088
Kc =
[HF] – + 0.011 0.011
[OAc–] 1.3 × 10–3 + 0.011 0.012
[ HF] OAc−
F− [ HOAc ] [ 0.011][0.012] 2.7 × 10−2 = F− [ 0.088] [F–] = 0.056 M Placing this value into the table as the equilibrium concentration of [F–], we find the initial [F–] must be 0.056 + 0.011 = 0.067 M. Therefore the amount of KF needed in the 175 mL solution is: 0.175 L(0.067 mol KF/L)(58.01 g KF/1 mol KF) = 0.68 g KF
457
Chapter 18
18.102 Step 1: Determine the [OH–] from the NH3 reaction with water: NH4+ + OH– NH3 + H2O
NH + OH – 4 Kb = NH3 NH + OH – 4 1.8 × 10−5 = NH 3 [NH3] 0.10 –x 0.10 – x
I C E Assume x << 0.10 [ x ][ x ] 1.8 × 10−5 = [ 0.10]
[OH–] – +x x
[NH4+] – +x x
x = 1.3 × 10–3 = [OH–] Step 2: Find the concentration of Mg2+ at the given concentration of OH–. The concentration of OH– from the NH3 is 1.3 × 10–3, there is an additional amount of OH– from the equilibrium of the Mg(OH)2, which makes the calculation: Ksp = 5.6 × 10–12 = [Mg2+][OH–]2 5.6 × 10–12 = (x)(1.3 × 10–3 + 2x)2 The additional amount of OH– can be ignored since it will be less than 1.3× 10–3: Using the molar solubility of Mg(OH)2 in distilled water: 5.6 × 10–12 = [Mg2+][OH–]2 s = [Mg2+] and 2s = [OH–] 4s3 = 5.6 × 10–12 s = 1.1 × 10–4 The solubility of Mg(OH)2 in distilled water is less than the amount of OH– supplied by the ammonia so we a re justified in ignoring its contribution. We may now solve for x x = 3.3 × 10–6 M = [Mg2+] The molar solubility of Mg(OH)2 = 3.3 × 10–6 M 18.103 At its simplest, this is only a Ksp problem. The concentration of Mn2+ is (0.400 L)(0.10 M Mn2+)/(0.500 L) = 0.080 M Mn2+
K sp = Mn 2+ OH −
2
1.6 × 10-13 = [ 0.080] OH − − -6 OH = 1.4 × 10
2
When [OH–] = 1.4 × 10–6 M, Mn(OH)2 precipitates. 0.100 L(2.0 mol/L) = 0.20 mol NH3 are added to 400 mL solution which would make an initial concentration of 0.20 mol/0.500 L = 0.40 M NH3. The following equilibrium is set up:
K b = 1.8 × 10-5
H2O + NH3 NH4+(aq) + OH–(aq)
The problem tells us that all of the Sn2+ is precipitated as Sn(OH)2:
458
Chapter 18
Sn2+(aq) + 2OH–(aq) Sn(OH)2(s) The concentration of the Sn2+ is (0.400 L)(0.10 M Sn2+)/(0.500 L) = 0.080 M Sn2+, just before it precipitates. The Sn2+ immediately uses the first 2(0.080 M) = 0.16 M OH– which is produced from the reaction of ammonia with water above, using 0.16 M NH3. This effectively brings our initial concentration of NH3 to 0.24 M. Now we determine how much NH3 will produce [OH–] = 1.3 × 10–6 M.
K b = 1.8 × 10-5
H2O + NH3 NH4+(aq) + OH–(aq)
I C E
Kb
[NH3] x –1.2 × 10–6 ?
[NH4+] – + 1.2 × 10–6 1.2 × 10–6
[OH–] 1.0 × 10–7 + 1.2 × 10–6 1.3 × 10–6
NH + OH − 4 = NH3
1.2 × 10-6 1.3 × 10-6 1.8 × 10 = NH3 [NH3] = 8.7 × 10–8 -5
Therefore the initial [NH3] should be 8.7 × 10–8 + 1.2 × 10–6 = 1.3 × 10–6. So we want to reduce [NH3] by 0.24 – 1.3 × 10–6 = 0.23999 M, essentially by 0.24 M. This would require adding equimolar amounts of HCl, or: 0.500 L(0.24 mol NH3/L)(1 mol HCl/1 mol NH3)(36.5 g HCl/1 mol HCl) = 4.4 g HCl. (The difficulty here arises from the fact that such a small amount of OH– is required to precipitate the Mg2+ from solution that even a minimal amount of NH3 produces enough hydroxide ion to do so.) 18.104 Three reactions are occurring in the solution: 1)
2)
Cu2+(aq) + 4NH3–(aq) Cu(NH3)42+–(aq)
+
–
NH3(aq) + H2O NH4 (aq) + OH (aq)
Cu(NH3 )4 Kform = 1.1 × 1013 = Cu 2+ NH 4 3 NH + OH – 4 Kb = 1.8 × 10 = NH3 –5
2
Ksp = 4.8 × 10–20 = Cu 2+ OH – Starting with the first reaction, using the Kform of Cu(NH3)42+, calculate the concentration of NH3. The concentration of Cu2+ before any reaction has occurred is 0.050 M, and the concentration of NH3 before any reaction has occurred is 0.50 M. Assume that all of the Cu2+ has reacted and then the reaction has come to equilibrium. Therefore, the initial concentration of Cu2+ is 0 and the initial concentration of NH3 is 0.50 M – (4 × 0.05 M) 3)
Cu(OH)2(s) Cu2+ + 2OH–
459
Chapter 18
I C E
[Cu2+] – +x + 2x
[NH3] 0.30 + 4x 0.30 + 4x
Cu(NH3)42+ 0.050 M –x 0.050 – x
Cu(NH3 )4 Kform = 1.1 × 1013 = Cu 2+ NH 4 3 Kform = 1.1 × 1013 =
[0.050 – x ] [ 2x ][0.30 + 4x ]4
assume x << 0.050
x = 2.8 × 10–13 [NH3] = 0.30 [Cu2+] = 5.6 × 10–13 Next, determine the [OH–] from the reaction of NH3 with H2O +
–
NH3(aq) + H2O NH4 (aq) + OH (aq)
I C E
[NH3] 0.30 –x 0.30 – x
[NH4+] – +x x
NH + OH – 4 Kb = 1.8 × 10 = NH3 –5
OH– – +x x
NH + OH – 4 Kb = 1.8 × 10–5 = NH3 [ x ][ x ] Kb = 1.8 × 10–5 = [0.30 – x ]
assume x << 0.30
x = 2.3 × 10–3 = [OH–] Finally, with the concentration of OH– and the concentration of Cu2+, determine if any precipitate has formed: Cu(OH)2 Cu2+ + 2OH– Ksp = 4.8 × 10–20 = Cu 2+ OH – [OH–] = 2.3 × 10–3 [Cu2+] = 5.6 × 10–13
2
2
Q = Cu 2+ OH – = (5.6 × 10–13)(2.3 × 10–3)2 = 3.0 × 10–18 Q > Ksp therefore a precipitate forms. Assume that all of the Cu2+ has reacted with the OH– and then the solution returns to equilibrium
I C E
[Cu2+] – +x +x
[OH–] 2.3 × 10–3 + 2x 2.3 × 10–3 + 2x
Ksp = 4.8 × 10–20 = Cu 2+ OH –
2
Ksp = 4.8 × 10–20 = [ x ] 2.3 ×10 –3 + 2x x = 9.1 × 10–15 M = [Cu2+]
2
x << 2.3 × 10–3
18.105 Al(OH)3 has such an exceedingly small Ksp, there is almost no dissociation in pure water. Therefore, the pH would be expected to be about 7. 460
Chapter 18
In doing the calculations, this is borne out:
K sp = Al3+ OH −
3
3
3 × 10−34 = [ x ][ 3x ] x = 1.8 × 10–9 [OH–] = 5.5 × 10–9 [OH–] = (1 × 10–7) + (5.5 × 10–9) = 1.055 × 10–7 pOH = 6.98 pH = 7.02 18.106 The value of K is very large, therfore, all of the EDTA4− will be consumed to produce the complex, PbEDTA2−. The reaction of Pb2+ with the EDTA4− is a one-to-one stoichiometric reaction, thus, Pb2+ is in excess. After the reaction we will have the following: [Pb2+] = 0.15 M – 0.10 M = 0.05 M [EDTA4−] = 0 [PbEDTA2+] = 0.10 M In reality, the system in in equilibirum so a very small amount of EDTA4− will exist in solution but the amount is insignificant. The hydroxide concentration can be determined from the pH of the solution. [OH−] = 10(14-12.5) = 3.2 x 10−2 Q = [Pb2+][OH−]2 = 0.05 x (3.2 x 10-2)2 = 5.1 x 10−5 This value is greater than Ksp for Pb(OH)2 which is 1.4 x 10−20 (CRC Handbook of Chemistry and Physics, 91st ed.,2010-2011), therefore, Pb(OH)2 will precipitate from the reaction mixture. 18.107 To answer this question we can use cation groupings from qualitative analysis. Ag+ and Pb2+ are group I ions. These ions can be separated from the others by adding Cl− to the solution. AgCl and PbCl2 will precipitate and can be filtered off from the solution. AgCl can be separated from PbCl2 by adding 6 M NH3. AgCl will react forming Ag(NH3)2+ while PbCl2 will remain a solid. Cu2+ and Bi3+ can be separated by adding 0.1 M H2S and 6M HCl. This will precipitate CuS and Bi2S3 which can be filtered The solids can be converted to soluble species by adding HNO3. Then, base can be added to precipitate hydroxides; Cu(OH)2 and Bi(OH)3. The addition of 6M NH3 will convert the Cu(OH)2 to Cu(NH3)42+ leaving the solid Bi(OH)3. Then add base to the remaining solution along with H2S to precipitate CoS and MnS. Filter the solution and dissolve the solid be adding acid. Cobalt can be separated by adding KNO2 which forms a yellow precipitate while Mn2+ remains in solution. Finally, Ba2+ and Ca+ can be separated by adding acid to make a weakly acidic solution and then adding Na2CrO4. BaCrO4 will precipitate while Ca2+ will remain in solution. 18.108 (1) (2)
In order, Figure C Figure B Figure E Figures A and D are excluded because PbBr2 will precipitate before PbCl2.
461
Chapter 18 18.109 The volume of a cone is given by V=(1/3)π r2h 3
1 2 12 in 2.54 cm x 3.1415 x ( 5 ft ) x 16ft x x 3 ft in 7 3 V = 1.19 x 10 cm
3
V=
The mass of the cone is: 1.19 x 107 cm3 x 2.71 g cm-3 = 3.22 x 107 g CaCO3 Saturated CaCO3 has a molar concentration of Ca2+ equal to: [Ca2+] = [CO32-] = Ksp1/2 = (3.4 x 10-9)1/2 = 5.8 x 10-5 M Mass of CaCO3 dissolved in a liter of saturated solution is: 5.8 x 10-5 mol L-1 x (100.09 g/ mol CaCO3) = 5.8 x 10-3 g L-1 The volume of solution required to build the cone would be equal to: 3.22 x 107 g CaCO3 x (1 L/5.8 x 10-3 g) = 5.6 x 109 L 10 drops have a volume of 1 mL. (10 drops/1 mL) x (1000 mL/L) = 10000 drops/L The number of drops required to deposit enough CaCO3 is given by: (10000 drops/L) x 5.6 x 109L = 5.6 x 1013 drops The age of the stalagmite equals: 5.6 x 1013 drops x (1s/drop) x (1 hr/3600 s) x (1 day/24 hrs) x (1yr/365 days) = 1.8 x 106 yrs 18.110 First, we must calculate the mass of CaSO4 dissolved. The volume is a cylinder, with V = hπr2: h = 0.50 in (2.54 cm/1 inch) = 1.27 cm r = 1/2 diameter = 0.50 cm Therefore: V = 1.0 cm3, and mass = 1.0 cm3(0.97 g/cm3) = 0.97 g However, because it is a hydrate (CaSO4·2H2O), plaster is only (136.2/172.2 = 0.79) 79% calcium sulfate, therefore the true mass of CaSO4 is: Mass = 0.97(.79) = 0.77 g CaSO4 In moles, this is 0.77 g CaSO4 x (1 mol CaSO4 /136.2g) = 0.0056 mol CaSO4 Now we must calculate the volume of water necessary to dissolve 0.77 g of CaSO4. We start by finding its molar solubility:
K sp = Ca 2+ SO42− = 4.9 × 10–5 K sp = [ x ][ x ] = 4.9 × 10–5 x = 7.0 × 10–3 mol/L
462
Chapter 18
So the volume of water needed is: 0.0056 mol(1 L/7.0 × 10–3 mol) = 0.8 L Finally, we find the amount of time needed to produce this much water: 0.8 L (1 day/2.00 L) = 0.40 days, or about 9.6 hours. 18.111 Start by determining the [Cl–] if the [Pb2+] = 0.0050 M. K sp
Cl− =
=
1.7 × 10−5 = 5.8 × 10−2 0.0050
Pb 2+ Then determine the concentration of chloride in a saturated solution, Ksp = 4x3 where x = [Pb2+] and [Cl–] =
Ksp
= 1.6 × 10−2 M, Cl− = 3.2 × 10−2 M 4 The volume of 0.10 M HCl which needs to be added is: M 1V 1 = M 2V 2 (x mL)(0.10 M HCl) = (3.2 × 10–2 M HCl)(x + 100 mL) 0.1 x = 3.2× 10–2x + 3.2 6.8 × 10–2 x = 3.2 x = 47 mL 47 mL of 0.10 M HCl need to be added. 2x, x = 3
18.112 CaCO3(s) + H+(aq) Ca2+(aq) + HCO3− (aq) HCO3−(aq) + H+(aq) H2CO3(aq) Combining these equations and solving for [Ca2+] we obtain,
2+
[Ca ] =
K sp H +
2
K a1K a 2
pH = 5.6 [Ca2+] =
3.4 x 10−9 10−5.6
(
4.3 x 10 −7
)(
2
5.6 x 10 −11
)
= 8.9 x 10-4 M
Therefore, the solubility of CaCO3 equals 8.9 x 10-4 M pH = 4.2 2+
[Ca ] =
3.4 x 10−9 10−4.2
2
( 4.3 x 10 )( 5.6 x 10 ) −7
−11
= 0.56 M
Therefore, the solubility of CaCO3 equals 0.56 M 18.113 Mg(OH)2(s) Mg2+(aq) + 2OH−(aq) Ksp = [Mg2+][OH−]2 [OH−] = 10−(14.0−9.8) = 6.3 x 10-5 M [Mg2+] = (6.3 x 10-5 mol OH−/L) x (1 mol Mg2+/2 mol OH−) = 3.2 x 10-5 M
463
Chapter 18
Ksp = 3.2 x 10-5 x (6.3 x 10-5)2 = 1.3 x 10-13 18.114 Π = iMRT
M = Π/(iRT)
i = 2 for NiCO3
1 atm 13 torr x 760 torr M= = 3.5 x 10− L atm x 298 K 2 x 0.082 K mol
4
Ksp = (3.5 x 10-4)2 = 1.2 x 10− 7
464
Chapter 19
Practice Exercises 19.1
w = –P∆V = –(14.0 atm)(12.0 L – 1.0 L) = –154 L atm ∆E = w + q = 0 0 = –154 + q Therefore, q = +154 L atm. (The energy is converted into heat; since the heat does not leave the system the temperature increases.)
19.2
∆E = q – P∆V since q = 0 ∆E = –P∆V but ∆V is negative for a compression so ∆E increases and T increases. Energy is added to the system in the form of work.
19.3
∆H =
∆H0f [NO2(g)] – { ∆H0f N2O (g)] + ∆H0f [O2(g)]}
∆H = {4 mol NO2(g) × (34 kJ/mol)} – {2 mol N2O × (81.5 kJ/mol) + 3 mol O2 × (0 kJ/mol)} ∆H = –27 kJ ∆E – ∆H = ∆nRT = (–1 mol)(8.314 J mol–1 K–1)(318 K) = –2.64 × 104 J ∆E – ∆H = –2.64 kJ ∆E is more exothermic. 19.4
∆E° = ∆H° – ∆nRT = –217.1 kJ – (–1 mol)(8.314 J mol–1 K–1)(298 K) = –217.1 kJ + 2.48 kJ = –214.6 kJ % Difference = (2.48/217) × 100 = 1 %
19.5
∆S should be negative since the reaction moves from great movement of ions in solution to less movement in the solid.
19.6
(a) (b)
∆S is negative since the products have a lower entropy, i.e. a lower freedom of movement. ∆S is positive since the products have a higher entropy, i.e. a higher freedom of movement.
19.7
(a)
∆S is negative since there are less gas molecules. (The product is also more complex, indicating an increase in order.) ∆S is negative since there are less gas molecules. (The product is also more complex, indicating an increase in order.)
(b)
19.8
(a) (b) (c)
19.9
∆S is negative since there is a change from a gas phase to a liquid phase. (The product is also more complex, indicating an increase in order.) ∆S is negative since there are less gas molecules. (The product is also more complex, indicating an increase in order.) ∆S is positive since the particles go from an ordered, crystalline state to a more disordered, aqueous state.
1 3 N2(g) + H2(g) NH3(g) 2 2 1 3 o ∆Sof = [ SNH3 (g) ] – [ SoN2 (g) + SoH2 (g) ] 2 2
465
Chapter 19
192.5 J 1 191.5 J 3 130.6 J ∆Sof = (1 mol NH 3 ) × − mol N 2 × + mol H 2 × mol K 2 mol K 2 mol K ∆Sof = –99.1 J K–1 19.10
19.11
∆S° = (sum S°[products]) – (sum S°[reactants]) a)
∆S° = {S°[H2O(l)] + S°[CaCl2(s)]} – {S°[CaO(s)] + 2S°[HCl(g)]} ∆S° = {1 mol × (69.96 J mol–1 K–1) + 1 mol × (114 J mol–1 K–1)} – {1 mol × (40 J mol–1 K–1) + 2 mol × (186.7 J mol–1 K–1)} ∆S° = –229 J/K
b)
∆S° = {S°[C2H6(g)]} – {S°[H2(g)] + S°[C2H4(g)]} ∆S° = {1 mol × (229.5 J mol–1 K–1)} – {1 mol × (130.6 J mol–1 K–1) + 1 mol × (219.8 J mol–1 K–1)} ∆S° = –120.9 J/K
N2(g) + 2O2(g) N2O4(g) ∆H oN2 O4 (g) = 9.67 kJ mol–1
∆SoN2 O4 (g) = (1 mol N2O4)(304 J mol–1 K–1) – (1 mol N2)(191.5 J mol–1 K–1) – (2 mol O2)(205.0 J mol–1 K–1) = –297.5 J K–1 ∆G = ∆H – T ∆Sof o f
o f
∆G of = 9.67 kJ mol–1 – (298 K)(–0.2975 kJ mol–1 K–1) ∆G of = 98.3 kJ mol–1
19.12
First, we calculate ∆S°, using the data: ∆S° = {2S°[Fe2O3(s)]} – {3S°[O2(g)] + 4S°[Fe(s)]} ∆S° = {2 mol × (90.0 J mol–1 K–1)} – {3 mol × (205.0 J mol–1 K–1) + 4 mol × (27 J mol–1 K–1)} ∆S° = –543 J/K = –0.543 kJ/mol Next, we calculate ∆H° using the data : ∆H° = (sum ∆Hf°[products]) – (sum ∆Hf°[reactants]) ∆H° = {2∆Hf°[Fe2O3(s)]} – {3∆Hf°[O2(g)] + 4∆Hf°[Fe(s)]} ∆H° = {2 mol × (–822.2 kJ/mol)} – {3 mol × (0.0 kJ/mol) + 4 × (0.0 kJ/mol)} ∆H° = –1644 kJ The temperature is 25.0 + 273.15 = 298.15 K, and the calculation of ∆G° is as follows: ∆G° = ∆H° – T∆S° = –1644 kJ – (298.15 K)(–0.543 kJ/K) = –1482 kJ
19.13
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) ∆Go = {2 mol Fe(s) × ∆Gf°[Fe(s)] + 3 mol CO2(g) × ∆Gf°[CO2(g)]} – {1 mol Fe2O3(s) × ∆Gf°[Fe2O3(s)] + 3 mol CO(g) × ∆Gf°[CO(g)]} ∆Go = {2 mol Fe(s) × 0 kJ mol–1 + 3 mol CO2(g) × –394.4 kJ mol–1} – {1 mol Fe2O3(s) × –741.0 kJ mol–1 + 3 mol CO(g) × –137.3 kJ mol–1} ∆Go = –30.3 kJ
466
Chapter 19
19.14
19.15
We calculate ∆G°rxn, using the data from Table 19.2: a)
∆G°rxn = {2∆Gf°[NO2(g)]} – {2∆Gf°[NO(g)] + ∆Gf°[ O2(g)]} ∆G°rxn = {2 mol × (+51.84 kJ mol–1)} – {2 mol × (+86.69 kJ mol–1) + 1 mol × (0 kJ mol–1)} ∆G°rxn = –69.7 kJ/mol
b)
∆G°rxn = {∆Gf°[CaCl2(s)] + 2∆Gf°[H2O(g)] } – {∆Gf°[Ca(OH)2(s)] + 2∆Gf°[ HCl(g)]} ∆G°rxn = {1 mol × (–750.2 kJ mol–1) + 2 mol × (–228.6 kJ mol–1)} – {1 mol × (–896.76 kJ mol–1) + 2 mol × (–95.27 kJ mol–1)} ∆G°rxn = –120.1 kJ/mol
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ∆G°rxn = {2∆Gf°[CO2(g)] + 3∆Gf°[H2O(l)]} – {∆Gf°[C2H5OH(l)] + 3∆Gf°[O2(g)]} ∆G°rxn = {2 mol × (–394.4 kJ mol–1) + 3 mol × (–237.2 kJ mol–1)} – {1 mol × (–174.8 kJ mol–1) + 3 mol × (0 kJ mol–1)} G°rxn = –1325.6 kJ For 125 g C2H5OH: 1 mol C 2 H 5 OH mol C2H5OH = 125 g C2H5OH = 2.71 mol C2H5OH 46.08 g C 2 H 5 OH Maximum work: (1325.6 kJ mol–1)(2.71 mol C2H5OH) = 3596 kJ C8H18(l) +
25 O2(g) 8CO2(g) + 9H2O(l) 2
∆G°rxn = {8∆Gf°[CO2(g)] + 9∆Gf°[H2O(l)]} – {∆Gf°[C8H18(l)] +
25 ∆Gf°[O2(g)]} 2
∆G°rxn = {8 mol × (–394.4 kJ mol–1) + 9 mol × (–237.2 kJ mol–1)} – {1 mol × (+17.3 kJ mol–1) + (0 kJ mol–1)} G°rxn = –5304.1 kJ For 125 g C8H18:
25 mol × 2
1 mol C8 H18 mol C8H18 = 125 g C8H18 = 1.09 mol C8H18 114.26 g C8 H18 Maximum work: (5304.1 kJ mol–1)(1.09 mol C8H18) = 5803 kJ
The octane is a better fuel on both a per gram and per mole basis. 19.16
The maximum amount of work that is available is the free energy change for the process, in this case, the standard free energy change, ∆G°, since the process occurs at 25 °C. 4Al(s) + 3O2(g) 2Al2O3(s) ∆G° = (sum ∆Gf°[products]) – (sum ∆Gf°[reactants]) ∆G° = 2∆Gf°[Al2O3(s)] – {3∆Gf°[O2(g)] + 4∆Gf°[Al(s)]} ∆G° = 2 mol × (–1576.4 kJ/mol) – {3 mol × (0.0 kJ/mol) + 4 mol × (0.0 kJ/mol)} ∆G° = –3152.8 kJ, for the reaction as written.
467
Chapter 19
This calculation conforms to the reaction as written. This means that the above value of ∆G° applies to the equation involving 4 mol of Al. The conversion to give energy per mole of aluminum is then: –3152.8 kJ/4 mol Al = –788 kJ/mol The maximum amount of energy that may be obtained is thus 788 kJ. 19.17
T ≈
∆H o ∆So
1000 J 3 ∆Ho = 21.7 kJ mol–1 = 21.7 × 10 J 1 kJ 21.7 × 103 J 239.9 K ≈ ∆So o –1 ∆S = 90.5 J K 19.18
For the vaporization process in particular, and for any process in general, we have: ∆G = ∆H – T∆S If the temperature is taken to be that at which equilibrium is obtained, that is the temperature of the boiling point (where liquid and vapor are in equilibrium with one another), then we also have the result that ∆G is equal to zero: ∆G = 0 = ∆H – T∆S, or Teq = ∆H/∆S We know ∆H to be 60.7 kJ/mol; we need the value for ∆S in units kJ mol–1 K–1: ∆S° = (sum S°[products]) – (sum S°[reactants]) ∆S° = S°[Hg(g)] – S°[Hg(l)] ∆S° = (175 × 10–3 kJ mol–1 K–1) – (76.1 × 10–3 kJ mol–1 K–1) ∆S° = 98.9 × 10–3 kJ mol–1 K–1 Teq = 60.7 kJ/mol ÷ 98.9 × 10–3 kJ/mol K = 614 K (341 °C)
19.19
∆G° = (sum ∆Gf°[products]) – (sum ∆Gf°[reactants]) 1 ∆G° = ∆Gf°[SO3(g)] – {∆Gf°[SO2(g)] + ∆Gf°[ O2(g)]} 2 ∆G° = 1 mol × (–370.4 kJ/mol) – {1 mol × (–300.4 kJ/mol) +
1 (0.0 kJ/mol)} 2
∆G° = –70.0 kJ/mol Since the sign of ∆G° is negative, the reaction should be spontaneous. 19.20
∆G° = ∆H° – T∆S° ∆G° = {2∆Gf°[HCl(g)] + ∆Gf°[CaCO3(s)]} – {∆Gf°[CaCl2(s)] + ∆Gf°[H2O(g)] + ∆Gf°[CO2(g)]} ∆G° = {2 mol × (–95.27 kJ/mol) + 1 mol × (–1128.8 kJ/mol)} – {1 mol × (–750.2 kJ/mol) + 1 mol × (–228.6 kJ/mol) + 1 mol × (–394.4 kJ/mol)} ∆G° = +53.9 kJ ∆G° is positive, the reaction is not spontaneous, and we do not expect to see products formed from reactants.
19.21
T = 75 °C + 273.15 K = 348 K ∆G° = 119.2 kJ – (348 K)(0.3548 kJ/K) –4.3 kJ 468
Chapter 19
19.22
∆G° = {∆Gf°[Na2CO3(s)] + ∆Gf°[CO2(g)] + ∆Gf°[H2O(g)]} – {2∆Gf°[NaHCO3(s)]} ∆G° = {1 mol × (–1048 kJ/mol) + 1 mol × (–394.4 kJ/mol) + 1 mol × (–228.6 kJ/mol)} – {2 mol × (–851.9 kJ/mol)} ∆G° = +32.8 kJ ∆H° = {∆Hf°[Na2CO3(s)] + ∆Hf°[CO2(g)] + ∆Hf°[H2O(g)]} – {2∆Hf°[NaHCO3(s)]} ∆H° = {1 mol × (–1131 kJ/mol) + 1 mol × (–393.5 kJ/mol) + 1 mol × (–241.8 kJ/mol)} – {2 mol × (–947.7 kJ/mol)} ∆H° = +129.1 kJ ∆S° = {Sf°[Na2CO3(s)] + Sf°[CO2(g)] + Sf°[H2O(g)]} – {2Sf°[NaHCO3(s)]} ∆S° = {1 mol × (136 J/mol K) + 1 mol × (213.6 J/mol K) + 1 mol × (188.7 J/mol K)} – {2 mol × (102 J/mol K)} ∆S° = +334.3 J K–1 = 0.3343 kJ K–1 ∆G o490 = ∆H° – T∆S° ∆G o490 = 129.1 kJ – (490 K)(0.3343 kJ K–1) = –34.7 kJ The equilibrium shifts to products.
19.23
PN O ∆G = ∆G o298 + RT ln 2 42 PNO 2
0.598 atm ∆G = −5.40 × 103 J mol −1 + 8.314 J mol−1 K −1 ( 298 K ) ln ( 0.260 atm )2 ∆G = 0 J mol–1 = 0 kJ mol–1 The system is at equilibrium, so that the reaction will not move.
(
19.24
)
Using the data provided we may write: PN O ∆G = ∆G o + RT ln 2 4 PNO 2
0.25 atm = − 5.40 × 103 J mol−1 + 8.314 J mol−1 K −1 ( 298 K ) ln ( 0.60 atm )2
(
)
(
= − 5.40 × 103 J mol−1 + −9.03 × 10 2 J mol−1 3
= − 6.30 × 10 J mol
)
−1
Since ∆G is negative, the forward reaction is spontaneous and the reaction will proceed to the right. 19.25
∆G° = –RT ln Kp ∆G° = –(8.314 J K–1 mol–1)(25 + 273 K) × ln(6.9 × 105) = –33 × 103 J ∆G° = –33 kJ
469
Chapter 19
19.26
∆G° = –RT ln Kp 3.3 × 103 J mol–1 = –(8.314 J K–1 mol–1)(25.0 + 273.15 K) × ln(Kp) Kp = 0.26
19.27
3C(g) + 8H(g) + O(g)
∆ H (Bond Formation) 0 f
∆ H (atoms) 3C(graphite) + 4H2(g) + ½ O2(g)
→
CH3CH2CH2OH(g)
∆ H 0f
The heat of formation of the gaseous reactants is endothermic, and therefore positive. 3 mol C x 716.7 kJ mol-1 = 2150.1 kJ 8 mol H x 217.9 kJ mol-1 = 1743.2 kJ 1 mol O x 249.2 kJ mol-1 = 249.2 kJ Total energy input is 4142.5 kJ Now determine the energy released during bond formation. 2 C-C bonds formed
2 mol x 348 kJ mol-1 = 696 kJ
7 C-H bonds formed
7 mol x 412 kJ mol-1 = 2884 kJ
1 C-O bond formed
1 mol x 360 kJ mol-1 = 360 kJ
1 O-H bond formed
1 mol x 463 kJ mol-1 = 463 kJ
Total energy released is 4403 kJ The heat of formation is: 4142.5 kJ + (– 4403 kJ) = -260.5 kJ 4C(s) + 9/2H2(g) + Br2(l)
→
CH3CHBrCH2CH3(g)
∆ H0f
The heat of formation of the gaseous reactants is endothermic, and therefore positive. 4 mol C x 716.7 kJ mol-1 = 2866.8 kJ 9 mol H x 217.9 kJ mmol-1 = 1961.1 kJ 1 mol Br x 112.4 kJ mol-1 = 112.4 kJ Total energy input is 4940.3 kJ Now determine the energy released during bond formation. 3 C-C bonds formed
3 mol x 348 kJ mol-1 = 1044 kJ
470
Chapter 19
9 C-H bonds formed
9 mol x 412 kJ mol-1 = 3708 kJ
1 C-Br bond formed
1 mol x 276 kJ mol-1 = 276 kJ
Total energy released is 5028 kJ The heat of formation is: 4940.3 kJ + (– 5028 kJ) = -88 kJ 19.28
6C(s) + 6H2(g)
→
C6H12(g)
∆ H0f
The heat of formation of the gaseous reactants is endothermic, and therefore positive. 6 mol C x 716.7 kJ mol-1 = 4300.2 kJ 12 mol H x 217.9 kJ mol-1 = 2614.8 kJ Total energy input is 6915 kJ Now determine the energy released during bond formation. 6 C-C bonds formed
6 mol x 348 kJ mol-1 = 2088 kJ
12 C-H bonds formed
12 mol x 412 kJ mol-1 = 4944 kJ
1 C-Br bond formed
1 mol x 276 kJ mol-1 = 276 kJ
Total energy released is 7032 kJ The heat of formation is: 6915 kJ + (– 7032 kJ) = -117 kJ 6C(g) + 6H(g)
→
C6H6(g)
The heat of formation of the gaseous reactants is endothermic, and therefore positive. 6 mol C x 716.7 kJ mol-1 = 4300.2 kJ 6 mol H x 217.9 kJ mmol-1 = 1307.4 kJ Total energy input is 5607.6 kJ Now determine the energy released during bond formation. Benzene bonding is neither single nor double bonding between carbon atoms but we will approximate the bonding as 3 single C-C bonds and 3 double C-C bonds. 3 C-C bonds formed
3 mol x 348 kJ mol-1 = 1044 kJ
3 C=C bonds formed
3 mol x 612 kJ mol-1 = 1836
6 C-H bonds formed
6 mol x 412 kJ mol-1 = 2472 kJ
Total energy released is 5352 kJ The heat of formation is: 5607.6 kJ + (– 5352 kJ) = +255.6 kJ
471
Chapter 19
Review Questions 19.1
The term thermodynamics is meant to convey the two ideas of thermo (heat) and dynamics (motion), namely movement or transfer of heat.
