Download CHAPTER 12 | The Chemistry of Solids

CHAPTER 12 | The Chemistry of Solids
12.1. Collect and Organize
From the drawings shown in Figure P12.1, we are to choose which represents a crystalline solid and which
represents an amorphous solid.
Analyze
In crystalline solids, atoms or molecules arrange themselves in regular, repeating three-dimensional patterns.
In an amorphous solid, the atoms or molecules are arranged randomly, with no defined repeating pattern.
Solve
Drawings b and d are analogous to crystalline solids because they show a definite pattern while a and c are
amorphous.
Think about It
In drawings b and d there are two kinds of atoms; if the drawings represent metals, these two substances would
be alloys.
12.3. Collect and Organize
Using Figure P12.3, we are to determine the unit cell and write the chemical formula for the compound where
element A is represented as red spheres and element B is represented as blue spheres.
Analyze
Once the unit cell is determined, the chemical formula can be deduced. Spheres on the corner are shared by
eight unit cells and so are counted as 18 in the unit cell. Likewise, spheres on the edges are shared by four unit
cells and are counted as 14 in the unit cell, and spheres on the faces of the unit cell are shared by two unit cells
and are counted as 12 in the unit cell. Any atom completely inside a unit cell belongs entirely to it and counts
as 1.
Solve
There are 8 ( 18 ) + 6 ( 12 ) = 4 red spheres or 4A atoms in the unit cell. There are 12 ( 14 ) + 1(1) = 4 blue spheres or
4B atoms in the unit cell. The chemical formula is A4B4 or AB.
Think about It
The empirical formula for this compound would be AB because that is the lowest whole-number ratio of
elements in the substance.
12.5. Collect and Organize
From Figure P12.5 in which a portion of the unit cell has 4 corner and 6 face A atoms and has 4 corner B
atoms, we are to determine the number of equivalent atoms of A and B.
Analyze
Atoms on the corners contribute
volume to the unit cell.
1
8
of their volume to the unit cell and atoms on the faces contribute
1
2
of their
1
2 | Chapter 12
Solve
4 ( 18 ) + 6 ( 12 ) = 3 12 A atoms
4 ( 18 ) =
1
2
B atom
Think about It
The empirical formula for this substance based on the unit cell is BA7.
12.7. Collect and Organize
From the portion of unit cell shown in Figure P12.7, we are to deduce the chemical formula of the ionic
compound.
Analyze
Atoms on the corners contribute 18 of their volume to the unit cell and atoms on the faces contribute 12 of their
volume to the unit cell. Atoms completely inside the unit cell contribute all of their volume to the unit cell.
Solve
8 ( 18 ) = 1 A cation
6 ( 12 ) = 3 B cations
1(1) = 1 X anion
The chemical formula is AB3X.
Think about It
In this unit cell notice that the X anion is located in the center, occupying the body-centered position, and the B
cations occupy face-centered positions.
12.9. Collect and Organize
Knowing the distance between the phosphorus atoms in its cubic form, we are to calculate the density.
Analyze
Density is mass per unit volume. The mass of phosphorus in one unit cell will be the number of atoms of P in
one unit cell multiplied by the molar mass of B and divided by Avogadro’s number. The volume of the unit
cell is the cube of the length of the side of the unit cell. Density is usually expressed as grams per cubic
centimeter (g/cm3) so we have to convert picometers to centimeters.
Solve
Mass of phosphorus in one unit cell
8 ( 18 ) ×
Volume of the unit cell
30.97 g
1 mol
×
= 5.143 × 10–23 g
1 mol 6.022 × 1023 atoms
3
⎛
1 × 10 –10 cm ⎞
238
pm
×
= 1.35 × 10 –23 cm 3
⎜⎝
1 pm ⎟⎠
Density of cubic phosphorus
5.143 × 10–23 g
= 3.81 g/cm3
–23
3
1.35 × 10 cm
Think about It
The density of phosphorus in another form would be different. If we measure the density of a crystal of known
composition, we may be able to determine what its unit cell looks like.
12.11. Collect and Organize
We are to determine the formula of a compound in which the oxide ions form an fcc unit cell and where half
of the octahedral holes are occupied by Al3+ and one-eighth of the tetrahedral holes are occupied by Mg2+.
The Chemistry of Solids | 3
Analyze
In a face-centered cubic (cubic closest-packed) array of oxide ions, there are 8 tetrahedral holes and 4
octahedral holes.
Solve
The number of oxide ions in the fcc unit cell is
(8 × 18 ) + (6 × 12 ) = 4
If one-half of the octahedral holes contain Al3+, then there are 2 Al3+ in the unit cell. If one-eighth of the
tetrahedral holes contain Mg2+, then there is 1 Mg2+ in the unit cell. The chemical formula for this compound
is MgAl2O4.
Think about It
This compound is a neutral salt; the total positive charge from the Al3+ and Mg2+ ions is 8+. The four oxide
ions each carry a 2– charge for a total negative charge of 8–.
12.13. Collect and Organize
We are to determine the formula of a lithium–sulfur compound in which the sulfide ions form an fcc
arrangement and the lithium ions occupy all of the tetrahedral holes.
Analyze
In an fcc arrangement, there are 8 tetrahedral holes in the unit cell.
Solve
The number of sulfide anions in the unit cell is
(8 × 18 ) + (6 × 12 ) = 4
The number of lithium cations in the unit cell is 8. The formula for the compound is Li2S.
Think about It
This salt is neutral because the charge on two lithium ions (1+) in the formula balances the charge on one
sulfide anion (2–).
12.15. Collect and Organize
We are to identify the elements highlighted in the periodic table in Figure P12.15 that do not adopt a rock salt
structure for their chloride salt.
Analyze
In the rock salt structure the cations must fit into the octahedral holes between the closest-packed layers of Cl–.
The radius ratio of cation to anion should be between 0.41 and 0.73. The radii of the cations of the highlighted
elements from Figure 10.2 are Li+ (76 pm), Na+ (102 pm), K+ (138 pm), Cs+ (170 pm), and Sr2+ (118 pm). The
radius of Cl– is 181 pm.
Solve
76 pm
= 0.42
181 pm
102 pm
For NaCl, the radius ratio is
= 0.564
181 pm
138 pm
For KCl, the radius ratio is
= 0.762
181 pm
170 pm
For CsCl, the radius ratio is
= 0.939
181 pm
118 pm
For SrCl2, the radius ratio is
= 0.652
181 pm
For LiCl, the radius ratio is
4 | Chapter 12
By the radius ratio rule, the rock salt structure is not probable for KCl (K+ would occupy cubic holes) nor for
CsCl (Cs+ would occupy cubic holes) or SrCl2 (because it does not have a 1-to-1 cation-to-anion
stoichiometry). LiCl is borderline. In nature, all of these adopt a rock salt structure except Cs (blue) and Sr
(purple).
Think about It
In the rock salt structure of SrCl2, Sr2+ would occupy one-half of the octahedral holes. This ionic compound
adopts a fluorite structure in nature.
12.17. Collect and Organize
From Figure P12.17 showing the unit cell of magnesium boride, we are to determine its formula.
Analyze
We are given that one B atom is inside the trigonal prismatic unit cell. Each corner of the unit cell has an Mg
atom. Because the unit cell is trigonal prismatic, each corner atom is shared with 11 other unit cells and so has
one-twelfth of its volume inside the unit cell.
Solve
The number of B atoms in the unit cell is 1. The number of Mg atoms in the unit cell is 6 ( 121 ) =
formula from the unit cell is MgB2 (in the lowest whole-number ratio).
1
2
. The
Think about It
It is tricky to see that the Mg atoms here are shared between 12 unit cells. Consider the center atom in the
figure on the right hand side. That atom is shared with 6 unit cells as shown, but there are also 6 more unit
cells stacked on top of it that are not shown in the diagram.
12.19. Collect and Organize
We are to explain how the electron-sea model accounts for the electrical conductivity of gold.
Analyze
In the electron-sea model the metal atoms exist in the structure as cations with their ionized electrons not
associated with particular metal ions.
Solve
Because the electrons are free to distribute themselves throughout the metal, the application of an electrical
potential across the metal causes the electrons to travel freely toward the positive terminal. This movement of
electrons in the structure is electrical conductivity.
Think about It
Metals have low ionization energies. This makes the electron sea somewhat easy to achieve in the lattice of
metals.
12.21. Collect and Organize
Given that the melting and boiling points of Na are lower than those of NaCl, we are to predict the relative
strengths of metallic and ionic bonds.
Analyze
Solid sodium is held together by metallic bonds, whereas sodium chloride is held together by ionic bonds. The
higher the melting point or boiling point of a substance, the stronger the forces between particles of that
substance.
Solve
Because the melting point and boiling point of NaCl are higher than those of Na, ionic bonds are stronger than
metallic bonds.
The Chemistry of Solids | 5
Think about It
Some metallic bonds are quite strong as evidenced by the melting points of some metals. Whereas Na has a
melting point of 97.72˚C, tungsten melts at 3422˚C.
12.23. Collect and Organize
We are to consider whether band theory can explain why hydrogen at very low temperatures and high
pressures might act like a metal and conduct electricity.
Analyze
Band theory is a model of bonding in which orbitals on many atoms are combined, as in molecular orbital
theory, to form a fully or partially filled valence band and an empty conduction band.
Solve
Yes, by combining the 1s orbitals on many hydrogen atoms at very low temperatures and at high pressures,
we could construct a molecular orbital diagram that would show a partially filled valence band (as for copper
in Figure 12.2) in which the electrons can move into the unfilled portion of the band where they can migrate
freely.
Think about It
Metallic liquid hydrogen was discovered with hydrogen at very high temperatures and very high pressures in
1996 at the Lawrence Livermore Laboratory quite by accident.
12.25. Collect and Organize
We are asked to identify which groups in the periodic table contain metals with filled valence bands.
Analyze
The metals in the periodic table are in groups 1–12 with some in group 13 (Al, Ga, In, TI), group 14 (Si, Ge,
Sn, Pb), group 15 (As, Sb, Bi), group 16 (Te, Po), and group 17 (At). Filled valence bands for metals occur for
electron configurations of ns2 and nd10.
Solve
Metals of groups 2 and 12 have filled valence bands.
Think about It
The orbital energy diagram for these metals will be similar to that of zinc in Figure 12.3.
12.27. Collect and Organize
We are asked why it might be important to exclude phosphorus contaminants in the manufacture of silicon
chips.
Analyze
The difference between silicon and phosphorus is that phosphorus has 5 valence electrons and silicon has 4
valence electrons.
Solve
Phosphorus with one more valence electron than silicon would give pure silicon a higher conductivity and so
would change the electrical nature of the silicon chips.
Think about It
Phosphorus, with one more electron than silicon, would form an n-type semiconductor if it were present as a
dopant in silicon.
12.29. Collect and Organize
We are asked to identify in which group of the periodic table we might find elements that would be useful to
form a p-type semiconductor with Sb2S3.
6 | Chapter 12
Analyze
Antimony has 5 valence electrons, and sulfur has 6 valence electrons. To form a p-doped semiconductor we
would choose elements with fewer than 5 valence electrons.
Solve
Group 14 elements with 4 valence electrons would form p-type semiconductors when doped into Sb2S3.
Think about It
Similarly, group 13 elements, with 3 valence electrons, might also be used to form p-doped Sb2S3
semiconductors.
12.31. Collect and Organize
We consider the properties of nitrogen-doped diamond.
Analyze
(a) Nitrogen, with 5 valence electrons, has one more electron than carbon, with 4 valence electrons.
(b) In diamond, the valence band and the conduction band are separated by a large band gap, making diamond
an insulator. Adding nitrogen to the diamond structure adds a partially filled band above diamond’s valence
band and below its conduction band.
(c) Because E = hc/λ, we can calculate the energy of the band gap when λ = 4.25 × 10–7 m.
Solve
(a) Because nitrogen has one more valence electron than carbon, nitrogen serves as an n-type dopant in
diamond.
(b)
(c) Eg =
6.626 × 10–34 J ⋅ s × 3.00 × 108 m/s
= 4.68 × 10–19 J
4.25 × 10–7 m
Think about It
Because nitrogen-doped diamonds absorb violet light, we observe a yellow color (the complement of violet)
in the diamond.
12.33. Collect and Organize
By calculating the wavelength associated with the band gap energies for AlN, GaN, and InN, we can
determine which energies correspond to radiation in the visible part of the electromagnetic spectrum.
Analyze
The wavelength of light absorbed or emitted by a semiconductor is inversely related to the band gap energy:
The Chemistry of Solids | 7
Eg =
hc
λ
To calculate the wavelength we can rearrange this equation to
hc
λ=
Eg
The wavelength is expressed in terms of a single photon, so we have to convert energy units (from kJ/mol to
J/photon).
Solve
For AlN λ =
For GaN λ =
6.626 × 10 –34 J ⋅ s × 3.00 × 108 m/s
= 2.06 × 10 –7 m or 206 nm
⎛ 580.6 kJ 1000 J
⎞
1 mol
⎜⎝ mol × kJ × 6.022 × 1023 photons ⎠⎟
6.626 × 10 –34 J ⋅ s × 3.00 × 108 m/s
= 3.72 × 10 –7 m or 372 nm
⎛ 322.1 kJ 1000 J
⎞
1 mol
⎜⎝ mol × kJ × 6.022 × 1023 photons ⎟⎠
6.626 × 10 –34 J ⋅ s × 3.00 × 108 m/s
= 6.21 × 10 –7 m or 621 nm
⎛ 192.9 kJ 1000 J
⎞
1 mol
⎜⎝ mol × kJ × 6.022 × 1023 photons ⎠⎟
Only InN (Eg = 192.9 kJ/mol) emits in the visible region of the spectrum.
For InN λ =
Think about It
The color of InN’s emission is red.
12.35. Collect and Organize
We are asked to differentiate between cubic closest-packed (ccp) and hexagonal closest-packed (hcp)
structures.
Analyze
Both structures contain layers of close-packed atoms and differ only in how the layers are stacked.
Solve
Cubic closest-packed structures have an abcabc . . . pattern, and hexagonal closest-packed structures have an
abab… pattern.
Think about It
The unit cell for ccp is face-centered cubic (Figure 12.11), and the unit cell for hcp is hexagonal (Figure 12.8).
12.37. Collect and Organize
We are asked which has the greater packing efficiency, the simple cubic or the body-centered cubic structure.
Analyze
Packing efficiency is the fraction of space within a unit cell that is occupied by the atoms.
Solve
We read in Section 12.3 of the textbook that the simple cubic cell has the lowest packing efficiency of all the
unit cells, so the body-centered cubic structure has a greater packing efficiency than the simple cubic
structure.
Think about It
The packing efficiency of the unit cell structures can be calculated. For simple cubic, the packing efficiency is
only 52%, whereas for body-centered cubic it is 68% and for face-centered and hexagonal it is 74%.
8 | Chapter 12
12.39. Collect and Organize
Iron can adapt either the bcc unit cell structure (at room temperature) or the fcc unit cell structure (at 1070˚C).
We are asked whether these two forms are allotropes.
Analyze
Allotropes are defined as different molecular forms of an element.
Solve
Iron is not molecular and the bcc and fcc unit cell structures describe only a difference in atom packing in the
metal. Therefore, these structural forms are not allotropes.
Think about It
Elements that do have allotropes include phosphorus, sulfur, and carbon.
12.41. Collect and Organize
Using the information in Figure 12.16 we are to derive the expression for the edge length (l ) of the bcc and
fcc unit cells in terms of r, the radius of the atoms.
Analyze
For the bcc unit cell, the atoms touch along the body diagonal for a distance of l 3 . For the fcc unit cell, the
atoms touch along the face diagonal for a distance of l 2 .
Solve
In the bcc unit cell, the body diagonal is 4r. From the Pythagorean theorem, the body diagonal is
4r =
( edge length )2 + ( face diagonal)2
where the edge length = l and the face diagonal is
l 2 + l 2 = 2l 2 = l 2
Therefore,
(
4r = l 2 + l 2
)
2
= l 2 + 2l 2 = 3l 2
4r = l 3
4r
l =
= 2.309r
3
In the fcc unit cell, the face diagonal is 4r:
l 2 + l 2 = 2l 2 = l 2 = 4r
l =
4r
2
= 2.828r
Think about It
For a given atom, the edge length of the bcc unit cell is less than that of atoms packed in an fcc unit cell.
12.43. Collect and Organize
Knowing that l = 240.6 pm for the bcc structure of europium, we are to calculate the radius of one atom of
europium.
Analyze
For the bcc structure the body diagonal contains 2 atoms of europium or 4r and the body diagonal is equal to
l 3 (Figure 12.16) where l is the unit cell edge length, 240.6 pm.
The Chemistry of Solids | 9
Solve
4r = l 3
r=
l 3 240.6 pm × 3
=
= 104.2 pm
4
4
Think about It
If the structure of the unit cell were simple cubic the radius of the europium atom would simply be
2r = l
r=
l
= 120.0 pm
2
12.45. Collect and Organize
We are to calculate the edge length of the unit cell of Ba knowing that it crystallizes in a bcc unit cell and that
rBa = 222 pm.
Analyze
The body diagonal of a bcc unit cell has the relationship l 3 = 4r. An edge has the length l , so all we need
to do is rearrange the body diagonal expression to solve for l :
4r
l =
3
Solve
l =
4r
3
=
4 × 222 pm
3
= 513 pm
Think about It
Be careful to not assume that the edge length of every unit cell is l = 2r as in a simple cubic.
12.47. Collect and Organize
We are to determine the type of unit cell for a form of copper using the density of the crystal (8.95 g/cm3) and
the radius of the Cu atom (127.8 pm).
Analyze
For each type of unit cell (simple cubic, body-centered cubic, and face-centered cubic), we can compare the
calculated density to that of the actual density given for the crystalline form of copper. Density is mass per
volume. The mass of each unit cell is the mass of the copper atoms contained in each unit cell. To find this,
we first have to determine the number of atoms of Cu in each unit cell based on its structure. Then, we
multiply by the mass of 1 atom of Cu:
1 Cu atom ×
1 mol
63.55 g
×
= 1.055 × 10–22 g
23
6.022 × 10 atoms 1 mol
The volume of each unit cell is l 3 , which is related to r in a way that depends on the type of unit cell.
Solve
For a simple cubic unit cell where all the Cu atoms are at the corners of the cube:
Number of Cu atoms = 8 × 18 = 1 Cu atom
l = 2r = 2 × 127.8 pm = 255.6 pm
Converting to centimeters for the calculation of density (g/cm3) gives
1 × 10–10 cm
255.6 pm ×
= 2.556 × 10–8 cm
1 pm
Volume = l 3 = (2.556 × 10–8 cm)3 = 1.670 × 10–23 cm3
10 | Chapter 12
1 Cu atom × 1.055 × 10 –22 g/atom
= 6.32 g/cm3
1.670 × 10–23 cm3
For a body-centered cubic unit cell where there is one Cu atom in the center of the unit cell and 8 Cu atoms at
the corners of the cube:
Number of Cu atoms = (8 × 18 ) + 1 = 2 Cu atoms
Density =
From the body diagonal 4r = l 3 or l = 4r
3,
4 × 127.8 pm
l =
= 295.1 pm
3
Converting to centimeters for calculation of density (g/cm3)gives
1 × 10–10 cm
295.1 pm ×
= 2.951 × 10–8 cm
1 pm
Volume = l 3 = (2.951 × 10–8 cm)3 = 2.570 × 10–23 cm3
2 Cu atoms × 1.055 × 10–22 g/atom
Density =
= 8.21 g/cm3
2.570 × 10–23 cm3
For a face-centered cubic unit cell where there are 8 Cu atoms at the corners of the unit cell and 6 on the
faces:
Number of Cu atoms = (8 × 18 ) + (6 × 12 ) = 4 Cu atoms
From the face diagonal 4r = l 2 or l = 4r
2,
4 × 127.8 pm
l =
= 361.5 pm
2
Converting to centimeters for calculation of density (g/cm3) gives
1 × 10–10 cm
361.5 pm ×
= 3.615 × 10 –8 cm
1 pm
Volume = l 3 = (3.615 × 10–8 cm)3 = 4.724 × 10–23 cm3
4 Cu atoms × 1.055 × 10–22 g/atom
Density =
= 8.93 g/cm3
4.724 × 10–23 cm3
The fcc unit cell gives a density closest to the actual density, so we predict that in this crystalline form of Cu
the atoms pack in an fcc unit cell (c).
Think about It
Notice that even though the fcc unit cell has the largest edge length, the unit cell contains more atoms and so it
yields the densest structure.
12.49. Collect and Organize
We can use the definitions of solid solution and homogeneous alloy to determine if there is a difference
between these terms.
Analyze
A solid solution is a homogeneous mixture of solids. An alloy is a mixture of two or more metallic elements
in solution with each other.
Solve
There is no real difference in these two terms. Both are homogeneous mixtures of solids. There may be a
subtle difference, however. An alloy is a mixture of two or more metallic elements, and a solid solution is a
mixture of two or more solids, which may or may not be metallic elements.
Think about It
All alloys are solid solutions, but not all solid solutions are alloys.
The Chemistry of Solids | 11
12.51. Collect and Organize
We are to explain why an alloy of Cu and Ag melts at a lower temperature than either pure Ag or pure Cu.
Analyze
When other metals are alloyed into a pure metal, the atomic structure of the metal is changed so that the atoms
are not in so regular of an arrangement as in the pure metal. From Sample Exercise 12.4, we know that Cu and
Ag form substitutional alloys because their atomic radiuses are so similar.
Solve
When these two metals form alloys with each other, the metallic bonding is weakened because of their slightly
different sizes and different ionization energies. This weaker metallic bonding means that the melting points
of the alloys are lower than those of the pure metals.
Think about It
Lowering the melting point of an alloy allows for easier forming, processing, and workability compared to a
pure metal.
12.53. Collect and Organize
For the tungsten–carbon alloy, we are to predict which element (W or C) is the host and which occupies the
interstices in the host’s lattice.
Analyze
Large atoms form a closest-packed structure in which smaller atoms may occupy the interstices, or holes,
between the closest-packed layers.
Solve
Tungsten, the larger element, is the host and carbon, the smaller element, occupies the holes between the
tungsten closest-packed layers.
Think about It
The atomic radii of tungsten and carbon from Appendix 3 are 139 pm and 77 pm, respectively. From the
radius ratio rule (Table 12.4) and
rC
77 pm
=
= 0.554
rW 139 pm
we predict that carbon occupies the octahedral holes in the tungsten lattice.
12.55. Collect and Organize
Given that a unit cell of an alloy of X and Y consists of an fcc arrangement of X atoms at each corner and Y
atoms on each face, we are to write the formula of the alloy and then consider the formula if the two atom
positions were reversed.
Analyze
Atoms at the corner of a unit cell count for
unit cell.
1
8
and atoms on the faces count for
1
2
of the total volume of the
Solve
(a) For 8 X atoms at the corners and 6 Y atoms on the faces of the unit cell:
1
8
(8 X atoms) + 12 (6 Y atoms) = XY3
(b) If the positions were reversed, we would have
1
8
(8 Y atoms) + 12 (6 X atoms) = YX3
Think about It
The fcc lattice is closest-packed. If X and Y may be in either the corners or faces in the unit cell, this alloy is a
substitutional alloy.
12 | Chapter 12
12.57. Collect and Organize
Given the atomic radii of C and V (77 and 135 pm, respectively), we can calculate the radius ratio to
determine which holes, octahedral or tetrahedral, C occupies in the V closest-packed structure.
Analyze
The radius ratios for tetrahedral, octahedral, and cubic holes are given in Table 12.4 as 0.22– 0.41, 0.41– 0.73,
and 0.73–1.00, respectively.
Solve
rC
77 pm
=
= 0.57
rV 135 pm
This ratio fits the range for carbon occupying octahedral holes in the vanadium lattice.
Think about It
Carbon can also fit into the larger cubic holes (radius ratio 0.73–1.00) but would occupy the smallest hole it
can in the lattice so as to maximize its interactions with the host (V) atoms in the closest-packed structure.
12.59. Collect and Organize
By calculating the radius ratio of Sn to Ag, we can determine whether dental alloys of Sn and Ag are
substitutional or interstitial alloys.
Analyze
It would be possible for an alloy to be both interstitial and substitutional as long as the voids within the lattice
are about the same size as the metal atom. In particular, if the host metal has a simple cubic structure, the
alloying metal may fit in a cubic void (radius ratio > 0.73) as well as substitute for the host metal in the lattice
structure.
Solve
rSn 140 pm
=
= 0.972
rAg 144 pm
Because these radii are within 15% of each other, an alloy of silver and tin is a substitutional alloy.
Think about It
Remember that it is also important for the metals in substitutional alloys to crystallize in the same closestpacked (hcp, ccp, simple cubic, bcc) structures.
12.61. Collect and Organize
Using the radius ratio rule we can determine whether N atoms (radius 75 pm) fit into the octahedral holes of
closest-packed Fe atoms (radius 126 pm).
Analyze
If the radius ratio falls between 0.41 and 0.73, the nitrogen would fit into the octahedral holes.
Solve
rN
75 pm
=
= 0.60
rFe 126 pm
This ratio does lie in the range for octahedral holes. Yes, N occupies the octahedral holes in Fe.
Think about It
The radius ratio would need to drop below 0.41 before N could occupy the tetrahedral holes. This would occur
for metals with atomic radii greater than 183 pm such as in the lanthanides.
The Chemistry of Solids | 13
12.63. Collect and Organize
We are to predict the formula of alloys formed when atom B variously occupies octahedral and tetrahedral
holes in an fcc lattice of A atoms.
Analyze
The fcc arrangement of A atoms has 4 atoms of A in the unit cell. This fcc unit cell also contains 4 octahedral
holes and 8 tetrahedral holes.
Solve
(a) If all of the octahedral holes are occupied there are 4 B atoms and 4 A atoms in the unit cell. The formula
of the alloy is AB.
(b) If half the octahedral holes are occupied there are 2 B atoms and 4 A atoms in the unit cell. The formula of
the alloy is A2B.
(c) If half of the tetrahedral holes are occupied there are 4 B atoms and 4 A atoms in the unit cell. The formula
of the alloy is AB.
Think about It
If all of the tetrahedral holes were filled, the formula of the alloy would be AB2.
12.65. Collect and Organize
For an alloy in which the A atoms are arranged in an fcc structure and in which there is one B atom for every
5 A atoms, we are to determine what fraction of the octahedral holes in the structure are occupied by B atoms.
Analyze
An fcc unit cell of A atoms contains 4 A atoms and 4 octahedral holes. If there are 5 A atoms for every 1 B
atom, we can consider five unit cells to give a whole number of A atoms, B atoms, and unit cells.
Solve
Five unit cells contain 20 A atoms and 20 octahedral holes. From the 5:1 ratio of A to B atoms, the five unit
cells contain 4 B atoms. These 4 B atoms, therefore, occupy 4/20 or one-fifth of the octahedral holes in the fcc
structure.
Think about It
If the ratio of A to B atoms were 2:1, this would mean that half of the octahedral holes would be filled with B
atoms in the fcc lattice structure of A atoms.
12.67. Collect and Organize
We are to explain why S8 is not a flat molecule.
Analyze
S8 is a ring of 8 sulfur atoms. Each sulfur atom brings 6 valence electrons to the structure. Because each sulfur
is bound to two other S atoms, each S atom has an octet of electrons.
Solve
Each S atom has two bonding pairs and two lone pairs in the structure. The sp3 hybridization on S and the
presence of the lone pairs give the S—S—S bonds a bent geometry and therefore the ring is not flat.
Think about It
The hybridization at each S atom is sp3, giving bond angles of ~109.5˚.
14 | Chapter 12
12.69. Collect and Organize
We are to consider the structure of graphite where the C atoms have been replaced by B and N atoms and
decide if that replacement would pucker the rings in the structure (refer to Figure P12.8).
Analyze
Boron has 3 valence electrons and nitrogen has 5. These atoms are replacing two carbon atoms in graphite
(Figure P12.8b), where each carbon atom brings 4 valence electrons to the structure.
Solve
In order to pucker the ring, lone pairs would have to be introduced onto the atoms in the structure. Because B
and N with 8 total valence electrons replace 2 C atoms, also with 8 valence electrons, the ring of BN atoms
remains flat.
Think about It
BN is boron nitride, a very tough, hard, high-temperature ceramic material.
12.71. Collect and Organize
For the ionic form of ice where the O2– ions are bcc and the H+ ions occupy the holes, we are to determine the
number of H+ and O2– ions in each unit cell and then draw the Lewis structure for ionic ice.
Analyze
A body-centered cubic arrangement of O2– ions would have 4 O2– ions at the corners of a cube and one O2– ion
in the center of the unit cell.
Solve
(a) For the bcc unit cell, there are 2 O2– ions. To balance the charge, there must be 4 H+ ions in the unit cell.
(b)
Think about It
The bcc unit cell has twice as many holes as there are ions in the unit cell. Therefore, the bcc unit cell has 4
holes. In ionic ice all 4 of these holes are filled with H+ ions.
12.73. Collect and Organize
We are to determine the bond angles in cyclic S6.
Analyze
Each sulfur atom in S6 has two bonding pairs and two lone pairs and is sp3 hybridized.
Solve
Each sulfur atom has a bent geometry with an ideal bond angle of 109.5˚.
Think about It
The S6 ring is not flat but, rather, is puckered and is flexible.
12.75. Collect and Organize
We are to explain why the Cl– ions in the rock salt structure of LiCl touch in the unit cell, but those in KCl do
not.
The Chemistry of Solids | 15
Analyze
In the fcc rock salt lattice, the alkali metal, Li+ or K+, is placed into the octahedral holes of the unit cell. From
the radius ratio rule applied to LiCl and KCl (ionic radii are given in Figure 10.2),
rLi+
rK + 138 pm
76 pm
=
= 0.42
=
= 0.76
rCl– 181 pm
rCl– 181 pm
Solve
The radius ratios show that K+ is large and so does not fit well into the octahedral holes (radius ratio between
0.41 and 0.73). Therefore, the rock salt structure for KCl has Cl– ions that may not touch in order to
accommodate the large K+ ions in the fcc lattice.
Think about It
The radius ratio rule is simply a guide to predicting in which type of hole cations will fit within the anion
lattice. You will find that sometimes we would predict an octahedral hole for a cation when the actual crystal
places the cation into a cubic hole, for example.
12.77. Collect and Organize
We are to describe how CsCl could be viewed as both a simple cubic and a body-centered cubic structure.
Analyze
A simple cubic structure consists of atoms located only at the corners of a cube. A body-centered cubic
structure consists of an atom inside a cube of atoms.
Solve
The radius of Cl– is 181 pm and the radius of Cs+ is 170 pm and so their radii are very similar. The Cs+ ion at
the center of Figure P12.77 occupies the center of the cubic cell, so CsCl could be viewed as a body-centered
cubic structure when taking into account the ions’ slight difference in size. However, if we look at the ions as
roughly equal in size, the unit cell becomes two interpenetrating simple cubic unit cells.
Think about It
In the body-centered cubic unit cell, the cubic hole in the simple cubic unit cell is filled.
12.79. Collect and Organize
We consider the rock salt structure to determine whether we could describe it as an fcc array of Na+ with Cl–
in octahedral holes.
Analyze
In order for Cl– to fit into the octahedral holes the radius ratio
rCl– 181 pm
=
rNa + 102 pm
would have to be between 0.41 and 0.73.
Solve
Although the Na+ ions, when viewed without the Cl– ions, are in an fcc array, the Na+ ions are not closestpacked. Also, the radius ratio rCl– rNa+ is greater than 1, indicating that the Cl– does not fit well into the
octahedral holes in the Na+ lattice. It is not accurate, then, to describe NaCl as an fcc array of Na+ with Cl– in
the octahedral holes.
Think about It
It is because of the large difference in radii that the “fcc array” of Na+ ions is not closest-packed. The Cl–
radius is so large in comparison that it has greatly expanded the fcc lattice of Na+ ions.
16 | Chapter 12
12.81. Collect and Organize
Given that the unit cell edge length for an alloy of Cu and Sn is the same as that for pure Cu, we are to
determine whether the alloy would be more dense than pure Cu.
Analyze
Density is mass per volume. If the unit cell edge length is the same for both the alloy and pure copper, then
the volumes of the unit cells of both substances are the same. The density would be greater only if the mass of
the unit cell alloy is greater than for the pure copper.
Solve
Because tin has a greater molar mass than copper, the mass of the unit cell in the alloy would be greater and
therefore, yes, the density of the alloy would be greater than the density of pure copper.
Think about It
The unit cell volume does not change much for substitutional alloys because the alloying atoms are close in
size (within 15%) to the atoms being replaced.
12.83. Collect and Organize
We are to predict the effect on density of an ionic compound of rock salt structure as the cation–anion radius
ratio increases.
Analyze
When the cation–anion ratio increases, the cation is increasing relatively in size. If the anions remain in the
rock salt structure, they might have to expand the fcc structure to accommodate larger cations.
Solve
As the cation–anion radius ratio increases, the closest-packed anions would expand so as to maintain the
cations in the octahedral holes of the fcc lattice. The cell volume, therefore, increases and therefore the
calculated density would be less than the measured density.
Think about It
Another factor in density is the molar mass of the ions of the structure.
12.85. Collect and Organize
For a ccp array of O2– ions with 14 of the octahedral holes containing Fe3+, 18 of the tetrahedral holes
containing Fe3+, and 14 of the octahedral holes containing Mg2+, we are asked to write the formula of the ionic
compound.
Analyze
In a ccp array of ions there are 4 octahedral holes and 8 tetrahedral holes. The ccp array has an fcc unit cell
which has 4 closest-packed anions.
Solve
The unit cell will have the following:
4 O2– anions in the fcc unit cell
1 Fe3+ in octahedral holes
1 Fe3+ in tetrahedral holes
1 Mg2+ in octahedral holes
The formula is MgFe2O4.
Think about It
The salt has charge balance as well: 4 O2– gives a charge of 8–, which is balanced by 2 Fe3+ plus 1 Mg2+ or a
charge of 8+.
12.87. Collect and Organize
The Chemistry of Solids | 17
We consider the structure of anatase, a form of TiO2, found on a map of Vinland believed to date from the
1400s.
Analyze
We can predict the type of hole Ti4+ is likely to occupy by calculating the radius ratio rTi4+ rO2– . The radius
ratios for tetrahedral, octahedral, and cubic holes are 0.22– 41, 0.41– 0.73, and 0.73–1.00, respectively (Table
12.4). In the ccp structure the unit cell is fcc. This unit cell contains 4 octahedral and 8 tetrahedral holes.
Solve
(a) From the radius ratio,
rTi4+
rO2–
=
60.5 pm
= 0.432
140 pm
Ti4+ is expected to occupy octahedral holes.
(b) To give charge balance there is one Ti4+ ion for every two O2– ions in the lattice. Because the unit cell
contains 4 O2– ions, there must be 2 Ti4+ ions in the unit cell. Since there are 4 octahedral holes in the unit cell,
this must mean that half of the octahedral holes in the unit cell are occupied.
Think about It
If the Ti4+ were small enough to fit into the tetrahedral holes, then one-fourth of the tetrahedral holes would be
occupied.
12.89. Collect and Organize
For two different forms of CdS we consider why the rock salt structure would be more dense than the
sphalerite form.
Analyze
In the rock salt structure, the S2– ions would be in an fcc array with Cd2+ occupying all of the octahedral holes.
In the sphalerite structure the S2– ions are also in an fcc array, but with Cd2+ ions occupying half of the
tetrahedral holes. From the radii of S2– and Cd2+ given, we can calculate the radius ratio to determine which
type of hole Cd2+ fits into.
rCd 2+
95 pm
=
= 0.52
rS2–
184 pm
Solve
The radius ratio predicts that Cd2+ fits into the octahedral holes in the ccp lattices of S2– ions. This rock salt
arrangement is more dense than the sphalerite arrangement because in sphalerite the lattice of S2– ions must
expand to accommodate the Cd2+ ions in the smaller tetrahedral holes.
Think about It
This example clearly demonstrates that the radius ratio rule is only a guide to the structure of ionic solids.
Here we predict CdS to be like rock salt, but in nature it is found as sphalerite.
12.91. Collect and Organize
Knowing the structure of the unit cell of ReO3 and the radii of Re and O atoms (given as 137 pm and 73 pm,
respectively), we are asked to calculate the density of ReO3.
Analyze
Given that there are Re atoms at each corner of the cubic unit cell, there is (8 × 18 ) or 1 Re atom in the unit
cell. Likewise, given that there are 12 edge O atoms, there are (12 × 14 ) = 3 O atoms in the unit cell. We are
told that the atoms touch (Re–O–Re) along an edge of the unit cell, therefore the cell edge length is
137 + 73 + 73 + 137 = 420 pm
To calculate the density of ReO3 we use
d=
mass of atoms in unit cell
volume of unit cell
18 | Chapter 12
Solve
The mass, m, of atoms in a unit cell of ReO3 is
186.21 g
15.999 g
1 Re atom ×
+ 3 O atoms ×
mol
mol = 3.889 × 10 –22 g
m=
6.022 × 1023 atoms mol
The volume, V, of the unit cell (in cm3) is
3
⎛
1 × 10 –10 cm ⎞
V = ⎜ 420 pm ×
= 7.409 × 10 –23 cm3
⎟
1 pm ⎠
⎝
The density, d, of the solid is then
3.889 × 10–22 g
d=
= 5.25 g/cm3
7.409 × 10–23 cm3
Think about It
Remember that this is a calculated, or theoretical, density. The actual density, even if the structure is correctly
described, may differ slightly from the theoretical value.
12.93. Collect and Organize
Given the structure (rock salt) and density (3.60 g/cm3) of MgO, we are asked to calculate the length of the
edge of the unit cell (l ) .
Analyze
The volume of the unit cell is simply l 3 . From the density we can calculate the volume, V, as
mass of unit cell
V=
density
In the fcc unit cell there are 4 Mg atoms and 4 O atoms to give a mass, m, of
24.305 g
15.999 g
4 Mg atoms ×
+ 4 O atoms ×
mol
mol = 2.677 × 10–22 g
m=
6.022 × 1023 atoms mol
Solve
V=
2.677 × 10 –22 g
= 7.436 × 10 –23 cm3
3.60 g/cm3
l = 3 V = 3 7.436 × 10–23 cm3 = 4.21 × 10 –8 cm or 421 pm
Think about It
This value seems reasonable. The radius of an oxygen atom is 73 pm and that of a manganese atom is 127 pm
(Appendix 3). Assuming that the O — Mn — O atoms touch along the fcc unit edge length (a good first
approximation), the edge length would be
73 + 127 + 127 + 73 = 400 pm
12.95. Collect and Organize
From a list of properties we are to identify which describe ceramics and which are associated with metals.
Analyze
Metallic bonding in metals allows atoms to slip past each other and be easily deformed. Ceramics have either
covalent or ionic bonding (or sometimes a mixture of both) that does not allow the atoms to easily slip past
each other when stress is applied. Metals have either partially filled valence bands or overlapping valence and
conduction bands that allow electrons to flow through the bands. Ceramics, however, have large band gaps
between their valence and conduction bands that render them thermally insulative and nonconductive to the
flow of electrons.
The Chemistry of Solids | 19
Solve
Ductility, electrical and thermal conductivity, and malleability describe metals. Ceramics can be characterized
as being electrically and thermally insulative and brittle.
Think about It
The bonding differences between metals and ceramics account for their very different properties and,
therefore, unique applications.
12.97. Collect and Organize
We are to write the formula for the mineral formed when Mg2+ replaces Al3+ in kaolinite [Al2(Si2O5)(OH)4].
Analyze
When replacing Al3+ with Mg2+ we must be careful to put in as many Mg2+ as needed to balance the charge of the
Al3+ removed. In this case, since we replace two Al3+ ions (6+ charge) we need 3 Mg2+ to balance the charge.
Solve
Mg3(Si2O5)(OH)4
Think about It
Several other ions (Fe3+, Be2+, etc.) could replace the Al3+ in kaolinite.
12.99. Collect and Organize
For the reaction of KAlSi3O8 with water and carbon dioxide, we are asked to determine whether it is a redox
reaction and to balance the reaction to give Al2(Si2O5)(OH)4, SiO2, and K2CO3 as products.
Analyze
In a redox reaction the oxidation state of atoms must change. All species for this reaction contain O2–, K+,
Si4+, H+, Al3+, and C4+.
Solve
2 KAlSi3O8(s) + 2 H2O (l ) + CO2(aq) → Al2(Si2O5)(OH)4(s) + 4 SiO2(s) + K2CO3(aq)
Because none of the atoms change oxidation state, this is not a redox reaction.
Think about It
In this weathering process the larger silicate anion Si3O84– is broken down by water to Si2O52– and SiO2.
12.101. Collect and Organize
For the transformation of anorthite (CaAl2Si2O8) to a mixture of grossular [Ca3Al2(SiO4)3], kyanite
(Al2SiO5), and quartz (SiO2) under high pressure, we are asked to write the balanced equation and to
determine the charges on the silicate anions.
Analyze
The charge on the silicate ions is found knowing that the cations are Ca2+ and Al3+.
Solve
(a) 3 CaAl2Si2O8(s) → Ca3Al2(SiO4)3(s) + 2 Al2SiO5(s) + SiO2(s)
(b) In anorthite, the silicate anion is Si2O88–.
In grossular, the silicate anion is SiO44–.
In kyanite, the silicate anion is SiO56–.
Think about It
There are many silicate minerals in nature.
12.103. Collect and Organize
Given the radii of Ba2+, Ti4+, and O2– (135, 60.5, and 140 pm, respectively), we can use the radius ratio rule
to determine which holes Ba2+ and Ti4+ occupy in a closest-packed arrangement of O2– anions.
20 | Chapter 12
Analyze
The radius ratio can help us predict whether the smaller ion would fit into a cubic (radius ratio 0.73–1.00),
octahedral (0.41– 0.73), or tetrahedral (0.22– 0.41) hole in the crystal lattice.
Solve
Cubic holes can accommodate Ba2+:
rBa 2+
rO2–
=
135 pm
= 0.964
140 pm
Octahedral holes (maybe tetrahedral) can accommodate Ti4+:
rTi4+ 60.5 pm
=
= 0.432
rO2–
140 pm
Think about It
Ti4+ has a radius ratio at the edge of the tetrahedral–octahedral ranges, so it is difficult to predict which of
these holes it will occupy.
12.105. Collect and Organize
We are asked to explain why an amorphous solid does not give sharp peaks when scanned by X-ray
diffraction.
Analyze
X-ray diffraction gives sharp peaks for crystalline materials that have a regular repeating array of atoms.
Amorphous materials have no long-range order in their arrangement of atoms.
Solve
An amorphous solid has no regular repeating lattice to diffract the X-rays and therefore cannot give rise to
the distinct constructive and destructive interference needed to produce sharp peaks in the X-ray diffraction
scan.
Think about It
Amorphous materials in X-ray diffraction either give no signal or may show very broad peaks.
12.107. Collect and Organize
X-rays and microwaves are both electromagnetic radiation. We are to consider why X-rays, not microwaves,
are used for structure determination.
Analyze
X-rays have wavelengths of the order of 10,000 to 10 pm. Microwaves have wavelengths of the order of 1 m
to 1 mm.
Solve
The separation of atoms in crystal lattices is of the order of 10–10 m or 100 pm. X-rays have wavelengths of
the order of the separation of atoms in crystals so constructive and destructive interference occurs.
Microwaves have wavelengths too long to be diffracted by crystal lattices.
Think about It
X-ray diffraction is a useful technique, but it has difficulty “seeing” light elements, like H atoms.
Complementary information on structure can be obtained from neutron-diffraction experiments.
12.109. Collect and Organize
We are to consider why different wavelengths of X-rays might be used to determine crystal structures.
The Chemistry of Solids | 21
Analyze
From the Bragg equation
nλ = 2d sin θ
we see that if λ is smaller, θ is smaller.
Solve
If a crystallographer uses a shorter λ wavelength, the data set can be collected over a smaller scanning range.
Think about It
In this way the crystallographer can see more data over a smaller scanning range so that if the camera or
source can only obtain 2θ = 30˚, for example, a shorter λ would reveal more peaks than a longer λ.
12.111. Collect and Organize
Given that the lattice spacing in sylvite (KCl) is larger than in halite (NaCl), we can use the Bragg equation
to predict which crystal will diffract X-rays of a particular wavelength through higher 2θ values.
Analyze
Rearranging the Bragg equation to solve for d gives
d=
This shows that sin θ is inversely proportional to d.
nλ
2sin θ
Solve
The smaller the distance d (the lattice spacing) the larger the sin θ and the larger the 2θ. Therefore, halite
with smaller lattice spacing diffracts X-rays through larger 2θ values.
Think about It
This means that the pattern for NaCl is more spread out than that of KCl.
12.113. Collect and Organize
For galena, which shows reflections in X-ray diffraction (λ = 71.2 pm) at 2θ = 13.98˚ and 21.25˚, we are
asked to determine the value of n for these reflections and to calculate the spacing between the layers of the
lattice (d).
Analyze
To determine n we must find a pattern in the angles of reflection. We notice here that for θ = 6.99˚ and
θ = 10.62˚ we have a factor of 1.52. The spacing of the layers in the lattice is calculated from the Bragg
equation:
d=
nλ
2sin θ
Solve
Because the ratio of the reflection angles (θ ) is 10.62˚/6.99˚ = 1.52, the values of n are 2 (θ = 6.99˚) and 3
(θ = 10.62˚). The lattice spacings are
2 × 71.2 pm
d=
= 585 pm
2sin 6.99 o
(
d=
)
3 × 71.2 pm
(
2sin 10.62 o
Average lattice spacing (585 + 580)/2 = 582 pm.
)
= 580 pm
Think about It
The calculated lattice spacings from the reflections may not be exactly equal, but they will be close as in the
example above.
22 | Chapter 12
12.115. Collect and Organize
For a lattice distance of 1855 pm, we are to calculate the smallest angle of diffraction (2θ ) for 154 pm
wavelength X-rays diffracted in pyrophyllite.
Analyze
We can rearrange the Bragg equation to solve for sin θ :
nλ
2d
where n = 1 for the smallest angle of diffraction. The 2θ value is twice the calculated value of θ.
sin θ =
Solve
sin θ =
1 × 154 pm
= 0.04151
2 × 1855 pm
θ = 2.38 o
2θ = 4.76 o
Think about It
For n = 3, the reflection would appear at
sin θ =
3 × 154 pm
= 0.1245
2 × 1855 pm
θ = 7.15 o
2θ = 14.3 o
12.117. Collect and Organize
For a unit cell with X at the 8 corners of the cubic unit cell, Y at the center of the cube, and Z at the center of
each face, we are to write the formula of the compound.
Analyze
Each atom at the corner of a cube counts as 18 in the unit cell. Each atom on a face counts as
cell. Each atom in the center of the unit cell counts as one.
1
2
in the unit
The Chemistry of Solids | 23
Solve
The formula for the compound is XYZ3.
Element X = 18 × 8 = 1 X
Element Y = 1 × 1 = 1 Y
Element Z = 6 × 12 = 3 Z
Think about It
An example of this kind of structure is perovskite, CaTiO3.
12.119. Collect and Organize
Given the density and radius of silicon (2.33 g/mL and 117 pm, respectively) we can determine the volume
of a silicon atom and the mass of 1.00 cm3 of silicon to calculate the packing efficiency of the atoms in the
solid using the equation
Packing efficiency (%) =
volume occupied Si atoms
× 100
volume of unit cell
Analyze
Assuming 1 cm3 of silicon that weighs 2.33 g, we can calculate the number of atoms in that 1 cm3. From the
radius of one silicon atom, we can calculate the volume of that silicon atom and multiply by the number of
atoms in the 1 cm3 sample. This gives the volume of space occupied by all the atoms in 1.00 cm3 of silicon.
The percentage of space occupied then is the volume of Si divided by 1 cm3.
Solve
Number of Si atoms in 1 cm3
2.33 g ×
1 mol 6.022 × 1023 atoms
×
= 4.995 × 1022 atoms
28.09 g
1 mol
Volume of 1 Si atom
3
4
4 ⎛
1×10–10 cm ⎞
–24
3
V = π r 3 = π ⎜117 pm ×
⎟ = 6.709 ×10 cm
3
3 ⎝
1 pm ⎠
Volume of Si atoms in 1.00 cm3 of Si
cm3
6.709 × 10−24
× 4.995 × 1022 atoms = 0.3351 cm3
atom
Packing efficiency
0.3351 cm3
× 100 = 33.5%
1.00 cm3
Think about It
This means that 66.5% of the silicon block is empty space!
12.121. Collect and Organize
We consider nanocubes of Mo (4.8 nm on each side) and are asked to calculate the radius of Mo in the
nanocubes, the density, and the number of Mo atoms in each nanocube.
Analyze
(a) We are given that the unit cells are body-centered cubic. In this arrangement the unit-cell edge length (l )
is 4r 3 where r is the radius of the atoms. The length of a unit cell is 4.8 nm/15 = 0.32 nm or 320 pm
because we are given that the nanocube is 15 unit cells on an edge.
(b) The density of the nanocube can be calculated knowing that in each bcc unit cell there are 2 Mo atoms.
Once we calculate the mass of these 2 atoms of Mo we divide by the volume (l 3 ) of the unit cell.
(c) Because we know that there are 15 unit cells on an edge for the nanocubes and that there are 2 Mo atoms
in each unit cell, we can calculate the number of atoms in each nanocube.
24 | Chapter 12
Solve
(a) Rearranging the equation for the unit edge length for a bcc unit cell gives the effective radius for Mo:
l 3 320 pm 3
=
= 139 pm
4
4
(b) The density of the bcc array of Mo in the nanocubes is
95.96 g
1 mol
⎛
⎞
×
⎜ 2 atoms ×
⎟
23
g
mol
6.022 ×10 atoms ⎠
d = ⎝
= 9.96
3
3
–10
cm
⎛
1×10 cm ⎞
⎜ 320 pm ×
⎟
1 pm ⎠
⎝
(c) The number of unit cells and Mo atoms in a nanocube are
r=
(15 unit cells on an edge)3 = 3375 unit cells
3375 unit cells ×
2 Mo atoms
= 6750 Mo atoms
unit cell
Think about It
Even though nanocubes are very small, they still contain thousands of Mo atoms.
12.123. Collect and Organize
We consider different possible crystal structures of Fe (radius 126 pm) and compare the densities of bcc and
hcp iron and calculate the density of a crystal of 96% Fe and 4% Si.
Analyze
For each structure the density is the mass of atoms in the unit cell divided by the volume of the cubic unit
cell. For the bcc unit cell there are 2 atoms per unit cell and the edge length is l = 4r 3 . For the hcp unit
cell there are 2 atoms per unit cell.
Solve
(a) The edge length of a bcc unit cell is
4r 4 × 126 pm
l =
=
= 291.0 pm or 2.91 × 10 –8 cm
3
3
The volume of the unit cell is
(
)
3
V = 291.0 × 10 –8 cm = 2.464 × 10 –23 cm3
The mass of the unit cell is
2 atoms Fe ×
The density is
55.85 g
1 mol
×
= 1.855 × 10–22 g
mol
6.022 × 1023 atoms
1.855 × 10–22 g
= 7.53 g/cm3
–23
3
2.464 × 10 cm
(b) In the hcp unit cell there are also 2 atoms in the unit cell. We are given that the volume of the unit cell is
5.414 × 10–23 cm3, so the density is
1.855 × 10–22 g
d=
= 3.42 g/cm3
–23
3
5.414 × 10 cm
(c) If 4% of the mass is due to Si replacing Fe, the molar mass of the alloy would be
28.09 g
55.85 g
0.04 ×
+ 0.96 ×
= 54.74 g/mol
mol
mol
The mass of 2 atoms in the unit cell would be
54.74 g
1 mol
2 atoms ×
×
= 1.818 × 10–22 g
mol
6.022 × 1023 atoms
The density is
d=
The Chemistry of Solids | 25
d=
1.818 × 10–22 g
= 3.36 g/cm3
5.414 × 10–23 cm3
Think about It
Replacing Fe with an element of lower molar mass (silicon) gives a lower density alloy.
12.125. Collect and Organize
For the same crystalline structure of AuZn with l = 319 pm and AgZn with l = 316 pm, we can use the
different molar masses of gold and silver and the length of the unit-cell edges to calculate the densities and
predict which alloy is more dense.
Analyze
Because gold has a much higher molar mass than silver, we can easily predict that the density of the AuZn
alloy would be greater than that of the AgZn alloy, despite the slightly smaller unit-cell edge length and
therefore smaller volume.
Solve
For AuZn
d=
( 262.38 g/mol × 1 mol/6.022 × 10 ) = 13.4 g/cm
(3.19 × 10 cm)
d=
(173.28 g/mol × 1 mol/6.022 × 10 ) = 9.12 g/cm
(3.16 × 10 cm)
23
For AgZn
–8
3
3
23
–8
3
3
The AuZn alloy is more dense than the AgZn alloy.
Think about It
Indeed, the prediction was correct and AuZn is substantially more dense than AgZn.
12.127. Collect and Organize
Using Equation 12.1 for packing efficiency we can calculate the value for a simple cubic unit cell.
Analyze
To find the packing efficiency we need to first calculate the volume of space occupied by the atoms as
spheres. The volume of a sphere is 4π r 3 3 . The volume of the atoms in the unit cell is the number of
atoms in the unit cell (1 for simple cubic) times the volume computed from the atom’s radius. Next, we
need to compute the volume of the unit cell from the edge length. For a simple cubic unit cell V = (2r)3
because 2r = l .
Solve
Using r as the radius for an atom, the packing efficiency of a simple cubic unit cell is
4 3
πr
packing efficiency = 3 3 × 100 = 52.4%
( 2r )
Think about It
Compare this to the packing efficiency of a body-centered unit cell (68%) and a cubic closest-packed or
hexagonal closest-packed unit cell (both 74%).
12.129. Collect and Organize
We are asked how Mn would incorporate into an iron lattice, either in the holes or as a substitute for iron.
26 | Chapter 12
Analyze
To occupy holes in a structure, the Mn would have to have a distinctly smaller radius than the Fe radius.
From Appendix 3, we find that the Mn radius is 127 pm and the Fe radius is 126 pm.
Solve
The radii of these metals are very close to each other, so Mn likely forms a substitutional alloy with iron.
Think about It
The radius of carbon (77 pm) is small enough to fit into the interstices of the iron–manganese lattice. The
radius ratio of carbon to iron (77/126) gives a value of 0.61. This indicates that carbon would fit into the
octahedral holes of the iron lattice.
12.131. Collect and Organize
From the formula Cu3Al for the alloy with a bcc structure, we are to determine how the copper and
aluminum atoms are distributed between the unit cells.
Analyze
The bcc unit cell contains 2 atoms. The radii of Al and Cu are 143 and 128, respectively. The radius ratio is
0.895.
Solve
The radius ratio indicates that Cu atoms fit into the cubic holes of the Al lattice. Another way to look at this
is as a simple cubic arrangement of Cu atoms with an Al atom in the center of the unit cell. To be consistent
with the formula (Cu3Al), we would have to consider three unit cells, one of which would have one Al atom
in the center.
Think about It
It is probably also possible to have an alloy of formula Al3Cu since their atomic radii are so similar.
12.133. Collect and Organize
For the reactions of Cl2 with Sn and Pb and for the thermal decomposition of SiCl4, we are asked to write
balanced chemical reactions.
Analyze
In the balanced equations we should include the physical state of the reactants and products. In the reaction
of gaseous Cl2 with solid Pb or Sn we form solid PbCl2 and liquid SnCl4, respectively. In the thermal
decomposition of liquid SiCl4 we form solid Si and gaseous Cl2.
Solve
(a) Sn(s) + 2 Cl2(g) → SnCl4 (l )
(b) Pb(s) + Cl2(g) → PbCl2(s)
o
>700 C
(c) SiCl4 (l ) ⎯⎯⎯⎯
→ Si(s) + 2 Cl2(g)
Think about It
Lead might also form PbCl4 (l ) which would require 2 mol of Cl2 for every mole of Pb in the balanced
chemical equation.
12.135. Collect and Organize
We consider how the “average” value of the ionization energies and electron affinities of group 14 elements
like C, Si, Ge, and Sn explains their tendency to form covalent bonds rather than ionic bonds.
Analyze
Ionization energy is the energy required (usually endothermic) to remove an electron from a gaseous species.
Electron affinity is the energy change (may be endo- or exothermic) when a species gains an electron.
The Chemistry of Solids | 27
Solve
Group 14 elements do not have low ionization energies or electron affinities so they tend to neither easily
lose nor easily gain an electron. As a result, they form bonds by sharing electrons, not by giving up or
accepting electrons.
Think about It
Two of these elements (Si, Ge) are also quite good as semiconductors whose band gap is intermediate
between metals and insulators.
CHAPTER 13 | Organic Chemistry: Fuels, Pharmaceuticals, and Materials
13.1. Collect and Organize
Given the carbon-skeleton structures in Figure P13.1 of four hydrocarbons, we are to determine the degrees of
unsaturation present in each.
Analyze
An unsaturated hydrocarbon contains one or more carbon–carbon double or carbon–carbon triple bonds. These
compounds have less than the maximum amount of hydrogen for each carbon atom. A double bond has one
degree of unsaturation and a triple bond has two degrees of unsaturation.
Solve
(a) This structure has one double bond. It has one degree of unsaturation.
(b) This structure has two double bonds. It has two degrees of unsaturation.
(c) This structure has neither double nor triple bonds. It is a saturated hydrocarbon with no degrees of
unsaturation.
(d) This structure has one double and one triple bond. It has three degrees of unsaturation.
Think about It
Because the structure in (d) has three degrees of unsaturation, it will combine with 3 molecules of hydrogen
(H2) to form a saturated hydrocarbon.
13.3. Collect and Organize
From the structures of fragrant oils in Figure P13.3, we are to identify those that contain the alkene functional
group.
Analyze
An alkene functional group is a carbon–carbon double bond.
Solve
Both pine oil and oil of celery contain a C
C bond and so are classified as alkenes.
Think about It
Camphor is not an alkene but it does have a C
O double bond. This functional group is a ketone.
13.5. Collect and Organize
Of the four hydrocarbons shown in Figure P13.5, we are to identify those that are aromatic.
Analyze
Aromatic compounds have planar, hexagonal rings of 6 sp2-hybridized carbon atoms with alternating single
and double bonds.
Solve
Compounds b and d are aromatic.
Think about It
Aromatic compounds have a special stability due to resonance.
13.7. Collect and Organize
For the molecules in Problem 13.6 we are asked to identify the other functional groups aside from the aromatic
rings.
55
56 | Chapter 13
Analyze
The other typical functional groups include alkenes (C
(R— O —R), aldehydes [R(C O)H], ketones [R(C
[R(C O)OR'], and amides [R(C O)NH2].
C), alkynes (C C), alcohols (R— OH), ethers
O)R'], carboxylic acids [R(C O)OH], esters
Solve
Benzyl acetate contains an ester group. Carvone contains two alkene groups and a ketone group.
Cinnamaldehyde contains an alkene and an aldehyde group.
Think about It
The wide variety of functional groups possible in organic compounds means that there are many combinations
of functional groups in organic chemistry. This gives rise to a seemingly endless array of different substances
with potentially useful properties.
13.9. Collect and Organize
From the structure of dihydroxydimethylsilane (Figure P13.9) we are to draw the condensed structure for the
repeating monomeric unit in Silly Putty.
Analyze
The condensation reaction of dihydroxydimethylsilane combines the monomers of Si(CH3)2(OH)2 to create a
larger molecule, losing water as a by-product. The –OH groups on the monomer will combine to produce the
– O — Si — O – linkage for the polymer backbone.
Solve
CH3
HO
Si
CH3
OH + HO
CH3
Si
CH3
OH
HO
CH3
Si
CH3
O
CH3
Si
OH
CH3
Continuing this reaction produces a long-chain polymer with the condensed structure
Think about It
These types of siloxane polymers have found uses in soft contact lenses, oils, and greases.
13.11. Collect and Organize
We are asked to draw monomeric units of cis- and trans-polyisoprene.
Analyze
The structures of the two polyisoprenes are shown in Figure P13.11. These polymers form through addition
reactions of an alkene. The monomeric unit is the smallest repeating unit in the polymer.
Solve
For cis-polyisoprene the monomeric unit is
For trans-polyisoprene the monomeric unit is
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 57
Think about It
The difference in these polymers is the orientation of how the monomeric units are joined together, which
gives a different orientation of the C C double bond in the polymer backbone.
13.13. Collect and Organize
Using the structure of Mevacor in Figure P13.13, we are to determine the number of chiral carbons this
cholesterol drug contains.
Analyze
A carbon atom is chiral when it is bonded to four different groups. Carbons that are sp2 or sp hybridized in a
structure cannot be chiral because they have only three or two groups bonded to them, respectively.
Solve
Mevacor has 8 chiral carbon atoms.
Think about It
The more chiral carbons there are in an organic molecule, the more difficult it is, in general, to synthesize and
purify it as a single enantiomer.
13.15. Collect and Organize
Carbon can have three different hybridized sets of s and p orbitals. We are asked to describe three ways in
which carbon– carbon bonds form.
Analyze
There are three 2p orbitals on carbon that may hybridize with the one 2s orbital.
Solve
The sp hybrid orbital on carbon can form triple bonds; the sp2 hybrid orbitals can form double bonds; the sp3
hybrid orbitals can form single bonds.
Think about It
Carbon does not form sp3d orbitals because there are no 2d orbitals (and the 3d orbitals are too high in energy
to mix with the 2s and 2p orbitals); it does not expand its valence beyond four.
13.17. Collect and Organize
Tungsten carbide contains carbon. We are asked whether this compound is considered to be an organic
compound.
58 | Chapter 13
Analyze
Organic compounds are compounds in which there is a carbon– element bond.
Solve
The carbon atoms in WC occupy the interstices of the closest-packed tungsten lattice and therefore are not
formally bonded to the tungsten at all. No, tungsten carbide is not considered to be an organic compound.
Think about It
We generally consider organic compounds to contain carbon covalently bonded to carbon, hydrogen, oxygen,
nitrogen, phosphorus, sulfur, and other atoms.
13.19. Collect and Organize
Looking back at Chapter 8, we are to identify the functional groups introduced there.
Analyze
A functional group is a unit of a compound that has characteristic properties for the organic compound.
Solve
The functional groups introduced in Chapter 8 include carboxylic acids, aldehydes, alkenes, alcohols, alkynes,
and aromatic rings.
Think about It
By identifying the functional groups in an organic compound, chemists can predict some of its properties and
reactivities.
13.21. Collect and Organize
Given that polyethylene is composed of C2H4 monomeric units, we are to calculate the number of monomers
of ethylene needed to give a polymer of molar mass 100,000 g/mol.
Analyze
We can find the number of monomers by dividing 100,000 g/mol by the molar mass of the C2H4 monomer
unit. C2H4 has a molar mass of 28.05 g/mol.
Solve
100, 000 g/mol
= 3565 monomer units
28.05 g/mol
Think about It
This is a relatively small polymer. Some polymers have tens of thousands or more monomeric units in their
polymer chains.
13.23. Collect and Organize
We are to compare the empirical formulas for linear and branched alkanes that have the same number of
carbon atoms.
Analyze
For any alkane, whether branched or linear, the empirical formula is CnH2n+2.
Solve
Yes, linear and branched alkanes with the same number of carbon atoms have the same empirical formula.
Think about It
An example of this is the following alkanes with 4 carbon atoms and the empirical formula C4H10:
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 59
n-Butane
2-Methylpropane
13.25. Collect and Organize
By considering the number of covalent bonds formed by carbon in alkanes, we can determine the
hybridization of the carbon atoms.
Analyze
In alkanes, carbon is singly bonded to four other atoms, either to other carbon atoms or to hydrogen atoms.
Solve
Carbon forms four single bonds when it undergoes sp3 hybridization.
Think about It
When the carbon is sp2 hybridized it may form double bonds to other carbon atoms (alkenes) or to oxygen
(ketones, aldehydes, carboxylic acids).
13.27. Collect and Organize
Given that cyclohexane (C6H12) is not planar we can draw the Lewis structure to help explain why.
Analyze
The Lewis structure for cyclohexane is
H
H
H
H
H
H
C
H
C
C
C
C
C
H
H
H
H
H
Solve
Each carbon atom in the six-membered ring is sp3 hybridized with ideal bond angles of 109.5˚. The ring,
therefore, is not planar.
Think about It
The cyclohexane ring can adopt two structures; the chair structure is more stable.
13.29. Collect and Organize
By considering the definition of a saturated hydrocarbon, we can determine whether cycloalkanes are
saturated.
Analyze
By the textbook definition (Section 13.2), a saturated hydrocarbon has the maximum ratio of hydrogen to
carbon atoms in its structure and has an empirical formula of CnH2n+2.
60 | Chapter 13
Solve
No. A cycloalkane has a formula of CnH2n and therefore is not a saturated hydrocarbon.
Think about It
This is a tricky definition, however. Another definition of a saturated hydrocarbon is when every carbon in the
molecule is bonded to 4 other atoms. That definition would classify cycloalkanes as saturated hydrocarbons.
13.31. Collect and Organize
By considering the definition of structural isomers we can determine if they always have the same chemical
properties.
Analyze
Two compounds are structural isomers if they have the same formula but different arrangements of the atoms.
Solve
No. Structural isomers have different chemical properties due to their different arrangement of atoms.
Think about It
Structural isomers also have different physical properties such as different melting points and vapor pressures.
13.33. Collect and Organize
For the molecular formula C5H12 we are asked to draw and name all the structural isomers.
Analyze
Structural isomers all have the same molecular formula so all of our structures must contain 5 C atoms and 12 H
atoms, just arranged differently. Naming of alkanes is described in Section 13.2 with examples in Table 13.4
for the linear alkanes.
Solve
n-Pentane
2-Methylbutane
2,2-Dimethylpropane
Think about It
The naming convention used for these compounds uniquely describes the structure of each isomer.
13.35. Collect and Organize
Of the five molecules shown in Figure P13.35, we are to identify and name those that are structural isomers of
n-octane.
Analyze
Any structural isomer of n-octane has to have the molecular formula C8H18.
Solve
(a) This molecule is C8H18 and is therefore a structural isomer.
(b) This molecule is C9H20 and is not a structural isomer of n-octane.
(c) This molecule is C8H18 and is therefore a structural isomer.
(d) This molecule is C8H18 and is therefore a structural isomer.
(e) This molecule is C9H20 and is not a structural isomer of n-octane.
The structural isomers of n-octane are (a) 2,3-dimethylhexane, (c) 2-methylheptane, and (d) 2-methylheptane.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 61
Think about It
Molecules c and d are not structural isomers of each other because they have the same arrangement of atoms
(just drawn differently), but molecules b and e are structural isomers of each other.
13.37. Collect and Organize
Using the examples for hydrocarbons in Section 13.2, we can convert the line drawings of each molecule in
Problem 13.35 to chemical formulas.
Analyze
In line drawings, only the carbon skeleton is shown. The end of a line is understood to be the –CH3 group and
the intersection of two lines is understood to be a – CH2 – group. If more than two lines intersect, then the
number of hydrogen atoms at that carbon is 4 minus the number of intersecting lines. This is because carbon
makes bonds to four atoms to satisfy its octet.
Solve
(a) C8H18
(b) C9H20
(c) C8H18
(d) C8H18
(e) C9H20
Think about It
The names of the isomers of C9H20 shown in Problem 13.35 are (b) 2-methyloctane and (e)
2,3,4-trimethylhexane.
13.39. Collect and Organize
By adding the energy of the bonds that form (–∆H) and the energy of the bonds broken, (+∆H) in the reaction
of C2H4 with H2 to give C2H6, we can estimate ΔHhydrogenation.
Analyze
The relevant bond strengths from Appendix 4 for the reactants and products are as follows:
C—H
413 kJ/mol
C C
614 kJ/mol
H—H
436 kJ/mol
C— C
348 kJ/mol
In the reactants a C C bond and a H—H bond are broken. In forming the products, a C— C bond and two
C—H bonds are formed.
Solve
ΔH rxn = ∑ ΔH bond breaking + ΔH bond forming
= [614 + 436] + ⎡⎣ –348 + ( 2 × –413)⎤⎦ = –124 kJ
Think about It
Because this reaction is exothermic, it is favored by enthalpy.
13.41. Collect and Organize
By examining the degree of dispersion forces in the nonpolar molecules C3H8, C14H30, and cyclooctane
(C8H16), we can put them in order of increasing boiling point.
Analyze
The larger the dispersion forces, the higher the boiling point. Dispersion forces are greater for nonpolar
molecules with more atoms and for those that are less branched or those that have longer chains.
62 | Chapter 13
Solve
Because C14H30 has the greatest number of atoms its boiling point is the highest. Next is cyclooctane, and
C3H8 has the lowest boiling point. In order of increasing boiling point, C3H8 < C8H16 < C14H30.
Think about It
The branched isomers of C14H30 have lower boiling points than the linear isomer.
13.43. Collect and Organize
By defining structural and geometric isomers we can differentiate between them.
Analyze
Both structural and geometric isomers have the same molecular formulas.
Solve
Structural isomers have different connectivity of the atoms; geometric isomers have the same connectivity of
the atoms but a different spatial arrangement.
Think about It
As examples 1-butene, 2-butene, and 2-methylpropene are structural isomers, whereas trans- and cis-2-butene
are geometric isomers.
H
H
C
H
C
CH3
C
H
CH2CH3
1-Butene
H3C
H
C
CH3
2-Methylpropene
H
C
C
H
CH3
trans-2-Butene
cis-2-Butene
13.45. Collect and Organize
By considering the empirical formula of an alkene and a cycloalkane with the same number of carbon atoms,
we can determine whether combustion analysis could distinguish between the two.
Analyze
Both cycloalkanes and alkenes have the formula CnH2n.
Solve
Since there is no difference in the number of either C atoms or H atoms between the alkene and the
cycloalkane, no, we cannot distinguish these by combustion analysis.
Think about It
We could, however, distinguish between an alkane of n C atoms (CnH2n+2) and an alkene (CnH2n).
13.47. Collect and Organize
We are to explain why alkenes in which the double bond involves the first (or last) carbon do not exhibit cis–
and trans isomerism.
Analyze
A cis isomer has two like groups on the same side of a line drawn through the double bond. A trans isomer
has two like groups on opposite sides of a line drawn through the double bond.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 63
Solve
When the double bond is “terminal” (occurs at the end or beginning of the carbon chain) there are two like
groups (H) so no cis and trans isomers are possible.
Think about It
No cis and trans isomers would likewise be possible for a double bond where two of the substituents on a
single carbon atom are the same, as shown in the following example:
H3C
H
C
H3C
C
CH3
2-Methyl-2-butene
13.49. Collect and Organize
From the structure of carvone we are asked to explain why this molecule does not have cis and trans isomers.
Analyze
A cis isomer has two like groups on the same side of a line drawn through the double bond. A trans isomer
has two like groups on opposite sides of a line drawn through the double bond.
Solve
The C C double bond outside of the ring does not show cis–trans isomerism because there are not two
dissimilar groups on the terminal carbon atom. The C C double bond in the ring of carbon atoms is cis in
the structure of carvone. This bond cannot be trans or the ring of 6 carbon atoms would not be possible.
Think about It
A related molecule that would show trans and cis isomers would be
Trans
Cis
13.51. Collect and Organize
By comparing the structures of ethylene and polyethylene, we can explain why ethylene reacts with HBr but
polyethylene does not.
Analyze
The structure of ethylene and polyethylene are as follows:
Ethylene
Solve
Ethylene has a C
Polyethylene
C bond with which HBr is reactive but polyethylene has only saturated C— C bonds.
Think about It
Polyethylene is produced from ethylene under high temperature and pressure.
64 | Chapter 13
13.53. Collect and Organize
For the two structures shown in Figure P13.53, we are asked to label the isomers as cis or trans and E or Z.
Analyze
A cis or Z isomer has two like groups on the same side of a line drawn through the double bond. A trans or E
isomer has two like groups on opposite sides of a line drawn through the double bond.
Solve
Isomer a is trans, E and isomer b is cis, Z.
Think about It
In designating these isomers we more often encounter cis and trans rather than E and Z, but knowing both
designations is helpful in studying organic chemistry.
13.55. Collect and Organize
o
Using ΔH fo values for the reactants and products we can calculate the ΔH rxn
for the controlled combustion of
methane.
Analyze
o
To calculate ΔH rxn
we use
o
ΔH rxn
= ∑ n ΔH f,o products – ∑ m ΔH f,o reactants
Solve
o
ΔH rxn
= ⎡⎣( 2 mol C2 H 2 × 226.7 kJ/mol) + ( 2 mol CO × –110.5 kJ/mol) + (10 mol H 2 × 0 kJ/mol)⎤⎦
– ⎡⎣( 6 mol CH 4 × –74.8 kJ/mol) + (1 mol O 2 × 0 kJ/mol)⎤⎦
= 681.2 kJ
This reaction, a controlled combustion of methane, is endothermic.
Think about It
The uncontrolled combustion of methane to give CO2 and H2O, however, is exothermic.
13.57. Collect and Organize
From the carbon-skeleton structure of vinyl acetate (Figure P13.57), we are to draw the structure of poly(vinyl
acetate).
Analyze
Poly(vinyl acetate) is an addition polymer in which the C C double bonds link together to form the polymer
backbone. In this polymer the acetate group is a side chain off the polymer chain.
Solve
Think about It
Poly(vinyl acetate), or PVA, is used in bookbinding due to its flexibility and, as an emulsion in water, as an
adhesive for wood, paper, and cloth.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 65
13.59. Collect and Organize
By looking at the Lewis structure of benzene we can explain why it is a planar molecule.
Analyze
Benzene has a six-membered carbon ring with alternating single and double bonds.
Solve
The line structure of benzene is
In benzene, each C atom is sp2 hybridized with bond angles of 120˚. This geometry at each of the carbon
atoms in the ring makes benzene a planar molecule.
Think about It
Benzene’s π electrons are delocalized by resonance (Figure 13.25), lending benzene a special stability.
13.61. Collect and Organize
By drawing the line structures of tetramethylbenzene and pentamethylbenzene, we can determine if these
molecules have any structural isomers.
Analyze
The methyl groups on these compounds are ring substitutents. That is, they take the place of H atoms in the
benzene structure.
Solve
Tetramethylbenzene has three structural isomers:
Pentamethylbenzene has no structural isomers:
Think about It
Remember that structural isomers have distinct chemical and physical properties.
13.63. Collect and Organize
By examining pyridine’s structure (Figure P13.63) we can determine if this compound is aromatic.
Analyze
Aromatic structures have alternating single and double bonds through which resonance delocalizes the
electrons.
66 | Chapter 13
Solve
Yes. Pyridine is an aromatic compound because the π electrons are delocalized over all the atoms in the ring
through resonance.
N
N
Think about It
In this aromatic compound, the alternating single and double bonds include C—N bonds.
13.65. Collect and Organize
We are to draw all the structural isomers of trimethylbenzene.
Analyze
The methyl groups on the benzene ring are substituents that have replaced the H atoms on benzene.
Solve
Think about It
The structures below are not additional structural isomers. They are the same as the first isomer above. All we
need to do is rotate each of the ones below to give the isomer above.
13.67. Collect and Organize
For benzene and ethylene we are to calculate the fuel values. We are also asked if 1 mol of benzene has a
higher or lower fuel value than 3 mol of ethylene.
Analyze
The fuel value is (–∆Hcomb/mass of fuel). For the two fuels the combustion reactions are
2 C6H6 (l ) + 15 O2(g) → 12 CO2(g) + 6 H2O (l )
C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O (l )
Solve
o
The ΔH comb
for 1 mol benzene:
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 67
Fuel value:
o
ΔH comb
= ⎡⎣(12 × –393.5) + ( 6 × –285.8)⎤⎦ − ⎡⎣( 2 × 49.0) + (15 × 0 )⎤⎦
= – 6534.8 kJ for 2 mol benzene (from balanced equation)
= –3267.4 kJ for 1 mol benzene
3267.4 kJ/mol
= 41.83 kJ/g
78.11 g/mol
o
The ΔH comb
for 3 mol ethylene:
o
ΔH comb
= ⎡⎣( 2 × –393.5) + ( 2 × –285.8)⎤⎦ − ⎡⎣(1 × 52.3) + ( 3 × 0)⎤⎦
= –1410.9 kJ for 1 mol ethylene (from balanced equation)
Fuel value:
1410.9 kJ/mol
= 50.30 kJ/g
28.05 g/mol
Looking at this carefully for the comparison:
1 mol benzene has an energy content of 3267.4 kJ
3 mol ethylene have an energy content of 3 × 1410.9 = 4232.7 kJ
Therefore, 1 mol benzene has a lower energy content than 3 mol ethylene.
=
Think about It
Ethylene has 6 additional C—H in 3 mol compared to 1 mol benzene. Breaking of these bonds and the
formation of additional H— O bonds in water must account for the difference in energy content.
13.69. Collect and Organize
By comparing the structure of methylamine to that of n-butylamine, we can explain why methylamine is more
soluble in water.
Analyze
The line structures of these two amines are
Solve
Methylamine has a smaller nonpolar hydrocarbon chain compared to n-butylamine and so it is more soluble in
water.
Think about It
Both amines are fairly soluble in water and indeed can react with water to form basic solutions.
CH3 NH 2 (aq) + H 2 O(l ) → CH3 NH3+ (aq) + OH – (aq)
13.71. Collect and Organize
In the structures of amphetamine and serotonin (Figure P13.71), we are to identify the primary and secondary
amine functional groups.
Analyze
Primary amines (1˚) have one R group bonded to the nitrogen atom, whereas secondary amines (2˚) have two
R groups bonded to the nitrogen atom.
Solve
68 | Chapter 13
Think about It
By analogy, tertiary amines (3˚) have three R groups bonded to the nitrogen atom and quaternary amines (4˚)
have four R groups bonded to the nitrogen atom.
13.73. Collect and Organize
We can use the ΔH fo values in Appendix 4 and the given ΔH fo for methylamine (–23.0 kJ/mol) to calculate
the enthalpy of the reaction
4 CH3NH2(g) + 2 H2O (l ) → 3 CH4(g) + CO2(g) + 4 NH3(g)
Analyze
The heat of a reaction may be calculated using ΔH fo for the reactants and products according to the equation
o
ΔH rxn
= ∑ n ΔH f,o products – ∑ m ΔH f,o reactants
Solve
o
ΔH rxn
= ⎡⎣( 3 mol CH 4 ) ( –74.8 kJ/mol) + (1 mol CO2 ) ( −393.5 kJ/mol) + ( 4 mol NH3 ) ( – 46.1 kJ/mol)⎤⎦
– ⎡⎣( 4 mol CH3 NH 2 ) ( –23.0 kJ/mol) + ( 2 mol H 2 O) ( –285.8 kJ/mol)⎤⎦
o
ΔH rxn
= –138.7 kJ
Think about It
Be careful to use the correct enthalpy of formation for the phases shown in the equation. In this problem
ammonia and carbon dioxide are in the gas phase and water is in the liquid phase.
13.75. Collect and Organize
We are asked why the fuel values of ethanol and diethyl ether are lower than that of ethane. In comparing the
fuel values we must take into account the oxygenation of the compounds.
Analyze
The more oxygenated a fuel, the lower is its fuel value. All three compounds have 2 carbons and 6 hydrogens.
Solve
Because dimethyl ether and ethanol both contain oxygen in their structures but ethane does not, the fuel values
of both dimethyl ether and ethanol are lower than that of ethane.
Think about It
We might expect, though, that the fuel values of dimethyl ether and ethanol will not differ much from each
other since they are isomers.
13.77. Collect and Organize
Ethers have the general structure R— O —Rʹ′ and alcohols have the general structure R— OH. We are asked to
explain their general difference in boiling point.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 69
Analyze
The lower the boiling point, the weaker are the intermolecular forces between the molecules. Since ethers boil
at lower temperatures than isomeric alcohols, as given in the problem statement, ethers must have weaker
intermolecular forces than alcohols.
Solve
Ethers have lower boiling points compared to alcohols because they have weaker dipole–dipole forces
compared to the alcohols, which have hydrogen bonding between the molecules.
Think about It
Both ethers and alcohols also have dispersion forces that attract their molecules to each other. These get
stronger as the length of the R groups on the ether or alcohol increases.
13.79. Collect and Organize
We need to consider the evaporation of ethanol to explain why our skin feels cold after wiping with ethanol.
Analyze
When ethanol comes in contact with your skin it begins to evaporate.
Solve
Evaporation of ethanol from the skin is an endothermic process (phase change from liquid to vapor). The heat
transfers from the skin to the ethanol so the skin feels cold.
Think about It
The reverse process, condensation, is an exothermic process, as you learned from Chapter 5.
13.81. Collect and Organize
From the structures shown in Figure P13.81 we are to identify which are ethers and which are alcohols and
place them in order of increasing boiling point.
Analyze
Ethers have the general formula R— O —Rʹ′ and alcohols have the general formula R— OH. The greater the
intermolecular forces between molecules, the higher is the boiling point. Because of hydrogen bonding,
alcohols generally have higher boiling points than ethers. The larger the molecule (greater number of atoms)
and the less branching it has, the greater is the boiling point.
Solve
Compounds a and d are alcohols; compounds b and c are ethers. In order of increasing boiling point, b < c < a < d.
Think about It
Our prediction for boiling point order is nearly correct: (b) diethyl ether, 35˚C < (c) isobutyl methyl ether,
59˚C < (a) 3-methyl-4-heptanol, 174˚C < (d) 2,5-dimethylcyclohexanol, 170˚C. Notice that compounds a and
d have nearly equal boiling points.
13.83. Collect and Organize
We are to calculate the fuel values of butanol and diethyl ether and indicate which has the higher fuel value.
Analyze
o
The fuel value is equal to – ΔH comb
divided by the mass of fuel. For these fuels the combustion reactions are
(C2H5)2O (l ) + 6 O2(g) → 4 CO2(g) + 5 H2O (l )
C4H9OH (l ) + 6 O2(g) → 4 CO2(g) + 5 H2O (l )
Solve
The fuel value for diethyl ether:
70 | Chapter 13
o
ΔH comb
= ⎡⎣( 4 × –393.5) + (5 × –285.8)⎤⎦ − ⎡⎣(1 × –279.6 ) + (6 × 0 )⎤⎦
= –2723.4 kJ/mol
2723 kJ/mol
Fuel value =
= 36.74 kJ/g
74.12 g/mol
The fuel value for n-butanol:
o
ΔH comb
= ⎡⎣( 4 × –393.5) + (5 × –285.8)⎤⎦ − ⎡⎣(1 × −327.3) + (6 × 0 )⎤⎦
= –2675.7 kJ
2675.7 kJ/mol
= 36.10 kJ/g
74.12 g/mol
Diethyl ether has a slightly higher fuel value than n-butanol.
Fuel value =
Think about It
We would expect these compounds to have very similar fuel values because they are isomers.
13.85. Collect and Organize
After calculating the fuel values of ethanol (C2H5OH) and methanol (CH3OH), we can decide the validity of
our prediction (Problem 13.76) that the fuel value increases as carbon atoms are added to the alcohol alkyl
chain.
Analyze
o
Fuel values are computed from ΔH comb
. For methanol and ethanol the balanced combustion reactions are
2 CH3OH (l ) + 3 O2(g) → 2 CO2(g) + 4 H2O (l )
C2H5OH (l ) + 3 O2(g) → 2 CO2(g) + 3 H2O (l )
o
f
o
We need ΔH values from the textbook (Appendix 4, Table A4.3) to compute ΔH comb
.
Solve
For methanol
o
ΔH comb
= ⎡⎣( 2 mol CO 2 ) ( –393.5 kJ/mol) + ( 4 mol H 2 O) ( −285.8 kJ/mol)⎤⎦
– ⎡⎣( 2 mol CH3OH ) ( –238.7 kJ/mol) + ( 3 mol O2 ) ( 0.0 kJ/mol)⎤⎦
ΔH
o
comb
= –1452.8 kJ for 2 mol CH3 OH burned
1452.8 kJ
Fuel value =
= 22.67 kJ/g
2 mol × 32.04 g/mol
For ethanol
o
ΔH comb
= ⎡⎣( 2 mol CO 2 ) ( –393.5 kJ/mol) + ( 3 mol H 2 O ) ( −285.8 kJ/mol)⎤⎦
– ⎣⎡(1 mol C2 H 5 OH ) ( –277.7 kJ/mol) + ( 3 mol O2 ) ( 0.0 kJ/mol)⎦⎤
o
ΔH comb
= –1366.7 kJ for 1 mol CH3 OH burned
1366.7 kJ
Fuel value =
= 29.67 kJ/g
1 mol × 46.07 g/mol
Yes, the answer supports the prediction made in Problem 13.76 that fuel values of alcohols increase as the
number of C atoms increases.
Think about It
When comparing alcohols to alkanes, however, alcohols have lower fuel values.
13.87. Collect and Organize
By looking at the structures of carboxylic acids and aldehydes, we can explain why carboxylic acids are
generally more soluble in water.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 71
Analyze
Carboxylic acids have the general molecular structure
Aldehydes have the general molecular structure
Solve
Both carboxylic acids and aldehydes have polar functional groups. Carboxylic acids, however, are more
soluble in water because they form strong hydrogen bonds with water.
Think about It
Remember that hydrogen bonds between a species and water are stronger than dipole–dipole interactions
between a species and water.
13.89. Collect and Organize
Given the structures for butanal (an aldehye) and 2-butanone (a ketone) (Figure P13.89), we are asked
whether these compounds are structural isomers.
Analyze
Structural isomers have the same molecular formula but different connectivity of their atoms. The molecular
formula of butanal is C4H8O and so is the formula of 2-butanone, C4H8O.
Solve
Yes, these compounds are structural isomers.
Think about It
n-Butanol (C4H10O), however, is not a structural isomer of these compounds because it has a different
molecular formula.
13.91. Collect and Organize
We can compare the molecular formulas of aldehydes and ketones to determine whether we could distinguish
them by combustion analysis.
Analyze
The molecular formula for aldehydes is CnH2nO and for ketones it is also CnH2nO.
Solve
Because the empirical formulas for a ketone and for an aldehyde with the same number of carbon atoms are
the same, no, we cannot distinguish between these compounds by combustion analysis.
Think about It
We would be able, however, to distinguish between a ketone or aldehyde and an alcohol, which has the
molecular formula CnH2n+2O.
13.93. Collect and Organize
By assigning formal charges to the atoms in the two resonance structures for acetic acid (Figure P13.93), we
can determine which form contributes more to the bonding.
72 | Chapter 13
Analyze
The resonance form that has the lowest formal charges and/or the negative formal charges on the most
electronegative atoms (oxygen for acetic acid) contributes the most to the structure.
Solve
Structure a contributes more to the bonding in acetic acid because all of the formal charges are zero.
Think about It
Once deprotonated, acetic acid forms the acetate anion, which has two equivalent resonance forms:
13.95. Collect and Organize
By comparing the general structures of an amine and an amide, we can distinguish between the two functional
groups.
Analyze
An amine has the general structure
An amide has the general structure
O
R
Solve
An amide includes a carbonyl (C
C
NH2
O) as part of its functional group in addition to the –NH2 group.
Think about It
Amides are somewhat related to carboxylic acids in structure:
O
R
C
OH
13.97. Collect and Organize
Given the molecular formula of an aldehyde (C5H10O), we are to choose which of four structures (Figure
P13.97) are structural isomers of it.
Analyze
Structural isomers have the same molecular formula but different connectivity of their atoms.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 73
Solve
Structure a has a formula of C5H10O, so it is a structural isomer of the given aldehyde. Structure b has a
formula of C5H10O, so it is a structural isomer. Structure c has a formula of C4H8O, so it is not a structural
isomer of aldehyde. Structure d has a formula of C5H10O so it is also a structural isomer of the given aldehyde.
Think about It
The names of the three structural isomers shown are (a) 3-methylbutanal, (b) 2-methylbutanal, and (d)
pentanal.
13.99. Collect and Organize
From the structures shown in Figure P13.99, we are to determine which compound is a ketone.
Analyze
Ketones have the general structure
Solve
Compound b is a ketone.
Think about It
Structure a represents an aldehyde, structure c represents a carboxylic acid, and structure d represents an
alcohol.
13.101. Collect and Organize
After plotting the carbon–hydrogen ratio in aldehydes as a function of the number of carbon atoms and
comparing it to that of alkanes and alkenes, we can find the better correlation for aldehydes.
Analyze
Aldehydes have the general formula CnH2nO so the C:H ratio is always 0.5. Alkanes have the general
formula CnH2n+2 so the C:H ratio changes as the number of C atoms increases. Alkenes have the general
formula CnH2n so the C:H ratio is always 0.5.
Solve
The plot of C:H ratio versus number of C atoms for aldehydes correlates exactly to that of alkenes and
poorly to that of alkanes.
74 | Chapter 13
Think about It
Although this correlation exists, aldehydes are not structural isomers of alkenes. Aldehydes have an oxygen
in their structure; alkenes do not.
13.103. Collect and Organize
Esters are formed in condensation reactions between carboxylic acids and alcohols. For each of the ester
structures shown in Figure P13.103, we can write an equation to identify the carboxylic acid and alcohol
used to synthesize them.
Analyze
The general reaction for the formation of esters is
Solve
(a) Pineapples
(b) Bananas
(c) Apples
Think about It
These are all condensation reactions because they give a small molecule, H2O, as the other product.
13.105. Collect and Organize
After calculating the fuel values of formaldehyde and formic acid (Figure P13.105), we can determine which
has the higher fuel value.
Analyze
Generally the more oxygenated the fuel, the lower the fuel value. We predict that formaldehyde has the
o
higher fuel value since it has less oxygen in its structure. The fuel value is equal to – ΔH comb
divided by mass
of fuel. For these two fuels the combustion reactions are
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 75
CH2O(g) + O2(g) → CO2(g) + H2O (l )
CH2O2(g) + 12 O2(g) → CO2(g) + H2O (l )
Solve
For formaldehyde
o
ΔH comb
= ⎡⎣( –393.5) + ( –285.8)⎤⎦ − ⎡⎣( –108.6) + ( 0)⎤⎦
= –570.7 kJ/mol
570.7 kJ/mol
Fuel value =
= 19.00 kJ/g
30.03 g/mol
For formic acid
o
ΔH comb
= ⎡⎣( –393.5) + ( –285.8)⎤⎦ − ⎡⎣( –378.7 ) + ( 0)⎤⎦
= –300.6 kJ
300.6 kJ/mol
Fuel value =
= 6.531 kJ/g
46.03 g/mol
Formaldehyde has a significantly higher fuel value than formic acid.
Think about It
Our prediction, based on the level of oxygenation of the two fuels, was correct.
13.107. Collect and Organize
o
For two reactions involving methanogenic bacteria we are to calculate ΔH rxn
.
Analyze
We determine the enthalpy of each reaction using the given value of the enthalpy of formation of formic
acid and the values in Appendix 4.
o
ΔH rxn
= ∑ n ΔH f,o products – ∑ m ΔH f,o reactants
Solve
For reaction 1
o
ΔH rxn
= ⎡⎣(1 mol CH4 ) ( –74.8 kJ/mol) + (1 mol CO2 ) ( −393.5 kJ/mol)⎤⎦ – ⎡⎣(1 mol CH3COOH) ( – 485.8 kJ/mol)⎤⎦
o
For
reaction
ΔH rxn
= 17.52kJ
o
ΔH rxn
= ⎡⎣(1 mol CH 4 ) ( –74.8 kJ/mol) + ( 3 mol CO2 ) ( −393.5 kJ/mol) + ( 2 mol H 2 O) ( −285.8 kJ/mol)⎤⎦
– ⎡⎣( 4 mol HCOOH)( –378.7 kJ/mol)⎤⎦
ΔH
o
rxn
= –312.1 kJ
Think about It
The breakdown of acetic acid to methane and carbon dioxide is endothermic and therefore not favored by
enthalpy but the process that gives methanol, carbon dioxide, and water is exothermic and is favored by
enthalpy.
13.109. Collect and Organize
For each of the polymers shown in Figure P13.109, which were synthesized through the condensation
reaction of H2N(CH2)6NH2 with HO2C(CH2)nCO2H, we are asked to determine the number of carbon atoms
in the chain (n) of the dicarboxylic acids used.
76 | Chapter 13
Analyze
The number of carbon atoms in the chain equals the value of n, which is the number of – (CH2) – units in the
chain of the dicarboxylic acid.
Solve
(a) 4
(b) 6
(c) 8
Think about It
Because n is defined in the problem as the number of carbon atoms in the dicarboxylic acid formula
HO2C(CH2)nCO2H, we do not count the carboxylic acid carbon atoms.
13.111. Collect and Organize
Given the reaction between dimethyl terephthalate and 1,4-di(hydroxymethyl)cyclohexane to form Kodel
(Figure P13.111), we are asked to classify the reaction as either a condensation or addition reaction and to
compare the properties of Kodel to those of Dacron, which is prepared using ethylene glycol.
Analyze
(a) In addition polymerization reactions, the atoms are joined in the monomers to form the polymeric
backbone without loss of atoms. In condensation polymerization reactions, the two monomers react to form
the polymeric backbone while a small molecule like water is formed.
(b) The structure of ethylene glycol is
OH
OH
Solve
(a) Kodel is a condensation polymer. Methanol (CH3OH) is the by-product of the polymerization reaction.
(b) Kodel has in its backbone a carbon six-membered ring, not a straight chain as in Dacron. Therefore,
Kodel might be better able to accept organic dyes that are nonpolar.
Think about It
Kodel, being fairly polar because of the saturated six-membered carbon ring, is fairly resistant to water and
has been used to make clothing.
13.113. Collect and Organize
Using the definitions of enantiomer, achiral, and optically active we can determine whether all three terms
may be applied to a single compound.
Analyze
An enantiomer is a molecule that is a nonsuperimposable mirror image of another molecule and is optically
active. An achiral molecule is superimposable on its mirror image.
Solve
No, all three of these terms cannot describe a single compound. Whereas enantiomer and optically active
describe the same chiral molecule, achiral does not.
Think about It
Many biomolecules are chiral.
13.115. Collect and Organize
We are to determine whether a racemic mixture is a homogeneous or heterogeneous mixture.
Analyze
For a mixture to be heterogeneous we must be able to discern by eye (or with a microscope) the different
components of the mixture. A racemic mixture is a mixture of two enantiomers.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 77
Solve
A racemic mixture is mixed at the molecular level, so it is a homogeneous mixture.
Think about It
When successfully separated the components of a racemic mixture rotate plane-polarized light in opposite
directions.
13.117. Collect and Organize
By examining the structure of glycine, we can explain why it is achiral.
Analyze
The structure of glycine from Figure P13.117 is
A carbon atom is chiral when it is bonded to four different groups. Carbons that are sp2 or sp hybridized in a
structure cannot be chiral because they have only three and two groups bonded to them, respectively.
Solve
Glycine has no chiral carbon centers. Glycine’s carbons are either sp2 hybridized (on the carboxylic acid
group) or have two of the same atoms or groups (H atoms in this case) bonded to the sp3-hybridized carbon
atom, so glycine is not chiral.
Think about It
All of the other common 20 amino acids in Table 20.1 are chiral.
13.119. Collect and Organize
We are asked which type of orbital hybridization on a carbon atom can give rise to a chiral center.
Analyze
A carbon atom is chiral when it is bonded to four different groups.
Solve
A carbon is bonded to four groups, and thus can give rise to enantiomers, only when it has sp3 hybridization.
Think about It
The mirror images of all the other hybridizations of carbon are all superimposable.
13.121. Collect and Organize
Given some ordinary objects, we are to determine which are chiral.
Analyze
A chiral object is not superimposable on its mirror image.
Solve
A tennis racket (b) (if we ignore the winding of the tape on the handle) is superimposable on its mirror
image, but (a) a golf club, (c) a glove, and (d) a shoe are not and so they are chiral objects.
Think about It
You might also learn later that an object is not chiral if it contains a plane of symmetry or has an inversion
center.
78 | Chapter 13
Solve
A screwdriver (a), a light bulb (b) (if we ignore the screw base), and a baseball (even considering the seams)
are all superimposable on their mirror images. A key, however, considering its grooves, is chiral because it is
not superimposable on its mirror image.
Think about It
You might also learn later that an object is not chiral if it contains a plane of symmetry or has an inversion
center.
13.123. Collect and Organize
From the line drawings of three carboxylic acids (Figure P13.123), we are to determine which are chiral.
Analyze
A molecule is chiral if it has at least one carbon atom bonded to four different groups.
Solve
Only molecule a has a chiral carbon center and so is the only molecule shown that is chiral:
Think about It
Molecule a would be achiral if the – OH group were replaced by a – CH3 group.
13.125. Collect and Organize
In each structure in Figure P13.125, we are to circle the chiral centers.
Analyze
There is a chiral center wherever in the molecule a carbon is bonded to four different groups.
Solve
Think about It
Because the ring of carbon atoms in sodium cyclamate is symmetrical, the carbon to which the –NHSO3–
group is bound is not chiral.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 79
13.127. Collect and Organize
Optical isomers are nonsuperimposable mirror images that contain a chiral carbon center. We are asked to
identify the chiral center and to draw the mirror image of the compound shown in Figure P13.127.
Analyze
There is a chiral center wherever in the molecule a carbon is bonded to four different groups. To draw the
mirror image we imagine a mirror reflecting the atoms and bonds on the page as done in Figure 13.51 for
limonene.
Solve
Think about It
These two enantiomers rotate plane-polarized light in opposite directions.
13.129. Collect and Organize
For the highlighted N atoms in nicotine and Valium (Figure P13.129), we are to identify the associated
functional group as either an amine or an amide.
Analyze
An amine has the following possible structures:
R—NH2
Primary
An amide has these possible structures:
R2—NH
Secondary
R3N
Tertiary
Solve
(a) Nicotine’s highlighted N atom is a tertiary amine group.
(b) Valium’s highlighted N atom, as it has an adjacent C O double bond, is an amide group.
Think about It
Both nicotine and Valium also contain aromatic rings in their structures.
13.131. Collect and Organize
For the combustion reaction of methanol
CH3OH (l ) + 32 O2(g) → CO2(g) + 2 H2O (l )
we are to calculate how many grams of methanol would be needed to raise the temperature of 454 g of water,
from 20.0˚C to 50.0˚C. We are also to calculate the mass of CO2 produced in this reaction.
Analyze
If all of the heat from burning the methanol is used to heat the water, then
80 | Chapter 13
qwater = –qcomb
The heat needed to raise the temperature of the water is given by
qwater = mcs ∆T
where m is the mass of the water, cs is the specific heat capacity of water (4.184 J/g ⋅ ˚C), and ∆T is the
change in temperature of the water (30.0˚C). We can use ΔH fo values from Appendix 4 to calculate the
enthalpy of combustion of one mole of methanol. The molar amount of methanol needed to heat the water,
then, can be calculated through
qcomb
= mol CH3OH to heat the water
ΔH comb
Once we know the moles of methanol required for the reaction, we can calculate the mass of CH3OH needed
and the mass of CO2 produced using the balanced equation.
Solve
The heat generated by the combustion reaction is
qwater = –qcomb = 454 g × 4.184 J/g ˚C × 30.0 ˚C = 57,000 J or 57.0 kJ
o
ΔH rxn
= ⎣⎡(1 mol CO 2 ) ( –393.5 kJ/mol) + ( 2 mol H 2 O ) ( −285.8 kJ/mol)⎦⎤
– ⎡⎣(1 mol CH3 OH ) ( –238.7 kJ/mol) + ( 32 mol O2 ) ( 0.0 kJ/mol)⎤⎦
o
ΔH rxn
= –726.4 kJ for 1 mol methanol
Moles of methanol to heat the water
–57.0 kJ
= 7.85 × 10–2 mol methanol
–726.4 kJ/mol
Mass of methanol needed to heat the water
32.04 g
7.85 × 10–2 mol ×
= 2.52 g methanol
mol
Mass of CO2 produced
1 mol CO2
44.01 g
7.85 × 10–2 mol CH3OH ×
×
= 3.45 g CO2
1 mol CH3OH
mol
Think about It
This mass of CO2 would occupy 1.92 L at 25˚C and 1 atm pressure:
nRT 7.85 × 10–2 mol × 0.0821 L ⋅ atm/mol ⋅ K × 298 K
V=
=
= 1.92 L
P
1 atm
13.133. Collect and Organize
We are to use bomb calorimetry combustion reactions and the rise in temperature for two compounds to
determine which is butanol (C4H9OH) and which is diethyl ether (C2H5OC2H5), both of which have a
molecular formula of C4H10O.
Analyze
To calculate the energy released during the combustion of a compound we multiply the heat capacity of the
calorimeter by the change in the temperature for the combustion of the compound:
∆Hcomb = Ccalorimeter ∆T
This answer is in kilojoules per sample. To compare the two heats we need to convert to kilojoules per gram
through
ΔH comb
= kJ/g
mass of sample
To identify for which compounds the ∆Hcomb values match the experimental values, we can use enthalpy of
o
formation values (Appendix 3) to calculate ΔH rxn
for the combustion reactions:
C4H9OH (l ) + 6 O2(g) → 4 CO2(g) + 5 H2O (l )
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 81
C2H5OC2H5 (l ) + 6 O2(g) → 4 CO2(g) + 5 H2O (l )
Solve
From calorimetry experiments for compound A:
3.640 kJ
× 10.33o C = –37.60 kJ
o
C
–37.60 kJ
per gram =
= –38.20 kJ/g
0.9842 g
ΔH comb =
ΔH comb
Compound B:
3.640 kJ
× 11.03o C = – 40.15 kJ
o
C
– 40.15 kJ
per gram =
= –36.17 kJ/g
1.110 g
ΔH comb =
ΔH comb
For butanol the standard enthalpy of combustion is
o
ΔH rxn
= ⎣⎡( 4 mol CO 2 ) ( –393.5 kJ/mol ) + ( 5 mol H 2 O ) ( −285.8 kJ/mol)⎦⎤
– ⎡⎣(1 mol C 4 H 9 OH ) ( –327.3 kJ/mol ) + ( 6 mol O 2 ) ( 0.0 kJ/mol)⎤⎦
o
ΔH rxn
= –2676 kJ/mol of butanol
–2676 kJ/mol
= −36.10 kJ/g
74.12 g/mol
For diethyl ether the standard enthalpy of combustion is
o
ΔH rxn
= ⎡⎣( 4 mol CO 2 ) ( –393.5 kJ/mol ) + ( 5 mol H 2 O ) ( −285.8 kJ/mol)⎤⎦
o
In kJ/g, this is ΔH rxn
=
– ⎡⎣(1 mol C 2 H 5 OC 2 H 5 ) ( –279.6 kJ/mol ) + ( 6 mol O 2 ) ( 0.0 kJ/mol )⎤⎦
ΔH
o
rxn
= –2723 kJ/mol of diethyl ether
–2723 kJ/mol
= −36.74 kJ/g
74.12 g/mol
From the enthalpy of combustion reaction calculations, we see that ΔH rxn of diethyl ether is slightly more
exothermic than that of butanol. Matching this result with the calorimetry results, compound A is diethyl
ether (more exothermic) and compound B is butanol (less exothermic).
o
In kJ/g, this is ΔH rxn
=
Think about It
The calculated ΔH rxn values for butanol and diethyl ether are close to each other because these two
compounds are structural isomers.
13.135. Collect and Organize
Using the line formula for dodecenal (Figure P13.135) we are to count the carbons the molecule contains,
identify its functional groups, and determine what types of isomerism are possible.
Analyze
(a) In line formulas, the end of a line represents a – CH3 group. Carbon atoms are also present where two
lines meet.
(b) Functional groups are portions of the structure that impact distinct chemical and physical properties to the
compound.
(c) Isomers may be either structural or geometric.
Solve
(a) There are 12 carbon atoms in dodecenal.
(b) Dodecenal contains an alkene and an aldehyde functional group.
82 | Chapter 13
(c) Geometric isomers (cis and trans) are possible around the C C bond. Structural isomers are also
possible. For example, the chain may be branched, the double bond can be shifted, and the aldehyde
functional group may be changed to a ketone.
O
H
O
Think about It
From the examples above, can you draw more structural isomers of dodecenal?
13.137. Collect and Organize
By writing the chemical formulas of naphthalene and anthracene (Figure P13.137), we can determine
whether these compounds can be distinguished by combustion analysis.
Analyze
If the ratio of carbon to hydrogen atoms is different between naphthalene and anthracene, then the
compounds are distinguishable by combustion analysis.
Solve
Naphthalene’s chemical formula is C10H8 or CnHn–2. Anthracene’s chemical formula is C14H10 or CnHn – 4. The
C:H ratio is different; therefore, yes, these compounds are distinguishable by combustion analysis.
Think about It
The balanced equation for the combustion reactions also show that we can distinguish between these
compounds.
C10H8(s) + 12 O2(g) → 10 CO2(g) + 4 H2O(g)
C14H10(s) + 332 O2(g) → 14 CO2(g) + 5 H2O(g)
13.139. Collect and Organize
From the structure of nylon we can identify functional groups that make the polymer hydrophilic.
Analyze
Hydrophilic means “water loving.” The polar functional groups on nylon are hydrophilic. The structure of
nylon is shown in Figure 13.46.
Solve
The amide groups [– C(
O)NH–] in nylon are hydrophilic.
Think about It
The alkyl chain – (CH2) 6– on the polymer backbone of nylon is hydrophobic, “water avoiding.”
13.141. Collect and Organize
We are to draw all the possible trimers resulting from the condensation reaction of putrescine
[H2N(CH2)4NH2] with adipic acid and terephthalic acid (Figure P13.141). In addition we are asked what ratio
of monomers would be necessary to prepare a polymer with a 1:1 ratio of the two carboxylic acids in the
polymer chain.
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 83
Analyze
In the condensation reaction the difunctional amine can react with two carboxylic acid molecules to form a
trimer.
Solve
(a)
O
H
O
N
OH
HO
N
O
H
O
O
H
O
N
HO
N
O
H
OH
O
O
HO
H
O
N
N
O
H
OH
O
(b) 1 mol adipic acid : 1 mol terephthalic acid : 2 mol putrescine.
Think about It
The polymer from part b would have the backbone structure
13.143. Collect and Organize
Using the given structures of maleic anhydride and styrene (Figure P13.143) we are to draw two structural
repeating units of the polymer formed from these two monomers. By comparing the structure to that of
polystyrene we can predict how the two polymers’ properties might differ.
Analyze
Both polystyrene and the polymer formed from maleic anhydride and styrene are addition polymers. The
structure of polystyrene is
84 | Chapter 13
Solve
(a)
(b) The polymer made from maleic anhydride and styrene is expected to be more hydrophilic because of the
presence of the oxygen atoms and to be less rigid.
Think about It
This type of polymer, which is derived from two different monomers, is a copolymer.
13.145. Collect and Organize
We consider the synthesis and properties of silicones [R2SiO]n (Figure P13.145).
Analyze
(a) The two equations are (1) the reaction of R2SiCl2 with water to give a new monomer, whose formula is
R2SiCl(OH), and HCl and (2) the reaction of two R2SiCl(OH) molecules eliminating water to give
ClR2Si—O—SiR2Cl.
(b) Water-repellent polymers are nonpolar.
Solve
(a) R2SiCl2(aq) + H2O( l ) → R2SiCl(OH)(aq) + HCl(aq)
R2SiCl(OH)(aq) + (HO)SiClR2(aq) → R2ClSi— O —SiClR2(aq) + H2O (l )
(b) Silicones are water repellent because the side chains (R groups) are nonpolar.
Think about It
Silicones have many applications including cookware (because they are non-stick and heat resistant), contact
lenses (because of their oxygen permeability), and lubricants (because of their clean application).
13.147. Collect and Organize
We are asked to explain why the enthalpy of combustion using bond energies for n-butane and
2-methylpropane, both with molecular formulas of C4H10, are the same, but when the enthalpy of combustion
for these compounds is determined experimentally their values are different.
Analyze
Experimentally determined heats of combustion are measured using bomb calorimetry, which accurately
measures the heat released by the compound when burned in oxygen. The heat of combustion calculated
Organic Chemistry: Fuels, Pharmaceuticals, and Materials | 85
using bond energies uses average bond energies (Table 8.2). The average bond energies are derived from
many examples of that type of bond.
Solve
The heat of combustion determined using experimental means is different from that calculated from average
bond energies because the bond energy of a particular bond depends on the structure of the rest of the
molecule.
Think about It
When using average bond energies to estimate the enthalpy of combustion for n-butane and
2-methylpropane, the calculated values will be equal because in each 10 C—H bonds and 4 C — C bonds are
broken in the combustion reaction to give 8 C O bonds (in CO2) and 10 H — O bonds (in H2O).
as hydrocarbons are non-renewable energy sources.
CHAPTER 14 | Thermodynamics: Spontaneous Processes, Entropy, and
Free Energy
14.1. Collect and Organize
As we add air to a balloon we are to consider whether the balloon and its contents increase or decrease in
entropy.
Analyze
Entropy is a measure of the distribution of energy at a specific temperature. It is related to the number of
microstates:
S = k lnW
where S is entropy, k is the Boltzmann constant, and W is the number of microstates. A microstate is a unique
distribution of particles among energy levels. As the number of particles in a system changes, the number of
microstates changes.
Solve
After adding more air to the balloon, there are more microstates available in the balloon because there are more
particles in the balloon. Therefore, the entropy of the balloon and its contents increases.
Think about It
We can think of entropy as the “disorder of a system.” When more particles are introduced into a system, there
is more disorder due to those particles’ motions.
14.3. Collect and Organize
For the system in Figure P14.3 showing a random distribution of two gases, we are to assess the probability
that one bulb will collect gas A and the other gas B.
Analyze
The system is already at high entropy because both gases can randomly move throughout the entire volume.
Solve
The probability that gases A and B will separate in the apparatus so as to occupy separate bulbs is very low.
Each gas would then be confined to a smaller volume. This change would involve a decrease in entropy.
Think about It
Systems tend to move spontaneously to maximum randomness. In that way we can think of spontaneous
changes as tending toward increasing entropy.
14.5. Collect and Organize
From the plot of ΔH and TΔS versus temperature in Figure P14.5, we are to consider the significance of the
point where the two lines meet and determine above what temperature the reaction is nonspontaneous.
Analyze
Entropy and enthalpy are related by the Gibbs free energy equation:
ΔG = ΔH – TΔS
On the plot ΔH does not change with temperature but the quantity TΔS decreases sharply with an increase in
temperature.
Solve
At the point of intersection, ΔH = TΔS, and it is here that ΔG = 0 and the reaction is at equilibrium. Because
TΔS decreases with increasing temperature on the graph, ΔS must be negative. For ΔG to be positive
(nonspontaneous) when TΔS is negative, ΔH would have to be more positive (less negative) than TΔS. This
occurs where the ΔH line is above the TΔS line, so this reaction is nonspontaneous above 60˚C.
116
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 117
Think about It
If ΔH is positive and TΔS is negative (ΔS is negative) for a reaction, according to the Gibbs free energy
equation the reaction will never be spontaneous. A plot describing that reaction would have ΔH and TΔS lines
that never cross.
14.7. Collect and Organize
We consider what happens to the sign of ΔS when we reverse a process.
Analyze
When the sign of ΔS is negative the process is not favored by entropy, and entropy is decreasing. When the
sign of ΔS is positive, the process is favored by entropy, and entropy is increasing.
Solve
If a process that is favored by entropy (+ΔS) is reversed, the reverse process will not be favored (–ΔS). Therefore, the
ΔS when a process is reversed has its sign reversed.
Think about It
In terms of order and disorder in a chemical or physical process, a process that has increasing disorder (∆S
positive) has increasing order (ΔS negative ) when the process is reversed.
14.9. Collect and Organize
By analyzing the possible outcomes in flipping three coins, we can determine the total number of possible
microstates and which of those microstates is most likely.
Analyze
Flipping a coin has two possibilities: heads (H) or tails (T). Because there are two possible outcomes for each
coin, for n coins there will be 2n different possible outcomes. For three coins there are 23 = 8 possibilities.
Solve
The eight possible microstates and their values where H = +1 and T = –1 are as follows:
HHH
HTH
HHT
THH
(+1 +1 +1) = +3
(+1 –1 +1) = +1
(+1 +1 –1) = +1
(–1 +1 +1) = +1
TTT
THT
TTH
HTT
(–1 –1 –1) = –3
(–1 +1 –1) = –1
(–1 –1 +1) = –1
(+1 –1 –1) = –1
The most likely microstates have sums of +1 and –1.
Think about It
As the number of coins increases, the number of possible microstates increases significantly. For
n = 4, there are 16 possibilities, for n = 5 there are 32, and for n = 6 there are 64.
14.11. Collect and Organize
Given ice cubes (the system) in a glass of lemonade (the surroundings), we are to determine the signs of ∆S
for the system and for the surroundings as the ice cools the lemonade from 10.0˚C to 0.0˚C.
Analyze
Cooling is the result of decreased molecular motion, which is correlated to a decrease in entropy.
Solve
The sign of ΔSsurr is negative because the lemonade is cooling, which decreases the motion of the lemonade
molecules. The sign of ΔSsys is positive because the ice is melting. When a phase changes from solid to liquid
occurs in the lemonade, disorder, and therefore entropy, of the substance increases.
118 | Chapter 14
Think about It
In this case, as long as ΔSsys > –ΔSsurr, the process is spontaneous.
14.13. Collect and Organize
Given three combinations of signs for changes in entropy for the system, the surroundings, and the universe,
we are to determine which combinations are possible.
Analyze
In order for a process to occur, the second law of thermodynamics must be obeyed. This law states that the
combination of the change in entropy for a system and the change in entropy for the surroundings must be
greater than zero. Therefore, ΔSuniv, which is equal to ΔSsys + ΔSsurr, must be greater than zero for a process to
occur and
ΔSsys + ΔSsurr > 0
Solve
(a) If ΔSsys < 0 and ΔSsurr > 0, then ΔSuniv > 0 when ΔSsurr > –ΔSsys. This combination of entropy changes is
possible.
(b) If ΔSsys > 0 and ΔSsurr < 0, then ΔSuniv > 0 when ΔSsys > –ΔSsurr. This combination of entropy changes is
possible.
(c) If ΔSsys > 0 and ΔSsurr > 0, then ΔSuniv must be > 0. ∆Suniv cannot be < 0 for these changes in entropy for the
system and the surroundings, so this combination of entropy changes is not possible.