19.2
∆E = Efinal – Einitial The first law of thermodynamics states that the internal energy may be transferred as heat or work, but it cannot be created or destroyed. The change in internal energy of a system is the sum of two terms, the amount of energy the system gains from heat transfer and the amount of energy the system gains from work transfer: ∆E = q + w. ∆E is positive is energy flows into a system and negative if energy flows out of a system.
19.3
If we refer to the equation in the answer to Review Question 19.2, we can see that if Efinal is larger than Einitial, then ∆E must be positive, by definition. Thus, ∆E for an endothermic process is positive.
19.4
∆E is the change in internal energy, which is positive for an endothermic process. Heat, q, is the absorbed by the system during the process, and it has a positive sign if the system absorbs heat during the process. Work, w, is done on the system by the surroundings during the process, and it has a negative sign if the system does work on the surroundings, whereas it has a positive sign if the surroundings do work on the system during the process.
19.5
A state function is a thermodynamic quantity whose value is determined only by the state of the system currently, and is not determined by a system's prior condition or history. A change in a state function is the same regardless of the path that is used to arrive at the final state from the initial state. That is, changes in state function quantities are path–independent. In the statement of the first law, only ∆E is a state function.
19.6
∆E is the heat of reaction at constant volume, and applies, for instance, to reactions performed in closed vessels such as bomb calorimeters. ∆H is the heat of reaction at constant pressure.
19.7
A pressure in Pascals, by definition, has the units Newtons/m2, or N/m2. Thus, P in pascals times ∆V in m3 is: N/m2 × m3 = N · m The last result is the Newton · meter, or a force times a distance. By definition, a force of 1 N operating over a distance of 1 meter is the quantity 1 joule. Thus, the quantity P × ∆V has the units of energy, namely joules.
19.8
Since the quantity P∆V is negative (corresponding to a decrease in volume), ∆E must be a smaller negative quantity than ∆H. ∆E is a smaller negative quantity than ∆H because, if the change is carried out at constant pressure, some energy is gained by the system as volume contracts.
19.9
The most negative value of ∆E will have the most energy released. (a) Compressing the spring: energy is added to the system: w > 0 Heating the spring: energy is added to the system: q > 0 (b) Expanding the spring: energy is removed from the system: w < 0 Cooling the spring: energy is removed from the system: q < 0 (c) Compressing the spring: energy is added to the system: w > 0 Cooling the spring: energy is removed from the system: q < 0 (d) Expanding the spring: energy is removed from the system: w < 0 Heating the spring: energy is added to the system: q > 0 In case (b), q < 0 and w < 0 so ∆E is the most negative
472
Chapter 19
19.10
19.11
A spontaneous change, in thermodynamic terms, is one for which the sign of ∆G is negative. It is a process that occurs by itself, without continued outside assistance. Kinetics plays no role in determining spontaneity. Student dependent answer. Examples of spontaneous changes may include: leaves falling from a tree, open soda cans losing their carbonation, food molding in the refrigerator, dorm rooms becoming messy. Examples of non–spontaneous changes may include: dorm rooms becoming neat and tidy, class notes becoming organized, leaves being raked into a pile.
19.12
Student dependent answer. For the examples given above all of the non–spontaneous changes are endothermic, they all require the surroundings expenditure of energy. The spontaneous changes are all exothermic except, perhaps, the dorm room becoming messy. This change may be either exothermic or endothermic.
19.13
A change that is characterized by a negative value for ∆H tends to be spontaneous. The only situation where this is not true is that in which the value of the product T∆S is sufficiently negative to make the quantity ∆G = ∆H – T∆S become positive.
19.14
Spontaneous processes tend to proceed from states of low probability to states of higher probability.
19.15
The ammonium and nitrate atoms are in a highly ordered geometry in the crystalline NH4NO3 sample. When NH4NO3 dissolves, the NH4+ and NO3– ions become randomly dispersed throughout the solvent. The increase in randomness that attends the dissolving of the solid is responsible for the process being favorable, or spontaneous, in spite of the fact that the enthalpy change is endothermic.
19.16
Entropy is a measure of the number of microstates of a system. An equivalent statement is that entropy is a measure of the statistical probability of a system.
19.17
Entropy is a measure of the randomness of a system, an increase in randomness corresponds to an increase in entropy. (a) An increase in temperature allows the atoms and molecules to move more rapidly, therefore there is an increase in entropy. (b) A decrease in volume limits the space that the molecules and atoms have to move in, therefore there is a decrease in entropy. (c) Changing a liquid to a solid limits the motion of the molecules; therefore there is a decrease in entropy. (d) Dissociating into individual atoms increases the entropy because there are more particles.
19.18
(a) (d)
19.19
The statistical probability of a system in a given state, relative to all the other states, is the same regardless of how the system happened to have been formed.
19.20
The entropy of the universe increases when a spontaneous event occurs.
19.21
A spontaneous event occurs when ∆G is negative. Since ∆G = ∆H – T∆S, even if the entropy is negative for a process, the enthalpy factor, ∆H may be negative enough at a given temperature to allow the overall change in free energy to be negative.
19.22
This is the statement that the entropy of a perfect crystalline solid at 0 K is equal to zero: S = 0 at 0 K.
19.23
The entropy of a mixture must be higher than that of two separate pure materials, because the mixture is guaranteed to have a higher degree of disorder. Said another way, a mixture is more disordered than either of its two separate components. Only a pure substance can have an entropy of zero, and then only at 0 K.
negative negative
(b) (e)
negative negative
(c) (f)
473
positive positive
Chapter 19
19.24
Entropy increases with increasing temperature because vibrations and movements within a solid lead to greater disorder at the higher temperatures. Melting especially produces more disorder and vaporization even more.
19.25
No, glass is an amorphous solid that is really a mixture of different substances so it is not a perfect crystalline solid and does not have an entropy value of 0 at 0K.
19.26
∆G = ∆H – T∆S
19.27
(a) (b) (c)
A change is spontaneous at all temperatures only if ∆H is negative and ∆S is positive. A change is spontaneous at low temperatures but not at high temperatures only if ∆H is negative and ∆S is negative. A change is spontaneous at high temperatures but not at low temperatures only if ∆H is positive and ∆S is positive.
19.28
A change is nonspontaneous regardless of the temperature, if ∆H is positive and ∆S is negative.
19.29
The value of ∆G is equal to the maximum amount of work that may be obtained from any process.
19.30
A reversible process is one in which the driving force of the process is nearly completely balanced by an opposing force. The situation is a bit esoteric, since no process can be run in a completely reversible manner. Nevertheless, the closer one obtains to reversibility, the more efficient the system becomes as a source of the maximum amount of useful work that can be achieved with the process.
19.31
The slower that the energy extraction is performed, the greater is the total amount of energy that can be obtained. This is the same as saying that the most energy is available from a process that occurs reversibly.
19.32
As with other processes that are not carried out in a reversible fashion, the energy is lost to the environment and becomes unavailable for use. In addition, much of the energy is lost as heat which is used to maintain our body temperature.
19.33
A truly reversible process would take forever to occur. Thus, if we can observe an event happening, it cannot be a truly reversible one.
19.34
At equilibrium, the value of ∆G is zero.
19.35
The process of bond breaking always has an associated value for ∆H that is positive. Also, since bond breaking increases disorder, it always has an associated value for ∆S that is positive. A process such as this, with a positive value for ∆H and a positive value for ∆S, becomes spontaneous at high temperatures.
19.36
Before the heat transfer, the molecules in the hot object vibrate and move more violently than do those of the cooler object. When in contact with one another, the objects transfer heat through collisions, and eventually, some of the kinetic energy of the molecules in the hot object is transferred to the molecules of the cool object. This process of energy transfer continues until the objects have the same temperature. The heat transfer is spontaneous because the scattering of kinetic energy among the molecules of both objects is a process with a positive value for its associated ∆S.
19.37
See Figure 19.15.
19.38
Although a reaction may have a favorable ∆G, and therefore be a spontaneous reaction in the thermodynamic sense of the word, the rate of reaction may be too slow at normal temperatures to be observed.
19.39
As the temperature is raised, °G' will become less negative, if ∆H° is negative and ∆S° is negative. Accordingly, less product will be present at equilibrium. 474
Chapter 19
19.40
∆G = ∆G° + RT lnQ
19.41
∆G° = –RT lnK
19.42
The natural log of 1 is zero so ∆G° = 0.
19.43
The amount of energy needed to break all the chemical bonds in one mole of gaseous molecules to give gaseous atoms.
19.44
It is easy to see why the conversion of a solid or liquid element to gaseous atoms is an endothermic process. In the case of elements which exist naturally as gases, most are polyatomic (the exception are the noble gases). To convert these elements to gaseous atoms will require an input of energy as bonds need to be broken.
19.45
Heat of formation is defined as the amount of energy needed to form the compound from its elements in their most stable state. C2(g) is not the naturally occurring state of carbon.
Review Problems 19.46
∆E = q + w = 0.450 kJ + 0.750 kJ = +1.200 kJ The overall process is endothermic, meaning that the internal energy of the system increases. Notice that both terms, q and w, contribute to the increase in internal energy of the system; the system gains heat (+q) and has work done on it (+w).
19.47
∆E = q + w –1785 J = 945 J + w w = –2730 J Since w is defined to be the work done on the system by the surroundings, then in this case, a negative amount of work is done on the system by the surroundings. The system, in fact, does work on the surroundings.
19.48
work = P × ∆V The total pressure caused by the hand pump: P = 30.0 lb/in2 Converting to atmospheres we get: P = 30.0 lb/in2 × 1 atm/14.7 lb/in2 = 2.04 atm Next we convert the volume change in units in3 to units L: 24.0 in3 × (2.54 cm/in)3 × 1 L/1000 cm3 = 0.393 L Hence P × ∆V = (2.04 atm)(0.393 L) = 0.802 L·atm 0.802 L·atm × 101.3 J/L·atm = 81 J
19.49
This is a reaction that produces 1 mol of a single gaseous product, CO2. Furthermore, this mole of gaseous product forms from nongaseous materials. The volume that will be occupied by this gas, once it is formed, can be found by application of Charles' Law: 22.4 L × 298 K/273 K = 24.5 L The work of gas expansion on forming the products is thus: W = –P∆V = –(1.00 atm)(24.5 L) = –24.5 L atm
19.50
We use the data supplied in Appendix C. 475
Chapter 19
(a)
3PbO(s) + 2NH3(g) 3Pb(s) + N2(g) + 3H2O(g) ∆H° = { ∆Hof [Pb(s)] + ∆Hof [N2(g)] + 3 ∆Hof [H2O(g)]}
– {3 ∆Hof [PbO(s)] + 2 ∆Hof NH3(g)]} ∆H° = {3 mol × (0 kJ/mol) + 1 mol × (0 kJ/mol) + 3 mol × (–241.8 kJ/mol)} – {3 mol × (–219.2 kJ/mol) + 2 mol × (–46.19 kJ/mol)} ∆H° = + 24.58 kJ ∆E = ∆H° – ∆nRT ∆E = 24.58 kJ – (+2 mol)(8.314 J/mol K)(10–3 kJ/J)(298 K) = 19.6 kJ (b)
NaOH(s) + HCl(g) NaCl(s) + H2O(l) ∆H° = { ∆Hof [NaCl(s)] + ∆Hf [H2O(l)]} – { ∆Hof [NaOH(s)] + ∆Hof [HCl(g)]} ∆H° = {1 mol × (–411.0 kJ/mol) + 1 mol × (–285.9 kJ/mol)} – {1 mol × (–426.8 kJ/mol) + 1 mol × (–92.3)} ∆H° = –178 kJ ∆E = ∆H° – ∆nRT ∆E = –178 kJ – (–1)(8.314 J/mol K)(10–3 kJ/J)(298 K) = –175 kJ
(c)
Al2O3(s) + 2Fe(s) Fe2O3(s) + 2Al(s) ∆H° = { ∆Hof [Fe2O3(s)] + 2 ∆Hof [Al(s)]} – { ∆Hof [ Al2O3(s)] + 2 ∆Hof [Fe(s)]} ∆H° = {1 mol × (–822.2 kJ/mol) + 2 mol ×(0 kJ/mol)} – {1 mol × (–1669.8 kJ/mol) + 2 mol × (0 kJ/mol)} ∆H° = 847.6 kJ ∆E = ∆H°, since the value of ∆n for this reaction is zero.
(d)
2CH4(g) C2H6(g) + H2(g) ∆H° = { ∆Hof [C2H6(g)] + ∆Hof [H2(g)]} – {2 ∆Hof [CH4(g)]} ∆H° = {1 mol × (–84.667 kJ/mol) + 1 mol × (0.0 kJ/mol)} – {2 mol × (–74.848 kJ/mol)} ∆H° = 65.029 kJ ∆E = ∆H°, since the value of ∆n for this reaction is zero.
19.51
We proceed as in the answer to Review Problem 19.50, using the data supplied in Appendix C. (a) 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) ∆H° = {2 ∆Hof [H2O(g)] + 4 ∆Hof [CO2(g)]} – {5 ∆Hof [O2(g)] + 2 ∆Hof [C2H2(g)]} ∆H° = {2 mol × (–241.8 kJ/mol) + 4 mol × (–393.5 kJ/mol)} – {5 mol × (0.0 kJ/mol) + 2 mol × (226.75 kJ/mol)} ∆H° = –2511.1 kJ
(b)
∆E = ∆H° – ∆nRT = –2511.1 kJ – (–1 mol)(8.314 J/mol K)(10–3 kJ/J)(298 K) ∆E = –2509 kJ C2H2(g) + 5N2O(g) 5N2(g) + H2O(g) + 2CO2(g) ∆H° = {5 ∆Hof [N2(g)] + ∆Hof [H2O(g)] + 2 ∆Hof [CO2(g)]} – {5 ∆Hof [N2O(g)] + ∆Hof [C2H2(g)]} ∆H° = {5 mol × (0.0 kJ/mol) + 1 mol × (–241.8 kJ/mol) + 2 mol × (–393.5 kJ/mol)} – {5 mol × (81.57 kJ/mol) + 1 mol × (226.75 kJ/mol)} ∆H° = –1663.4 kJ
476
Chapter 19
∆E = ∆H° – ∆nRT ∆E = –1663.4 kJ – (2 mol)(8.314 J/mol K)(10–3 kJ/J)(298 K) ∆E = –1668 kJ (c)
NH4Cl(s) NH3(g) + HCl(g) ∆H° = { ∆Hof [HCl(g)] + ∆Hof [NH3(g)]} – { ∆Hof [NH4Cl(s)]} ∆H° = {1 mol × (–92.3 kJ/mol) + 1 mol × (–46.19 kJ/mol)} – {1 mol × (–315.4 kJ/mol)} ∆H° = 176.9 kJ ∆E = ∆H° – ∆nRT ∆E = 176.9 kJ – (2 mol)(8.314 J/mol K)(10–3 kJ/J)(298 K) ∆E = 171.9 kJ
(d)
(CH3)2CO(l) + 4O2(g) 3CO2(g) + 3H2O(g) ∆H° = {3 ∆Hof [CO2(g)] + 3 ∆Hof [H2O(g)]} – { ∆Hof [(CH3)2CO(l)] + 4 ∆Hof [O2(g)]} ∆H° = {3 mol × (–393.5 kJ/mol) + 3 mol × (–241.8 kJ/mol)} – {1 mol × (–248.1 kJ/mol) + 4 × (0 kJ/mol)} ∆H° = –1657.8 kJ ∆E = ∆H° – ∆nRT ∆E = –1657.8 kJ – (2 mol)(8.314 J/mol K)(10–3 kJ/J)(298 K) ∆E = –1662.8 kJ
19.52
∆H = ∆E + ∆ngasRT ∆E = ∆H – ∆ngasRT ∆H = –163.14 kJ ∆ngas = 3 mol – 2 mol = 1 mol R = 8.314 J mol–1 K–1 T = 25 °C + 273 K = 298 K For 195 g N2O, first calculate the number of moles of N2O and then how many kJ of energy will be released for that many moles of N2O ∆E = –163.14 kJ – (1 mol)(8.314 × 10–3 kJ mol–1 K–1)(298 K) ∆E = –165.62 kJ 1mol N 2 O mol N2O = 195 g = 4.43 mol N2O 44.02 g N 2 O −165.62 kJ Amount of Energy = 4.43 mol N2O = –367 kJ 2 mol N 2 O For T = 212 °C T = 212 °C + 273 K = 485 K ∆E = –163.14 kJ – (1 mol)(8.314 × 10–3 kJ mol–1 K–1)(485 K) ∆E = –163.14 kJ –4.03 kJ ∆E217 °C = –167.17 kJ for 2 moles of N2O
For 195 g N2O, 4.43 mol N2O: −167.17 kJ Amount of Energy at 212 °C = 4.43 mol N2O = –370 kJ 2 mol N 2 O
19.53
C4H10(g) +
13 O2(g) 4CO2(g) + 5H2O(g) 2 477
Chapter 19
Find the moles of C4H10 at the first set of pressure, volume, and temperature data, then find the moles of the CO2 and H2O resulting from the reaction. Finally, find the new volume of the products. PV = nRT ( 2.00 atm )(10.0 L ) mol C4H10 = = 0.826 mol C4H10 ( 0.0821 L atm mol−1 K −1 ) ( 295 K ) 4 mol CO 2 mol CO2 = ( 0.826 mol C 4 H10 ) = 3.30 mol CO2 1 mol C 4 H10 5 mol H 2 O mol H2O = ( 0.826 mol C 4 H10 ) = 4.13 mol H2O 1 mol C 4 H10 total moles of product = 3.30 mol CO2 + 4.13 mol H2O = 7.43 mol ( 0.826 mol) 0.0821 L atm mol−1 K −1 ( 305 K ) initial volume = = 20.7 L 1.00 atm ( 7.43 mol) 0.0821 L atm mol−1 K −1 ( 305 K ) = 186 L final volume = 1.00 atm w = –P∆V w = –(1.00 atm)(186 L – 20.7 L) = 165 L atm 101.325 J 1 kJ In kilojoules: 165 L atm = 16.7 kJ 1 L atm 1000 J
(
)
(
19.54
19.55
)
In general, we have the equation: ∆H° = (sum ∆Hf [products]) – (sum ∆Hf [reactants]) (a)
∆H° = { ∆Hof [CaCO3(s)]} – { ∆Hof [CO2(g)] + ∆Hof [CaO(s)]} ∆H° = {1 mol × (–1207 kJ/mol)} – {1 mol × (–393.5 kJ/mol) + 1 mol × (–635.5 kJ/mol)} ∆H° = –178 kJ ∴ favored.
(b)
∆H° ={ ∆Hof [C2H6(g)]} – { ∆Hof [C2H2(g)] + 2 ∆Hof [H2(g)]} ∆H° = {1 mol × (–84.5 kJ/mol)} – {1 mol × (226.75 kJ/mol) + 2 mol × (0.0 kJ/mol)} ∆H° = –311 kJ ∴ favored.
(c)
∆H° = { ∆Hof [Fe2O3(s)] + 3 ∆Hof [Ca(s)]} – {2 ∆Hof [Fe(s)] + 3 ∆Hof [CaO(s)]} ∆H° = {1 mol × (–822.2 kJ/mol) + 3 mol × (0.0 kJ/mol)} – {2 mol × (0.0 kJ/mol) + 3 mol × (–635.5 kJ/mol)} ∆H° = +1084.3 kJ ∴ not favorable from the standpoint of enthalpy alone.
a)
∆H° = { ∆Hof [HCl(g)] + ∆Hof [NH3(g)]} – { ∆Hof [NH4Cl(s)]} ∆H° = {1 mol × (–92.3 kJ/mol) + 1 mol × (–46.19 kJ/mol)} – {1 mol × (–315.4 kJ/mol)} ∆H° = 176.9 kJ ∴ not favorable from the standpoint of enthalpy.
(b)
∆H° = {2 ∆Hof [H2O(g)] + 4 ∆Hof [CO2(g)]} – {5 ∆Hof [O2(g)] + 2 ∆Hof [C2H2(g)]} ∆H° = {2 mol × (–241.8 kJ/mol) + 4 mol × (–393.5 kJ/mol)} – {5 mol × (0.0 kJ/mol) + 2 mol × (226.75 kJ/mol)} ∆H° = –2512 kJ ∴ favored from the standpoint of enthalpy alone.
(c)
∆H° = {5 ∆Hof [N2(g)] + ∆Hof [H2O(g)] + 2 ∆Hof [CO2(g)]} – {5 ∆Hof [N2O(g)] + ∆Hof [C2H2(g)]} ∆H° = {5 mol × (0.0 kJ/mol) + 1 mol × (–241.8 kJ/mol) + 2 mol × (–393.5 kJ/mol)} – {5 mol × (81.57 kJ/mol) 478
Chapter 19
+ 1 mol × (226.75 kJ/mol)} ∆H° = –1663 kJ ∴ favorable from the standpoint of enthalpy. 19.56
2N2O(g) 2N2(g) + O2(g) The factors needed to be consider in order to determine the sign of ∆S are (1) the number of moles of products versus reactants in this case the number of moles increases (2) the state of the products versus reactants both products and reactants are gases (3) the complexity of the molecules N2O is more complex than either N2 or O2 ∆S is expected to be positive.
19.57
2HI(g) H2(g) + I2(g) The factors needed to be consider in order to determine the sign of ∆S are (1) the number of moles of products versus reactants in this case the number of moles are the same (2) the state of the products versus reactants both products and reactants are gases (3) the complexity of the molecules HI is more complex than either H2 or I2 ∆S is expected to be negative.
19.58
(a) (b) (c) (d)
negative – since the number of moles of gaseous material decreases. negative – since the number of moles of gaseous material decreases. negative – since the number of moles of gas decreases. positive – since a gas appears where there formerly was none.
19.59
(a) (b) (c)
∆S is positive since randomness in a gas is higher than that in a solid. ∆S is negative. There are fewer moles of gases among the products. ∆S is negative since gaseous material (which is highly random) is replaced by a solid (which is highly ordered). ∆S is negative since the relatively random liquid reactant disappears in a process that makes only a solid.
(d)
19.60
∆S° = (sum S°[products]) – (sum S°[reactants]) (a)
∆S° = {2S°[NH3(g)]} – {3S°[H2(g)] + S°[N2(g)]} ∆S° = {2 mol × (192.5 J mol–1 K–1)} – {3 mol × (130.6 J mol–1 K–1) + 1 mol × (191.5 J mol–1 K–1)} ∆S° = –198.3 J/K ∴ not spontaneous from the standpoint of entropy.
(b)
∆S° = {S°[CH3OH(l)]} – {2S°[H2(g)] + S°[CO(g)]} ∆S° = {1 mol × (126.8 J mol–1 K–1)} – {2 mol × (130.6 J mol–1 K–1) + 1 mol × (197.9 J mol–1 K–1)} ∆S° = –332.3 J/K ∴ not favored from the standpoint of entropy alone.
(c)
∆S° = {6S°[H2O(g)] + 4S°[CO2(g)]} – {7S°[O2(g)] + 2S°[C2H6(g)]} ∆S° = {6 mol × (188.7 J mol–1 K–1) + 4 mol × (213.6 J mol–1 K–1)} – {7 mol × (205.0 J mol–1 K–1) + 2 mol × (229.5 J mol–1 K–1)} ∆S° = +92.6 J/K ∴ favorable from the standpoint of entropy alone.
(d)
∆S° = {2S°[H2O(l)] + S°[CaSO4(s)]} – {S°[H2SO4(l)] + S°[Ca(OH)2(s)]}
479
Chapter 19
∆S° = {2 mol × (69.96 J mol–1 K–1) + 1 mol × (107 J mol–1 K–1)} – {1 mol × (157 J mol–1 K–1) + 1 mol × (76.1 J mol–1 K–1)} ∆S° = +14 J/K ∴ favorable from the standpoint of entropy alone. (e)
19.61
19.62
∆S° = {2S°[N2(g)] + S°[SO2(g)]} – {2S°[N2O(g)] + S°[S(s)]} ∆S° = {2 mol × (191.5 J mol–1 K–1) + 1 mol × (248 J mol–1 K–1)} – {2 mol × (220.0 J mol–1 K–1) + 1 mol × (31.9 J mol–1 K–1)} ∆S° = +159 J/K ∴ favorable from the standpoint of entropy alone.
∆S° = (sum S°[products]) – (sum S°[reactants]) (a)
∆S° = {S°[AgCl(s)]} – {1/2S°[Cl2(g)] + S°[Ag(s)]} ∆S° = {1 mol × (96.2 J mol–1 K–1)} – {1/2 mol × (223.0 J mol–1 K–1) + 1 mol × (42.55 J mol–1 K–1)} ∆S° = –57.9 J/K
(b)
∆S° = {S°[H2O(g)]} – {1/2S°[O2(g)] + S°[H2(g)]} ∆S° = {1 mol × (188.7 J mol–1 K–1)} – {1/2 mol × (205.0 J mol–1 K–1) + 1 mol × (130.6 J mol–1 K–1)} ∆S° = –44.4 J/K
(c)
∆S° = {S°[H2O(l)]} – {1/2S°[O2(g)] + S°[H2(g)]} ∆S° = {1 mol × (69.96 J mol–1 K–1)} – {1/2 mol × (205.0 J mol–1 K–1) + 1 mol × (130.6 J mol–1 K–1)} ∆S° = –163.1 J/K
(d)
∆S° = {S°[CO2(g)] + S°[H2O(g)] + S°[CaSO4(s)]} – {S°[CaCO3(s)] + S°[H2SO4(l)]} ∆S° = {1 mol × (213.6 J mol–1 K–1) + 1 mol × (188.7 J mol–1 K–1) + 1 mol × (107 J mol–1 K–1)} – {1 mol × (92.9 J mol–1 K–1) + 1 mol × (157 J mol–1 K–1)} ∆S° = +259 J/K
(e)
∆S° = {S°[NH4Cl(s)]} – {S°[HCl(g)] + S°[NH3(g)]} ∆S° = {1 mol × (94.6 J mol–1 K–1)} – {1 mol × (186.7 J mol–1 K–1) + 1 mol × (192.5 J mol–1 K–1)} ∆S° = –284.6 J/K
The entropy change that is designated ∆Sf is that which corresponds to the reaction in which one mole of a substance is formed from elements in their standard states. Since the value is understood to correspond to the reaction forming one mole of a single pure substance, the units may be written either J K–1 or J mol–1 K–1.
(a)
2C(s) + 3H2(g) C2H6(g) 5.69 130.6 229.5 ∆S° = 229.5 – [2(5.69) + 3(130.6)] 229.5 – 11.38 – 391.8 =– 173.7 J/K.mol
(b)
3C(s) + ½ O2(g) + 3 H2(g) → (CH3)2CO(l) 5.69 205.0 130.6 200.4 ∆S° = 200.4 – [3(5.69) + ½ (205.0) + 3 (130.6)] 200.4 – 17.07 – 102.5 = 391.8 480
Chapter 19
= –311.0 J/K.mol (c)
19.63
19.64
19.65
19.66
Cu(s) + S(s) + 9 2 O2(g) + 5 H2(g) → CuSO4.5H2O 33.15 31.9 205.0 130.6 300.4 ∆S° = 300.4 – [1(33.15) + 1(31.9) + 9 2 (205.0) +5(130.6)] 300.4 – 33.15 –31.9 – 922.5 – 653.0 ∆S° = –1340.2 J/K.mol
The entropy change that is designated ∆Sf is that which corresponds to the reaction in which one mole of a substance is formed from elements in their standard states. Since the value is understood to correspond to the reaction forming one mole of a single pure substance, the units may be written either J K–1 or J mol–1 K–1. (a)
Na(s) + ½ O2(g) + ½ H2(g) → NaOH(s) 51.0 205.0 130.6 64.18 ∆S° = 64.18 – [51.0 + (½)(205.0) + (½) 130.6] 64.18 – 51.0 – 102.5 – 65.3 ∆S° = – 154.62 J/K.mol
(b)
2Fe(s) + 3 2 O2(g) → Fe2O3(s) 27 205.0 90.0 ∆S° = 90.0 – [(2)(27) + ( 3 2 ) (205)] 90.0 – 54 – 307.5 ∆S° = –271.5 J/K.mol
(c)
Ag(s) × ½ N2(g) + 3 2 O2(g) → AgNo3(s) 42.55 191.5 205.0 141.0 ∆S° = 141.0 – [42.55 + (½)(191.5) + ( 3 2 )(205.0)] = 141.0 – 42.55 – 95.8 – 307.5 ∆S° = – 304.85 J/K.mol
∆S° = (sum S°[products]) – (sum S°[reactants]) ∆S° = {2S°[HNO3(l)] + S°[NO(g)]} – {3S°[NO2(g)] + S°[H2O(l)]} ∆S° = {2 mol × (155.6 J mol–1 K–1) + 1 mol × (210.6 J mol–1 K–1)} – {3 mol × (240.5 J mol–1 K–1) + 1 mol × (69.96 J mol–1 K–1)} ∆S° = –269.7 J/K ∆S° = (sum S°[products]) – (sum S°[reactants]) ∆S° = {S°[HC2H3O2(l)] + S°[H2O(l)]} – {S°[C2H5OH(l)] + S°[O2(g)]} ∆S° = {1 mol × (160 J mol–1 K–1) + 1 mol × (69.96 J mol–1 K–1)} – {1 mol × (161 J mol–1 K–1) + 1 mol × (205.0 J mol–1 K–1)} ∆S° = –136 J/K The quantity ∆G f applies to the equation in which one mole of pure phosgene is produced from the naturally occurring forms of the elements: C(s) + 1/2O2(g) + Cl2(g) COCl2(g), ∆G f = ? We can determine if we can find values for ∆Hf and ∆Sf , because: s ∆G° = ∆H° – T∆S° The value of ∆Sf is determined using S° for phosgene in the following way: ∆Sof = {S°[COCl2(g)]} – {S°[C(s)] + 1/2S°[O2(g)] + S°[Cl2(g)]}
481
Chapter 19
∆Sof = {1 mol × (284 J mol–1 K–1)} – {1 mol × (5.69 J mol–1 K–1) + 1/2 mol × (205.0 J mol–1 K–1) + 1 mol × (223.0 J mol–1 K–1)} o ∆Sf = –47 J mol–1 K–1 or –47 J/K ∆G of = ∆Hof – T ∆Sof = –223 kJ/mol – (298 K)(–0.047 kJ/mol K) = –209 kJ/mol 19.67
∆S° = (sum S°[products]) – (sum S°[reactants]) 2Al(s) + 3/2O2(g) Al2O3(s) ∆Sof = {S°[Al2O3(s)]} – {2S°[Al(s)] + 3/2S°[O2(g)]} ∆Sof = {1 mol × (51.0 J mol–1 K–1)} – {2 mol × (28.3 J mol–1 K–1) + 3/2 mol × (205.0 J mol–1 K–1)} ∆Sof = –313.1 J/K or –313.1 J mol–1 K–1 = –0.3131 kJ mol–1 K–1 ∆G f = ∆Hf – T ∆Sof = –1669.8 kJ/mol – (298 K)(–0.3131 kJ mol–1 K–1) ∆G f = –1576.5 kJ/mol This value agrees well with the value listed in Appendix C2.
19.68
∆G° = (sum ∆G of [products]) – (sum ∆G of [reactants]) (a)
∆G° = { ∆G of [H2SO4(l)]} – { ∆G of [H2O(l)] + ∆G of [SO3(g)]} ∆G° = {1 mol × (–689.9 kJ/mol)} – {1 mol × (–237.2 kJ/mol) + 1 mol × (–370 kJ/mol)} ∆G° = –82.3 kJ
(b)
∆G° = {2 ∆G of [NH3(g)] + ∆G of [H2O(l)] + ∆G of [CaCl2(s)]} – { ∆G of [CaO(s)] + 2 ∆G of [NH4Cl(s)]} ∆G° = {2 mol × (–16.7 kJ/mol) + 1 mol × (–237.2 kJ/mol) + 1 mol × (–750.2 kJ/mol)} – {1 mol × (–604.2 kJ/mol) + 2 mol × (–203.9 kJ/mol)} ∆G° = –8.8 kJ
(c)
19.69
∆G° = { ∆G of [H2SO4(l)] + ∆G of [CaCl2(s)]} – { ∆G of [CaSO4(s)] + ∆G of [HCl(g)]} ∆G° = {1 mol × (–689.9 kJ/mol) + 1 mol × (–750.2 kJ/mol)} – {1 mol × (–1320.3 kJ/mol) + 2 mol × (–95.27 kJ/mol)} ∆G° = +70.7 kJ
∆G° = (sum ∆G of [products]) – (sum ∆G of [reactants]) (a)
∆G° = { ∆G of [H2O(g)] + ∆G of [CaCl2(s)]} – {2 ∆G of [HCl(g)] + ∆G of [CaO(s)]} ∆G° = {1 mol × (–228.6 kJ/mol) + 1 mol × (–750.2 kJ/mol)} – {2 mol × (–95.27 kJ/mol) + 1 mol × (–604.2 kJ/mol)} ∆G° = –184.1 kJ
482
Chapter 19
19.70
(b)
∆G° = {2 ∆G of [Ag(s)] + ∆G of [CaCl2(s)]} – {2 ∆G of [AgCl(s)] + ∆G of [Ca(s)]} ∆G° = {2 mol × (0.0 kJ/mol) + 1 mol × (–750.2 kJ/mol)} – {2 mol × (–109.7 kJ/mol) + 1 mol × (0.0 kJ/mol)} ∆G° = –530.8 kJ
(c)
∆G° = { ∆G of [NO(l)] + 2 ∆G of [HNO3(l)]} – {3 ∆G of [NO2(l)] + ∆G of [H2O(l)]} ∆G° = {1 mol × (86.69 kJ/mol) + 2 mol × (–79.91 kJ/mol)} – {3 mol × (51.84 kJ/mol) + 1 mol × (–237.2 kJ/mol)} ∆G° = +8.6 kJ
Multiply the reverse of the second equation by 2 (remembering to multiply the associated free energy change by –2), and add the result to the first equation: 4NO(g) 2N2O(g) + O2(g), 2NO(g) + O2(g) 2NO2(g),
∆G° = –139.56 kJ ∆G° = -69.70 kJ
(1) (2)
Reverse reaction (1) double reaction (2) ∆G° = 139.56 kJ + 2 x (-69.70) kJ 2N2O(g) + 3O2(g) 4NO2(g), 19.71
19.72
∆G° = +0.16 kJ
Add the reverse of the first equation to the second equation plus twice the third equation: CO(NH2)2(s) + 2NH4Cl(s) COCl2(g) + 4NH3(g), COCl2(g) + H2O(l) CO2(g) + 2HCl(g), 2NH3(g) + 2HCl(g) 2NH4Cl(s),
∆G° = +332.0 kJ ∆G° = –141.8 kJ ∆G° = –183.9 kJ
CO(NH2)2(s) + H2O(l) 2NH3(g) + CO2(g),
∆G° = +6.3 kJ
The maximum work obtainable from a reaction is equal in magnitude to the value of ∆G for the reaction. Thus, we need only determine ∆G° for the process: ∆G° = (sum ∆G of [products]) – (sum ∆G of [reactants]) ∆G° = {3 ∆G of [H2O(g)] + 2 ∆G of [CO2(g)]} – {3 ∆G of [O2(g)] + ∆G of [C2H5OH(l)]} ∆G° = {3 mol × (–228.6 kJ/mol) + 2 mol × (–394.4 kJ/mol)} – {3 mol × (0.0 kJ/mol) + 1 mol × (–174.8 kJ/mol)} ∆G° = –1299.8 Kj Since this is for 2 mols C2H5OH we must multiply the ∆G° value by 2. –1299.8 kJ (2) = –2599.6 kJ
19.73
We must first determine ∆G° for the reaction: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ∆G° = (sum ∆G of [products]) – (sum ∆G of [reactants]) ∆G° = { ∆G of [CO2(g)] + 2 ∆G of [H2O(g)]} – { ∆G of [CH4(g)] + 2 ∆G of [O2(g)]} ∆G° = {1 mol × (–394.4 kJ/mol) + 2 mol × (–228.6 kJ/mol)} – {1 mol × (–50.79 kJ/mol) + 2 mol × (0.0 kJ/mol)} ∆G° = –800.8 kJ/mol Next, we determine the amount of work available from the combustion of 64.0 g of CH4: 483
Chapter 19
19.74
1 mol CH 4 −800.8 kJ 3 kJ = ( 64.0 g CH 4 ) = − 3.19 × 10 kJ 16.043 g CH 1 mol CH 4 4 At equilibrium, ∆G = 0 = ∆H – T∆S Teq = ∆H/∆S, and assuming that ∆S is independent of temperature, we have: Teq = (31.4 × 103 J mol–1) ÷ (94.2 J mol–1 K–1) = 333 K
19.75
At equilibrium, ∆G = 0 = ∆H – T∆S Teq = ∆H/∆S, and assuming that ∆S and ∆H are independent of temperature, we have: Teq = (10.0 × 103 J mol–1) ÷ (9.50 J mol–1 K–1) = 1.05 × 103 K or 780 °C
19.76
At equilibrium, ∆G = 0 = ∆H – T∆S Thus ∆H = T∆S, and if we assume that both ∆H and ∆S are independent of temperature, we have: ∆S = ∆H/Teq = (37.7 × 103 J/mol) ÷ (99.3 + 273.15 K) ∆S = 101 J mol–1 K–1
19.77
We proceed as in the answer to Review Problem 19.76: ∆S = ∆H/Teq = (31.9 × 103 J/mol) ÷ (56.2 + 273.15 K) ∆S = 96.9 J mol–1 K–1
19.78
The reaction is spontaneous if its associated value for ∆G° is negative. ∆G° = (sum ∆G of [products]) – (sum ∆G of [reactants]) ∆G° = { ∆G of [HC2H3O2(l)] + ∆G of [H2O(l)] + ∆G of [NO(g)] + ∆G of [NO2(g)]} – { ∆G of [C2H4(g)] + ∆G of [HNO3(l)]} ∆G° = {1 mol × (–392.5 kJ/mol) + 1 mol × (–237.2 kJ/mol) + 1 mol × (86.69 kJ/mol) + 1 mol × (51.84 kJ/mol)} – {1 mol × (68.12 kJ/mol) + 1 mol × (–79.91 kJ/mol)} ∆G° = –479.4 kJ Yes, the reaction is spontaneous.
19.79 We first balance each equation, and then calculate a value of ∆G°. If ∆G° is a negative number, then the reaction is spontaneous. (a)
3PbO(s) + 2NH3(g) 3Pb(s) + N2(g) + 3H2O(g) ∆G° = {3 ∆G of [Pb(s)] + ∆G of [N2(g)] + 3 ∆G of [H2O(g)]} – {3 ∆G of [PbO(s)] + 2 ∆G of [NH3(g)]} ∆G° = {3 mol × (0.0 kJ/mol) + 1 mol × (0.0 kJ/mol) + 3 mol × (–228.6 kJ/mol)} – {3 mol × (–189.3 kJ/mol) + 2 mol × (–16.7 kJ/mol)} ∆G° = –84.5 kJ ∴ the reaction is spontaneous.
(b)
NaOH(s) + HCl(g) NaCl(s) + H2O(l) ∆G° = { ∆G f [NaCl(s)] + ∆G f [H2O(l)]} – { ∆G f [NaOH(s)]+ ∆G f [HCl(g)]} ∆G° = {1 mol × (–384.0 kJ/mol) + 1 mol × (–237.2 kJ/mol)} – {1 mol × (–382 kJ/mol) + 1 mol × (–95.27)} ∆G° = –144 kJ ∴ the reaction is spontaneous.
484
Chapter 19
19.80
19.81
(c)
Al2O3(s) + 2Fe(s) Fe2O3(s) + 2Al(s) ∆G° = { ∆G of [Fe2O3(s)] + 2 ∆G of [Al(s)]} – { ∆G of [Al2O3(s)] + 2 ∆G of [Fe(s)]} ∆G° = {1 mol × (–741.0 kJ/mol) + 2 mol × (0.0 kJ/mol)} – {1 mol × (–1576.4 kJ/mol) + 2 mol × (0.0 kJ/mol)} ∆G° = +835.4 k ∴ the reaction is not spontaneous.
(d)
2CH4(g) C2H6(g) + H2(g) ∆G° = { ∆G of [C2H6(g)] + ∆G of [H2(g)]} – {2 ∆G of [CH4(g)]} ∆G° = {1 mol × (–32.9 kJ/mol) + 1 mol × (0.0 kJ/mol)} – {2 mol × (–50.79 kJ/mol)} ∆G° = +68.7 kJ ∴ the reaction is not spontaneous.
PCl3(g) +
1 O 2( g ) 2
→ POCl3(g)
(a)
∆G° = {1 mol × ∆G of [POCl3(g)]} – {1 mol × ∆G of [PCl3(g)] + 1/2mol× ∆G of [O2(g)]} ∆G° = {1 mol × (–512.9kJ/mol)} – {1 mol × (–267.8 kJ/mol) + 1/2 mol × (0 kJ/mol)} ∆G° = –245.1 kJ = –2.45 × 105 J –2.45 × 105 J = –RTlnKp = –(8.314 J/K mol)(298 K) × lnKp lnKp = 98.9 and Kp = 9.20 x 1042
(b)
∆G° = {1 mol × ∆G of [SO2(g)] + 1mol × ∆G of [O2(g)]} – {1 mol × ∆G of [SO3(g)]} ∆G° = {1 mol × (–300.4 kJ/mol) + 1 mol × (0 kJ/mol)} – {1 mol × (–370.4 kJ/mol)} ∆G° = 70 kJ = 7. 0 × 104 J 7.0 × 104 J = –RTlnKp = –(8.314 J/K mol)(298 K) × lnKp lnKp = 28.25 and Kp = 5.37 × 10–13
(a)
∆G° = {2 × ∆G f [NO(g)] + 2 × ∆G f [H2O(g)]} – {1 × ∆G f [N2H4(g)] + 2 × ∆G f [O2(g)]} ∆G° = {2 mol × (86.69 kJ/mol) + 2 mol × (–228.6 kJ/mol)} – {1 mol × (159.3 kJ/mol) + 2 mol × (0.0 kJ/mol)} ∆G° = – 443.1 kJ = – 4.431 × 105 J − 4.431× 105 J = –RTlnKp = –(8.314 J/K mol)(298 K) × lnKp lnKp = 178.8 and Kp = 4.728 × 1077
(b)
The ∆G of for H2O2(g) is not given in Appendix C. The value was found in the Handbook of Chemistry and Physics, 84th Edition, 2003 – 2004, David R. Lide, Ed., CRC Press, New York. ∆G° = {2 × ∆G of [NO2(g)] + 8 × ∆G of [H2O(g)]} – {1 × ∆G of [N2H4(g)] + 6 × ∆G of [H2O2(g)]} ∆G° = [2 mol × (51.84 kJ/mol) + 8 mol × (–228.6 kJ/mol)] – [1 mol × (159.3 kJ/mol) + 6 mol × (–105.6 kJ/mol)] ∆G° = –1250.8 kJ = –1.2508 × 106 J –1.2508 × 106 J = –RTlnKp = –(8.314 J/K mol)(298 K) × lnKp lnKp = 5.048 × 102 ∴ Kp = 1.70 × 10219 This is rocket fuel!
485
Chapter 19
19.82
∆G° = –RT ln Kp –9.67 × 103 J = –(8.314 J/K mol)(1273 K) × ln Kp ln Kp = 0.914 ∴ Kp = 2.49 N 2 O O 2 = ( 0.150 )( 0.350 ) = 0.66 Q= NO 2 [ NO ] ( 0.200)( 0.40) Since the value of Q is less than the value of K, the system is not at equilibrium and must shift to the right to reach equilibrium.
19.83
79.8 × 103 J = –(8.314 J/K mol)(673 K) × ln Kp ln Kp = –14.3 ∴ Kp = 6.40 × 10–7 We start by calculating the reaction quotient, Q. Be sure to determine the pressure of the gases using the ideal gas law.
(
)(
)
3.8 × 10−3 mol 0.0821 L atm ( 673 K ) mol K 1.50 L Q= L atm L atm ( 0.040 mol ) 0.0821 673 K ) ( 0.022 mol ) 0.0821 mol mol K ( K 1.50 L 1.50 L
(
)
(
) ( 673 K )
= 0.117
Since the value of Q is larger than the value of Kp, the system must shift to the left in order to reach equilibrium.
19.84
∆G° = –RT ln Kp –50.79 × 103 J = –(8.314 J K–1 mol–1)(298 K) × ln Kp ln Kp = 20.50 Taking the exponential of both sides of this equation gives: Kp = 8.000 × 108 This is a favorable reaction, since the equilibrium lies far to the side favoring products and is worth studying as a method for methane production.
19.85
T = 37 + 273 = 310 K ∆G°T = –RT ln Kc –33 × 103 J = –(8.314 J K–1 mol–1)(310 K) × ln Kc ln Kc = 13 Kc = 3.6 × 105
19.86
If ∆G° = 0, Kc = 1. If we start with pure products, the value of Q will be infinite (there are zero reactants) and, since Q > Kc, the equilibrium will shift towards the reactants, i.e., the pure products will decompose to their reactants.
19.87
∆G°500 = –RT ln Kp ∆G°500 = –(8.314 J K–1 mol–1)(500 K) ln (6.25 × 10–3) ∆G°500 = 2.11 × 104 J = 21.1 kJ
19.88
This requires the breaking of three N–H single bonds: NH3 N + 3H
486
Chapter 19
The enthalpy of atomization of NH3 is thus three times the average N–H single bond energy: 3 × 388 kJ/mol = 1.16 × 103 kJ/mol 19.89
The energy released during the formation of 1 mol of acetone is equal to the sum of all of the bond energies in the molecule: for the eight C—H bonds: 412 kJ/mol × 8 mol C—H bonds for the two C—C bonds: 348 kJ/mol × 2 mol C—C bonds for the one C=O bond: 743 kJ/mol × 1 mol C=O bonds for the two C–O bonds: 360 kJ/mol × 2 mol C–O bonds Adding the above contributions we get 5.46 × 104 kJ released per mole of acetone formed.
19.90
The heat of formation for ethanol vapor describes the following change: 2C(s) + 3H2(g) +1/2O2(g) C2H5OH(g) This can be arrived at by adding the following thermochemical equations, using data from Table 19.3: 3H2(g) 6H(g) ∆H1° = (6)217.89 kJ = 1307.34 kJ ∆H2° = (2)716.67 kJ = 1,433.34 kJ 2C(s) 2C(g) 1/2O2(g) O(g) ∆H3° = (1)249.17 kJ = 249.17 kJ 6H(g) + 2C(g) + O(g) C2H5OH(g) ∆H4° = x 3H2(g) + 2C(s) + 1/2O2(g) C2H5OH(g)
∆Hf° = (2989.85 + x) kJ
Since ∆Hf° is given as –235.3 kJ… –235.3 = 2989.85 + x x = –3225.2 kJ ∆H°atom is the reverse reaction, so the sign will change: ∆H°atom = 3225.2 kJ The sum of all the bond energies in the molecule should be equal to the atomization energy: ∆H°atom = 1(C–C bond) + 5(C–H bonds) + 1(O–H bond) + 1(C–O bond) We use bond energy values from Table 19.4: 3225.2 kJ = 1(348 kJ) + 5(412 kJ) + 1(463 kJ) + 1(C–O bond) C–O bond energy = 354 kJ/mol 19.91
∆Hf [C2H4(g)] refers to the enthalpy change under standard conditions for the following reaction: 2Cgraphite + 2H2(g) C2H4(g),
∆Hf [C2H4(g)] = 52.284 kJ/mol
We can arrive at this net reaction in an equivalent way, namely, by vaporizing all of the necessary elements to give gaseous atoms, and then allowing the gaseous atoms to form all of the appropriate bonds. The overall enthalpy of formation by this route is numerically equal to that for the above reaction, and, conveniently, the enthalpy changes for each step are available in either Table 19.3 or Table 19.4: ∆Hf = sum( ∆Hf [gaseous atoms]) – sum(average bond energies in the molecule) ∆Hf [C2H4(g)] = 52.284 kJ/mol = [2 × 716.7 + 4 × 218.0] – [4 × 412 + C=C] from which we can calculate the C=C bond energy: 605 kJ/mol.
487
Chapter 19
19.92
There are two C=S double bonds to be considered: ∆Hf = sum( ∆Hf [gaseous atoms]) – sum(average bond energies in the molecule) ∆Hf [CS2(g)] = 115.3 kJ/mol = [716.67 + 2 × 276.98] – [2 × C=S] The C=S double bond energy is therefore given by the equation: C=S = –(115.3 – 716.67 – 2 × 276.98) ÷ 2 = 577.7 kJ/mol
19.93
∆Hf = sum( ∆Hf [gaseous atoms]) – sum(average bond energies in the molecule) ∆Hf [H2S(g)] = –20.15 kJ/mol = [277.0 + 2 × 218.0] – [2 × H–S] H–S = (20.15 + 277.0 + 2 × 218.0) ÷ 2 = 366.6 kJ/mol
19.94
There are six S—F bonds in the molecule: ∆Hf = sum( ∆Hf [gaseous atoms]) – sum(average bond energies in the molecule) ∆Hf [SF6(g)] = –1096 kJ/mol = [277.0 + 6 × 79.14] – [6 × S—F] S—F = (1096 + 277.0 + 6 × 79.14) ÷ 6 = 308.0 kJ/mol
19.95
∆Hf = sum( ∆Hf [gaseous atoms]) – sum(average bond energies in the molecule) ∆Hf [SF4(g)] = [277.0 + 4 × 79.14] – [4 × 308.0] = –638.4 kJ/mol The % difference is [(718.4 – 638.4) ÷ 718.4] × 100% = 11 %
19.96
∆Hf = sum( ∆Hf [gaseous atoms]) – sum(average bond energies in the molecule) ∆Hf [C2H2(g)] = [2 × 716.7 + 2 × 218.0] – [2 × 412 + 960] = 85 kJ/mol
19.97
∆Hf = sum( ∆Hf [gaseous atoms]) – sum(average bond energies in the molecule) ∆Hf [CCl4(g)] = [716.67 + 4 × 121.47] – [4 × 338] = –149 kJ/mol
19.98
The heat of formation of CF4 should be more exothermic than that of CCl4 because more energy is released on formation of a C—F bond than on formation of a C—Cl bond. Also, less energy is needed to form gaseous F atoms than to form gaseous Cl atoms.
19.99
The computed value for ∆Hf for benzene is likely to be larger than the experimentally measured value. The reason for the difference is that the computation neglects the stabilization provided by the high degree of conjugation/resonance.
Additional Exercises 19.100 P(atm) × ∆V(L) = 1 L atm of work Substituting the given information: P × ∆V = 101,325 Pa × 1 dm3 Now it is true that 1 dm = 0.1 m, so: 1 dm3 = (0.1 m)3 = 1 × 10–3 m3 Also, 1 Pa = 1 N/m2 Thus we have: P(Pa) × ∆V(dm3) = (101,325 N/m2)(1 × 10–3 m3) = 101.325 N·m 101.325 N·m × 1 J/1 N·m = 101.325 J or 1 L atm ×
8.314 J mol−1 K −1 0.0821 L atm mol−1 K −1
= 101.325 J
488
Chapter 19
19.101 w = –P∆V P = 1 atm ∆V = 8.00 L – 4.00 L = 4.00 L w = –(1 atm)(4.00 L) = –4.00 L atm In joules: 101.325 J –4.00 L atm = –405 J 1 L atm 19.102 Under reversible conditions, the expansion of an ideal gas produces the maximum amount of work that can be obtained from the change. Such a situation cannot be accomplished by any real process. In an ideal gas, the gas particles are non-interacting. In a real gas, the interaction between the individual particles must be overcome so that the gas may expand. It requires an input of energy to overcome these attractive forces. 19.103 The expansion of a real gas is not reversible, and not all of the free energy change for the process is converted to work as the gas expands. Some of the energy must be used to overcome the intermolecular forces of attraction that exist between molecules. In an ideal gas, the particles are assumed to be noninteracting and this energy loss is not a concern. See 19.102 19.104 See 19.102 and 19.101 w= –P∆V = (–2.0 atm)(30 L – 12 L) = –36 L atm If the temperature is constant, then heat must be absorbed to balance the amount of energy expended as work and the amount of heat absorbed must be equal to the amount of work expended. q = 36 L atm 19.105 We can calculate the work of the expanding gas (P∆V) if we can calculate the change in volume ∆V. Since the initial volume is given (4.50 L), we need only to calculate the final volume. For this, it is first necessary to determine the value of n, the number of moles of gas.
n=
PV (3.50 atm)(4.50 L) = = 0.644 mol L atm RT 0.0821 mol (298 K) K
V2 =
(
)
(
(0.644 mol) 0.0821 nRT = P2 1 atm
L atm mol K
) (298 K) = 15.8 L
∆V = (15.8 L − 4.50 L) = 11.3 L The work of gas expansion against a constant pressure of 1 atm is then given by the quantity –P∆V: w = – (1 atm)(11.3 L) = –11.3 L atm 19.106 As in exercise 19.105, we solve for the volume after the first expansion to a pressure of 2 atm:
V2 =
(
(0.817 mol) 0.0821 nRT = P2 2 atm
L atm mol K
) (298 K)
= 10.0 L
The work after the first stage of expansion is therefore: w1 = –(2 atm)(10.0 L – 5.00 L) = –10.0 L atm. The work performed during the second stage of expansion is: w2 = –(1 atm)(20.0 L – 10.0 L) = –10.0 L atm. The sum for the whole stepwise process is w1 + w2 = –20.0 L atm. Thus, we see that more work is performed in this multi–step expansion than in a single expansion. To obtain more work, the expansion needs to be done in smaller steps 19.107 First, review the information provided. Since the salt dissolved, the process is spontaneous, and the sign for ∆G is negative. The temperature went down indicating that this is an endothermic process and ∆H must 489
Chapter 19
be positive. Dissolving the solid salt increases the disorder of the system which indicates ∆S is positive. The general equation from which we must work is ∆G = ∆H – T∆S. Using this equation, the magnitude of T∆S must be larger than the magnitude of ∆H in order to obtain a negative value for ∆G. However, the magnitude of ∆H is almost certainly larger than that of ∆S as can be appreciated from the fact that ∆S˚ values are given in J and those of ∆H˚f are given in kJ. 19.108 (a) (b)
( c)
(d)
(e)
H2C2O4(g) + 1/2O2(g) H2O(l) + 2CO2(g) ∆H°comb = –246.05 kJ H2(g) + 2C(s) + 2O2(g) H2C2O4(s) ∆Hf = ? kJ mol–1 ∆H°comb = { ∆Hof (H2O(l)) + 2 ∆Hof (CO2(g))} – { ∆Hof (H2C2O4(s)) + 1/2 ∆Hof (O2(g))} Rearranging; ∆Hf (H2C2O4(s)) = { ∆Hof (H2O(l)) + 2 ∆Hof (CO2(g))} –∆H°comb = –285.8 kJ + 2(–393.5 kJ) – (–246.05 kJ) = –826.8 kJ ∆Sf = S°(H2C2O4(s)) – {Sf°(H2(g)) + 2S°(C(g))+ 2S°(O2(g))} = 120.1 J/K – {130.6J/K + 2(5.69 J/K) + 2(205.0 J/K)} = –431.9J/mol K ∆S°comb = {S°(H2O(l)) + 2S°(CO2(g))} – {S°(H2C2O4(g)) + 1/2S°(O2(g))} = {69.96 J/K + 2(213.6 J/K)} – {120.1 J/K + 1/2(205.0 J/K) = 274.6 J/mol K ∆G f = ∆Hf – T ∆Sf = –826.8 kJ – (298 K)(–0.4319 kJ/K) = –698.1 kJ ∆Gcomb° = ∆Hcomb° – T∆Scomb° = –246.05 kJ – (298K)(0.2746 kJ/K) = –327.9 kJ
19.109 glucose + phosphate glucose–6–phosphate + H2O ATP + H2O ADP + phosphate
∆G = 13.13 kJ ∆G = –32.22 kJ
Add these two reactions together to get: glucose + ATP glucose–6–phosphate + ADP
∆G = –19.09 kJ
19.110 The value for ∆G° is given in the problem, and values for the free energy of formation of all other reactants and products can be obtained in Appendix C2.
∆G = (sum of ∆G of of the products) − (sum of ∆G of of the reactants) ∆G = {8 × ∆G of [CO 2 (g)] + 10 × ∆G fo [H 2 O(g)]} − {2 × ∆G of [C4 H10 (g)] + 13 × ∆G of [O 2 (g)]} ∆G of [C4 H10 (g)] =1/2[ {8 × ∆G of [CO 2 (g)] + 10 × ∆G fo [H 2 O(g)]} − {13 × ∆G of [O 2 (g)]} − ∆G ] 1 [{(8 mol × –394.4 kJ mol–1) + (10 mol × –228.6 kJ mol–1)} – (13 mol × 0 kJ mol–1) – 2 (–5407 kJ)] = –17 kJ ∆G° = –RTlnK –17,000 J = –(8.314 J mol–1 K–1)(298 K)lnK ln K = 6.86 Kc = 954 ∆G of [C 4 H10 (g)] =
19.111 1500 °C = 1773 K o = –RTlnK ∆G1173 o = –(8.314 J mol–1 K–1)(1773 K)(ln 5.67) ∆G1173
490
Chapter 19
o = –25,600 J = –25.6 kJ ∆G1173
19.112 2HNO3(l) N2O5(g) + H2O(l) ∆G° = 37.6 kJ N2(g) + 3O2(g) + H2(g) 2HNO3(l) H2O(l) H2(g) + 1/2O2(g)
∆G° = –159.82 kJ ∆G° = 237.2 kJ ∆G° = sum of the other equations = 115.0 kJ
N2(g) + 5/2O2(g) N2O5(g) 19.113 2ClNO(g) Cl2(g) + 2NO(g)
[ClNO] 1.00 –2x 1.00 – 2x
I C E
[Cl2] – +x +x
[NO] – +2x +2x
2x = 5% of 1.00 = 0.05 x = 0.025 [ClNO] = 0.95 [Cl2] = 0.025 [NO] = 0.05
[Cl2 ][ NO] 2 [ ClNO]
2
K=
[0.025][0.05] 2 [0.95]
2
=
= 6.9 × 10–5
∆G o298 = –RTlnK ∆G o298 = –(8.314 J mol–1 K–1)(298 K)(ln 6.9 × 10–5) ∆G o298 = 23,700 J = 23.7 kJ
19.114 N2O(g) + O2(g) NO2(g) + NO(g) ∆G = ∆H – T∆S ∆G = –42.9 kJ – (773 K)(–0.0261 kJ K–1) = –22.7 kJ ∆G o773 = –RTlnK –22,700 J = –(8.314 J mol–1)(773 K)lnK ln K = 3.53 K = 34.2
[N2O] 0.06 –x 0.06 –x
I C E K=
[O2] 0.06 –x 0.06 – x
( x )( x ) [ NO ] [ NO ] = [ N O ][ O ] ( 0.06 − x )( 0.06 − x ) 2
2
[NO2] – +x +x
= 34.2
2
Take the square root of both sides of the equation ( x) 5.85 = ( 0.06 − x ) 0.351 – 5.85x = x 0.351 = 6.85x x = 0.0512 The percentage of N2O reacted is:
491
[NO] – +x +x
Chapter 19
0.0512 × 100% = 85.4% 0.06 19.115 The reaction is N2(g) 2N(g), ∆H° = 2 ∆Hf [N(g)] – ∆Hf [N2(g)]. Since the enthalpy of formation of molecular nitrogen is defined as zero, the enthalpy change for this reaction and, consequently, the bond energy of a nitrogen molecule, is simply two times the enthalpy of formation of atomic nitrogen. 472.68 kJ Bond Energy = ( 2 mol N atoms ) = 945.36 kJ mol N atoms A similar argument holds true for oxygen: 249.17 kJ Bond Energy = ( 2 mol O atoms ) = 498.34 kJ mol N atoms 19.116 C(s) ↓(1) C(g)
+
2Cl2(g) ↓(2) 4Cl(g)
+
(3)
CCl4(l) ↑(4) CCl4(g)
∆H = 717 kJ ∆H = 4(121.5 kJ) = 486 kJ ∆Hf°(CCl4(g)) = 4(BE(C–Cl)) = 4(338 kJ) = –1352 kJ ∆Hcond = –29.9 kJ
Step 1: Step 2: Step 3: Step 4:
∆H = the sum of the four steps = –179 kJ 19.117 2N2O + 3O2 4NO2
[N2O] 0.157 –2x 0.157 –2x
I C E
[O2] 0.0968 –3x 0.0968 – 3x
[NO2] – +4x 4x
4x = 0.0743 M x = 0.0186 M [N2O] = 0.157 – 2(0.0186) = 0.120 M [O2] = 0.0968–3(0.186) = 0.041 M [NO2] = 0.0743 M
[ NO ] [ N O] [ O ] 4
K=
2
2
2
3
2
=
( 0.0743) 4 ( 0.120) 2 ( 0.041) 3
= 31
∆G° = –RTlnK ∆G° = –(8.314 J mol–1K–1)(298 K)(ln 31) ∆G° = –8.51 × 103 J mol–1 ∆G° = –8.51 kJ mol–1 19.118 S(s) + O2(g) + F2(g) SF2O2(g) 2O(g) (2)
+
O 2( g )
+
∆H of = –858 kJ mol–1
S(g) (1)
+
2F(g) (3)
S(s)
+
F2( g )
(1)
∆H1o = ∆H of S(g) = 1 mol × 276.98 kJ mol–1 = 276.98 kJ
(2)
∆H o2 = ∆H of O(g) = 2 mol × 249.17 kJ mol–1 = 498.34 kJ
SF2O2(g)
492
Chapter 19
(3)
∆H 3o = ∆H of F(g) = 2 mol × 79.14 kJ mol–1 = 158.28 kJ
The sum of the three steps is the total energy input: 933.6 kJ The sum of the bond energies for SF2O2(g) less the total energy input to form the gaseous atoms is the enthalpy of formation. The sum of the bond energies: 2 S―F 2 mol × 308.0 kJ/mol = 616 kJ 2 S═O 2 mol × x kJ/mol = 2x kJ ∆H of SF2 O2 (g) = 933.6 kJ + [–616 kJ + 2x kJ] –858 kJ = 317 kJ + 2x kJ x = –588 kJ The bond energy for S═O is 588 kJ
Multi-Concept Problems 19.119 We need to calculate the amount of energy produced when one gallon of each of these fuels is burned; 0.7893 g ethanol 1 mole ethanol mol ethanol = 3.78 × 103 mL ethanol = 64.8 moles ethanol 1 mL ethanol 46.07 g ethanol
(
)
−1299.8 kJ 4 kJ = (64.8 moles ethanol) = 8.42 × 10 kJ 1 mole ethanol 0.7025 g octane 1 mole octane mol octane = 3.78 × 103 mL octane = 23.2 moles octane 1 mL octane 114.23 g octane
(
)
−5307 kJ 5 kJ = (23.2 moles octane) = 1.23 × 10 kJ 1 mole octane In spite of the large number of moles of ethanol in one gallon of liquid, the energy produced from the combustion of a gallon of octane is greater than the amount produced when one gallon of ethanol is burned. 19.120 Strong acids and bases are strong electrolytes. Therefore, the net ionic equation for the reaction of a strong acid and strong base is given by: H+(aq) + OH−(aq)
→
∆Ho = -55.86 kJ
H 2O
(1)
Propionic acid is a weak acid, therefore the net ionic equation for its reaction with a strong base is: HC3H5O2(aq) + OH−(aq)
→
C3H5O2− (aq) + H2O
∆Ho = -49.23 kJ
(2)
If we reverse the first reaction (1) and combine it with the second reaction (2) we obtain the desired ionization reaction and the enthalpy of ionization. H 2O → H+(aq) + OH−(aq) HC3H5O2(aq) + OH−(aq) → HC3H5O2(aq)
→
∆Ho = +55.86 kJ mol−1 C3H5O2 (aq) + H2O ∆Ho = -49.23 kJ mol−1 −
C3H5O2− (aq) + H+(aq)
∆Ho = +6.63 kJ mol−1
The entropy change for the reaction can be determined from the enthalpy of the reaction and the equilibrium constant for propionic acid.