Think about It
In part c, ΔSuniv would be calculated as < 0 if ΔSsys and ΔSsurr were both < 0 or if ΔSsys + ΔSsurr < 0. This
process, however, would not be spontaneous.
14.15. Collect and Organize
Given that a chemical reaction has ΔSsys = – 48.0 J/K, we are to determine the maximum value of ΔSsurr needed
in order for the reaction to be nonspontaneous.
Analyze
For a spontaneous process,
ΔSuniv = ΔSsys + ΔSsurr > 0
Solve
In order for the reaction to be nonspontaneous, ΔSsurr must be less (more negative) than +48.0 J/K.
Think about It
If ΔSsurr is more positive than 48.0 J/K, then ΔSuniv > 0 and the reaction would be spontaneous.
14.17. Collect and Organize
Given pairs of systems we are asked to determine which of the pair has the greater entropy.
Analyze
The system with the most randomness in its molecular motion is the system with the most entropy.
Solve
(a) Propene has more entropy than cyclopropane because it has more degrees of freedom as its atoms are not
tied together in a ring structure.
(b) Wet paint has more entropy than dry paint.
(c) One mol of SO3 gas has more entropy than 1 mol of SO2 gas.
(d) An aquarium with fish has more entropy than an aquarium without fish.
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 119
Think about It
In part c, because SO3 has more atoms in its structure than SO2, it has more bond vibrations possible, thus
leading to higher entropy.
14.19. Collect and Organize
For a particular substance we are asked which state (solid, liquid, or gas) has the most positive standard molar
entropy.
Analyze
Entropy is the measure of the disorder, or randomness, of a system. The more disorder, the more entropy.
Solve
The disorder of molecules between the three physical states decreases in the order gas > liquid > solid because
as the gas condenses and the liquid solidifies the motions of the molecules become more fixed. Therefore, a
substance in the gaseous state has the highest standard molar entropy.
Think about It
Heating up a substance, even without a phase change occurring, also increases the entropy of the substance
because of increased molecular motion.
14.21. Collect and Organize
We are to predict which has the higher standard molar entropy, diamond or the fullerenes.
Analyze
Diamond is a three-dimensional, highly ordered network of covalently bonded carbon atoms (see the section,
Carbon: Diamonds, Graphite, and the Molecules of Life, Chapter 5 in the textbook). Fullerenes are also made
up of covalently bonded carbon atoms, but they form discrete structures instead of an extended network.
Solve
Fullerenes with less extensive bonding have a higher standard molar entropy than diamond.
Think about It
We would expect the standard molar entropy of graphite to be in between that of diamond and the fullerenes
because of its intermediate structure.
14.23. Collect and Organize
By considering the phase and size of a compound, we can rank the compounds in each series in order of
increasing standard molar entropy.
Analyze
Generally speaking, compounds that are larger in size have greater S˚.
Solve
(a) CH4(g) < CF4(g) < CCl4(g)
(b) CH3OH (l ) < CH3CH2OH (l ) < CH3CH2CH2OH (l )
(c) HF(g) < H2O(g) < NH3(g)
Think about It
Larger molecules have greater S˚ because they have more opportunities for internal motion.
14.25. Collect and Organize
For each process described, we are to predict the sign of the change in entropy for the system.
120 | Chapter 14
Analyze
ΔSsys is positive when a process results in less order, an increase in temperature, or an increase in volume, or if
the number of moving molecules increases.
Solve
(a) The sign of ΔSsys is negative for a bricklayer building a wall out of a pile of bricks (the system).
(b) ΔSsys is negative for the leaves (the system) being raked into a pile.
(c) ΔSsys is negative for the formation of AgCl(s) from Ag+(aq) and Cl–(aq) (the system).
(d) ΔSsys is positive for the oxidation of Zn(s) by HC1(aq) to form H2(g) and ZnC12(aq).
Think about It
Notice that not all processes have a positive ΔSsys. It is the case, however, that these processes have a positive
ΔSuniv through the equation
ΔSsys + ΔSsurr = ΔSuniv > 0
14.27. Collect and Organize
When the products of a process have a lower entropy than the reactants, we are to determine the sign of the
entropy change for the overall process.
Analyze
The entropy change for a process is calculated by
ΔSrxn = ∑ nSproducts – ∑ mSreactants
Solve
If nSproducts < mSreactants, then ∆Srxn for the process is negative.
Think about It
A negative ΔSrxn means that the process resulted in more order in going to products.
14.29. Collect and Organize
For the reaction
A(s) → B (l )
we are asked to compare the entropy of this reaction to that where the temperature is raised above the melting
point of A.
Analyze
The new reaction would be
A (l ) → B (l )
The entropy of A (l ) is substantially greater than that of A(s) because the molecular motion in a liquid is
greater than in a solid.
Solve
Because the entropy for a reaction is
ΔSrxn = ∑ nSproducts – ∑ mSreactants
the use of A (l ) in the reaction, which has a greater entropy than A(s), means that
therefore, yes, the ΔSrxn changes.
∑ mS
reactants
is larger and
Think about It
Without knowing the relative magnitudes of S˚ for A (l ) , A(s), and B (l ) , however, we cannot definitively
o
determine if ΔSrxn
is positive or negative.
14.31. Collect and Organize
o
Using values of S˚ for the reactants and products for four atmospheric reactions, we are to calculate ΔS rxn
.
Standard molar entropies are in Appendix 4.
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 121
Analyze
A change in entropy for a reaction is
o
o
o
ΔSrxn
= ∑ nSproducts
– ∑ mSreactants
Solve
o
(a) ΔSrxn
= ( 2 mol NO × 210.7 J/mol ⋅ K )
– ⎡⎣(1 mol N 2 × 191.5 J/mol ⋅ K ) + (1 mol O2 × 205.0 J/mol ⋅ K )⎤⎦
= 24.9 J/K
o
(b) ΔSrxn
= ( 2 mol NO2 × 240.0 J/mol ⋅ K )
– ⎡⎣( 2 mol NO × 210.7 J/mol ⋅ K ) + (1 mol O2 × 205.0 J/mol ⋅ K )⎤⎦
= −146.4 J/K
o
(c) ΔSrxn
= (1 mol NO2 × 240.0 J/mol ⋅ K )
– ⎡⎣(1 mol NO × 210.7 J/mol ⋅ K ) + ( 12 mol O 2 × 205.0 J/mol ⋅ K )⎤⎦
(d) ΔS
o
rxn
= −73.2 J/K
= (1 mol N2 O4 × 304.2 J/mol ⋅ K ) − ( 2 mol NO2 × 240.0 J/mol ⋅ K )
= −175.8 J/K
Think about It
Notice that because the balanced equation in part b is twice that in part c the value of the entropy change for
the reaction is doubled.
14.33. Collect and Organize
o
Given the ΔSrxn
for the conversion of Cl and O3 into ClO and O2 and the S˚ for Cl, O3, and O2 from Appendix
4, we are to calculate S˚ for ClO.
Analyze
o
To calculate ΔS rxn
, we need SOo3 = 238.8 J/mol ⋅ K , SClo = 165.2 J/mol ⋅ K, and SOo 2 = 205.0 J/mol ⋅ K along
with the balanced equation:
Cl(g) + O3(g) → ClO(g) + O2(g)
The change in entropy for a reaction is
o
o
o
ΔSrxn
= ∑ nSproducts
– ∑ mSreactants
Solve
(
)
o
o
ΔSrxn
= 19.9 J/K = ⎡⎣ 1 mol ClO × SClO
+ (1 mol O2 × 205.0 J/mol ⋅ K )⎤⎦
– ⎡⎣(1 mol Cl × 165.2 J/mol ⋅ K ) + (1 mol O3 × 238.8 J/mol ⋅ K )⎤⎦
o
19.9J/K = –199.0 J/K + SClO
o
SClO
= 218.9 J/mol ⋅ K
Think about It
o
Because ΔSrxn
is positive, the entropy for the system (the reaction) increases when Cl and O3 react to form
ClO and O2.
14.35. Collect and Organize
We are to assess whether it is correct to say that all exothermic reactions are spontaneous.
Analyze
A reaction is spontaneous when its free-energy change is negative according to the equation
ΔG = ΔH – S
When H is negative, the reaction is exothermic.
122 | Chapter 14
Solve
No. The assumption that all exothermic reactions are spontaneous is incorrect. If the value of T∆S is more
negative than the value of H, then ΔG would be positive and the reaction would be nonspontaneous.
Think about It
A reaction that is exothermic is nonspontaneous when its entropy change is negative and the reaction is run at
“high temperature.”
14.37. Collect and Organize
We assess whether a reaction that has a negative ΔG is spontaneous.
Analyze
By definition a reaction is spontaneous when its value of free-energy change is negative.
Solve
Yes, a reaction with negative ∆G is spontaneous.
Think about It
A spontaneous reaction may be either exothermic or endothermic according to the equation
ΔG = ΔH – T ΔS
14.39. Collect and Organize
For a reaction that has a positive ΔG we are to state what we can say about the reaction.
Analyze
The Gibbs free-energy, ΔG, which can be negative, positive, or zero, relates the free energies of the products
and the reactants:
o
ΔGrxn
= ∑ n ΔGf,o products – ∑ m ΔGf,o reactants
Solve
In a reaction in which ΔG is positive, the products have a higher free energy than the reactants. The
reaction is nonspontaneous and proceeds spontaneously in the reverse direction (product to reactants).
Think about It
Spontaneous reactions have ∆G values that are negative, in which case the free energy of the products is lower
than the free energy of the reactants.
14.41. Collect and Organize
We consider whether exothermic reactions are spontaneous only at low temperature.
Analyze
Spontaneity is shown by a negative free-energy value (ΔG) in the equation
ΔG = ΔH – TΔS
For exothermic reactions, we also know that ΔH is negative.
Solve
No. As we can see from the equation, if ΔS is positive then the value of T∆S would be positive. In that case
ΔG would always be negative, regardless of the temperature, and the exothermic reaction would be
spontaneous at all temperatures.
Think about It
If ∆S is negative for an exothermic reaction, then it is true that the reaction would be spontaneous only at low
temperatures.
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 123
14.43. Collect and Organize
For the sublimation of dry ice at room temperature, we are to determine the signs of ΔS, ΔH, and ΔG.
Analyze
The reaction describing the sublimation is
CO2(s) → CO2(g)
Solve
ΔS is positive because the solid is subliming to a gas, which has much greater entropy.
ΔH is positive because heat is required to effect the phase change.
ΔG is negative because at room temperature (25˚C) this process spontaneously occurs.
Think about It
The spontaneity of this process is temperature dependent. At lower temperatures the value of T∆S in the
equation
ΔG = ΔH – T∆S
would not be great enough to give a negative value for ΔG.
14.45. Collect and Organize
For the processes described, we are to determine whether they are spontaneous.
Analyze
Spontaneous processes occur without outside intervention once they are started.
Solve
(a) Fragrance spontaneously spreads throughout a room.
(b) A broken clock does not spontaneously mend itself.
(c) An iron fence spontaneously rusts.
(d) An ice cube spontaneously melts in water.
Think about It
Spontaneous processes are favored either by enthalpy, or entropy, or both according to the equation
ΔG = ΔH – T∆S
14.47. Collect and Organize
o
For the dissolution of NaBr and NaI in water, we are to calculate the value of ∆G˚ knowing the ΔH sol
and
o
ΔS sol for each of these soluble salts.
Analyze
We can calculate the value of ΔG o from the standard state enthalpy and entropy of the reaction using
o
o
o
ΔGrxn
= ΔH rxn
− T ΔSrxn
where T = 298 K.
Solve
For NaBr
0.057 kJ ⎞
⎛
o
ΔGrxn
= –1 kJ/mol – ⎜ 298 K ×
⎟ = –18 kJ/mol
⎝
mol ⋅ K ⎠
For NaI
0.074 kJ ⎞
⎛
o
ΔGrxn
= –7 kJ/mol – ⎜ 298 K ×
⎟ = –29 kJ/mol
⎝
mol ⋅ K ⎠
124 | Chapter 14
Think about It
Be sure to use consistent units for ΔH˚ and ΔS˚ in calculations. In this problem, we have to change the ΔS˚
values given in J/mol ⋅ K to kJ/mol ⋅ K.
14.49. Collect and Organize
o
For the reaction of C(s) with H2O(g) to produce H2(g) and CO(g), we are to calculate the value of ΔGrxn
from
the data listed in Appendix 4:
o
ΔGrxn
= ∑ n ΔGf,o products – ∑ m ΔGf,o reactants
Using
o
o
o
ΔGrxn
= ΔH rxn
− T ΔSrxn
we can predict the lowest temperature at which the reaction is spontaneous.
Analyze
To calculate the lowest temperature at which CO and H2 form from C and steam, we must first calculate
o
o
and ΔSrxn
using Appendix 4. For a spontaneous reaction ∆G is negative. Therefore, if we set ΔG = 0
ΔH rxn
and solve for T, that would give the temperature at which the reaction changes spontaneity.
ΔG = 0 = ΔH – T∆S
ΔH
T=
ΔS
Be sure to have consistent units for ΔH and ΔS for this calculation.
Solve
o
ΔGrxn
= ⎣⎡(1 mol H 2 × 0.0 kJ/mol ) + (1 mol CO × –137.2 kJ/mol )⎦⎤
– ⎡⎣(1 mol H 2 O × –228.6 kJ/mol ) + (1 mol C × 0.0 kJ/mol)⎤⎦
= 91.4 kJ
To determine the lowest temperature at which this reaction is spontaneous we must first calculate ∆H˚ and ∆S˚
o
ΔH rxn
= ⎡⎣(1 mol H 2 × 0.0 kJ/mol) + (1 mol CO × –110.5 kJ/mol)⎤⎦
– ⎣⎡(1 mol H 2 O × –241.8 kJ/mol) + (1 mol C × 0.0 kJ/mol)⎦⎤
= 131.3 kJ
ΔS
o
rxn
= ⎡⎣(1 mol H 2 × 130.6 J/mol ⋅ K ) + (1 mol CO × 197.7 J/mol ⋅ K )⎤⎦
– ⎡⎣(1 mol H 2 O × 188.8 J/mol ⋅ K ) + (1 mol C × 5.7 J/mol ⋅ K )⎤⎦
= 133.8 J/K
o
Solving for T when ΔGrxn
= 0 gives
T=
131.3 kJ
= 981.3 K or 708.2oC
0.1338 kJ/K
The lowest temperature at which this reaction is spontaneous is just above 981.3 K or 708.2˚C.
Think about It
For this calculation we used carbon in the form of graphite. We would not want to use diamond as a reactant!
14.51. Collect and Organize
For the vaporization of water at 1.00 atm we are to calculate ΔH˚ and ΔS˚. Assuming that calculated quantities
(using data listed in Appendix 4 at 298 K) are not temperature dependent, we are to calculate the boiling point
of water.
Analyze
To calculate both ΔH˚ and ΔS˚ for the phase change
H2O (l ) → H2O(g)
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 125
we need the following (from Appendix 4): ΔH fo of H2O (l ) = –285.8 kJ/mol, ΔH fo of H2O(g) = –241.8 kJ/mol,
S˚ of H2O (l ) = 69.9 J/mol ⋅ K, and S˚ of H2O(g) = 188.8 J/mol ⋅ K. To calculate the boiling point we can
rearrange the Gibbs free-energy equation to solve for the temperature where ΔG = 0 because at the boiling
point H2O (l ) and H2O(g) are in equilibrium
o
o
ΔGrxn = 0 = ΔH rxn
− T ΔS rxn
Tb =
o
ΔH rxn
o
ΔSrxn
Solve
o
ΔH rxn
= (1 mol H 2 O(g ) × –241.8 kJ/mol) – [1 mol H 2 O(l ) × –285.8 kJ/mol]
= 44.0 kJ
ΔS
o
rxn
= [1 mol H 2 O(g ) × –188.8 J/mol ⋅ K ] – [1 mol H 2 O(l ) × 69.9 kJ/mol ⋅ K ]
= 118.9 J/K or 0.1189 kJ/K
Tb =
44.0 kJ
= 370.1 K or 96.9o C
0.1189 kJ/K
Think about It
Since the calculated value of the boiling point temperature is a little different from the actual
Tb (100.0˚C), we can see that there is some temperature dependence of ΔH and ΔS.
14.53. Collect and Organize
o
o
For the reaction of SO2(g) with H2S(g) to form S(s) and H2O(g), we are to calculate ΔH rxn
and ΔSrxn
. Using
the Gibbs free-energy equation we can predict the temperatures over which the reaction is spontaneous.
Analyze
For a reaction to be spontaneous, ΔG must be negative. The temperature at which a reaction changes
spontaneity is where ΔGrxn = 0. Rearranging the free-energy equation gives
o
o
o
ΔGrxn
= ΔH rxn
− T ΔS rxn
=0
T=
o
ΔH rxn
o
ΔS rxn
Solve
o
ΔH rxn
= ⎡⎣( 3 mol S × 0.0 kJ/mol) + ( 2 mol H 2 O × –241.8 kJ/mol)⎤⎦
– ⎡⎣(1 mol SO2 × –296.8 kJ/mol) + ( 2 mol H 2S × –20.17 kJ/mol)⎤⎦
= –146.5 kJ
ΔS
o
rxn
= ⎡⎣( 3 mol S × 32.1 J/mol ⋅ K ) + ( 2 mol H 2 O × 188.8 J/mol ⋅ K )⎤⎦
– ⎡⎣(1 mol SO2 × 248.2 J/mol ⋅ K ) + ( 2 mol H 2S × 205.6 J/mol ⋅ K )⎤⎦
= –185.5 J/K
–146.5 kJ
= 789.8 K or 516.6oC
–0.1855 kJ/K
Because this reaction is favored by enthalpy (–ΔH) but not by entropy (–ΔS), it is spontaneous at low
temperatures. Therefore, the reaction is spontaneous for temperatures below 789.8 K.
Tb =
Think about It
The equation specifies that H2O(g) is produced in the reaction so the useful range for the reaction is
373–789.8 K. Below 373 K, the water is in liquid form.
14.55. Collect and Organize
126 | Chapter 14
o
For each reaction we are to calculate ΔGrxn
and determine whether the reactions are spontaneous at 355 K.
Analyze
The standard free energy of the reaction can be calculated using Appendix 4 values for ΔGfo for the reactants
and products:
o
ΔGrxn
= ∑ n ΔGf,o products – ∑ m ΔGf,o reactants
o
o
To determine the spontaneity of the reactions at 355 K we first must calculate ΔH rxn
and ΔSrxn
and then
o
calculate ΔGrxn from
o
o
o
ΔGrxn
= ΔH rxn
− T ΔSrxn
o
where T = 355 K. If ΔGrxn
is negative, the reaction is spontaneous.
Solve
(a)
o
ΔGrxn
= ( 2 mol NO × 86.6 kJ/mol)
– ⎡⎣(1 mol N 2 × 0.0 kJ/mol) + (1 mol O 2 × 0.0 kJ/mol)⎤⎦
= 173.2 kJ
ΔH
o
rxn
= ( 2 mol NO × 90.3 kJ/mol)
– ⎡⎣(1 mol N 2 × 0.0 kJ/mol) + (1 mol O 2 × 0.0 kJ/mol)⎤⎦
= 180.6 kJ
ΔS
o
rxn
= ( 2 mol NO × 210.7 J/mol ⋅ K )
– ⎡⎣(1 mol N 2 × 191.5 J/mol ⋅ K ) + (1 mol O 2 × 205.0 J/mol ⋅ K )⎤⎦
= 24.9 J/K
ΔG355K = 180.6 kJ – 355 K × 0.0249 kJ/K = 171.8 kJ
This reaction is nonspontaneous at 355 K.
o
= ( 2 mol NO 2 × 51.3 kJ/mol)
(b) ΔGrxn
– ⎡⎣( 2 mol NO × 86.6 kJ/mol) + (1 mol O 2 × 0.0 kJ/mol)⎤⎦
= –70.6 kJ
ΔH
o
rxn
= ( 2 mol NO 2 × 33.2 kJ/mol)
– ⎡⎣( 2 mol NO × 90.3 kJ/mol) + (1 mol O 2 × 0.0 kJ/mol)⎤⎦
= –114.2 kJ
ΔS
o
rxn
= ( 2 mol NO 2 × 240.0 J/mol ⋅ K )
– ⎡⎣( 2 mol NO × 210.7 J/mol ⋅ K ) + (1 mol O 2 × 205.0 J/mol ⋅ K )⎤⎦
= –146.4 J/K
ΔG355K = –114.2 kJ – 355 K × – 0.1464 kJ/K = – 62.2 kJ
This reaction is spontaneous at 355 K.
o
(c) ΔGrxn
= (1 mol NO 2 × 51.3 kJ/mol)
– ⎡⎣(1 mol NO × 86.6 kJ/mol) + ( 12 mol O 2 × 0.0 kJ/mol )⎤⎦
= –35.3 kJ
ΔH
o
rxn
= (1 mol NO 2 × 33.2 kJ/mol)
– ⎡⎣(1 mol NO × 90.3 kJ/mol) + ( 12 mol O 2 × 0.0 kJ/mol )⎤⎦
= –57.1 kJ
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 127
o
ΔSrxn
= (1 mol NO2 × 240.0 J/mol ⋅ K )
– ⎡⎣(1 mol NO × 210.7 J/mol ⋅ K ) + ( 12 mol O 2 × 205.0 J/mol ⋅ K )⎤⎦
= –73.2 J/K
ΔG355 K = –57.1 kJ – 355 K × – 0.0732 kJ/K = –31.1 kJ
This reaction is spontaneous at 355 K.
o
(d) ΔGrxn
= (1 mol N 2 O4 × 97.8 kJ/mol) – ( 2 mol NO2 × 51.3 kJ/mol)
= – 4.8 kJ
ΔH
o
rxn
ΔS
o
rxn
= (1 mol N 2 O4 × 9.2 kJ/mol) – ( 2 mol NO2 × 33.2 kJ/mol)
= –57.2 kJ
= (1 mol N 2 O4 × 304.2 J/mol ⋅ K ) – ( 2 mol NO2 × 240.0 J/mol ⋅ K )
= –175.8 J/K
ΔG355 K = –57.2 kJ – 355 K × – 0.1758 kJ/K = 5.2 kJ
This reaction is not spontaneous at 355 K.
Think about It
When calculating ΔG through ΔH and ΔS, always be sure to use consistent units for enthalpy and entropy.
14.57. Collect and Organize
For the reactions in Problem 14.55 we are to determine which are spontaneous at high temperatures, low
temperatures, or all temperatures.
Analyze
Spontaneity is indicated by a negative free-energy value through the equation
ΔG = ΔH – T∆S
We can see that an exothermic reaction will be spontaneous at all temperatures if entropy is positive and at
low temperatures if entropy is negative. An endothermic reaction will never be spontaneous at any
temperature if the entropy is negative but will be spontaneous at high temperature if entropy is positive.
Solve
(a) Because Reaction (a) in Problem 14.55 is an endothermic reaction with positive entropy, this reaction
will be spontaneous at high temperature.
(b) Because Reaction (b) in Problem 14.55 is an exothermic reaction with negative entropy, this reaction will
be spontaneous at low temperature.
(c) Because Reaction (c) in Problem 14.55 is an exothermic reaction with negative entropy, this reaction
will be spontaneous at low temperature.
(d) Because Reaction (d) in Problem 14.55 is an exothermic reaction with negative entropy, this reaction will
be spontaneous at low temperature.
Think about It
A reaction must be favored either by enthalpy or entropy in order to be spontaneous at any temperature.
14.59. Collect and Organize
For the decomposition of gaseous ammonia into nitrogen and hydrogen, we are asked if the reaction is
spontaneous at standard conditions.
Analyze
o
We can use the values in Appendix 4 for ΔGfo of the products and reactants to calculate ΔGrxn
:
o
ΔGrxn
= ∑ n ΔGf,o products – ∑ m ΔGf,o reactants
o
If ΔGrxn
is negative, then the reaction is spontaneous.
128 | Chapter 14
Solve
o
ΔGrxn
= ⎡⎣(1 mol N 2 × 0.0 kJ/mol) + (1 mol H 2 × 0.0 kJ/mol)⎤⎦
− ( 2 mol NH 3 × –16.5 kJ/mol)
= 33.0 kJ
No, this reaction is not spontaneous at standard conditions.
Think about It
The reverse reaction to form NH3 from N2 and H2 is spontaneous.
14.61. Collect and Organize
o
o
We can use the given thermodynamic quantities of entropy ( ΔS rxn
= 80.0 J/mol ⋅ K) and enthalpy ( ΔH rxn
=
40.0 kJ/mol) to assess the heat of reaction, change in entropy, and spontaneity of the reaction
A(g) → B(g) + C(g)
We are also to calculate the temperature at which the reaction becomes spontaneous.
Analyze
To determine whether the reaction is spontaneous, we can use the expression for the Gibbs free energy:
o
o
o
ΔGrxn
= ΔH rxn
− T ΔSrxn
To calculate the temperature at which the reaction becomes nonspontaneous, we can solve the equation for T
when ΔG = 0.
o
o
o
ΔGrxn
= ΔH rxn
− T ΔS rxn
=0
T=
o
ΔH rxn
o
ΔS rxn
Solve
(a) The reaction is endothermic because ΔH is positive.
(b) Yes. The positive entropy value makes sense because the reaction is producing two moles of gaseous
products from one mole of gaseous reactants.
(c) No. Because the reaction is not favored by enthalpy but is favored by entropy, the ΔGrxn is negative (and
the reaction is spontaneous) only at high temperatures, not at all temperatures.
(d)
T=
40.0 kJ/mol
= 500 K
0.0800 kJ/mol ⋅ K
At temperatures higher than 500 K, this reaction is spontaneous.
Think about It
Only a reaction that has –ΔH and +ΔS is spontaneous at all temperatures.
14.63. Collect and Organize
By examining the structures of glucose 6-phosphate and fructose 6-phosphate in Figure P14.63 we can explain
why the conversion between these two sugars has a ΔG o close to zero.
Analyze
Changes in free energy result from changes in enthalpy (related to the number and types of chemical bonds)
and changes in entropy (disorder in the system).
Solve
In the conversion of glucose 6-phosphate to fructose 6-phosphate the six-membered ring becomes a fivemembered ring. The bond arrangements are only slightly different between the two structures and so the
enthalpies and entropies of the product and reactant are very close in value.
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 129
Think about It
Glucose 6-phosphate and fructose 6-phosphate are structural isomers of each other.
14.65. Collect and Organize
For a process that has two steps we are to explain how to calculate the overall value of ΔG.
Analyze
Free energy, like entropy and enthalpy, are state functions. Therefore, we can apply Hess’s law to the
calculation of ΔG for an overall chemical equation.
Solve
To calculate the overall free-energy change for a two-step process we need only add the ΔG values of the
individual steps.
Think about It
Remember that Hess’s law may only be applied to state functions.
14.67. Collect and Organize
o
Given ΔGfo values for maltose, glucose, and water we are to calculate ΔGrxn
for the hydrolysis reaction that
converts 1 mole of maltose into 2 moles of glucose.
Analyze
o
is calculated from ΔGfo values by
ΔGrxn
o
ΔGrxn
= ∑ n ΔGf,o products – ∑ m ΔGf,o reactants
Solve
o
ΔGrxn
= ( 2 mol glucose × –910.1 kJ/mol)
– ⎡⎣(1 mol maltose × –695.65 kJ/mol) + (1 mol H2 O × –237.2 kJ/mol)⎤⎦
= –887.4 kJ
Think about It
This reaction is spontaneous at standard conditions.
14.69. Collect and Organize
We examine the behavior and entropy change when a substance cools and freezes.
Analyze
Substances when cooled have less molecular motion.
Solve
The statement is wrong in that entropy of a system does not increase when a substance is cooled. The second
sentence should read, “At some point of cooling they freeze, and continuing to cool the sample decreases its
entropy.”
Think about It
An exception to the generally true statement that substances contract when they cool is water. At 4˚C, water is
most dense but becomes less dense as it cools down to 0˚C because of hydrogen bonding (Figure 10.29).
14.71. Collect and Organize
For the decomposition of solid NH4Cl into gaseous NH3 and HCl, we are to calculate the temperature at which
o
= 0.
ΔGrxn
130 | Chapter 14
Analyze
o
o
For this we need to first calculate ΔH rxn
and ΔSrxn
using values in Appendix 4. Then we can calculate T by
rearranging the free-energy equation:
o
o
o
ΔGrxn
= ΔH rxn
− T ΔS rxn
=0
T=
o
ΔH rxn
o
ΔS rxn
Solve
o
ΔH rxn
= ⎡⎣(1 mol NH3 × – 46.1 kJ/mol) + (1 mol HCl × –92.3 kJ/mol)⎤⎦
− (1 mol NH 4 Cl × –314.4 kJ/mol)
= 176.0 kJ
ΔS = ⎡⎣(1 mol NH3 × 192.5 J/mol ⋅ K ) + (1 mol HCl × 186.9 J/mol ⋅ K )⎤⎦
o
rxn
− (1 mol NH 4 Cl × 94.6 J/mol ⋅ K )
= 284.8 J/K
T=
176.0 kJ
= 618.0 K or 344.8oC
0.2848 kJ/K
Think about It
This reaction is favored by entropy but not by enthalpy. It is spontaneous at high temperature.
14.73. Collect and Organize
For the gas-phase reaction of NO with H2 to form N2 and H2O we can use the values of ΔGfo in Appendix 4 to
o
calculate ΔGrxn
and then determine if the reaction is spontaneous.
Analyze
o
We can use the values for ΔGfo of the products and reactants to calculate ΔGrxn
:
o
ΔGrxn
= ∑ n ΔGf,o products – ∑ m ΔGf,o reactants
o
If ΔGrxn
is negative then the reaction is spontaneous.
Solve
o
ΔGrxn
= ⎣⎡(1 mol N 2 × 0.0 kJ/mol) + ( 2 mol H 2 O × –228.6 kJ/mol )⎦⎤
− ⎡⎣( 2 mol NO × 86.6 kJ/mol ) + ( 2 mol H 2 × 0.0 kJ/mol )⎤⎦
= – 630.4 kJ
Yes, the reaction is spontaneous at standard temperature and pressure.
Think about It
o
Because 4 mol of gas combine as reactants and form 3 mol of gas as products, we predict that ΔSrxn
is
negative. From the data in Appendix 4, we see that this is indeed the case:
14.75. Collect and Organize
For a reaction that is favored by enthalpy (ΔH negative) and by entropy (ΔS positive) we are to determine
whether the reaction is spontaneous at any temperature.
Analyze
A reaction is spontaneous when ΔG is negative according to the Gibbs free-energy equation:
ΔG = ΔH – T∆S
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 131
Solve
If ΔS is positive then the term TΔS is positive. Subtracting T∆S from a negative value of ΔH means that ΔG is
always negative and the reaction is spontaneous at all temperatures.
Think about It
If a reaction is not favored by both enthalpy (ΔH positive) and entropy (ΔS negative), it will not become
spontaneous by changing the temperature.
14.76. Collect and Organize
We can use ΔH fo and S o values in Appendix 4 to estimate the ΔG at 225˚C (498 K) for the reaction
C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g)
Analyze
o
o
After computing ΔH rxn
and ΔSrxn
we can calculate ΔGrxn with the equation
o
o
ΔGrxn = ΔH rxn
– T ΔSrxn
Solve
o
ΔH rxn
= ⎡⎣( 2 mol CO 2 × –393.5 kJ/mol) + ( 2 mol H 2 O × –241.8 kJ/mol)⎤⎦
− ⎡⎣(1 mol C2 H 4 × 52.3 kJ/mol) + ( 3 mol O 2 × 0.0 kJ/mol)⎤⎦
ΔS
o
rxn
= –1322.9 kJ
= ⎡⎣( 2 mol CO 2 × 213.6 J/mol ⋅ K ) + ( 2 mol H 2 O × 188.8 J/mol ⋅ K )⎤⎦
− ⎡⎣(1 mol C 2 H 4 × 219.5 J/mol ⋅ K ) + (3 mol O 2 × 205.0 J/mol ⋅ K )⎤⎦
= –29.7 J/K
ΔGrxn = –1322.9 kJ – ( 498 K × – 0.0297 kJ/K) = –1308.1 kJ
Think about It
Even though this reaction is not favored by entropy (–ΔS), it is favored by enthalpy (–ΔH) and is spontaneous
at low temperatures. As the temperature increases, this reaction becomes less spontaneous and ΔG becomes
more positive.
14.77. Collect and Organize
We are to explain why the elements at standard conditions have nonzero molar entropy values.
Analyze
Entropy is a measure of disorder in a system. The third law of thermodynamics states that the entropy of a
perfect crystal at 0 K is zero.
Solve
Because standard conditions are not defined as 0 K, all substances, even the elements, above this temperature
have positive entropy.
Think about It
Be careful in calculating ΔS for a reaction because the S˚ for an element, unlike ΔH fo and ΔGfo, is not zero.
14.79. Collect and Organize
For HCN we are to calculate the normal boiling point (Tb) given ΔH fo and S o values for HCN(l ) and HCN(g).
Analyze
At the boiling point ΔG = 0 because the system is at equilibrium. Therefore,
132 | Chapter 14
ΔG = 0 = ΔH vap − Tb ΔS vap
T=
ΔH vap
ΔS vap
Solve
For the vaporization process
HCN (l ) → HCN(g)
ΔH vap = (1 mol HCN(g ) × 135.1 kJ/mol ) − (1 mol HCN(l ) × 108.9 kJ/mol)
= 26.2 kJ
ΔSvap = (1 mol HCN(g ) × 202 J/mol ⋅ K ) − (1 mol HCN(l ) × 113 J/mol ⋅ K )
= 89 J/K
Tb =
26.2 kJ
= 294 K
0.089 kJ/K
Think about It
The actual boiling point of HCN(l ) is 299 K, just about room temperature, so our calculation is
approximately correct.
14.81. Collect and Organize
For CCl4 we are to estimate ΔSvap from ΔH vap (32.5 kJ/mol) and Tb (76.7˚C) and compare that value to our
prediction of whether ΔSvap should be positive or negative.
Analyze
The reaction is
CCI4 (l ) → CCl4(g)
At the boiling point ΔG = 0 because the system is at equilibrium. Therefore,
ΔG = 0 = ΔH vap − Tb ΔS vap
Tb =
ΔH vap
ΔS vap
Solve
Because we have a phase change from liquid to gas, we expect an increase in the entropy for the system and
we expect ΔSvap to be positive.
o
ΔSvap
=
32.5 kJ/mol
= 0.0929 kJ/mol ⋅ K or 92.9 J/mol ⋅ K
349.85 K
Think about It
We can check our calculation using S˚ values from Appendix 4:
ΔSvap = [1 mol CCl4 ( g ) × 309.7 J/mol ⋅ K ] − [1 mol CCl 4 (l ) × 216.4 J/mol ⋅ K ]
= 93.3 J/mol ⋅ K
The values differ slightly because of the temperature dependence of ΔH and ΔS.
14.83. Collect and Organize
Given the melting point Tm (3422˚C) and ΔHfus (35.4 kJ/mol) of tungsten, we are to calculate ΔSfus.
Analyze
The reaction for this process is
W(s) → W(l )
At the melting point ΔG = 0 because the system is at equilibrium. Therefore,
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 133
ΔG = 0 = ΔH fus − Tm ΔSfus
ΔSfus =
ΔH fus
Tm
Solve
ΔSfus =
35.4 kJ/mol
= 0.00958 kJ/mol ⋅ K or 9.58 J/mol ⋅ K
3695 K
Think about It
Tungsten has the highest melting point of all the metals, making it useful as a filament in incandescent
lightbulbs.
14.85. Collect and Organize
Given the transition temperature (369 K) and the enthalpy change for the interconversion of two allotropes of
S8 (297 J/mol) we are to calculate the entropy change for the transition.
Analyze
At the transition temperature, ΔG = 0 because the system is at equilibrium. Therefore,
ΔG = 0 = ΔH trans − T ΔStrans
ΔStrans =
ΔH trans
T
Solve
ΔStrans =
297 J/mol
= 0.805 J/mol ⋅ K
369 K
Think about It
This transition, not favored by enthalpy (ΔH positive), is favored by entropy (ΔS positive).
14.87. Collect and Organize
Using the values given for ΔH foand S˚ for CaCO3, CaO, and CO2 we are to explain why S˚ of CaCO3 is higher
than that of CaO and calculate the temperature at which the pressure of CO2 over CaCO3 is 1.0 atm.
Analyze
By considering the phase and size of each compound, we can rank the compounds in order of increasing
standard molar entropy. The reaction involved is
CaCO3(s) → CaO(s) + CO2(g)
To calculate the temperature at which the partial pressure of CO2 is 1.0 atm, we must recognize that at that
temperature the reaction will be at equilibrium, ΔG = 0.
Solve
S˚ for CaCO3 is greater than S˚ for CaO because there are more atoms in CaCO3.
o
o
To calculate the temperature at which the pressure of CO2 is 1.0 atm we must first calculate ΔH rxn
and ΔS rxn
.
o
⎡
⎤
ΔH rxn = ⎣(1 mol CaO × –636 kJ/mol) + (1 mol CO2 × –394 kJ/mol)⎦
− (1 mol CaCO3 × –1207 kJ/mol)
= 177 kJ
ΔS
o
rxn
= ⎡⎣(1 mol CaO × 40 J/mol ⋅ K ) + (1 mol CO2 × 214 J/mol ⋅ K )⎤⎦
− (1 mol CaCO3 × 93 J/mol ⋅ K )
= 161 J/K
ΔG = 0 = 177 kJ – T × 0.161 kJ/K
134 | Chapter 14
T = 1099 K or 826oC
Think about It
Although this reaction is endothermic it is favored by entropy and so is spontaneous at high temperature.
14.89. Collect and Organize
We consider what changes occur in ΔS for the heating and cooling of DNA and then we are to write an
equation that relates the melting temperature of DNA to ΔH and ΔS.
Analyze
(a and b) Entropy is a measure of disorder in the system. As DNA unwinds into its two single strands, more
disorder is present in the system.
(c) At the melting point ΔG is equal to 0. We can rearrange the free-energy equation
ΔG = ΔH – T∆S = 0
to solve for T.
Solve
(a) The sign of ΔS for the process of DNA separating into two strands through heating is positive.
(b) ΔS for the re-formation of the double helix of DNA is negative.
(c) ΔH – T∆S = 0
ΔH
T=
ΔS
Think about It
When ΔG = 0, the system is at equilibrium, as we will see in the study of chemical equilibria in Chapter 16.
14.91. Collect and Organize
For the hydrogen and water we are to draw the possible vibrational modes for comparison to those of
ammonia.
Analyze
Figure 8.9 shows the vibrational modes for CO2. These modes include both symmetric and asymmetric
stretching and bending modes. We can use the CO2 example to depict, count, and compare the possible
vibrational modes for H2, H2O, and NH3.
Solve
Hydrogen has only one possible vibrational mode, a symmetric stretching mode.
H
H
Symmetrical stretch
Water has three possible vibrational modes: a symmetric stretch, an asymmetric stretch, and a bending mode.
H
O
H
H
O
H
H
H
H
H
Symmetrical stretch
H
N
H
H
Asymmetrical stretch
H
Bend
Symmetrical stretch
Asymmetrical stretch
Ammonia has six possible vibrational modes.
N
O
H
N
H
H
Asymmetrical stretch
Thermodynamics: Spontaneous Processes, Entropy, and Free Energy | 135
H
H
N
H
H
N
H
H
H
N
H
H
Deformation
Bend
Bend
Both water and hydrogen, with fewer atoms in their structures, have fewer possible vibrational modes than
ammonia, with more atoms in its structure.
Think about It
The number of possible vibrational modes for a molecule with N atoms is predictable. For linear molecules,
the number of possible vibrational modes is 3N – 5; for nonlinear molecules, the number of possible
vibrational modes is 3N – 6.
CHAPTER 15 | Chemical Kinetics
15.1. Collect and Organize
For the plot of Figure P15.1 we are to identify which curves represent [N2O] and [O2] over time for the
conversion of N2O to N2 and O2 according to the equation
2 N2O(g) → N2(g) + O2(g)
Analyze
As the reaction proceeds, the concentration of the reactant, N2O, decreases and the concentration of the
product, O2, increases. The rate at which N2O is used up in the reaction is twice the rate at which O2 is
produced.
Solve
The [N2O] is represented by the green line and [O2] is represented by the red line.
Think about It
Notice that [N2], represented by the blue line, increases twice as fast as [O2] because there are 2 N2 molecules
produced for every 1 O2 molecule in the reaction.
15.3. Collect and Organize
For three different initial concentrations of reactant A shown in Figure P15.3, we are to choose which would
have the fastest rate for the conversion 2 A→B.
Analyze
We are given that the reaction is second order in A. The rate law is written as follows:
Rate = k[A]2
As the concentration of A increases, the rate increases.
Solve
Figure 15.3b has the fastest reaction rate because it has the highest concentration of A.
Think about It
The higher the concentration of reactant molecules, the more often they collide, which increases the rate of the
reaction.
15.5. Collect and Organize
For the three reaction profiles in Figure P15.5, we are to choose the one that has the slowest reaction rate.
Analyze
The rate of a reaction is determined by the activation energy (Ea) of the slowest step. All of the reactions
shown consist of a single step. The Ea is the energy difference between the reactants (on the left-hand side of
the graph) and the transition state (the highest point on the reaction profile curve).
Solve
Reaction profile b has the largest Ea and therefore is the slowest reaction.
Think about It
Be careful not to assume that reaction b is slow because it is nonspontaneous. The rate of a reaction does not
depend on the thermodynamics of the reactants and the products.
15.7. Collect and Organize
For the reaction profile in Figure P15.7 we are to match it with the possible reactions given.
Analyze
156
Chemical Kinetics | 157
The reaction profile shows a two-step reaction that has a slightly larger activation energy for its second step
compared to its first step.
Solve
Reaction c is correct because it is a two-step reaction with the first step faster than the second.
Think about It
Reaction b, which occurs in a single step, would show only one transition state and one activation energy in
its reaction profile.
15.9. Collect and Organize
Given the reaction profile of an uncatalyzed reaction (Figure P15.9), we are to choose the reaction profile
corresponding to the catalyzed reaction.
Analyze
A catalyst increases the rate of reaction by decreasing the activation energy of the reaction through an
alternate pathway to the products. This alternate pathway usually involves more steps.
Solve
Reaction profile b correctly shows the catalyzed reaction.
Think about It
Reaction profiles a and c cannot be correct because the initial uncatalyzed nonspontaneous reaction is
represented as spontaneous. A catalyst cannot change a nonspontaneous reaction into a spontaneous reaction.
15.11. Collect and Organize
Of the highlighted elements in Figure P15.11 we are to choose which forms gaseous oxides associated with
photochemical smog.
Analyze
Photochemical smog is the result of sunlight interacting with NOx produced by automobile emissions and
volatile organic compounds (VOCs) released into the atmosphere.
Solve
Nitrogen (light blue) forms the volatile oxides that are components of photochemical smog.
Think about It
Sunlight causes a reaction of NOx and VOCs to produce peroxyacyl nitrates that are very irritating to the
lungs.
15.13. Collect and Organize
Of the highlighted elements in Figure P15.13 we are to identify which are widely used as heterogeneous
catalysts.
Analyze
Heterogeneous catalysts have a different phase than the reactants. We read in Section 15.6 about the specific
metals that are used in catalytic converters.
Solve
Both of the transition metals, palladium (blue) and platinum (orange), can serve as heterogeneous catalysts,
and were specifically identified in the chapter as catalysts.
Think about It
Because catalytic converters contain precious metals such as rhodium, platinum, and palladium, there is great
interest in recycling the metals from catalytic converters.
158 | Chapter 15
15.15. Collect and Organize
By considering the levels of O3 during the day as seen in Figure 15.2, we are to explain why [O3]max occurs
later in the day than [NO]max and [NO2]max.
Analyze
Ozone in the troposphere (the lowest portion of Earth’s atmosphere) is due to the reaction of O2 with O
generated from the interaction of UV light with NO2.
Solve
The presence of NO2 in the atmosphere and ample sunlight allows the O atoms to react with O2 to generate
O3. The reactant NO2 is present in the atmosphere due to automobile exhausts, which build up during the day.
The buildup of O3 lags behind until later in the day, until [NO2] increases and the sunlight becomes stronger
as midday approaches.
Think about It
Ozone is a very reactive gas and is irritating to lung tissues.
15.17. Collect and Organize
We are to explain why there isn’t an increase in NO concentration after the evening rush hour.
Analyze
The reaction in the troposphere (lower atmosphere) that produces NO is
sunlight
NO2 ( g ) ⎯⎯⎯
→ NO(g ) + O(g)
Solve
In the evening the sunlight (and UV radiation) is less intense, so the photochemical breakdown of NO2 does
not occur to as great an extent as after the morning rush hour.
Think about It
The use of catalytic converters to reduce the NOx to N2 and O2 in automobile exhaust has greatly helped to
reduce photochemical smog in large urban and suburban centers.
15.19. Collect and Organize
o
Using ΔH fo for NO, O2, and NO2 in Appendix 4, we can calculate ΔH rxn
for
2 NO(g) + O2(g) → 2 NO2(g)
Analyze
o
The ΔH rxn
can be calculated using
o
ΔH rxn
= ∑ n ΔH f,oproducts – ∑ m ΔH f,o reactants
Solve
o
ΔH rxn
= ( 2 mol NO2 × 33.2 kJ/mol) − ⎡⎣( 2 mol NO × 90.3 kJ/mol) + (1 mol O2 × 0.0 kJ/mol)⎤⎦
= −114.2 kJ
Think about It
This reaction is exothermic and favored by enthalpy.
15.21. Collect and Organize / Analyze
For the reaction of N2 with O2 to produce N2O and N2O5 we are to write balanced chemical equations.
Chemical Kinetics | 159
Solve
(a)
N 2 (g ) + 12 O 2 ( g ) → N 2 O(g )
or 2 N 2 (g ) + O 2 (g ) → 2 N 2 O(g )
(b)
N 2 (g ) + 52 O 2 ( g ) → N 2 O5 (g )
or 2 N 2 (g ) + 5 O 2 (g ) → 2 N 2 O5 (g )
Think about It
Balanced chemical equations usually are written with whole-number coefficients.
15.23. Collect and Organize
We are to explain the difference between the average rate and the instantaneous rate of a reaction.
Analyze
The rate of reaction is measured by the change in concentration of a reactant or product over time. The
difference between the average and instantaneous rates is the length of the period of time over which the
change in concentration is measured.
Solve
The average rate is the rate averaged over a fairly long period of time, whereas the instantaneous rate is the
rate at a specific moment in time (or over a very, very short period of time).
Think about It
The rate of a reaction is always positive.
15.25. Collect and Organize
We are to explain why the average rates of most reactions change over time.
Analyze
The forward rate of a reaction as measured by the average rate depends on the concentration of reactants.
Solve
As the reaction proceeds, the concentrations of the reactants decrease. Because most reactions depend on the
availability (i.e., concentration) of reactants to proceed, the decrease in reactant concentrations lowers the
reaction rate.
Think about It
Reactions that have no dependence on the concentrations of the reactants, although rare, do not show a change
in rate as the reaction proceeds.
15.27. Collect and Organize
Given the balanced equation for the reaction of ammonia with oxygen to produce H+, NO2–, and H2O, we are
asked to relate the formation of products and consumption of reactants.
Analyze
The coefficients in the balanced chemical equation tell us how the rate of formation of products and
consumption of reactants are related to each other. For the generic chemical equation where A, B, C, and D
are the reactants and products and a, b, c, and d are their stoichiometric coefficients
aA+bB→cC+dD
the relationship of the rates is given by
1 Δ [ A]
1 Δ [ B] 1 Δ [ C] 1 Δ [ D]
Rate = −
=−
=
=
a Δt
b Δt
c Δt
d Δt
160 | Chapter 15
Solve
For the reaction given
+
–
1 Δ [ NH3 ]
1 Δ [ O2 ] 1 Δ ⎡⎣ H ⎤⎦ 1 Δ ⎡⎣ NO2 ⎤⎦ 1 Δ [ H 2 O]
Rate = –
=–
=
=
=
2 Δt
3 Δt
2 Δt
2
Δt
2 Δt
+
–
Δ [ NH3 ] Δ ⎡⎣ H ⎤⎦ Δ ⎡⎣ NO2 ⎤⎦
(a) −
, the rate of consumption of NH3 is equal to the rate of formation of H+
=
=
Δt
Δt
Δt
and NO2–.
Δ ⎡ NO2 – ⎤⎦
2 Δ [ O2 ]
(b) ⎣
, the rate of formation of NO2– is two-thirds the rate of consumption of O2.
=−
Δt
3 Δt
Δ [ NH 3 ] 2 Δ [ O 2 ]
(c)
, the rate of consumption of NH3 is two-thirds the rate of consumption of O2.
=
Δt
3 Δt
Think about It
The negative sign is used in front of the expressions involving the consumption of a reactant to give a positive
reaction rate because the change in concentration, [X]f – [X]i or Δ[X], is negative for reactants because
[X]f < [X]i.
15.29. Collect and Organize
For each of three reactions we are to write expressions for the rate of formation of products and consumption
of reactants.
Analyze
The coefficients in the balanced chemical equation tell us how the rate of formation of products and
consumption of reactants are related to each other. For the generic chemical equation where A, B, C, and D
are the reactants and products and a, b, c, and d are their stoichiometric coefficients
aA+bB→cC+dD
the relationship of the rates is given by
Rate = −
1 Δ [ A]
1 Δ [ B] 1 Δ [ C] 1 Δ [ D]
=−
=
=
a Δt
b Δt
c Δt
d Δt
Solve
Δ [ H 2 O2 ]
1 Δ [ OH ]
2 Δt
Δ [ O2 ] Δ [ ClO3 ]
(b) Rate = −
=−
=
Δt
Δt
Δt
Δ [ N 2 O5 ]
Δ [ H 2 O ] 1 Δ [ HNO3 ]
(c) Rate = −
=−
=
Δt
Δt
2
Δt
(a) Rate = −
Δt
Δ [ ClO]
=
Think about It
The negative sign is used in front of the expressions involving the consumption of a reactant to give a positive
reaction rate because the change in concentration, [X]f – [X]i or Δ[X], is negative for reactants because
[X]f < [X]i.
15.31. Collect and Organize
Using the balanced equation describing the reaction of SO2 with CO, we are to write expressions to compare
the rates of formation of products and the rates of consumption of reactants.
Analyze
From the balanced equation we see that the reaction may be expressed as
Chemical Kinetics | 161
Rate = −
Δ [SO 2 ]
Δt
=−
1 Δ [ CO ]
1 Δ [ CO 2 ] Δ [ COS]
=
=
3 Δt
2 Δt
Δt
Solve
(a) Rate =
(b) Rate =
(c) Rate =
Δ [ CO 2 ]
Δt
Δ [ COS]
Δt
Δ [ CO ]
Δt
=−
=−
=3
2 Δ [ CO]
3 Δt
Δ [SO 2 ]
Δt
Δ [SO 2 ]
Δt
Think about It
These relative rates make sense based on the stoichiometry of the reaction. For every 1 mol of SO 2 used in the
reaction, 2 mol of CO2 and 1 mol of COS are produced. So, for example, the concentration of CO2 will
increase twice as fast as the concentration of SO2 decreases.
15.33. Collect and Organize
Using the relative rate expressions and the rate of the consumption of ClO in two reactions, we are to calculate
the rate of change in the formation of the products of the two reactions.
Analyze
(a) For this reaction
Rate = −
(b) For this reaction
Rate = −
1 Δ [ ClO] Δ [ Cl 2 ] Δ [ O 2 ]
=
=
2 Δt
Δt
Δt
Δ [ ClO]
=–
Δ [ O3 ]
=
Δ [ O2 ]
=
Δ [ ClO 2 ]
Δt
Δt
Δt
Δt
For each reaction we are given ∆[ClO]/∆t and we can use this value in the relationships above to calculate the
rate of change in (a) the concentration of Cl2 and O2 and (b) the concentration of O2 and ClO2.
Solve
Δ [Cl2 ] Δ [O2 ]
1 Δ [ClO]
1
(a)
=
=−
= – × –2.3 ×107 M /s
Δt
Δt
2 Δt
2
= 1.2 ×107 M /s
(b)
Δ [O 2 ]
Δt
=
Δ [ClO2 ]
Δt
=−
Δ [ClO]
Δt
= – ( –2.9 ×104 M /s )
= 2.9 ×104 M /s
Think about It
The rates of the formation of products is positive because [X]f > [X]i so [X]f – [X]i = Δ[X] is positive.
15.35. Collect and Organize
Given the [O3] over time when it reacts with NO2– we are to calculate the average reaction rate for two time
intervals.
Analyze
The average rate of reaction can be found according to
Δ [ O3 ] [ O3 ]f – [ O3 ]i
=
Δt
tf – ti
162 | Chapter 15
Solve
Between 0 and 100 µs:
–
Δ [ O3 ]
Δt
(9.93 × 10
=
–3
)
– 1.13 × 10 –2 M
(100 – 0) µs
= 1.4 × 10 –5 M /µs
Between 200 and 300 µs:
–
Δ [ O3 ]
Δt
=
(8.15 × 10
–3
)
– 8.70 × 10 –3 M
(300 – 200) µs
= 5.50 × 10 –6 M /µs
Think about It
Notice that as the reaction proceeds the rate of consumption of ozone decreases. This is due to the decreasing
reactant concentrations.
15.37. Collect and Organize
After we plot [ClO] versus time and [Cl2O2] versus time we can determine the instantaneous rate of change at
1 s for each compound.
Analyze
The instantaneous rate is the slope of the line that is tangent to the curve at the time we are interested in. We
can estimate fairly well the instantaneous rate at 1 s from the plots by choosing two points that are close to 1 s
to calculate the slope. Since we are given only values for [ClO], we need to calculate [Cl2O2] for each time.
[Cl2 O2 ] t =
([ClO] – [ClO] )
0
t
2
where [ClO]0 = 2.60 × 1011 M
Time
[ClO]t
[Cl2O2]t
s
molecules/cm3
molecules/cm3
0
0.00
2.60 × 1011
1
1.08 × 1011
7.60 × 1010
2
6.83 × 1010
9.59 × 1010
10
3
4.99 × 10
1.05 × 1011
10
4
3.93 × 10
1.10 × 1011
10
5
3.24 × 10
1.14 × 1011
10
6
2.76 × 10
1.16 × 1011
Note that initially there is no Cl2O2 present, so [Cl2O2]0 = 0 molecules/cm3.
Solve
For the change in concentration of ClO versus time we obtain the following plot.
Chemical Kinetics | 163
To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from
the data set that surround the data point of interest and calculate ∆[ClO]/∆t.
For t = 1 s, we can use the points t = 0 s and t = 2 s
Δ[ClO] (6.83 × 1010 − 2.60 × 1011 ) molecules/cm3
–
=
= 9.59 × 1010 molecules ⋅ cm –3 ⋅ s –1
Δt
(2 – 0) s
Using a graphing program which calculates the slope of the tangent to the line at t = 1 s, we get an
instantaneous rate of 8.28 × 1010 molecules ⋅ cm –3 ⋅ s.
For the change in concentration of Cl2O2 versus time we obtain the following plot.
To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from
the data set that surround the data point of interest and calculate ∆[Cl2O2]/∆t.
For t = 1 s, we can use the points t = 0 s and t = 2 s
Δ[Cl2 O2 ] (9.59 × 1010 − 0) molecules/cm3
=
= 4.80 × 1010 molecules ⋅ cm–3 ⋅ s –1
Δt
(2 – 0) s
Using a graphing program which calculates the slope of the tangent to the line at t = 1 s, we get an
instantaneous rate of 4.13 × 1010 molecules ⋅ cm –3 ⋅ s.
Think about It
Because we expect the rate of disappearance of ClO to be twice the rate of appearance of Cl2O2 from the
balanced equation
2 ClO(g) → Cl2O2(g)
our answers make sense
164 | Chapter 15
Δ[ClO]
9.59 × 1010 molecules ⋅ cm –3 ⋅ s −1
Δt
=
= 2.0
Δ[Cl2 O2 ] 4.80 × 1010 molecules ⋅ cm –3 ⋅ s −1
Δt
15.39. Collect and Organize
We consider whether two different chemical reactions can have the same rate-law expression.
Analyze
The rate-law expression is of the form
Rate = k[A]x
Where k is the rate constant, [A] is the concentration of the reactant(s) on which the rate depends, and x is the
order of the reaction for that concentration of reactant.
Solve
Yes, two different reactions can have the same form of the rate law. They both may have the same
dependence on the concentration of reactants, yet yield different products. For example, A may decompose by
two routes:
A→B+C
A→D+E
If both reactions are first order (or even second order) in [A] they would have the same rate law:
Rate = k[A]
Think about It
The value of the rate constant, however, is expected to be different for the two reactions.
15.41. Collect and Organize
We are asked if the units of the half-life for a second-order reaction are the same as those of the half-life for a
first-order reaction.
Analyze
The half-life is the time it takes for the amount of reactant originally present to decrease by one-half.
Solve
Yes. Because the half-life is a time measurement, the units, no matter what the order of the reaction, are
always in units of time (s, min, hr, yr, etc.).