∆Go = ∆Ho − T∆S = −RT ln K Thus, ∆S = (−RT ln K − ∆Ho)/T = {− 8.314 J mol−1 K−1 x 298 K x ln (1.3 x 10−5) – 6,663 J mol−1}/298 K = 71.2 J mol−1 K−1 493
Chapter 19
19.121 The proposed reaction is: CH4(g) + H2O →
CH3OH(l) + H2(g)
Some of the claims that need to be explored are: (1) Does a catalyst shift equilibrium? (2) Is the mass of product consistent with the mass of methane in 117 m3 at STP? (3) Does thermodynamics favor the products? (4) Is the equilibrium constant large or small?
A catalyst alters the kinetics of a reaction, increasing the rate so that equilibrium is achieved more quickly, but does not shift the equilibrium. Therefore, this claim is false. The number of moles of methane in 117 m3 at STP is: n = PV/(RT) = (1.00 atm x 117 m3 x 1000 L m-3)/(0.0821 L atm K−1 mol−1 x 273 K) = 5220 mol CH4 The number of moles of product, assuming 100% yield, is:
5220 mol CH 4 x
1 mol CH3 OH 32.04 g x = 1.67 x 105 mol CH3 OH mol CH 4 mol CH3OH
The reaction should produce 167 kg of methane so the claim that it can product 134 kg is valid. The Gibb’s Free energy of the reaction is: ∆G orxn = ∆G of [H 2 ( g )] + ∆G of [CH 3OH(l )] − {∆G of [H 2 O( g )] + ∆G of [CH 4 (g )] = 1 mol CH3OH x(−166.2 kJ mol−1) –{1 mol H2O x (-237.2 kJ mol−1) + 1 mol CH4 x (-50.79 kJ mol−1) = 121.8 kJ The reaction is not favored. Let’s look at the entropy change. ∆Sorxn = So [H 2 ( g )] + So [CH 3OH(l )] − {So [H 2 O( g )] + So [CH 4 (g )] = {1 mol H2 x (130.6 J mol−1 K−1 + 1 mol CH3OH x(126.8 kJ mol−1 K−1)} –{1 mol H2O x 69.96 J mol−1 K−1 ) + 1 mol CH4 x (186.2 J mol−1 K−1) = 1.24 J K−1 Entropy favors the reaction, but only slightly. Therefore, the enthalpy is large and positive. To drive the reaction forward would require an increase in the temperature to a value large enough so that T∆S was greater than ∆H .
∆G = ∆H − T ∆S
T = (∆G − ∆H)/∆S
The equilibrium constant is very small so the reactants are favored. ∆G = − RT ln K 121.8 kJ x 1000 J kJ−1 = −8.314 J K−1 mol−1 x 298 K ln K lnK = -49.66
K = 4.5 x 10−22
494
Chapter 19
The overall assessment is that this project is not one in which you should not invest the funds. 19.122 The equilibrium constant can be used to determine ∆Go and the reaction data can be used to determine the enthalpy of the neutralization reaction. Then we need to see how these two results can be used along with our knowledge of equilibrium and thermodynamic relationships to calculate the entropy of the reaction.
∆Go = − RT lnK =RT pK = 8.314 J mol−1 K−1 x 298 K x 5.83 = 1.44 x 104 J or 14.4 kJ mol−1 The neutralization reaction is a 1:1 mole ratio. Also, you have two different volumes at two different initial temperatures. Finally, you have an excess of on reactant so this is a limiting reagent problem. mmol of acid = 100.0 mL x 0.00525 M = 0.525 mmol mmol of base = 45.6 mL x 0.00634 M = 0.289 mmol Therefore, 0.289 mmol of acid react and the heat of reaction will be for this many mmoles reacting. qrxn = mCP∆T = 100.0 g x 4.184 J g−1 oC−1 x (21.26 – 24.88) oC + 45.6 g x 4.184 J g−1 oC−1 x (22.18 – 24.88) oC = -2029 J This is the amount of heat released when 0.289 mmol of acid react. Therefore, the molar enthalpy of the reaction is: -2029 J/(0.289 mmol x 1 mol/1000 mmol) = −7.02 x 106 J mol−1 or -702 kJ mol−1 To summarize, HA A− + H+
∆G o = 14.4 kJ mol−1
HA + OH− → A− + H2O
∆Ho = −702 kJ mol−1
Using Hess’s Law we can add together a series of reactions and their enthalpies to achieve the desired equilibrium reaction and its enthalpy. H2O H+(aq) + OH−(aq)
∆Ho = 1 mol OH− x (-230.0 kJ mol−1) − 1 mol H2O x (-285.9 kJ mol−1) = 55.9 kJ HA + OH− → A− + H2O H2O H+(aq) + OH−(aq) HA A− + H+
∆ G o = ∆ H o − T ∆ So
∆Ho = −702 kJ mol−1 ∆Ho = 55.9 kJ mol−1
∆Ho = −646.1 kJ mol−1 ∆So = (∆Go − ∆Ho)/T
∆So = {14.4 kJ – (-702 kJ)}/298 K = 2.40 kJ mol−1 K−1
495
or 2.4 x 103 J mol−1 K−1
Chapter 20
Practice Exercises Mg(s) Mg2+(aq) + 2e– Fe2+(aq) + 2e– Fe(s) Mg(s)|Mg2+(aq)||Fe2+(aq)|Fe(s)
20.1
anode: cathode: cell notation:
20.2
anode: cathode: overall:
20.3
E°cell = E°substance reduced – E°substance oxidized 2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)
Al(s) Al3+(aq) + 3e– Ni2+(aq) + 2e– Ni(s) 3Ni2+(aq) + 2Al(s) 2Al3+(aq) + 3Ni(s)
E ocell = E oAg+ − E oCu 2+ E°cell = 0.80 V – 0.34 V = 0.46 V 2Ag+(aq) + Zn(s) 2Ag(s) + Zn2+(aq)
E ocell = E oAg+ − E oZn 2+ E°cell = 0.80 V – (–0.76 V) = 1.56 V Zinc will have the larger value for E°cell. 20.4
E°cell = E°substance reduced – E°substance oxidized Fe2+(aq) + Mg(s) Mg2+(aq) + Fe(s) 1.93 V = E°Fe2+ – (–2.37 V) E°Fe2+ = 1.93 V + (–2.37 V) = –0.44 V This agrees exactly with Table 20.1.
20.5
The half–reaction with the more positive value of E° (listed higher in Table 20.1) will occur as a reduction. The half–reaction having the less positive (more negative) value of E° (listed lower in Table 20.1) will be reversed and occur as an oxidation. (a) I2(aq) + 2e– 2I–(aq) oxidation Fe3+(aq) + e– Fe2+(aq) reduction 496
Chapter 20
(b)
(c)
20.6
2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq) Mg2+(aq) + 2e– Mg(s) oxidation Cr3+(aq) + 3e– Cr(s) reduction 3Mg(s) + 2Cr3+(aq) 3Mg2+(aq) + 2Cr(s) Co2+(aq) + 2e– Co(s) oxidation SO42–(aq) + 4H+(aq) + 2e– H2SO3(aq) + H2O reduction SO42–(aq)+ 4H+(aq) + Co(s) H2SO3(aq) + H2O + Co2+(aq)
The half–reaction with the more positive value of E° (listed higher in Table 20.1) will occur as a reduction. The half–reaction having the less positive (more negative) value of E° (listed lower in Table 20.1) will be reversed and occur as an oxidation. Br2(aq) + 2e– 2Br–(aq) SO42–(aq) + 4H+(aq) + 2e– H2SO3(aq) + H2O Br2(aq) + H2SO3(aq) + H2O 2Br–(aq) + SO42–(aq) + 4H+(aq)
reduction oxidation
20.7
Either nickel(II) or iron(III) will be reduced, depending on which way the reaction proceeds. Iron(III) is listed higher than nickel(II) in Table 20.1 (it has a greater reduction potential), so we would expect that the reaction would not be spontaneous in the direction shown. The spontaneous reaction is: Ni(s) + 2Fe3+(aq) Ni2+(aq) + 2Fe2+(aq)
20.8
The half–reaction having the more positive value for E° will occur as a reduction. The other half–reaction should be reversed, so as to appear as an oxidation. NiO2(s) + 2H2O + 2e– Ni(OH)2(s) + 2OH–(aq) Fe(s) + 2OH–(aq) 2e– + Fe(OH)2(s) NiO2(s) + Fe(s) + 2H2O Ni(OH)2(s) + Fe(OH)2(s)
reduction oxidation net reaction
E°cell = E°substance reduced – E°substance oxidized E°cell = E°NiO2 – E°Fe E°cell = 0.49 – (–0.88) = 1.37 V 20.9
5{Cr(s) -
Cr3+(aq) + 3e-} +
-
oxidation 2+
3{MnO4 (aq) + 8H (aq) + 5e Mn (aq) + 4H2O(l)} reduction Net reaction 5Cr(s) + 3MnO4−(aq) + 24H+(aq) 3Mn2+(aq) + 12H2O(l) + 5Cr3+(aq) E°cell = E°substance reduced – E°substance oxidized E ocell = E oMnO− − E oCr3+ 4
E°cell = 1.51 – (–0.74) = 2.25 V 20.10
The half–reaction having the more positive value for E° will occur as a reduction. The other half–reaction should be reversed, so as to appear as an oxidation. 3Cu2+(aq) + 2Cr(s) 3Cu(s) + 2Cr3+(aq) E°cell = E°substance reduced – E°substance oxidized o o E ocell = E Cu 2+ − E Cr 3+ E°cell = 0.34 – (–0.74) = 1.08 V
497
Chapter 20
20.11
A reaction will occur spontaneously in the forward direction if the value of E° is positive. We therefore evaluate E° for each reaction using: E°cell = E°substance reduced – E°substance oxidized (a) Br2(aq) + 2e– 2Br–(aq) reduction I2(s) + 2e– 2I–(aq) oxidation, when reversed E°cell = E°Br2 – E°I2 E°cell = 1.07 V – (0.54 V) = 0.53 V, ∴ spontaneous (b)
MnO4–(aq) + 8H+ + 5e– Mn2+(aq) + 4H2O reduction 5Ag+(aq) + 5e– 5Ag(s) oxidation, when reversed
E ocell = E oMnO− − E oAg+ 4
E°cell = 1.51 V – (0.80V) = +0.71 V, ∴ spontaneous 20.12
A reaction will occur spontaneously in the forward direction if the value of E° is positive. We therefore evaluate E° for each reaction using: E°cell = E°substance reduced – E°substance oxidized (a)
Br2(aq) + 2e– 2Br–(aq) reduction Cl2(aq) + 2H2O 2HOCl(aq) + 2H+(aq) + 2e–
oxidation
E°cell = E°Br2 – E°HOCl E°cell = 1.07 V – (1.63 V) = –0.56 V, (b)
∴ non-spontaneous
2Cr3+(aq) + 6e– 2Cr(s) reduction 3Zn(s) 3Zn2+(aq) + 6e– oxidation E°cell = E°Cr3+ – E°Zn2+ E°cell = –0.74 V – (–0.76 V) = +0.02 V,
∴ spontaneous
20.13
From the equation ∆G° = –nFE°cell ∆G° = –30.9 kJ or –30,900 J F = 96,500 C mol–1 E°cell = 0.107 V –30,900 J = –n(96,500 C mol–1)(0.107 V) n = 2.99 which rounds to 3 Therefore, 3 moles of electrons are transferred in the reaction.
20.14
∆G° = –n F E°cell From Practice Exercise 11 (a): n = 2 e–, E°cell = 0.53 V ∆G° = –n F E°cell = –(2 e–)(96,500 F)(0.53 V) = –102,000 J = –102 kJ From Practice Exercise 11 (b): n = 5 e–, E°cell = 0.71 V ∆G° = –n F E°cell = –(5 e–)(96,500 F)(0.71 V) = –342,600 J = –343 kJ From Practice Exercise 12 (a): n = 2 e–, E°cell = –0.56 V ∆G° = –n F E°cell = –(2 e–)(96,500 F)(–0.56 V) = 108080 J = 108 kJ From Practice Exercise 12 (b): n = 6 e–, E°cell = 0.02 V ∆G° = –n F E°cell = –(6 e–)(96,500 F)(0.02 V) = 11,600 J = –11.6 kJ 498
Chapter 20
20.15
Using Equation 20.7,
RT Eocell = n F ln K c
−0.46 V =
(8.314 J mol −1K −1 )(298 K)
2(96,500 C mol−1 ) ln K c = −35.83 Taking the antilog (ex) of both sides of the above equation gives Kc = 2.7 × 10–16.
ln K c
This very small value for the equilibrium constant means that the products of the reaction are not formed spontaneously. The equilibrium lies far to the left, favoring reactants, and we do not expect much product to form. The reverse reaction will be spontaneous, therefore, the value for Kc for the spontaneous reaction will be: Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) 1 1 K c' = = = 3.7 × 1015 K c 2.7 × 10−16 20.16
Ag+(aq) + e– Ag(s) E° = 0.80 V E° = 0.07 V AgBr(s) + e– Ag(s) + Br–(aq) Equation for the spontaneous reaction: Ag+(aq) + Br–(aq) AgBr(s) E°cell = 0.73 V 1 K= Ag + Br − ln Kc =
( )(
)
− −1 E ocell n F ( 0.73 V ) 1 e 96,500 C mol = = 28.43 RT 8.314 J mol −1 K −1 ( 298 K )
(
Kc = 2.23 × 10
)
12
Ksp for AgBr is 5.4 × 10–13 1 1 = = 4.5 × 10–13 K c 2.23 × 1012 The Kc is the inverse of the Ksp. So the cell potential value agrees with Ksp table. 20.17
Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq) Mg 2 + RT Ecell = E ocell − ln nF Cu 2 +
E°cell = 0.34 – (–2.37 V) = 2.71 V
(8.314 J mol K ) ( 298 K ) Ecell = 2.71 V – ( 2 ) ( 96,500 C mol ) −1
−1
−1
20.18
Ecell = Eocell
−
2.2 × 10−6 = 2.82 V ln [0.015]
[1] RT ln 2 nF H+
499
Chapter 20
(8.314 J mol K ) ( 298 K ) ( 2 ) ( 96,500 C mol ) −1
0.00 = 0.14 V –
−1
−1
ln
[1] H+
2
0.14 V = -2.57 x 10-2 ln[H+] 5.45 = -ln[H+] = -2.303 log[H+] 2.37 = -log[H+] = pH
20.19
Zn(s) Zn2+(aq) + 2e– Cu2+(aq) + 2e– 2Cu(s)
oxidation reduction
E°cell = E°Cu2+ – E°Zn2+ E°cell = +0.34 V – (–0.76 V) = +1.10 V The Nernst equation for this cell is:
Zn 2+ RT Ecell = E cell − ln nF Cu 2 + (8.314 J mol−1K −1 )(298 K) E cell = 1.10 V − ln 2(96,500 C mol−1 ) =1.10 V – 0.01284(4.605) = 1.04 V o
20.20
Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq) Mg 2 + RT Ecell = Eocell − ln nF Cu 2 +
[ 0.010]
E°cell = 0.34 – (–2.37 V) = 2.71 V
(8.314 J mol K ) ( 298 K ) 2.79 = 2.71 V – ( 2 ) ( 96,500 C mol ) −1
[1.0]
−1
−1
Mg 2 + ln [ 0.015]
Mg 2 + Mg 2 + ln 0.08 V = –0.0128 ln [ 0.015] [ 0.015] Mg 2 + –6.25 = ln [ 0.015] Mg 2 + e–6.25 = 0.015 [ ] [Mg2+] = 2.95 × 10–5 M
20.21
Cu(s) Cu2+(aq) + 2e– Ag+(aq) + e– Ag(s)
oxidation reduction
E°cell = E°Ag+ – E°Cu2+ E°cell = +0.80 V – (+0.34 V) = +0.46 V
500
Chapter 20
Ecell = Eocell −
Cu 2+ RT ln 2 nF + Ag
Cu 2+ = (Eo − E )/ RT ln cell cell 2 + nF Ag
=
( 0.46 V
− 0.57 V )
0.01284 = − 8.5670
Cu 2+ = e −8.5670 = 1.9 × 10−4 2 + Ag Since the [Ag+] = 0.225 M, [Cu2+] = 9.6 × 10–6 M Substituting the second value of 0.82 V into the same expression gives [Cu2+] = 3.4 × 10–14 M
20.22
We are told that, in this galvanic cell, the chromium electrode is the anode, meaning that oxidation occurs at the chromium electrode. Now in general, we have the equation: E°cell = E°reduction – E°oxidation which becomes, in particular for this case: E°cell = E°Ni2+ – E°Cr3+ The net cell reaction is given by the sum of the reduction and the oxidation half–reactions, multiplied in each case so as to eliminate electrons from the result: 3 × [Ni2+(aq) + 2e– Ni(s)] reduction 2 × [Cr(s) Cr3+(aq) + 3e–] oxidation 3Ni2+(aq) + 2Cr(s) 2Cr3+(aq) + 3Ni(s) net reaction In this reaction, n = 6, and the Nernst equation becomes: 2
Cr 3+ RT o Ecell = E cell − ln 3 nF 2 + Ni 2
Cr 3+ = (Eo − E )/ RT ln cell cell 3 nF Ni 2 +
501
Chapter 20
2
Cr 3+ = ( 0.487 V − 0.552 V ) ln 3 0.004279 Ni 2+ 2
Cr 3+ = − 15.190 ln 3 2 + Ni Cr 3+
2
2+
3
Ni
= e −15.190 = 2.5 × 10−7
Substituting [Ni2+] = 1.20 M, we solve for [Cr3+] and get: [Cr3+] = 6.6 × 10–4 M.
20.23
Fe3+(aq) + e– Fe2+(aq) 2I2(s) + e– 2I–(aq) O2(g) + 4H+(aq) + 4e– 2H2O
Eo = 0.77 V Eo = +0.54 V Eo = +1.23 V
The reaction with the least positive reduction potential will be the easiest to oxidize, and its product will be the product at the anode. I2 will be produced. 20.24
The cathode is always where reduction occurs. We must consider which species could be candidates for reduction, then choose the species with the highest reduction potential from Table 20.1. Cd2+(aq) + 2e– 2Cd(s) Sn2+(aq) + 2e– 2Sn(s) 2H2O + 2e– H2(g) + 2OH–(aq)
E° = –0.40 V E° = –0.14 V E° = –0.83 V
Tin(II) has the highest reduction potential, so we would expect it to be reduced in this environment. We expect Sn(s) at the cathode. 20.25
The number of Coulombs is: 4.00 A × 180 s = 720 C The number of moles is:
mol OH − = 720 C × 20.26
The number of moles of Au to be deposited is: 3.00 g Au ÷ 197 g/mol = 0.0152 mol Au. The number of Coulombs (A × s) is: 3F 96,500 C Coulombs = 0.0152 mol Au × × = 4.40 × 103 C 1 mol Au 1F The number of minutes is:
min =
20.27
1F 1 mol OH − × = 7.46 × 10–3 mol OH− 96,500 C 1F
4.40 × 103 A • s 1 min = 7.33 min × 10.0 A 60 s
As in Practice Exercise 26 above, the number of Coulombs is 4.40 × 103 C. This corresponds to a current of:
A=
4.40 × 103 A • s 1 min × = 3.67 A 20.0 min 60 s 502
Chapter 20
20.28
The number of Coulombs is: 0.100 A × 1.25 hr(3600 s/hr) = 450 C The number of moles of copper ions produced is:
mol Cu 2 + = 450 C ×
1 mol e− 1 mol Cu 2 + × = 0.00233 mol Cu2+ 96,500 C 2 mol e −
Therefore, the increase in concentration is: M = mol/L = (0.00233 mol Cu2+)/(0.125 L) = + 0.0187 M Review Questions 20.1
A galvanic cell is one in which a spontaneous redox reaction occurs, producing electricity. A half–cell is one of either the cathode or the anode, together with the accompanying electrolyte.
20.2
The salt bridge connects two half–cells, and allows for electrical neutrality to be maintained by a flow of appropriate ions.
20.3
These must be kept separate, because otherwise Ag+ ions would be reduced directly by Cu metal, and no external current would be produced.
20.4
The anode is the electrode at which oxidation takes place, and the cathode is the electrode at which reduction takes place. The charges of the electrodes in a galvanic cell are opposite to those in the electrolysis cell; the cathode is positive and the anode is negative.
20.5
In both the galvanic and the electrolysis cells, the electrons move away from the anode and toward the cathode.
20.6
The anions move away from the cathode toward the anode, and the cations move away from the anode toward the cathode.
20.7
Aluminum metal constitutes the anode: Al(s) Al3+(aq) + 3e– The cathode is tin metal: Sn2+(aq) + 2e– Sn(s)
20.8
A cell potential is a standard potential only if the temperature is 25 °C, the pressure is 1 atm, and all ions have a concentration of 1 M.
20.9
The cell potential for the anode half–reaction is subtracted from the cell potential for the cathode half–cell: o o E ocell = E substance reduced – E substance oxidized
E ocell = E oreduction – E ooxidation 20.10
The standard hydrogen electrode is diagramed in Figure 20.5 of the text. It consists of a platinum wire in contact with a solution having [H+] equal to 1 M, and hydrogen gas at a pressure of 1 atm is placed over the system. The half–cell potential is 0 V.
20.11
A positive reduction potential indicates that the substance is more easily reduced than the hydrogen ion. Conversely, a negative reduction potential indicates that the substance comprising the half–cell is less easily reduced than the hydrogen ion.
20.12
The difference between the reduction potentials for hydrogen and copper is a constant that is independent of the choice for the reference potential. In other words, the reduction half–cell potential for copper is to be 0.34 units higher for copper than for hydrogen, regardless of the chosen point of reference. If E° for copper is taken to be 0 V, then E° for hydrogen must be –0.34 V. 503
Chapter 20
20.13
The negative terminal of the voltmeter must be connected to the anode in order to obtain correct readings of the voltage that is generated by the cell.
20.14
The metals are placed into the activity series based on their values of standard reduction potentials.
20.15
The silver will be reduced, according to the standard reduction potentials.
External circuit electron flow
(+)
(–)
Salt Bridge
Ag
Ag+ (aq)
Fe
Fe3+ (aq)
Cathode
Anode
The net cell reaction is: Fe(s) + 3Ag+(aq)
→
Fe3+(aq) + 3Ag(s)
20.16
External circuit (–)
electron flow
(+)
Salt Bridge
Pt
Fe2+(aq) Fe3+(aq) Anode
Pt
Br2(aq) Br–(aq) Cathode
2Fe2+(aq) + Br2(aq) 2 Fe3+(aq) + 2Br–(aq) 20.17
∆G° = –nF E cell
20.18
E ocell =
0.0592 log K c n
504
Chapter 20
20.19
The Nernst equation: RT Ecell = E ocell − ln Q nF –1 R = 8.314 J mol K–1 T = 25 °C = 298 K F = 96,494 C mol–1 RT = 0.02567 J C–1 F RT ln Q Ecell = E ocell − nF If the system is at equilibrium, Q = Kc. And Ecell = 0 RT ln K c 0 = E ocell − nF RT E ocell = ln K c nF
20.20
We begin by separating the reaction into its two half–reactions, in order to obtain the value of n. Pb(s) + SO42–(aq) PbSO4(s) + 2e– PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– PbSO4(s) + 2H2O Thus, n is equal to 2, and the equation that we are to use is:
0.0592 log Q n 0.0592 1 = 2.05 V − log 4 2 2 H + SO 4 2 −
E cell = E ocell − E cell
20.21
A concentration cell consists of two almost identical half–cells, the two half–cells are composed of the same substances, but have difference concentrations of the solute species.
[ion ]dilute Ecell = E ocell − RT ln nF [ion ]conc E ocell = 0 because the standard cell potential is the reduction potential of the substance being reduce less the reduction potential of the substance being oxidized, and the two are equal to each other because they are the same substance. 20.22
Pb(s) + SO42–(aq) PbSO4(s) + 2e– PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– PbSO4(s) + 2H2O Connecting six cells in series produces 12 volts.
anode cathode
20.23
PbSO4(s) + 2e– Pb(s) + SO42–(aq) PbSO4(s) + 2H2O 2e– + PbO2(s) + 4H+(aq) + SO42–(aq)
cathode anode
20.24
This is diagramed in Figure 20.11. The float inside the hydrometer sinks to a level that is inversely proportional to the density of the liquid that is drawn into it. This works because the concentration of the sulfuric acid (and hence the state of charge of the battery) is proportional to the concentration of sulfuric acid in the battery.
20.25
The anodic reaction is: Zn(s) Zn2+(aq) + 2e–
505
Chapter 20
Several reactions take place at the cathode; one of the important ones is: 2MnO2(s) + 2NH4+(aq) + 2e– Mn2O3(s) + 2NH3(aq) + H2O 20.26
Zn(s) + 2OH–(aq) ZnO(s) + H2O + 2e– 2MnO2(s) + H2O + 2e– Mn2O3(s) + 2OH–(aq)
anode cathode
20.27
Cd(s) + 2OH–(aq) Cd(OH)2(s) + 2e– 2e– + NiO2(s) + 2H2O Ni(OH)2(s) + 2OH–(aq)
anode cathode
The overall cell reaction on discharge of the battery is: Cd(s) + NiO2(s) + 2H2O Cd(OH)2(s) + Ni(OH)2(s) On charging the battery, the reactions are reversed. 20.28
The hydrogen is held in a metal alloy, Mg2Ni, which has the ability to absorb and hold substantial amounts of hydrogen. The electrolyte is KOH.
20.29
MH(s) + OH–(aq) M(s) +H2O + e– NiO(OH)(s) + H2O + e– Ni(OH)2(s) + OH–(aq) MH(s) + NiO(OH)(s) M(s) + Ni(OH)2(s)
anode cathode overall reaction
The reactions are reversed upon charging. 20.30
Lithium has the most negative reduction potential of any metal, so it is very easy to oxidize making it an excellent material for an anode, and it is a very lightweight metal. The major problem with lithium in a cell is that it reacts vigorously with water. Also, lithium batteries often have a large negative ∆H.
20.31
In a typical primary lithium cell, the electrodes are lithium as the anode and manganese(IV) oxide as the cathode. Li Li+ + e– anode MnO2 + Li+ + e– LiMnO2 cathode Li + MnO2 LiMnO2 net cell reaction
20.32
The electrode materials in a typical lithium ion cell are graphite and cobalt oxide. When the cell is charged, Li+ ions leave LiCoO2 and travel through the electrolyte to the graphite. When the cell discharges, the Li+ ions move back through the electrolyte to the cobalt oxide while electrons move through the external circuit to keep the charge in balance.
20.33
O2(g) + 2H2O + 4e– 4OH–(aq) H2(g) + 2OH–(aq) 2H2O + 2e– 2H2(g) + O2(g) 2H2O
20.34
Fuel cells are more efficient thermodynamically, and more of the energy of the reaction can be made available for useful work provided that the supply of reactants is maintained. The only product formed by the cell is water.
20.35
In an electrolytic cell, the cathode is negative, and the anode is positive. The opposite is true of a galvanic cell. An inert electrode is an electrode which does not chemically react in the measurement of electrochemical data.
20.36
The flow of electrons in the external circuit must be accompanied by the electrolysis reaction. Otherwise the electrodes would accumulate charge, and the system would cease to function.
20.37
In solid NaCl, the ions are held in place and cannot move about. In molten NaCl, the crystal lattice of the solid has been destroyed; the ions are free to move, and consequently to conduct current by migrating either to the anode or to the cathode. Anode: 2Cl–(l) Cl2(g) + 2e–
cathode anode net cell reaction
506
Chapter 20
Cathode: Net:
Na+(l) + e– Na(l) 2Na+(l) + 2Cl–(l) 2Na(l) + Cl2(g)
20.38
oxidation: reduction:
2H2O 4H+(aq) + 4e– + O2(g) 2H2O + 2e– H2(g) + 2OH–(aq)
20.39
It is reduction that occurs at the cathode, and near it, the pH increases due to the formation of OH–(aq). At the anode, where the oxidation of water occurs, the pH decreases due to the production of H+(aq). See the equations given in the answer to Review Question 20.38. The overall change in pH is 0 since the amount of H+ formed and the amount of OH– formed are equal. K2SO4 serves as charge carriers to balance the charge that occurs upon electrolysis of the K2SO4 solution.
20.40
One Faraday (F) is equivalent to one mole of electrons. Also, one Faraday is equal to 96,500 Coulombs, and a Coulomb is equivalent to an Ampere⋅second: 1 F = 96,500 C and 1 C = 1 A⋅s
20.41
The deposition of 0.10 mol of Cr from a Cr3+ solution will take longer than the deposition of 0.10 mol Cu from a Cu2+ solution because the Cr3+ requires 1.5 times as many electrons for deposition than Cu2+. This is due to the difference in charges on the two ions.
20.42
The Ag+ solution will give more metal deposited since it is in the +1 state while the Cu2+ solution will give half as much since the copper is in the +2 state.
20.43
Copper has a larger atomic mass than iron; therefore, the copper will deposit a greater mass of metal. Both metals are in the same +2 state.
20.44
Electroplating is a procedure by which a metal is deposited on another conducting surface.
20.45
Al2O3(s) is dissolved in molten cryolite, Na3AlF6. The liquid mixture is electrolyzed to drive the following reaction: 2Al2O3(l) 4Al(l) + 3O2(g) The two half–reactions are: anode: cathode: overall cell reaction
2O2– O2(g) + 4e– Al3+(l) + 3e– Al(l) 6O2– + 4Al3+ 3O2(g) + 4Al(l) 507
Chapter 20
20.46
The oxygen that is formed slowly decomposes the anode.