Think about It
The half-life of a reaction depends on the value of the reaction’s rate constant and, except for first-order
reactions, on the initial concentration. The larger the rate constant, the faster is the reaction and the shorter is
the half-life of the reaction.
15.43. Collect and Organize
For a second-order reaction we are to predict the effect of doubling [A]0 on the half-life.
Analyze
For a second-order reaction,
t1/ 2 =
1
k [ A ]0
Solve
From the equation for the half-life of a second-order reaction, we see that doubling [A]0 halves the half-life.
Chemical Kinetics | 165
Think about It
The half-life of a second-order reaction, like that of a first-order reaction, is inversely related to the rate
constant.
15.45. Collect and Organize
For each rate-law expression, we are to determine the order of the reaction with respect to each reactant and
the overall reaction order.
Analyze
The order of a reaction is the experimentally determined dependence of the rate of a reaction on the
concentration of the reactants involved in the reaction. In the rate-law expression the order is shown as the
power to which the concentration of a particular reactant is raised. The overall reaction order is the sum of the
powers of the reactants in the rate-law expression.
Solve
(a) For the rate-law expression Rate = k[A] [B], the reaction is first order in both A and B and second order
overall.
(b) For Rate = k[A]2 [B], the reaction is second order in A, first order in B, and third order overall.
(c) For Rate = k[A] [B]3, the reaction is first order in A, third order in B, and fourth order overall.
Think about It
The higher the order of the reaction for a particular reactant, the greater is the effect of a change in
concentration of that reactant on the reaction rate.
15.47. Collect and Organize
For each of the reactions described we are to write the rate law and determine the units for k, using the units M
for concentration and s for time.
Analyze
The general form of the rate law is
Rate = k[A]x[B]y
where k is the rate constant, A and B are the reactants, and x and y are the orders of the reaction with respect
to each reactant as determined by experiment.
Solve
(a) Rate = k[O][NO2]
Because rate has units of M/s and each concentration has units of M,
M s
k=
= M −1 s −1
M2
(b) Rate = k[NO]2[Cl2]
Because rate has units of M/s and each concentration has units of M,
M s
k=
= M −2 s −1
M3
(c) Rate = k [CHCl3 ][Cl2 ]1/ 2
Because rate has units of M/s and each concentration has units of M,
M s
1
k = 3 / 2 = M − 2 s −1
M
(d) Rate = k[O3]2[O]–1
Because rate has units of M/s and each concentration has units of M,
k=
M s
= s −1
2
M M
Think about It
The units of the rate constant clearly depend on the overall order of the reaction.
166 | Chapter 15
15.49. Collect and Organize
Given the changes in rate of the decomposition of BrO to Br2 and O2 when [BrO] is changed, we are to predict
the rate law in each case.
Analyze
The general form of the rate law for the reaction is
Rate = k[BrO]x
where x is an experimentally determined exponent.
Solve
(a) If the rate doubles when [BrO] doubles, then x = 1 and the rate law is
Rate = k[BrO]
(b) If the rate quadruples when [BrO] doubles, then x = 2 and the rate law is
Rate = k[BrO]2
(c) If the rate is halved when [BrO] is halved, then x = 1 and the rate law is
Rate = k[BrO]
(c) If the rate is unchanged when [BrO] is doubled, then x = 0 and the rate law is
Rate = k[BrO]0 = k
Think about It
For this reaction the relationship is straightforward between the change in rate when the concentration of the
reactant was changed to determine x. To determine x for more complicated reactions, use
rate 2 ⎛ [ A ]2
= ⎜
rate1 ⎜⎝ [ A ]1
⎛ rate 2
ln ⎜
⎝ rate1
⎞
⎟⎟
⎠
x
⎛ [ A ]2
⎞
⎟ = x ln ⎜⎜
⎠
⎝ [ A ]1
⎞
⎟⎟
⎠
⎛ rate 2 ⎞
ln ⎜
⎟
rate1 ⎠
x = ⎝
⎛ [ A ]2 ⎞
ln ⎜
⎜ [ A ] ⎟⎟
⎝
1 ⎠
15.51. Collect and Organize
Given that the rate of the reaction quadruples when both [NO] and [ClO] are doubled, we are to identify what
additional information we would need to write the rate law for the reaction
NO(g) + ClO(g) → NO2(g) + Cl(g)
Analyze
The general form of the rate law for this reaction is
Rate = k[NO]x[ClO]y
Solve
If the rate quadruples when both [NO] and [ClO] are doubled, the rate law could be any of the following:
Rate = k[NO][ClO]
Rate = k[NO]2
Rate = k[ClO]2
In order to differentiate among these we need to determine the change in the rate when only [NO] or [ClO] is
changed.
Think about It
Chemical Kinetics | 167
If the rate is only doubled when [NO] and [ClO] are independently changed, then the rate law is
Rate = k[NO][ClO]
If the rate is quadrupled when [NO] is doubled but remains constant if [ClO] is doubled, then the rate law is
Rate = k[NO]2
If the rate is quadrupled when [ClO] is doubled but remains constant if [NO] is doubled, then the rate law is
Rate = k[ClO]2
15.53. Collect and Organize
For the reaction of NO2 with O3 to produce NO3 and O2 we are to write the rate law given that the reaction is
first order in both NO2 and O3. From the rate law and given the rate constant we can calculate the rate of the
reaction for a given [NO2] and [O3]. From this we can calculate the rate of appearance of NO3 and the rate of
the reaction when [O3] is doubled.
Analyze
The general form of the rate law for this reaction is
Rate = k[NO2]x[O3]y
The rate of consumption of reactants and formation of products is
Rate = −
Δ [ NO2 ]
Δt
=–
Δ [ O3 ]
Δt
=
Δ [ NO3 ]
Δt
=
Δ [ O2 ]
Δt
Solve
(a) Rate = k[NO2][O3]
1.93 × 104
(b) Rate =
× 1.8 × 10–8 M × 1.4 × 10–7 M = 4.9 × 10–11M /s
M ⋅s
Δ [ NO3 ]
(c) Rate =
= 4.9 × 10 –11 M /s
Δt
(d) When [O3] is doubled, the rate of the reaction doubles.
Think about It
When [O3] = 2.8 × 10–7 M (double that in part b) the rate of reaction is 9.73 × 10–11 M/s, which is twice that
calculated in part b, so our prediction in part d is correct.
15.55. Collect and Organize
By comparing the rate constants for four reactions that are all second order, we can determine which reaction
is the fastest if all the initial concentrations are the same.
Analyze
The reaction with the largest rate constant has the fastest reaction rate.
Solve
Reaction c has the largest value of k, so it proceeds the fastest.
Think about It
The slowest reaction is a reaction with the smallest value of k.
15.57. Collect and Organize
Given the information that the rate of the reaction between NO and NO2 with water doubles when either [NO]
or [NO2] doubles and there is no dependence of the rate on [H2O], we can write the rate law for the reaction.
Analyze
The general form of the rate law for this reaction is
Rate = k[NO]x[NO2]y[H2O]z
168 | Chapter 15
Doubling of the reaction rate with doubling of [NO] or [NO2] means the reaction is first order in those
reactants. Because the rate does not depend on [H2O], the reaction is zero order in that reactant.
Solve
The rate law for this reaction is
Rate = k[NO]1[NO2]1[H2O]0 = k[NO][NO2]
Think about It
If [NO] and [NO2] are doubled simultaneously, the rate of reaction quadruples.
15.59. Collect and Organize
In the reaction of ClO2 with OH– the rate of the reaction was measured for various concentrations of both
reactants. From the data we are to determine the rate law and calculate the rate constant, k.
Analyze
To determine the dependence of the rate on a change in the concentration of a particular reactant, we can
compare the reaction rates for two experiments in which the concentration of that reactant changes but the
concentrations of the other reactants remain constant. Once we have the order of the reaction for each
reactant, we can write the rate-law expression. To calculate the rate constant for the reaction, we can rearrange
the rate law to solve for k and use the data from any of the experiments.
Solve
Using experiments 1 and 2 we find that the order of the reaction with respect to ClO2 is 1:
rate1 ⎛ [ClO 2 ]0,1 ⎞
= ⎜
⎟
rate 2 ⎜⎝ [ClO 2 ]0,2 ⎟⎠
x
0.0248 M /s ⎛ 0.060 M ⎞
= ⎜
⎟
0.00827 M /s ⎝ 0.020 M ⎠
x
3.00 = 3.01x
x =1
Using experiments 2 and 3 we find that the order of the reaction with respect to [OH–] is also 1:
–
rate3 ⎛ [OH ]0,3 ⎞
= ⎜
⎟
rate 2 ⎜⎝ [OH – ]0,2 ⎟⎠
x
0.0247 M /s ⎛ 0.090 M ⎞
= ⎜
⎟
0.00827 M /s ⎝ 0.030 M ⎠
x
2.99 = 3.0 x
x =1
The rate law for this reaction is Rate = k[ClO2][OH–].
Rearranging the rate-law expression to solve for k and using the data from experiment 1 gives
rate
0.0248 M /s
k=
=
= 14 M –1 s –1
–
[ClO2 ][OH ] 0.060 M × 0.030 M
Think about It
We may use any of the experiments in the table to calculate k. Each experiment’s data give the same value of
k as long as the experiments were all run at the same temperature.
15.61. Collect and Organize
In the reaction of H2 with NO the rate of the reaction was measured for various concentrations of both
reactants. From the data we are to determine the rate law and calculate the rate constant, k.
Chemical Kinetics | 169
Analyze
To determine the dependence of the rate on a change in the concentration of a particular reactant, we can
compare the reaction rates for two experiments in which the concentration of that reactant changes, but the
concentrations of the other reactants remain constant. Once we have the order of the reaction for each
reactant, we can write the rate-law expression. To calculate the rate constant for the reaction, we can rearrange
the rate law to solve for k and use the data from any of the experiments.
Solve
Using experiments 1 and 2 we find that the order of the reaction with respect to NO is 2:
rate 2 ⎛ [NO]0,2 ⎞
= ⎜
⎟
rate1 ⎜⎝ [NO 2 – ]0,1 ⎟⎠
x
0.0991 M /s ⎛ 0.272 M ⎞
= ⎜
⎟
0.0248 M /s ⎝ 0.136 M ⎠
x
4.00 = 2.00 x
x=2
Using experiments 3 and 4 we find that the order of the reaction with respect to H2 is 1:
rate 4 ⎛ [H 2 ]0,4
= ⎜
rate3 ⎜⎝ [H 2 ]0,3
⎞
⎟⎟
⎠
x
1.59 M /s ⎛ 0.848 M ⎞
= ⎜
⎟
0.793 M /s ⎝ 0.424 M ⎠
x
2.01 = 2.00 x
x =1
The rate law for this reaction is Rate = k[NO]2[H2].
Rearranging the rate-law expression to solve for k and using the data from experiment 1 gives
0.0248 M /s
k=
= 6.32 M –2 s –1
2
( 0.136 M ) × 0.212 M
170 | Chapter 15
Think about It
We may use any of the experiments in the table to calculate k. Each experiment’s data give the same value of
k as long as the experiments were all run at the same temperature.
15.63. Collect and Organize
From the data given for the concentration of NO3 over time as it decomposes to NO2 and O2, we are to
calculate the value of k for this reaction.
Analyze
We are given a single data set and the fact that the reaction is second order. For this second-order reaction the
plot of 1/[NO3] versus time gives a straight line with a slope equal to k, the reaction rate constant.
Solve
The slope = k = 0.32 µM–1 ⋅ min–1
Think about It
The half-life of this second-order reaction is
1
1
t1 2 =
=
–1
–1
k[NO3 ]0
0.3202 µ M ⋅ min × 1.470 ×10−3 µ M
(
) (
)
= 2124 min or 35.4 hr
15.65. Collect and Organize
From the data given for concentration of NH3 over time as it decomposes to N2 and H2, we are to determine
the rate law and calculate the value for k for this reaction.
Analyze
We are given a single data set. If a plot of [NH3] versus time yields a straight line, then the reaction is zero
order. If a plot of ln[NH3] versus time yields a straight line, then the reaction is first order. If a plot of 1/[NH3]
versus time yields a straight line, then the reaction is second order. The slope of the line on the appropriate
graph gives the rate constant k for the reaction (slope = –k for a zero- or first-order reaction and slope = +k for
a second-order reaction).
Chemical Kinetics | 171
Solve
The first-order plot is linear so that the rate law is Rate = k[NH3]. The slope = –k = –0.003 s–1 and thus
k = 0.003 s–1.
Think about It
The half-life of this first-order reaction is
0.693
0.693
t1 2 =
=
= 231 s or 3.9 min
k
0.0030 s –1
15.67. Collect and Organize
For the decomposition reaction of N2O to N2 and O2, we are given that the plot of ln[N2O] versus time is
linear. We are to write the rate law and then determine the number of half-lives it would take for the
concentration of N2O to become 6.25% of its original concentration.
Analyze
The integrated rate laws for zero-, first-, and second-order reactions with their half-life equations are as
follows:
[A] = –kt + [A]0
zero order
t1/2 = [A]0/2k
ln[A] = –kt + ln[A]0
first order
t1/2 = 0.693/k
1/[A] = kt + 1/[A]0
second order
t1/2 = 1/k[A]0
Solve
(a) Since we are given that the plot of ln[N2O] versus time is linear, the reaction is first order and the rate law
is Rate = k[N2O].
(b) The number of half-lives needed to reduce the concentration to 6.25% would be, where n = number of
half-lives,
172 | Chapter 15
6.25
= 0.50n
100
0.0625 = 0.50 n
ln 0.0625 = n ln 0.50
n=4
Think about It
The number of half-lives needed to reduce the concentration of a reactant to products is not dependent on the
order of the reaction nor on the magnitude of the rate constant.
15.69. Collect and Organize
From the data given for the concentration of 32P over time, we are to determine the rate law and calculate the
value for k for this radioactive decay.
Analyze
Radioactive decay follows first-order kinetics. This is confirmed in part b, where we are told to determine the
first-order rate constant. A plot of ln[32P] versus time for this decay gives a straight line with slope = –k. The
half-life of a first-order decay is given by
0.693
t1/ 2 =
k
Solve
(a)
First-order plot for Problem 15.69
The rate law for this radioactive decay is Rate = k[32P].
(b) k = –slope = 0.0485 day–1
0.693
(c) t1/ 2 =
= 14.3 days
0.0485 day –1
Think about It
The time needed for [32P] to reduce to 1.00% of its original concentration, where n = number of half-lives, is
1.00
= 0.50n
100
0.0100 = 0.50n
ln 0.0100 = n ln 0.50
n = 6.64 half-lives
The number of days is n × t1/ 2 = 6.64 × 14.3 days = 95 days.
15.71. Collect and Organize
Given that the dimerization of ClO to Cl2O2 is second order we are to determine the value of the rate constant
and to calculate the half-life of this reaction.
Chemical Kinetics | 173
Analyze
We are given a single data set and knowledge that the reaction is second order. For this second-order reaction
a plot of 1/[ClO] versus time gives a straight line with a slope equal to k, the reaction rate constant. The halflife of this second-order reaction is
1
t1/ 2 =
k[ClO] 0
where [ClO]0 is the initial concentration of ClO used in the reaction.
Solve
The value of k for this reaction is k = slope = 5.40 × 10–12 cm3 molecules–1 s–1, and the half-life of the reaction
is as follows:
1
1
t1/ 2 =
=
= 0.712 s
−12
3
k[ClO] 0 5.40 × 10 cm 2.60 × 1011 molecules
×
molecules ⋅ s
cm3
Think about It
For a second-order reaction as we increase the initial concentration of the reactant, the half-life gets shorter.
15.73. Collect and Organize
From the data for the pseudo-first-order hydrolysis of sucrose provided, we are to write the rate law and
determine the value of the pseudo-first-order rate constant, kʹ′.
Analyze
To obtain the value of kʹ′ we plot ln[sucrose] over time. The slope of the line is equal to –k.
Solve
The rate law is Rate = k[C12H22O11][H2O] = kʹ′[C12H22O11].
The pseudo-first-order plot gives kʹ′ = –slope = 6.21 × 10–5 s–1.
174 | Chapter 15
Think about It
If we knew the concentration of water in the hydrolysis reaction we could calculate the value of k using
k=
k ʹ′
[H2 O]0
15.75. Collect and Organize
We are asked why some spontaneous reactions are slow.
Analyze
Spontaneous reactions have negative free energy values, which tells us that the free energy of the products is
lower than that of the reactants.
Solve
A spontaneous reaction is slow if it has a large activation energy.
Think about It
Likewise a nonspontaneous reaction may be fast to reach equilibrium.
15.77. Collect and Organize
Of the four statements concerning the relationship between ΔH, ΔG ,and ΔS and the rate of a reaction, we are
to choose which are true. (a) Are exothermic reactions (–ΔH) always fast? (b) Are reactions with ΔG > 0
(nonspontaneous) slow? (c) Are endothermic reactions (+ΔH) always slow? (d) Are reactions in which
entropy increases (+ΔS) fast?
Analyze
Thermodynamics tells us about the relative enthalpy, entropy, and free energy of the reactant(s) and products.
Kinetics tells us how fast a reaction proceeds from reactants to products.
Solve
Because thermodynamic properties do not involve the consideration of the pathways that reactants take to get
to products (i.e., they are state functions), neither the sign nor the magnitude of ΔG, ΔH, or ΔS are related to
the speed of a reaction. Therefore, all of the statements are false.
Think about It
Just because a reaction is favored thermodynamically does not necessarily mean that it is fast.
15.79. Collect and Organize
We are to explain why the order of a reaction is independent of temperature, yet the value of k changes with
temperature.
Analyze
We need to consider how temperature affects the motion and collision of the reactants.
Solve
An increase in temperature increases the frequency and the kinetic energy at which the reactants collide. This
speeds up the reaction, changing the value of k. The activation energy of the slowest step in the reaction,
however, is not affected by a change in temperature and, therefore, the order of the reaction is unaffected.
Chemical Kinetics | 175
Think about It
As a general empirical observation, heating a reaction by 10˚C doubles the rate of reaction.
15.81. Collect and Organize
In comparing two first-order reactions with different activation energies, we are to decide which would show
a larger increase in its rate as the reaction temperature is increased.
Analyze
We can use the Arrhenius equation to mathematically determine which reaction would be most accelerated by
an increase in temperature:
ln k = ln A –
Ea
RT
Solve
Let’s assume that T2 = 2T1. For either reaction the difference in their rate constants is as follows:
E
E
ln kT1 = ln A – a
ln kT2 = ln A – a
RT1
RT2
ln kT2 – ln kT1 =
But since T2 = 2T1,
– Ea Ea
E ⎛ 1 1 ⎞
+
= a ⎜ − ⎟
RT2 RT1 R ⎝ T1 T2 ⎠
ln kT2 – ln kT1 =
Ea ⎛ 1
Ea
1 ⎞
⎜ −
⎟ =
R ⎝ T1 2T1 ⎠ 2RT1
For Ea = 150 kJ/mol,
ln kT2 – ln kT1 =
150 kJ/mol
2 RT1
For Ea = 15 kJ/mol,
ln kT2 – ln kT1 =
15 kJ/mol
2RT1
Comparing these as a ratio,
150 kJ/mol
2 RT1
=
= 10
15 kJ/mol
ln kT2 – ln kT1 for Ea = 15 kJ/mol
2 RT1
Therefore, the reaction with the larger activation energy (150 kJ/mol) would be accelerated more than the
reaction with the lower activation energy (15 kJ/mol) when heated.
ln kT2 – ln kT1 for Ea = 150 kJ/mol
Think about It
Our derivation demonstrates that different reactions with different activation energies will not accelerate in the
same way when they are heated.
15.83. Collect and Organize
We can use an Arrhenius plot of rate constant versus temperature for the reaction of O and O3 to determine the
activation energy (Ea) and the value of the frequency factor (A).
Analyze
The Arrhenius equation is
– Ea ⎛ 1 ⎞
⎜ ⎟ + ln A
R ⎝ T ⎠
If we plot ln k (y-axis) versus 1/T (x-axis) we obtain a straight line with slope m = –Ea /R. The activation
energy is therefore calculated by
Ea = –slope × R
ln k =
176 | Chapter 15
where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor by
b = ln A or eb = A.
Solve
The Arrhenius plot gives a slope of m = –2060 and a y-intercept of b = 0.00160.
Ea = – (–2060 K × 8.314 J/mol ⋅ K) = 1.713 × 104 J/mol or 17.1 kJ/mol
b = 0.00160 = ln A
A = e0.00160 = 1.002
Think about It
Once we have the values of Ea and A from the plot, we can calculate the value of k at any temperature.
15.85. Collect and Organize
We can use an Arrhenius plot of rate constant versus temperature for the reaction of N2 with O2 to form NO to
determine the activation energy (Ea), the frequency factor (A), and the rate constant for the reaction at 300 K.
Analyze
The Arrhenius equation is
– Ea ⎛ 1 ⎞
⎜ ⎟ + ln A
R ⎝ T ⎠
If we plot ln k (y-axis) versus 1/T (x-axis) we obtain a straight line with slope m = –Ea/R. The activation energy
is therefore calculated by
Ea = –slope × R
where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor by
b = ln A or eb = A. Once Ea and A are known, we may use another form of the Arrhenius equation to calculate
k at any temperature:
⎛ E ⎞
k = A exp ⎜ – a ⎟
⎝ RT ⎠
ln k =
Solve
The Arrhenius plot gives a slope of m = –37758 and a y-intercept of b = 24.641.
Chemical Kinetics | 177
(a) Ea = – ( –37758 K × 8.314 J/mol ⋅ K) = 3.14 × 105 J/mol or 314 kJ/mol
(b) A = e24.641 = 5.03 × 1010
⎛
⎞
3.14 ×105 J/mol
–44
–1/2 –1
(c) k = 5.03 ×1010 exp ⎜ −
s
⎟ = 1.06 ×10 M
8.314
J/mol
⋅
K
×
300
K
⎝
⎠
Think about It
Alternatively, k can be calculated from the original Arrhenius equation:
– Ea
–3.14 × 105 J/mol
ln k =
+ ln A =
+ 24.641
RT
8.314 J/mol ⋅ K × 300 K
= −101.25
k = e −101.25 = 1.07 × 10−44 M –1/ 2 s –1
15.87. Collect and Organize
We can use an Arrhenius plot of rate constant versus temperature for the reaction of ClO2 and O3 to determine
the activation energy (Ea) and the value of the frequency factor (A).
Analyze
The Arrhenius equation is
– Ea ⎛ 1 ⎞
⎜ ⎟ + ln A
R ⎝ T ⎠
If we plot ln k (y-axis) versus 1/T (x-axis) we obtain a straight line with slope m = –Ea/R. The activation energy
is therefore calculated by
Ea = –slope × R
where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor by b =
ln A or eb = A.
ln k =
Solve
The Arrhenius plot gives a slope of m = – 4698.7 and a y-intercept of b = 27.872.
178 | Chapter 15
Ea = – ( – 4698.7 K × 8.314 J/mol ⋅ K) = 3.91 × 104 J/mol or 39.1 kJ/mol
b = 27.872 = ln A
A = e27.872 = 1.27 × 1012
Think about It
Once we have the values of Ea and A from the plot, we can calculate the value of k at any temperature.
15.89. Collect and Organize
By comparing the rate laws for two reactions, we can determine whether their mechanisms are similar.
Analyze
For the reaction between NO and H2,
For the reaction between NO and Cl2,
Rate = k[NO]2[H2]
Rate = k[NO][Cl2]
Solve
No. The different rate laws for the two reactions indicate different mechanisms.
Think about It
However, just because two rate laws are similar, it does not mean that their reaction mechanisms are similar.
15.91. Collect and Organize / Analyze
We are to identify the conditions under which a bimolecular reaction shows behavior that is pseudo-firstorder.
Solve
Pseudo-first-order kinetics occurs when one of the reactants is in sufficiently high concentration that it does
not change appreciably over the course of the reaction.
Think about It
We solved problems relating to pseudo-first-order reactions earlier in this chapter (Problems 15.73 and 15.74).
15.93. Collect and Organize
We are asked to draw reaction profiles that fit the reaction A → B (which has Ea = 50.0 kJ/mol) for three
different mechanisms: (a) a single elementary step, (b) a two-step reaction with Ea,step 2 = 15 kJ/mol, and (c) a
two-step reaction in which the second step is rate determining.
Analyze
The reaction profile plots energy versus reaction progress and shows the presence of an activated complex at
highest energy, the reaction intermediates in the “valleys” between reactant and products, and the relative
activation energies in multistep reactions. The rate-determining step is the slowest step in the mechanism and
has the largest activation energy in the reaction profile.
Chemical Kinetics | 179
Solve
Think about It
These reaction profiles could also be drawn for a nonspontaneous reaction. In that case the reactant A would
be lower in energy than product B.
15.95. Collect and Organize
For each of the elementary steps given we are to write the rate law and determine whether the step is uni-, bi-,
or termolecular.
Analyze
The rate law for an elementary step in a mechanism is written in the form
Rate = k[A]x[B]y[C]z
where A, B, and C are the reactants involved in the elementary reaction and x, y, and z are the stoichiometric
coefficients for the respective reactants in the elementary reaction.
Solve
(a) Rate = k[SO2Cl2]. Because this elementary step involves only a molecule of SO2Cl2, it is unimolecular.
(b) Rate = k[NO2][CO]. Because this elementary step involves a molecule of NO2 and a molecule of CO, it is
bimolecular.
(c) Rate = k[NO2]2. Because this elementary step involves two molecules of NO2, it is bimolecular.
Think about It
Termolecular elementary reactions are rare.
15.97. Collect and Organize
From three elementary steps that describe a reaction mechanism, we are to write the overall chemical
equation.
Analyze
To write the overall chemical reaction we need to add the elementary steps, being sure to cancel the
intermediates in the reaction.
Solve
180 | Chapter 15
N 2 O5 ( g ) → NO3 (g ) + NO2 (g )
NO3 ( g ) → NO2 ( g ) + O( g )
2 O( g ) → O2 ( g )
N 2 O5 ( g ) + O( g ) → 2 NO 2 ( g ) + O 2 ( g )
Think about It
In this reaction, NO3 is a reaction intermediate. It is generated in the reaction but consumed in a subsequent
step in the mechanism.
15.99. Collect and Organize
We are given the mechanism by which N2 reacts with O2 to form NO. For a given rate law of
Rate = k[N2][O2]1/2
we are to determine which step in the mechanism is the rate-determining step.
Analyze
To determine which step in the proposed mechanism might be the slowest, we can write the rate law for the
mechanism when the first, second, or third step is slow and then match the theoretical rate law to the
experimental rate law.
Solve
If the first step is slow, the rate law is
Rate = k1[O2]
This does not match the experimental rate law, so the first step is not the slowest step in the mechanism.
If the second step is slow then the rate law is
Rate = k2[O][N2]
Because O is an intermediate we use the first step to express its concentration in terms of concentrations of the
reactants. For a fast step occurring before a slow step in a mechanism,
Rateforward = ratereverse
k1[O2] = k–1[O]2
Rearranging to solve for [O],
1/ 2
⎛ k
⎞
[O] = ⎜ 1 [O2 ] ⎟
⎝ k–1
⎠
Substituting this into the rate law from the second step gives
1/ 2
⎛ k
⎞
1/ 2
Rate = k2 ⎜ 1 [O2 ] ⎟ [ N2 ] = k [O2 ] [ N 2 ]
⎝ k–1
⎠
This rate law matches the experimental rate law, so the rate-determining step is the second step.
We should check to see if the mechanism of the third step is the slow step and might also give the
experimental rate law. Following the logic above,
Rate = k3[N][O]
From the second fast step in the mechanism,
k2[O][N2] = k–2[NO][N]
Solving for [N], an intermediate, gives
k [ O] [ N 2 ]
[N] = 2
k –2 [ NO ]
Chemical Kinetics | 181
From the first fast step in the mechanism,
k1[O2] = k–1[O]2
2
solving for [O] gives
k1
[ O2 ]
k–1
Substituting these expressions into the rate law from the third step in the mechanism gives
k k [O 2 ][N 2 ]
Rate = k3 2 1
k –2 k –1
[NO]
k[N 2 ][O 2 ]
=
[NO]
This rate law does not match the experimental rate law.
[O]2 =
Think about It
Just because the rate law for a mechanism matches the experimental rate law does not mean that the
mechanism is the mechanism. Another mechanism might also give the same experimental rate law.
15.101. Collect and Organize
We are given the mechanism by which NO reacts with Cl2 to produce NOCl2. For a given rate law of
Rate = k[NO][Cl2]
we are to determine which step in the mechanism is the rate-determining step.
Analyze
To determine which step in the proposed mechanism might be the slowest, we can write the rate law for the
mechanism when the first, second, or third step is slow and then match the rate law to the experimental rate law.
Solve
If the first step is slow, the rate law is
Rate = k1[NO][Cl2]
This matches the experimental rate law, so the first step is the rate-determining step.
We should check to see if the rate law for the mechanism with the second step as the slow step,
Rate = k2[NOCl2][NO]
might also give the experimental rate law. If the second step is slow, then
Rate1 = rate–1
k1[NO][Cl2] = k–1[NOCl2]
k
[NOCl2 ] = 1 [ NO] [ Cl2 ]
k–1
Substituting this into the rate-law expression gives
⎛ k
⎞
2
Rate = k2 ⎜ 1 [ NO][Cl2 ] ⎟ [ NO] = k [ NO] [Cl2 ]
k
⎝ –1
⎠
This does not match the experimental rate law, so the second step in the mechanism is not the ratedetermining step.
Think about It
Just because the rate law for a mechanism matches the experimental rate law does not mean the mechanism
is the mechanism. Another possible mechanism might also give the same experimental rate law.
15.103. Collect and Organize
From the mechanisms given, we are to determine which are possible for the thermal decomposition and
which are possible for the photochemical decomposition of NO2. We are given the rate laws: for the thermal
decomposition reaction, Rate = k[NO2]2; for the photochemical decomposition, Rate = k[NO2].
182 | Chapter 15
Analyze
Using the slowest elementary step in the mechanism, we can write the rate-law expression for each of the
mechanisms and then determine which is consistent with the order of the reaction given for each process.
Solve
For mechanism a, the first step in the mechanism is slow, so the rate law is
Rate = k[NO2]
For mechanism b, the second step in the mechanism is slow, so the rate law is
Rate = k2[N2O4]
Using the first step to express [N2O4] in terms of the concentrations of the reactant NO2 gives
k1[NO2]2 = k–1[N2O4]
k1
2
NO2 ]
[
k–1
Substituting into the rate expression from the second step,
kk
2
2
Rate = 2 1 [ NO2 ] = k [ NO2 ]
k–1
For mechanism c, the first step in the mechanism is slow so the rate law is
Rate = k[NO2]2
Therefore, mechanisms b and c are consistent with the thermal decomposition of NO2 and mechanism a is
consistent with the photochemical decomposition of NO2.
[N 2 O4 ] =
Think about It
To distinguish between the two possible mechanisms for thermal decomposition, we might try to detect the
different intermediates formed in each. Detection of the formation of N2O4 would support mechanism b over
mechanism c.
15.105. Collect and Organize
We are asked if a catalyst affects both the rate and the rate constant of a reaction.
Analyze
A catalyst speeds up a reaction by providing an alternate pathway (mechanism) to the products having a
lower activation energy.
Solve
Yes. Because the rate of the reaction is faster (affecting the rate) and the activation energy is lowered
(affecting the value of k), a catalyst affects both the rate of the reaction and the value of the rate constant.
Think about It
A “negative” catalyst that slows down a reaction would increase Ea and decrease k for a reaction. We call
these “negative catalysts” inhibitors.
15.107. Collect and Organize
We were asked if a substance (catalyst) that increases the rate of a reaction also increases the rate of the
reverse reaction.
Analyze
A catalyst speeds up a reaction by providing an alternate pathway (mechanism) to the products having a
lower activation energy.
Solve
Yes, both the reverse and forward reaction rates are increased when a catalyst is added to a reaction. The
activation energies of both processes are lowered by the different pathway the catalyst provides for the
reaction.
Chemical Kinetics | 183
Think about It
Likewise, an inhibitor would decrease the rates of both forward and reverse reactions.
15.109. Collect and Organize
We are to explain why the concentration of a homogeneous catalyst does not appear in the rate law.
Analyze
A catalyst is used in a reaction and later regenerated.
Solve
The concentration of a homogeneous catalyst may not appear in the rate law because the catalyst itself is not
involved in the rate-limiting step.
Think about It
If the catalyst is involved in the slowest step of the mechanism, however, it is involved in the rate law.
15.111. Collect and Organize
Given the mechanism for the reaction of NO and N2O to form N2 and O2, we are to determine whether NO or
N2O is used in the reaction as the catalyst.
Analyze
A catalyst is used in a reaction and later regenerated and provides a lower energy pathway to the products by
lowering the activation energy of the reaction, thereby speeding up the reaction.
Solve
We can assume that the presence of either NO or N2O, if either is a catalyst, increases the rate of the
reaction. In examining the mechanism we see that N2O is a reactant, not a catalyst, but that NO is a catalyst
because it is used in the reaction and then regenerated. Thus, NO is a catalyst for the decomposition of N2O.
Think about It
If the slow step of this mechanism were the first step, the rate law would be
Rate = k[NO][N2O]
If the second step were slow, the rate law would be
2
NO] [ N 2 O]
[
Rate = k
2
[ N2 ]
2
15.113. Collect and Organize
Using the Arrhenius equation we can compute and compare the rate constants for the reaction of O3 with O
versus the reaction of O3 with Cl.
Analyze
We are given values A and Ea for each reaction at 298 K. The Arrhenius equation is
ln k =
– Ea
+ ln A
RT
Solve
For the reaction of O3 with O,
–17.1 × 103 J/mol
+ ln 8.0 × 10 –12 cm3 /molecule ⋅ s
8.314 J/mol ⋅ K × 298 K
ln k = –32.45
ln k =
k = 8.05 × 10 –15 cm3 /molecule ⋅ s
For the reaction of O3 with Cl,
(
)
184 | Chapter 15
–2.16 × 103 J/mol
+ ln 2.9 × 10 –11 cm3 /molecule ⋅ s
8.314 J/mol ⋅ K × 298 K
ln k = –25.14
ln k =
(
)
k = 1.21 × 10 –11 cm3 /molecule ⋅ s
Therefore, the reaction of O3 with Cl has the larger rate constant.
Think about It
Our answer is consistent with a qualitative look at the activation energies and frequency factors for the two
reactions. The higher activation energy and lower frequency factor for the reaction of O3 with O give a
smaller reaction rate constant.
15.115. Collect and Organize
We are to explain why a glowing wood splint burns faster in a test tube filled with O2 than in air.
Analyze
Air is composed of about 21% O2.
Solve
When the concentration of a reactant (O2 for the combustion reaction) increases, the rate of reaction also
increases. As we place the glowing wood in pure O2, the rate of combustion increases.
Think about It
If the wood splint were placed in a test tube filled with argon, the combustion reaction would stop.
15.117. Collect and Organize
We are to explain why a person submerged in cold water is less likely to have a lack of oxygen for a given
period of time than a person submerged in a warm pool.
Analyze
Chemical reactions are slower at colder temperatures than at warmer temperatures.
Solve
The bodily reactions that use O2 are slower at colder temperatures, so the person submerged in an icecovered lake uses less of the already dissolved oxygen in their system than the person in a warm pool.
Think about It
Rapid cooling technology is being investigated at Argonne National Laboratory for use in surgery patients
and heart attack victims to reduce the damage done to cells by lack of oxygen in the blood.
15.119. Collect and Organize
In the case where ratereverse << rateforward we are to consider whether the method to determine the rate law
(initial concentrations and initial rates) would work at other times, not just at the start of the reaction. If so,
we are to specify which concentrations might be used to determine the rate law.
Analyze
The method that uses initial rates and concentrations to determine the rate law is under the condition where
no reverse reaction is occurring.
Solve
Yes, we could use this method at other times, not just t = 0, to determine the rate law if the rate of the reverse
reaction is much slower than the forward reaction as long as [products] << [reactants] at the time so that no
appreciable reverse reaction is occurring.
Think about It
We will see in Chapter 16 that when the rate of the reverse reaction equals the rate of the forward reaction,
the reaction is at equiibrium.
Chemical Kinetics | 185
15.121. Collect and Organize
In the plot of 1/[X]–1/[X]0 as a function of time, t, we are asked how the rate constant, k, can be determined.
Analyze
The plot of 1/[X]–1/[X]0 as a function of time, t, is the plot for a second-order rate equation.
Rate = k [ X ]
2
1
1
= kt +
[ X]
[ X ]0
Solve
In this plot 1/[X]–1/[X]0 divided by t – t0 is the slope of the line which corresponds to k, the reaction rate
constant. All we need to do to determine k from this plot is to determine the slope of the line.
Think about It
The integrated form of the rate law allows us to obtain the value of k from the concentration vs. time data
from a single experiment.
15.123. Collect and Organize
We are asked to explain why an elementary step may not have a rate law that is zero order.
Analyze
An elementary step in a reaction mechanism describes the collisions of molecular or atomic species taking
place in a reaction.
Solve
For an elementary step to take place, there must be some involvement from a molecular or atomic species.
Therefore, there cannot be no dependence (or zero order) of the reactant in an elementary step of a reaction
mechanism.
Think about It
Because most elementary steps are either unimolecular or bimolecular, the rate expressions of elementary
steps are usually first or second order.
15.125. Collect and Organize
Given the balanced equation for the reaction between NO2 and O3 to produce N2O5 and O2, we are asked to
relate the rates of change in [NO2], [O3], [N2O5], and [O2].
Analyze
From the balanced equation the rate of formation of products and consumption of reactants is
Δ [ O3 ] Δ [ N 2 O5 ] Δ [ O 2 ]
1 Δ [ NO 2 ]
Rate = −
=−
=
=
2 Δt
Δt
Δt
Δt
Solve
The rate of consumption of O3 is the same as the rate of formation of N2O5 and O2 and one-half the rate of
consumption of NO2.
Think about It
The rate of consumption of N2O5 is half the rate of formation of NO2 and twice the rate of formation of O2.
15.127. Collect and Organize
We can write the rate law from the order of the decomposition reaction determined in Problem 15.126. From
that we are to calculate the value of the rate constant at the experimental temperature and write the complete
rate-law expression.
186 | Chapter 15
Analyze
From Problem 15.126 we know that the reaction is first order in [N2O5].
Solve
Because this reaction is a first-order decomposition reaction, the rate-law expression is
Rate = k[N2O5]
Using the data in experiment 1 in Problem 15.126,
1.8 × 10–5 M/s = k × 0.050 M
k = 3.6 × 10–4 s–1
The complete rate-law expression is then
Rate = (3.6 × 10–4 s–1) × [N2O5]
Think about It
The data from experiment 2 in Problem 15.126 would give the same value of k.
15.129. Collect and Organize
For the reaction of NO with O3 to produce NO2 and O2 we can use the information that the reaction is first
order in both NO and O3 along with the values of the rate constants at two different temperatures to
determine whether the reaction occurs in a single or many steps and to calculate the activation energy, the
rate of the reaction at another concentration of the reactants, and the rate constants at two other temperatures.
Analyze
To answer the questions we need to use the Arrhenius equation
ln k =
– Ea
+ ln A
RT
and the rate-law expression, which states that the reaction is first order in both NO and O3:
Rate = k[NO][O3]
Solve
(a) Because the rate law where the reaction is first order in both NO and O3 is consistent with that in which
the reaction would occur in a single step, this reaction might indeed occur in a single step.
(b) We can calculate the activation energy for the reaction by comparing the rate constant at the two
temperatures:
– Ea ⎛ 1 ⎞
ln k25o C =
⎜
⎟ + ln A
R ⎝ 298 K ⎠
ln k75o C =
Subtracting ln k at 25˚C from ln k at 75˚C gives
ln k75o C − ln k25o C =
– Ea
R
⎛ 1 ⎞
⎜
⎟ + ln A
⎝ 348 K ⎠
⎛ k75o C
– Ea ⎛ 1
1 ⎞
−
⎜
⎟ = ln ⎜⎜
R ⎝ 348 K 298 K ⎠
⎝ k25o C
⎞
⎟⎟
⎠
⎛ 3000 M –1 s –1 ⎞
– Ea
1 ⎞
⎛ 1
ln ⎜
=
−
⎜
⎟
–1 –1 ⎟
⎝ 80 M s ⎠ 8.314 J/mol ⋅ K ⎝ 348 K 298 K ⎠
Ea = 6.25 × 104 J/mol or 62.5 kJ/mol
(c) We can use the rate-law expression to calculate the rate of the reaction at 25˚C when [NO] = 3 × 10–6 M
and [O3] = 5 × 10–9 M. We are given in the statement of the problem that k at 25˚C is 80 M–1 s–1.
Rate = 80 M–1 s–1 × (3 × 10–6 M) × (5 × 10–9 M) = 1.2 × 10–12 M/s
(d) To use the Arrhenius equation to calculate the values of k at 10˚C and 35˚C we have to first determine the
value of the frequency factor A. To do this we can use the value for Ea and k for 25˚C:
Chemical Kinetics | 187
(
–6.25 × 104 J/mol ⎛ 1 ⎞
⎜
⎟ + ln A
8.314 J/mol ⋅ K ⎝ 298 K ⎠
ln A = 29.608
)
ln 80 M –1 s –1 =
A = 7.22 × 1012
So the value of k at 10˚C (283 K) is
–6.25 × 104 J/mol ⎛ 1 ⎞
12
⎜
⎟ + ln(7.22 × 10 )
8.314 J/mol ⋅ K ⎝ 283 K ⎠
ln k = 3.045
ln k =
k = 21 M –1 s –1
The value of k at 35˚C (308 K) is
–6.25 × 104 J/mol ⎛ 1 ⎞
12
ln k =
⎜
⎟ + ln(7.22 × 10 )
8.314 J/mol ⋅ K ⎝ 308 K ⎠
ln k = 5.20
k = 1.8 × 102 M –1 s –1
Think about It
The values of k at 10˚C and 35˚C calculated above make sense because they are a little lower and a little
higher, respectively, than the value of k at 25˚C. Also, an Arrhenius plot using the two k values at 25˚C and
75˚C could be used to determine the activation energy and the frequency factor values in this problem.
15.131. Collect and Organize
We consider the mechanism for the exchange of H216O around a Na+ cation for H218O in order to write the
rate law for the rate-determining step. We also need to think about the relative energies of the reactants and
products if we are to sketch the reaction-energy profile.
Analyze
(a) We are given that the first step of the reaction is rate determining so this is the step from which we write
the rate-law expression.
(b) In deciding which has the higher energy, the reactants or products, for the reaction profile, we need to
consider the relative strength of the Na+–16OH2 interaction versus that of the Na+–18OH2 interaction.
Solve
(a) Rate = k[Na(H2O)6+]
(b) Neither. The ion–dipole interaction should not be significantly different for H218O versus H216O, so the
energy of the reactants and the products in the reaction profile will be about the same.
Think about It
In reality, there is an isotope effect in which the H218O–Na+ interaction is slightly stronger than the
H216O–Na+ interaction, so the energy of the product in this reaction is slightly lower than the energy of the
reactants.
15.133. Collect and Organize
For the reaction of NO with ONOO– we are to use the provided data to determine the rate law for the
reaction. We are also to draw the Lewis structure with resonance forms to describe the bonding in the
o
ONOO– anions and then use bond energies to estimate ΔH rxn
.
Analyze
(a) To determine the rate law we can compare the effects of changing the concentrations of NO and ONOO–
on the rate of the reaction.
(b) After drawing the resonance forms for ONOO– we can determine which is the preferred structure by
assigning formal charges to the atoms in each resonance form.
188 | Chapter 15
o
(c) The ΔH rxn
may be estimated by bond energies using
o
ΔH rxn
= ∑ ΔH bond breaking +∑ ΔH bond forming
Solve
(a) For the order of reaction with respect to ONOO– we can compare experiments 1 and 2:
–
rate2 ⎛ [ONOO ]0,2 ⎞
= ⎜
⎟
rate1 ⎜⎝ [ONOO – ]0,1 ⎟⎠
x
x
1.02 ×10 –11 M /s ⎛ 0.625 ×10 –4 M ⎞
= ⎜
⎟
2.03 ×10 –11 M /s ⎝ 1.25 ×10 –4 M ⎠
0.502 = 0.500 x
x =1
Notice that there are not two experiments in the data table for which the concentration of ONOO– stays the
same. For the order of reaction with respect to NO, we can compare experiments 2 and 3 as long as we take
into account the knowledge that the reaction is first order in ONOO–. Between these two experiments we see
that as the ONOO– concentration is quadrupled we would expect that the rate would be quadrupled. The rate
of the reaction, however, when the NO concentration is halved simultaneously with the quadrupling of the
rate on changing the ONOO–, doubles:
rate3 ⎛ 2.03 ×10–11 M /s ⎞
= ⎜
⎟ = 1.99
rate2 ⎝ 1.02 ×10–11 M /s ⎠
Because the ∆[ONOO–] = 4, which would quadruple the rate of the reaction, the ∆[NO] must affect the rate
by 1.99/4 = 0.498. Therefore,
x
⎛ 0.625 ×10–4 M ⎞
0.498 = ⎜
⎟
–4
⎝ 1.25 ×10 M ⎠
0.498 = 0.500 x
x =1
The rate law for this reaction is Rate = k[NO][ONOO–].
We can use any experiment to calculate the value of k. Using the data from experiment 1:
2.03 × 10–11 M/s = k × (1.25 × 10–4 M) × (1.25 × 10–4 M)
k = 1.30 × 10–3 M–1s–1
(b)
(c)
Bonds broken = O — O (146 kJ/mol)
Bonds formed = N— O (201 kJ/mol)
ΔHrxn = [(1 mol × 146 kJ/mol) + (1 mol × –201 kJ/mol)] = –55 kJ
Think about It
To solve this problem you had to draw on several concepts you have learned so far in this course. You had to
review not only how to write a rate law given kinetic data, but also how to draw resonance structures and
how bond energies might be used to estimate the enthalpy of a reaction.
15.135. Collect and Organize
Using data for [HNO2] over time, we can determine the order of the isotopic exchange reaction with respect
to [HNO2]. We are also asked whether the rate of the reaction will depend on [H218O].
Chemical Kinetics | 189
Analyze
We are given a single data set. If the plot of [HNO2] versus time yields a straight line, then the reaction is
zero order. If the plot of ln[HNO2] versus time yields a straight line, then the reaction is first order. If the plot
of 1/[HNO2] versus time yields a straight line, then the reaction is second order.
Solve
(a)
This reaction is second order in [HNO2].
(b) Because the reaction mixture has a very large [H218O], we cannot observe a change in [H218O] over time.
The rate might be dependent on [H218O], but we cannot tell from the information given.
Think about It
If it were possible, we could place the reaction in a nonreactive solvent and then vary the [H218O] over time
to determine the rate’s dependence on its concentration.
15.137. Collect and Organize
Using the raw data obtained for four experiments in which the concentrations of NH2 and NO were varied,
we are to write the rate law and determine the value of k for the reaction between NH2 and NO at 1200 K.
Analyze
To determine the rate law we can compare the effects of changing the concentrations of NH2 and NO on the
rate of the reaction. Once we have determined the order of the reaction with respect to each reactant, we can
write the rate-law expression and use any of the experiments to calculate the value of the rate constant, k.
Solve
(a) For the order of the reaction with respect to NH2, we can use the data from experiments 1 and 2:
190 | Chapter 15
rate2 ⎛ [NH 2 ]0,2
= ⎜
rate1 ⎜⎝ [NH 2 ]0,1
⎞
⎟⎟
⎠
x
x
0.24 M /s ⎛ 2.00 ×10 –5 M ⎞
= ⎜
⎟
0.12 M /s ⎝ 1.00 ×10 –5 M ⎠
2.0 = 2.00 x
x =1
For the order of the reaction with respect to NO, we can use the data from experiments 2 and 3:
rate3 ⎛ [NO]0,3
= ⎜
rate2 ⎜⎝ [NO]0,2
⎞
⎟⎟
⎠
x
x
0.36 M /s ⎛ 1.50 ×10 –5 M ⎞
= ⎜
⎟
0.24 M /s ⎝ 1.00 × 10 –5 M ⎠
1.5 = 1.50 x
x =1
The rate-law expression is Rate = k[NH2][NO].
(b) Using experiment 1 to calculate the value of k,
0.12 M/s = k × (1.00 × 10–5 M) × (1.00 × 10–5 M)
k = 1.2 × 109 M–1 s–1
Think about It
Any of the experiments listed in the data table would give us the same value of k in the calculation in part b.
CHAPTER 16 | Chemical Equilibrium
16.1. Collect and Organize
For two reversible reactions we are given the reaction profiles (Figure P16.1). The profile for the conversion of
A to B shows that reactant A has a lower free energy than product B. The profile for the conversion of C to D
shows that C has a higher free energy than D. From the profiles we are to determine which reaction has the
larger kf, the smaller kr, and the larger value of Kc.
Analyze
The magnitude of the rate constant is inversely related to the magnitude of the activation energy, Ea. The
reaction with the larger kf, smaller kr, and the larger Kc is that reaction with the lowest Ea for the forward
reaction and where kf > kr.
Solve
The reaction C É D has the larger kf, the smaller kr, and the larger Kc (because it has the lowest activation
energy for the forward direction).
Think about It
Remember that a large k (rate constant) means that the reaction is fast and therefore the reaction has a low
activation energy.
16.3. Collect and Organize
From Figure P16.3 showing 26 blue spheres (product B) and 13 red spheres (reactant A), we are to write the
chemical equation of the equilibrium reaction and calculate the value of Kc.
Analyze
(a) In this reaction A is transformed into B. We represent a system at equilibrium using double-headed reaction
arrows between the reactants and products. We will assume that one molecule of A produces one molecule of
B in the reaction.
(b) The value of Kc is the ratio of the concentration of the products (number of B spheres) and the
concentration of the reactants (number of A spheres) raised to their respective stoichiometric coefficients from
the balanced chemical equation.
Solve
(a) A  B
26
(b) K c =
= 2.0
13
Think about It
If the chemical equation were written as 2A É B, then Kc would be
Kc =
[B]
26
=
= 0.15
[A]2 (13) 2
16.5. Collect and Organize
By comparing the relative distributions of reactants A and B with product AB at two different temperatures as
shown in Figure P16.5, we can determine whether the reaction is endothermic or exothermic.
Analyze
From the equation
ΔG = – RT ln K = ΔH – T ΔS
we see that as temperature rises for an exothermic reaction, ΔG becomes less negative and therefore K
decreases. If, however, the temperature is raised on an endothermic reaction ΔG becomes more negative and K
224
Chemical Equilibrium | 225
increases. Therefore, if products increase upon raising, the temperature the reaction is endothermic; if products
decrease, the reaction is exothermic.
Solve
At 300 K the equilibrium mixture is 6 A, 10 B, and 5 AB. This gives an equilibrium constant of
K300 K =
5
= 0.083
6 × 10
At 400 K, the equilibrium mixture is 3 A, 7 B, and 8 AB. This gives an equilibrium constant of
8
K 400 K =
= 0.38
3× 7
As temperature increases for this reaction, K increases, indicating that more products form at higher
temperatures, so this reaction is endothermic.
Think about It
We assume in this problem that the difference in entropy for the two temperatures at which this reaction is run
is minimal, so only ∆H contributes to the difference in ∆G at the two temperatures.
16.7. Collect and Organize
We are asked to define chemical equilibrium in terms of the rates of the forward and reverse reactions.
Analyze
Equilibrium is a state where the composition of the reaction is not changing.
Solve
A reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. For
this system, the composition of the reaction is unchanging.
Think about It
Chemical equilibrium is a dynamic process. At the molecular level the forward and reverse reactions are still
occurring. Because they occur at the same rate, however, we observe no macroscopic changes in the
concentrations of the reactants and products in the mixture.
16.9. Collect and Organize
From the graph showing the concentrations of both B and A over time (Figure P16.9), we are to determine if
the reaction has reached equilibrium at 20 µs.
Analyze
At equilibrium the rate of the reaction to produce B from A is equal to the rate of the reaction to produce A
from B. The concentrations of A and B at equilibrium are constant.
Solve
No. At 20 µs the concentrations of A and B in Figure P16.9 are still changing (there is a nonzero
instantaneous rate of reaction) so at this time the reaction has not yet reached equilibrium.
Think about It
The concentrations of both A and B level off at about 50 µs. After this time, the reaction is at equilibrium.
16.11. Collect and Organize
For a reaction where kf > kr we are to determine whether K is greater than, less than, or equal to 1.
Analyze
The equilibrium constant defined in terms of the rate constants for a reaction is
K=
kf
kr
226 | Chapter 16
Solve
When kf > kr, K is greater than 1.
Think about It
When K > 1 more products are present at equilibrium than reactants.
16.13. Collect and Organize
For the decomposition of N2O to N2 and O2, we are to identify the species present after one day from the
given molar masses. The initial reaction mixture contains 15N2O, N2, and O2.
Analyze
The only isotope in the reaction for oxygen is 16O but for nitrogen there is both 15N and 14N present at the
beginning of the reaction. After one day, the 14N will be incorporated into N2O and 15N will be incorporated
into N2.
Solve
Molar Mass
Compound
How Present
28
29
30
32
44
45
46
14
Originally present
From decomposition of 15N14NO
From decomposition of 15N2O
Originally present
From combination of 14N2 and O2
From combination of 15N14N and O2
Originally present
N2
N14N
15
N2
O2
14
N 2O
15 14
N NO
15
N 2O
15
Think about It
The redistribution of 15N from N2O to N2 and 14N from N2 to N2O demonstrates that both forward and reverse
reactions occur in a dynamic equilibrium process.
16.15. Collect and Organize
From the rate laws and rate constants for the forward and reverse reactions for
AB
we are to calculate the value of the equilibrium constant.
Analyze
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium
constant is the ratio of the concentrations of the products raised to their coefficients to the concentrations of
the reactants raised to their respective coefficients.
Solve
For the reaction A É B we are given that the forward reaction is first order in A:
Ratef = k1[A], where k1 = 1.50 × 10–2 s–1
and that the reverse reaction is first order in B:
Rater = k–1[B] where k–1 = 4.50 × 10–2 s–1
At equilibrium, ratef = rater so
k1[A] = k–1[B]
Rearranging this to give the equilibrium constant expression allows us to solve for K:
[ B] = k1 = 1.50 × 10–2 s –1 = 0.333
Kc =
[ A] k–1 4.50 × 10–2 s –1
Think about It
For the reverse equilibrium
BA
Chemical Equilibrium | 227
the equilibrium constant is 1/0.333 or 3.00.
16.17. Collect and Organize
From the equation relating Kc and Kp we can determine under what conditions Kc = Kp.
Analyze
The equation relating Kc and Kp is
Kp = Kc (RT )Δn
Solve
Kp equals Kc when Δn = 0. This is true when the number of moles of gaseous products equals the number of
moles of gaseous reactants in the balanced chemical equation.
Think about It
The value of Kp may also be less than Kc (for Δn < 0) or greater than Kc (for Δn > 0).
16.19. Collect and Organize
Given three reactions involving nitrogen oxides, we are to write Kc and Kp expressions for each reaction.
Analyze
The Kc expression uses concentration (molarity) units for the reactants and the products while the Kp
expression uses partial pressure. The K expression for a balanced chemical reaction takes the general form
wA + xB  yC + zD
[C] y [D]z
Kc =
[A]w [B]x
y
z
PC ) ( PD )
(
Kp =
( PA )w ( PB ) x
and
Solve
(
)
(a) Kc =
PN2 O4
[N 2 O4 ]
and K P =
2
[N 2 ][O2 ]
PN2 PO2
(b) K c =
[NO 2 ][N 2 O]
and
[NO]3
2
( )( )
( P )( P )
K =
NO 2
p
( PNO )
N2O
3
2
(c) K c =
( )( )
( )
PN2 PO2
[N 2 ]2 [O 2 ]
and K p =
2
2
[N 2 O]
PN2 O
Think about It
Kc uses concentration units (usually in mol/L), whereas Kp uses partial pressure units (usually in atm).
16.21. Collect and Organize
Given a plot of the concentration versus time for the decomposition of N2O to N2 and O2 (Figure P16.21), we
are to estimate the value of K.
Analyze
The amounts of N2O, N2, and O2 are given in concentration units and the form of the Kc expression is
1/ 2
Kc =
[ N ][ O ]
[ N O]
2
2
2
The concentrations of each species at equilibrium can be read from the graph as those concentrations that are
no longer changing with time. This occurs after 5 s and gives [N2] = 0.00030 M, [O2] = 0.00014 M, and
[N2O] = 7.10 × 10–6 M.
228 | Chapter 16
Solve
1/ 2
Kc
[ N ][ O ]
=
[ N O]
2
2
=
2
(0.00030 M )(0.00014 M )1/ 2
= 0.50
(7.10 × 10−6 M )
Think about It
Because this equilibrium constant is neither high nor low, at equilibrium the distribution of the reactants and
products is expected to be roughly equal. A general rule is that if K < 0.01, the reactants are favored in the
equilibrium and if K > 100 then the products are favored in the equilibrium.
16.23. Collect and Organize
Given the balanced chemical equation and the equilibrium partial pressures of all species in the decomposition
reaction of BrCl to give Br2 and Cl2, we are to calculate the value of Kp for the reaction.
Analyze
The form of the Kp expression for this reaction is
Kp =
( P )( P )
Br2
Cl2
( PBrCl )2
Solve
Kp =
( P )( P ) = (0.33 atm)(0.33 atm) = 4.8
Br2
Cl2
( PBrCl )2
(0.15 atm) 2
Think about It
Remember that equilibrium constants, K, do not have any units.
16.25. Collect and Organize
Given the equilibrium molar concentrations of N2, O2, and NO we are to calculate Kc for
N 2 (g) + O 2 (g)  2 NO(g)
Analyze
The equilibrium constant expression for this reaction is
Kc =
[ NO]2
[ N 2 ][ O2 ]
Solve
–3
Kc
2
(3.1 × 10 M )
=
(3.3 × 10 M )(5.8 × 10
−3
−3
M
)
= 0.50
Think about It
Be careful to account for the coefficients in the mass action equation when calculating Kc. In this problem, we
must be sure to square the equilibrium concentration of NO.
16.27. Collect and Organize
Given the initial moles of H2O and CO and the moles of CO2 present at equilibrium, we are to determine Kc for
H 2O(g) + CO(g)  H 2 (g) + CO 2 (g)
Chemical Equilibrium | 229
Analyze
Because the ratios of reactants and products are 1:1:1:1 we do not need the volume of the vessel because that
volume would cancel in the final Kc expression. We can therefore simply use the mole amounts. Furthermore,
we can calculate the amounts of reactants and products present at equilibrium from the initial amounts of the
reactants, the given moles of the product CO2 formed (8.3 × 10–3mol), and the stoichiometry of the reaction.
Solve
Initial (mol)
Change (mol)
Equilibrium (mol)
H 2O
0.0150
–x
0.0150 – x
CO
0.0150
–x
0.0150 – x
H2
0
+x
x
CO2
0
+x
x
At equilibrium, we know that [CO2] = x = 8.3 × 10–3 mol. This gives
H2O = CO = 0.0150 – x = 0.0067 mol
H2 = CO2 = x = 8.3 × 10–3 mol
[ H ][ CO ] = (8.3 × 10 mol)(8.3 × 10 mol) = 1.5
[ H O][ CO] ( 0.0067 mol)(0.0067 mol)
–3
Kc =
2
–3
2
2
Think about It
We must have a balanced chemical reaction so that we can stoichiometrically relate the quantity of products to
the reactants and correctly calculate the value of the equilibrium constant.
16.29. Collect and Organize
We are given Kp = 32 for the following reaction at 298 K:
A(g) + B(g)  AB(g)
We are to calculate the value of Kc, which we can do using the relationship
K p = K c ( RT )
Δn
Analyze
The change in the number of moles of gas for this reaction (Δn) is
1 mol AB – (1 mol A + 1 mol B) = –1
Solve
L ⋅ atm
⎛
⎞
32 = K c × ⎜ 0.08206
× 298 K⎟
⎝
⎠
mol ⋅ K
−1
K c = 780
Think about It
When Δn is positive, Kp is greater than Kc, but when Δn is negative, Kp is less than Kc. When Δn is zero, Kp
equals Kc.
16.31. Collect and Organize
We are given Kp = 1.45 × 10–5 for the following reaction at 500˚C:
N 2 (g) + 3 H 2 (g)  2 NH 3 (g)
We are to calculate the value of Kc.
Analyze
Rearranging the equation relating Kp and Kc to solve for Kc gives
Kp
Kc =
(RT ) Δn
The change in the number of moles is
230 | Chapter 16
(2 mol NH3) – (1 mol N2 + 3 mol H2) = –2
The temperature in kelvins is 500 + 273 = 773 K.
Solve
Kc =
1.45 × 10–5
L ⋅ atm
⎛
⎞
× 773 K⎟
⎜⎝ 0.08206
⎠
mol ⋅ K
–2
= 0.0583
Think about It
Here, Kp < Kc because Δn is negative. When Δn = 0, Kp = Kc.
16.33. Collect and Organize
We are asked to determine in which of the three reactions the values of Kc and Kp are equal.
Analyze
Kp and Kc are equal when Δn = 0 in the relationship
K p = K c ( RT )
Δn
where Δn is the change in the number of moles of gas in the reaction:
Δn = (moles of gaseous products) – (moles of gaseous reactants)
Solve
(a) Δn = 2 – 3 = –1
(b) Δn = 1 – 1 = 0
(c) Δn = 2 – 2 = 0
Reactions b and c have Kp = Kc. Only reaction a has Kp ≠ Kc.
Think about It
Remember that Δn is the difference in the moles of gaseous products and gaseous reactants. In reaction b
Fe(s) and FeO(s) are not considered.
16.35. Collect and Organize
Given Kc = 5.0 for the following reaction at 327˚C (600 K)
Cl 2 (g) + CO(g)  COCl 2 (g)
we are to calculate the value of Kp at 325˚C. We can use the relationship
K p = K c ( RT )
Δn
Analyze
For this reaction
Δn = (1 mol COCl2) – (1 mol Cl2 + 1 mol CO) = –1
Solve
L ⋅ atm
⎛
⎞
K p = 5.0 × ⎜ 0.08206
× 598 K ⎟
⎝
⎠
mol ⋅ K
–1
= 0.10
Think about It
Because Δn is negative, the value of Kp is less than that of Kc.
16.37. Collect and Organize
We consider how K changes when the coefficients of a balanced reaction are scaled up or down.
Analyze
To answer the question let’s consider the following reactions:
Chemical Equilibrium | 231
(1) 2 A + B  C
(2) 4 A + 2 B  2 C
The equilibrium constant expressions for the reactions are
[C]
[C]2
K1 =
and
K
=
2
[A]2 [B]
[A]4 [B]2
These are related by
K2 = (K1)2
Solve
When scaling the coefficients of a reaction up or down the new value of the equilibrium constant is the first K
raised to the power of the scaling constant.
Think about It
Scaling reaction 1 by 13 gives
2
1
1
A+ B  C
3
3
3
13
[C]
or (K1 ) 1
K=
3
[A]1 3 [B]1 3
16.39. Collect and Organize
Given the equilibrium constant for the reaction of 1 mol I2(g) with 1 mol Br2(g) to give 2 mol IBr(g), we are to
calculate the value of the equilibrium constant for the reaction of 12 mol of I2 and Br2 to give 1 mol IBr.
Analyze
The Kc expressions for these reactions are
2
IBr ]
[
K c1 =
[ I2 ][ Br2 ]
and
K c2 =
[IBr ]
12
[I2 ] [Br2 ]
12
Kc2 for the second reaction is, therefore, related to that of the first by
12
K c2 = ( K c1 )
Solve
12
K c2 = ( K c1 )
= (120)1 2 = 11.0
Think about It
When we multiply a chemical reaction by a number, the new value of the equilibrium constant is the first
equilibrium constant raised to the number.
16.41. Collect and Organize
For the reaction of NO with NO3 we are to explain how Kc for the reverse reaction relates to Kc of the forward
reaction by writing their equilibrium constant expressions.
Analyze
The form of the equilibrium constant expression for the forward reaction is
[NO2 ]2
K c,forward =
[NO][NO3 ]
For the reverse reaction, the equilibrium constant expression is
[NO][NO3 ]
Kc,reverse =
[NO2 ]2
Solve
Examining these two expressions we see that
232 | Chapter 16
K c,reverse =
1
K c,forward
Think about It
Another way to think about this relationship is
Kc,forward × Kc,reverse = 1
16.43. Collect and Organize
For two reactions of different stoichiometry for the reaction of SO2 with O2 to produce SO3, we are to write
the equilibrium constant expressions and explain how they are related.
Analyze
Equilibrium constant expressions take the form
wA + xB  yC + zD
Kc =
Solve
The equilibrium expressions for the reactions are
[SO3 ]
Kc =
[SO2 ][O2 ]1 2
These expressions are related by
[C]y [D]z
[A]w [B]x
and
K cʹ′ =
K cʹ′ = ( K c )
[SO3 ]2
[SO 2 ]2 [O 2 ]
2
Think about It
If the second reaction were
1
2
then
SO 2 (g) + 14 O 2 (g)  12 SO3 (g)
K cʹ′ʹ′ =
[SO3 ]1 2
[SO 2 ]1 2 [O 2 ]1 4
16.45. Collect and Organize
Given the value of Kc for the reaction
2 SO 2 (g) + O 2 (g)  2 SO3 (g)
–3
as 2.4 × 10 , we are to calculate Kc for three other forms of this reaction at the same temperature.
Analyze
When a reaction is multiplied by a factor x, the new equilibrium constant is (Kc)x. When a reaction is reversed,
the new equilibrium constant is 1/Kc.
Solve
(a) This reaction is the original equation multiplied by 12 . The new Kc is
–3 1 2
( 2.4 ×10 )
= 4.9 ×10–2
(b) This reaction equation is the reverse of the original equation. The new Kc is
1 2.4 × 10 –3 = 420
(
)
Chemical Equilibrium | 233
(c) This reaction equation is the reverse of the original equation multiplied by 12 . The new Kc is
12
1 ( 2.4 ×10–3 )
= 20
Think about It
Because we can relate the equilibrium constants for different forms of a reaction, we need to only tabulate one
of the equilibrium constants. All others can be calculated from that value.
16.47. Collect and Organize
We are asked to calculate K for a reaction given Kc values for two other reactions. We can add the two
equations together and reverse the resulting equation and then calculate the equilibrium constant for this new
chemical equation.
Analyze
The Kc expressions for the reactions and the overall reaction are
A+2 B C
K1 =
!C#
" $
= 3.3
2
!A#!B#
" $" $
2
C2D
! D#
" $
K2 =
= 0.041
!C#
" $
2
! D#
" $
A+2 B 2 D
K3 =
2
!A#!B#
" $" $
where K3 = K1 × K2. The equilibrium constant for the reaction
2 D  A+2 B
is the inverse of K3.
Solve
K=
1
= 7.4
0.041 × 3.3
Think about It
When adding reactions, the new equilibrium constant is the product of the equilibrium constants of the
reactions that were combined.
16.49. Collect and Organize / Analyze
We are to define the term reaction quotient.
Solve
The reaction quotient is the ratio of the concentrations of the products raised to their stoichiometric
coefficients to the concentrations of reactants raised to their stoichiometric coefficients. The reaction quotient
has the same form as the equilibrium constant expression, but the reaction concentrations (or partial pressures)
are not necessarily at their equilibrium values.
Think about It
If the reaction quotient, Q, is greater than K, then the reaction mixture must reduce its concentration of
products to attain equilibrium. When Q is less than K, the reaction mixture must increase its concentration of
products to attain equilibrium.
16.51. Collect and Organize
We are asked what it means when Q = K.
234 | Chapter 16
Analyze
Both K and Q take the form of the ratio of the concentrations of products raised to their stoichiometric
coefficients to the concentrations of reactants raised to their stoichiometric coefficients.
Solve
When Q = K the system is at equilibrium.
Think about It
Whenever Q ≠ K, the reaction is not at equilibrium and it adjusts its relative concentrations of reactants and
products so that Q = K.
16.53. Collect and Organize
We are asked whether the reaction
A(aq)  B(aq)
where [A(aq)] = 0.10 M, [B(aq)] = 2.0 M, and K = 22 is at equilibrium. If the reaction is not at equilibrium,
we are to state in which direction the reaction will proceed to reach equilibrium.
Analyze
The reaction quotient (Q) is
[ B] = Q where [B] and [A] are the concentrations of A and B in the reaction mixture.
[ A]
If Q > K, the reaction will proceed to the left to reach equilibrium. If Q < K, the reaction will proceed to the
right. If Q = K, the reaction is at equilibrium.
Solve
2.0 M
= 20
0.10 M
No, the reaction is not at equilibrium. Q < K, so this reaction proceeds to the right to reach equilibrium.
Q=
Think about It
In this reaction, more B forms as the reaction proceeds to equilibrium.
16.55. Collect and Organize
Given two sets of reactant and product concentrations for the reaction of A and B to form C (where Kp =
1.00), we are to determine whether either reaction mixture is at equilibrium. The temperature is 300 K.
Analyze
These systems are at equilibrium when Q = Kp = 1.00. For the reaction where A, B, and C are expressed in
terms of molarity, we must convert Kp to Kc using
K p = K c ( RT )
Δn
where for this reaction Δn = –1.
Solve
1.0 atm
= 1.0
1.0
atm
(
)(1.0 atm)
This reaction mixture is at equilibrium.
(b) Rearranging the equation to solve for Kc gives
Kp
1.00
Kc =
=
= 24.6
Δn
–1
( RT )
⎡⎛
L ⋅ atm ⎞
⎤
⎢⎜⎝ 0.08206 mol ⋅ K ⎟⎠ × 300 K ⎥
⎣
⎦
(a) Qp =
Chemical Equilibrium | 235
1.0 M
= 1.0
(1.0 M )(1.0 M )
Because Qc < Kc this reaction mixture is not at equilibrium and shifts to the right to attain equilibrium.
Qc =
Think about It
Both Kp and Kc are close to 1 and so these reaction mixtures have roughly equal proportions of reactants and
products when they reach equilibrium.
16.57. Collect and Organize
By comparing Q versus K for the reaction of N2 with O2 to form NO, we can determine in which direction the
reaction will proceed to attain equilibrium.
Analyze
The form of the reaction quotient for this reaction is
Qp =
( PNO )2
( P )( P )
N2
O2
–3
Because ∆n = 0 for this reaction Kp = Kc = 1.5 × 10 .
Solve
–3 2
Qp
(1.00 × 10 )
=
= 1.00
(1.00 × 10 )(1.00 × 10 )
–3
–3
Because Q > K, the system is not at equilibrium and proceeds to the left.
Think about It
Notice that for this problem Kp = Kc. This is not always the case, so be careful to notice which value is
provided for K and in what units the amounts of the reactants and products are expressed.
16.59. Collect and Organize
Given initial concentrations of the reactants X and Y and product Z and the value of the equilibrium constant
(Kc = 1.00 at 350 K), we can use the reaction quotient to determine in which direction the reaction will shift to
reach equilibrium.
Analyze
The reaction quotient for this reaction has the form
Qc =
[Z]
[X][Y]
Solve
[Z]
0.2 M
=
=5
[X][Y] 0.2 M × 0.2 M
Because Qc > Kc the reaction proceeds to the left (a) producing more X and Y.
Qc =
Think about It
To calculate the equilibrium concentration of each reactant and product, we solve for x in the equation
0.2 − x
1.00 =
( 0.2 + x )2
16.61. Collect and Organize
For the dissolution of CuS to give Cu2+ and S2– in aqueous solution, we are asked to write the Kc expression.
236 | Chapter 16
Analyze
Because any pure solids or liquids do not change in concentration over the course of a reaction, their
“concentrations” do not appear in the mass action (equilibrium constant) expression.
Solve
Kc = [Cu2+][S2–]
Think about It
Because the reactant in the dissolution reaction is a pure solid, only the concentrations of the products
influence the position of the equilibrium.
16.63. Collect and Organize
For the decomposition reaction of CaCO3 to CO2 and CaO, we are to explain why [CaCO3] and [CaO] do not
appear in the Kc expression.
Analyze
The Kc expression includes the concentrations of the reactants and products that may change during the course
of the reaction attaining equilibrium.
Solve
The concentrations (expressed as densities) of pure solids (CaCO3 and CaO) do not change during the reaction
to attain equilibrium and so they do not appear in the equilibrium constant expression.
Think about It
The form of Kc for this reaction is simply Kc = [CO2].
16.65. Collect and Organize
We are asked whether the value of K increases when more reactants are added to a reaction already at
equilibrium.
Analyze
The general form of the equilibrium constant expression for a reaction is
wA + xB  yC + zD
Kc =
[C] y [D]z
[A]w [B]x
Solve
No, the equilibrium constant is not changed when the concentration of the reactants is increased. The relative
concentrations of the reactants and products in that case adjust until they achieve the value of K. The value of
the equilibrium constant is only affected by temperature.
Think about It
The value of the reaction quotient Q decreases below the value of K when reactants are added to a system
previously at equilibrium, and the reaction shifts to the right.
16.67. Collect and Organize
Given that the K for the binding of CO to hemoglobin is larger than that for the binding of O2 to hemoglobin,
we are to explain how the treatment of CO poisoning by administering pure O2 to a patient works.
Analyze
By giving a patient pure O2 to breathe we increase the partial pressure of O2 to which the hemoglobin in the
patient’s blood is exposed. This oxygen can displace the CO bound to the hemoglobin through application of
Le Châtelier’s principle.
Chemical Equilibrium | 237
Solve
Combining the two equations, we can write the expression for the displacement of CO bound to hemoglobin
by O2:
Hb(CO) 4  Hb + 4 CO(g)
Hb + 4 O 2 (g)  Hb(O 2 ) 4
Hb(CO) 4 + 4 O 2 (g)  Hb(O 2 ) 4 + 4 CO(g)
As the concentration of O2 increases the reaction shifts to the right and the CO on the hemoglobin is
displaced.
Think about It
If we were given the values of the equilibrium constants for the reactions, we could calculate the new
equilibrium constant for the overall reaction.
16.69. Collect and Organize
We are to interpret Henry’s law through Le Châtelier’s principle.
Analyze
The dissolution of a gas (let’s use oxygen in this example) in a liquid (let’s use water) can be written as a
chemical equation:
O 2 (g)  O 2 (aq)
Solve
According to Le Châtelier’s principle an increase in the partial pressure (or concentration) of O2 above the
water shifts the equilibrium to the right so that more oxygen becomes dissolved in the water. This is consistent
with Henry’s law.
Think about It
The solubilities of different gases in a liquid are different from each other, but all gases are more soluble in a
liquid when present at higher partial pressures.
16.71. Collect and Organize
Of four reactions we are to determine which will shift its equilibrium so as to form more products when the
mixture is compressed to half of its original volume.
Analyze
Decreasing the volume by half doubles the partial pressures of all the gaseous reactants and products in the
reactions. This would cause a shift in the position of the equilibrium toward the side of the reaction with the
fewer number of moles of gas.
Solve
(a) This equilibrium shifts to the left, forming more reactants.
(b) This equilibrium shifts to the right, forming more products.
(c) This equilibrium shifts neither to the left nor to the right since the number of moles of gas is the same on
both sides of the equation.
(d) This equilibrium shifts to the right, forming more products.
Reactions b and d will form more products when the volume of the mixture is decreased by half.
238 | Chapter 16
Think about It
The opposite shifts occur in the position of the equilibrium when the volumes of the reaction mixtures are
increased.
16.73. Collect and Organize
We are to predict the effect on the position of the equilibrium
2 O3 (g)  3 O 2 (g)
with various changes in concentration and volume.
Analyze
Increasing the concentration of a reactant shifts the equilibrium to the right whereas increasing the
concentration of a product shifts the equilibrium to the left. An increase in volume decreases the pressure and
shifts the equilibrium toward the side of the reaction with the greater moles of gas.
Solve
(a) Increasing the concentration of the reactant O3 shifts the equilibrium to the right, increasing the
concentration of the product O2.
(b) Increasing the concentration of the product O2 shifts the equilibrium to the left, increasing the
concentration of the reactant O3.
(c) Decreasing the volume of the reaction to 1/10 its original volume shifts the equilibrium to the left,
increasing the concentration of the reactant O3.
Think about It
Adding O2 and decreasing the volume have the same shift in the position of the equilibrium.
16.75. Collect and Organize
We are to determine how decreasing the partial pressure of O2 affects the equilibrium
2 SO 2 (g) + O 2 (g)  2 SO3 (g)
Analyze
According to Le Châtelier’s principle, increasing the partial pressure or concentration of a reactant shifts the
equilibrium to the right. Decreasing the partial pressure of a reactant, then, shifts the equilibrium to the left.
Solve
Decreasing the partial pressure of O2 in this reaction shifts the equilibrium to the left.
Think about It
At equilibrium then, we have less SO3 product when we reduce the partial pressure of O2 in the reaction
mixture.
16.77. Collect and Organize
For three processes at equilibrium we are to determine which shows an increased yield of product C at
increasing temperatures.
Analyze
Endothermic processes have heat as a reactant. When the temperature is raised for these reactions the
equilibrium shifts to the right, increasing the amount of product C formed.
Chemical Equilibrium | 239
Solve
Reaction a is the only endothermic process (∆H > 0), so it is the only process for which the product yield
increases with increasing temperature.
Think about It
Temperature changes do not affect the amount of product formed for reaction b, for which ∆H = 0.
16.79. Collect and Organize
We are to explain why equilibrium calculations are simpler when the value of the equilibrium constant, K, is
very small.
Analyze
When K is very small, the reaction does not proceed very far to the right. The concentrations of products
formed once the reaction achieves equilibrium are very small.
Solve
When K is small, the count of reactants that are transformed into products is small and so at equilibrium the
concentration of the reactants is approximately equal to the initial concentrations. This means that the
approximation
x
x
K=
≈
(A − x)(B − x) (A)(B)
for the reaction
A+B C
is valid. This makes our calculations easier and avoids having to use the quadratic equation.
Think about It
The assumption is considered valid if x is < 5% for the value of both A and B.
16.81. Collect and Organize
For the decomposition of PCl5 to PCl3 and Cl2 (Kp = 23.6 at 500 K), we are to calculate the equilibrium partial
pressures given the initial partial pressures of PCl5 and PCl3. We are also to determine how the concentration
of PCl3 and PCl5 change when more Cl2 is added to the system already at equilibrium.
Analyze
(a) To calculate the partial pressures of all the species present we set up an ICE table. Because the initial
partial pressure of Cl2 is 0.0 atm, we know that the reaction proceeds to the right to attain equilibrium.
(b) To determine how the equilibrium shifts when Cl2 is added, we apply Le Châtelier’s principle.
Solve
(a)
Initial
Change
Equilibrium
PPC15
PPC13
PC12
0.560 atm
–x
0.560 – x
0.500 atm
+x
0.500 + x
0.00 atm
+x
x
After placing these values into the equilibrium constant expression, we can solve for x using the quadratic
formula:
( 0.500 + x)( x)
23.6 =
(0.560 − x)
x 2 + 24.1x –13.216 = 0
x = 0.536 or – 24.6
Because 0.500 + x would be negative if x = –24.6, the actual root for this problem is x = 0.536. The
equilibrium partial pressures of PCl5, PCl3, and Cl2 are
240 | Chapter 16
PPCl5 = 0.560 − 0.536 = 0.024 atm
PPCl3 = 0.500 + 0.536 = 1.036 atm
PCl2 = 0.536 atm
(b) When the partial pressure of Cl2 is increased the partial pressure (or concentration) of PCl3 decreases and
the partial pressure of PCl5 increases.
Think about It
Because K > 1, the products of this reaction are favored.
16.83. Collect and Organize
For the initial concentrations of H2O and Cl2O as 0.00432 M in the equilibrium reaction
H 2O(g) + Cl 2O(g)  2 HOCl(g)
where Kc = 0.0900 at 25˚C, we are to calculate the equilibrium concentrations of H2O, Cl2O, and HOCl.
Analyze
We first set up an ICE table to solve this problem. We assume here that the initial concentration of
HOCl = 0.00 M. The equilibrium constant expression for this reaction is
[HOCl]2
Kc =
[H 2 O][Cl2 O]
Solve
Initial
Change
Equilibrium
[H2O]
0.00432 M
–x
0.00432 – x
[Cl2O]
0.00432 M
–x
0.00432 – x
HOCl
0
+2x
2x
After placing these values into the equilibrium constant expression, we can solve for x by taking the square
root of both sides:
0.0900 =
( 2 x )2
(0.00432 − x) 2
2x
0.300 =
0.00432 − x
1.296 × 10 –3 − 0.30 x = 2 x
x = 5.63 × 10 –4
The concentration of all the gases at equilibrium are
[H2O] = [Cl2O] = 0.00432 – x = 3.76 × 10–3 M
[HOCl] = 2x = 1.13 × 10–3 M
Think about It
This reaction, with its equilibrium constant less than 1, favors reactants over product at equilibrium.
16.85. Collect and Organize
Given Kp = 2 × 106 at 25˚C for the reaction of 1 mol of NO with 12 mol of O2 to give 1 mol of NO2, we are to
calculate the equilibrium ratio of the partial pressure of NO2 to that of NO in air where the partial pressure of
oxygen is 0.21 atm.
Chemical Equilibrium | 241
Analyze
The Kp expression for this reaction is
Kp =
PNO2
12
PNO × ( PO2 )
Solve
When PO = 0.21 atm,
2
PNO2
= 2 ×106
12
PNO × ( 0.21)
Rearranging this equation to solve for the ratio of the partial pressure of NO2 to the partial pressure of NO gives
PNO2
12
= 2 × 106 × ( 0.21) = 9 × 105
PNO
Kp =
Think about It
The high value of K indicates that the product NO2 is highly favored. This is consistent with the high value we
calculated for the ratio of the partial pressures.
16.87. Collect and Organize
Given that Kp = 1.5 at 700˚C, we are to calculate PCO and PCO at equilibrium for the reaction
2
CO 2 (g) + C(s)  2 CO(g)
where the initial partial pressures of CO2 and CO are 5.0 atm and 0.0 atm, respectively.
Analyze
Because carbon is a pure solid, it does not appear in the equilibrium constant expression:
Kp
2
PCO )
(
=
PCO
2
Solve
Initial
Change
Equilibrium
PCO2
PCO
5.0 atm
–x
5.0 – x
0.0 atm
+2x
2x
After placing these values into the equilibrium constant expression, we can solve for x using the quadratic
formula:
1.5 =
( 2 x )2
(5.0 – x)
2
4 x + 1.5 x − 7.5 = 0
x = 1.20 or − 1.57
The value of x = –1.57 would give a negative partial pressure for CO, so x = 1.19. The partial pressures of the
gases at equilibrium are
PCO = 2x = 2.4 atm
PCO 2 = 5.0 – x = 3.8 atm
Think about It
Checking our results should give a value close to 1.5, the equilibrium constant value:
242 | Chapter 16
( 2.4)2
3.8
= 1.5
16.89. Collect and Organize
For the decomposition of NO2 to NO and O2 when PO = 0.136 atm at equilibrium, we are asked to calculate
2
the partial pressures of NO and NO2 at equilibrium at 1000 K, where K = 158, and to calculate the total
pressure in the flask at equilibrium.
Analyze
(a) To calculate the partial pressures of NO and NO2 we set up an ICE table where we start with pure NO2 (A)
and end with PO = 0.136 atm = x at equilibrium.
2
(b) The total pressure is
PT = PNO2 + PNO + PO2
Solve
(a)
PNO 2
Initial
Change
Equilibrium
PO 2
PNO
0.00 atm
+2x
2x
A atm
–2x
A – 2x
0.00 atm
+x
x
We know that at equilibrium x = 0.136 atm, so
K p = 158 =
2
( P ) = ( 2 x ) ( x ) = ( 0.272) ( 0.136)
(P )
(P )
(P )
( PNO )
2
2
2
NO2
2
( PNO2 ) =
2
O2
NO2
2
NO2
2
( 0.272 ) × ( 0.136 )
158
12
PNO2 = ( 6.37 × 10 –5 )
= 6.37 × 10 –5
= 7.98 × 10 –3 atm
At equilibrium,
PNO = 2x = 0.272 atm
PO 2 = x = 0.136 atm
PNO 2 = 7.98 × 10–3 atm
(b) PT = 7.98 × 10–3 atm + 0.272 atm + 0.136 atm = 0.416 atm
Think about It
The amount of NO2 initially present can also be calculated:
A − 2 x = 7.98 × 10 –3 atm
Because x = 0.136 atm
A = 7.98 × 10 –3 + 2 × 0.136 = 0.280 atm
16.91. Collect and Organize
For the equilibrium reaction
N 2 (g) + O 2 (g)  2 NO(g)
the value of Kp is 0.050 at 2200˚C. We are to calculate the partial pressures of N2, O2, and NO at equilibrium
given that the initial partial pressures of these gases are 0.79 atm N2, 0.21 atm O2, and 0.00 atm NO.
Analyze
We set up an ICE table to solve this problem. The equilibrium constant expression for this reaction is
Chemical Equilibrium | 243
Kp =
( PNO )2
( P )( P )
O2
N2
Solve
PN 2
PO 2
PNO
Initial
0.79 atm
0.21 atm
0.00 atm
Change
–x
–x
+2x
Equilibrium
0.79 – x
0.21 – x
2x
After placing these values into the equilibrium constant expression, we can solve for x using the quadratic
formula:
0.050 =
0.050 =
(2x)
2
(0.79 − x)(0.21 − x)
4x2
( 0.1659 − x + x )
2
3.95 x 2 + 0.050 x – 0.008295 = 0
x = 0.03993 or –0.05260
The value of x = –0.05260 would give a negative partial pressure for NO, so x = 0.03993. The partial
pressures of the gases at equilibrium are
PO = 0.21 – x = 0.17 atm
2
PN 2 = 0.79 – x = 0.75 atm
PNO = 2x = 0.080 atm
Think about It
Using these equilibrium partial pressures we can check our answers, which should give Kp = 0.050:
( 0.080)2 = 0.050
( 0.75)( 0.17)
16.93. Collect and Organize
For the following reaction
2 H 2S(g)  2H 2(g) + S2(g)
the value of Kc is 2.2 × 10
the initial [H2S] is 6.00 M.
–4
at 1400 K. We are to calculate the equilibrium concentration of H2S, given that
Analyze
We are told to assume that the initial concentrations of H2 and S2 are 0.00 M. Because the equilibrium
constant is small we may be able to make a simplifying assumption. The equilibrium constant expression for
this reaction is
[H ]2 [S ]
K c = 2 22
[H 2S]
244 | Chapter 16
Solve
[H2S]
6.00 M
–2x
6.00 – 2x
Initial
Change
Equilibrium
[H2]
0.00 M
+2x
2x
[S2]
0.00 M
+x
x
Since the equilibrium constant is small, this reaction does not proceed very far to the right and we may be able
to make a simplifying assumption. After placing these values into the equilibrium constant expression, we can
solve for x:
2
2.2 ×10– 4 =
(2x ) ( x )
(6.00 − 2 x) 2
x = 0.126
2
≈
(2x ) ( x )
(6.00) 2
Checking this assumption shows that it is valid:
2 × 0.126
× 100 = 4%
6.00
The equilibrium [H2S] is 6.00 – 2x = 5.75 M.
Think about It
Without the simplifying assumption we would have to solve a cubic equation.
16.95. Collect and Organize
For the following reaction Kc = 5.0 at 600 K:
CO(g) + Cl 2 (g)  COCl 2(g)
We are asked to calculate the partial pressures of the gases at equilibrium given that the initial partial
pressures of CO and Cl2 are 0.265 atm and that of COCl2 is 0.000 atm.
Analyze
We have to first calculate Kp from Kc where ∆n = –1:
5.0
= 0.102
L ⋅ atm ⎞
⎛
⎜⎝ 0.08206
⎟ × 600 K
mol ⋅ K ⎠
We can set up an ICE table to solve for the equilibrium partial pressures of the gases.
K p = Kc ( RT ) Δn =
Solve
Initial
Change
Equilibrium
PCO
PCl 2
PCOCl 2
0.265 atm
–x
0.265 – x
0.265 atm
–x
0.265 – x
0.000 atm
+x
x
After placing these values into the equilibrium constant expression, we can solve for x using the quadratic formula:
x
0.102 =
(0.265 − x) 2
x
0.0702 − 0.530 x + x 2
0.102 x 2 − 1.05406 x + 0.00716 = 0
x = 10.33 or 0.00680
The value of x = 10.33 would give negative partial pressures for CO and Cl2, so x = 0.00680. The partial
pressures of the gases at equilibrium are
PCO = PCl = 0.265 – x = 0.258 atm
2
0.102 =
PCOCl 2 = x = 0.00680 atm
Chemical Equilibrium | 245
Think about It
Checking our result, we get the value of Kp:
0.00680
= 0.102
(0.258) 2
16.97. Collect and Organize
We are to calculate the concentrations of all the gases at equilibrium for the reaction
CO(g) + H 2O(g)  CO 2 (g) + H 2 (g)
given that the initial concentrations of all the gases are 0.050 M and that Kc = 5.1 at 700 K.
Analyze
The equilibrium constant expression for this reaction is
[CO2 ][H 2 ]
Kc =
[CO][H 2 O]
We have to use Q, the reaction quotient, to determine the direction in which the reaction goes to reach
equilibrium:
Q=
(0.050)(0.050)
= 1.0
(0.050)(0.050)
Q < K so the reaction proceeds to the right.
Solve
Initial
Change
Equilibrium
[CO]
[H2O]
[CO2]
[H2]
0.050 M
–x
0.050 – x
0.050 M
–x
0.050 – x
0.050 M
+x
0.050 + x
0.050 M
+x
0.050 + x
5.1 =
(0.050 + x) 2
(0.050 – x) 2
Taking the square root of both sides:
(0.050 + x)
(0.050 – x)
0.1129 − 2.258 x = 0.050 + x
0.0629 = 3.258 x
0.0193 = x
The concentrations of the gases at equilibrium are
[CO] = [H2O] = 0.050 – x = 0.031 M
[CO2] = [H2] = 0.050 + x = 0.069 M
2.258 =
Think about It
We did not need to use the quadratic formula here because with the concentrations of the reactants equal to
each other and the concentrations of the products equal to each other we could simplify the math by taking the
square root of both sides of the equation to solve for x.
16.99. Collect and Organize
We are to relate the sign of ΔG˚ to the magnitude of K.
Analyze
The equation relating ΔG˚ to K is
ΔG o = – RT ln K
Solve
246 | Chapter 16
Yes, the value of ΔG˚ is positive when K < 1. Because ln K is negative for K < 1 and because ΔG˚ =
–RT ln K, the value of ΔG˚ is positive.
Think about It
This means that reactions with K < 1 are not spontaneous.
16.101. Collect and Organize
If a reaction starts with 100% reactants we are to determine in which direction the reaction shifts when it is
spontaneous (ΔG˚ < 0).
Analyze
The equation relating ΔG˚ to K is
ΔG o = – RT ln K
Solve
When ΔG˚ < 0 then ln K of the reaction is positive, giving K > 1. The reaction favors the formation of
products, so the reaction will shift to the right.
Think about It
The reaction will proceed to the right until the mixture of reactants and products as expressed in the reaction
quotient Q matches the value of the equilibrium constant K.
16.103. Collect and Organize
Given three reactions with associated ΔG˚ values, we are asked to determine which reaction has the largest
value of K at 25˚C.
Analyze
The equilibrium constant of a reaction is related to the Gibbs free energy through the equation
ΔG˚ = –RT ln K
Rearranging this equation gives
−ΔG o
ln K =
RT
From this we see that the more negative the value of ΔG, the larger K will be.
Solve
Two of the reactions have negative values of ΔG˚ and are spontaneous (b and c); they will have K > 1. The
reaction with the more negative free energy (reaction c) has the largest equilibrium constant.
Think about It
The value of K for reaction c is
ln K =
−ΔGo
27.9 kJ
=
= 11.26
–3
RT
8.314 × 10 kJ/mol ⋅ K × 298 K
(
)
11.26
K =e
= 7.8 × 10
4
16.105. Collect and Organize
We can use the equation
ΔG˚ = –RT ln K
to calculate the value of K at 298 K for the phosphorylation of glucose, which has ΔG˚ = 113.8 kJ/mol.
Analyze
Rearranging the equation to solve for ln K gives
ln K =
−ΔG o
RT
Chemical Equilibrium | 247
where R = 8.314 × 10–3 kJ/mol ⋅ K.
Solve
ln K =
−13.8 kJ/mol
(
)
8.314 × 10 –3 kJ/mol ⋅ K × 298 K
= –5.57
K = e –5.57 = 3.81 × 10 –3
Think about It
When ΔG is positive the value of K is less than 1 and the reactants are favored at equilibrium over the
products.
16.107. Collect and Organize
From the value of Kc for the hydrolysis of sucrose (Kc = 5.3 × 1012 at 298 K), we are to calculate the value of
ΔG˚.
Analyze
The value of ΔG˚ is calculated from K using the equation
ΔG˚ = –RT ln K
where R = 8.314 J/mol ⋅ K and T is the temperature in kelvins.
Solve
ΔGo = −8.314
J
× 298 K × ln(5.3 × 1012 ) = –7.3 × 104 J/mol or –73 kJ/mol
mol ⋅ K
Think about It
When the value of K is larger than 1, the Gibbs free energy of the reaction is negative.
16.109. Collect and Organize
In this problem we can add two equations to calculate ΔG˚ and Kp at 298 K for the reaction
N 2 (g) + 2 O 2 (g)  2 NO 2 (g)
Analyze
To obtain the overall reaction we simply add the two reactions given and add their ΔG˚ values. We can
calculate the value of Kc using
ΔG˚ = –RT ln K
and from Kc we can calculate Kp using
K p = K c ( RT )
Δn
where Δn for this reaction is –1.
Solve
Adding the equations gives the overall equation and ΔG˚:
N 2 (g) + O 2 (g)  2 NO(g)
ΔG  = 173.2 kJ
2 NO(g) + O 2 (g)  2 NO 2 (g)
ΔG  = –69.7 kJ
N 2 (g) + 2 O 2 (g)  2 NO 2 (g)
ΔG  = 103.5 kJ
Rearranging to calculate Kc,
ln K =
– ΔG o
–103.5 kJ/mol
=
= –41.77
RT
(8.314 ×10–3 kJ/mol ⋅ K ) × 298 K
K = e – 41.77 = 7.24 × 10−19
The value of Kp can now be calculated
248 | Chapter 16
–1
K p = 7.24 × 10 –19 × [ (0.08206 L ⋅ atm/mol ⋅ K) × 298] = 2.96 × 10 –20
Think about It
Because the free-energy change is large and positive, K is much less than 1 and the reactants are greatly
favored over the product at equilibrium.
16.111. Collect and Organize
Using the relationship
⎛ K ⎞
ΔH o ⎛ 1 1 ⎞
ln ⎜ 2 ⎟ = −
⎜ − ⎟
R ⎝ T2 T1 ⎠
⎝ K1 ⎠
we can determine whether a reaction is endothermic or exothermic if the value of K decreases with
increasing T.
Analyze
In the equation, K2 < K1 so ln(K2/K1) < 0 and T2 > T1 so (1/T2 – 1/T1) < 0.
Solve
In order for ln(K2/K1) to be negative when (1/T2 – 1/T1) is also negative, the value of ∆H must be negative
and the reaction must be exothermic.
Think about It
An endothermic reaction, on the other hand, shows an increased K with increasing T because ∆H > 0 and
thus ln(K2/K1) > 0 so K2/K1 > 0.
16.113. Collect and Organize
Given that Kp for the reaction of CO with H2O increases as the temperature decreases, we can use the
relationship
⎛ K 2 ⎞
ΔH o ⎛ 1 1 ⎞
ln ⎜
=−
⎟
⎜ − ⎟
⎜ K1 ⎟
R ⎝ T2 T1 ⎠
⎝
⎠
to determine whether the reaction is endothermic or exothermic.
Analyze
In the equation K2 > K1 so ln(K2/K1) > 0. Since T2 < T1, (1/T2 – 1/T1) > 0.
Solve
In order for ln(K2/K1) to be positive when (1/T2 – 1/T1) is also positive, ∆H must be negative and the reaction
must be exothermic.
Think about It
An endothermic reaction, on the other hand, shows a decreased K with a decrease in temperature.
16.115. Collect and Organize
Given ∆H˚ = 180.6 kJ and Kc = 4.10 × 10– 4 for the conversion of N2 and O2 to NO at 2000˚C, we are to
calculate the value of Kc at 25˚C.
Analyze
The relationship between equilibrium constants at two different temperatures is given by
⎛ K ⎞
ΔH o ⎛ 1 1 ⎞
ln ⎜ 2 ⎟ = −
⎜ − ⎟
R ⎝ T2 T1 ⎠
⎝ K1 ⎠
Chemical Equilibrium | 249
where ∆H˚ is the enthalpy of the reaction (in J/mol), R is the gas constant (in J/mol ⋅ K), and T1 and T2 are
the temperatures (in kelvins).
Solve
⎛ 4.1× 10 – 4
ln ⎜
⎜
K2
⎝
⎞
1.806 × 105 J/mol ⎛ 1
1 ⎞
−
⎟⎟ = −
⎜
⎟
8.314
J/mol
⋅
K
2273
K
298
K ⎠
⎝
⎠
⎛ 4.1× 10 – 4 ⎞
ln ⎜
⎟⎟ = 63.337
⎜
K2
⎝
⎠
–4
4.1× 10
= e63.337 = 3.214 × 1027
K2
K 2 = 1.3 × 10 –31
Think about It
As we would expect, the equilibrium constant for this endothermic reaction decreases as the reaction is
cooled.
16.117. Collect and Organize
Given that K1 = 1.5 × 105 at 430˚C and K2 = 23 at 1000˚C, we are to calculate the standard enthalpy of the
reaction between NO and O2 to produce NO2.
Analyze
We can use the following equation to solve this problem:
⎛ K ⎞
ΔH o ⎛ 1 1 ⎞
ln ⎜ 2 ⎟ = −
⎜ − ⎟
R ⎝ T2 T1 ⎠
⎝ K1 ⎠
Solve
ΔH o
1 ⎞
⎛ 23 ⎞
⎛ 1
ln ⎜
=
−
−
⎟
⎜⎝
⎟
⎝ 1.5 × 105 ⎠
8.314 J/mol ⋅ K 1273 K 703 K ⎠
ΔH o = –1.15 × 105 J/mol or –115 kJ/mol
Think about It
The result that this reaction, where K decreases with increasing T, is exothermic is consistent with our
answer to Problem 16.112.
16.119. Collect and Organize
Given the percentage of decomposition of CO2 at three temperatures and that each equilibrium mixture
begins with 1 atm of CO2 we are to determine whether the reaction is endothermic and then calculate Kp at
each temperature. We are also to comment on the decomposition reaction as a remedy for global warming.
Analyze
The amount of CO2 decomposed increases with increasing temperature. This means that K2 > K1 in the
equation
⎛ K ⎞
ΔH o ⎛ 1 1 ⎞
ln ⎜ 2 ⎟ = −
⎜ − ⎟
R ⎝ T2 T1 ⎠
⎝ K1 ⎠
where T2 > T1 so (1/T2 – 1/T1) < 0. To calculate Kp for each temperature, we set up an ICE table:
250 | Chapter 16
PCO 2
PCO
PO 2
Initial
1 atm
0 atm
0 atm
Change
–2x
+2x
+x
Equilibrium
1 – 2x
2x
x
We know that 2x is equal to the percent decomposition divided by 100. The form of the equilibrium
expression is
Kp =
( PCO )2 ( PO
(P )
2
)
2
CO2
Because the percent decomposition increases with temperature, the value of Kp is expected to increase as T
increases.
Solve
Because the value of K increases with increasing temperature, ln(K2/K1) > 0. With (1/T2 – 1/T1) < 0 the value
of ∆H must be positive, so this reaction is endothermic.
At 1500 K, 2x = 0.00048, x = 0.00024, and 1 – 2x = 0.99952:
(0.00048)2 (0.00024)
Kp =
= 5.5 × 10–11
(0.99952)2
At 2500 K, 2x = 0.176, x = 0.088, and 1 – 2x = 0.824:
(0.176)2 (0.088)
Kp =
= 4.0 × 10–3
(0.824)2
At 3000 K, 2x = 0.548, x = 0.274, and 1 – 2x = 0.452:
(0.548)2 (0.274)
Kp =
= 0.40
(0.452)2
As predicted, the value of the equilibrium constant increases with increasing temperature. This reaction,
however, does not favor products even at very high temperature, so this is not a viable source of CO and is
not a remedy to decrease CO2 as a contributor to global warming. Also, the process produces poisonous CO
gas.
Think about It
By using values in Appendix 4 we can confirm that the reaction is endothermic.
16.121. Collect and Organize
By combining the two equations given we can calculate the overall equilibrium constant, K, for the reaction
and then calculate the concentration of X2– in an equilibrium mixture where [H2X]eq = 0.1 M and [HCl]eq =
[H3O+]eq = 0.3 M.
Analyze
When we add equations the overall equilibrium constant is the product of the individual equilibrium
constants. From the overall reaction we can write the equilibrium constant expression and from that solve for
[X2–] at equilibrium.
Solve
H 2 X(aq) + H 2O()  HX – (aq) + H 3O+ (aq)
K1 = 8.3×10 –8
HX – (aq) + H 2O()  X 2– (aq) + H 3O+ (aq)
K 2 = 1×10 –14
H 2 X(aq) + 2 H 2O()  X 2– (aq) + 2 H 3O+ (aq)
⎡ X 2– ⎤⎦ ⎡⎣ H 3 O + ⎤⎦
K c = ⎣
[ H2X]
2
K 3 = K1 × K 2 = 8.3×10 –22
Chemical Equilibrium | 251
We know that Kc = 8.3 × 10–22, [H3O+] = 0.3 M, and [H2X] = 0.1 M for the saturated solution. Rearranging
the equilibrium constant expression and solving for [X2–] gives
K c × [ H 2 X ] 8.3 × 10–22 × 0.1
⎡⎣ X 2– ⎤⎦ =
=
= 9 × 10–22 M
2
+ 2
(0.3)
⎡⎣ H3O ⎤⎦
Think about It
Our answer makes sense because with a very small overall equilibrium constant, we expect a very small
concentration of product, X2–.
16.123. Collect and Organize
For the reaction of CH4 with S8 to form CS2 and H2 and the reaction, we are to calculate the value of the
equilibrium constant using thermodynamic data from Appendix 4 for the reaction at both 25˚C and 500˚C.
Analyze
We have to calculate the value of Kp from ∆G at 500˚C. To do so, we must first determine ∆H˚ and ∆S˚ from
tabulated values in the appendix:
∆H˚ = [(4 × –115.3) + (8 × 0.0)] – [(4 × –74.8) – (1 × 0.0)] = 760.4 kJ
∆S˚ = [(4 × 237.8) + (8 × 130.6)] – [(4 × 186.2) – (1 × 32.1)] = 1219 J/K
We can then calculate the value of ∆G25˚C and ∆G500˚C through
∆G25˚C = ∆H˚ – T ∆S˚ = 760.4 kJ – (298 K × 1.219 kJ/K) = 397.1 kJ
∆G500˚C = ∆H˚ – T ∆S˚ = 760.4 kJ – (773 K × 1.219 kJ/K) = –181.9 kJ
The value of Kp for each temperature is calculated though the value of ∆G˚
ΔG o
ln K p = –
RT
Solve
At 25˚C (298 K)
ln K p = −
397.1 kJ/mol
= −160.3
⎛ 8.314 × 10−3 kJ ⎞
⎜⎝
⎟⎠ × 298 K
mol ⋅ K
K p = e –160.3 = 2.47 × 10−70
At 500˚C (773 K)
ln K p = −
−181.9 kJ/mol
= 28.30
⎛ 8.314 × 10−3 kJ ⎞
⎜⎝
⎟⎠ × 773 K
mol ⋅ K
K p = e28.30 = 1.96 × 1012
Think about It
This reaction changed its spontaneity upon heating because it is an endothermic reaction with a favorable
entropy change.
16.125. Collect and Organize
Given ∆G˚ = – 418.6 kJ for the reaction in which CaO reacts with SO2 in the presence of O2 to form solid
CaSO4, we can calculate K and then determine the PSO in this reaction when the partial pressure of oxygen
2
is 0.21 atm.
Analyze
To calculate K from ∆G˚ we use
252 | Chapter 16
ln K = –
The equilibrium constant expression for the reaction is
Kp =
ΔG o
RT
1
1/ 2
( P )( P )
SO2
O2
Solve
ln K = –
–418.6 kJ/mol
= 168.956
8.314 ×10–3 kJ/mol ⋅ K × 298K
K = e168.956 = 2.38 ×1073
2.38 × 1073 =
1
PSO2 × (0.21)1/ 2
PSO2 = 9.2 × 10–74 atm
Think about It
Essentially all of the SO2 is “scrubbed” by the CaO in this reaction, making this an efficient method to
remove SO2 from exhaust gases.
CHAPTER 17 | Equilibrium in the Aqueous Phase
17.1. Collect and Organize
In Figure P17.1 four lines are shown to describe the possible dependence of percent ionization of acetic acid
with concentration. We are to choose the one that best represents the trend for this weak acid.
Analyze
The ionization of acetic acid is described by the following chemical reaction:
CH3COOH(aq)  CH3COO–(aq) + H+
The degree of ionization is the ratio of the quantity of a substance that is ionized to the concentration of the
substance before ionization.
Solve
According to Figure 17.8, the change in degree of ionization of a weak acid with concentration is not linear
and is best described by the red line in Figure P17.1. The degree of ionization increases with decreasing acetic
acid concentration.
Think about It
The percent ionization could be calculated for each concentration if we knew the equilibrium concentration of
the acetate ion in solution and the initial concentration of acetic acid dissolved.
⎡⎣ H + ⎤⎦
equilibrium
% ionization =
× 100
[ acetic acid]initial
17.3. Collect and Organize
From Figure P17.3 we are to choose which titration curve represents a strong acid and which represents a weak
acid, each of 1 M concentration at the start of the titration.
Analyze
A strong acid is completely ionized in solution and has a lower initial pH than the weak acid, which is only
partially ionized in solution.
Solve
The blue titration curve represents the titration of a 1 M solution of strong acid. The red titration curve
represents the titration of a 1 M solution of weak acid. This is because the pH of the strong acid is expected to
be much lower than that of the weak acid at the start of the titration (where no base has yet been added).
Think about It
Notice that the equivalence point of the titration of the strong acid (pH 7) does not equal that of the weak acid
(pH 10).
17.5. Collect and Organize
For the red titration curve in Figure P17.3, we are to choose the indicator, according to its pKa, that would be
best for the titration.
Analyze
The best indicator is the one with a pKa that is nearest to the end point of the titration.
Solve
The end point for the red curve is at approximately pH 10. Therefore, the best indicator is the one with a pKa
of 9.0.
281
282 | Chapter 17
Think about It
The lower pKa indicators would show a color change before the end point of the titration was reached. Using
these would therefore underestimate the concentration of the weak acid in the original solution.
17.7. Collect and Organize
We are shown two titration curves in Figure P17.7. The blue curve has one equivalence point and the red curve
has two equivalence points. We are to assign each of these curves to either Na2CO3 or NaHCO3.
Analyze
Both of the bases being titrated are soluble sodium salts. The equation describing the titration of CO32– (Na+ is
a spectator ion) shows CO32– to be “dibasic”; it reacts in two steps to form H2CO3.
CO32–(aq) + H+(aq) → HCO3–(aq)
HCO3–( aq) + H+(aq) → H2CO3(aq)
HCO3– , however, reacts with acid in one step; it is “monobasic”:
HCO3–(aq) + H+(aq) → H2CO3(aq)
Solve
The red titration curve represents the titration of Na2CO3 because it shows two equivalence points. The blue
titration curve represents the titration of NaHCO3 because it shows one equivalence point.
Think about It
Notice that both titration curves start at high pH. This is due to the hydrolysis of CO32– and HCO3– in water:
CO32–(aq) + H2O (l ) → HCO3–(aq) + OH–(aq)
HCO3–(aq) + H2O (l ) → H2CO3(aq) + OH–(aq)
17.9. Collect and Organize
For HBr(aq) we are to identify the Brønsted–Lowry acid and base.
Analyze
A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor.
Solve
HBr is a strong acid in water. It acts as a Brønsted–Lowry acid, donating its proton to H2O, the Brønsted–
Lowry base:
HBr(aq) + H2O (l ) → H3O+(aq) + Br–(aq)
Think about It
Hydrobromic acid is a strong acid in water. It completely dissociates in water to H3O+ and Br–.
17.11. Collect and Organize
For NaOH(aq) we are to identify the Brønsted–Lowry acid and base.
Analyze
NaOH is a soluble salt that forms Na+ and OH– in water. Na+ does not react with water so it is a spectator ion.
We need then to consider the behavior of OH– in water. A Brønsted–Lowry acid is a proton donor. A
Brønsted–Lowry base is a proton acceptor.
Solve
OH– is a strong base in water. It acts as a Brønsted–Lowry base, removing a proton from H2O, the Brønsted–
Lowry acid:
OH–(aq) + H2O (l )  H2O (l ) + OH–(aq)
Equilibrium in the Aqueous Phase | 283
Think about It
In Problems 17.9 and 17.10, water acted as a Brønsted–Lowry base. In this problem, it acts as an acid. This
dual acid–base behavior makes water amphoteric.
17.13. Collect and Organize
For three acid–base reactions we are to identify which reactant is the acid and which reactant is the base.
Analyze
For all of these reactions that involve the transfer of a proton between the acid and base, we can apply the
Brønsted–Lowry definitions of acid and base. A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry
base is a proton acceptor.
Solve
(a) HNO3 is the acid. It transfers H+ to the base NaOH.
(b) HCl is the acid. It transfers H+ to the base CaCO3.
(c) HCN is the acid. It transfers H+ to the base NH3.
Think about It
In reactions a and b, Na+, Ca2+, and Cl– do not get involved in the reaction. They are spectator ions. The net
ionic equations are
HNO3(aq) + OH–(aq) → NO3–(aq) + H2O (l )
CO32–(aq) + 2 H+(aq) → CO2(g) + H2O (l )
17.15. Collect and Organize
For each species listed we are to write the formula for the conjugate base.
Analyze
The conjugate base form of a species has H+ removed from its formula.
Solve
The conjugate base of HNO2 is NO2–.
The conjugate base of HOCl is OCl–.
The conjugate base of H3PO4 is H2PO4–.
The conjugate base of NH3 is NH2–.
Think about It
Be sure to account for the change in charge when H+ is removed to form the conjugate base.
17.17. Collect and Organize
Given that the concentration of a nitric acid solution is 1.50 M, we are to calculate the concentration of H+
ions in the solution.
Analyze
Nitric acid is a strong acid that completely dissociates in water, so the concentration of H+ ions is
stoichiometrically related to the concentration of HNO3:
HNO3(aq) → H+(aq) + NO3–(aq)
Solve
[H+] = 1.50 M
Think about It
The concentration of OH– in strong base solutions, likewise, is the same as the concentration of the strong
base dissolved into the solution.
284 | Chapter 17
17.19. Collect and Organize
Given that a solution is 0.0800 M in the strong base Sr(OH)2, we are asked to calculate the concentration of
OH–.
Analyze
Sr(OH)2, being a strong base, completely dissociates according to the equation
Sr(OH)2(aq) → Sr2+(aq) + 2 OH–(aq)
Therefore, [OH–] = 2 × [Sr(OH)2].
Solve
2 × 0.0800 M = 0.160 M = [OH–]
Think about It
Be sure to account for both OH– ions in this strong base.
17.21. Collect and Organize
We are asked how to prepare 2.50 L of 0.70 M OH– using NaOH(s).
Analyze
From the desired volume and concentration we first calculate the moles of OH– required for the solution.
Since 1 mol of OH– is produced for every 1 mol of NaOH dissolved, this is also the moles of NaOH required.
To calculate the mass of NaOH needed we multiply the number of moles by the molar mass of NaOH
(40.00 g/mol).
Solve
Mass of NaOH needed
0.70 mol OH – 1 mol NaOH 40.00 g NaOH
Mass NaOH = 2.50 L ×
×
×
= 70.0 g NaOH
L
1 mol NaOH
1 mol OH –
Dissolve 70.0 g of NaOH(s) in water and dilute to a total volume of 2.50 L.
Think about It
Remember that volume times concentration for solutions gives us the moles of that substance in solution.
17.22. Collect and Organize
We are to calculate the volume of a 1.00 M NaOH solution that is needed to prepare 250 mL of a solution that
is 0.0200 M in OH– concentration.
Analyze
Since NaOH is a strong base, 1.00 M NaOH is 1.00 M in OH–. We can use the relationship from Chapter 4 for
dilutions to solve this problem for Vinitial:
Vinitial × Minitial = Vfinal × Mfinal
Solve
Vinitial × 1.00 M = 250 mL × 0.0200 M
Vinitial = 5.00 mL
Think about It
Be careful in this type of calculation if the strong base is one like Ba(OH)2. In that case, a 1.00 M Ba(OH)2
solution would be 2.00 M in OH– concentration.
17.23. Collect and Organize
We are to explain why the pH value decreases for solutions as acidity increases.
Analyze
The pH of a solution is calculated through
Equilibrium in the Aqueous Phase | 285
pH = –log[H+]
Solve
Because the pH function is a –log function, as [H+] increases, the value of –log[H+] decreases.
Think about It
The pH scale is typically 0 –14 for concentrations of H+ from 1 M to 1 × 10–14 M, but values of pH may be
negative or greater than 14.
17.25. Collect and Organize
We are asked under what conditions the pH of a solution may be negative.
Analyze
The pH scale is often seen as 0 –14. This occurs when [H+] is between 1 M and 1 × 10–14 M.
Solve
When [H+] is greater than 1 M, the pH of the solution is negative.
Think about It
For example, a 3.00 M solution of HCl has a pH of
pH = –log[H+] = –log(3.00 M) = – 0.48
17.27. Collect and Organize
Given either the [OH–] or [H+] for a solution, we are asked to calculate the pH and pOH and determine
whether the solution is acidic, basic, or neutral.
Analyze
To calculate the pH or the pOH from the [H+] or [OH–], respectively, we use
pH = –log[H+]
pOH = –log[OH–]
To find the pOH from the pH and vice versa we use the relationship
pH + pOH = 14
If the pH of a solution is less than 7, the solution is acidic. If the pH is equal to 7, the solution is neutral. If the
pH is greater than 7, the solution is basic.