20.47
Sodium is obtained from electrolysis of molten NaCl using the Downs cell. Some of the uses of sodium are to make tetraethyl lead, for sodium vapor lamps, and as a coolant in nuclear reactors. Na+ + e– Na(l) Cl– 1/2Cl2(g) + e– NaCl(l) Na(l) + 1/2Cl2(g)
20.48
cathode anode net reaction
This is shown in a photo and a diagram (Figure 20.24) of the text. Impure copper is the anode, which dissolves during the process. Pure copper is deposited at the cathode. Anode sludge contains precious metals, whose value makes the process cost effective. Typical reactions occurring at the anode are: Cu(s) Cu2+(aq) + 2e– Zn(s) Zn2+(aq) + 2e– Fe(s) Fe2+(aq) + 2e– The reaction that occurs at the cathode is: Cu2+(aq) + 2e– Cu(s)
20.49
One of the various methods arise diagramed in Figure 20.25 of the text. The physical apparatus influences the products that are obtained. The cathode reaction is the same in stirred and unstirred cells: 2H2O + 2e– H2(g) + 2OH–(aq) The unstirred anode reaction is: 2Cl–(aq) Cl2(g) + 2e– The net reaction is: 2NaCl(aq) + 2H2O 2NaOH(aq) + Cl2(g) + H2(g) In a stirred cell, the Cl2(g) that is produced reacts with the OH–(aq) present forming Cl–(aq), OCl–(aq) and water. The anode reaction in the stirred cell is therefore, Cl–(aq) + 2OH–(aq) OCl–(aq) + H2O + 2e– The net reaction in a stirred cell is: NaCl(aq) + H2O NaOCl(aq) + H2(g)
Review Problems 20.50
20.51
(a)
anode: cathode: cell:
Fe2+ → Fe3+ + e− 3e− + 4H+ + NO3− → NO + 2H2O 4H+ + NO3− + 3Fe2+ → 3Fe3+ + NO + 2H2O
(b)
anode: cathode: cell:
2Br− → Br2 + 2e− 2e− + Cl2 → 2Cl− Cl2 + 2Br− → Br2 + 2Cl−
(c)
anode: cathode: cell:
Ag → Ag+ + e− Au3+ + 3e− → Au 3Ag + Au3+ → 3Ag+ + Au
(a)
anode:
Fe → Fe2+ + 2e− 508
Chapter 20
cathode: cell:
Cd2+ + 2e− → Cd Fe + Cd2+ → Fe2+ + Cd
(b)
anode: cathode: cell:
Ag → Ag+ + e− 2e−+ 4H+ + NiO2 → Ni2+ + 2H2O 4H+ + NiO2 + 2Ag → 2Ag+ + Ni2+ +2H2O
(c)
anode: cathode: cell:
Mg → Mg2+ + 2e– Cd2+ + 2e− → Cd Mg + Cd2+ → Mg2+ + Cd
20.52
(a) (b) (c)
Cd(s)|Cd2+(aq)|| Au3+(aq)|Au(s) Fe(s)|Fe2+(aq)|| Br−(aq), Br2(aq)| Pt(s) Cr(s)|Cr3+(aq)|| Cu2+(aq)|Cu(s)
20.53
(a) (b) (c)
Zn(s)|Zn2+(aq)|| Cr3+(aq)|Cr(s) Pb(s), PbSO4(s)|HSO4−(aq)||H+(aq), HSO4−(aq)|PbO2(s), PbSO4(s) Mg(s)|Mg2+(aq)||Sn2+(aq)|Sn(s)
20.54
(a)
Sn(s)
20.55
a)
MnO4–(aq)
20.56
(a) (b) (c)
0.40 + 1.42 = 1.82 V 1.07 + 0.44 = 1.51 V 0.34 + 0.74 = 1.08 V
20.57
(a) (b) (c)
0.76 – 0.74 = 0.02 V 0.36 + 1.69 = 2.05 V 2.37 – 0.14 = 2.23 V
20.58
The reactions are spontaneous if the overall cell potential is positive. Eocell = Esubstance reduced − Esubstance oxidized
20.59
Br–(aq) (c) (b)
Zn(s)
Au3+(aq)
(d) (c)
I–(aq) PbO2(s)
(a)
Eocell = 1.42 V – (0.54 V) = 0.88 V
spontaneous
(b)
Eocell = 1.07 V – (0.17 V) = 0.90 V
spontaneous
(c)
Eocell = –0.74 V – (–2.76 V) = 2.02 V
spontaneous
(d)
HOCl(aq)
A reaction is spontaneous if its net cell potential is positive: o o E ocell = E substance reduced – E substance oxidized (a)
E ocell = 1.07 V – (1.36 V) = –0.29 V
not spontaneous
(b)
E ocell E ocell
= –0.44 V – (0.96 V) = –1.40 V
not spontaneous
= –0.25 V – (–0.44 V) = +0.19 V
spontaneous
(c) 20.60
(b)
The half–cell with the more positive E ocell will appear as a reduction, and the other half–reaction is reversed, to appear as an oxidation: BrO3–(aq) + 6H+(aq) + 6e– Br–(aq) + 3H2O reduction 3 × (2I–(aq) I2(s)+ 2e–) oxidation BrO3–(aq) + 6I–(aq) + 6H+(aq) 3I2(s) + Br–(aq) + 3H2O net reaction 509
Chapter 20
Eocell = Esubstance reduced − Esubstance oxidized or Eocell = Eoreduction − Eooxidation = 1.44 V – (0.54 V) = 0.90 V 20.61
The half–reaction having the more positive reduction potential is the reduction half–reaction, and the other is reversed to become the oxidation half–reaction: MnO2(s) + 4H+(aq) + 2e– Mn2+(aq) + 2H2O reduction Pb(s) + 2Cl–(aq) PbCl2(s) + 2e– oxidation MnO2(s) + 4H+(aq) + Pb(s) + 2Cl–(aq) Mn2+(aq) + 2H2O + PbCl2(aq) net reaction o o E ocell = E substance reduced – E substance oxidized = 1.23 – (–0.27) = 1.50 V
20.62
The half–reaction having the more positive standard reduction potential is the one that occurs as a reduction, and the other one is written as an oxidation: 2 × (2HOCl(aq) + 2H+(aq) + 2e– Cl2(g) + 2H2O) reduction 3H2O + S2O32–(aq) 2H2SO3(aq) + 2H+(aq) + 4e– oxidation 4HOCl(aq) + 4H+(aq) + 3H2O + S2O32–(aq) 2Cl2(g) + 4H2O + 2H2SO3(aq) + 2H+(aq) which simplifies to give the following net reaction: 4HOCl(aq) + 2H+(aq) + S2O32–(aq) 2Cl2(g) + H2O+ 2H2SO3(aq)
20.63
Br2(aq) + 2e– 2Br–(aq)
E cell = 1.07 V
Cl2(g) + 2e– 2Cl–(aq)
E cell = 1.36
Since the second of these has the larger reduction half–cell potential, it occurs as a reduction, and the first is reversed to become an oxidation: 2Br−(aq) + Cl2(g) → 2Cl−(aq) + Br2(aq) 20.64
The two half–reactions are: SO42–(aq) + 2e–+ 4H+(aq) H2SO3(aq) + H2O() 2I–(aq) I2(s) + 2e–
reduction oxidation
Eocell = Eoreduction − Eooxidation = 0.17 V – (0.54 V) = –0.37 V Since the overall cell potential is negative, we conclude that the reaction is not spontaneous in the direction written. 20.65
The two half–reactions are: S2O82– + 2e– 2SO42– Ni(OH)2 + 2OH– NiO2 + 2H2O + 2e–
reduction oxidation
o o E ocell = E substance reduced – E substance oxidized = 2.01 V – (0.49 V) = 1.52 V
Since the overall cell potential is positive, we conclude that the reaction is spontaneous in the direction written.
20.66
First, separate the overall reaction into its two half–reactions: 2Br–(aq) Br2(aq) + 2e– oxidation I2(s) + 2e– 2I–(aq) reduction
510
Chapter 20
Eocell = Eoreduction − Eooxidation = 0.54 V – (1.07 V) = –0.53 V The value of n is 2: ∆G° = –n F Eocell = –(2)(96,500 C)(–0.53 J/C) = 1.0 × 105 J = 1.0 × 102 kJ 20.67
Using the equation ∆G° = –n F E ocell , we have ∆G° = –n(96,500 C/mol e–)(1.69 V) for which we need n. Upon writing the two half–reactions, i.e., MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O reduction HCHO2(l) CO2(g) + 2H+(aq) + 2e– oxidation we see that we need to multiply the reduction half–reaction by 2 and the oxidation reaction by 5 in order to balance the eqaution: 2MnO4–(aq) + 16H+(aq) + 10e– 2Mn2+(aq) + 8H2O reduction 5HCHO2(l) 5CO2(g) + 10H+(aq) + 10e– oxidation The net reaction has n = 10. So, ∆G° = –(10 mol e–)(96,500 C/mol e–)(1.69 V) = –1.63 × 103 kJ.
20.68
(a) (b)
(c)
Eocell = Eoreduction − Eooxidation = 2.01 V – (1.47 V) = 0.54 V Since n = 10, ∆G° = –n F Eocell = –(10)(96,500 C)(0.54 J/C) = –5.2 × 105 J ∆G° = –5.2 × 102 kJ
RT Eocell = nF ln K c 0.54 V =
(8.314 J mol-1K -1 )(298 K)
ln K c 10(96,500 C mol-1 ) ln K c = 210.3 Taking the exponential of both sides of this equation: Kc = 2.1 × 1091 20.69
Ni2+ is reduced by two electrons and Co is oxidized by two electrons. E ocell = E oreduction − E ooxidation = –0.25 – (–0.28) = +0.03 V
E ocell =
0.0592 log K c n
+0.03 = (0.0592/2) × log Kc log Kc = 1 and Kc = 101 = 10 20.70
Sn is oxidized by two electrons and Ag is reduced by two electrons:
0.0592 log K c n –0.015 V = (0.0592 V/2) × log Kc log Kc = –0.51 Kc = antilog(–0.51) = 0.31 Eocell =
20.71
First, separate the overall reaction into two half–reactions: 2H2O 4H+ + 4e– + O2 oxidation reduction 2 × (Cl2 + 2e– 2Cl–)
E ocell = E oreduction − E ooxidation = 1.36 – (1.23) = 0.13 V
511
Chapter 20
0.0592 log K c n +0.13 V = (0.0592 V/4) log Kc log Kc = 8.78 Kc = antilog(8.78) = 6.08 × 108. E ocell =
20.72
This reaction involves the oxidation of Ag by two electrons and the reduction of Ni by two electrons. The concentration of the hydrogen ion is derived from the pH of the solution: [H+] = antilog (–pH) = antilog (–2.10) = 1 × 10–2.10 M 2
E cell
Ag + Ni 2 + 0.0592 V = 2.48 V − log 4 2 + H 2
3.5 × 10−2 3.5 × 10−2 0.0592 V = 2.48 V − log 4 2 − 3 7.9 × 10 Ecell = 2.48 V – 0.120 V = 2.36 V 20.73
The following half–reactions indicate that the value of n is 30: 5Cr2O72– + 70H+ + 30e– 10Cr3+ + 35H2O 3I2 + 18H2O 6IO3– + 36H+ + 30e– 6
E cell
10
IO3− Cr 3+ 0.0592 V = 0.135 V − log 34 5 30 + 2 H Cr2 O7 − 6
10
[0.00017] [0.0015] 0.0592 V = 0.135 V − log 30 [0.11]34 [ 0.012]5 = 0.135 V −
0.0592 V log 2.19 ×10−9 30
= 0.152 V
20.74
2+ RT Mg Ecell = E cell − ln n F Cd 2 + o
E cell = 1.97 −
(8.314 J mol-1K -1 )(298 K) -1
2(96,500 C mol )
1.45 V = 1.97 V − 0.01284 ln
ln
[1.00] Cd 2+
1 Cd 2+
ln(1/[Cd2+]) = 40.498 Taking (ex) of both sides: 1/[Cd2+] = 3.87 × 1017 [Cd2+] = 2.58 × 10–18 M
20.75
Since the copper half–cell is the cathode; this is the half–cell in which reduction takes place. The silver half–cell is therefore the anode, where oxidation of silver occurs. The standard cell potential is:
512
Chapter 20
E ocell = E oreduction − E ooxidation = 0.3419 V – 0.2223 V = 0.1196 V. The overall cell reaction is: Cu2+(aq) + 2Ag(s) + 2Cl–(aq) Cu(s) + 2AgCl(s), and the Nernst equation becomes:
E cell = 0.1196 V −
0.0592 V 1 log 2 2 Cu 2+ Cl−
If we use the values given in the exercise, we arrive at: 0.0895 V = 0.1196 V – 0.0296 V × log(1/[Cl–]2), which rearranges to give: log(1/[Cl–]2) = 1.017, [Cl–] = 0.310 M 20.76
In the iron half–cell, we are initially given: 0.0500 L × 0.100 mol/L = 5.00 × 10–3 mol Fe2+(aq) The precipitation of Fe(OH)2(s) consumes some of the added hydroxide ion, as well as some of the iron ion: Fe2+(aq) + 2OH–(aq) Fe(OH)2(s). The number of moles of OH– that have been added to the iron half–cell is: 0.500 mol/L × 0.0500 L = 2.50 × 10–2 mol OH– The stoichiometry of the precipitation reaction requires that the following number of moles of OH– be consumed on precipitation of 5.00 × 10–3 mol of Fe(OH)2(s): 5.00 × 10–3 mol Fe(OH)2 × (2 mol OH–/mol Fe(OH)2) = 1.00 × 10–2 mol OH– The number of moles of OH– that are unprecipitated in the iron half–cell is: 2.50 × 10–2 mol – 1.00 × 10–2 mol = 1.50 × 10–2 mol OH– Since the resulting volume is 50.0 mL + 50.0 mL, the concentration of hydroxide ion in the iron half–cell becomes, upon precipitation of the Fe(OH)2: [OH–] = 1.50 × 10–2 mol/0.100 L = 0.150 M OH– The standard cell potential is: Eocell = Eoreduction − Eooxidation = 0.3419 V – (–0.447 V) = 0.7889 V The Nernst equation is: 2+ RT Fe Ecell = E cell − ln n F Cu 2 + o
1.175 = 0.7889 −
(8.314 J mol-1K -1 )(298 K) 2(96,500 C mol-1 )
Fe 2+ ln [1.00]
1.175 = 0.7889 − 0.01284 ln Fe 2+ ln[Fe2+] = –30.08 [Fe2+] = 8.66 × 10–14 M 20.77
The half–cell reactions and the overall cell reaction are: Cu2+(aq) + 2e– Cu(s) E°red = +0.3419 V H2(g) 2H+(aq) + 2e– E°ox = 0.0000 V Cu2+(aq) + H2(g) Cu(s) + 2H+(aq) (a)
The standard cell potential is: 513
Chapter 20
E ocell = E oreduction − E ooxidation = 0.3419 V – 0 V = +0.3419 V The Nernst equation for this system is: 2
E cell =
E ocell
H + 0.0592 V − log 2 Cu 2 +
which becomes, under the circumstances defined in the problem:
E cell = E ocell −
2 0.0592 V log H + 2
Rearranging the last equation gives:
( 2 ) ( Ecell − Eocell ) 0.0592 V
= − log H +
2
which becomes the desired relationship:
(E
cell
o − E cell
0.0592 V (b)
)=
− log H + = pH
The equation derived in the answer to part (a) of this question is conveniently rearranged to give:
E cell = (0.0592 V)(pH) + E ocell = (0.0592 V)(5.15) + 0.3419 V = 0.647 V (c)
The equation that was derived in the answer to part (a) of this question may be used directly:
pH =
20.78
o
E =
E ocell
(E
cell
o − E cell
0.0592 V
( 0.645 V
− 0.3419 V )
0.0592 V
+ RT Ag dilute − ln n F Ag + conc
(8.314 J mol K ) ( 298 K ) ln [0.015] (1 mol ) ( 9.65 ×10 C mol ) [0.50] −1
Eo = 0 −
)=
−1 4
−1
Eo = 0.090 V
o
E =
E ocell
+ RT Ag dilute − ln n F Ag + conc
(8.314 J mol K ) ( 348 K ) ln [0.015] (1 mol ) ( 9.65 ×10 C mol ) [ 0.50] −1
Eo = 0 −
−1
4
−1
Eo = 0.105 V
514
= 5.12
Chapter 20
20.79
At 25 °C (298 K) o
E =
E ocell
2+ RT Cu dilute ln − n F Cu 2 + conc
(8.314 J mol K ) ( 298 K ) ln [0.015] (1 mol ) ( 9.65 ×10 C mol ) [0.50] −1
Eo = 0 −
−1 4
−1
Eo = 0.090 V At 75 °C (348 K)
(8.314 J mol K ) ( 348 K ) ln [0.015] (1 mol ) ( 9.65 ×10 C mol ) [ 0.50] −1
Eo = 0 −
−1
4
−1
Eo = 0.10 V 20.80
(a) (b) (c) (d)
20.81
20.82
Fe2+(aq) + 2e– Fe(s) 0.35 mol Fe2+ × 2 mol e–/mol Fe2+ = 0.70 mol e– Cl–(aq) 1/2Cl2(g) + e– 0.85 mol Cl– × 1 mol e–/mol Cl– = 0.85 mol e– Cr3+(aq) + 3e– Cr(s) 1.25 mol Cr3+ × 3 mol e–/mol Cr3+ = 3.75 mol e– Mn2+(aq) + 4H2O(l) MnO4–(aq) + 8H+(aq) + 5e– 2.5 × 10–2 mol Mn2+ × 5 mol e–/mol Mn2+ = 0.125 mol e–
(a)
Mg2+(aq) + 2e– Mg(s)
(b)
1 mol Mg 2 mol e − − mol e− = ( 4.75 g Mg ) = 0.391 mol e 24.31 g Mg 1 mol Mg Cu2+(aq) + 2e– Cu(s) 1 mol Cu 2 mol e − − mol e− = ( 45.0 g Cu ) = 1.42 mol e 63.55 g Cu 1 mol Cu
Fe(s) + 2OH–(aq) Fe(OH)2(s) + 2e– The number of Coulombs is: 15.0 min × 60 s/min × 8.00 C/s = 7.20 × 103 C. The number of grams of Fe(OH)2 is: 1 mol Fe(OH)2 89.86 g Fe(OH) 2 g Fe(OH) 2 = 7.20 × 103 C 1 mol Fe(OH) − 2 mol e 2 = 3.35 g Fe(OH)2
(
20.83
2Cl–(l) Cl2(g) + 2e– The number of Coulombs is: 4.50 A × 45.0 min × 60 s/min = 1.22 × 104 C The number of grams of Cl2 that will be produced is: 1 mol e − 1 mol Cl2 70.91 g Cl2 g Cl2 = 1.22 × 104 C = 4.46 g Cl2 96,500 C 2 mol e − 1 mol Cl 2 Cr3+(aq) + 3e– Cr(s) The number of Coulombs that will be required is:
(
20.84
)
)
515
Chapter 20
1 mol Cr 3 mol e − 96,500 C Coulombs = ( 85.0 g Cr ) = 4.73 × 106 C − 52.00 g Cr 1 mol Cr 1 mol e The time that will be required is: 1 s 1 hr hr = 4.73 × 106 C = 52.6 hr 2.50 C 3600 s
(
20.85
)
The number of Coulombs that will be required is: 1 mol Pb 2 mol e − 96,500 C Coulombs = ( 35.0 g Pb ) = 3.26 × 104 C − 207.2 g Pb 1 mol Pb 1 mol e The time that will be required is: 1 s 1 hr hr = 3.26 × 104 C = 6.04 hr 1.50 C 3600 s Mg2+ + 2e– Mg(l) The number of Coulombs that will be required is: 1 mol Mg 2 mol e− 96,500 C Coulombs = ( 45.0 g Mg ) = 3.57 × 105 C − 24.31 g Mg 1 mol Mg 1 mol e The number of amperes is: 3.57 × 105 C ÷ 7200 s = 49.6 amp
(
20.86
20.87
)
Al3+ + 3e– Al(s) The number of Coulombs that are required is:
1 mol Al 3 mol e − 96,500 C Coulombs = 409 × 103 g Al = 4.39 × 109 C − 26.98 g Al 1 mol Al 1 mol e
(
)
The number of amperes is: 4.39 × 109 C ÷ 8.64 × 104 s = 5.08 × 104 A (Note: There are 8.64 × 104 s in 24.0 hr.)
20.88
The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of Coulombs is: 2.00 A × 20.0 min × 60 s/min = 2.40 × 103 C. The number of moles of OH– is: 1 mol e − 2 mol OH − mol OH − = 2.40 × 103 C = 0.0249 mol OH − 96,500 C 2 mol e − [OH-] = 0.0249 mol/0.250 L = 0.0996 M
(
20.89
)
The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of seconds is: 25.0 min × 60 s/min = 1.50 × 103 s. The number of moles of OH– is: − − 1.45 C 1 mol e 2 mol OH mol OH − = 1.50 x 103 s − s 96,500 C 2 mol e
(
)
2.25 × 10−3 mol OH − −2 − OH - = 2.25 x 10 mol OH 0.100 L − 2 Cl → Cl2 + 2e−
= 0.225 M
516
=
Chapter 20
1 mol Cl2 mol OH − ) = 1.125 x 10−2 mol Cl2 − 2 mol OH Assuming Cl2 and OH- react to make OCl−
( 2.25 x 10
OCl- = 20.90
−2
1.125 x 10−2 mol = 0.122 M 0.100 L
Possible cathode reactions: Al3+ + 3e– Al(s) –
E° = –1.66 V –
2H2O + 2e H2(g) + 2OH (aq) E° = –0.83 V
20.91
Possible anode reactions: S2O82– + 2e– 2SO42–
E° = +2.05 V
O2 + 4H+ + 4e– 2H2O
E° = +1.23 V
Cathode reaction:
2H2O + 2e– H2(g) + 2OH–(aq)
E° = –0.83 V
Anode reactions:
2H2O O2 + 4H+ + 4e–
E° = –1.23 V
Net cell reaction:
2H2O 2H2(g) + O2(g)
Possible cathode reactions: Cd2+ + 2e– Cd(s)
E° = –0.40 V
2H2O + 2e– H2(g) + 2OH–(aq)
E° = –0.83 V
Possible anode reactions: O2 + 4H+ + 4e– 2H2O
E° = +1.23 V
–
I2(s) + 2e 2I
20.92
–
E° = –2.06 V
E° = +0.54 V
Cathode reaction:
Cd2+ + 2e– Cd(s)
Anode reaction:
2I– I2(s) + 2e–
E° = –0.54 V
Net reaction:
Cd2+ +2I– I2(s) + Cd(s)
E° = –0.94 V
E° = –0.40 V
The answers to the previous Review Problems guide us here: Possible cathode reactions: K+ + e– K(s) E° = –2.92 V Cu2+ + 2e– Cu(s) –
E° = +0.34 V –
2H2O + 2e H2(g) + 2OH (aq)
E° = –0.83 V
Cathode reaction: Cu2+ + 2e– Cu(s) Possible anode reactions:
517
Chapter 20
2SO42– S2O82– + 2e–
E° = –2.01 V
2Br– Br2 + 2e–
E° = –1.07 V +
2H2O O2(g) + 4H (aq) + 4e
–
E° = –1.23 V
Anode reaction: 2Br– Br2 + 2e– Overall reaction: Cu2+ + 2Br– Br2 + Cu(s) 20.93
At the cathode, where reduction occurs, we expect Cu(s). At the anode, where oxidation occurs, we expect I2(aq). The net cell reaction would be Cu2+(aq) + 2I–(aq) Cu(s) + I2(aq)
20.94
In aqueous solution the following reduction of water can occur: 2H2O(l) + 2e−
→
H2(g) + 2OH-(aq)
Eo = − 0.83 V
Reactions that are less positive than this cannot occur at the cathode. Therefore, Al3+, Mg2+, Na+, Ca2+, K+, and Li+ would not be reduced at the cathode. 20.95
In aqueous solution the following oxidation of water can occur: 2H2O(l)
→
O2(g) + 4H+(aq) + 4e−
E(oxidation) = −1.23 V
Reactions that are more negative than this cannot occur at the anode. Therefore, Cl−, Au, Br−, Pb2+, Mn2+, Cl2, PbSO4, SO42−, and F− would not be oxidized at the anode. Additional Exercises 20.96
∆G° = –n F E cell , E cell = 1.34 V = 1.34 J/C, and n = 2 ∆G° = –(2)(96,500 C)(1.34 J/C) = –2.59 × 105 J per mol of HgO The maximum amount of work that can be derived from this cell, on using 1.10 g of HgO, is thus:
1 mol HgO 2.59 × 105 J 3 J = (1.10 g HgO ) = 1.32 × 10 J 216.6 g HgO 1 mol HgO Now, since 1 watt = 1 J s–1, then 2 × 10–3 watt = 2 × 10–3 J s–1, and the time required for this process is:
1s 1 min 1 hr hr = 1.32 × 103 J = 183 hr −3 60 s 60 min 2 × 10 J
(
20.97
)
The initial numbers of moles of Ag+ and Zn2+ are: 1.00 mol/L × 0.100 L = 0.100 mol. The number of Coulombs (A × s) that have been employed is: 0.10 C/s × 18.00 hr × 3600 s/hr = 6.5 × 103 C. The number of moles of electrons is: 6.5 × 103 C ÷ 96,500 C/mol = 6.7 × 10–2 mol electrons. For Ag+, there is 1 mol per mole of electrons, and for Zn2+, there are two moles of electrons per mol of Zn. This means that the number of moles of the two ions that have been consumed or formed is given by: 6.7 × 10–2 mol e– × 1 mol Ag+/1 mol e– = 6.7 × 10–2 mol Ag+ reacted. 6.7 × 10–2 mol e– × 1 mol Zn2+/2 mol e– = 3.4 × 10–2 mol Zn2+ formed 518
Chapter 20
The number of moles of Ag+ that remain is: 0.100 – 0.067 = 0.033 mol of Ag+ The final concentration of silver ion is: [Ag+] = 0.033 mol/0.100 L = 0.33 M The number of moles of Zn2+ that are present is: 0.100 + 0.034 = 0.134 mol Zn2+ The final concentration of zinc ion is: [Zn2+] = 0.134 mol/0.100 L = 1.34 M The standard cell potential should be: E ocell = E oreduction − E ooxidation = 0.80 – (–0.76) = 1.56 V We now apply the Nernst equation:
E cell = E ocell −
0.0592 V 1.34 log 2 ( 0.33)2
Ecell = 1.56 V – 0.032 V = 1.53 V 20.98
The concentration of Br– in solution will be 0.10 M, since the Ksp of AgBr is so small, the amount of dissociation of AgBr from the electrode will be negligible. The reduction reaction will be AgBr(s) + e– Ag(s) + Br–(aq) E° = 0.070 V The oxidation reaction will occur at the standard hydrogen electrode. The net cell reaction is: 2AgBr(s) + H2 2Ag(s) + 2Br–(aq) + 2H+(aq) E cell = 0.070 V – 0.000 V = 0.070 V The potential for the constructed cell is calculated using the Nernst equation: 2 2 0.0592 V log Br − H + n 0.0592 2 2 E cell = 0.070 V − log [ 0.1] [1] 2 Ecell = 0.070 V – (–0.0592) Ecell = 0.129 V
E cell = E ocell −
20.99
Our strategy will be thus: Step A: Pressure H2 (wet) partial pressure H2 mol H2 # e– used Step B: Find total charge used = (current)(time) Step C: The charge per electron can be arrived at by: Charge per e– = total charge/total #e– used = Step A/Step B Step A: Pressure H2 (wet) partial pressure H2 mol H2 # e– used The total pressure of wet hydrogen is 767 torr, but some of this is provided by water vapor. Consulting the water vapor pressure table in the appendices, we find that at 27 ˚C, the vapor pressure of water is 26.7 torr. Therefore pressure solely due to hydrogen gas (the partial pressure of hydrogen gas) is: 519
Chapter 20
PH2 = 767 – 26.7 = 740 torr 740 torr(1 atm/760 torr) = 0.974 atm Using the ideal gas law, PV = nRT (0.974 atm)(0.288 L) = n(0.0821 L·atm mol–1·K–1)(27 + 273K) n = 0.0114 mol H2 According to the electrolysis equation 2H–+ + 2e– H2(g), 2 moles of electrons are required per mol of H2 gas formed. Therefore, electrons = 0.0114 mol H2(2 mol e–/1 mol H2)(6.022 × 1023 electrons/mol) = 1.37 × 1022 electrons Step B: Total charge used = (1.22 A)(1800 s) = 2200 C Step C: Charge per e– = total charge/total # e– used = 2200 C/1.37 × 1022 electrons = 1.61 × 10–19 C per electron (This is a very good estimate; the accepted value is 1.60 × 10–19 C.) 20.100 The oxidation reaction is: Fe2+ + 2e– Fe The reduction reaction is: 2H+ + 2e– H2 The overall cell reaction is 2H+ + Fe H2 + Fe2+
Ecell =
E cell
E° = –0.44 V E° = 0.00 V
E cell = 0.44 V
Fe2 + 0.0592 V log – 2 n H+
[Fe2+] = 0.12 M [H+] is from the ionization of acetic acid H + C 2 H 3O 2 − = 1.8× 10–5 Ka = HC2 H3 O 2 [H+] = x [C2H3O2–] = x [ x ][ x ] 1.8 × 10–5 = [0.18 − x ]
[HC2H3O2] = 0.10 – x
Assume x is small compared to the concentration of HC2H3O2 x = 1.8 × 10–3 M = [H+] = [C2H3O2–] [HC2H3O2] = 0.18 M n=2 Ecell = 0.44 V –
[0.12] 0.0592 V log 2 2 1.8 × 10−3 520
Chapter 20
Ecell = 0.44 V – 0.14 V Ecell = 0.30 V 20.101
The approach is as follows: Area x thickness Cr volume Cr mass Cr moles Cr # e– needed current VCr = (area)(thickness) = (1.00 m2)(4.0 × 10–5 m) = 4.0 × 10–5 m3 3
100 cm 7.19 g Cr 1 mol Cr 6 F 96,500 C e − = 4.0 × 10−5 m3 1 m 1 cm3 52.0 g Cr 1 mol Cr 1 F = 3.20 × 106 C
(
)
This is done in 4.50 hr (16,200 s). So the current must be: Current = charge/time = 3.20 × 106 C/16,200 s = 198 A 20.102 (a)
First, we calculate the number of Coulombs: 1.50 A × 30.0 min × 60 s/min = 2.70 × 103 A·s = 2.70 × 103 C Then we determine the number of moles of electrons: 1 mol e− mole e– = (2.70 × 103 C) = 0.0280 mol e– 96,500 C
(b) (c)
0.475 g ÷ 50.9 g/mol = 9.33 × 10–3 mol V (2.80 × 10–2 mol e–)/(9.33 × 10–3 mol V) = 3.00 mol e–/mol V The original oxidation state was V3+.
20.103
F2 Cl2 Br2 I2
+2.87 +1.36 +1.07 +0.54
Electron Affinitiy (kJ.mol) -328 -348 -325 -295
Li+ Na+ K+ Rb+ Cs+
-3.05 -2.71 -2.92 -2.93 -2.92
-60 -53 -48 -47 -45
520 496 419 403 376
1.0 1.0 0.9 0.9 0.9
Mg2+ Ca2+ Sr2+ Ba2+
-2.32 -2.76 -2.89 -2.90
+230 +155 +176 +50
738 590 549 503
1.3 1.1 1.0 0.9
Eo (V)
521
Ionization Energy1 (kJ/mol) 1682 1251 1140 1008
Electronegativity
4.1 2.9 2.8 2.2
Chapter 20
(1) wiki.chemeddl.org
(a) Reduction potentials decrease in the same order as electronegativity and ionization energy. Electron affinity follows no pattern. (b) Reduction potentials and the properties decrease going down the family, though Li is out of order due to its small size and larger effective nuclear charge compared to the other members of the family. (c) Reduction potentials and the properties decrease going down the family though the electron affinity value for Sr2+ is out of order.