Solve
(a) pH = –log(3.45 × 10–8) = 7.462
pOH = 14 – pH = 6.538
This solution is basic.
(b) pH = –log(2.0 × 10–5) = 4.70
pOH = 14 – pH = 9.30
This solution is acidic.
(c) pH = –log(7.0 × 10–8) = 7.15
pOH = 14 – pH = 6.85
This solution is basic.
(d) pOH = –log(8.56 × 10– 4) = 3.068
pH = 14 – pOH = 10.932
This solution is basic.
Think about It
When determining how many significant figures to include in your answers when computing the pH or pOH,
remember that the first number in the pH or pOH gives the location of the decimal point. The significant
digits, therefore, follow after the decimal point.
286 | Chapter 17
17.29. Collect and Organize
Given that stomach acid is 0.155 M HCl, we are to calculate its pH.
Analyze
The pH is defined as pH = –log[H+]. Since HCl is a strong acid, we know that 0.155 M HCl is also
0.155 M H+.
Solve
pH = –log(0.155) = 0.810
Think about It
The acid in your stomach is fairly strong and the HCl is produced by parietal cells to break down food. Your
stomach is protected in turn by epithelial cells that secrete a bicarbonate solution that neutralizes the acid and
forms a coating to protect the stomach’s tissues.
17.31. Collect and Organize
Given that the concentration of NaOH in a solution is 0.0450 M, we are to calculate its pH and pOH.
Analyze
Because NaOH is a strong base, we know that the [OH–] in this solution is 0.0450 M. The pOH is calculated
from pOH = –log[OH–], with the pH being found from the pOH through the relationship pH + pOH = 14.
Solve
pOH = –log (0.0450) = 1.347
pH = 14 – pOH = 12.653
Think about It
This solution is as alkaline (basic) as drain cleaner (Figure 17.6).
17.33. Collect and Organize
Given a solution that is 1.33 M in HNO3, we are to calculate its pH.
Analyze
Since HNO3 is a strong acid, we know that for this solution the [H+] from the HNO3 is 1.33 M. The pH is
calculated from pH = –log[H+].
Solve
pH = –log (1.33) = –0.124
Think about It
This pH is negative because the concentration of acid is greater than 1.00 M, which corresponds to
pH = 0.000.
17.35. Collect and Organize
For one-molar solutions of CH3COOH, HNO2, HClO, and HCl we are to rank these in order of decreasing
concentration of H+. For this we need the Ka values for each acid.
Analyze
The Ka values for these acids are listed in Tables 17.1 and 17.2. The greater the value of the Ka, the greater the
concentration of H+ in the solution of the acid. We note that HCl is a strong acid.
Solve
In order of largest Ka (strongest acid) to smallest Ka (weakest acid), HCl > HNO2 > CH3COOH > HClO.
Think about It
For the weak acids in this series, there is a wide range in their acidities (about 10,000-fold comparing their Ka
values).
Equilibrium in the Aqueous Phase | 287
17.37. Collect and Organize
We are to explain why the electrical conductivity of 1.0 M NaNO2 is much better than that of 1.0 M HNO2.
Analyze
Solutions with a larger concentration of dissolved ions conduct electricity better than those with lower
concentrations of dissolved ions.
Solve
NaNO2 is completely soluble in water, separating into Na+ and NO2– ions, each in 1.0 M concentration for a
total ion concentration of 2.0 M. HNO2, however, only weakly dissociates in water:
HNO2(aq)  NO2–(aq) + H+(aq)
and so produces just slightly greater than 1.0 M ions in solution. NaNO2, therefore, with more dissolved ions
in solution, is a better conductor of electricity.
Think about It
Later in this chapter, you will learn that NO2– reacts with water to a small extent:
NO2–(aq) + H2O (l )  HNO2(aq) + OH–(aq)
However, this produces another ion, OH–, and so our analysis that a solution of 1.0 M NaNO2 is 2.0 M in ions
is still correct.
17.39. Collect and Organize
The formula for hydrofluoric acid is HF. From this we can write the mass action expression for this weak
acid.
Analyze
The general form of the mass action expression for weak acids, based on HA  A– + H+, is
Ka =
Solve
[H+ ][A – ]
[HA]
HF(aq)  F–(aq) + H+(aq)
Ka =
[H+ ][F– ]
[HF]
Think about It
The hydrofluoric acid is transferring a proton to water in this equation, so an equivalent expression is
HF(aq) + H2O (l )  F–(aq) + H3O+(aq)
Ka =
[H3O+ ][F– ]
[HF]
17.41. Collect and Organize
Given that the Ka of alanine is less when it is dissolved in ethanol than it is when dissolved in water, we are to
determine which solvent ionizes alanine to the larger extent and which solvent is the stronger Brønsted–
Lowry base.
Analyze
The larger the value of Ka, the greater the extent to which a substance has ionized. The solvent in which an
acid ionizes the most must be the strongest Brønsted–Lowry base toward that acid.
Solve
(a) Because Ka for alanine is greater in water than in ethanol, alanine in water is ionized to a greater extent
than in ethanol.
288 | Chapter 17
(b) Water is the stronger base for alanine compared to ethanol because alanine is ionized to a greater extent in
water.
Think about It
This question demonstrates that acid–base strengths can depend on the basicity of the solvent.
17.43. Collect and Organize
Given that CH3NH2 is slightly basic in water we can write the equation describing its reaction with water to
identify which species in the reaction is the Brønsted–Lowry acid and which is the base.
Analyze
Acting as a base, CH3NH2 accepts H+ from surrounding water molecules.
Solve
The reaction describing the basicity of CH3NH2 is
CH3NH2(aq) + H2O (l )  CH3NH3+(aq) + OH–(aq)
In this reaction, H2O is the acid and CH3NH2 is the base.
Think about It
In another solvent, such as diethylamine, methylamine may act as an acid:
CH3NH2 + (CH3CH2)2NH  CH3NH– + (CH3CH2)2NH2+
This occurs because diethylamine (Kb = 8.6 × 10– 4) is a stronger base than methylamine (Kb = 4.4 × 10– 4).
17.45. Collect and Organize
Using the measured percent ionization of 1.00 M lactic acid of 2.94%, we are to calculate the Ka of this weak
acid.
Analyze
The equilibrium equation from Appendix 5 that describes the ionization of lactic acid is
CH3CHOHCOOH(aq)  H+(aq) + CH3CHOHCOO–(aq)
We can set up an ICE table to solve this problem, where x = 1.00 M × 0.0294 = 0.0294. The Ka expression is
[H + ][CH3CHOHCOO – ]
Ka =
[CH3CHOHCOOH]
Solve
Initial
Change
Equilibrium
[CH3CHOHCOOH]
1.00
–x = – 0.0294
0.9706
Ka =
[H+]
0
+x = + 0.0294
0.0294
[CH3CHOHCOO–]
0
+x = + 0.0294
0.0294
(0.0294) 2
= 8.91× 10 – 4
0.9706
Think about It
Compare this with the Ka at 25˚C value of 1.4 × 10 – 4 listed in Appendix 5. The difference may be attributable
to a temperature difference because body temperature is 37˚C.
17.47. Collect and Organize
We are given the [H+] for an equilibrium solution of an unknown acid with an initial concentration of
0.250 M. From this information we can calculate the degree of ionization and Ka for this weak acid.
Analyze
Equilibrium in the Aqueous Phase | 289
The equilibrium expression for the unknown acid is
HA(aq)  H+(aq) + A–(aq)
with a Ka expression of
[H+ ][A – ]
Ka =
[HA]
We can set up an ICE table to show how this weak acid ionizes; for this acid, [H+] at equilibrium is
4.07 × 10–3 M.
Solve
[HA]
Initial
0.250
Change
–x = –0.00407
Equilibrium
0.24593
(0.00407)2
Ka =
0.24593
[H + ]eq
Degree of ionization =
=
[HA]initial
[H+]
0
+x = +0.00407
0.00407
[A–]
0
+x = +0.00407
0.00407
= 6.74 × 10–5
0.00407 M
= 0.0163 or 1.63%
0.250 M
Think about It
In this problem we can use the [H+] as equivalent to x in the ICE table, enabling us to calculate both the Ka
and the degree of ionization for the acid.
17.49. Collect and Organize
Given that formic acid has Ka = 1.8 × 10– 4, we can calculate the pH of a 0.060 M aqueous solution of this
weak acid.
Analyze
To solve this problem we set up an ICE table where the initial concentration of formic acid, HCOOH, is
0.060 M. We let x be the amount of formic acid that ionizes. The equilibrium and Ka expressions are
HCOOH(aq)  H+(aq) + HCOO–(aq)
[H+ ][HCOO– ]
= 1.8 × 10–4
[HCOOH]
The pH can be calculated through the equation pH = –log[H+].
Ka =
Solve
[H+]
[HCOO–]
Initial
0
0
Change
+x
+x
Equilibrium
x
x
2
2
x
x
1.8 × 10–4 =
≈
0.060 − x 0.060
x = 3.29 × 10–3
We should first check the simplifying assumption we made above before calculating the pH.
3.29 × 10–3
× 100 = 5.5%
0.060
This is a little over 5%, so technically we should solve this by the quadratic equation (which we do below).
However, if we allow this simplifying assumption to be valid, the pH of the solution would be calculated by
[HCOOH]
0.060
–x
0.060 – x
pH = –log 3.29 × 10–3 = 2.48
Now, solving by the quadratic equation gives
1.8 × 10– 4(0.060 – x) = x2
2
x + 1.8 × 10– 4x – 1.08 × 10–5 = 0
290 | Chapter 17
x = 3.20 × 10–3
pH = –log 3.20 × 10–3 = 2.49
Think about It
The difference between the pH values when making the simplifying assumption and solving the equation
exactly using the quadratic equation is 2.49 – 2.48 = 0.01, which is fairly small.
17.51. Collect and Organize
By comparing the pH of rain in a weather system in the Midwest of 5.02 with the pH of the rain in that same
system when it reached New England of 4.66, we can calculate how much more acidic the rain in New
England was.
Analyze
We want to find the ratio
[H + ]New England
[H + ]Midwest
Because pH = –log[H ], the [H ] = 1 × 10 . Therefore,
[H + ]New England 1 × 10 –4.66
=
[H + ]Midwest
1 × 10 –5.02
+
+
–pH
Solve
[H + ]New England
1 × 10 –4.66 2.19 × 10 –5 M
=
= 2.3
[H + ]Midwest
1 × 10 –5.02 9.55 × 10 –6 M
The rain in New England in this weather system was 2.3 times more acidic than the rain in the Midwest.
=
Think about It
Among the causes of the acidity of the rain in New England are the coal-burning electricity power plants in
the Midwest, which expel SO2 and SO3 into the air which make H2SO3 and H2SO4.
17.53. Collect and Organize
Given the Kb for dimethylamine of 5.9 × 10– 4, we are to find the pH of a solution of dimethylamine that is
1.20 × 10–3 M.
Analyze
The equilibrium and Kb expressions for dimethylamine are
(CH3)2NH(aq) + H2O (l )  (CH3)2NH2+(aq) + OH–(aq)
[OH – ][(CH3 )2 NH 2 + ]
[(CH3 ) 2 NH]
We can set up an ICE table to solve this problem, where x is the amount of dimethylamine that ionizes. By
solving for x we can calculate [OH–], from which we can determine the pOH and pH:
pOH = –log[OH–]
pH = 14 – pOH
Kb =
Solve
Initial
Change
Equilibrium
[(CH3)2NH]
[OH–]
[(CH3)2NH2+]
1.20 × 10–3
–x
1.20 × 10–3 – x
0
+x
x
0
+x
x
x2
x2
≈
–3
1.2 × 10 − x 1.2 × 10–3
x = 8.41 × 10 –4
5.9 × 10–4 =
Equilibrium in the Aqueous Phase | 291
Checking the simplifying assumption shows that our simplifying assumption is not valid:
8.41 × 10–4
× 100 = 70%
1.20 × 10–3
so we must solve this by the quadratic equation:
5.9 × 10– 4 (1.20 × 10–3 – x) = x2
x2 + 5.9 × 10– 4x – 7.08 × 10–7 = 0
x = 5.966 × 10– 4
pOH = –log 5.97 × 10– 4 = 3.22
pH = 14 – 3.22 = 10.77
Think about It
Our answer of a pH > 7 is consistent with dimethylamine’s behavior as a base in aqueous solution.
.
17.55. Collect and Organize
Given the pKa of the conjugate acid of codeine (pKa = 8.21), we are to calculate the pH of a 3.42 × 10– 4 M
solution of codeine, a weak base.
Analyze
We first need to determine the value of the Kb from the pKa for codeine:
Ka = 1 × 10– pKa = 1 × 10–8.21 = 6.17 × 10–9
Kw
1× 10 –14
= 1.62 × 10 – 6
K a 6.17 × 10 –9
The equilibrium and Kb expressions for the ionization of codeine are
Codeine(aq) + H2O (l )  codeineH+(aq) + OH–(aq)
Kb =
=
[OH – ][codeineH + ]
= 1.62 ×10– 6
[codeine]
We can set up an ICE table to solve this problem, where x is the amount of codeine that ionizes. By solving
for x we can calculate [OH–], from which we can determine the pOH and pH:
pOH = –log[OH–]
pH = 14 – pOH
Kb =
Solve
Initial
Change
Equilibrium
[Codeine]
3.42 × 10– 4
–x
3.42 × 10– 4 – x
1.62 × 10 – 6 =
[OH–]
0
+x
x
[CodeineH+]
0
+x
x
x2
x2
≈
3.42 × 10– 4 − x 3.42 × 10 – 4
x = 2.35 × 10 –5
Checking the simplifying assumption shows that our simplifying assumption is not valid:
2.35 × 10–5
× 100 = 6.9%
3.42 × 10–4
so we must solve this by the quadratic equation:
1.62 × 10–6(3.42 × 10–4 – x) = x2
2
x + 1.62 × 10–6x – 5.54 × 10–10 = 0
x = 2.27 × 10–5
pOH = –log 2.27 × 10–5 = 4.644
pH = 14 – 4.644 = 9.356
292 | Chapter 17
Think about It
Comparing the pKb of codeine to that of morphine in Problem 17.54, we see that codeine and morphine are
identical in their basicity.
17.57. Collect and Organize
We are to explain why for H3PO4 Ka > Ka > Ka .
1
2
3
Analyze
The equations describing these acid dissociation constants are as follows:
H3PO4(aq)  H2PO4–(aq) + H+(aq)
K a1
H2PO4–(aq)  HPO42–(aq) + H+(aq)
Ka2
HPO42–(aq)  PO43–(aq) + H+(aq)
Ka
3
Solve
With each successive ionization, it becomes more difficult to remove H+ from a species that is negatively
charged. Therefore it is harder to remove H+ from HPO42– than from H2PO4– than from H3PO4. This is
reflected in decreasing Ka values as H3PO4 is ionized.
Think about It
From Appendix 5 we can compare the Ka values for phosphoric acid: K a = 7.11 × 10–3, K a = 6.32 × 10–8,
1
2
K a = 4.5 × 10–13. These span ten orders of magnitude.
3
17.59. Collect and Organize
We have to use the K a of H2SO4 to calculate the pH of a solution of 0.300 M H2SO4.
2
Analyze
The first H+ is completely removed from the H2SO4 and the initial concentrations of the species in solution
before the second ionization are [H+] = 0.300 M, [H2SO4] = 0.0 M, and [HSO4–] = 0.300 M. The equation
describing the second ionization is
HSO4–(aq)  SO42–(aq) + H+(aq)
Ka =
Solve
Initial
Change
Equilibrium
[H + ][SO4 2– ]
= 1.2 × 10–2
–
[HSO4 ]
[HSO4–]
0.300
–x
0.300 – x
[H+]
0.300
+x
0.300 + x
[SO42–]
0
+x
x
Plugging equilibrium concentrations into the Ka expression gives
1.2 × 10–2 =
x(0.300 + x)
0.300 − x
Solving this by the quadratic equation gives
x2 + 0.312x – 0.0036 = 0
x = 0.0111
[H+] = 0.300 + 0.0111 = 0.311
pH = –log 0.311 = 0.51
Think about It
If we did not take into consideration the second ionization of H2SO4 we would have underestimated the
acidity of the solution by 0.016 pH units.
Equilibrium in the Aqueous Phase | 293
17.61. Collect and Organize
Given a 0.250 M solution of ascorbic acid and the K a and K a for this weak diprotic acid (1.0 × 10–5 and
2
1
5 × 10–12, respectively), we are to calculate the pH.
Analyze
Because the K a is so much smaller than the K a for ascorbic acid, we can say that the second ionization
2
1
contributes little to the [H+] in the solution. Therefore, we can solve this by examining only the first ionization.
Solve
Initial
Change
Equilibrium
[Ascorbic acid]
0.250
–x
0.250 – x
[Ascorbate–]
0
+x
x
[H+]
0
+x
x
x2
x2
≈
0.250 − x 0.250
x = 1.58 × 10–3
Checking the simplifying assumption shows that it is valid:
1.58 × 10–3
× 100 = 0.63%
0.250
The pH is found from the [H+]:
pH = –log 1.58 × 10–3 = 2.80
1.0 × 10–5 =
Think about It
This solution is about as acidic as vinegar.
17.63. Collect and Organize
Given a 0.00100 M solution of nicotine and the K b and K b (from Appendix 5) for this weak dibasic
1
2
compound, we are to calculate the pH.
Analyze
The K b (1.3 × 10–11) is so much smaller than the K b (1.0 × 10–6) for nicotine that we may ignore the
2
1
contribution of the second ionization of nicotine to the [OH–] in the solution. We can therefore solve this
problem by examining only the first ionization.
Solve
[NicotineH+]
Initial
0
Change
+x
Equilibrium
x
2
x
x2
1.0 ×10 – 6 =
≈
0.00100 − x 0.00100
x = 3.16 ×10 –5
Checking the simplifying assumption shows that it is valid:
3.16 × 10–5
× 100 = 3.2%
0.00100
[Nicotine]
0.00100
–x
0.00100 – x
We can calculate the pOH and pH from the [OH–]:
pOH = –log 3.16 × 10–5 = 4.500
pH = 14 – 4.500 = 9.500
Think about It
[OH–]
0
+x
x
294 | Chapter 17
The relatively high pH of this dilute solution of nicotine shows that this compound is a fairly strong weak
base.
17.65. Collect and Organize
Given a 0.01050 M solution of quinine and the K b and K b for this weak dibasic compound, we are to
1
2
calculate the pH.
Analyze
The K b (1.4 × 10–9) is so much smaller than K b (3.3 × 10–6) for quinine that we may ignore the contribution
2
1
of the second ionization of quinine to the [OH–] in the solution. We can therefore solve this problem by
examining only the first ionization.
Solve
Initial
Change
Equilibrium
[Quinine]
0.01050
–x
0.01050 – x
[QuinineH+]
0
+x
x
[OH–]
0
+x
x
x2
x2
≈
0.01050 − x 0.01050
x = 1.86 × 10– 4
Checking the simplifying assumption shows that it is valid:
1.86 × 10 – 4
× 100 = 1.8%
0.01050
The pOH and pH can be calculated from the [OH–]:
pOH = –log 1.86 × 10– 4 = 3.730
pH = 14 – 3.730 = 10.270
3.3 × 10– 6 =
Think about It
Quinine has a very complicated molecular structure that includes aromatic rings, an alcohol, an amine, an
alkene, and an ether as functional groups.
17.67. Collect and Organize
We are asked to explain why H2SO4 is a stronger acid (greater Ka ) than H2SeO4.
1
Analyze
The only difference in these acids is the central atom. Sulfur and selenium belong to the same group in the
periodic table. These elements differ in size and electronegativity.
Solve
Sulfur is more electronegative than selenium. This higher electronegativity on the sulfur atom stabilizes the
anion HSO4– more than the anion HSeO4–. Therefore, H2SO4 is a stronger acid.
Think about It
We would expect this trend to continue, so we predict that H2TeO4 is a weaker acid than H2SeO4.
17.69. Collect and Organize
We are to predict which acid of a pair is stronger.
Equilibrium in the Aqueous Phase | 295
Analyze
The more oxygen atoms bound to the central atom and the more electronegative the central atom (X) in the
acid, the more acidic is the compound.
Solve
(a) H2SO3 is a stronger acid than H2SeO3.
(b) H2SeO4 is a stronger acid than H2SeO3.
Think about It
The presence of oxygen atoms bound to the central atom in an oxyacid can have a dramatic effect on acidity.
17.71. Collect and Organize
Given the pKa values of conjugate acids of three pyridine derivatives where methyl groups are added, we are
to determine if more methyl groups increase or decrease the basicity of pyridine.
Analyze
The pKa is a measure of the conjugate acid’s acidity. From the equation
pKa = –log Ka
we see that the larger the pKa, the weaker is the acid. The weaker the acid, the stronger is the conjugate base.
Solve
As methyl groups are added the pKa increases so the acidity decreases. If the acidity decreases, the basicity of
the conjugate base increases. Therefore, more methyl groups on the parent pyridine increases the base
strength.
Think about It
Our prediction is true. The Kb values of the pyridine bases show that more methyl groups lead to increased
basicity.
N
Kb = 1.51 × 10–9
N
Kb = 9.77 × 10–8
N
Kb = 2.69 × 10–7
17.73. Collect and Organize
Of the three salts given, we are to determine which, when dissolved in water, gives an acidic solution.
Analyze
To give an acidic solution, the cation of the salt must donate a proton to water without the anion reacting with
water, or if the anion hydrolyzes, then the pKa of the cation must be lower than the pKb of the anion.
Solve
Both NH4+ and CH3COO– of ammonium acetate hydrolyze:
NH4+(aq) + H2O (l )  NH3(aq) + H3O+(aq)
pKa = 9.25
CH3COO–(aq) + H2O (l )  CH3COOH(aq) + OH–(aq)
pKb = 9.25
Because pKa = pKb, this salt’s solution is neutral.
Only NH4+ of ammonium nitrate hydrolyzes. This gives an acidic solution:
NH4+(aq) + H2O (l )  NH3(aq) + H3O+(aq)
–
Only HCOO of sodium formate hydrolyzes. This gives a basic solution:
HCOO–(aq) + H2O (l )  HCOOH(aq) + OH–(aq)
Therefore, of the three salts, only ammonium nitrate dissolves to give an acidic solution.
296 | Chapter 17
Think about It
Remember that neither Na+ nor NO3– hydrolyze because they would form either a strong base or a strong acid
which are always 100% ionized.
Na+(aq) + H2O (l ) ⎯⎯
→ NaOH(aq) + H+(aq)
NO3–(aq) + H2O (l ) ⎯⎯
→ HNO3(aq) + OH–(aq)
17.75. Collect and Organize
We consider why lemon juice is used to reduce the fishy odor due to the presence of (CH3)3N in not-so-fresh
seafood.
Analyze
Trimethylamine is a weak base, and lemon juice contains citric acid that is a weak acid.
Solve
The citric acid in the lemon juice neutralizes the volatile trimethylamine to make a nonvolatile dissolved salt
that neutralizes the fishy odor:
HOC(CH2)2(COOH)3(aq) + (CH3)3N(aq)  HOC(CH2)2(COOH)2COO–(aq) + (CH3)3NH+(aq)
Think about It
Because the pKb of trimethylamine of 4.19 and the pKa of citric acid of 3.13 are lower than the pKa of
trimethylammonium (9.81) and the pKb of citrate (10.87), this equilibrium lies to the right favoring the
products.
17.77. Collect and Organize
Given Ka = 2.1 × 10–11 for the conjugate acid of saccharin, we are asked to calculate the value of pKb for
saccharin.
Analyze
From the Ka we can calculate the pKa:
pKa = –log Ka
From the pKa we can calculate the pKb because pKw = 14.00 at 25˚C
pKb = 14.00 – pKa
Solve
pKa = –log(2.1 × 10–11) = 10.68
pKb = 14.00 – 10.68 = 3.32
Think about It
Alternatively, we could calculate the Kb from Ka using
Kb =
and then calculate pKb using
Kw
Ka
pKb = –log Kb
17.79. Collect and Organize
From Appendix 5 we know that the Ka of HF is 6.8 × 10– 4. Using this we are to calculate the pH of a solution
that is 0.00339 M in NaF.
Analyze
When NaF dissolves in water the F– ion hydrolyzes to give a basic solution:
F–(aq) + H2O (l )  HF(aq) + OH–(aq)
Equilibrium in the Aqueous Phase | 297
The Kb for this reaction is
Kw
1× 10 –14
= 1.47 × 10 –11
K a 6.8 × 10 – 4
We can solve for [OH–] using an ICE table and then compute the pH.
Kb =
Solve
=
[F–]
0.00339
–x
0.00339 – x
[HF]
Initial
0
Change
+x
Equilibrium
x
2
x
x2
1.47 × 10–11 =
≈
0.00339 − x 0.00339
x = 2.23 × 10–7
Checking the simplifying assumption shows that it is valid:
2.23 × 10–7
× 100 = 0.0066%
0.00339
The pH is calculated from the [OH–]:
pOH = –log 2.23 × 10–7 = 6.65
pH = 14 – 6.65 = 7.35
[OH–]
0
+x
x
Think about It
Because HF is a moderately strong weak acid, F–, its conjugate base, is a fairly weak base.
17.81. Collect and Organize
We are to explain why a solution of CH3COOH with CH3COONa is a better pH buffer than a solution
containing NaCl and HCl.
Analyze
A buffer is composed of a weak acid and its conjugate base.
Solve
Because the weak acid CH3COOH is combined with its weak conjugate base, CH3COO–, this buffer can
absorb added H+ or OH–. The other mixture, HCl with Cl–, is a strong acid paired with its very, very weak
conjugate base. This pairing cannot absorb added H+ or OH–.
Think about It
It is a key idea that for the acid–base pair in a buffer system both conjugates be weak.
17.83. Collect and Organize
For a buffer that is 0.244 M in acetic acid and 0.122 M in sodium acetate, we can use the Henderson–
Hasselbalch equation to calculate the pH of this buffer at 25˚C (Ka = 1.76 × 10–5) and at 0˚C (Ka = 1.64 × 10–5).
Analyze
The Henderson–Hasselbalch equation is
[base]
[acid]
For this problem, [base] = [acetate] = 0.122 M and [acid] = [acetic acid] = 0.244 M. Because the Ka values are
different for the two temperatures, the pH of these solutions will differ.
pH = pK a + log
Solve
At 25˚C, pH = –log(1.76 × 10–5 ) + log
0.122
= 4.453
0.244
298 | Chapter 17
At 0˚C, pH = –log(1.64 × 10–5 ) + log
0.122
= 4.484
0.244
Think about It
At the lower temperature the pH of this buffer is less acidic than at the higher temperature.
17.85. Collect and Organize
For a buffer that is 0.225 M in both HPO42– and PO43– (with the Ka of HPO42– equal to 4.5 × 10–13), we can use
the Henderson–Hasselbalch equation to calculate the pH of the buffer. From the pH we can also calculate the
pOH.
Analyze
The Henderson–Hasselbalch equation is
pH = pK a + log
[base]
[acid]
For this problem, [base] = [acid] = 0.225 M.
Solve
0.225M
= 12.34
0.225M
pOH = 14 – pH = 14 –12.34 = 1.65
pH = –log(4.5 × 10–13 ) + log
Think about It
Notice that when the concentrations of the acid and its conjugate base are equal, pH = pKa because log 1 = 0.
17.87. Collect and Organize
Given the pH of an acetic acid–acetate buffer solution (pH = 3.56) where the Ka of acetic acid is 1.76 × 10–5,
we can use the Henderson–Hasselbalch equation to calculate the ratio of acetate ion to acetic acid in the
solution.
Analyze
Rearranging the Henderson–Hasselbalch equation to solve for the ratio gives
[acetate]
pH = pK a + log
[acetic acid]
[acetate]
= pH – pK a
[acetic acid]
[acetate]
( pH – pK a )
= 1× 10
[acetic acid]
log
Solve
The pKa = –log(1.76 × 10–5) = 4.754, so pH – pKa = 3.56 – 4.754 = –1.194 and the ratio of acetate to acetic
acid is
[acetate]
= 1 × 10 –1.194 = 0.064
[acetic acid]
Think about It
Because [acetic acid] > [acetate], the pH of this buffer is less than the pKa of acetic acid.
17.89. Collect and Organize
We can use the Henderson–Hasselbalch equation to determine the pH of a solution prepared by mixing a
volume of 0.05 M NH3 with an equal volume of 0.025 M HCl.
Equilibrium in the Aqueous Phase | 299
Analyze
Mixing equal volumes of solutions dilutes them both. Therefore, after mixing and before reaction, [NH3] =
0.025 M and [HCl] = 0.0125 M in the combined solution. When HCl reacts with NH3, NH4+ is produced
according to the equation
NH3(aq) + H+(aq) → NH4+(aq)
Stoichiometrically, this would give [NH3] = 0.0125 M and [NH4+] = 0.0125 M after complete reaction with
H+. Because this is a solution of an acid (NH4+) and its conjugate base (NH3), we can use the Henderson–
Hasselbalch equation to calculate the pH of the solution. To do so we need the pKa of NH4+ (9.25) from
Appendix 5 (Table A5.1).
Solve
pH = pKa + log
[NH3 ]
0.0125
= 9.25 + log
= 9.25
0.0125
[NH4 + ]
Think about It
Because the HCl added to the NH3 in this solution converts exactly half of the NH3 to NH4+, the ratio of the
concentrations equals 1 and the pH of the solution equals the pKa of NH4+.
17.91. Collect and Organize
We are to calculate how much of the strong acid HNO3 (10 M) we would add to a buffer solution that is
0.010 M acetic acid and 0.10 M sodium acetate to give a solution that is pH 5.00.
Analyze
We can use the Henderson–Hasselbalch equation to first calculate the ratio [acetate]/[acetic acid] needed to
give a solution of pH 5.00. We then use that ratio to determine how much HNO3 to add in order to convert the
acetate in solution to acetic acid so as to give that ratio. The pKa of acetic acid is 4.75.
Solve
[acetate]
[acetic acid]
[acetate]
5.00 = 4.75 + log
[acetic acid]
[acetate]
= 1.78
[acetic acid]
When we add HNO3 to the solution, acetate is converted to acetic acid and the amounts of each are
[acetate] = 0.10 – x
[acetic acid] = 0.10 + x
The ratio of these is 1.78. Solving for x gives
[acetate]
0.10 − x
= 1.78 =
[acetic acid]
0.10 + x
0.10 − x = 0.0178 + 1.78 x
pH = pK a + log
x = 0.0296
The value of x is the moles of HNO3 that we need to add to the solution. In mL of 10 M HNO3, then
0.0296 mol ×
1000 mL
= 3.0 mL for 1.00 liter of the buffer solution
10 mol
Think about It
The small volume addition of HNO3 does not appreciably change the concentration of the acid and the base,
so we do not need to account for it.
17.93. Collect and Organize
300 | Chapter 17
We are to compare the pH of 1.00 L of a buffer that is 0.120 M in HNO2 and 0.150 M in NaNO2 before and
after 1.00 mL of 12.0 M HCl is added.
Analyze
We can use the Henderson–Hasselbalch equation in both cases. After the addition of HCl, however, the
amounts (calculated in moles) of HNO2 and NO2– have to be adjusted before using the Henderson–
Hasselbalch equation. The pKa of HNO2 is 3.40.
Solve
Without added HCl the pH of the buffer solution is
pH = 3.40 + log
0.150
= 3.50
0.120
Because we have 1.00 L of the buffer solution, we originally have 0.120 mol HNO2 and 0.150 mol NO2– in
the solution. Adding 1.00 mL of 12.0 M HCl adds
1.00 mL ×
12.0 mol
= 0.0120 mol H+
1000 mL
This will increase the moles of HNO2 to 0.120 mol + 0.0120 mol = 0.132 mol and decrease the moles of NO2–
to 0.150 mol – 0.0120 mol = 0.138 mol. Because the volume of the solution is 1.00 L, the concentrations of
these species are [HNO2] = 0.132 M and [NO2–] = 0.138 M. Using the Henderson–Hasselbalch equation to
calculate the pH gives
pH = 3.40 + log
0.138
= 3.42
0.132
Think about It
The pH of the buffer changed very little. The change in pH of 1.00 L of water after adding 0.0120 mol of H+
would be from 7.00 to 1.92.
17.95. Collect and Organize / Analyze
We compare the terms molar solubility and solubility product.
Solve
Molar solubility is the mole of a substance that dissolves in a solvent. The solubility product is the equilibrium
constant for the dissolution of a substance.
Think about It
Solubility has units of grams or moles per volume of solution, but, like other equilibrium constants, the
solubility product is unitless.
17.97. Collect and Organize
By comparing the Ksp values of MgCO3, CaCO3, and SrCO3 we can identify which cation (Mg2+, Ca2+, or
Sr2+) precipitates first as carbonate mineral.
Analyze
From Appendix 5, the Ksp values are
MgCO3
Ksp = 6.8 × 10–6
CaCO3
Ksp = 5.0 × 10–9
SrCO3
Ksp = 5.6 × 10–10
Equilibrium in the Aqueous Phase | 301
Solve
Because SrCO3 has the lowest Ksp, the cation Sr2+ precipitates first as a carbonate mineral.
Think about It
The order of solubility from least to most soluble for these carbonates is SrCO3 < CaCO3 < MgCO3.
17.99. Collect and Organize
For the case of SrSO4 whose Ksp increases as the temperature increases, we are to determine whether the
dissolution is exothermic or endothermic.
Analyze
We can include heat as a reactant (for an endothermic reaction) or as a product (for an exothermic reaction)
and apply Le Châtelier’s principle:
SrSO4(s) + heat  Sr2+(aq) + SO42–(aq)
SrSO4(s)  Sr2+(aq) + SO42–(aq) + heat
Solve
Applying Le Châtelier’s principle, we see that the reaction shifts to the right and more SrSO4 dissolves as the
temperature is increased. The dissolution is endothermic.
Think about It
The opposite effect of temperature occurs for an exothermic dissolution: less solid dissolves as the
temperature is increased.
17.101. Collect and Organize
By writing the equation for the dissolution of hydroxyapatite, we can explain why acidic substances erode
tooth enamel.
Analyze
The solubility of hydroxyapatite is described by
Ca5(PO4)3OH(s)  OH–(aq) + 3 PO43–(aq) + 5 Ca2+(aq)
Solve
Acidic substances react with the OH– released upon dissolution of hydroxyapatite. The equilibrium is shifted
to the right, dissolving more hydroxyapatite.
Think about It
The equilibrium would be shifted in the opposite direction (to the left) in an alkaline environment.
17.103. Collect and Organize
Given the [Ba2+] in a saturated solution of BaSO4 (1.04 × 10–5 M Ba2+), we are to calculate the value of Ksp
for BaSO4:
BaSO 4 (s)  Ba 2+ (aq) + SO 2−
(aq)
4
Analyze
The Ksp expression is
Ksp = [Ba2+][SO42–]
Because for every mole of BaSO4 that dissolves we get 1 mol of Ba2+ and 1 mol of SO42–, the molarities of
Ba2+ and SO42– are the same for this saturated solution of BaSO4.
Solve
Ksp = (1.04 × 10–5)(1.04 × 10–5) = 1.08 × 10–10
Think about It
302 | Chapter 17
Because BaSO4 is quite insoluble, we can add SO42– to a solution of dissolved Ba2+ to quantitatively
precipitate the barium out of solution. After weighing the dried precipitate we can then calculate how much
Ba2+ was present in the original solution.
17.105. Collect and Organize
Given that Ksp = 1.02 × 10–6, we are to calculate [Cu+] and [Cl–] for a saturated solution of CuCl.
Analyze
The solubility equation and Ksp expression for CuCl are
CuCl(s)  Cu + (aq) + Cl − (aq)
K sp [Cu + ][Cl – ]
The [Cu+] = [Cl–] in this solution because for every particle of CuCl that dissolves we get one particle of Cu+
and one particle of Cl–.
Solve
Let [Cu+] = [Cl–] = x. The Ksp expression becomes
Ksp = 1.02 × 10–6 = (x)(x)
x = 1.01 × 10–3
+
–
–3
Therefore, [Cu ] = [Cl ] = 1.01 × 10 M.
Think about It
We do not need to know how much CuCl is originally placed into the solution because it does not appear in
the Ksp expression as a pure solid.
17.107. Collect and Organize
Given the Ksp of CaCO3 (9.9 × 10–9), we are to calculate the solubility of this substance in units of grams per
milliliter.
Analyze
The solubility equation and Ksp expression for CaCO3 are
CaCO3 (s)  Ca 2+ (aq) + CO32− (aq)
K sp = [Ca 2+ ][CO32– ]
In this solution, [Ca2+] = [CO32–] because for every CaCO3 that dissolves one Ca2+ and one CO32– are
produced. We can then say that
Ksp = x2
and we can solve for x, which is the molar solubility (mol/L) of CaCO3. To convert this to grams per
milliliter we multiply by the molar mass of CaCO3 (100.09 g/mol) and divide by 1000 mL/L.
Solve
K sp = 9.9 × 10 –9 = x 2
x = 9.95 × 10 –5
9.95 × 10 –5 mol 100.09 g
1L
×
×
= 9.96 × 10 –6 g/mL
L
1 mol
1000 mL
Think about It
The value of x that we calculate in the Ksp expression is the molar solubility of the solid because it is that
amount (“x”) that dissolves into the solution.
calculate in the Ksp expression is the molar solubility of the solid because it is that amount (“x”) that
dissolves into the solution.
17.109. Collect and Organize
Given the Ksp for the dissolution of AgOH (1.52 × 10–8) in Appendix 5, we are to calculate the pH of a
saturated solution.
Equilibrium in the Aqueous Phase | 303
Analyze
The solubility equation and Ksp expression for AgOH are
AgOH(s)  Ag + (aq) + OH − (aq)
K sp = [Ag + ][OH − ]
Letting [Ag+] = [OH–] = x (because the stoichiometry is 1:1), we can solve for x using the Ksp expression.
The pH of the solution will then be
pH = 14 – (–log x)
Solve
Ksp = 1.52 ×10–8 = x 2
x = 1.233 ×10– 4
pH = 14 – ( – log 1.233 × 10– 4 ) = 10.091
Think about It
Even though the Ksp of AgOH is not high, this solution is quite basic.
17.111. Collect and Organize
Using the common-ion effect, we can determine in which 0.1 M solution (NaCl, Na2CO3, NaOH, or HCl) the
most CaCO3 dissolves.
Analyze
Whenever a common ion is already present in the solution, the CaCO3 is less soluble. Any solution,
therefore, with Ca2+ or CO32– would have lower solubility of CaCO3 compared to water. We also should look
for solutions that might react with either Ca2+ or CO32– and shift the solubility equilibrium to the right.
Solve
(a) NaCl(aq) has neither an ion common to CaCO3 nor ions that react with either Ca2+ or CO32–. CaCO3 has
the same solubility in this NaCl solution as in water.
(b) The 0.1 M Na2CO3 solution is 0.1 M in CO32–. This decreases the solubility of CaCO3.
(c) NaOH(aq) has neither an ion common to CaCO3 nor reacts with either Ca2+ or CO32–. CaCO3 has the
same solubility in this NaOH solution as water.
(d) A solution of HCl reacts with CO32– to form H2CO3 which then decomposes to H2O and CO2. This
reaction shifts the solubility equilibrium to the right so more CaCO3 dissolves.
The solution of (d) 0.1 M HCl dissolves the most CaCO3.
Think about It
A higher concentration of acid dissolves even more CaCO3 as the equilibrium shifts to the right:
CaCO3(s) + 2 H+(aq)  Ca2+(aq) + H2CO3(aq)
17.113. Collect and Organize
Given the average concentrations of SO4–2 and Sr2+ in seawater (0.028 M and 9 × 10–5 M, respectively) and
the Ksp of SrSO4 (3.4 × 10–7), we are to determine if the concentration of Sr2+ is controlled by the relative
insolubility of SrSO4.
Analyze
The solubility equation and Ksp expression for SrSO4 are
SrSO 4 (s)  Sr 2+ (aq) + SO 2–
(aq)
4
K sp = [Sr 2+][SO 2–
]
4
Solve
2–
K sp = 3.4 × 10 –7 = [Sr 2+ ][SO 4 ]
= [Sr 2+ ] ( 0.028 M )
[Sr 2 + ] = 1.21× 10 –5 M
304 | Chapter 17
This is the expected concentration of Sr2+ in seawater with a known [SO42–] of 0.028 M. This [Sr2+] is lower
than the average [Sr2+] of 9 × 10– M, so some other process must be controlling the [Sr2+].
5
Think about It
As the [SO42–] increases, the solubility of SrSO4 decreases because of the common-ion effect.
17.115. Collect and Organize
Given 125 mL solution that is 0.375M in Ca(NO3)2 we are asked whether CaF2 will precipitate when
245 mL of a 0.255 M NaF solution is added.
Analyze
The Ksp for CaF2 from Appendix 5 is 3.9 × 10–11. When the two solutions are mixed, the total volume is
370 mL and the [Ca2+] and [F –] in the mixed solution is
125 mL × 0.375M = 370 mL × [Ca 2 + ]
[Ca 2 + ] = 0.127 M
245 mL × 0.255M = 370 mL × [F− ]
[F− ] = 0.169 M
From the Ksp expression
Ksp = [Ca2+][F–]2 = 3.9 × 10–11
If the [Ca2+]initial × [F–]2initial > Ksp, then CaF2 will precipitate.
Solve
[Ca2+]initial × [F–]2initial = 0.127 × (0.169)2 = 3.63 × 10–3. This is greater than the value of Ksp for CaF2, so it will
preciptate from the mixed solution.
Think about It
Because CaF2 has a small solubility product constant, we expect that most of the ions will precipitate as
CaF2. In this solution the Ca2+ ions are in excess (0.0469 mol) compared to that of F– (0.0625 mol) so the
final solution will have only a small amount of F– in solution.
17.117. Collect and Organize
When a 0.250 M solution of Pb2+(aq) is added to a solution that is 0.010 M in Br– and SO42– we can use the
values of Ksp for PbBr2 and PbSO4 to determine which anion is the first to precipitate. We are then asked to
calculate the concentration of the anion that precipitates first at the moment that the second ion starts to
precipitate. This will be when the solution is saturated in the lead salt of the first anion to precipitate.
Analyze
The Ksp value shows that PbSO4 has a smaller solubility product constant (1.8 × 10–8) compared to that of
PbBr2 (6.6 × 10– 6).
Solve
(a) PbSO4 with a smaller solubility product constant will precipitate first from the solution, so the SO42–
anion will precipitate first.
(b) When Br– begins to precipitate, the maximum amount of Pb2+ that could be present that will not cause
Br– to precipitate is
Ksp = 6.6 × 10−6 = [Pb 2+ ] [Br – ]2
[Pb 2+ ] =
6.6 × 10−6
= 6.6 × 10−2 M
(0.010)2
The [SO42–] in the solution when [Pb2+] = 6.6 × 10–2 M is
Equilibrium in the Aqueous Phase | 305
Ksp = 1.8 × 10−8 = [Pb2+ ] [SO4 2 − ]
[Pb 2+ ] =
1.8 × 10−8
= 2.7 × 10−7 M
6.6 × 10−2
Think about It
We do not need an ICE table to solve this problem.
17.119. Collect and Organize
We are to describe the difference between a titration curve for a strong acid and that of a weak acid.
Analyze
A strong acid is completely ionized in aqueous solution whereas a weak acid is only partially hydrolyzed.
This affects the pH of the solution at the start of the titration for equal concentrations of the acids. The
equivalence or end point of the titration is where equal moles of OH– have been added to the acid. The
species formed at the end point for a strong acid and weak acid differ. This affects the pH of the solution at
the equivalence point.
Solve
The weak acid titration curve has an initial pH that is higher (less acidic) than that of an equimolar solution
of a strong acid (lower pH, more acidic). The pH at the equivalence point in the titration of a strong acid is
7.00 because the species formed in the titration are water and a nonhydrolyzing anion, such as Cl–:
HCl(aq) + OH–(aq) → H2O (l ) + Cl–(aq)
The pH at the equivalence point for a weak acid is basic because of the formation of a hydrolyzing anion,
such as NO2– in the titration of HNO2:
HNO2(aq) + OH–(aq) → NO2–(aq) + H2O (l )
NO2–(aq) + H2O (l )  HNO2(aq) + OH–(aq)
Think about It
Titration of a weak acid always gives an end point of pH > 7.
17.121. Collect and Organize
We are asked whether the pH at the equivalence point is the same for the titration of all weak acids with
strong base.
Analyze
When a weak acid is titrated, the species present in solution at the equivalence point is the conjugate base of
the weak acid. This conjugate base is a weak base and will hydrolyze in water according to the equation
A–(aq) + H2O (l )  HA(aq) + OH–(aq)
Solve
No. Because the extent to which A– (the conjugate base) hydrolyzes depends on the base strength of A–, the
pH at the equivalence point in the titration for weak acids is not expected to be the same.
Think about It
Likewise, the pH values at the equivalence point for weak bases differ based on the strength of the conjugate
acid.
17.123. Collect and Organize
We are to calculate the pH along various points of the titration curve when 25.0 mL of 0.100 M acetic acid
(Ka = 1.76 × 10–5) is titrated with 0.125 M NaOH.
Analyze
For each step of the titration we have to consider the moles of NaOH added that react with the moles of
CH3COOH initially present:
306 | Chapter 17
25.0 mL ×
0.100 mol
= 0.00250 mol CH3 COOH
1000 mL
For each point in the titration curve the moles of OH– (from NaOH) added are as follows:
0.125 mol
10.0 mL ×
= 0.00125 mol OH –
1000 mL
0.125 mol
20.0 mL ×
= 0.00250 mol OH –
1000 mL
0.125 mol
30.0 mL ×
= 0.00375 mol OH –
1000 mL
The OH– reacts with the CH3COOH in solution to give CH3COO–, the conjugate base of acetic acid. Our
strategy for the problem is to first react as much of the added OH– with acetic acid as possible and then use
the equilibrium Ka expression to calculate the pH:
CH3COOH(aq)  CH3COO–(aq) + H+(aq)
[CH3 COO – ][H + ]
[CH3 COOH]
Or we can use the equivalent equilibrium Kb expression:
CH3COO–(aq) + H2O (l )  CH3COOH(aq) + OH–(aq)
Ka =
[CH3COOH][OH – ]
[CH3COO – ]
For that calculation we have to be careful to determine the molarity of the species in solution by remembering
that the volume in a titration increases through the addition of the titrant.
Kb =
Solve
When 10.0 mL of OH– are added, the 0.00125 mol of OH– reacts with the 0.00250 mol of CH3COOH to
produce 0.00125 mol of CH3COO– and leave 0.00125 mol of CH3COOH unreacted. Since the total volume
of the solution is now 25.0 + 10.0 = 35.0 mL the molarity of these species is
0.00125 mol
= 0.0357 M
0.0350 L
Using an ICE table to calculate the pH of this solution gives the following:
[CH3COOH]
[CH3COO–]
[H+]
Initial
0.0357
0.0357
0
Change
–x
+x
+x
Equilibrium
0.0357 – x
0.0357 + x
x
x(0.0357 + x) x(0.0357)
≈
0.0357 − x
0.0357
–5
x = 1.76 × 10
The simplifying assumption is valid (0.05%) so
pH = –log 1.76 × 10–5 = 4.754
–
When 20.0 mL of OH are added we have added an equal number of moles of OH– as there are moles of
CH3COOH initially present. This reaction produces 0.00250 mol of CH3COO– so it makes sense here to use
the Kb expression to calculate the pH of the solution. Since the total volume of the solution is now 45.0 mL,
the molarity of these species is
0.00250 mol
= 0.0556 M
0.0450 L
Using an ICE table to calculate the pH of this solution gives the following:
[CH3COO–]
[CH3COOH]
[OH–]
Initial
0.0556
0
0
Change
–x
+x
+x
Equilibrium
0.0556 – x
x
x
1.76 × 10 –5 =
Equilibrium in the Aqueous Phase | 307
Kb =
Kw
Ka
=
1× 10−14
x2
x2
–10
=
5.68
×
10
=
≈
0.0556 − x 0.0556
1.76 × 10 –5
x = 5.62 × 10 – 6
The simplifying assumption is valid (0.01%) so
pOH = –log 5.62 × 10– 6 = 5.250
pH = 14 –5.250 = 8.750
When 30.0 mL of OH– are added we convert all of the CH3COOH to 0.00250 mol CH3COO– and have
0.00125 mol OH– remaining. Since the total volume of the solution is now 55.0 mL, the molarity of these
species is
0.00250 mol
= 0.0455 M CH3COO –
0.0550 L
0.00125 mol
= 0.0227 M OH –
0.0550 L
Using an ICE table to calculate the pH of this solution gives the following:
[CH3COO–]
[CH3COOH]
[OH–]
Initial
0.0455
0
0.0227
Change
–x
+x
+x
Equilibrium
0.0455 – x
x
0.0227 + x
x(0.0227 + x) x(0.0227)
5.68 × 10 –10 =
≈
0.0455 − x
0.0455
x = 1.14 × 10 –9
The simplifying assumption is valid so
pOH = –log 0.0227 = 1.644
pH = 14 –1.644 = 12.356
Think about It
When exactly half the moles of strong base are added in the titration of a weak acid (as in this problem
where 10 mL of the titrant were added), this point is the midpoint of the titration. Notice that at this point
pH = pKa of the weak acid.
17.125. Collect and Organize
In the titration of a 100.00 mL NH3 solution with 0.1145 M HCl, it takes 22.35 mL to reach the equivalence
point. From this information we are to calculate the concentration of ammonia in the solution.
Analyze
Because at the equivalence point the moles of HCl added as a titrant equal the moles of NH3 in the solution,
we can calculate the amount of NH3 through
Moles NH3 = mL HCl used as titrant × molarity of HCl solution ×
1 mol NH3
1 mol HCl
Because we know the volume of the original solution of NH3, the molarity of the sample is
mol NH3
volume of sample in L
Solve
0.1145 mol 1 mol NH3
×
= 2.559 × 10–3 mol
1000 mL 1 mol HCl
2.559 × 10–3 mol
Molarity of NH3 =
= 0.02559 M
0.100 L
Moles NH3 = 22.35 mL HCl ×
Think about It
Remember that at the equivalence point, what is equal is the moles, not the volume nor the concentration of
acid and base.
308 | Chapter 17
17.127. Collect and Organize
We are to calculate the volume of 0.0100 M HCl required to titrate 250 mL of 0.0100 M Na2CO3 to the first
equivalence point.
Analyze
Before setting out to do a lot of calculations here, let’s stop and think. The concentration of the titrant (HCl)
is equal to the concentration of the base we are titrating! We don’t need, therefore, to do any calculations
because the same volume of HCl is needed to neutralize the base.
Solve
Titration of 250 mL of 0.0100 M Na2CO3 to the first equivalence point requires 250 mL of HCl.
Think about It
To reach the second end point for the Na2CO3 solution, 500 mL of the HCl titrant would be required.
17.129. Collect and Organize
In comparing the titration of a weak acid in which the amount of NaOH titrant needed to reach the
equivalence point is double that for another titration, we are asked what the pH halfway to the equivalence
point is for the second titration if the pH at that point for the first titration is 4.44.
Analyze
The midpoint is where half of the weak acid has been converted into its conjugate base. It is at this point
where [acid] = [conjugate base] that the pH = pKa.
Solve
The pH at the midpoint in the second titration would be the same as it is in the first titration, 4.44.
Think about It
Certainly the volume of the added titrant at which the midpoint is reached for the second titration would be
twice that as in the first titration, but the pH at those points would be the same.
17.131. Collect and Organize
To sketch the titration curve for the titration of 50.0 mL of a solution of the weak acid HNO2 with 1.00 M
NaOH, we must use the information provided to calculate the pH of the HNO2 solution before any NaOH is
added and at the equivalence point.
Analyze
The initial pH of 0.250 M HNO2 is calculated from the value of Ka = 4.0 × 10– 4 after setting up an ICE table
where the initial concentrations of H+ and NO2– are 0.00 M. To calculate the pH of the solution at the
equivalence point we need to recognize that at that point all of the HNO2 is converted to NO2–. We use Kb,
therefore, for the ICE table calculation. We also need to take into account the added volume of the solution
when NaOH titrant is added.
Solve
The initial pH of 0.250 M HNO2 is found as follows:
[HNO2]
Initial
0.250
Change
–x
Equilibrium
0.250 – x
[H+]
0.00
+x
x
[NO2–]
0.00
+x
x
x2
x2
≈
0.250 − x 0.250
x = 0.0100
pH = –log 0.0100 = 2.00
–
At the equivalence point, moles of OH added = moles of HNO2 present in the initial solution:
K a = 4.0 ×10– 4 =
Equilibrium in the Aqueous Phase | 309
Moles HNO2 = 50.0 mL ×
0.250 mol
= 0.0125 mol
1000 mL
Because all the HNO2 is converted to NO2– at the equivalence point, the amount of NO2– at equivalence point
is initially 0.0125 mol. The NO2– produced hydrolyzes by the equation
NO2–(aq) + H2O (l )  HNO2(aq) + OH–(aq)
to give a slightly basic solution at the equivalence point.
We need to determine the volume of NaOH titrant added so that we can use the [NO2–] in an ICE table to
calculate the pH at the equivalence point. The 0.0125 mol of HNO2 requires 0.0125 mol of OH– to neutralize
it. The volume of NaOH needed to provide this 0.0125 mol OH– is
1 mol NaOH – 1000 mL
0.0125 mol OH – ×
×
= 12.5 mL
1.00 mol
1 mol OH –
The total volume of the solution at the equivalence point is 12.5 + 50.0 mL = 62.5 mL. The molarity of NO2–
at the equivalence point is
0.0125 mol NO2
= 0.200 M
0.0625 L
–
[NO2 ]
[HNO2]
[OH–]
Initial
0.200
0
0
Change
–x
+x
+x
Equilibrium
0.200 – x
x
x
This is a Kb expression so we must calculate Kb from Ka:
Kw
1× 10−14
Kb =
=
= 2.5 × 10 –11
K a 4.0 × 10 – 4
Solving the equilibrium constant expression for x gives
x2
x2
≈
0.200 − x 0.200
x = 2.24 × 10− 6
pOH = –log 2.24 × 10– 6 = 5.65
pH = 14 – 5.65 = 8.35
K b = 2.5 ×10–11 =
Think about It
Note in the titration curve that there is a relatively flat region before the equivalence point. Here the pH does
not change much despite the addition of more and more titrant. This is often called the buffer region.
17.133. Collect and Organize
For the titration of quinine, a dibasic malaria drug, we need to calculate the initial pH and the position of
each of the equivalence points in order to sketch the titration curve.
310 | Chapter 17
Analyze
Because the second base ionization constant (Kb = 1.35 × 10–9) is much smaller than the first (K b =
2
1
3.31 × 10 – 6), we can ignore it in calculating the initial pH of the solution. Because the concentration of the
quinine and HCl titrant are equal (both solutions are 0.100 M), we know that the first equivalence point will
be at the point where 40.0 mL of HCl have been added and the second equivalence point is at 80.0 mL of
HCl added.
Solve
The initial pH is calculated as follows:
Initial
Change
Equilibrium
[QuinineH+]
0
+x
x
[Quinine]
0.100
–x
0.100 – x
[OH–]
0
+x
x
x2
x2
≈
0.100 − x 0.100
x = 5.75 × 10− 4
K b = 3.31×10– 6 =
1
pOH = –log 5.75 ×10− 4 = 3.240
pH = 14 – 3.240 = 10.760
The first equivalence point is where 40.0 mL of HCl have been added to give
0.100 mol quinine 1 mol quinineH +
40.0 mL HCl ×
×
= 4.00 × 10–3 mol
1000 mL
1 mol quinine
The molarity of quinineH+ is
4.00 × 10–3 mol
= 0.0500 M
0.080 L
At this point quinineH+ can act both as an acid and as a base:
QuinineH+(aq) + H2O (l )  quinineH22+(aq) + OH–(aq)
K b1 =1.35 × 10–9
Kw
1× 10 –14
= 3.02 × 10 –9
K b2 3.31× 10 – 6
This calculation is beyond the scope of general chemistry study, but we can see that Ka > Kb and so we
2
1
expect this equivalence point to be slightly acidic. In fact, it is, at approximately pH 6.8.
The second equivalence point is where 80.0 mL of HCl have been added. This solution contains 4.00 ×
10–3 mol quinineH22+ with a molarity of
4.00 × 10–3 mol
= 0.0333 M
0.120 L
because the total volume is now 40.0 mL quinine solution + 80.0 mL HCl titrant. The pH at this equivalence
point is dependent on the equilibrium
Kw
1× 10 –14
QuinineH2+(aq)  quinineH+(aq) + H+(aq) K a1 =
=
= 7.41× 10 – 6
K b2 1.35 × 10 –9
The pH at this equivalence point is as follows:
[QuinineH2+]
[QuinineH+]
[H+]
Initial
0.0333
0
0
Change
–x
+x
+x
Equilibrium
0.0333– x
x
x
QuinineH+(aq)  quinine(aq) + H+(aq)
Ka2 =
=
Equilibrium in the Aqueous Phase | 311
x2
x2
≈
0.0333 − x 0.0333
x = 4.97 × 10− 4
pH = –log 4.97 ×10− 4 = 3.304
Ka = 7.41×10– 6 =
1
Think about It
Notice that at the two midpoints (20 mL and 60 mL HCl added) that the pH equals pK a (8.52) and pKa
2
1
(5.13).
17.135. Collect and Organize
We are asked to describe the changes in bonding and intermolecular forces when the weak base CH3NH2 is
dissolved in water.
Analyze
Methylamine is a gas and when dissolved in water the individual methylamine molecules are surrounded
with water. CH3NH2 also reacts with water according to the equation
CH 3 NH 2 (aq) + H 2O()  CH 3 NH +3 (aq) + OH – (aq)
Solve
The hydrogen bonds between some of the water molecules must break and re-form around the species
CH3NH2. Also, the amine hydrolyzes and forms CH3NH3+ and OH–; thus, ion–dipole forces are added when
these ions are surrounded by water molecules.
Think about It
Depending on the strength of the forces formed versus those broken, this dissolution may be either
exothermic or endothermic.
17.137. Collect and Organize
We are to compare the structures of H3PO3 and H3PO4 and explain why the K a1 values for these acids are similar.
Analyze
After drawing the Lewis structure and identifying which H atom is ionizable on H3PO3, we can compare that
to the structure of H3PO4.
312 | Chapter 17
Solve
(a) and (b) Phosphorous acid has the Lewis structure
(c) Phosphoric acid has a similar structure with its ionizable H atoms also bonded to oxygen atoms, so it is
not surprising that these two acids have similar values for Ka .
1
Think about It
Notice that the H atom bonded to P in H3PO3 is not ionizable. The electronegativity difference between P
and H is not great enough to make this H atom acidic.
17.139. Collect and Organize
We are asked whether the pH of the solution changes when a cook adds more baking soda to water used in a
recipe and to explain why or why not.
Analyze
Baking soda is a soluble sodium salt with the formula NaHCO3. In solution this salt forms Na+(aq) + HCO3–.
Na+ does not react with water but HCO3– does, which changes the pH of the solution.
Solve
Yes, the pH of the solution increases due to the increase in hydrolysis of HCO3– according to the equation
HCO3– (aq) + H 2O()  H 2CO3 (aq) + OH – (aq)
Think about It
H2CO3 decomposes at baking temperatures to give CO2(g) and H2 O(l ).
17.141. Collect and Organize
Given the change in pH of a lake from 6.1 to 4.7 when 400 gallons of 18 M H2SO4 were added to the lake,
we are to calculate the volume of the lake.
Analyze
To make this calculation easier, we can assume that we only ionize the first proton on H2SO4. First, we must
calculate the moles of H2SO4 added to the lake and then determine the increase in the concentration of H+ in
the lake in going from pH 6.1 to pH 4.7. Knowing that we simply divide the moles of acid added by the
molarity change of H+ in the lake, we can obtain the size of the lake.
Solve
The amount of H2SO4 added is
400 gal ×
3.78 L 18 mol
×
= 27, 216 mol
1 gal
L
The change in lake [H+] is
The size of the lake is
Initial [H+] = 1 × 10– 6.1 = 7.94 × 10–7 M
Final [H+] = 1 × 10– 4.7 = 2.00 × 10–5 M
∆[H+] = 1.92 × 10–5 M increase
27, 216 mol ×
1L
= 1.4 × 109 L
1.92 × 10–5 mol
Equilibrium in the Aqueous Phase | 313
Think about It
This volume is equivalent to 1.4 × 106 m3. If the lake were 10 m deep, it would cover an area of 1.4 × 105 m2.
If thought of as a square, that is 374 m on a side.
17.143. Collect and Organize
For the drug Zoloft we are to use Figure P17.143 to determine which form is for the acid salt and whether
solutions of Zoloft are acidic or basic.
Analyze
It is important to remember that this drug is sold as the HCl salt.
Solve
(a) The acid salt form has H on the amine (R2NH) moiety with Cl– as a counterion. This structure is shown
on the right of Figure P17.143.
(b) Because this drug is sold as the HCl salt, solutions of this drug are acidic.
Think about It
Many drugs are sold as HCl salts to render the drugs more soluble in aqueous solution.
17.145. Collect and Organize
By examining the equilibrium reactions of HF in water and aqueous F–, we are to determine the major
species present at pH 7.00, calculate the equilibrium constant for the combination of the two equilibrium
equations, and, finally, calculate the pH and [HF2–]eq when [HF] is 0.150 M.
Analyze
(a) By examining the values of the two equilibrium constants, and considering the concentrations of species
in solution, we can predict which species is more likely to be present at pH 7.00, F– or HF2–.
(b) The overall equation is the sum of the two equilibrium reactions, so the K for the combined reaction is
the product of the two K values for the individual reactions.
(c) We can use an ICE table and the value of the overall K calculated in part b to determine the pH and [HF2–]eq.
Solve
(a) Because the equilibrium constant of the reaction of F– with HF is larger than the dissociation of HF, we
might expect the most likely species to be HF2–. However, because HF is weak, the [F–] is low compared to
that of water and so HF reacts with H2O to form F– as the major anionic species.
(b) Koverall = Ka × K = (1.1 × 10–3) × (2.6 × 10–1) = 2.86 × 10– 4
(c) We must tackle this problem in two steps. First, we consider the hydrolysis of HF (aq).
[HF]
[H3O+]
[F–]
Initial
0.150
0
0
Change
–x
+x
+x
Equilibrium
0.150 – x
x
x
1.1×10 –3 =
x2
( 0.150 − x )
x 2 + 1.1×10 –3 x − 1.65 ×10 – 4 = 0
x = 0.0123
–
So [F ] = 0.0123 M and [HF] = 0.138 M
Now, we consider the second equilibrium
[F–]
Initial
0.0123
Change
–x
Equilibrium
0.0123 – x
[HF]
0.138
–x
0.138 – x
[HF2–]
0
+x
x
314 | Chapter 17
0.26 =
x
x
≈
(0.0123 − x)(0.138 − x) (0.0123)(0.138)
x = 4.4 ×10– 4
Therefore,
pH = –log (0.0123) = 1.91
[HF2–] = 4.4 × 10– 4 M
Think about It
Be careful in making simplifying assumptions. For the first equilibrium, we must solve using the quadratic
equation.
17.147. Collect and Organize
Given the structure of Naproxen (Figure P17.147), we are to draw the structure of the sodium salt, explain
whether a solution of the salt is acidic or basic, and explain why the salt is more soluble than Naproxen itself.
Analyze
The ionizable functional group in Naproxen is the carboxylic acid (– COOH) group.
Solve
(a)
(b) A solution of the salt of Naproxen is basic because the ionized – COO– group reacts with water, giving
– COOH + OH–:
(c) The salt is more soluble because it is charged and water molecules form stronger ion – dipole forces
around the molecule compared to the dipole–induced dipole forces between the neutral molecule and water.
Think about It
Being soluble in water also helps to deliver the drug to the body.
17.149. Collect and Organize
For three reactions of nitrogen and sulfur compounds, we are to complete the equations.
Analyze
All three reactants are covalent compounds. Reactions of covalent compounds are discussed on page 865 of
Chapter 17.
Solve
(a) SO3(g) + H2 O(l ) → H2SO4 (l )
(b) 3 NO2(g) + H2 O(l ) → 2 HNO3 (l ) + NO(g)
(c) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Think about It
The first two reactions show how acids are produced from nonmetal oxides.
Equilibrium in the Aqueous Phase | 315
17.151. Collect and Organize
We are asked which steps of the Ostwald synthesis of nitric acid would have a higher yield at higher
temperature.
Analyze
By Le Châtelier’s principle the yield of a reaction increases as temperature increases for endothermic
reactions.
Solve
The steps in the Ostwald process with their ∆H values calculated from the data in Appendix 4 are as follows:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
ΔH = [(4 × 90.3) + (6 × –241.8)] – [(4 × – 46.1) + (5 × 0)]
= –905.2 kJ
2 NO(g) + O2(g)  2 NO2(g)
ΔH = (2 × 33.2) – [(2 × 90.3) + (1 × 0)]
= –114.2 kJ
3 NO2(g) + H2 O(l )  2 HNO3 (l ) + NO(g)
ΔH = [(2 × –174.1) + (1 × 90.3)] – [(3 × 33.2) + (1 × –285.8)]
= –71.7 kJ
All of these reactions are exothermic, so none of the steps has a higher yield at a higher temperature.
Think about It
The reactions, however, are run at 900˚C, which enhances the rate of the reaction.
17.153. Collect and Organize
o
o
After writing the equations corresponding to ΔH f,SO
and ΔH f,SO
, we are to apply Hess’s law to show how to
3
2
o
calculate ΔH rxn for
2 SO2(g) + O2(g) → 2 SO3(g)
Analyze
The chemical reactions corresponding to ΔH fo involve preparing the compound from the elements.
Solve
The balanced formation reactions for SO2 and SO3 are
1
8 S8 ( s ) + O 2 (g ) → SO 2 (g )
o
ΔH f,SO
2
o
S8 ( s) + 32 O 2 (g ) → SO3 (g )
ΔH f,SO
3
Reversing the first reaction, multiplying it by 2, and adding to the second reaction (also multiplied by 2)
o
allow us to calculate ΔH rxn
.
o
2 SO2(g) → 14 S8(s) + 2 O2(g)
–2 ΔH f,SO
2
1
8
1
4
S8(g) + 3 O2(g) → 2 SO3(g)
2 SO2(g) + O2(g) → 2 SO3(g)
o
2 ΔH f,SO
2
o
o
o
2 ΔH f,SO
– 2 ΔH f,SO
= ΔH rxn
2
3
Think about It
Remember that Hess’s law is applicable to other state functions, such as S and G.
17.155. Collect and Organize
We are asked to compare the solubility in water of SO3 and NO2 with CO2 and whether the Henry’s law
constants for SO3 and NO3 are greater than that for CO2 (which is 3.5 × 10–2 M/atm).
Analyze
316 | Chapter 17
Carbon dioxide is a linear nonpolar molecule that forms dipole–induced dipole intermolecular interactions
when dissolved in water. To determine whether SO3 and NO2 are more or less soluble in water than CO2 we
must draw their Lewis structures and determine their molecular polarity.
Solve
SO3 is trigonal planar and therefore nonpolar like CO2:
SO3, like CO2, is nonpolar and forms dipole–induced
dipole interactions with water. These intermolecular
forces, however, are stronger for SO3 because SO3 is a
larger molecule. SO3, therefore, is slightly more soluble in
water than CO2.
NO2 is bent and therefore polar.
NO2 forms dipole – dipole interactions with water. These
are stronger than the water – CO2 intermolecular forces, so
NO2 is significantly more soluble in water compared to
CO2.
Yes, the corresponding constants for SO3 and NO2 are expected to be greater than the Henry’s law constant
for CO2.
Think about It
Of the three substances, NO2 is the most soluble in water.
17.157. Collect and Organize
After drawing the Lewis structure for H2S2O3, we are to consider the acid properties of thiosulfuric acid
compared to H2SO4.
Analyze
We are given that H2S2O3 is isostructural with H2SO4. This means that they have their atoms arranged in the
same way with an S atom in H2S2O3 taking the place of one of the O atoms in H2SO4.
Solve
(a)
(b) When a less electronegative sulfur atom replaces an oxygen atom in the acid, the acidity decreases.
Therefore, H2S2O3 is less acidic than H2SO4.
Think about It
Thiosulfuric acid is indeed less acidic than sulfuric acid. H2S2O3 has pKa = 0.6 and pK a = 1.6 whereas
1
2
H2SO4 has pKa < 0 and pK a = 1.92. While pK a for H2S2O3 is higher than that of H2SO4, remember that the
1
2
2
first ionization constant in diprotic acids dominates, so H2SO4 is stronger as an acid than H2S2O3.
CHAPTER 18 | The Colorful Chemistry of Metals
18.1. Collect and Organize
From the highlighted elements in Figure P18.1, we are to choose those whose chlorides are colored.
Analyze
Chloride compounds are colored for the transition elements that have incomplete d shells. The chlorides of the
highlighted elements and the electron configurations of the transition metal ions are
CaCl2 Ca2+
[Ar]
CrCl2 Cr2+
[Ar]3d 4
3+
CrCl3 Cr
[Ar]3d 3
2+
CoCl2 Co
[Ar]3d 7
3+
CoCl3 Co
[Ar]3d 6
2+
ZnCl2 Zn
[Ar]3d 10
Solve
Chromium (green) and cobalt (yellow) have colored chloride salts.
Think about It
Remember that we remove the s electrons first in forming transition metal cations.
18.3. Collect and Organize
Of the elements highlighted in Figure P18.3, we are to identify which have M2+ ions that form colorless
tetrahedral complex ions.
Analyze
Transition metal ions that are colorless have either filled or empty d orbitals. The electron configurations for
the M2+ cations are
Red
V2+
[Ar]d 3
2+
Purple
Mn
[Ar]d 5
2+
Yellow
Co
[Ar]d 7
2+
Blue
Zn
[Ar]d 10
Solve
Zinc (blue) forms colorless tetrahedral complex ions.
Think about It
Because nearly all tetrahedral complex ions are high spin with the d orbital splitting diagram shown below,
the numbers of unpaired electrons for the other ions in this problem are as follows: V2+, 3 unpaired e–; Mn2+,
5 unpaired e–; and Co2+, 3 unpaired e–.
18.5. Collect and Organize
For the structure of a chelating ligand shown in Figure P18.5, we are to count the number of electron-pair
donor groups for the ligand when the – COOH groups are ionized.
Analyze
The electron-pair donor groups on the neutral ligand include the –SH groups. When the carboxylic acid groups
are ionized, the – OH group becomes – O– and so these are added to the –SH electron-donor groups.
Solve
353
354 | Chapter 18
When the two carboxylic acid groups are ionized, two more electron-donor groups are added to give a total of
4 electron-pair donors.
Think about It
When ionized this ligand goes from being a bidentate to a tetradentate ligand.
18.7. Collect and Organize
In Figure P18.7 three solutions of different colors for three Co3+ ions are shown. Using the spectrochemical
series (Table 18.5) we can identify which solution is yellow, blue, and orange. The three ligands are F–, NH3,
and CN–.
Analyze
First, we have to recognize that the observed color is the complementary color of the light absorbed. The
yellow solution, therefore, absorbs violet light, the blue solution absorbs orange light, and the orange solution
absorbs blue light. The order of these wavelengths in order of increasing energy is
Orange < blue < violet
–
–
The higher the ligand (F , NH3, CN ) on the spectrochemical series, the higher is the energy absorbed because
of increased d–d splitting on the transition metal ion.
Solve
From the spectrochemical series, the order of splitting of the d orbitals on Co3+ for these ligands is
F– < NH3 < CN–
The solutions therefore are
(a) Yellow solution absorbing violet = [Co(CN)6]3–
(b) Blue solution absorbing orange = [CoF6]3–
(c) Orange solution absorbing blue = [Co(NH3)6]3+
Think about It
Ligands high on the spectrochemical series are called strong field ligands and are usually low-spin.
18.9. Collect and Organize
We are asked whether a substance may be a Lewis base (an electron-pair donator) but not a Brønsted–Lowry
base (proton acceptor).
Analyze
The difference between a Lewis base and a Brønsted–Lowry base hinges on what is being donated or accepted
by the base. For Lewis bases, an electron pair is donated to the acid; for Brønsted–Lowry bases, a proton is
accepted by the base.
Solve
Yes. A substance can be a Lewis base without being a Brønsted–Lowry base if it can donate an electron pair
but does not accept a proton.
Think about It
Cl–(aq) is an example of a Lewis base that is not a Brønsted–Lowry base. It may share an electron pair with a
transition metal, such as Co2+, but it does not accept a proton. (If it did accept a proton it would form HCl,
which is 100% ionized in solution.)
18.11. Collect and Organize
We are given that BF3 is a Lewis acid. We are to explain why it is not a Brønsted–Lowry acid.
Analyze
A Brønsted–Lowry acid is a proton donor. A Lewis acid is an electron-pair acceptor.
The Colorful Chemistry of Metals | 355
Solve
BF3 is a Lewis acid because it accepts electron pairs from Lewis bases to complete the octet around boron. It
is not a Brønsted–Lowry acid, however, because is does not have any H atoms to donate as H+.
Think about It
The reaction of NH3 with BF3 is typical of the Lewis acid–base reactions of BF3:
18.13. Collect and Organize
After drawing the Lewis structures for F– and BF3, we can show how the electron pairs move and bonds form
in the formation of BF4–, and identify the Lewis acid and Lewis base in the reaction.
Analyze
F– has a full octet in its Lewis structure. BF3 has only 6 electrons around B in its Lewis structure.
Solve
The Lewis base is F– and the Lewis acid is BF3.
Think about It
BF3 is often called an electron-deficient compound and is a strong Lewis acid.
18.15. Collect and Organize
After drawing the Lewis structures for CO2, H2O, and H2CO3, we can show how the electron pairs move and
bonds form and break in the formation of H2CO3, and identify the Lewis acid and Lewis base in the reaction.
Analyze
All the compounds have covalent bonding. A Lewis acid accepts an electron pair in an acid–base reaction,
whereas a Lewis base donates an electron pair.
Solve
In this reaction CO2 both accepts an electron pair (Lewis acid) and accepts a proton (Brønsted–Lowry base
and Lewis base). Likewise, H2O both donates an electron pair (Lewis base) and donates a proton (Brønsted–
Lowry acid and Lewis acid). Both CO2 and H2O in this reaction act as both Lewis acids and Lewis bases.
356 | Chapter 18
Think about It
Notice that one of the O —H bonds in water breaks in this reaction so that two O —H bonds are present in the
product, H2CO3.
18.17. Collect and Organize
After drawing the Lewis structures for B(OH)3, H2O, B(OH)4–, and H+, we can show how the electron pairs
move and bonds form and break in the formation of B(OH)4–, and identify the Lewis acid and Lewis base in
the reaction.
Analyze
All the compounds have covalent bonding. A Lewis acid accepts an electron pair in an acid–base reaction,
whereas a Lewis base donates an electron pair.
Solve
B(OH)3 is the Lewis acid, and H2O is the Lewis base.
Think about It
Notice that one of the O —H bonds in water breaks in this reaction so that an additional – OH group on B is
formed.
18.19. Collect and Organize
We are asked which molecules or ions (H2O or Cl–) surround Na+ when NaCl is dissolved in water.
Analyze
NaCl is completely soluble in water; the Na+ and Cl– ions are 100% dissociated.
Solve
Water molecules occupy the inner coordination sphere of Na+ ions.
Think about It
The oxygen atoms, which carry partial negative charge, are pointed toward the Na+ ion in the coordination
sphere.
18.21. Collect and Organize
We are asked which species surround Ni2+ when Ni(NO3)2 dissolves in water.
Analyze
The compound that is dissolved in water consists of Ni2+ and NO3– ions. We also have a lot of H2O that may
bind to the metal cation. The cation attracts species that are negative or that are polar. This means that
potentially Ni2+ might be surrounded by NO3– or H2O. Which one of these surrounds the Ni2+ depends, then,
on their Lewis basicity toward Ni2+. From our previous study of acid–base behavior, we know that in order of
basicity, NO3– < H2O.
The Colorful Chemistry of Metals | 357
Solve
H2O molecules in solution surround the Ni2+ ion because H2O is the strongest base.
Think about It
This means that the NO3– ions are spectator ions in the solution of Ni(NO3)2.
18.23. Collect and Organize
We are to identify the counter ion present in Na2[Zn(CN)4].
Analyze
A counter ion is the ion of opposite charge to the complex ion and is not directly bonded to the metal cation.
Solve
Na+, sodium ion
Think about It
The counter ion balances the charge to give a neutral salt. In this case we know that the charge on the complex
ion is 2– because there are two Na+ counter ions.
18.25. Collect and Organize
We are to explain why a solution of Cu2+ in the presence of EDTA is not toxic to phytoplankton.
Analyze
EDTA is a chelating ligand that forms very stable complexes by bonding its six Lewis base groups to a cation.
Solve
The Cu2+ in a solution containing EDTA is not toxic to phytoplankton because the Cu2+ ion is fully complexed
by the EDTA.
Think about It
How EDTA complexes with cations is shown in Figure 18.10.
18.27. Collect and Organize
We are to explain why AgCl dissolves easily in a solution of NH3 but not in water.
Analyze
The Ksp reaction for AgCl is
AgCl(s)  Ag + (aq) + Cl – (aq)
K sp = 1.8 ×10 –10
The Kf reaction for Ag(NH3)2+ is
Ag + (aq) + 2 NH 3 (aq)  Ag(NH 3 )+2 (aq)
K f = 1.7 ×107
Solve
Ag+ forms a soluble complex with NH3, removing Ag+ from solution and shifting the equilibrium for the
dissolution of AgCl to the right.
Think about It
The overall reaction for the dissolution of AgCl in aqueous ammonia is
AgCl(s) + 2 NH 3 (aq)  Ag(NH 3 )+2 (aq) + Cl – (aq)
K overall = K sp × K f = 3.1×10 –3
under standard conditions, 1 M NH3(aq).
18.29. Collect and Organize
Given the amount of Ni(NO3)2 (1 mmol) that is dissolved in 250 mL of a 0.500 M aqueous solution of NH3 we
can calculate the concentration of Ni(NO3)2 dissolved. Using the formation constant of Ni(NH3)62+ (Kf = 5.5 ×
358 | Chapter 18
108) we can then calculate the concentration of uncomplexed Ni2+(aq) in the solution at equilibrium. The
relevant equation is
Ni 2+ (aq) + 6 NH 3 (aq)  Ni(NH 3 )62+ (aq)
Analyze
Because the Kf for this complex is very large we can expect that all of the Ni2+ dissolved [from 1 mmol of
Ni(NO3)2] is converted into Ni(NH3)62+. We can use the Kf to calculate the amount of Ni2+ that remains in
solution. The amount of Ni(NO3)2 and Ni2+ initially in the solution is
1.0 × 10–3 mol
= 4.0 × 10–3 M
0.250 L
At equilibrium then, if [Ni2+] = x, the [Ni(NH3)62+] = 4.0 × 10–3 – x. The [NH3] at equilibrium will be [0.500 –
(6 × 4.0 × 10–3)] + 6x.
Solve
(a) The amount of Ni(NO3)2 initially dissolved in the solution is
1.0 × 10–3 mol
= 4.0 × 10–3 M
0.250 L
(b) To determine the concentration of free Ni2+ in the solution, we set up an ICE table and use the Kf
expression:
[Ni2+]
[NH3]
[Ni(NH3)62+]
Initial
Change
Equilibrium
4.0 × 10–3
– (4.0 × 10–3 – x)
x
5.5 × 108 =
0.500
–6(4.0 × 10–3 – x)
0.476 + 6x
0
4.0 × 10–3 – x
4.0 × 10–3 – x
4.0 × 10–3 – x 4.0 × 10 –3
≈
x(0.476 + 6x)6 x(0.476)6
x = 6.3 × 10–10 M
Think about It
Indeed, only a very small amount of nickel is present as uncomplexed Ni2+ in this solution.
18.31. Collect and Organize
Given the initial molar amounts of Co(NO3)2, NH3, and ethylenediamine (en) in a 500 mL solution
(1.00 mmol, 100 mmol, and 100 mmol, respectively), we can calculate the concentration of Co2+(aq) in the
solution at equilibrium.
Analyze
The complex ion formation equations are
Co 2+ (aq) + 6 NH 3 (aq)  Co(NH 3 )62+
K f = 7.7 ×104
Co 2+ (aq) + 3 en(aq)  Co(en)32+
K f = 8.7 ×1013
We can reverse the first equation and add it to the second equation to obtain an equilibrium equation for the
conversion of Co(NH3)62+ into Co(en)32+:
Co(NH 3 )62+ (aq) + 3 en(aq)  Co(en)32+ (aq) + 6 NH 3 (aq)
8.7 ×1013
= 1.13×109
7.7 ×104
Because the Kf for Co(en)32+ is so large we can assume that initially all Co2+ is converted into Co(en)32+.
Therefore [Co(en)32+]initial = 2.00 × 10–3 M, [en]initial = (0.200 M – 3 × 2.00 × 10–3), and [Co(NH3)62+]initial = 0.
Once this equilibrium is established we can calculate [Co2+] knowing [Co(NH3)62+] and using the first
equilibrium equation above.
K=
The Colorful Chemistry of Metals | 359
Solve
[Co(NH3)62+]
Initial
Change
Equilibrium
1.13 × 10
0
+x
x
9
[Co(en)3]2+
[en]
0.194
+3x
0.194 + 3x
( 2.00 × 10
=
−3
)
− x ( 0.200 − 6 x )
x ( 0.194 + 3 x )
[NH3]
–3
0.200
– 6x
0.200 – 6x
2.00 × 10
–x
2.00 × 10–3 – x
6
3
( 2.00 × 10 ) (0.200)
≈
−3
x ( 0.194 )
6
3
x = 1.55 × 10 −14
This result can now be used in the first equilibrium equation where [Co(NH3)62+] = 2.02 × 10–19 M, [NH3] ⊕
0.200 M, and [Co2+] = 0 initially:
[Co2+]
[NH3]
[Co(NH3)62+]
Initial
Change
Equilibrium
0
x
x
7.7 × 104 =
1.55 × 10–14
–x
1.55 × 10–14 – x
0.200
+6x
0.200 + 6x
(1.55 × 10
−14
−x
x ( 0.200 + 6 x )
(
6
) ≈ (1.55 × 10
4.928 x = 1.55 × 10
−14
x ( 0.200)
−14
−x
−x
)
6
)
5.928 x = 1.55 × 10 −14
x = 2.6 × 10−15 M = [Co 2 + ]
Think about It
In this 500 mL solution there are only
2.6 × 10–15 mol 6.022 × 1023 Co2+
0.500 L ×
×
= 7.8 × 108 Co2 + ions
L
1 mol
compared to the initial Co2+ dissolved in solution
6.022 × 1023 Co2+
1.00 × 10–3 mol ×
= 6.022 × 1020 Co2+ ions
1 mol
for a % of Co2+ remaining at equilibrium of
7.8 × 108
× 100 = 1.3 × 10–10 %
6.022 × 1020
18.33. Collect and Organize
We can use the conventions for naming given in Section 18.4 to name three transition-metal complex ions:
Cr(NH3)63+, Co(H2O)63+, and [Fe(NH3)5Cl]2+.
Analyze
For each cation, we first name the ligands in alphabetical order, indicating with a prefix how many of each
ligand are bonded to the metal ion. Then, we add the name of the metal, indicating the charge on the metal ion
with a Roman numeral.
Solve
(a) Hexaamminechromium(III)
(b) Hexaaquacobalt(III)
(c) Pentaamminechloroiron(III)
360 | Chapter 18
Think about It
Be sure to correctly account for the charge on the metal ion by considering the charge on the ligands and the
charge on the overall complex. In part c the chloro ligand has a 1– charge. With an overall charge on the
complex of 2+, the iron ion must have a 3+ charge.
18.35. Collect and Organize
We can use the conventions for naming given in Section 18.4 to name three transition metal complex ions:
CoBr42–, Zn(H2O)(OH)3–, and Ni(CN)53–.
Analyze
For each anion, we first name the ligands in alphabetical order, indicating with a prefix how many of each
ligand are bonded to the metal ion. Then we add the name of the metal, using -ate as the ending and indicate
the charge on the metal ion with a Roman numeral.
Solve
(a) Tetrabromocolbaltate(II)
(b) Aquatrihydroxozincate(II)
(c) Pentacyanonickelate(II)
Think about It
Be sure to correctly account for the charge on the metal ion by considering the charge on the ligands and the
charge on the overall complex. In part c the cyanide ligands have a 1– charge each. With an overall charge on
the complex of 3–, the Ni metal ion must have a 2+ charge.
18.37. Collect and Organize
We can use the conventions for naming given in Section 18.4 to name three transition metal coordination
compounds: [Zn(en)]SO4, [Ni(NH3)5(H2O)]Cl2, and K4Fe(CN)6.
Analyze
To name these coordination compounds, we separately name the cation and the anion, with the name of the
cation being written first.
Solve
(a) Ethylenediaminezinc(II) sulfate
(b) Pentaammineaquanickel(II) chloride
(c) Potassium hexacyanoferrate(II)
Think about It
We need not indicate the number of sulfate, chloride, or potassium ions in the name. They are understood to
be counter ions. When writing the formulas we would indicate how many are needed to balance the charge so
as to make a neutral compound.
18.39. Collect and Organize
For four chloride salts we are to identify which would give an acidic solution.
Analyze
All the salts are soluble. The Cl– does not hydrolyze but the cations might.
Solve
(a) Ca2+(aq) does not hydrolyze because it would form the strong base Ca(OH)2, which is 100% ionized in
solution.
(b) Cr3+(aq) does hydrolyze to form an acidic solution according to the equation
Cr3+(aq) + 3 H2O (l ) É
Cr(OH)3(s) + 3 H+(aq)
(c) Na+(aq) does not hydrolyze because it would form the strong base NaOH, which is 100% ionized in
solution.
The Colorful Chemistry of Metals | 361
(d) Fe2+(aq) does hydrolyze to form an acidic solution according to the equation
Fe2+(aq) + 2 H2O (l )  Fe(OH)2(s) + 2 H+(aq)
Both (b) CrCl3 and (d) FeCl2 produce an acidic solution.
Think about It
According to Table 18.2, Cr3+ is among the most acidic of the hydrated metal ions.
18.41. Collect and Organize
We are asked how the oxidation of Fe2+ to Fe3+ affects the acidity of the ozone solution. To answer this we are
to compare the acidity of Fe2+(aq) versus Fe3+(aq).
Analyze
Both Fe2+(aq) and Fe3+(aq) hydrolyze according to the equations
Fe2+(aq) + 2 H 2O()  Fe(OH)2(s) + 2 H+(aq)
Fe3+(aq) + 3 H 2O()  Fe(OH)3(s) + 3 H+(aq)
Solve
An aqueous solution of Fe3+ is more acidic and has a lower pH because Fe3+ hydrolyzes to a greater extent
than Fe2+. The higher ionic charge of Fe3+ polarizes the O —H bonds in water bound to Fe3+ to a greater extent
than Fe2+ does.
Think about It
Notice too that Fe3+ could hydrolyze to produce 3 mol of H+ rather than 2 mol of H+ from Fe2+ hydrolysis.
18.43. Collect and Organize
For the amphiprotic Cr(OH)3 we are to write chemical equations to describe this property.
Analyze
Amphiprotic compounds react with both acids and bases.
Solve
In basic solution Cr(OH)3 adds OH– to form the soluble Cr(OH)4– ion:
Cr(OH)3(s) + OH–(aq)  Cr(OH)4–(aq)
In acidic solution Cr(OH)3 reacts with H+ to form Cr3+ and water:
Cr(OH)3(s) + 3 H+(aq)  Cr3+(aq) + 3 H2 O(l )
Think about It
Other amphiprotic hydroxide compounds that behave similarly to Cr(OH)3 are Al(OH)3 and Zn(OH)2.
18.45. Collect and Organize
We are to explain why Ca2+, Mg2+, and Fe3+ combined with strong base (OH–) are insoluble, but Al3+ is
soluble.
Analyze
All of these metal cations form insoluble hydroxide compounds (see Appendix 5 for Ksp values ranging from
1.9 × 10–33 for Al3+ to 4.7 × 10–6 for Ca2+). Al(OH)3, however, is amphoteric and reacts with OH– to form a
complex ion.
Solve
Al(OH)3 reacts with OH– in solution to form soluble Al(OH)4–. The other ions do not form this type of soluble
complex ion and therefore remain insoluble as Mg(OH)2, Ca(OH)2, and Fe(OH)3 in strongly basic solution.
362 | Chapter 18
Think about It
If either Cr3+ or Zn2+ contaminated the ore, they too would be soluble as the complex ions Cr(OH)4– and
Zn(OH)42–.
18.47. Collect and Organize
From the Ka of Al3+(aq) we can calculate the pH of a solution that is 0.50 M in Al(NO3)3.
Analyze
The nitrate ions of Al(NO3)3 do not react with water. The reaction of Al3+ with water, however, gives an acidic
solution with Ka = 1 × 10–5:
Al 3+ (aq) + H 2O()  Al(OH) 2+ (aq) + H 3O+ (aq)
Solve
Initial
Change
Equilibrium
[Al3+]
[Al(OH)2+]
[H3O+]
0.50
–x
0.50 – x
0
+x
x
0
+x
x
1 × 10–5 =
The assumption that x << 0.500 is valid:
The pH of the solution is, therefore
( x)( x)
( 0.50 – x)
≈
x2
0.50
x = 2.24 × 10–3
2.24 × 10–3
× 100 = 0.45%
0.50
pH = –log(2.24 × 10–3) = 2.65
Think about It
Aluminum with its small size and high positive charge as Al3+ gives fairly acidic aqueous solutions.
18.49. Collect and Organize
From the Ka of Fe3+(aq) we can calculate the pH of a solution that is 0.100 M in Fe(NO3)3.
Analyze
The nitrate ions of Fe(NO3)3 do not react with water. The reaction of Fe3+ with water, however, gives an acidic
solution with Ka = 3 × 10–3:
Fe3+ (aq) + H 2O()  Fe(OH) 2+ (aq) + H 3O+ (aq)
Solve
Initial
Change
Equilibrium
[Fe3+]
[Fe(OH)2+]
[H3O+]
0.100
–x
0.100 – x
0
+x
x
0
+x
x
3 × 10–3 =
( x)( x)
( 0.100 – x)
≈
x = 1.73 × 10–2
x2
0.100
The assumption that x << 0.100 is not valid:
1.73 × 10–2
× 100 = 17%
0.100
so we must solve by the quadratic equation:
The Colorful Chemistry of Metals | 363
3 ×10 –3 (0.100 – x) = x 2
3 × 10 – 4 – 3 × 10–3 x = x 2
x 2 + 3 × 10 –3 x – 3 × 10 – 4 = 0
x = 0.0159 or – 0.0189
The concentration of H3O+ and Fe(OH)2+ must be positive, so x = 0.0159 and the pH of the solution is
pH = –log(0.0159) = 1.80
Think about It
Fe3+ is the most acidic hydrated metal ion listed in Table 18.3, so we expect this solution to be quite acidic.
18.51. Collect and Organize
We are to sketch the titration curve for the titration of 25 mL of 0.5 M FeCl3 with 0.50 M NaOH.
Analyze
The Fe3+ ions react with OH– to form, in steps, Fe(OH)3:
Fe3+ (aq) + OH – (aq)  Fe(OH) 2+ (aq)
Fe(OH) 2+ (aq) + OH – (aq)  Fe(OH)+2 (aq)
Fe(OH)+2 (aq) + OH – (aq)  Fe(OH)3 (s)
The titration curve shows three equivalence points. Notice that the concentration of OH– in the titrant equals
the concentration of Fe3+ in the solution. The equivalence points will, therefore, occur at 25 mL, 50 mL, and
75 mL.
Solve
Think about It
Remember that Fe(OH)3 does not further react with OH–.
18.53. Collect and Organize / Analyze
After first defining sequestering agent, we are to describe the properties that make a compound an effective
sequestering agent.
Solve
A sequestering agent is a multidentate ligand that separates metal ions from other substances so that they can
no longer react. Properties that make a sequestering agent effective include strong bonds formed between the
metal and the ligand and large formation constants.
Think about It
EDTA is an example of a good sequestering agent.
18.55. Collect and Organize
364 | Chapter 18
By examining the structure of an aminocarboxylate ligand, we can predict how its chelating ability changes as
pH changes.
Analyze
The general structure of an aminocarboxylate is
O
HO
NH2
Solve
+
At low pH the aminocarboxylate ligand contains – COOH and –NH3 groups. As the pH increases the chelating
ability increases because OH– removes the H on the amine and carboxylic acid groups, providing additional
sites for binding to the metal cation.
Think about It
EDTA is an example of an aminocarboxylate ligand.
18.57. Collect and Organize
We are to explain why most compounds of Sc through Zn are colored.
Analyze
Most of the first-row transition metals form ions with electrons in 3d orbitals.
Solve
When the transition metals bond to ligands, the d orbitals split in energy. If there is a d to d transition possible
for the ion, the compound is likely to be colored.
Think about It
Not all the first-row transition elements give colored complexes. For example, if there are no electrons in the d
orbitals as in Sc3+ or if the d orbitals are filled as in Zn2+, the compounds are not colored.
18.59. Collect and Organize
For a square-planar crystal field geometry, we are to explain why dxy is higher in energy than either the dxz or
dyz orbitals.
Analyze
In a crystal field d orbitals that are pointed directly at the ligands are raised in energy and those that are not
pointed directly at ligands are lowered in energy.
Solve
The repulsions due to the ligands in a square-planar crystal field are highest for the dxy orbital and so it is
raised in energy because this orbital lies in the plane of the ligands. The dxz and dyz orbitals, however, are
perpendicular to the plane of the four ligands, and therefore these orbitals are lower in energy compared to the
energy of the dxy orbital.
Think about It
In a square-planar geometry the d x2 – y2 orbital points directly at the four ligands. It has the most repulsions and
therefore is the highest orbital in energy.
18.61. Collect and Organize
The Colorful Chemistry of Metals | 365
Given yellow and violet aqueous solutions of Cr3+ complex ions, we are to determine which solution contains
Cr(H2O)63+ and which contains Cr(NH3)63+.
Analyze
From the spectrochemical series we know that NH3 splits the d orbitals to a greater extent than H2O (Table
18.5). Cr(NH3)63+ therefore absorbs a higher energy (shorter wavelength) of light than Cr(H2O)63+.
Solve
The wavelength (color) absorbed is complementary to the wavelength (color) observed. The yellow aqueous
solution absorbs violet light and the violet aqueous solution absorbs yellow light. Because violet light has a
higher energy and shorter wavelength than yellow light, the yellow solution contains (b) Cr(NH3)63+. The
violet solution contains (a) Cr(H2O)63+.
Think about It
A Cr3+ compound giving a red solution would have a d-orbital splitting energy between that of Cr(NH3)63+ and
Cr(H2O)63+.
18.63. Collect and Organize
Given the octahedral crystal field splitting for Co(phen)33+ (5.21 × 10–19 J/ion), we can determine the color of
the solution of Co(phen)33+ by calculating the wavelength of light absorbed and correlating that to the color of
light reflected or transmitted.
Analyze
To calculate the wavelength of light absorbed we rearrange E = hc/λ to solve for the wavelength of light:
hc
λ=
E
We can use the visible spectrum (400–700 nm) to determine the color of the absorbed wavelength and then
the color wheel (Figure 18.15) to choose the complementary color (the color we observe).
Solve
6.626 × 10–34 J ⋅ s × 3.00 × 108 m/s
= 3.82 × 10–7 m or 382 nm
5.21 × 10–19 J/ion
This wavelength is in the UV region, so the solution is colorless.
λ=
Think about It
If a complex absorbs in the UV, sometimes it appears slightly yellow because its absorption “tails” into the
violet region of the visible spectrum.
18.65. Collect and Organize
For the complexes NiCl42– and NiBr42–, whose solutions absorb light at 702 and 756 nm, respectively, we are
asked which ion has the greater d-orbital energy splitting.
Analyze
The shorter the wavelength absorbed, the larger the split of the d-orbital energies, ∆o.
Solve
Because NiCl42– absorbs at a shorter wavelength than NiBr42–, NiCl42– has a greater split of the d-orbital
energies.
Think about It
A solution of NiCl42– appears blue-green and a solution of NiBr42– appears yellow-green.
18.67. Collect and Organize / Analyze
We can use the crystal field model to explain how a transition metal may be either high spin or low spin.
366 | Chapter 18
Solve
The magnitude of the crystal field splitting energy compared to the pairing energy of the electrons in a lower
energy d orbital determines whether a transition metal cation is high spin or low spin.
Think about It
For a high-spin complex the pairing energy is greater than the crystal field splitting energy, whereas for a lowspin complex the pairing energy is less than the crystal field splitting energy.
18.69. Collect and Organize
For high-spin tetrahedral complexes of Fe2+, Cu2+, Co2+, and Mn3+, we are to determine the number of
unpaired electrons.
Analyze
The tetrahedral crystal field diagram is
Since all these complexes are high spin, the pairing energy is greater than the crystal field splitting, ∆t.
Solve
Fe2+ has a d 6 electron configuration, giving 4 unpaired electrons.
Cu2+ has a d 9 electron configuration, giving 1 unpaired electron.
Co2+ has a d 7 electron configuration, giving 3 unpaired electrons.
Mn3+ has a d 4 electron configuration, giving 4 unpaired electrons.
Think about It
Because Δt ≈ 4/9Δo, nearly all tetrahedral complexes are high spin.
18.71. Collect and Organize
Of the transition metal cations Co2+, Cr3+, Ni2+, and Zn2+, we are to determine which of their electron
configurations could result in either a high-spin or low-spin complex in a tetrahedral field.
Analyze
The tetrahedral field diagram is
and it may hold 10 e–. The possibility for high-spin or low-spin configurations occurs when, depending on the
magnitude of Δt, electrons may be placed in either the low-lying orbital set or the high-lying orbital set after
the d 2 configuration. These situations may occur for metal ions that have 3, 4, 5, or 6 d electrons.
Solve
Co2+ has the electron configuration [Ar]4s03d 7, so it has 7 electrons in its 3d orbitals.
Cr3+ has the electron configuration [Ar]4s03d 3, so it has 3 electrons in its 3d orbitals.
Ni2+ has the electron configuration [Ar]4s03d 8, so it has 8 electrons in its 3d orbitals.
Zn2+ has the electron configuration [Ar]4s03d 10, so it has 10 electrons in its 3d orbitals.
Only Cr3+ may have either high-spin or low-spin configurations.
Think about It
Because Δt ≈ 4/9Δo, nearly all tetrahedral complexes are high spin.
The Colorful Chemistry of Metals | 367
18.73. Collect and Organize
For the minerals MnO2 and Mn3O4, where the Mn ions are surrounded by six O2– ions (and are therefore in an
octahedral crystal field), we are to determine the charges of the Mn ions in each mineral and which mineral
might have a possibility of high-spin and low-spin Mn ions.
Analyze
High-spin and low-spin complexes are possible in an octahedral field when there are 4, 5, 6, or 7 electrons
occupying the d orbitals.
Solve
(a) Mn4+ in MnO2
2 Mn3+ and 1 Mn2+ in Mn3O4
(b) Both low-spin and high-spin configurations are possible in Mn3O4 (d 4 and d 5) but not in MnO2 (d 3).
Think about It
It is not unusual for minerals to have a metal ion present in two different oxidation states as is the case here
for Mn3O4.
18.75. Collect and Organize
We are asked whether tetrahedral CoCl42– is paramagnetic or diamagnetic.
Analyze
The electron configuration of Co2+ in CoCl42– is [Ar]4s03d7.
Solve
In a tetrahedral field the Co2+ ions in CoCl42– have a d7 configuration and have 3 unpaired electrons in its dxy,
dxz, and dyz orbitals. This complex is paramagnetic.
Think about It
Because of its partially filled d orbitals, we also expect this compound to be colored.
18.77. Collect and Organize / Analyze
We are asked to differentiate between cis- and trans- for an octahedral complex ion.
Solve
For an octahedral geometry, cis- means that two ligands are side by side and have a 90˚ bond angle between
them. Ligands that are trans- to each other have a 180˚ bond angle between them.
A
A
M
M
A
Cis
A
Trans
Think about It
Complexes that have cis- and trans-placed ligands are isomers of each other.
18.79. Collect and Organize
In considering geometric isomers of a square-planar complex, we are to determine how many different ligands
are necessary to give geometric isomers.
Analyze
There are four coordination sites on a square-planar complex.
Solve
To have geometric isomers we must have at least two ligands that are different from the others.
368 | Chapter 18
Think about It
Some geometric isomer possibilities for square-planar complexes are as follows:
MA2B2
B
A
B
M
MA2BC
B
C
A
A
A
C
M
MABCD
B
D
B
A
M
A
A
A
D
M
C
A
M
B
A
C
M
B
B
A
M
C
D
B
18.81. Collect and Organize
By looking at the structure of [Co(en)(H2O)2Cl2]2+, we can determine if it can have geometric isomers.
Analyze
Ethylenediamine (en) must bind in a cis fashion to the Co4+ metal ion.
Solve
Yes, [Co(en)(H2O)2Cl2]2+ has geometric isomers because the H2O (or Cl) ligands may be either cis or trans to
each other and relative to the ethylenediamine ligand:
Think about It
Isomer a also has an optical isomer (nonsuperimposable mirror image) as a possibility.
18.83. Collect and Organize
For square-planar CuCl2Br22– we are asked to sketch the possible geometric isomers and are asked if any of
the isomers drawn are chiral.
Analyze
The possible arrangements of the two different ligands on a square-planar metal cation are cis and trans.
Solve
Cis
Trans
No, neither isomer is chiral, because both are superimposable on their mirror images.
Think about It
If this complex were tetrahedral, it also would not have any geometric isomers and would not be chiral.
The Colorful Chemistry of Metals | 369
18.85. Collect and Organize
Given that the Kf for the following equation is 5 × 1013:
Ag + (aq) + 2 S2O32– (aq)  Ag(S2O3 )3–
(aq)
2
we are to calculate the ratio [Ag+]/[Ag(S2O3)23–], where [S2O32–] = 0.233 M.
Analyze
The equilibrium constant expression for the reaction is
⎡⎣ Ag(S2 O3 )23– ⎤⎦
Kf =
2
⎡⎣ Ag + ⎤⎦ ⎡⎣S2 O32 − ⎤⎦
rearranging this to solve for the desired concentration ratio gives
⎡⎣ Ag + ⎤⎦
1
=
⎡⎣ Ag(S2 O3 )23– ⎤⎦ Kf × ⎡S2 O32 − ⎤ 2
⎣
⎦
Solve
⎡⎣ Ag + ⎤⎦
1
=
= 4 × 10 −13
2
13
⎡⎣ Ag(S2 O3 ) 23– ⎤⎦
5 × 10 × ( 0.233)
(
)
Think about It
Our answer, which indicates that there is very little Ag+ compared to Ag(S2O3)23– in solution, is consistent
with the large Kf value for this reaction.
18.87. Collect and Organize
Given the observed colors and magnetic properties of two cobalt complexes, we are asked which has the
largest ∆o.
Analyze
We can presume that both complexes have octahedral geometry. Oxidation converts Co2+ to Co3+. Co2+ has a
d7 configuration. Because this complex is observed to be purple, it is absorbing relatively low-energy yellow
light. For this complex to have 3 unpaired e– (high spin), Δo must be small. Co3+ has a d 6 configuration.
Because this complex is observed to be yellow, it is absorbing relatively high-energy purple light. For this
complex to have no unpaired e– (low spin), Δo must be large.
Solve
The yellow complex containing Co3+ in aqueous ammonia has the larger Δo.
Think about It
Remember that the spin of a complex is a result of the magnitude of Δo.
18.89. Collect and Organize
We are to explain why square-planar Ag2+ is paramagnetic but square-planar Ag+ and Ag3+ are diamagnetic.
Analyze
The square-planar crystal field diagram is
370 | Chapter 18
Solve
Ag2+ has 9 d electrons, leaving an unpaired electron in the d x2 – y2 orbital to make it paramagnetic. Ag3+ has 8 d
electrons and Ag+ has 10 d electrons. Both have all electrons paired, so these silver ions are diamagnetic.
Think about It
Because AgO is diamagnetic we know that Ag2+ is not in the compound.
18.91. Collect and Organize
As we replace H2O ligands on Cu(H2O)62+ with NH3 ligands, we can use the ligand’s relative placement in the
spectrochemical series (Table 18.5) to predict whether the color of the series of complexes will shift to longer
or shorter wavelengths.
Analyze
In the spectrochemical series, NH3 is a stronger-field ligand than H2O. We must keep in mind that the color
we see is complementary to the color absorbed.
Solve
As we replace weak-field H2O ligands with stronger-field NH3 ligands, the ∆o for the complex increases. This
means that the complex absorbs at shorter wavelengths. The color we see, therefore, shifts to longer
wavelengths.
Think about It
Water and ammonia are not far apart in the spectrochemical series, however, so we may not expect the shift in
wavelength to be significant.
CHAPTER 19 | Electrochemistry and the Quest for Clean Energy
19.1. Collect and Organize
For the voltaic cell shown in Figure P19.1, we are to explain why a porous separator is not required.
Analyze
The porous separator serves to keep the reduction and oxidation half-reactions separate so that electrons are
passed through the external circuit.
Solve
Because of the careful layering, each half-cell has its metal in contact with its cation solution. The solutions are
not mixing, but nevertheless the layers allow the ions needed to balance the charge in each half-cell to pass.
Think about It
The half-reactions and overall reaction for this voltaic cell are
o
= – 0.7618 V
Eanode
Zn(s) → Zn2+(aq) + 2 e–
o
= 0.3419 V
Ecathode
Cu2+(aq) + 2 e– → Cu(s)
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
o
o
o
= Ecathode
– Eanode
Ecell
= 0.3419 – (– 0.7618) = 1.1037 V
19.3. Collect and Organize
For the voltaic cell shown in Figure P19.3 in which an Ag+/Ag cell is connected to a standard hydrogen
electrode (SHE), we are to determine which electrode is the anode and which is the cathode and indicate in
which direction the electrons flow in the outside circuit.
Analyze
A voltaic cell runs spontaneously when Ecell is positive. By comparing the reduction potentials of each halfcell, we can write the reaction that is spontaneous for the cell. The half-cell where reduction (gain of electrons)
occurs contains the cathode and the half-cell where oxidation occurs contains the anode. Electrons flow from
the anode, where they are produced by oxidation, toward the cathode, where they are required for reduction.
Solve
The spontaneous reaction for this cell is
2 × (Ag+ + e– → Ag)
H 2 → 2 H + + 2 e–
2 Ag+ + H2 → 2 H+ + Ag
o
= 0.7996 V
Ecathode
o
= 0.000 V
Eanode
o
o
o
= Ecathode
– Eanode
Ecell
= 0.7996 – 0.000 = 0.7996 V
Thus, Ag is the cathode, Pt in the SHE is the anode, and electrons flow from the SHE to Ag (to the left in the
circuit shown in Figure P19.3).
Think about It
The shorthand notation for this cell would be
Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s)
where Pt is an inert electrode used in the SHE.
19.5. Collect and Organize
The graph of cell potential versus [H2SO4] (Figure P19.5) shows four lines, some curved, some linear, some
increasing, and some decreasing, as the concentration of H2SO4 decreases. From the shape of the curves and
their trends we are to choose the line that best represents the trend in potential versus [H2SO4] in a lead-acid
battery.
390
Electrochemistry and the Quest for Clean Energy | 391
Analyze
The scale for [H2SO4] is logarithmic and voltage in the battery varies with the [H2SO4] according to the Nernst
equation:
0.0592
1
Ecell = 2.04 V –
log
2
2
[ H 2SO4 ]
Solve
From the Nernst equation we see that the cell potential drops as the log 1/[H2SO4]2 decreases. So as [H2SO4]
decreases, the cell potential also decreases. The red line on the graph shows the opposite trend: The voltage
increases as [H2SO4] decreases. In considering which of the remaining lines might describe the lead-acid
battery, we must consider that because the cell voltage drops as log 1/[H2SO4]2 we expect that the decrease in
potential is linear. Therefore, the blue line best describes the potential as a function of [H2SO4] concentration.
Think about It
Another characteristic of lead-acid batteries is that their cell voltage does not drop substantially until over 90%
of the battery has been discharged (see Figure 19.12).
19.7. Collect and Organize
For the electrolysis of water in which the two product gases, H2 and O2, are collected in burettes (Figure
P19.7), we are to write the half-reactions occurring at each electrode and discuss why a small amount of acid
was added to the water to speed up the reaction.
Analyze
In the electrolysis of water, electricity is supplied to make the nonspontaneous oxidation and reduction
reactions occur. From Figure P19.7 we notice that the left burette has collected twice the volume of gas
compared to the right burette. From the overall balanced equation
2 H2 O(l ) → 2 H2 (g ) + O2 (g )
o
we can identify the left burette as containing H2 and the right burette as containing O2. Ered
values are given in
Appendix 6.
Solve
(a) The left electrode is the cathode where reduction is occurring:
2 H 2 O(l ) + 2 e – → H 2 (g ) + 2 OH – (aq)
o
Ecathode
= –0.8277 V
The right electrode is the anode where oxidation is occurring:
2 H 2 O(l ) → O2 ( g ) + 4 H + (aq) + 4 e –
o
Eanode
= 1.229 V
(b) A small amount of H2SO4 is added to the water to increase the conductivity of the solution.
Think about It
The electrochemical potential for the overall process is
o
o
o
Ecell
= Ecathode
− Eanode
= –0.8277 V –1.229 V = –2.057 V
19.9. Collect and Organize
We are to explain the function of the porous separator in an electrochemical cell.
392 | Chapter 19
Analyze
The porous separator is situated between the half-reactions in the electrochemical cell.
Solve
The porous separator allows nonreactive ions to pass through the separator so that electrical neutrality is
maintained in each of the half-cells.
Think about It
Without the flow of ions through the separator the flow of electrons stops.
19.11. Collect and Organize
We are to explain why a metal wire cannot function as a porous separator in an electrochemical cell.
Analyze
The porous separator allows charged ions to pass through to maintain electrical neutrality in each cell as the
redox reaction progresses.
Solve
A wire can pass only electrons through it, not ions, and so it cannot function as a porous separator.
Think about It
A wire is used as the conduit for electrons in the external circuit of the voltaic cell.
19.13. Collect and Organize
For the Pb2+/Pb and Zn2+/Zn voltaic cell in which we are told that the products are Pb(s) and Zn2+(aq), we are
to write the appropriate half-reactions, write the balanced overall cell reaction, and diagram the cell.
Analyze
Because we know that Pb and Zn2+ are produced, Pb2+ is reduced and Zn is oxidized in this process. In
balancing the overall reaction, we must be sure to cancel all the electrons produced by oxidation with those
used in reduction. To do so, we might have to multiply either half-reaction, or both, by some factor. Finally, to
diagram the cell we use the following convention:
Anode | oxidation half-reaction species || reduction half-reaction species | cathode
making sure to indicate the phases of the species involved and to use vertical lines to separate phases.
Solve
(a) Pb2+(aq) + 2 e– → Pb(s)
cathode, reduction
Zn(s) → Zn2+(aq) + 2 e–
anode, oxidation
(b) We simply add the above reactions to obtain the overall reaction:
Pb2+(aq) + Zn(s) → Zn2+(aq) + Pb(s)
2+
2+
(c) Zn(s) | Zn (aq) || Pb (aq) | Pb(s)
Think about It
As this reaction proceeds, the lead electrode gets heavier and the zinc electrode loses mass.
19.15. Collect and Organize
For the Cd/Cd(OH)2 and MnO4–/MnO2 voltaic cell in which we are told that the products are Cd(OH)2(s) and
MnO2(s), we are to write the appropriate half-reactions, write the balanced overall cell reaction, and diagram
the cell. The electrolyte is an alkaline aqueous solution.
Analyze
Because we know that Cd(OH)2 and MnO2 are produced, MnO4– is reduced and Cd is oxidized in this process.
In balancing the overall reaction, we must be sure to cancel all the electrons produced by oxidation with those
used in reduction. To do so, we might have to multiply either half-reaction, or both, by some factor. Finally, to
diagram the cell we use the following convention:
Electrochemistry and the Quest for Clean Energy | 393
Anode | oxidation half-reaction species || reduction half-reaction species | cathode
making sure to indicate the phases of the species involved and to use vertical lines to separate phases.
Solve
(a) MnO4–(aq) + 2 H2 O(l ) + 3 e– → MnO2(s) + 4 OH–(aq)
cathode, reduction
–
–
Cd(s) + 2 OH (aq) → Cd(OH)2(s) + 2 e
anode, oxidation
(b) To cancel the electrons, we multiply the cathode reaction by 2 and the anode reaction by 3:
2 MnO4–(aq) + 4 H2 O(l ) + 6 e– → 2 MnO2(s) + 8 OH–(aq)
3 Cd(s) + 6 OH–(aq) → 3 Cd(OH)2(s) + 6 e–
2 MnO4–(aq) + 4 H2 O(l ) + 3 Cd(s) → 2 MnO2(s) + 3 Cd(OH)2(s) + 2 OH–(aq)
(c) Cd(s) | Cd(OH)2(s) || MnO4–(aq) | MnO2(s) | Pt(s)
Think about It
Because we would find it difficult to make an electrode out of MnO2(s), this cell requires the use of an inert
electrode such as Pt.
19.17. Collect and Organize
For the “super iron” battery, we are to determine the number of electrons transferred in the spontaneous
reaction and the oxidation states for Fe and Zn in the reactants and products. Finally, we are to diagram the
cell.
Analyze
(a) By writing the half-reactions involved in the redox reaction and balancing them, we can determine the
number of electrons transferred.
(b) Knowing the typical oxidation states of oxygen (–2) and potassium (+1) helps us determine the oxidation
states of Fe in K2FeO4 and Zn in ZnO and K2ZnO2.
(c) To diagram the cell we use the following cell notation:
Anode | oxidation half-reaction species || reduction half-reaction species | cathode
making sure to indicate the phases of the species involved and to use vertical lines to separate phases.