20.104
RT ln Cl − nF (8.314 J mol−1K −1 )(298 K) 0.0532 V = 0.2223 V − ln Cl − 1(96,500 C mol−1 ) E cell = Ecell −
−0.1691 V =
(8.314 J mol−1K −1 )(298 K) 1(96,500 C mol−1 )
ln Cl −
−6.586 = ln Cl −
1.379 × 10−3 = Cl− 20.105 The half–reactions diagrammed in this problem are: Ag(s) Ag+(aq) + e– anode E° = –0.80 V Fe3+(aq) + e– Fe2+(aq) cathode E° = 0.77 V E cell = E reduction – E oxidation = 0.77 V – 0.80 V = –0.03 V
Fe2 + Ag + 0.0592 V log 1 Fe3+ 5.5 × 10−4 3.8 × 10−2 0.0592 V = − 0.03 V − log 1 1.2 × 10−3 = + 0.07 V
E cell =E ocell − E cell
As stated, being a galvanic cell, and using conventions in cell notation, the left-side half–cell is the anode and negatively charged, and the right–side half–cell is the cathode and positively charged. The equation for the spontaneous cell reaction is Fe3+(1.2 × 10–3 M) + Ag(s) Ag+(5.5 × 10–4 M) + Fe2+(0.038 M). This is an example of a concentration cell. Multi-Concept Problems 20.106 The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of Coulombs is: 2.00 A × 20.0 min × 60 s/min = 2.40 × 103 C. The number of moles of OH– is: 1 mol e − 2 mol OH − mol OH − = 2.40 × 103 C = 0.0249 mol OH − 96,500 C 2 mol e −
(
)
522
Chapter 20 The volume of acid solution that will neutralize this much OH– is: 1 mol HCl 1000 mL HCl mL HCl = 0.0249 mol OH − = 40.1 mL HCl 1 mol OH − 0.620 mol HCl
(
)
20.107 The electrolysis of NaCl solution results in the reduction of water, together with the formation of chlorine gas and hydroxide ion: 2Cl–(aq) Cl2(g) + 2e–. The number of Coulombs is: 2.50 A × 15.0 min × 60 s/min = 2.25 × 103 C. The number of moles of Cl2 gas collected is: 1 mol e− 1 mol Cl 2 mol Cl2 = 2.25 × 103 C = 0.0117 mol Cl2 96,500 C 2 mol e − The volume of Cl2 gas that is collected is:
(
)
(
)
L atm 298 K ) ( 0.0117 mol ) 0.0821 mol nRT K ( = 0.299 L = 299 mL = P 1 atm ( 750 torr − 23.76 torr ) 760 torr 20.108 2H+(aq) + 2e– H2(g) The number of Coulombs is: 15.00 min × 60 s/min × 0.750 C/s = 675 C. The number of moles of H2 is:
V=
1 mol e − 1 mol H 2 mol H 2 = ( 675 C ) = 3.50 × 10−3 mol H 2 96,500 C 2 mol e− Finally, we calculate the volume of H2 gas:
(
)
L atm 293 K ) ( 0.00350 mol ) 0.0821 mol nRT K ( = 0.0870 L = 87.0 mL = P 1 atm 735 torr ( ) 760 torr 20.109 In the iron half–cell, we are initially given: 0.0500 L × 0.100 mol/L = 5.00 × 10–3 mol Fe2+(aq)
V=
The precipitation of Fe(OH)2(s) consumes some of the added hydroxide ion, as well as some of the iron ion: Fe2+(aq) + 2OH–(aq) Fe(OH)2(s). The number of moles of OH– that have been added to the iron half–cell is: 0.500 mol/L × 0.0500 L = 2.50 × 10–2 mol OH– The stoichiometry of the precipitation reaction requires that the following number of moles of OH– be consumed on precipitation of 5.00 × 10–3 mol of Fe(OH)2(s): 5.00 × 10–3 mol Fe(OH)2 × (2 mol OH–/mol Fe(OH)2) = 1.00 × 10–2 mol OH– The number of moles of OH– that are unprecipitated in the iron half–cell is: 2.50 × 10–2 mol – 1.00 × 10–2 mol = 1.50 × 10–2 mol OH– Since the resulting volume is 50.0 mL + 50.0 mL, the concentration of hydroxide ion in the iron half–cell becomes, upon precipitation of the Fe(OH)2: [OH–] = 1.50 × 10–2 mol/0.100 L = 0.150 M OH– We have assumed that the iron hydroxide that forms in the above precipitation reaction is completely insoluble. This is not accurate, though, because some small amount does dissolve in water according to the following equilibrium: Fe(OH)2(s) Fe2+(aq) + 2OH–(aq)
523
Chapter 20
This means that the true [OH–] is slightly higher than 0.150 M as calculated above. Thus we must set up the usual equilibrium table, in order to analyze the extent to which Fe(OH)2(s) dissolves in 0.150 M OH– solution:
I C E
[Fe2+] – +x +x
[OH–] 0.150 +2x 0.150+2x
The quantity x in the above table is the molar solubility of Fe(OH)2 in the solution that is formed in the iron half–cell. Ksp = [Fe2+][OH–]2 = (x)(0.150 + 2x)2 The standard cell potential is: o E ocell = E oreduction − E oxidation = 0.3419 V – (–0.447 V) = 0.7889 V The Nernst equation is:
E cell =
E ocell
Fe2+ RT − ln nF Cu 2+
1.175 = 0.7889 −
(8.314 J mol-1K -1 )(298 K) 2(96,500 C mol-1 )
Fe 2+ ln 1.00 [ ]
1.175 = 0.7889 − 0.01284 ln Fe 2+ ln[Fe2+] = –30.07 [Fe2+] = 8.72 × 10–14 M This is the concentration of Fe2+ in the saturated solution, and it is the value to be used for x in the above expression for Ksp. Ksp = (x)(0.150 + 2x)2 = (8.72 × 10–14)[0.150 + (2)(8.72 × 10–14)]2 Ksp = 1.96 × 10–15 20.110 The desired reaction is: AgBr(s) Ag+(aq) + Br−(aq) While this does not look like an oxidation-reduction reaction, we can use Hess’s law and the half-reaction provided in the problem to construct the desired reaction. Reverse the direction of the first reaction and keep the second reaction in the given direction. Be sure to change the sign of the cell potential when the reaction direction is changed. Ag
→
Ag+ + e−
Eo = − 0.80 V
AgBr
→
Ag + Br− + e−
Eo = + 0.07 V
Sum the two reactions and their cell potentials:
524
Chapter 20
AgBr
→ Ag+ + Br− Eocell = − 0.73 V
Now use the Nernst equation to determine the equilibrium constant for the reaction.
At equilibrium, Ecell = 0
0.0592 V log Ag + Br − 1 - 0.73 V = −12.3 log Ag + Br − = log K sp ( AgBr ) = 0.0592V K sp = 4.7 x 10−13 E cell =E ocell −
20.111 The range of mmoles of Cl− is: 0.096 M x 3.00 mL = 0.288 mmol Cl− 0.106 M x 3.00 mL = 0.318 mmol Cl− − 1 mol 1 mol e Coulombs = 0.288 mmol x 1000 mmol 1 mol Cl→ The time that will be required is: 1s seconds = ( 27.8 C ) = 55.6 s 0.500 C
96,500 C = 27.8 C − 1 mol e
− 1 mol 1 mol e Coulombs = 0.318 mmol x 1000 mmol 1 mol Cl→
96,500 C = 30.7 C − 1 mol e
1s seconds = ( 30.7 C ) = 61.4 s 0.500 C The range of time is thus, 55.6 to 61.4 seconds to precipitate the chloride ions as AgCl. The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide 20.112 ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of seconds is: 25.0 min × 60 s/min = 1.50 × 103 s. The number of moles of OH– is:
0.250 mol H + mol OH − = 15.5 mL H + 1000 mL H +
(
)
1 mol OH − 1 mol H +
−3 − = 3.87 × 10 mol OH
The number of Coulombs that will form this much OH– is:
2 mol e− Coulombs = 3.87 × 10−3 mol OH − 2 mol OH − The average current in amperes is: 3.74 × 102 C Current = = 0.250 A 1.50 × 103 s
(
)
525
96,500 C = 3.74 × 102 C − 1 mol e
Chapter 20
20.113
1 H 2 ( g ) + O 2 ( g ) → H 2 O(g) 2 o o ∆GT ≈ ∆H 289 - T∆So298 1000 J −1 −1 - 383 K{(1 mol H 2 O)(188.7 J mol K ) kJ
= (1 mol H 2 O )(-241.8 kJ mol −1 )
− [(1 mol H 2 )(130.6 J mol−1 K −1 ) + (
1 2
mol O 2 )(205.0 J mol −1 K −1 )]}
= − 2.248 x 105 J or 224.8 kJ The is the maximum amount of work obtained by the reaction of one mole of hydrogen gas.
( 2.248 x 10 J ) 1Jwatt = 2.248 x 10 s 5
−1
5
watt s
Therefore, one mole of hydrogen gas will produce 224.8 kW s. We want enough hydrogen gas to produce 1 kW s. The mass of H2, at 100% efficiency would be:
1 mol H 2 2.02 g −3 = 8.98 x 10 g H 2 224.8 kW s mol H 2
(1 kW s )
Since the efficiency is only 70% we will need:
8.98 x 10−3 g = 1.28 x 10−2 g H 2 0.70
526
Chapter 21
Practice Exercises 21.1
m0
m=
1 −
(vc )
2
45.9g
m=
(
−1 1 − 75 m s
3.00x108 m s −1
)
2
= 45.9 g
The golf ball is too large and travelling too slow to have any relativistic effect on its mass. 21.2
45.9g
m=
(
6 −1 1 − 7.5x10 m s
3.00x108 m s −1
)
2
= 46.5 g
A slight relativistic effect on mass occurs at the higher velocity. 21.3
20 10
Ne has 10 protons and 10 neutrons.
10 protons have a mass of: 10 neutrons have a mass of:
10 x 1.0072764669 u = 10.072764669 u 10 x 1.0086649156 u = 10.086649156 u
The total rest mass of the isotope is: 20.159413825 u The mass defect is: 21.4
10 5
20.159413825 u – 19.992440 u = 0.166974 u
B has 5 protons and 5 neutrons.
5 protons have a mass of: 10 neutrons have a mass of:
5 x 1.0072764669 u = 5.0363823345 u 5 x 1.0086649156 u = 5.043324578 u
The total rest mass of the isotope is: 10.079706912 u The mass defect is:
10.079706912 u – 10.012937 u = 0.066770 u E = mc2
The binding energy is given by:
E = 0.066770 u x
1.66065389 x 10−27 kg x (2.9979 x108 m s −1 ) 2 u
= 9.965 x 10-12 J There are ten nucleons in this isotope. Therefore, the bonding per nucleon is given as: 9.965 x 10-12 J/10 nucleons = 9.965 x 10-13 J/nucleon 21.5
226 88 Ra
→
222 86 Rn
+
4 2 He
+
0 0γ
An alpha particle is emitted.
527
Chapter 21
21.6
90 38 Sr
21.7
11 6
C
21.8
13 4
Be
21.9
72 34
Se +
21.10
242 96
21.11
59 26
21.12
Using the value of k for Pu from Example 21.2: Activity = kN Activity = 6.22 × 1011 Bq = 6.22 × 1011 disintegrations/second k = 2.50 × 10–10 seconds–1 6.22 × 1011 disintegrations/second = (2.50 × 10–10 seconds–1)N N = 2.49 × 1021 atoms Pu Now find the mass of Pu 1 mole Pu 244 g Pu mass Pu = (2.49 × 1021 atoms Pu) 1 mole Pu = 1.01 g Pu 23 6.022 × 10 atoms Pu The percentage of Pu in the sample: 1.01 g Pu × 100% = 50.4% Pu 2.00 g sample
21.13
Half life for Rn-222 = 4 days x (24 h/day) x (3600 s/hr) = 345,500 s k = ln2/t1/2 = (0.6931)/(345,500 s) = 2.01 × 10–6 s–1
→
Cm
Fe
90 39Y
+
11 5
→
0 −1
B + 10 e + ν
12 4
→
Be + 72 33
→
e
238 94
→ →
0 −1 e
59 27
1 0
n
As + X ray + ν
Pu +
Co +
0 −1
4 2
Ηe
e
Activity = 4 pCi = 4 × 10–12 Ci (3.7 × 1010 dps/1Ci) = 0.1482 Bq = 0.148 disintegrations per second Activity = disintegrations/sec = kN 0.148 = (2.01 × 10–6 s–1)N N = 7.36 × 104 atoms Rn-222
21.14
We make use of the Inverse Square Law: I1 d 2 = 22 I2 d1 2
( x m) 4.8 units = 0.30 units ( 5.0 m )2 x m = 20 m
528
Chapter 21
21.15
We make use of the Inverse Square Law: I1 d 2 = 22 I2 d1 2
(1.2 m ) 1.4 units = = 100 units (assuming 1 significant figure) I2 (10 m )2 Review Questions 21.1
We do not normally, in the course of everyday life, encounter nuclear reactions. In typical chemical reactions, the mass of the various materials that are involved is unaffected by the relatively slow velocities that these materials possess.
21.2
If the velocity of an object approaches the speed of light, then the denominator on the right side of equation 21.1 would take on a value increasingly close to zero. This would require the mass m of the object to increase, and, in the limit, become infinity.
21.3
(a) (b)
21.4
The sum of all of the energy in the universe, plus all of the mass equivalent of energy, is constant. ∆E = ∆mo × c2
The rest mass of a particle is the mass when it is stationary. However, it is not possible to measure the mass of a stationary particle. Therefore, the mass of a particle is measured while it is in motion. The rest mass is determined by applying special relativity corrections to the mass. Mass appears to increase, relative to an observer, as a particle approaches the speed of light. We use rest mass since the speed of a particle is difficult to determine, and in most cases is not relevant for our calculations.
21.5
A small amount of mass is converted to energy, and lost from the system, when the nucleons assemble into a stable atomic nucleus. The amount of mass that is converted into energy corresponds to the gain in stability that the nucleus achieves by loss of this amount of energy.
21.6
Iron-56 has the largest binding energy per nucleon. If the binding energy is too small nuclear fission occurs, producing species that are more stable.
21.7
A substance is radioactive if it emits particles and/or electromagnetic radiation from an unstable atomic nucleus. Radioactive decay is used to describe the phenomenon because the rate at which radiation is emitted depends on the concentration of the radionuclides in the sample. As the radioactive nuclei decay, the number of radiocuclides remaining decreases and the intensity of the radiation drops, or decays.
21.8
Naturally occurring radionuclides emit alpha particles (alpha radiation), beta particles, and gamma radiation.
21.9
(a) (b) (c) (d)
21.10
The alpha particle is more massive than the other particles, and this, coupled with its high positive charge, makes it likely that an alpha particle will collide with something soon after it is ejected. This collision normally transforms the alpha particle into a helium atom, by gain of two electrons.
21.11
Gamma rays are emitted from the nucleus, whereas X rays arise from electronic transitions.
The alpha particle is composed of helium nuclei, and has a 2+ charge. Beta particles are electrons. Positrons are particles with a +1 charge, and have the same mass as the electron. A deuteron consists of one neutron and one proton, i.e. the nucleus of a deuterium atom.
529
Chapter 21
21.12
Electron capture of a low–level electron by the nucleus creates a low–level hole in the electron configuration that is filled by descent of an electron from an upper electronic level. The atom emits radiation corresponding to the difference in energies between the two electronic levels, and this radiation lies in the X ray region of the electromagnetic spectrum.
21.13
Beta decay originates in the nucleus. The weak force intereaction converts a neutron into a proton, emitting an electron and an electron antineutrino.
21.14
For each atom, the number of protons is plotted against the number of neutrons. The band is established by the location in this plot of all known stable nuclei.
21.15
Barium–140 should have the longer half-life because it has an even number of protons and neutrons.
21.16
Tin–112 has an even number of protons and neutrons. Both of these are odd for indium–112 (atomic number 49).
21.17
Lanthanum–139 is a stable nuclide because it has 82 neutrons, a magic number. Additionally, lanthanum– 140 has both an odd number of protons and an odd number of neutrons which implies instability.
21.18
There must be a sufficient number of neutrons to shield the protons from one another, especially so as the number of protons becomes increasingly large.
21.19
The loss of an alpha particle is the most effective way to move the nuclide toward the band of stability, if the unstable nuclide lies to the right and above the band of stability, because the alpha particle has a relatively large mass.
21.20
The neutron-to-proton ratio of lead–164 is too low. It has too few neutrons for the number of protons in the nucleus.
21.21
The loss of a beta particle is most likely, because the net nuclear effect is the conversion of a neutron into a proton, and this reduces the neutron–to–proton ratio.
21.22
The loss of a positron is most likely, because the net nuclear effect is the conversion of a proton into a neutron, and this increases the neutron–to–proton ratio.
21.23
The net effect of electron capture is the conversion of a proton into a neutron, and this increases the neutron–to–proton ratio. Radionuclides lying below the band of stability are more likely to undergo electron capture than those lying above the band of stability.
21.24
A compound nucleus is unstable because of excess energy.
21.25
Because the compound nucleus has no memory of how it was prepared, it may decay by any path which lowers the energy. The actual pathway depends on the amount of excess energy the compound nucleus has.
21.26
4 2 He
21.27
Neutrons have zero charge. Therefore, when they collide there is no charge repulsion. Alpha particles and protons are charged and thus experience coulombic repulsion.
21.28
The Geiger counter detects the ions that form in an enclosed gas through which radiation is passing. The radiation must have high enough energy to penetrate the tube holding the gas and the radiation must then be able to ionize the gas.
21.29
Radiation generates unstable and reactive ions within cells. These reactive ions can participate in reactions that can eventually lead to birth defects, tumors, mutations, and cancer.
+
14 7N
→
18 * 9F
→
1 0n
+
17 9F
530
Chapter 21
21.30
(a) (b) (c) (d)
The common unit of radioactivity is the curie (Ci), whereas the SI unit of radioactivity is the becquerel (Bq). The unit of energy for radiation is the electron–volt, or its multiples and the SI unit is the joule. The SI unit of absorbed dose is the gray (Gy), while the rad is an older common unit. The SI unit of dose equivalent is the sievert (Sv) and the rem, or roentgen equivalent for man, is the older common unit.
21.31
The sievert accounts for the type of radiation exposure and the type of tissue as well as the dose.
21.32
A short half-life radioactive compound means the body will be exposed to more events than from one with a long half-life. Therefore, the short half-life is more dangerous. The danger also depends on the type of emissions.
21.33
A short half–life is desirable in medical work in order to minimize the exposure that a patient receives from radioactive materials. This reduces radiation–caused damage to the body. If the half–life is too short, the radionuclide will not be present in the body long enough to permit useful measurements.
21.34
Alpha particles cannot reach a detector outside the body, because they cannot penetrate tissue. Furthermore, as tissues in the body capture alpha particles, they are damaged.
21.35
A sample possessing a specific element is bombarded with neutrons creating a neutron enriched sample for analysis. These samples decay by gamma emission at frequencies that are characteristic of the element in question. By analyzing the frequency of the emission, the identity of the emitting element may be determined. By analyzing the intensity of the gamma emission, the concentration may be determined.
21.36
The assumption is that the lead in a mixture being analyzed has all come from the disintegration of uranium only.
21.37
Contamination with air could introduce new sources of carbon–containing compounds which would completely invalidate any analysis.
21.38
Some of the forms of radiation that everybody experiences are cosmic rays, diagnostic X rays, radioactive pollutants, fallout from atmospheric tests of nuclear devices, and, in some cases, radiation used in medicine.
21.39
(a) (b)
(c) (d)
A thermal neutron is one whose kinetic energy is governed by the temperature of its surroundings, even at room temperature. Nuclear fission is the breakup of a heavy nucleus, normally caused by absorption of a neutron, to give two or more lighter nuclei, plus two or more high-energy neutrons (i.e. not simply thermal neutrons). A fissile isotope is one that is capable of undergoing fission following neutron capture. Nuclear fusion is the formation of atomic nuclei by the joining together of the nuclei of lighter atoms.
21.40
The naturally occurring fissile isotope is uranium–235.
21.41
The fission of each uranium–235 produces more than two new neutrons, which can react further to sustain a chain reaction. In other words, more neutrons are generated by each fission event than are needed to cause another such event.
21.42
In fission reactions, large nuclei are broken into smaller nuclei. In terms of stability of a nucleus, the necessary ratio of protons to neutrons decreases as the atomic number decreases. Thus there is an excess of neutrons in a fission reaction. For example, the initial isotopes of Kr and Ba that are produced by the fission of uranium–235 are neutron–rich materials. In other words, their neutron–to–proton ratios are too high, and they spontaneously emit the excess neutrons to form Kr–94 and Ba–139. 531
Chapter 21
21.43
The sub–critical mass is incapable of self–capturing all of the emitted neutrons, which are necessary for sustaining a chain reaction.
21.44
A moderator in a nuclear reactor is used to slow down the fast neutrons that are produced by fission.
21.45
No critical mass can form because the atoms of the fissionable isotopes are greatly diluted by those of the non–fissile isotopes.
21.46
2 3 1H + 1H
→ 42 He + 01 n Tritium must be synthesized for this reaction because it is radioactive and has a relatively short half–life, so it does not occur in nature in large quantities.
21.47
In order to construct a fusion reactor, the main issue is overcoming the electrostatic force. That is the repulsion between the protons in the two nuclei as they approach each other. In order to overcome the electrostatic force, the atoms need to be heated to over 10 million degrees Celsius and there needs to be a high concentration of the nuclei so that they can react with each other.
Review Problems 21.48
Equation 21.1 becomes: 1.00 m= 2 1 − v c
( )
On substituting the various velocities for the value of v in the above expression, and using the value c = 3.00 × 108 m s–1, we get: (a) m = 1.01 kg (b) 3.91 kg (c) 12.3 kg 21.49
m=
mo 1 −
(vc )
2
This equation can be rearranged to give: 1
1
mo2 2 1.000 kg 2 2 8 −1 v = c 1 − 3.00 x 10 m s 1 = − m 2 m2 On substituting the various masses for the value of m in the above expression we get: (a) v = 2.99 x 107 m s−1 (b) 1.25 x 108 m s−1 (c) 2.94 x 108 m s−1 21.50
Solve the Einstein equation for ∆m: ∆m = ∆E/c2 1 kJ = 1.25 × 103 J = 1.25 × 103 kg m2 s–2 ∆m = 1. 25 × 103 kg m2 s–2 ÷ (3. 00 × 108 m/s)2 = 1.39 × 10–14 kg = 1.39 × 10–11 g
21.51
∆E = ∆mc2 =(1.25 g x (1 kg/1000 g))(3.00 × 108 m/s)2 ∆E = 1.1 × 1014 kg m2/s2 = 1.1 × 1014 J
21.52
The joule is equal to one kg m2/s2, and this is employed directly in the Einstein equation: ∆m = ∆E/c2, where ∆E is the enthalpy of formation of liquid water, which is available in Table 7.2.
532
Chapter 21
H2(g) + O2(g) H2O(l), ∆H = –285.9 kJ/mol (2 mol) = − 571.8 kJ ∆m = (–571.8 × 103 kg m2/s2) ÷ (3.00 × 108 m/s2)2 = –6.35 × 10–12 kg (–6.35 × 10–12 kg) × 1000 g/kg × 109 ng/g = –6.35 ng The negative value for the mass implies that mass is lost in the reaction. The percent of mass that is lost is 6.35 × 10−12 kg –8 ×100% = 3.52 × 10 % −3 18.02 × 10 kg 21.53
Solve the Einstein equation for ∆m: ∆m = ∆E/c2 1780 kJ = 1780 × 103 J = 1780 × 103 kg m2 s–2 ∆m = 1780 × 103 kg m2 s–2 ÷ (3.00 × 108 m/s)2 = 1.98 × 10–11 kg = 1.98 × 10−8g = 19.8 ng Note: we have assumed three sig figs for this calculation in order to agree with the text.
21.54
The mass of the deuterium nucleus is the mass of the proton (1.0072764669 u) plus that of a neutron (1.0086649156 u), or 2.015941381 u. The difference between this calculated value and the observed value is equal to ∆m: ∆m = (2.015941 – 2.0135) = 2.44 × 10–3 u ∆E = ∆mc2 = (2.44 × 10–3 u)(1.6606 × 10–27 kg/u)(3.00 × 108 m/s)2 ∆E = 3.65 × 10–13 kg m2/s2 = 3.65 × 10–13 J Since there are two nucleons per deuterium nucleus, we have: ∆E = 3.65 × 10–13 J/2 nucleons = 1.8 × 10–13 J per nucleon
21.55
The mass of the tritium nucleus is the mass of one proton plus that of two neutrons: 1.007276470 + 2(1.008664904) = 3.024606 u. The difference between this calculated value and the observed value is equal to ∆m: ∆m = 3.024606 – 3.01550 = 9.11 × 10–3 u ∆E = ∆mc2 = (9.11 × 10–3 u)(1.6605 × 10–27 kg/u)(3.00 × 108 m/s)2 ∆E = 1.36 × 10–12 kg m2/s2 = 1.36 × 10–12 J Since there are three nucleons per tritium nucleus, the energy per nucleon is: 1.36 × 10–12 J/3 nucleons = 4.53 × 10–13 J per nucleon
21.56
63 29
Cu has 29 protons and 34 neutrons.
29 protons have a mass of: 34 neutrons have a mass of:
29 x 1.0072764669 u = 29.2110175401 u 34 x 1.0086649156 u = 34.02946071304 u
The total rest mass of the isotope is: 63.50562477 u The mass defect is:
63.50562477 u – 62.9529601 u = 0.5526646 u
The binding energy is given by:
E = 0.5526646 u x
E = mc2
1.66065389 x 10−27 kg x (2.9979 x108 m s −1 ) 2 u
= 8.2485 x 10-11 J 533
Chapter 21
21.57
The mass of the americium-241 nucleus is the mass of 95 proton2 (95 x 1.0072764669 u) plus that of 146 neutrons (146 x 1.0086649156 u), or 242.956342034 u. The difference between this calculated value and the observed value is equal to ∆m: ∆m = (242.956342034 – 241.05682) = 1.90 u ∆E = ∆mc2 = (1.90 u)(1.6606 × 10–27 kg/u)(3.00 × 108 m/s)2 ∆E = 2.84 × 10–10 kg m2/s2 = 2.84 × 10–10 J
21.58
(a)
211 83 Bi
(b)
177 72 Hf
(c)
216 84 Po
(d)
19 9F
21.59
(a)
241 94 Pu
(b)
146 57 La
(c)
58 30 Zn
(d)
68 31 Ga
21.60
(a)
242 4 238 94 Pu 2 He + 92 U 28 0 28 12 Mg −1 e + 13 Al 26 0 26 14 Si 1 e + 13 Al 37 0 37 18 Ar + −1 e 17 Cl
(b) (c) (d) 21.61
55 26 Fe 42 19 K 93 44 Ru 251 98 Cf
(a) (b) (c) (d)
0 −1 e −01 e + 01 e + 42 He
+
55 25 Mn 42 20 Ca 93 43Tc + 247 96 Cm
21.62
(a)
261 102 No
(b)
211 82 Pb
(c)
141 61 Pm
(d)
179 74W
21.63
(a)
80 38 Sr
(b)
121 50 Sn
(c)
50 25 Mn
(d)
257 100 Fm
21.64
87 36 Kr
21.65
58 26 Fe
21.66
The more likely process is positron emission, because this produces a product having a higher neutron–to– 38 proton ratio: 19 K 01 e + 38 18 Ar
21.67
Electron capture is the more likely event:
86 36 Kr
+
1 0n
37 18 Ar
+
0 −1 e
37 17 Cl
This produces chlorine–37, which has a magic number, i.e., 20 neutrons. Beta emission would give potassium–37, which is less stable. 21.68
The most probable decay is alpha emission. 209 84
21.69
Po
205 82
→
Pb +
4 2
He
Silicon-32 decays by beta emission. 32 14 Si
0 −1 e
+
32 15 P
534
Chapter 21
21.70
Six half–life periods correspond to the fraction 1/64 of the initial material. That is, one sixty–fourth of the initial material is left after 6 half lives: 4.00 mg × 1/64 = 0.0625 mg remaining.
21.71
After four half–life periods, one sixteenth of the original sample remains: (8.00 × 10–9 g) × 1/16 = 5.00 × 10–10 g remaining
21.72
53 * 24 Cr
21.73
19 9F
21.74
80 35 Br
21.75
115 48 Cd
+
21.76
55 26 Fe
55 25 Mn
21.77
23 11 Na
+
4 2 He
21.78
70 30 Zn
+
208 82 Pb
21.79
209 83 Bi
+
64 28 Ni
21.80
Radiation ∝
forms. The reactions are: 4 2 He
+
;
23 * 11 Na
1 0n
116 48 Cd
+
1 1p
27 13 Al
22 11 Na
2 1H
+
+
53 * 24 Cr
1 0n
+ γ 1 0n
+
55 26 Fe
1 0n
+ γ
278 112 Uub 273 111 Rg
51 23V
→
1 0n
+
+
277 112 Uub
272 111 Rg
1 d2
I1 d 2 = 22 I2 d1 d 2 = d1
21.81
I1 2.8 = 1.8m = 5.7 m I2 0.28
I1 d 2 = 22 I2 d1 d2 = d1
I1 = I2
100 = 1.05 90
Increase the distance by 5% to decrease exposure by 10%. 21.82
This calculation makes use of the Inverse Square Law: I1 d 2 = 22 I2 d1
d 22 8.4 rem = 0.50 rem (1.50 m )2 d2 = 6.1 m 21.83
This calculation makes use of the Inverse Square Law:
535
1 1p
+
52 23V
Chapter 21
I1 d 2 = 22 I2 d1 2
( 0.40 m ) 50 mrem = I2 ( 4.0 m )2 I2 = 5.0 × 103 mrem or 5.0 rm 21.84
This is one curie. It is also: 1.0 Ci × (3.7 × 1010 Bq/Ci) = 3.7 × 1010 Bq
21.85
( 3.5 x 10
Bq 8 Ci 3.7 x 1010 = 1.3 x 10 Bq Ci
)
disintegrations 8 Ci 3.7 x 1010 = 1.3 x 10 disintegrations Ci ln 2 ln 2 Activity = kN and t1 2 = or k= k t1 2
( 3.5 x 10
21.86
−3
)
−3
Activity =
ln 2 N t1 2
1 g 241 Am 1 mol N = 0.22 mg 1000 mg 241 g
Am 6.022 × 1023 atoms 241 Am 1 mol 241 Am
241
241
Am 17 241 = 5.5 × 10 atoms Am
ln 2 17 241 Activity = × (5.5 × 10 atoms Am) 1.70 × 105 d 2.2 × 1012 Am decays 1 d 1 h 7 –1 = 3600 s = 2.6 × 10 s d 24 h 1000 mCi 1 Ci –1 2.6 × 107 s–1 = 2.6 × 107 Bq = 2.6 × 107 Bq = 7.0 × 10 mCi 10 1 Ci 3.7 × 10 Bq = 6.5 × 102 µCi 21.87
Activity = kN Activity =
and
t1 2 =
ln 2 k
k=
or
ln 2 t1 2
ln 2 N t1 2
1 mol 90 Sr 6.022 × 1023 atoms N = 1.10 g 90 g 90 Sr 1 mol 90 Sr
90
Sr 21 90 = 7.4 × 10 atoms Sr
ln 2 21 290 Activity = × (7.4 × 10 atoms Sr) 1.00 × 104 d 5.1× 1017 Sr decays 1 d 1 h 12 –1 = 3600 s = 5.9 × 10 s d 24 h 1000 mCi 1 Ci 5 5.9 × 1012 s–1 = 5.9 × 1012 Bq = 5.9 × 1012 Bq = 1.6 × 10 mCi 10 1 Ci 3.7 × 10 Bq
536
Chapter 21
21.88
Activity = kN
k=
activity N
1 g 131 I 1 mol N = number of 131I atoms = 1.00 mg 131I 1000 mg 131 I 131 g 18 131 = 4.60 × 10 atoms I 12 1 s −1 4.6 × 10 Bq –6 –1 k= = 1.0 × 10 s 4.60 × 1018 atoms 131I 1 Bq t1 2 =
21.89
I 6.022 × 1023 g 131 I 1 mol 131 I
131
k=
activity N
1 g 201 Tl 1 mol N = number of 201Tl atoms = 10.0 mg 201Tl 1000 mg 201 Tl 201 g 19 201 = 3.00× 10 atoms Tl 1 s −1 7.9 × 1013 Bq –6 –1 k= = 2.64 × 10 s 3.00 × 1019 atoms 201Tl 1 Bq
21.90
I
ln 2 ln 2 = = 6.9 × 105 s (This is about 8 days.) −6 −1 k 1.0 × 10 s
Activity = kN
t1 2 =
131
Tl 6.022 × 1023 g 201 Tl 201 Tl 1 mol 201 Tl
201
ln 2 ln 2 = = 2.63 × 105 s −6 −1 k 2.64 × 10 s
The chemical product is BaCl2.Recall that for a first order process k =
0.693 t1 2
So k = 0.693/30.1 yr = 2.30 × 10–2/yr. Also, [A] 0 ln = kt [A] t
[A] t = [A] 0 exp( − kt) [A] 0 = exp[− (2.30 × 10−2 /yr)(160 yr)] [A] t [A] 0 = 2.52 × 10−2 [A] t so 2.5% of the original sample remains. 21.91
For a first–order process, the rate constant is related to the half–life by the equation: k = 0.693/t1/2 Hence, k = 0.693/28.1 yr = 2.47 × 10–2 yr–1 Also, for a first–order process, the concentration varies with time according to the equation: ln which allows us to solve for the time, t, for the activity to decrease by the specified amount:
t=
(
)
0.245 Ci g [A] 0 1 1 × ln = × ln = 409 yr k [ A ] t 2.47 × 10−2 yr −1 1.00 × 10−5 Ci g
(
)
537
[A] 0 [A ] t
= kt
Chapter 21
21.92
This calculation makes use of the first order rate equation, where knowing [A]t, we need to calculate [A]0: [A] 0 ln = kt . [A ] t k = 0.693/t1/2 = 0.693/8.07 d = 8.59 × 10–2 d–1 [A] 0 ln = 8.59 × 10−2 d −1 ( 29.0 d ) 25.6 × 10−5 Ci g
(
(
)
)
Taking the exponential of both sides of the above equation gives:
[A] 0
( 25.6 ×
10−5 Ci
g
)
= e 2.49 = 12.1
Solving for the value of [A]0 gives: [A]0 = 3.09 × 10–3 Ci/g 21.93
The rate constant for the first–order process is first determined: k = 0.693/t1/2 = 0.693/6.02 hr = 0.115 hr–1 Also, we know that: ln
( 4.96 × 10 ln
−6
Ci
[A] t
)=
[A] 0 [A ] t
= kt
( 0.115 hr ) (8.00 hr ) = 0.920 −1
Taking the exponential of both sides of this equation gives:
( 4.96 × 10
−6
Ci
[ A ]t
) = antiln (0.920) = 2.51
[A]t = 1.98 × 10–6 Ci, the activity after 8 hours. 21.94
In order to solve this problem, it must be assumed that all of the argon–40 that is found in the rock must have come from the potassium–40, i.e., that the rock contains no other source of argon–40. If the above assumption is valid, then any argon–40 that is found in the rock represents an equivalent amount of potassium–40, since the stoichiometry is 1:1. Since equal amounts of potassium–40 and argon–40 have been found, this indicates that the amount of potassium–40 that remains is exactly half the amount that was present originally. In other words, the potassium–40 has undergone one half–life of decay by the time of the analysis. The rock is thus seen to be 1.3 × 109 years old.
21.95
In a fashion similar to that outlined in the answer to Review Problem 21.94, we conclude that, in order for the rock to be one half–life old (1.3 × 109 yr), there must be equal amounts of the two isotopes, one having been formed by decay of the other. The answer is, thus, 1.16 × 10–7 mol of potassium–40.
21.96
14 C ln 12 = 1.2 × 10−4 t C Taking the natural log we determine:
(
1.2 × 10−12 ln 4.8 × 10−14
)
−4 = 1.2 × 10 t
(
)
538
Chapter 21
−12 1 1.2 × 10 t= ln = 2.7 × 104 yr −4 −14 1.2 × 10 4.8 × 10 The tree died 2.7 × 104 years ago. This is when the volcanic eruption occurred.
21.97
ln
r0 = kt rt
1.2 × 10−12 ln rt
−4 −1 9.0 × 103 y = 1.21 × 10 y
(
)(
)
1.2 × 10−12 rt
−4 −1 9.0 × 103 y = exp 1.21 × 10 y
1.2 × 10−12 2.97
−13 = rt = 4.0 × 10
21.98
235 92 U
21.99
Both will be beta emitters. Both lie above the band of stability, and they can mover closer to it by emitting beta particles. Some of the extra neutrons produced by the fission shown in Review Problem 21.98 may bombard additional U–235 nuclei to cause further fission products to form.
+
1 0n
(
94 38 Sr
)(
+
140 54 Xe
)
+ 2 01n
Additional Exercises 21.100 The half –life of tritium is 12.26 yrs. Assuming a sample of pure tritium, in 40 years Then 40 yrs/12.26 yrs = 3.26 half-lives have occurred.
1 2
3.26
= 0.104
Thus, 10.4 % of the original sample remains. The sum of the tritium and helium-3 would represent the original amount of tritium in the sample. 21.101 (a) (b) (c) (d)
30 0 30 13 Al -1 e + 14 Si 252 4 248 99 Es 2 He + 97 Bk 93 0 93 42 Mo + -1 e 41 Nb 28 0 28 15 P 1 e + 14 Si
21.102 The mass of the iron–56 nucleus is the mass of twenty–six protons plus the mass of thirty neutrons. The difference between this calculated value and the observed value is equal to m: (26 × 1.007276470 u) + (30 × 1.008664904 u) = 56.449135 u ∆m = (56.449135 u – 55.9349 u) = 0.5142 u ∆E = ∆mc2 = (0.5142 u)(1.6605 × 10–27 kg/u)( 2.9979 × 108 m/s)2 ∆E = 7.685 × 10–11 kg m2/s2 = 7.685 × 10–11 J Since there are fifty-six neucleons per iron–56 nucleus, we have: ∆E = 7.685 × 10–11 J/56 nucleons = 1.37 × 10–12 J per nucleon.
539
Chapter 21
This has to be the largest binding energy possible based upon the information provided in Figure 21.1. According to this Figure, iron–56 is the most stable isotope known. 21.103 (a) (b) (c) (d)
10 6C 243 96 Cm 49 23V + 20 8O
0 1e
+
10 5B
4 2 He
0 −1 e 0 −1 e
+
+
239 94 Pu
49 22Ti 20 9F
21.104 None. The parent mass may be exactly the same as the daughter mass since the positron has a mass of zero. 21.105 Using the masses of the proton and antiproton and converting it into energy, we can then determine the wavelength of the two photons. The mass of the antiproton is equal to the mass of the proton: 2 × 1.0072764669 u = 2.014552932 u 1.66065389 × 10−27 kg 8 –1 2 –10 E = (2.014552932 u) × × (2.99792458 × 10 m s ) = 3.00676313 × 10 J 1 u Each photon will have half the energy:
2.99792458 ×10−10 J = 1.503381564 × 10–10 J 2 The wavelength for the photon:
(
)(
6.6260690 ×10−34 J s 2.99792458 ×108 m s −1 E λ= = hc 1.50338164 × 10−10 J The decimal multiplier should be femto: 1.32131826 fm
) = 1.32131826 × 10
–15
m
21.106 In order to fuse, these two nuclei must overcome the repulsive force that exists between particles of like charge. This is only possible by adding huge amounts of thermal energy to a mixture of the particles or by using extreme pressure. 21.107
214 83 Bi 210 81Tl 210 82 Pb 210 83 Bi 210 84 Po
4 210 2 He + 81Tl 0 210 −1 e + 82 Pb 0 210 83 Bi −1 e + 0 210 −1 e + 84 Po 4 206 2 He + 82 Pb
Element E is 21.108 (a) (b)
206 82 Pb
15 7N
ln
[A] 0 [A ] t
= kt .
k = 0.693/t1/2 = 0.693/124 s = 5.59 × 10–3 s–1
ln
750 mg = 5.59 × 10−3 s −1 ( 300 s ) [A] t
(
)
[A]t = 140 mg
540
Chapter 21
21.109 Positron decay 21.110 Recall that for a first order process k = 0.693/t1/2 So k = 0.693/1.3 × 109 yr = 5.30 × 10–10/yr. Also, [A] 0 ln = kt [A] t
t =
[A] 0 1 1 3.22 × 10−5 mol ln ln = = 8.2 × 108 yrs − 10 k [A] t 5.30 × 10 /yr 2.07 × 10−5 mol
21.111 Recall that for a first order process k = 0.693/t1/2 So k = 0.693/5730 yr = 1.21 × 10–4/yr. Also, [A] 0 ln = kt [A] t
t =
[A] 0 1 1 ln = k [A] t 1.21 × 10
−4
/yr
ln
9 = 1.82 × 104 yrs 1
21.112 Because this is an equilibrium process, there is a chance that either the forward or the reverse reactions will occur. As NO reacts with NO2 we form ONO*NO. This compound can decompose and give either NO2 and *NO or NO and *NO2. After sufficient time, both *NO or NO will be present in the mixture. 21.113 As time passes, equal parts of CH3HgI and CH3*HgI will be produced. Since half the atoms are labeled, half the products will be labeled. 21.114 This problem is similar to a dilution problem, i.e., C1V1 = C2V2.We will use cpm as the concentration unit. First, we need to account for the density difference. 580 cpm 0.792 g cpm Methanol: concentration = = 459 g mL mL
29 cpm 0.884 g cpm Coolant: concentration = = 26 mL g mL Now we want the volume of the cooling system. Solve the following: 459 cpm mL (12.0 mL ) = 212 mL 26 cpm mL 21.115 Starting with the half-life of 131I, 8.07 d, calculate k. ln 2 t 12 = k ln 2 k= t 12
0.693 = 0.0859 d −1 8.07 d Then calculate the desired activity from the mass of the thyroid and the dose: 86 µCi A = (22 g thyroid) = 1890 µCi g k=
541
Chapter 21
3.7 × 1010 disintegrations/sec 1 µCi 7 A = 1890 µCi 6 = 7.00 × 10 disintegrations/sec 1 µ Ci 10 Ci Now calculate the number of 131I atoms that will give this activity: A = kN 1 d 1 h 7.00 × 107 disintegrations/sec = 0.0859 d–1 24 h 3600 s N 7.00 × 107 disintegrations/sec = 9.94 × 10–7 N N = 7.04 × 1013 131I atoms 1 mol 131I Mass of 131I = 7.04 × 1013 131I atoms 6.022 × 1023
131 g 131I =1.53 × 10–8 g 131I 1 mol 131I
21.116 A 235 92 U fission bomb could serve as a trigger for a fusion bomb because the reaction of neutrons and the neutrons could be used in the reaction of Li–6 to form He–4 and H–3.
238 92 U
produces
21.117 Cf–245 should be Cf–248: 248 48 294 1 98 Cf + 20 Ca 118 Uuo + 2 0 n 21.118 The activity is 4.1 pCi L–1. It must be converted to disintegrations per second and the decay constant needs to be determined from the half life. 24 h 3600 s 5 t 1 = 3.823 d 1 h = 3.303 × 10 s 2 1 d k=
ln 2 ln 2 = = 2.099 × 10–6 s–1 t1 3.303 × 105 s 2
Activity = kN Activity = 4.1 × 10
N=
–12
3.7 ×1010 dis s −1 = 1.5 × 10-1 dis s–1 L–1 Ci L Ci –1
Activity 0.15 dis s −1 L−1 = = 7.23 × 104 atoms 222Rn L–1 k 2.099 ×10−6 s −1
21.119 A gray (Gy) is one Joule of energy per kilogram of absorbing material, and a rad is 100 Gy. First we need to calculate the number of Gy first we need to calculate how many disintegrations are occurring, and then how much energy is released.
1g 145 Pr 1.2 mg 145Pr = 0.0012 g 145Pr 1000 mg 140 Bq Number of disintegrations = 0.0012 g 145Pr = 0.168 Bq g In 10 minutes (600 seconds) the number of disintegrations is: 0.168 Bq(600 s) = 101 disintegrations 1.80 MeV 106 eV 1.6022 × 10−19 J –11 101 disintegrations = 2.91 × 10 J di sin tegration 1 MeV eV If the person weighs 70 kg, then the Gy is 2.91× 10−11 J = 4.15 × 10–13 Gy 70 kg
542
Chapter 21
To convert to rad: 1 rad –15 4.15 × 10–13 Gy = 4.15 × 10 rad 100 Gy 21.120 Elements with an odd number of protons form stable isotopes when the number of neutrons is even. Thus, Cl-35 and Cl-36 are stable. Cl-36 has an odd number of protons and neutrons, an unstable configuration. The isotope formed is 36 17 Cl
→
36 18 Ar
+
36 18 Ar
0 −1 e
Multi-Concept Questions 21.121 Start by determining an activity per mol of Cr and mol C2O42– For Cr: 1 mol K 2 Cr2 O7 843 cpm 294.18 g 5 = 1.24 × 10 cpm/mol Cr g 1 mol K Cr O 2 mol Cr 2 2 7 For oxalate 345 cpm 90.04 g 4 2− = 3.1 × 10 cpm/mol C2 O 4 g 1 mol Again this is an equilibrium problem. 165 cpm Moles Cr = = 1.33 × 10−3 mole Cr 1.24 × 105 cpm/mol Cr
83 cpm
= 2.67 × 10−3 mole C 2 O 4 2 − 3.11 × 10 cpm/mol C 2 O 4 2 − It is easy to see by inspection that there are 2 moles C2O42– for each mole of Cr(III)
Mole C2O42– =
21.122
30
4
MBq 106 Bq 1 cm 1 Ir x x 140 mm x x =1.8 x 1015 Ir atoms cm MBq 10 mm Bq
30 MBq x
106 Bq 1 Ci x = 8.1 x 10−4 Ci MBq 3.7 x 1010 Bq
Lead will protect against beta emission.
543
Chapter 22
Practice Exercises 22.1
[Ag(S2O3)2]3-
(NH4)3Ag(S2O3)2
22.2
AlCl3•6H2O
[Al(H2O)6]3+
22.3
[CrCl2 (H2O)4]+
the counter ion would be a halide
22.4
(a) (b) (c)
[SnCl6]2– (NH4)2[Fe(CN)4(H2O)2] OsBr2(H2NCH2CH2NH2)2 or OsBr2(en)2
22.5
(a) (b) (c)
potassium hexacyanoferrate(III) dichlorobis(ethylenediamine)chromium(III) sulfate hexaaquacobalt(II) hexafluorochromate(II)
22.6
Since there are three ligands and C2O42– is a bidentate ligand, the coordination number is six.
22.7
(a) (b)
22.8
(c)
The coordination number is six. There are two bidentate ligands and two monodentate ligands. The coordination number is six. Both C2O42– and ethylenediamine are bidentate ligands. Since there are three bidentate ligands, the coordination number must be six. EDTA is a hexadentate ligand so the coordination number is six.
(a)
The coordination number is four. There are four monodentate ligands attached to the metal. The coordination geometry is square planar.
(b)
The coordination number is two. There are two monodentate ligands attached to the metal. The coordination geometry is linear with the thiosulfate ions attached to the silver ion through the sulfur atoms.
Review Questions 22.1
22.2
22.3
The metal ion is a Lewis acid, accepting a pair of electrons from the ligand, which serves as a Lewis base. (a) The Lewis acid is Cu2+, and the Lewis base is H2O. (b) The ligand is H2O. (c) Water provides the donor atom, oxygen. (d) Oxygen is the donor atom, because it is attached to the copper ion. (e) The copper ion is the acceptor. A ligand serves as a Lewis base because it donates the two electrons that are required for the formation of the coordinate covalent bond with the metal ion. They are formed by the use of coordinate covalent bonds.
22.4 NH3 H3N
2+ Cu
NH3
Cl
2+
NH 3
Cl
2+ Cu Cl
544
2-
Cl
Chapter 22
22.5
A bidentate ligand must have two atoms that can form a coordinate covalent bond with a metal to give a ring including the metal.
22.6
A chelate is a compound formed between a metal ion and a polydentate ligand. The multiple sites at which a coordinate covalent bond may form enable these ligands to “grab” the metal ion very strongly. The oxalate ion, C2O42–, is an example of a bidentate ligand that forms a chelate with a metal ion. The ligand is able to form coordinate covalent bonds using the lone pairs of electrons on each terminal oxygen atom. O
C C
C
O
O M
n+
22.7
EDTA has six donor atoms.
22.8
EDTA forms complexes with the sorts of metal ions that would otherwise promote spoilage.
22.9
EDTA forms complexes with the metal ions that are otherwise responsible for hardness.
22.10
The net charge is 1–, and the formula is [Co(EDTA)]–.
22.11
[Cr(en)3]3+ is more stable since the ligands are bidentate.
22.12
The coordination number of a metal complex is the number of ligand atoms that are attached to the metal. The common coordination geometries of complex ions having coordination number 4 are the square plane and the tetrahedron. The most common structure observed for coordination number six is octahedral
22.13
22.14 O O
N Co
O
N
CH 2 CH 2
O The curved lines represent –CH2–C=O
22.15
Isomers are compounds that have the same chemical formula, but that are also distinct and different substances, owing to differences in the arrangements of the atoms.
22.16
Stereoisomerism is the existence of isomers that differ only in the spatial orientation of their atoms. Geometric isomerism is the existence of isomers that differ because their molecules have different geometries. Chiral isomers are isomers that differ because one is the non-superimposable mirror image of the other. 545
Chapter 22
Enantiomers are chiral isomers of one another, i.e., they are non-superimposable mirror images of one another. 22.17
In a cis isomer, like groups are located adjacent to one another, whereas in a trans isomer, like groups are located opposite one another.
22.18
Chirality requires the nonsuperimposability of mirror images.
22.19
Optical isomers are chiral isomers, i.e. isomers that differ because they are non-superimposable mirror images on one another, and they rotate plane polarized light.
22.20
Geometric isomers have the same structural formula but differ in the arrangement of the atoms. They also have different physical properties. Optical isomers differ in that their mirror images are not superimposable. These isomers interact with light. Their physical properties such as boiling point and melting point are identical.
22.21
See Figure 22.13.
22.22
As shown in Figure 22.13 the orbitals that are oriented between the axes are: dxy, dxz, dyz, whereas those that are oriented along the axes are: d x 2 − y2 and d z2 .
22.23
The d x 2 − y2 and d z2 orbitals are located near the ligands, whereas the dxy, dxz, and dyz, orbitals are located between (farther from) the ligands.
22.24
See Figure 22.15
Energy
22.25
∆ ∆
Co 2+
Co3+
As indicated in the figure above, the oxidation of Co2+ to give Co3+ removes an electron from a higher energy level. The resulting electron configuration is more stable. Additionally, the energy separation, ∆, increases as the oxidation state increases. This also stabilizes the compound and further explains the ease of oxidation. 22.26
Various ligands may be bound to a metal, giving different complexes, all having their own characteristic value for ∆. As the value of ∆ changes, so too does the frequency of light that a complex absorbs.
22.27
(a)
22.28
The spectrochemical series is a list of ligands, arranged in their order of increasing ability to produce a large value of ∆. This series is determined by measuring the frequencies of light that are absorbed by a series of complexes having the various ligands.
Green-blue
(b)
Violet-blue
546
Chapter 22
22.29
The high–spin case is the one in which the maximum number of unpaired electrons is found. The converse is true of the low–spin case; it is the one having the fewest possible number of unpaired electrons.
22.30
Both high– and low–spin configurations are possible only for the following configurations: d4, d5, d6, and d 7.
22.31
As the charge of the metal ion increases, ∆ increases. Also, as the metal ion’s radius decreases ∆ increases. The following ligand factors decrease ∆, decreasing electronegativity (see the spectrochemical series p 1036. F− causes a larger splitting than I−);decreasing basicity (for example, CN− is a stronger base than F−); increasing ligand size which causes a decrease in orbital overlap between the metal d orbitals and the ligand orbitals ( for example, F− is smaller than I−). An increase in π-bonding ability (so called π-acceptor ligands increases ∆. CO and CN− are good examples of strong π-acceptors and the halide ions are good examples of strong π-donors; poor π-acceptors.
22.32
There is promotion of an electron into the upper energy level, by absorption of a photon having energy equal to E = ∆ = hν:
+ hν
22.33
This is a complex of Co3+. The complex is known to be diamagnetic, so no unpaired electrons are present. We conclude that it must be a low spin complex:
22.34
The d z2 , dxz, and dyz orbitals will become somewhat stabilized because the given change should result in less electron-electron repulsions along the axes. Conversely, the dx2–y2 and dxy orbitals will become destabilized, as electron repulsions along the x and y axes are increased:
22.35
It is the porphyrin ligand that is found in chlorophyll, heme and vitamin B12.
22.36
The porphyrin ring ligand by itself produces four-coordinate covalent bonds to a central metal ion having a coordination of 4.
22.37
The heme is the oxygen carrier in hemoglobin and myoglobin. 547
Chapter 22
22.38
Both have square planar structures. Co2+
22.39
The calcium ion is essential for the overall health and well being of the body. The most critical role calcium plays is that it is essential to bones. Calcium is necessary to maintain bone density. Insufficient levels of calcium may lead to loss in bone density and result in osteoporosis.
Review Problems 22.40
The net charge is –3, and the formula is [Fe(CN)6]3–. The IUPAC name for the complex is hexacyanoferrate(III) ion
22.41
This is the ion [Ag(NH3)2]+, which can exist as the chloride salt: [Ag(NH3)2]Cl. The IUPAC name for the complex is diamminesilver(I) ion.
22.42
[CoCl4(en)]−
22.43
[Cr(NH3)4(NO2)]+
22.44
(a) (c)
C2O42– oxalato Cl– chloro
(b) (d)
S2– (CH3)2NH
sulfido or thio dimethylamine
22.45
(a) (c)
NH3 SO42–
(b) (d)
N3– C2H3O2–
azido acetato
22.46
(a) (b) (c) (d) (e)
tetraaquadicyanoiron(III) ion tetraamineoxalatonickel(III) bisaquatetracyano iron(III) ion potassium hexathiocyanatomanganate(III) tetrachlorocuprate(II) ion
22.47
(a) (b) (c) (d) (e)
tetrachloroaurate(III) ion bis(ethylendiamine)dinitroiron(II) ion tetraaminedicarbonatocobalt(III) nitrate ethylendiaminetetraacetoferrate(II) ion diaminedichloroplatinum(II)
22.48
(a) (b) (c) (d) (e)
[Ni(NH3)6]2− [CrCl3(NH3)3]− [Co(NO2)6]3− Mn(CN)4(NH3)2]2− [Fe(C2O4)3]
22.49
(a) (b) (c) (d) (e)
[AgI2]– [SnS3]2− [Co(en)2(H2O)4]2(SO4)3 [CrCl(NH3)5]SO4 K3[Co(C2O4)3]
22.50
The coordination number is six, and the oxidation number of the iron atom is +2.
ammine sulfato
548
Chapter 22
22.51
C2O42– is bidentate × 2 NO2– is monodentate × 2 Coordination number = 6
22.52
(a)
(b)
The curved lines represent the backbone of the oxalate ion. 22.53
Co3+ compounds are 6 coordinated. Therefore, two waters are needed to complete the structure.
(a)
(b)
22.54 O N
O M
O
OH 2
OH2 The curved lines represent –CH2–C(=O)– groups
549
Chapter 22
22.55
(a)
The nitrogen atoms are the donor atoms.
(b)
This is 2 × 3 = 6.
(c) 3+
N N
N Co
N
N N
(d)
Co(dien)23+, due to the chelate effect
(e)
H2
H2 C
N
H2 C
H N
H2 C
H2 C
H N
H2 C
H2 C
N
H2
22.56
Since both are the cis isomer, they are identical. One can be superimposed on the other after simple rotation.
22.57
Yes, the isomers are chiral since the mirror image may not be superimposed on the original.
22.58
cis:
trans: Br Pt Cl
NH3
Br
NH 3
H3 N
Pt
NH3 Cl
22.59 Cl Cl
Cl
NH 3
H 3N
Co Cl
NH 3
H 3N
NH3
O
3-
O
3-
O
O
O
O
O
Co O
Cl Cl
22.60 O
NH 3 Co
O Co O O
550
Chapter 22
22.61
The cis isomer is chiral:
The trans isomer is nonchiral +
N N
Cl Co N
Cl N
[Cr(H2O)6]2+
(b)
[CrCl6]3−
22.62
(a)
22.63
[Cr(CN)6]3– < [Cr(NO2)6]3– < [Cr(en)3]3+ < [Cr(NH3)6]3+ < [Cr(H2O)6]3+ < [CrF6]3– < [CrCl6]3–
22.64
[Cr(H2O)6]3+
22.65
(a)
22.66
(a) The value of ∆ increases down a group. Therefore, we choose: [FeCl(NH3)5]3+ (b) The value of ∆ increases with oxidation state of the metal. Therefore, we choose: [Ru(NH3)6]2+
22.67
Ligand B produces the smaller splitting. The absorbed colors are the complements of the perceived colors: CoA63+ – absorbed color is green CoB63+ – absorbed color is red The absorbed color with the highest energy (shortest wavelength) is green. We conclude that ligand B is lower in the spectrochemical series.
22.68
This is the one with the strongest field ligand, since Co2+ is a d7 ion: CoA63+. You are removing a higher energy electron in CoA62+ than in CoB62+. A produces a larger ligand field splitting than does B.
22.69
The color of CoA63+ is red. The complex with the lower oxidation state should have a smaller value for ∆. Therefore, the 2+ ion should absorb in the red, and appear blue.
22.70
This is a weak field complex of Co2+, and it should be a high–spin d7 case. It cannot be diamagnetic; even if it were low spin, we would still have one unpaired electron.
22.71
For Fe(H2O)63+, we expect a relatively small value for ∆, and we predict the high–spin case having five unpaired electrons:
[Fe(CN)6]4−
(b)
[Mn(CN)6]3–
551
Chapter 22 Fe(H2O)63+
Fe(CN)63–
For Fe(CN)63–, we expect a relatively large value for ∆, and we predict the low–spin case having one unpaired electron:
Additional Exercises 22.72
The geometry of four coordinated nickel complexes is typically square planar while that of six coordinated nickel is octahedral. Ni2+ is a d8 system. In a square planar geometry the transition would be from the dxy to the d x 2 − y 2 . This separation is smaller for the same ligand type than the separation of the d-orbitals in a octahedral field. The smaller the transition energy the larger the wavelength for the transition. Therefore, square planar nickel should absorb in the red region and the solutions would appear blue. Octahedral nickel complexes would be more likely to appear red since they would absorb light in the blue region of the spectra
22.73
SCN− is linear
The complex is square planar, as shown above. The Pd has an oxidation state of +4. 22.74
The typical electron configuration for the first row transition elements is 4s2 3dn. They tend to lose the 2 electrons in the 4s orbital to form +2 ions.
22.75 -
O O
N Co
O
N
-
O
CH 2
H2C
CH2
H 2C
O
N
O Co
N
O O
The mirror images are not superimposable. 22.76
Two different substituents on a tetrahedron cannot be arranged in isomeric ways. That is, all possibilities are the same, being superimposable on each other. Tetrahedral form: Cl Pt Cl
NH 3 NH 3
552
Chapter 22
The square planar forms can form two isomers: Cl Pt Cl
NH 3
Cl
NH3
H 3N
Pt
NH 3 Cl
22.77
For a linear complex with the z axis as the intermolecular axis, the d z 2 orbital will experience the greatest repulsion and therefore have the greatest increase in energy. The two orbitals oriented between the x and z or y and z axes will experience the next greatest repulsion. Finally, the orbital orieented between the x and y axes and the orbital oriented along the x and y axes will experience the least repulsion.
22.78
(a)
The number of moles of chloride that have been precipitated is:
1 mol AgCl −3 mol AgCl = (0.814 g AgCl) = 5.68 × 10 mol AgCl 143.32 g AgCl The number of moles of Cr that were originally present is:
1 mol CrCl3 • 6H 2 O 1 mol Cr mol Cr = (0.757 g CrCl3 • 6H 2 O) 266.4 g CrCl3 • 6H 2 O 1 mol CrCl3 • 6H 2 O = 2.84 × 10−3 mol Cr The ratio of moles of Cl– per mole of Cr is therefore: 5.68/2.84 = 2.00. This means that there were 2 mol of Cl– that were free to precipitate. The other mole of chloride ion must have been bound as a ligand to the Cr. In other words, the complex ion was [Cr(Cl)(H2O)5]2+. (b)
[Cr(Cl)(H2O)5]Cl2⋅H2O
(c) 2+
Cl H 2O
OH2 Cr
H 2O
OH2 OH 2
(d)
There is only one isomer.
553
Chapter 22
22.79 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
fac–[Cr(NH3)3(H2O)3]3+ mer–[Cr(NH3)3(H2O)3]3+ cis-[CrCl(NH3)2(H2O)3]2+ trans-[CrCl(NH3)2(H2O)3]2+ cis-[Cr(NH3)3(H2O)2]2+ trans-[Cr(NH3)3(H2O)2]2+ cis-[CrCl2(NH3)3(H2O)]+ trans-[CrCl2(NH3)3(H2O)]+ cis-[CrCl2(NH3)3(H2O)]+ trans-[CrCl2(NH3)3(H2O)]+ chiral [CrCl2(NH3)2(H2O)2]+ chiral [CrCl2(NH3)2(H2O)2]+ chiral [CrCl2(NH3)2(H2O)2]+ chiral [CrCl2(NH3)2(H2O)2]+ cis-cis-trans-[CrCl2(NH3)2(H2O)2]+ cis-cis-trans-[CrCl2(NH3)2(H2O)2]+ cis-trans-cis-[CrCl2(NH3)2(H2O)2]+ cis-trans-cis-[CrCl2(NH3)2(H2O)2]+ trans- cis-cis-[CrCl2(NH3)2(H2O)2]+ trans- cis-cis-[CrCl2(NH3)2(H2O)2]+ trans- trans-trans-[CrCl2(NH3)2(H2O)2]+ trans- trans-trans-[CrCl2(NH3)2(H2O)2]+ cis-[CrCl2(NH3)(H2O)3]+ trans-[CrCl2(NH3)(H2O)3]+ cis-[CrCl2(NH3)(H2O)3]+ trans-CrCl2(NH3)(H2O)3]+
[CrCl6]3– [CrCl6]3– [CrCl5(NH3)]2– [CrCl5(NH3)]2– [CrCl5(H2O)]2– [CrCl5(H2O)]2– cis-[CrCl4(H2O)2]– cis-[CrCl4(H2O)2]– trans-[CrCl4(H2O)2]– trans-[CrCl4(H2O)2]– cis-[CrCl4(NH3)(H2O)]– cis-[CrCl4(NH3)(H2O)]– trans-[CrCl4(NH3)(H2O)]– trans-[CrCl4(NH3)(H2O)]– cis-[ [CrCl4(NH3)(H2O)]– trans-[ [CrCl4(NH3)(H2O)]– cis-[ [CrCl4(NH3)(H2O)]– trans-[ [CrCl4(NH3)(H2O)]– cis-[CrCl4(NH3)(H2O)]– trans-[CrCl4(NH3)(H2O)]– cis-[CrCl4(NH3)(H2O)]– trans-[CrCl4(NH3)(H2O)]– cis-[ [CrCl4(NH3)2]– cis-[CrCl4(NH3)2]– trans-[ [CrCl4(NH3)2]– trans-[ [CrCl4(NH3)2]–
22.80
The ligands that are along the z axis are pulled out, this lowers the energy of the dxz and dyz and dz2 orbitals. This in turns lowers the energy of the molecule since the dxz and dyz and dz2 orbitals will be filled orbitals.
22.81
The first two isomers are cis-cis and are optical isomers. The other two isomers are cis-trans.
554
Chapter 22
22.82
Copper(II) is a d9 system with one unpaired electron. The ligands cause the d-orbitals to split and because there is a vacant orbital in the upper d level, d-d transitions can occur causing copper(II) compounds to be colored. Zinc(II) is a d10 system with all d-orbitals filled. Therefore, no d-d transition can occur and zinc(II) compounds are colorless.
Multi-Concept Problems 22.83
Moles of CN− 4.0 L x 3.00 M = 12 moles CN− Moles of Ag+ 50 L x 0.2 M = 10 moles Ag+ The formation constant for Ag(CN)22− is Kform = 5.3 x 1018. Therefore, all of the cyanide will complex with silver to form the complex ion. The concentration of free silver will be from the silver ions left after the reaction. 10 mol Ag+ − 12 mol CN− x
1 mol Ag + 2 mol CN −
= 4 mol Ag+
4 mol Ag + = 7.4 x 10−2 M 54 L Silver is available for reduction from the free silver ions in solution as well as from the Ag(CN)22− complex. AgCl is insoluble, and therefore not readily available for reduction. [Ag+] =
22.84
The large charge of cobalt(III) compared to cobalt(II) and the smaller size of the Co3+ions results in more stable complexes cobalt(III) complexes. This is at least partial due to the higher electrostatic attraction between ligands and the metal ion when the metal ion is highly charged and small in size.
Co(NH3 )36+ Co3+ + 6NH3 K inst =
1 4.8 x 1048
Co2+ + 6NH3 Co(NH3 )62+ K form = 1.3 x 1014 Combining the above equations we can write:
Co(NH3 )36+ + Co2 + Co3+ + Co(NH3 )62 + K form = 2.71 x 10−35 An equilibrium constant can be used to determine the cell potential for the reaction.
∆G o = − RT ln K ∆G o = − n F E o At 298 K we can write: log K =
n Eo 0.0592 V
For the reaction of interest n = 1. Therefore, Eo = 0.0592 V x log K 555
Chapter 22
Eo = 0.0592 V x log 2.71 x 10−35 = -2.05 V Now we need to determine the cell potential for the reduction of Co(NH3)63+ to Co(NH3)62+. If we combine the reduction equation determined above with the reduction equation for Co3+ we will obtain the desired reduction equation.
Co(NH3 )36+ + Co2 + Co3+ + Co(NH3 )62+ E o = − 2.05 V Co3+ → Co 2+ E o = 1.82 V Co(NH3 )36+ Co(NH3 )62+ E o = − 0.226 V The reduction potential of the Co(H2O)63+ complex is +1.82 V and the reduction potential for the Co(NH3)63+ complex is − 0.226 V. Therefore, the aquated cobalt species is a much stronger oxidizing agent. 22.85
Using the osmotic pressure data we can determine the van Hoff factor.
i = i=
Π MRT 12.2 atm
( 0.1 M x 0.0821 L atm K−1 mol−1 x 298 K )
i = 4.96 Therefore, there would be 5 moles of ions in a solution that is 1.00 M. The compound is [Pt(NH3)6]Cl4. This is the only arrangement that would create 5 moles of ion per mole of complex. When this compound is dissolved in water, there will be 4 moles of free Cl− ions per mole of compound.
0.126 g x
1 mol PtCl4 (NH3 )6 4 mol Cl− 1 mol AgCl 143.3 g = x x x − 439.1 g mol PtCl4 (NH3 )6 mol AgCl mol Cl
0.164 g AgCl from the "free" Cl−
There are no isomers of this complex ion.
556
Chapter 23
Practice Exercises 23.1
2,2–dimethylpropane
23.2
(a) (b) (c)
23.3
OH │ CH3CH2–CH–CH3
23.4
(a)
3–methylhexane 4–ethyl–2,3–dimethylheptane 5–ethyl–2,4,6–trimethyloctane
(depending on the strength of the oxidizing agent) (b)
O || CH3CH2CCH2CH3
(c) Tertiary alcohols do not undergo oxidation in the same manner as primary and secondary alcohols.
23.5
(a)
(b) 23.6
Cl | CH3CHCH2CH3 CH3CH=CH2
O || CH3CH2COH + CH3OH
O H+ catalyst || CH3CH2COCH3 + H2O
The product is methylpropanoate
23.7
(a)
(b)
(c)
557
Chapter 23
23.8
The products are:
CH3CH2OH and
O || CH3CO −
23.9
O || H2NCH2CH2CH3 + CH3CH2COH →
23.10
(a)
O || CH3CH2CNHCH2CH2CH3 + H2O
O || CH3CH2–C–OH + CH3CH2NH2 (b)
O || CH3CH2–C–O– + HOCH(CH3)2
(c)
CH3CH2CH=CH2
23.11
23.12
558
Chapter 23
Review Questions 23.1
This is due to the high strength of covalent bonds carbon forms with itself and other nonmetals at the same time, and that carbon can make four bonds.
23.2
Since ethane is H3C–CH3, R is CH3, a methyl group.
23.3
(a) (b)
23.4
(c)
This is impossible, since there should be three hydrogen atoms attached to the first carbon atom. This is impossible, since there should be only two hydrogen atoms attached to the first carbon atom. This is possible.
(a)
CH3CH2CH2CH3
(b)
(c)
23.5
The functional group is the characteristic portion of a molecule that sets it into one of a number of families of similar substances. It is the focus of the reactivity of the molecule.
23.6
The functional groups typically impart polarity to what would be an otherwise nonpolar hydrocarbon, and they therefore give the molecule more of an opportunity for reaction than a simple hydrocarbon would have. In other words, since the functional group is the site of polarity, it is the likely site of attack by polar or ionic reactants. Amines, for instance, are Lewis and Brønsted bases, and they characteristically can be protonated.
559
Chapter 23
23.7 amine group
NH 2 H2 C
(a) H 3C
CH
CH 3 O
(b)
H2 C
H 3C
C H2
OH
ketone group-two R groups attached to the C=O
O (c)
H 3C
(d)
H2 C
C
H2 C
CH3
carboxylate group
C
C
CH 3
NH 2
amide group
O
(e)
HO
H2 C
H2 C
OH
alcohol groups
aldehyde group
O HC (f )
C H2
CH3
ester group
O
(g)
H 3C
O
C
560
CH2
CH3
Chapter 23
23.8
Isomers have identical chemical formulas but different structures.
23.9
Free rotation is prevented by the π–nature of the second of two bonds in the carbon–carbon double bond. Since the bond is rigid, by changing the order of the substituents, isomers can be formed.
23.10
CH3CH3CH2OH is more soluble in water than CH3CH2CH2CH3 because it can hydrogen bond with water.
23.11
Propanol, CH3CH3CH2OH, has the higher boiling point, due to its capacity for hydrogen bonding.
23.12
(a) (b)
C, Compound C has two OH groups, and it will have the greatest capacity for hydrogen bonding. A, Compound A has no polar groups, and has only the very weak London forces.
23.13
(b) is more water soluble because it can hydrogen bond with water and is also slightly polar. 23.14
Substance A has the higher boiling point. Since it is larger and has more branching, the London forces are stronger than in B.
23.15
Aldehydes and ketones do not have the capacity to form hydrogen bonds to themselves.
23.16
Ethanoic acid forms dimers containing two hydrogen bonds. O H3C
H
O
C
C O
H
CH3
O
23.17
The ester does not have the capacity to form hydrogen bonds to itself, whereas the acid readily forms a hydrogen-bonded dimer.
23.18
The amide can form hydrogen bonds and are quite polar compared to aminobutane.
23.19
The N–H bond is not as polar as the O–H bond, and the hydrogen bond of amines is therefore not as strong as that of alcohols. Also, there are more sites available for hydrogen bonding to an O–H group than to an N–H group.
561
Chapter 23
23.20 (a)
(b)
(c)
(d)
(e)
23.21 (a)
O
(b)
H3C
C
Cl
(c)
H3C
C H
CH3 C H2 O
C H
C
OH
C H2
C H2
562
C H2
CH3
Chapter 23
O
CH3
(d)
H3C (e)
H2N
C H O
O
C
CH3
C H2
CH3
CH3
C
C H
O
O
C
C
23.22 (a)
H3C
CH3
(b)
NH2 O
(c)
H3C
C H
C
H
(d)
CH3 CH3 O
(e)
H3C 23.23
C H
C H
C
O Na+
The proper IUPAC name is 2–butanol:
OH H3C
C H
C H2
CH3
23.24
For propene, both positions for the double bond are equivalent.
23.25
Carbon dioxide and water.
23.26
CH3CH=CH2 + H+ CH3CH+–CH3
563
Chapter 23
23.27
23.28
The hydrogenation adds across the double bond the product for both isomers is butane H H CH3 H
C
+
H 3C
H
H
H C
H 3C
H2
+
C
H 3C 23.29
C
H 3C
H2
CH3
C
C
H H
H H
C
C
H
H
CH3
CH3
The oxygen atom of an alcohol group acquires another proton when the alcohol is treated with a strong acid, and this weakens all of the bonds to oxygen, including the C –– O bond:
CH3 H
C
CH3
H +
O
H+
H
CH3 23.30
C CH3
H O H
The compound 1–butanol can be oxidized by the dichromate ion, and it is a specific isomer of butanol. The isomer of butanol that cannot be oxidized by the dichromate ion is 2–methyl–2–propanol.
CH3 H3C
C
H O
CH3 2-methyl-2-propanol 23.31
Carboxylic acid. The organic acid reacts with the base to form an ionic compound which is soluble in water. In the acid form, it is insoluble in water because it is not ionic.
23.32
Amine: As derivatives of ammonia, these compounds are weakly basic and will react or dissolve in acid but as a weak base, will not dissolve in a strongly basic solution.
23.33
The lone pair on the nitrogen atom of an amide is not basic due to the electronegativity of the amide oxygen atom.
564
Chapter 23
23.34
The carbonyl group in urea pulls electron density away from the nitrogen atoms making urea a neutral compound.
23.35
CH3CH2CH2NH2 + H2O CH3CH2CH2NH3+ + OH–
23.36
a, b, c, f
23.37
(a) (b) (c) (d)
CH3CH2CH2NH3+ Br– no reaction no reaction CH3CH2CH2NH2 + H2O +
23.38
H CH3CH2CO2H + CH3OH CH3CH2CO2CH3 + H2O
23.39
A macromolecule is a molecule whose molecular mass is very large. Examples are starch, cellulose, silk or hair.
23.40
A polymer is a substance formed by linking together many simpler units called monomers. Not all macromolecules are polymers because they are not made of monomers.
23.41
A monomer is a substance of relatively low formula mass that is used to make a polymer. Iyt provides the structural skeleton of the repeating unit. (a) H
CH3
C
C
H
H
F
F
C
C
F
F
H
Cl
C
C
H
H
(b)
(c)
23.42
The polymer backbone is the long chain of atoms that have substituents at regular intervals.
23.43
The repeat unit for polypropylene is
565
Chapter 23
H
CH3
C
C
H
H
The formula for the polypropylene polymer is [CH2CH(CH3)]n. Uses of polypropylene are dishwasher-safe food containers, indoor-outdoor carpeting, and artificial turf. 23.44
Propylene has a double bond between the two carbons, while the repeating unit of polypropylene has a single bond.
23.45
An addition polymer is the simple addition of one monomer unit to another. The condensation polymer eliminates a small molecule when the two monomer units become joined.
23.46
Three uses of polystyrene are clear plastic drinking glasses, molded car parts, housing of computers and kitchen appliances, Styrofoam cups and insulation materials. 23.47
A copolymer is formed by combining two different monomers.
23.48
(a) Nylon 6,6 O C
CH2CH 2CH 2CH2
566
O
H
C
N
H CH2CH2 CH 2CH 2CH2CH2
N
n
Chapter 23
(b)
O
O
O
C
C
O
H2 C
H2 C
n 23.49
O
O
C
C
N
N
H amide bond
H
n
23.50 carbonate O O
C
CH3 O
C CH3
n
ester linkage
23.51
Branching is when a polymer chain grows off the main backbone.
23.52
High density polyethylene is not branched; rather, it contains high molecular weight linear strands. These strands easily line up to form long fibers
23.53
Ultrahigh molecular weight polyethylene has molecular weights of 3 to 6 million. It results in an amorphous product with little structural strength.
23.54
The higher the polymer crystallinity, the stronger the polymer
23.55
Low-density polyethylene has short polymer chains and extensive branching which prevents the polymer form lining up in an orderly fashion.
23.56
Nylon forms strong fibers since the substituents on the chains can form hydrogen bonds between the polymer chains.
567
Chapter 23
23.57
C
C
N
N
O
O
H
H
H
H
O
O
N
N
C
C
n
n 23.58
HDPE UHDPE Kevlar
Lightweight, strong, tear and water resistant material like milk jugs Bulletproof vests, cut-resistant fabric, surgical gloves Bulletproof vests, thin yet strong hulls of racing boats.
23.59
Biochemistry is the science of living systems, emphasizing the study of cell chemistry and chemical interactions among cells.
23.60
These are materials (carbohydrates, lipids, proteins, nucleic acids, water, certain metal ions, and micronutrients, etc.), energy (supplied by lipids, carbohydrates, etc.), and information in the form of genetic codes (supplied by nucleic acids).
23.61
Carbohydrates are naturally occurring polyhydroxyaldehydes and polyhydroxyketones, or substances that give these on hydrolysis. They include the monosaccharides, the disaccharides and the polysaccharides. Examples are sucrose, starch and cellulose.
23.62
Each yields glucose on complete hydrolysis: (a) glucose (b) glucose (c) glucose
23.63
The three forms differ from one another only in the manner in which the ring is either opened or closed, as illustrated in Figure 23.6. Furthermore, the open form (which gives the substance the characteristics of an aldehyde) is readily regenerated by the two dynamic and reversible equilibria that convert it into either the α or the β form.
23.64
The digestion of sucrose yields glucose and fructose.
23.65
The hydrolysis of lactose gives glucose and galactose.
23.66
They each give glucose: (a) glucose
(b) glucose
23.67
Amylose constitutes 20 % of starch. It is one of two kinds of glucose polymer in starch, the other being amylopectin. Amylopectin is larger than amylose, because it is composed of numerous amylose molecules linked by oxygen bridges. See page 1079 of text.
23.68
Humans lack the digestive enzyme required to hydrolyze cellulose.
568
Chapter 23
23.69
Glycogen stores glucose units.
23.70
Lipids are substances that are found in living systems and that are soluble in nonpolar solvents such as ethers and benzene. The lipids include a diverse range of substances such as cholesterol, and hormones, as well as fats, oils, fatty acids, and other materials known as triacylglycerols.
23.71
Large portions of lipids are made up of nonpolar hydrocarbon-like segments.
23.72
It is soluble in nonpolar solvents, and it is a substance found in living things.
23.73
This particular type of unsaturation refers only to the presence of the C=C double bond.
23.74
No, because it has an odd number of carbon atoms.
23.75
(a) (b)
The monomers of polypeptides are alike in having the carbon – nitrogen – carbonyl repeating backbone, one linked to another by the peptide bond. The monomers of polypeptides have different side groups, since each of the various amino acids could be employed at any point in the polymer.
23.76
The peptide bond is like that found in the amide bond, a carbonyl – nitrogen bond. They are the same type of bonds.
23.77
Two isomeric polypeptides can differ in the amino acid sequence.
23.78
Two dissimilar polypeptides can differ in (1) the number of amino acids, (2) the identities of the amino acids, and (3) the sequence of amino acids that comprise the chain.
23.79
Although some proteins are made up of a single polypeptide, others may contain two or more associated polypeptides, an organic group, as well as a metal ion.
23.80
The final shape is determined by the exact amino acid sequence, because the side chains are of different sizes and some are hydrophillic and others are hydrophobic.
23.81
Proteins
23.82
These deactivate enzymes. Examples include nerve poisons, botulism toxin, and heavy metals.
23.83
A particular amino acid sequence in a polypeptide is obtained by following a sort of "direction" from the gene that is responsible for a particular polypeptide synthesis. There is a one-to-one relationship between the series of side chain bases (taken in groups of three) in the DNA of a gene and the series of amino acids that are to be assembled in the polypeptide that is to be made under the "direction" of the gene. The gene directs the synthesis of hnRNA, which the cell possesses to make mRNA. It is mRNA that is the bearer of the genetic message from the nucleus of the cell to a site outside the nucleus, where the polypeptide is to be assembled. This is the process of transcription. Next, the mRNA accumulates at the ribosomes, awaiting the arrival of the correct amino acids, which are delivered by tRNA. The process of polypeptides synthesis next involves translation of the information in mRNA into a specific amino acid sequence in the growing polypeptide.
23.84
The genetic messages are carried as a sequence of side chain bases on a DNA segment.
23.85
The two strands in DNA are held together by hydrogen bonds between specific base pairs, as shown in Figures 23.10 and 23.11.
23.86
This is shown in Figures 23.10 and 23.11. Only one particular base can be found opposite another in the double helix. Both structures also have sugar–phosphate backbones.
569
Chapter 23
23.87 23.88
DNA and RNA differ in the identity of the monosaccharide. In RNA it is ribose. In DNA it is deoxyribose. Also, DNA uses the bases A, T, G, and C, whereas RNA uses the bases A, U, G, and C. (a) In DNA, A pairs with T. (b) In RNA, A pairs with U. (c) C pairs with G.
23.89
Transcription begins with DNA and ends with the synthesis of mRNA.
23.90
Translation begins with mRNA, uses tRNA, and ends with a specific polypeptide.
Review Problems 23.91 (a)
H
H
H
C
N
H
(b)
H
Br
H
H
C
Br
H
Cl
(c)
C
H
(d)
H
Cl
N
O
H
H
N
N
H
Cl (e)
H
C
C
H
(f)
H
H
23.92 (a)
H
H
C
C
H
H
H
C
H
(d)
H
H
C
H
C
O
H
C
N
C
N
C
or
H
H
(f)
C
C
H
H
(a) (b) (c)
alkene alcohol ester
C
H
(d) carboxylic acid (e) amine (f) alcohol 570
H
N
H 23.93
H
H
H C
O
(b)
O
(c)
(e)
H
O
Chapter 23
23.94
(a) alkyne (b) aldehyde (c) ketone
23.95
The saturated compounds are b, e, and f.
23.96
The saturated compounds are d and e; b, c, and f are saturated in the sense that they have no C–C multiple bonds.
23.97
(a)
amine
23.98
(a) (b) (c) (d)
ether, amine ether, amide ester, aldehyde ester, carboxylic acid
23.99
(a) (b) (c) (d) (e) (f) (g)
These are identical, being oriented differently only. These are identical, being drawn differently only. These are unrelated, being alcohols with different numbers of carbon atoms. These are isomers, since they have the same molecular formula, but different structures. These are identical, being oriented differently only. These are identical, being drawn differently only. These are isomers, since they have the same molecular formula, but different structures.
23.100 (a) (b) (c) (d) (e) (f) (g)
These are identical, being oriented differently only. These are isomers, since they have the same molecular formula, but different structures. These are isomers, since they have the same molecular formula, but different structures. These are identical, being oriented differently only. These are identical, being drawn differently only. These are isomers, since they have the same molecular formula, but different structures. These are unrelated.
23.101 (a) (b) (c)
butane 2–methylbutane 2,5–dimethylheptane
23.102 (a) (b) (c)
2,4–dimethylheptane 3–heptane 5–methyl–2–hexene
23.103 (a)
(d) ether (e) amine (f) amide
(b)
amine
(c)
amide
No isomers
(b) H
CH2CH2CH3 C
H
C
H3C
C H
trans
H C
H3C
CH2CH2CH3 cis
571
(d)
amine, ketone
Chapter 23
(c) CH3
CH3
C
H
CH3
C
Br
C H
Br
CH3
cis
23.104
(a)
C
trans
No isomers
(b)
CH3
H C Br
H2C
H
C
C
H2C
CH3
trans
Br
CH3 cis
(c)
(c)
23.105 (a) (b) (c) (d) (e) (f)
CH3CH3 ClCH2CH2Cl BrCH2CH2Br CH3CH2Cl CH3CH2Br CH3CH2OH
23.106 (a) (b) (c)
CH3CHClCH3 CH3CHICH3 CH3CHOHCH3
23.107 (a)
CH3CH2CH2CH3
(b)
H3C
Cl
Cl
C
C
H
H
C
CH3
572
CH3
Chapter 23
(c)
H3C
Br
Br
C
C
H
H
H
Cl
C
C
H
H
H
Br
C
C
H
H
H
OH
C
C
H
H
CH3
(d)
H3C
CH3
(e)
H3C
CH3
(f)
H3C
CH3
23.108 (a)
(b)
Cl
I
(c)
OH 23.109 This sort of reaction would disrupt the π delocalization of the benzene ring. The subsequent loss of resonance energy would not be favorable.
573
Chapter 23
23.110 Benzene does not "add" Br2, if it did, it would add across a double bond.
Br
Br This does not occur because the loss of resonance energy would not be favorable. Rather, in the presence of a catalyst, it gives bromobenzene: C6H6 + Br2 → C6H5Br + HBr (FeBr3 catalyst) 23.111 CH3OH IUPAC name = methanol; CH3CH2OH IUPAC name = ethanol; CH3CH2CH2OH IUPAC name = 1–propanol; (CH3)2COH or as shown below
common name = methyl alcohol common name = ethyl alcohol common name = propyl alcohol
CH3 H3C
C H
OH IUPAC name = 2–propanol;
23.112 CH3CH2CH2CH2OH CH3CH2CHOHCH3 (CH3)2CHCH2OH (CH3)3COH
1–butanol 2–butanol 2–methyl–1–propanol 2–methyl–2–propanol
23.113 CH3CH2CH2–O–CH3 CH3CH2–O–CH2CH3 (CH3)2CH–O–CH3
methyl propyl ether diethyl ether methyl 2–propyl ether
23.114 Those that can be oxidized to aldehydes are: CH3CH2CH2CH2OH and (CH3)2CHCH2OH One can be oxidized to a ketone: CH3CH2CHOHCH3 23.115 (a)
(b)
C H
CH2
574
common name = isopropyl alcohol
Chapter 23
(c)
C H
CH2
23.116
23.117 (a)
O
(b)
O C
CH3
(c)
O C H2
CH
575
Chapter 23
23.118 The product is propene.
H2C
C H
CH3
The mechanism is an addition of H+ to the oxygen followed by loss of water to form a carbocation and finally the formation of the C-C double bond.
23.119 The elimination of water can result in a C=C double bond in two locations: CH3CH=CHCH3 CH2=CHCH2CH3 1–butene 2–butene 23.120 The ratio should be 3:2, since there are three hydrogen atoms on the first carbon of 2–butanol, whereas there are only two hydrogen atoms on the third carbon atom of 2–butanol. In fact , the product is 90% 2butene and 10% 1-butene. Obviously, other factors are at work that alter the predicted outcome. 23.121 The aldehyde is more easily oxidized. The product is:
23.122 (a) This is C, giving:
O CH3
C
OH
(b) This is B, giving:
O CH3
C
O Na+
(c) This is B, giving:
O CH3
C
O
CH3 576
Chapter 23
(d) This is A, giving:
O CH3
C
H CH2=CH2
(e) This is A, giving:
23.123 (a) CH3CH2CO2H (b) CH3CH2CO2H + CH3OH (c) Na+ + CH3CH2CH2CO2– + H2O 23.124 (a) CH3CH2CONH2 + H2O (b) CH3CH2CH2CO2H + NH3 (c) Na+ + CH3CH2CO2– + CH3OH 23.125 CH3CO2H + CH3CH2NHCH2CH3 23.126 (a)
The amine neutralizes HCl: CH3CH2CH2NH2 + HCl CH3CH2CH2NH3+ + Cl–
(b)
The amide is hydrolyzed: O H 3C
(c)
CH2
C
NH 2
+ H2O CH3CH2COH + NH3
The alkylammonium cation neutralizes sodium hydroxide: CH3CH2NH3+ + OH– CH3CH2NH2 + H2O
23.127
23.128
O C
O H2 C
H2 C
C
O
H2 C
H2 C
O n
577
Chapter 23
23.129
23.130 O
O
H
C
C
N
H H2 C
H2 C
H2 C
H2 C
N n
23.131 O H2C
O
C
(CH2)7
C H
CH(CH2)7CH3
O HC
O
C
(CH2)7 C H
H2 C
C H
C H
CH(CH2)4CH3
O H2C
O
C
(CH2)14CH3
23.132 These are, first, the tri-alcohol:
OH H2C
OH
OH
C H
CH2
and the three carboxylic acids: linoleic acid:
O H2 C
C HO
C H2
H2 C C H2
H2 C C H2
H2 C
H C C H2 578
C H
H2 C
H C C H
C H2
H2 C C H2
CH3
Chapter 23
oleic acid O
H2 C
C HO
H2 C
C H2
C H2
H2 C C H2
C H2
H2 C
H2 C
H C C H
C H2
H2 C C H2
H2 C C H2
CH3
myristic acid
O H2 C
C HO
H2 C
C H2
C H2
H2 C C H2
H2 C C H2
H2 C C H2
H2 C C H2
CH3
23.133 O H2C
O
C
(CH2)16CH3
O HC
O
C
(CH2)12CH3
O H2C
O
C
(CH2)16CH3
23.134 First, we have glycerol:
OH H2C
OH
OH
C H
CH2
and then three anions, as follows. (1) linoleate anion:
O -
H2 C
C O
C H2
H2 C C H2
H2 C C H2
H2 C
H C C H2
(2) myristate anion: 579
C H
H2 C
H C C H
C H2
H2 C C H2
CH3
Chapter 23
O H2 C
C -
O
C H2
H2 C
H2 C
C H2
C H2
H2 C C H2
H2 C C H2
H2 C C H2
CH3
(3) oleate anion: O H2 C
C -
O
C H2
H2 C C H2
H2 C C H2
C H2
H2 C
H2 C
H C
C H2
C H
H2 C
H2 C C H2
C H2
CH3
Hydrophobic sites are composed of fatty acid units. Hydrophilic sites are composed of charged
23.135 units.
23.136 See Figure 23.7and pages 1081 and 1082 of the text. 23.137
O +
H3N
C H2
C
O N H
C H2
C
O–
23.138
O +
H3N
CH C
O N H
CH3
CH C
O N H
CH3
CH C
O–
CH3
23.139
O +
H3N
C H2
C
O N H
CH C
O
O–
+
H3N
CH2
CH C CH2
580
O N H
C H2
C
O–
Chapter 23
23.140
O +
alanine
H3N
CH C
O–
CH3 O +
cysteine
H3N
CH C
O–
CH2SH O +
phenyl alanine
H3N
CH C
O–
CH2
Additional Exercises 23.141 (a) (b) (c)
C7H14O C10H22 C10H20N2O
23.142 (a) (b)
(CH3)2CHCH2OH and (CH3)3COH The one that cannot be oxidized is (CH3)3COH.
(c) CH3 H3C
(d)
C +
CH3 CH3
H3C
CH CH2+
The first one is more stable, since it is the one that leads to the observed product. It is a tertiary carbocation.
23.143 The dimethylamine has a higher boiling point, inspite of the lower formula mass due to the hydrogen bond between the nitrogen and the hydrogen.
23.144 (a)
+
Na –O
O
O
C
C
(b) 581
O– Na +
+ 2 H 2O
Chapter 23
OH H2 C
CH H3C
(c) (d) (e)
C CH3 H2 CH3NHCH2CH2CH3 + H2O CH3CH2OCH2CH2CO2H + CH3OH
HO (f)
C H2
H C Br
23.145 There are six possibilities, Gly-Ala-Phe, Gly-Phe-Ala, Ala-Gly-Phe, Ala-Phe-Gly, Phe-Gly-Ala, and Phe-Ala-Gly (GAP, GPA, AGP, APG, PGA, and PAG).
Multi-Concept Problems 23.146 (a)
(b)
Only primary alcohols will be oxidized to organic acids. A is the only primary alcohol among the choices. C5H9COOH
(c)
(d)
36.8 g C6 H12 O x
1 mol C6 H12 O 2 mol Na 2 Cr2 O7 262.0 g x x = 100.2 g 3 mol C6 H12 O mol Na 2 Cr2 O7
64.1 g Na 2 Cr2 O7 (e)
36.8 g C6 H12 O x
1 mol C6 H12 O 1 mol C6 H10 O 2 1 mol NaOH 40.0 g x x x = 100.2 g 1 mol C6 H12 O 1 mol C6 H10 O 2 mol NaOH
14.7 g NaOH 23.147 Using the heats of formation of the gaseous atoms and the bond energies of the product, we can determine the enthalpy of formation of the product.
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Chapter 23
From App. C Tables C.3 and C.4 (Also, see pages 906-907 of text)
∆H of ( C6 H12 O ) = 6C(g) + 12H(g) + O − [6 C-C + 11 C-H + 1 C-O + 1 O-H] = 6 mol C (716.76 kJ mol−1) + 12 H mol (217.89 kJ mol−1) + 1 mol O (249.17 kJ mol−1) − [6 mol C-C (-348 kJ mol−1) + 11 mol C-H (−412 kJ mol−1) + 1 mol C-O (−360 kJ mol−1) + 1 mol O-H (-463 kJ mol−1)] = −279 kJ mol−1 C6H12O + 17/2 O2(g)
→
6CO2(g) + 6H2O
∆H rxn = 6 mol CO 2 x ( − 393.5 kJ mol−1 ) + 6mol H 2 O x ( − 285.9 kJ mol−1 ) − (−279 kJ mol−1 ) = −3797 kJ per mol reactant −3797 kJ 1 mol x x 11.35 g = − 430 kJ mol 100.2 g 23.148 We first find the number of grams of C, H, and O that were present in the original sample:
12.01 mg C g C = (12.44 mg CO2 ) = 3.395 mg C 44.01 mg CO 2 2.016 mg H g H = ( 5.091 mg H 2 O ) = 0.5696 mg H 18.02 mg H 2 O The mass of oxygen is given by: 6.225 mg – 3.395 mg – 0.5696 mg = 2.260 mg O The mole amounts are: 3.395 g C ÷ 12.01 mg/mmol = 0.2827 mmol C 0.5696 mg H ÷ 1.008 mg/mmol = 0.5651 mmol H 2.260 mg O ÷ 16.00 mg/mmol = 0.1413 mmol O The relative mole amounts are: for C: 0.2827 mmol/0.1413mmol = 2.00 for H: 0.5651mmol/0.1413 mmol = 4.00 for O: 0.1413 mmol/0.1413 mmol = 1.000 These relative mole amounts correspond to the empirical formula: C2H4O The molecular formula could be any number of these units but we know this is a monoprotic acid so need two oxygen atoms in the molecule. Therefore, the molecular formula is C4H8O2. The structure is:
O H2 C H3C 0.822 g x
C H2C
OH
1 mol C4 H8 O2 1 mol KOH 1L 1000 mL x x x = 88.1 g mol C4 H8 O2 0.224 mol KOH L
41.6 mL 583
Chapter 23
23.149 The original number of moles of hydroxide are: 0.1016 M × 0.05000 L = 5.080 × 10–3 mol OH– The moles of hydroxide not neutralized by the organic acid are: 0.1182 M × 0.02378 L = 2.811 × 10–3 mol OH– Therefore, the moles of hydroxide that were neutralized by the acid were: 5.080 × 10–3 mol – 2.811 × 10–3 mol = 2.269 × 10–3 mol OH– This is also equal to the number of moles of the organic acid. The formula mass is therefore: 0.2081 g/2.269 × 10–3 mol = 91.71 g/mol Alternate Method Mmoles of base = 5.08 mmol Mmoles of acid neutralized = 2.811 mmol Mmoles of acid = 5.08 – 2.811 = 2.269 mmol Molar mass = 208.1 mg/2.269 mmol = 91.71 mg/mmol This is equal to the molecular mass only if the unknown is a monoprotic acid. 23.150 CH3CH2CHSO + 5O2 1.225 g x
→
3CO2 + 3H2O + SO2
1 mol C3 H 6SO 5 mol O 2 32.0 g x x = 90.14 g mol C3 H 6SO mol O 2
2.17 g O2
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