Solve
(a) The half-reactions are as follows (K+ is a spectator ion):
5 H2 O(l ) + 3 Zn(s) → ZnO(s) + 2 ZnO22–(aq) + 1 OH+(aq) + 6 e–
6 e– + 10 H+(aq) + 2 FeO42–(aq) → Fe2O3(s) + 5 H2O (l )
Six electrons are transferred in this reaction.
(b) FeO42– has Fe6+
Fe2O3 has Fe3+
Zn has Zn0
ZnO and ZnO22– have Zn2+
(c) Zn(s) | ZnO(s) | ZnO22–(aq) || FeO42–(aq) | Fe2O3(s) | Pt(s)
Think about It
Because we would find it difficult to make an electrode out of Fe2O3(s) this cell requires the use of an inert
electrode such as Pt.
19.19. Collect and Organize / Analyze
We are to describe the function of platinum in the standard hydrogen electrode.
Solve
The platinum electrode transfers electrons to the half-cell and because it is inert it is not involved in the
reaction.
394 | Chapter 19
Think about It
The inert electrode simply serves as a place for the oxidation or reduction reaction to occur, and as a conduit
for electrons for the reaction.
19.21. Collect and Organize
o
We are asked to equate two different ways to calculate Ecell
.
Analyze
Equation 19.2 is
o
o
o
Ecell
= Ered
(cathode) − Ered
(anode)
o
o
Other textbooks express E cell
in terms of E red
only:
o
o
Ecell
= Ered
(cathode) + Eoxo (anode)
Solve
o
o
Because Ered
(anode) = – Eoxo (anode) we can substitute – Eoxo (anode) for Ered
(anode) in the expression
o
o
Ecell
= Ered
(cathode) + Eoxo (anode)
to obtain
(
o
o
o
Ecell
= Ered
(cathode) + – Ered
(anode)
o
red
o
cell
This is equal to the expression for E
)
o
red
= E (cathode) – E (anode)
in Equation 19.2.
Think about It
o
Either expression to calculate Ecell
is valid.
19.23. Collect and Organize
o
Using Ered
values from Appendix 6, we can determine whether O2 is a stronger oxidizing agent in acid or in
base.
Analyze
An oxidizing agent causes another substance to be oxidized and therefore it is itself reduced. The more
o
positive the Ered
for O2 in base or in acid, the better its ability to oxidize another substance.
Solve
o
o
O2 is a stronger oxidizing agent in acid because Ered
(acid, 1.229 V) is greater than Ered
(base, 0.401 V).
Think about It
o
o
This also means that OH– (with Eanode
= –0.0401 V) is more easily oxidized than H2O (with Eanode
= –1.229 V).
19.25. Collect and Organize
For each redox reaction given, we are to calculate the cell potential after calculating the value of free energy.
Analyze
We calculate ∆G˚ for each reaction using Gfo values for the products and reactants from Appendix 4. The cell
potential can then be calculated using
o
o
ΔGcell
= – nFEcell
o
Ecell
=
– ΔG o
nF
Electrochemistry and the Quest for Clean Energy | 395
Solve
(a) ΔG o = ⎡⎣(1 mol Cu 2+ × 65.5 kJ mol) + (1 mol Cu × 0.0 kJ mol )⎤⎦ – (2 mol Cu + × 50.0 kJ mol) = –34.5 kJ
+34,500 J
o
Ecell
=
= + 0.358 V
1 mol × 9.65 × 104 C/mol
(b)
ΔG o = ⎡⎣(1 mol Ag + × 77.1 kJ/mol) + (1 mol Fe2 + × −78.9 kJ/mol) ⎤⎦ −
o
Ecell
⎡⎣(1 mol Ag × 0.0 kJ/mol) + (1 mol Fe3+ × −4.7 kJ/mol)⎤⎦ = 2.9 kJ
−2900 J
=
= −0.030 V
1 mol × (9.65 × 104 C/mol)
Think about It
Reaction (a) is spontaneous and reaction (b) is nonspontaneous.
19.27. Collect and Organize
By calculating Ecell for the possible redox reaction between Ag and Cu2+, we can determine if the reaction is
spontaneous.
Analyze
Because we are told that [Ag+] = [Cu2+] = 1.00 M, this reaction occurs under standard conditions, and we can
o
o
o
calculate Ecell
from values of Ered
in Appendix 6. If Ecell
is calculated as positive, the reaction is spontaneous.
Solve
o
o
The half-reactions, Ecathode
and the overall reaction and cell potential are
, and Eanode
o
2 Ag(s) → 2 Ag+(aq) + 2 e–
= 0.7996 V
Eanode
Cu2+(aq) + 2 e– → Cu(s)
2 Ag(s) + Cu2+(aq) → 2 Ag+(aq) + Cu(s)
No, the reaction is not spontaneous.
o
= 0.3419 V
Ecathode
o
o
o
= Ecathode
– Eanode
= – 0.4577 V
Ecell
Think about It
The reverse reaction, Cu placed into Ag+ to dissolve copper and deposit silver, is spontaneous.
19.29. Collect and Organize
We are asked to write the overall cell reaction for the reaction of Zn with O2 in the zinc–air battery to form
Zn(OH)42–.
Analyze
For each half-reaction we need to have OH– or H2O as a reactant. We balance each for both atoms and charge
before adding the two half-reactions together.
Solve
Cathode
Anode
O2(g) + 2 H2O (l ) + 4 e– → 4 OH–(aq)
[Zn(s) + 4 OH–(aq) → Zn(OH)42–(aq) + 2 e–] × 2
O2(g) + 2 H2O (l ) + 2 Zn(s) + 4 OH–(aq) → 2 Zn(OH)42–(aq)
Think about It
This differs from the other overall reaction for the zinc–air battery shown in Figure 19.6 in that water and OH–
are reactants to give soluble Zn(OH)42– instead of solid ZnO.
19.31. Collect and Organize
o
o
For a Ni–Zn cell we are asked to compare its Ecell
with that of a Cu–Zn cell that has Ecell
= 1.10 V.
396 | Chapter 19
Analyze
o
o
o
Using Ered
values in Appendix 6, we can identify the spontaneous reaction (having positive Ecell
) and calculate Ecell
.
Solve
The spontaneous reaction in the Ni–Zn cell is
Ni2+(aq) + 2 e– → Ni(s)
2+
Zn(s) → Zn (aq) + 2 e
o
= – 0.257 V
Ecathode
–
o
= – 0.7618 V
Eanode
o
o
o
= Ecathode
– Eanode
= 0.505 V
Ecell
o
o
The Ecell
for a Ni–Zn cell is less than Ecell
for a Cu–Zn cell.
Think about It
o
Because Ecell
for the Ni–Zn cell is less positive, we also know that the cell reaction has a less negative ∆G.
19.33. Collect and Organize
o
For each reaction we can break it up into its appropriate half-reactions and, using E red
values from Appendix 6,
o
calculate Ecell . From this cell potential we can then calculate the free energy for each reaction.
Analyze
To calculate the cell potential we use
o
o
o
= Ecathode
– Eanode
Ecell
o
Once we have calculated E cell
we use
o
ΔG o = – nFEcell
to calculate the free energy of the reaction. Here, we have to remember that n is the number of moles of
electrons transferred in the overall balanced equation.
Solve
(a)
Cu(s) → Cu2+(aq) + 2 e–
2+
–
Sn (aq) + 2 e → Sn(s)
Cu(s) + Sn2+(aq) → Cu2+(aq) + Sn(s)
ΔGo = –2 mol ×
(b)
o
= – 0.136 V
Ecathode
o
o
o
= Ecathode
– Eanode
= – 0.478 V
Ecell
9.65 × 104 C –0.478 J 1 kJ
×
×
= 92.2 kJ
mol
C
1000 J
Zn(s) → Zn2+(aq) + 2 e–
Ni2+(aq) + 2 e– → Ni(s)
Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)
ΔGo = –2 mol ×
o
= 0.3419 V
Eanode
o
= – 0.7618 V
Eanode
o
= – 0.257 V
Ecathode
o
o
o
= Ecathode
– Eanode
= 0.548 V
Ecell
9.65 × 104 C 0.5048 J 1 kJ
×
×
= –97.4 kJ
mol
C
1000 J
Think about It
Reaction (b) with its positive cell potential is spontaneous. Reaction (a) is nonspontaneous, so copper metal in
contact with Sn2+ solution does not oxidize.
19.35. Collect and Organize
For each pair of reduction reactions, we can use the reduction potentials in Appendix 6 to write the equation
for the spontaneous cell reaction, identifying the reactions occurring at the anode and cathode.
Electrochemistry and the Quest for Clean Energy | 397
Analyze
o
o
The overall cell potential, Ecell
we
, must be positive for the spontaneous voltaic cell reaction. To calculate Ecell
use
o
o
o
= Ecathode
– Eanode
Ecell
The oxidation reaction occurs at the anode, and the reduction reaction occurs at the cathode.
Solve
(a)
Anode
Cathode
Zn(s) → Zn2+(aq) + 2 e–
2+
o
= – 0.7618 V
Eanode
–
Hg (aq) + 2 e → Hg(l )
2+
o
= 0.851 V
Ecathode
2+
Zn(s) + Hg (aq) → Zn (aq) + Hg(l )
o
o
o
= Ecathode
– Eanode
= 1.613 V
Ecell
(b)
Anode
Cathode
Zn(s) + 2 OH–(aq) → ZnO(s) + H2 O(l ) + 2 e–
–
–
Ag2O(s) + H2 O(l ) + 2 e → 2 Ag(s) + 2 OH (aq)
Zn(s) + Ag2O(s) → ZnO(s) + 2 Ag(s)
o
= –1.25 V
Eanode
o
= 0.342 V
Ecathode
o
o
o
= Ecathode
– Eanode
= 1.59 V
Ecell
(c)
Anode
Cathode
2 × [Ni(s) + 2 OH–(aq) → Ni(OH)2(s) + 2 e–]
o
= – 0.72 V
Eanode
–
o
= 0.401 V
Ecathode
–
O2(g) + 2 H2 O(l ) + 4 e → 4 OH (aq)
2 Ni(s) + O2(g) + 2 H2 O(l ) → 2 Ni(OH)2(s)
o
o
o
= Ecathode
– Eanode
= 1.12 V
Ecell
Think about It
Notice that when multiplying a half-reaction in order to obtain the balanced cell reaction, the oxidation or
reduction potential does not change. This is because cell potential is an intensive, not extensive, property.
19.37. Collect and Organize
We can write the overall cell reaction for the battery by adding the equations together. The overall cell
o
o
o
potential is Ecell
= Ecathode
– Eanode
.
Analyze
Because the reduction reaction and the oxidation reaction each involve the exchange of one electron, we do
not need to multiply either reaction in order to write the overall reaction.
Solve
(a) NiO(OH)(s) + TiZr2H(s) → TiZr2(s) + Ni(OH)2(s)
o
(b) Ecell
=1.32 V – 0.00 V = 1.32 V
Think about It
This reaction is spontaneous with a ∆G value of
9.65 × 104 C 1.32 J 1 kJ
ΔGo = –1 mol ×
×
×
= –127 kJ
mol
C
1000 J
19.39. Collect and Organize / Analyze
We are asked how a spontaneous electrochemical cell with a positive cell potential does negative work.
Solve
A positive Ecell indicates a spontaneous reaction that does electrical work on the surroundings, so the sign of w
is negative.
Think about It
A positive cell potential gives a negative free energy through the equation
∆G = –nFEcell
398 | Chapter 19
19.41. Collect and Organize
Using the relationship between free energy and cell potential
∆Gcell = –nFEcell
we can calculate ∆Gcell for a Zn–MnO2 battery generating 1.50 V.
Analyze
To use the equation we need the value of n, the number of electrons transferred in the overall balanced
equation. From the cell reaction provided, we see that Zn is oxidized to Zn2+ and 2 mol of Mn4+ (in MnO2) are
reduced Mn3+ (in Mn2O3), so 2 mol of electrons are transferred in the reaction. We should also be aware of the
units for the quantities in this equation. The value of F is 9.65 × 104 C/mol and E is in volts which, being
equivalent to joules per coulomb (J/C), means that ∆G is calculated in joules, which we can convert to
kilojoules, the usual units for free energy.
Solve
ΔGcell = –2 mol ×
9.65 × 104 C 1.50 J 1 kJ
×
×
= –290 kJ
mol
C
1000 J
Think about It
This reaction is spontaneous because the cell potential is positive, making ∆Gcell negative.
19.43. Collect and Organize
Using the relationship between free energy and cell potential
∆Gcell = –nFEcell
we can calculate ∆Gcell for a nickel–metal hydride battery cell generating 1.20 V.
Analyze
To use the equation we need the value of n, the number of electrons transferred in the overall balanced
equation. From the cell reaction provided we see that Ni3+ [in NiO(OH)] is reduced to Ni2+ [in Ni(OH)2] and
M– (in MH) is oxidized to M, so 1 mol of electrons are transferred in the reaction. We should also be aware of
the units for the quantities in this equation. The value of F is 9.65 × 104 C/mol and E is in volts which, being
equivalent to joules per coulomb (J/C), means that ∆G is calculated in joules, which we can convert to
kilojoules, the usual units for free energy.
Solve
ΔGcell = –1 mol ×
9.65 ×104 C 1.20 J 1 kJ
×
×
= –116 kJ
mol
C
1000 J
Think about It
This reaction is spontaneous because the cell potential is positive, making ∆Gcell negative.
19.45. Collect and Organize
We are to explain why the voltage of most batteries changes little until the battery is almost discharged, where
the voltage drops significantly.
Analyze
Voltage of a battery (a voltaic cell) is governed by the Nernst equation:
RT
o
Ecell = Ecell
–
ln Q
nF
As a battery discharges, the value of Q, the reaction quotient, changes:
[products]x
Q=
[reactants]y
Electrochemistry and the Quest for Clean Energy | 399
Solve
At the start of the reaction, Q is very small because [reactants] >> [products]. As the reaction proceeds
[products] grows and Q increases but does not increase significantly until significant amounts of products
form, that is, when the battery is nearly discharged.
Think about It
As an example, Figure 19.12 shows the cell potential of a lead–acid battery as a function of discharge. Notice
that the voltage is relatively constant until the battery is approximately 90% discharged.
19.47. Collect and Organize
We can use the Nernst equation to calculate the cell potential when Fe3+ is combined with Cr2+ at nonstandard
conditions.
Analyze
Because T = 298 K, we can use the following form of the Nernst equation
0.0592
o
Ecell = Ecell
–
log Q
n
o
o
where Ecell
is the potential of the cell under standard conditions (calculated from tabulated Ered
values), n is
the moles of electrons transferred in the overall balanced redox equation, and Q is the reaction quotient.
Solve
o
First, we need to calculate Ecell
and determine the value of n. The half-reactions and overall cell reaction are
Fe3+(aq) + e– → Fe2+(aq)
o
Ecathode
= 0.770 V
Cr2+(aq) → Cr3+(aq) + e–
o
Eanode
= −0.41 V
o
o
o
= Ecathode
– Eanode
Ecell
= 1.18 V
Fe3+(aq) + Cr2+(aq) → Fe2+(aq) + Cr3+(aq)
We see that n = 1. Now we can use the Nernst equation to calculate Ecell when [Fe3+] = [Cr2+] = 1.50 × 10–3 M
and [Fe2+] = [Cr3+] 2.5 × 10– 4 M.
Ecell
(
(
2
)
)
2.5 × 10− 4
0.0592
= 1.18 V –
log
n
1.50 × 10− 3
2
= 1.27 V
Think about It
This reaction became more spontaneous (higher cell potential) under these conditions.
19.49. Collect and Organize
We can use the following equation to calculate the equilibrium constant for the given redox reaction:
o
nEcell
log K =
0.0592
Analyze
o
o
First, we have to determine the standard cell potential, Ecell
values in Appendix 6 and also
, using the Ered
determine the value of n, the moles of electrons transferred in the overall balanced equation.
Solve
Fe3+(aq) + e– → Fe2+(aq)
Cr2+(aq) → Cr3+(aq) + e–
Fe3+(aq) + Cr2+(aq) → Fe2+(aq) + Cr3+(aq)
log K =
1 × 1.18 V
= 19.9324
0.0592
o
Ecathode
= 0.770 V
o
Eanode
= −0.41 V
o
o
o
= Ecathode
– Eanode
Ecell
= 1.18 V
n =1
400 | Chapter 19
K = 1 × 1019.9324 = 8.56 × 1019
Think about It
Because K for this reaction is very large, the reaction goes very far to the right.
19.51. Collect and Organize
We can use the Nernst equation to calculate the potential of the hydrogen electrode at pH = 7.00.
Analyze
We are reminded that under standard conditions [1 atm H2(g) and 1.00 M H+ = pH = 0.00] the voltage of the
o
hydrogen cell is zero. This is Ecell
. The overall reaction for the cell (against SHE) is
2 H+(aq) + 2 e– → H2(g)
(where [H+] = 1 × 10–7 M)
+
–
H2(g) → 2 H (aq) + 2 e
(SHE)
2 H+(aq) (1 × 10–7 M) + H2(g) (1 atm) → 2 H+(aq) (1.00 M) + H2(g) (1 atm)
The form of Q for the Nernst equation is
2
1.00 M ) × (1 atm )
(
1
Q=
=
2
2
–7
(1 × 10 M ) × (1 atm) (1 × 10–7 )
Solve
Ecell = 0.000 V –
0.0592
1
log
2
1 × 10 –7
(
)
2
= –0.414 V
Think about It
The spontaneous reaction actually is the reverse reaction:
2 H+(aq) (1.00 M) + H2(g) (1 atm) → 2 H+(aq) (1 × 10–7 M) + H2(g) (1 atm)
o
= 0.414 V
Ecell
In this redox cell, acid in the SHE will be reduced and H2 in the cell where pH = 7.00 will be oxidized.
19.53. Collect and Organize
We can use the Nernst equation to calculate the potential for the reduction of MnO4– to MnO2 in the presence
of SO32– when [MnO4–] = 0.150 M, [SO32– ] = 0.256 M, [SO42–] = 0.178 M, and [OH–] = 0.0100 M. We are
also to assess whether the potential increases or decreases as reactants become products in the reaction.
Analyze
o
To use the Nernst equation we need to know the value of Ecell
and n. We can determine both of these by
writing out the half-reactions and balancing the redox reaction.
Solve
2 MnO4–(aq) + 4 H2 O(l ) + 6 e– → 2 MnO2(s) + 8 OH–(aq)
o
Ecathode
= 0.59 V
3 SO32–(aq) + 6 OH–(aq) → 3 SO42–(aq) + 3 H 2 O(l ) + 6 e –
o
Eanode
= −0.92 V
o
o
o
Ecell
= Ecathode
− Eanode
= 1.51 V
2 MnO4–(aq) + 3 SO32–(aq) + H2 O(l ) → 2 MnO2 (s) + 3 SO42– (aq) 2 OH– (aq)
3
Ecell = 1.51 V –
Ecell = 1.54 V
n=6
2
( 0.178 M ) ( 0.0100 M )
0.0592
log
6
( 0.150 M )2( 0.256 M )3
Electrochemistry and the Quest for Clean Energy | 401
As the reaction proceeds, the concentrations of the reactants decrease and the concentrations of the products
increase, so Q increases and log Q becomes more positive. When a more positive log Q is multiplied by
o
0.0592/6 and then subtracted from the Ecell
, Ecell decreases.
Think about It
Be sure to use half-reactions to determine the correct value of n.
19.55. Collect and Organize
o
For the reaction of copper pennies with nitric acid, we are to calculate Ecell
and then Ecell when [H+] =
–
2+
0.100 M, [NO3 ] = 0.0250 M, [Cu ] = 0.0375 M, and PNO = 0.00150 atm.
Analyze
o
o
o
o
is calculated by adding Eanode
and Ecathode
must be
Ecell
. The reaction is spontaneous, so our calculated Ecell
positive. In balancing the reaction, we can also determine the value of n for the Nernst equation.
Solve
(a)
3 Cu(s) → 3 Cu2+(aq) + 6 e–
–
+
o
Eanode
= 0.3419 V
–
2 NO3 + 8 H (aq) + 6 e → 2 NO(g) + 4 H2 O(l )
o
Ecathode
= 0.96 V
3 Cu(s) + 2 NO3– + 8 H+(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2 O(l )
3
(b)
Ecell = 0.6181 V –
o
o
o
Ecell
= Ecathode
− Eanode
= 0.62 V
2
( 0.0375 M ) ( 0.00150 atm) = 0.61 V
0.0592
log
6
( 0.0250 M )2( 0.100 M )8
Think about It
We may mix concentration units of atmospheres and molarity as we do in this calculation of Q.
19.57. Collect and Organize
o
For the reaction of NH4+ with O2 in water we are to calculate Ecell
. We can then use the Nernst equation to
–
+
determine [NO3 ]/[NH4 ] for PO2 = 0.21 atm and pH = 5.60 at 298 K.
Analyze
o
o
o
o
is calculated by adding Eanode
and Ecathode
must be
Ecell
. The reaction is spontaneous, so our calculated Ecell
positive. In balancing the reaction we can determine the value of n. Because the system is at equilibrium in
part b, we know that Ecell = 0. Therefore, [NO3–]/[NH4+] can be determined through the equation
[NO3 – ][H + ]2
0.0592
o
0 = Ecell
–
log
2
n
[NH 4 + ] PO2
( )
Solve
(a)
NH4+(aq) + 3 H2 O(l ) → NO3–(aq) + 10 H+ + 8 e–
2 O2(g) + 8 H+(aq) + 8 e– → 4 H2 O(l )
NH4+(aq) + 2 O2(g) → NO3–(aq) + H 2 O(l ) + 2 H + (aq)
o
Eanode
= 0.88 V
o
Ecathode
= 1.229 V
o
o
o
Ecell
= Ecathode
− Eanode
= 0.349 V
402 | Chapter 19
(b)
[NO3 ] ( 2.512 × 10 – 6 M )
0.0592
0 = 0.349 V –
log
2
+
8
[NH 4 ] ( 0.21 atm )
−
[NO3 ] ( 2.512 × 10 – 6 M )
0.0592
−0.349 V = –
log
2
+
8
[NH 4 ] ( 0.21 atm )
−
2
2
−
⎛
[NO3 ] ⎞
47.16 = log ⎜⎜ (1.431× 10 –10 ) ×
+ ⎟
[NH 4 ] ⎟⎠
⎝
−
1× 1047.16 = 1.453 × 1047 = 1.43 × 10 –10 ×
[NO3 ]
+
[NH 4 ]
−
[NO3 ]
+
[NH 4 ]
= 1.02 × 1057
Think about It
o
This ratio is consistent with a spontaneous reaction as indicated by the positive Ecell
.
19.59. Collect and Organize
We are to compare a 12-volt lead–acid battery with one that has a lower ampere-hour rating.
Analyze
An ampere-hour is a unit of electrical charge and is defined as the electric charge transferred by 1 A of current
for 1 hr. It is used to describe the life of a battery.
Solve
The total masses of the electrode materials (c) and the combined surface areas of the electrodes (f) are likely
to be different.
Think about It
Both batteries use the same components (b and e) and have the same voltage (a and d).
19.61. Collect and Organize
We are to compare two voltaic cells to determine which produces more charge per gram of anode material.
Analyze
For each cell we must first identify which species is the anode and the number of electrons transferred when
1 mol of anode is consumed in the reaction. The charge generated by the reaction is
C = nF
where C = charge in coulombs, n = moles of electrons transferred in balanced equation, and F = 9.65 ×
104 C/mol. This gives the charge per mol of anode. To convert this into charge per gram we use the molar
mass:
charge
coulombs/mol
=
gram
molar mass of anode material
Solve
For the Ni– Cd voltaic cell, Cd is the anode material:
C 2 mol e– × 9.65 × 104 C / mol
=
= 1.72 × 103 C/g Cd
g
112.41 g/mol
For the Al– O2 voltaic cell, Al is the anode material:
C 12 mol e – × 9.65 × 104 C / mol
=
= 4.29 × 104 C/g Al
g
26.98 g/mol
Therefore, the Al– O2 cell produces a greater charge per gram.
Electrochemistry and the Quest for Clean Energy | 403
Think about It
Notice that the number electrons transferred in the oxidation of Al to Al3+ is 4 mol × 3 e–/mol = 12 e–.
19.63. Collect and Organize
We are to compare two voltaic cells to determine which produces more energy per gram of anode material.
Analyze
The energy of a voltaic cell is the force to move electrons from the anode to the cathode. The unit of volts is
energy per unit charge, so
Energy = volts × charge (in units of V⋅ C)
where 1 V = 1 J/C. The charge in the cell generated by 1 g of anode material is
mol e –
9.65 × 104 C
Charge = 1 g × molar mass of anode material ×
×
mol anode
mol
Solve
For the Zn–Ni(OH)2 cell, Zn is the anode material:
1 mol 2 mol e − 9.65 × 104 C
Charge = 1 g ×
×
×
= 2.952 × 103 C
65.38 g 1 mol Zn
mol e−
Energy = 1.20
J
× 2.952 × 103 C = 3.54 × 103 J
C
For the Li–MnO2 cell, Li is the anode material:
1 mol 1 mol e– 9.65 × 104 C
Charge = 1 g ×
×
×
= 1.390 × 104 C
6.941 g 1 mol Li
mol e–
Energy = 3.15
J
× 1.390 × 104 C = 4.38 × 104 J
C
Therefore, the Li–MnO2 cell generates more energy per gram of anode.
Think about It
Notice that although the charge generated per mole of Li versus that of a mole of Zn in these cells is lower,
the high voltage of the Li–MnO2 cell means that this cell generates more energy.
19.65. Collect and Organize
We are to explain the differences in the signs of the cathode in a voltaic versus an electrolytic cell.
Analyze
The signs of the electrodes in a cell indicate the direction of electron flow.
Solve
In a voltaic cell, the electrons are produced at the anode so a negative (–) charge builds up there; in an
electrolytic cell, electrons are being forced onto the cathode so that it builds up negative (–) charge. The flow
of electrons in the outside circuit is reversed in an electrolytic cell compared to the flow in a voltaic cell.
Think about It
An electrolytic cell uses an outside source of electrical energy to cause a nonspontaneous reaction to occur.
19.67. Collect and Organize
In a mixture of molten Br– and Cl– salts, we are to predict which product, Br2 or Cl2, forms first in an
electrolytic cell as the voltage is increased.
Analyze
The oxidations of Br– and Cl– are expressed as
2 Br – (l ) → Br2 (l ) + 2 e–
2 Cl – (l ) → Cl2(g) + 2 e
–
404 | Chapter 19
Solve
The halide that is first to be oxidized is the one with the lowest ionization energy. Br–, being larger and less
electronegative than Cl–, loses its electron more readily and therefore Br2 forms first in the cell as the voltage
is increased.
Think about It
If the molten salt also contains F–, F2 would form after Br2 and Cl2.
19.69. Collect and Organize
For the electrolysis of a 1.0 M Cu2+ solution, we are to determine whether the potential at the cathode where
the reduction of Cu2+ occurs needs to be more negative or less negative than 0.34 V in order to quantitatively
reduce the Cu2+ in solution to Cu.
Analyze
o
We are given that Ered
for Cu2+ is 0.34 V. This is under standard conditions when [Cu2+] = 1.0 M, the
concentration of the solution at the start of the electrolysis. As [Cu2+] decreases as Cu is deposited on the
electrode, Ecell can be calculated using the Nernst equation:
0.0592
1
Ecell = 0.34 V –
log
2
[Cu 2+ ]
Solve
As the reaction proceeds and [Cu2+] decreases the value of log (1/[Cu2+]) becomes more positive. As a result
Ecell decreases, so the cathode potential must be more negative than 0.34 V to complete the reduction.
Think about It
There might be a slight overpotential required to accomplish the electrolysis, however, because of a kinetic
barrier to the reduction reaction.
19.71. Collect and Organize
In an electroplating process, we are to calculate the mass of silver deposited on an object when 1.7 A ⋅ hr of
charge is delivered from a battery. To determine this we need to relate the ampere-hours to the total number of
electrons generated. Then we can relate the number of electrons to the mass of Ag deposited from a solution
of Ag+.
Analyze
We can convert ampere-hours to coulombs:
1 C 3600 s
×
A⋅s
hr
The moles of electrons used in the process can be calculated from this result knowing that 1 mol e– = 9.65 ×
104 C. To calculate the mass of Ag deposited we also need to know that the reduction of Ag+ to Ag is a oneelectron process.
A ⋅ hr ×
Solve
1 C 3600 s
×
= 6120 C
A⋅s
hr
1 mol e –
6120 C ×
= 6.342 × 10–2 mol e–
4
9.65 × 10 C
1 mol Ag 107.87 g Ag
–2
6.342 × 10 mol e– ×
×
= 6.8 g
mol
1 mol e–
1.7 A ⋅ hr ×
Think about It
The higher the amps for the battery, the faster an object can be electroplated.
Electrochemistry and the Quest for Clean Energy | 405
19.73. Collect and Organize
We are to calculate how long it will take to recharge a battery that contains 4.10 g of NiO(OH) and is 50%
discharged. This means that 2.05 g of NiO(OH) has been depleted from the battery. The charger for the
battery operates at 2.00 A and 1.3 V.
Analyze
We first need to calculate the moles of electrons needed to recover 2.05 g of NiO(OH), which in the NiMH
battery forms Ni(OH)2 in a 1-electron process. Next, we will convert the moles of electrons to coulombs.
Because 1 C = 1 A ⋅ s and we know the amperes at which the charger operates, we can then calculate the time
it takes the charger to deliver the electrons to recharge the battery.
Solve
2.05 g NiO(OH) ×
0.0224 mol e – ×
1 mol
1 mol e –
×
= 0.0224 mol e –
91.70 g 1 mol NiO(OH)
9.65 × 104 C A ⋅ s
1
×
×
= 1080 s
–
C
2.00 A
mol e
In minutes this is 1080 s × 1 min/60 s = 18.0 min.
Think about It
The larger the battery, the more of the reactant that is needed to be regenerated and so the longer it takes to
recharge it.
19.75. Collect and Organize
We are to calculate the amount of O2 that could be generated in one hour on a submarine using electrolysis
and then consider the practicality of using seawater as the source of oxygen for the submarine.
Analyze
We can calculate the moles of O2 produced by the electrolytic cell through
1 mol O2
1C
mol e –
Moles O2 = time in seconds × amperes ×
×
×
4
A ⋅ s 9.65 × 10 C 4 mol e–
–
Notice that the oxidation of water to O2 is a 4 e process. We can then use the ideal gas law to calculate the
volume of O2 produced.
Solve
(a)
1 mol O 2
3600 s
1C
1 mol e –
× 0.025 A ×
×
×
= 2.332 × 10 – 4 mol O 2
4
1 hr
A ⋅ s 9.65 × 10 C 4 mol e –
2.332 × 10 –4 mol × 0.08206 L ⋅ atm / mol ⋅ K × 298 K
V=
= 5.78 × 10 –3 L or 5.8 mL
0.98692 atm/1 bar
(b) Seawater contains a fairly high concentration of Cl– and Br– that can be oxidized, so the direct electrolysis
of seawater would not be useful as an oxygen source.
Moles O2 = 1 hr ×
Think about It
Submarines probably purify their water, perhaps through a reverse osmosis process, to remove the chloride
and bromide and other ions before the electrolysis process.
19.77. Collect and Organize
For the process that electroplates nickel, we are to calculate the lowest potential required to deposit Ni onto a
piece of iron using a 0.35 M Ni2+ solution.
Analyze
To solve this problem we need to use the Nernst equation:
406 | Chapter 19
0.0592
log Q
n
o
where Ecell
is the reduction potential of Ni2+ versus the SHE (– 0.257 V), n is the number of electrons needed
to reduce Ni2+ to Ni (2), and Q is 1/[Ni2+] based on the reduction reaction
Ni2+(aq) + 2 e– → Ni(s)
o
Ecell = Ecell
−
Solve
Ecell = –0.257 V −
0.0592
1
log
= –0.270 V
2
0.35 M
Think about It
For this electrolysis reaction, using a more dilute solution of the metal cation necessitates an increase in the
potential needed to cause Ni2+ to deposit on the iron.
19.79. Collect and Organize
We are to consider the advantages and disadvantages of hybrid power systems versus all-electric fuel-cell
systems.
Analyze
A parallel hybrid power system uses traditional petroleum-based fuel for high power demands and an electric
motor for lower power demands. An all-electric fuel cell system uses only electrochemical power based on
combustion half-reactions to supply power.
Solve
A hybrid vehicle uses a relatively inexpensive fuel (gasoline) in the internal combustion engine and has good
fuel economy but still gives off emissions. A fuel-cell vehicle does not give off emissions (the reaction
produces H2O) but requires a more expensive and explosive fuel (hydrogen); moreover, current battery
technologies incorporate materials that are still very expensive and bulky.
Think about It
The determining factor whether alternate fuels and power systems get used will be the cost of petroleum used
to produce gasoline, which traditionally has been much less expensive than alternative energy sources.
19.81. Collect and Organize
Given the information that methane may be used in fuel cells, we are to consider why these fuel cells are
likely to produce less CO2 emissions per mile than an internal combustion engine fueled by methane.
Analyze
Both the methane fuel cell and the combustion of methane in a combustion reaction have the balanced
equation
CH4(g) + 2 O2(g) → 2 H2O(g) + CO2(g)
Solve
Fuel cells burn methane fuel more efficiently. Electric engines are more efficient by converting more of the
energy into motion instead of losing it as heat. Therefore, less CO2 is produced per mile with fuel cells.
Think about It
The bulkiness and short range of fuel cells currently limit the use of fuel cells in transportation.
19.83. Collect and Organize
For the reactions of CH4 and CO with water, we are to assign oxidation numbers to the C and H atoms in all
o
the species in the reactions and calculate ΔGrxn
for each and for the overall reaction
CH4(g) + 2 H2O(g) → 4 H2(g) + CO2(g)
Electrochemistry and the Quest for Clean Energy | 407
Analyze
o
We can use the usual rules of assigning oxidation states from Chapter 4. To calculate ΔGrxn
we use ΔGfo
values from Appendix 4 in the following equation:
o
ΔGrxn
= ∑ nΔGf,o products − ∑ mΔGf,o reactants
Solve
−4 +1
+1
+2
0
(a) C H 4 ( g ) + H 2 O(g ) → C O(g ) + 3H 2 ( g )
+2
+1
0
+4
C O( g ) + H 2 O(g ) → H 2 (g ) + C O 2 ( g )
(b) For the reaction of CH4 with H2O
o
ΔGrxn
= ⎡⎣(1 mol CO × –137.2 kJ/mol ) + ( 3 mol H 2 × 0.0 kJ/mol )⎤⎦ –
⎡⎣(1 mol CH 4 × –50.8 kJ/mol ) + (1 mol H 2 O × −228.6 kJ/mol)⎤⎦ = 142.2 kJ
For the reaction of CO with H2O
o
ΔGrxn
= ⎡⎣(1 mol CO 2 × –394.4 kJ/mol) + (1 mol H 2 × 0.0 kJ/mol)⎤⎦ –
⎡⎣(1 mol CO × –137.2 kJ/mol) + (1 mol H 2 O × −228.6 kJ/mol)⎤⎦ = −28.6 kJ
For the overall reaction
o
o
o
ΔGoverall
= ΔGrxn
+ ΔGrxn
= 113.6 kJ
1
2
Think about It
The spontaneity of the second reaction is not enough to overcome the positive free energy of the first reaction,
so the overall reaction is nonspontaneous.
19.85. Collect and Organize
We consider electrolysis of a molten Mg2+ salt from evaporated seawater (so it may contain NaCl).
Analyze
The possible reactions are (with E˚ values when listed in Appendix 6)
Mg 2+ (l ) + 2 e– → Mg(s)
Na + (l ) + e– → Na(s)
2 H2 O(l ) + 2 e– → H2(g) + 2 OH–(aq)
o
Ered
= –0.8277 V
2 H2 O(l ) → O2(g) + 4 H+(aq) + 4 e–
o
Ered
= 1.229 V
Mg2+(aq) + 2 e– → Mg(s)
o
Ered
= –2.37 V
Na+(aq) + e– → Na(s)
o
Ered
= –2.71 V
Solve
(a) Mg2+ undergoes a reduction reaction that occurs at the cathode. Mg forms at the cathode.
(b) No. Mg2+, with a higher positive charge, has a lower (less negative) reduction potential than Na+, so the
Mg2+ would not need to be separated from the NaCl in seawater first.
(c) No. The electrolysis of MgCl2(aq) would not produce Mg(s) because water, with a less negative reduction
potential, would be electrolyzed.
(d) H2 and O2 gases would be produced.
Think about It
Because different components are reduced at different potentials, electrolysis of molten salts is one way to
separate components (e.g., metals) from each other.
408 | Chapter 19
19.87. Collect and Organize
For a Mg–Mo3S4 battery for which we are given the half-reaction potential of the anode reaction (2.37 V) and
o
the overall cell potential (1.50 V), we are to calculate Ered
of Mo3S4. We are also to consider why Mg2+ is
added to the battery’s electrolyte and determine the oxidation states and electron configurations of Mo in
Mo3S4 and MgMo3S4.
Analyze
o
o
o
o
o
Because Ecell
= Ecathode
− Eanode
, the reduction potential for Mo3S4 will be Ecell
+ Eanode
.
Solve
o
(a) Ecell
= 1.50 V + (– 2.37 V) = – 0.87 V
(b) Sulfur usually carries a 2– charge so each Mo atom in Mo3S4 has a calculated charge of 2.67+. This,
therefore, is a mixed oxidation state compound where it is likely that two of the Mo atoms have a 3+ charge
and one Mo atom has a 2+ charge. In MgMo3S4, the Mg atom has a 2+ charge, so the Mo atoms have a 2+
charge. The electron configurations for the two oxidation states of Mo are
Mo in +2 oxidation state [Kr]4d 4
Mo in +3 oxidation state [Kr]4d 3
2+
(c) Mg is added to the electrolyte to better carry the charge in the cell. This cation is produced at the anode
and consumed at the cathode.
Think about It
This battery resembles the lithium–ion battery in that the migration of a cation in the cell generates the
electrical current.
19.89. Collect and Organize
We consider the thermodynamic and electrochemical properties of the synthesis of F2 both chemically and in
an electrolysis reaction.
Analyze
(a) We can use the usual rules described in Chapter 4 to assign oxidation numbers to all the elements in the
reactants and products for the chemical synthesis of F2. From the change in oxidation numbers we can deduce
the number of electrons involved in the process.
o
(b) To calculate ΔH rxn
we use
(c) When we assume ∆S ≈ 0
o
ΔH rxn
= ∑ nΔH f,o products − ∑ mΔH f,o reactants
∆G = ∆H – T∆S ≈ ∆H
o
so we can use ∆H for ∆G in the equation to calculate Ecell
.
Solve
(a) In K2MnF6: K = +1, Mn = +4, F = –1
In SbF5: Sb = +5, F = –1
In KSbF6: K = +1, Sb = +5, F = –1
In MnF3: Mn = +3, F = –1
In F2: F = 0
This is a 1-electron process.
o
(b) ΔH rxn
= ⎡⎣( 2 mol KSbF6 × –2080 kJ/mol) + (1 mol MnF3 × –1579 kJ/mol) + ( 12 mol F2 × 0.0 kJ/mol)⎤⎦
– ⎡⎣(1 mol K 2 MnF6 × –2435 kJ/mol) + ( 2 mol SbF5 × –1324 kJ/mol)⎤⎦ = –656 kJ
6.56 × 105 J
= 6.80 V
1 mol e – × 9.65 × 104 C/mol
(d) If ∆S is positive then the ∆G estimate is too positive; ∆G would be more negative, giving a more positive
o
o
o
using only the ΔH rxn
value would be too low.
Ecell
. Therefore, our calculated Ecell
o
(c) Ecell
≈
Electrochemistry and the Quest for Clean Energy | 409
(e) In H2: H = 0
In F2: F = 0
In KF: K = +1, F = –1
In KHF2: K = +1, H = +1, F = –1
This is a 2-electron process.
Think about It
The chemical synthesis of F2 relies on the oxidation of F– by the strong oxidant MnF62–. This process,
however, is not very practical and fluorine is prepared industrially by the electrolysis reaction described in
part e.
19.91. Collect and Organize
For three reactions, we are to write balanced chemical equations.
Analyze
In each of the reactions either carbon or oxygen combines with the oxide or sulfide anion.
Solve
(a) 2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)
(b) 2 ZnO(s) + C(s) → 2 Zn(s) + CO2(g)
(c) HgS(s) + O2(g) → Hg(l ) + SO2(g)
Think about It
These are all redox reactions.
19.93. Collect and Organize
We consider the lattice structures of ZnS and CdS.
Analyze
First, we need to remember that both hcp and ccp lattices are closest-packed. Then, we have to consider the
size of the atoms that are closest packed (the anion S2–) in each structure to determine whether the tetrahedral
holes are the same size in each.
Solve
(a) Yes. Because both are closest-packed, both ZnS and CdS have the same packing efficiency.
(b) If the S2– are closest packed (as given in the statement of the problem) in both structures, yes, the size of
the tetrahedral holes is the same in both structures.
Think about It
From the radius ratios of these compounds, however, we see that Cd2+ would not fit well into the tetrahedral
holes and so might expand the ccp lattice to have larger tetrahedral holes:
410 | Chapter 19
rZn 2+
rS2–
rCd2+
rS2–
=
74 pm
= 0.402
184 pm
=
95 pm
= 0.516
184 pm
19.95. Collect and Organize
o
For three redox reactions we are to calculate Ecell
.
Analyze
o
To calculate Ecell
we add the reduction potential to the oxidation potential, using the appropriate halfreactions.
Solve
(a)
Zn(s) → Zn2+(aq) + 2 e–
Cd2+(aq) + 2 e– → Cd(s)
o
Eanode
= −0.7618 V
o
Ecathode
= –0.403 V
o
o
o
Ecell
= Ecathode
− Eanode
= 0.359 V
Hg(l ) → Hg2+(aq) + 2 e–
(b)
Zn2+(aq) + 2 e–→ Zn(s)
o
Eanode
= 0.851 V
o
Ecathode
= –0.7618 V
o
o
o
Ecell
= Ecathode
− Eanode
= −1.613 V
Cd(s) → Cd2+(aq) + 2 e–
(c)
2+
–
Hg2 (aq) + 2 e → 2 Hg(l )
o
Eanode
= −0.403 V
o
Ecathode
= 0.7973 V
o
o
o
Ecell
= Ecathode
− Eanode
= 1.200 V
Think about It
o
Reaction (b) is not spontaneous because of its negative Ecell
.
19.97. Collect and Organize
By comparing the oxidation potentials of Zn, Cd, and Hg we can determine which of these metals is the best
reducing agent.
Analyze
o
The best reducing agent has the least positive reduction potential, Ered
.
Solve
o
For Zn, Ered
= – 0.7618 V
o
For Cd, Ered
= – 0.403 V
o
For Hg, Ered
= +0.7973 V (to Hg22+) and +0.851 V (to Hg2+)
2+
Zn is the least easily reduced of these metal ions, and therefore Zn metal is the best reducing agent.
Think about It
The ease of zinc’s oxidation is the basis for its use in galvanized steel and as a sacrificial anode to prevent the
rusting of the iron.
19.99. Collect and Organize
For this question we explain why zinc and magnesium have similar chemistries.
Electrochemistry and the Quest for Clean Energy | 411
Analyze
The electron configurations of Zn and Mg are
Zn
Mg
[Ar]4s23d10
[Ne]3s2
Solve
Because the filled d orbitals in Zn are considered to be core electrons, both Zn and Mg have the same
number of valence electrons (two s e–) and therefore have the same oxidation state in compounds (+2).
Think about It
The chemistries of Li+ and Mg2+ are also similar but because they have similar charge densities.
19.101. Collect and Organize
We are asked why solutions of Zn2+(aq) are colorless.
Analyze
Aqueous solutions of many transition metal compounds are colored because of d-orbital splitting induced by
ligands. Although zinc is a transition metal, the Zn2+ ion has the electron configuration [Ar]3d10.
Solve
Because Zn2+ has a d 10 configuration there are no d– d transitions possible; therefore, solutions of Zn2+ are
colorless.
Think about It
Other colorless transition metal ions include d 0 ions such as Ti4+.
CHAPTER 21 | Nuclear Chemistry
21.1. Collect and Organize
Of the highlighted elements in the periodic table shown in Figure P21.1, we are to choose the one that has a
stable isotope containing no neutrons.
Analyze
Neutrons are always present when there is more than one proton in the nucleus.
Solve
Only the element highlighted in red (hydrogen) contains no neutrons in its nucleus.
Think about It
Neutrons are sometimes referred to as the “glue” that holds the protons together in the nucleus. It is the strong
nuclear force that binds together a neutron’s three quarks, and the residual strong force binds neutrons to
protons in the nucleus.
21.3. Collect and Organize
Of the highlighted elements in the periodic table shown in Figure P21.1, we are to choose the one that has no
stable isotopes.
Analyze
Elements beyond bismuth (Z = 83) have no stable isotopes.
Solve
The orange element (astatine) is the element with no stable nuclei.
Think about It
It was until recently believed that bismuth was the last stable isotope, but theory had predicted that indeed
bismuth is unstable. In 2003 researchers in France measured the half-life of bismuth-209 to be 1.9 × 1019 yr.
This is a billion times longer than the age of the universe, so scientists still treat 209Bi as the last stable element
in the periodic table.
21.5. Collect and Organize
From the two graphs shown in Figure P21.5, we are to determine which one describes β decay.
Analyze
In β decay a neutron is changed into a proton and an electron is produced. In β decay, then, the number of
neutrons decreases and the number of protons increases.
Solve
Because β decay results in the emission of an electron from the nucleus, which increases the atomic number of
the nucleus, graph a illustrates β decay.
Think about It
Figure P21.5(b) represents the process of proton emission from the nucleus, where the number of protons is
reduced by 1 but the number of neutrons remains the same.
21.7. Collect and Organize
Given the five curves shown in Figure P21.7, we are to choose the one that represents a nuclear process with
t1/2 = 2.0 days.
Analyze
462
Nuclear Chemistry | 463
The half-life of a nuclear decay process is the time it takes for half of the original concentration to decay.
Starting from the original isotope quantity of 100%, it is the time it takes for that concentration to decrease to
50%.
Solve
The blue line b represents a decay process with t1/2 = 2.0 days. The quantity of the isotope is 50% after 2 days
after starting at 100%.
Think about It
Notice that the value of t1/2 does not change. On line b the time it takes for the quantity of the isotope to drop
from 50% to 25% is also 2.0 days.
21.9. Collect and Organize
From the processes depicted in Figure P21.9, we are to assign each as either fission or fusion.
Analyze
In a fission process the nucleus splits into smaller nuclei. In a fusion process smaller nuclei combine to form a
heavier nucleus.
Solve
Process 1 represents fission where smaller nuclei are generated (along with some nuclear particles shown as
gray spheres). Process 2 represents fusion where lighter nuclei are fused into a heavier nucleus.
Think about It
Note that in fusion, some nuclear particles (neutrons, α particles, or β particles) may be released to stabilize the
larger nucleus formed.
21.11. Collect and Organize
Given a list of nuclear particles, we are to arrange them in order of increasing mass.
Analyze
For this problem we need to recognize that a β particle has the same mass and charge as an electron, that a
positron has the same mass but opposite charge as the electron, that a neutron is slightly more massive than a
proton, that an α particle is equivalent in mass to a helium nucleus with two protons and two neutrons, and
that a deuteron consists of one proton and one neutron.
Solve
In order of increasing mass: electron = β particle = positron < proton < neutron < deuteron < α particle.
Think about It
A positron is an antielectron. When it encounters an electron both are immediately annihilated.
21.13. Collect and Organize
We are to describe how antihydrogen differs from hydrogen.
Analyze
An antiparticle has the same mass but opposite charge of the particle it is partnered with.
Solve
Antihydrogen has the same mass as hydrogen but its nucleus has a negative charge with a positively charged
electron. It contains the antiproton in the nucleus and a positron in place of the electron in the 1s orbital.
Think about It
Antihydrogen is immediately destroyed when it encounters the walls of the reactor in which it is generated.
464 | Chapter 21
21.15. Collect and Organize
For the annihilation of a proton and an antiproton when they collide, we can use Einstein’s relation between
mass and energy to calculate the energy released and the relationship between energy and wavelength to
calculate the wavelength of the gamma rays emitted from this process.
Analyze
We first have to find ∆m for the annihilation. Since there is no particle mass left over after the collision, ∆m is
the sum of the mass of the proton and the antiproton. This is simply twice the mass of the proton since the
mass of the proton equals the mass of the antiproton. We can then use Einstein’s equation to calculate ∆E. The
wavelength of the gamma ray emitted is found by the relationship
hc
λ=
E
As shown by Equation 21.6, we will assume that 2 gamma rays are emitted.
Solve
Δm = 2 × 1.67262 × 10 –24 g ×
1 kg
= 3.34524 × 10 –27 kg
1000 g
(
)
2
ΔE = 3.34524 × 10 –27 kg × 3.00 × 108 m/s = 3.010716 × 10 –10 J
λ=
hc 6.626 ×10 –34 J ⋅ s × 3.00 ×108 m/s
=
= 1.32 ×10 –15 m or 1.32 ×10 – 6 nm
E
3.010716 ×10 –10 J/2 photons
Think about It
If a mole of protons collided with a mole of antiprotons, the energy released would be
6.022 × 1023
3.010716 × 10–10 J/photon ×
= 1.813 × 1014 J
1 mol
21.17. Collect and Organize
We are to determine what percentage of radioactivity of a sample remains after two half-lives. A half-life is
defined as the time at which half of the radioactivity remains.
Analyze
After each half-life, there is a 50% decrease in radioactivity. We can start with 100% and decrease by 50% for
each half-life.
Solve
first
second
100% ⎯⎯⎯
→ 50% ⎯⎯⎯
→ 25%
half-life
half-life
Think about It
Alternatively, we can use Equation 21.3
Nt
= 0.5n
N0
where N is the number of half-lives:
Nt
= 0.52 = 0.25 or 25%
N0
21.19. Collect and Organize / Analyze
We can use the definitions of mass defect and binding energy to discriminate between these two terms.
Solve
The mass defect is the difference between the mass of the nucleus of an isotope and the sum of the masses of
the individual nuclear particles that make up that isotope. The binding energy is the energy released when
individual nucleons combine to form the nucleus of an isotope.
Nuclear Chemistry | 465
Think about It
The mass defect and binding energy are related to each other through Einstein’s equation:
ΔE = Δmc 2
21.21. Collect and Organize
We are asked to calculate the binding energy of the nucleus in 51V.
Analyze
The binding energy is calculated using
ΔE = Δmc 2
where c is the speed of light and ∆m is the mass defect. Because the nucleus of 51V is made up of 23 protons
(1.67262 × 10–27 kg each) and 28 neutrons (1.67493 × 10–27 kg each), the mass defect is the sum of the masses
of those particles subtracted from the actual mass of 51V (given as 50.9440 amu).
1.6605402 × 10–27 kg
50.9440 amu ×
= 8.45946 × 10–26 kg
amu
Solve
Δm = 8.45946 × 10 –26 kg – ⎡⎣ 23 × 1.67262 × 10 –27 kg + 28 × 1.67493 × 10 –27 kg ⎤⎦ = –7.737 × 10 –28 kg
(
(
ΔE = 7.737 × 10 –28 kg × 3.00 × 108 m/s
) (
)
2
)
= 6.96 × 10 –11 J
Think about It
Vanadium-51 is a stable isotope and its binding energy is relatively high for an element. Iron-56 has the
highest binding energy per nucleon and is therefore the most stable of all the nuclides.
21.23. Collect and Organize
For the four fusion reactions given, we can use Einstein’s equation to calculate the energy released in each.
Analyze
In the equation
ΔE = Δmc 2
c is the speed of light and ∆m is the mass defect. The mass defect is the mass lost in the reaction and must be
expressed in kilograms. Since the masses in this problem are expressed in atomic mass units (amu) we need to
use the conversion factor
1 amu = 1.6605402 × 10–27 kg
Solve
(a) For the production of 28Si from 14N + 14N
Δm = 27.97693 amu – (2 ×14.00307 amu) = –0.02921 amu
0.02921 amu ×
1.6605402 × 10–27 kg
= 4.8504379 × 10–29 kg
1 amu
(
)
2
ΔE = 4.8504379 × 10–29 kg × 3.00 × 108 m/s = 4.37 × 10–12 J
(b) For the production of 28Si from 10B + 16O + 2H
Δm = 27.97693 amu – (10.0129 amu +15.99491 amu + 2.0146 amu ) = –0.04548 amu
0.04548 amu ×
1.6605402 × 10–27 kg
= 7.552 × 10–29 kg
1 amu
(
)
2
ΔE = 7.552 × 10–29 kg × 3.00 × 108 m/s = 6.80 × 10 –12 J
466 | Chapter 21
(c) For the production of 28Si from 16O + 12C
Δm = 27.97693 amu – (15.994915 amu + 12.000 amu ) = –0.01798 amu
0.01798 amu ×
1.6605402 × 10–27 kg
= 2.98565 × 10–29 kg
1 amu
(
)
2
ΔE = 2.98565 × 10 –29 kg × 3.00 × 108 m/s = 2.69 × 10 –12 J
(d) For the production of 28Si from 24Mg + 4He
Δm = 27.97693 amu – (23.98504 amu + 4.00260 amu ) = –0.01071 amu
0.01071 amu ×
1.6605402 × 10 –27 kg
= 1.77844 × 10 –29 kg
1 amu
(
)
2
ΔE = 1.77843 × 10–29 kg × 3.00 × 108 m/s = 1.60 × 10 –12 J
Think about It
These reactions all release energy (there is a negative mass defect in proceeding from reactants to products).
Notice that in solving for the energy released we have used the positive value of the mass defect.
21.25. Collect and Organize
We are asked to compute the binding energy (BE) per nucleon in a nucleus of 4He.
Analyze
We can calculate the binding energy from Einstein’s equation:
BE = ΔE = Δmc 2
where ∆m is the difference in the mass between the 4He nucleus and the sum of the masses of the 2 protons
and 2 neutrons that compose the nucleus. Because there are 4 total nucleons, we divide the calculated value of
∆E by 4.
Solve
Δm = 4.00260 amu – ⎡⎣( 2 × 1.00728 amu ) + ( 2 × 1.00867 amu )⎤⎦ = –0.0293 amu
2
⎛
1.6605402 × 10–27 kg ⎞
8
⎜⎝ 0.0293 amu ×
⎟⎠ × 3.00 × 10 m/s
amu
BE per nucleon =
= 1.09 × 10–12 J/nucleon
4
(
Think about It
As shown in Figure 21.3,
nuclear stability.
56
)
Fe has the highest binding energy per nucleon and therefore has the highest
21.27. Collect and Organize
We are to describe how the belt of stability (shown in Figure 21.4) can be used to predict the possible decay
modes of radioactive nuclides.
Analyze
The belt of stability in Figure 21.4 plots the number of neutrons versus the number of protons.
Nuclear Chemistry | 467
Solve
If the nuclide lies in the belt of stability (green dots on the plot in Figure 21.4), it is not radioactive and is
stable. If it lies above the belt of stability, then it is neutron rich and tends to undergo β decay to increase the
number of protons and reduce the number of neutrons in its nucleus. If it lies below the belt of stability, it is
neutron poor and tends to undergo positron emission or electron capture to increase the number of neutrons
and reduce the number of protons in its nucleus.
Think about It
From Figure 21.4 we can see that there are often several known isotopes for an element that are radioactive
(orange dots).
21.29. Collect and Organize
Using the decay series for 238U shown in Figure 21.7, we can explain why several α decays are often followed
by β decay.
Analyze
We know, too, from the statement of the problem that the neutron-to-proton ratio of decay products must
decrease to form stable fission products.
Solve
Alpha decay increases the neutron-to-proton ratio to produce less stable isotopes, which can then be made
more stable through β emission to decrease the neutron-to-proton ratio.
Think about It
Ultimately, 238U decays to form stable 206Pb.
21.31. Collect and Organize
For the nuclear decay reactions described, we are to identify the modes of decay.
Analyze
By writing balanced equations for each decay, we can identify the mode of decay.
Solve
137
53
137
54
0
I → 137
54 Xe + –1 β
0
Xe → 137
55 Cs + –1 β
Both of these processes are β decays.
Think about It
The effect of the β -emission process is that the atomic number increases by 1, leaving the mass number
unchanged.
21.33. Collect and Organize
We consider the neutron-to-proton ratio for an isotope with a mass number, A, more than two times the atomic
number, Z.
Analyze
When the neutron-to-proton ratio is 1.00, there are equal numbers of protons and neutrons in the nucleus.
When the ratio is greater than 1, there are more neutrons than protons. When the ratio is less than 1, there are
more protons than neutrons.
Solve
When A > 2Z, the neutron-to-proton ratio is greater than 1.
468 | Chapter 21
Think about It
For light elements the neutron-to-proton ratio is about 1.0, but for heavier stable nuclei the ratio is greater than
1.0 and up to 1.5 for the heaviest elements.
21.35. Collect and Organize
For the decay of 26Al to 26Mg, we are to write a balanced nuclear equation.
Analyze
Since the mass number in this nuclear process stays the same but the atomic number decreases,
undergoes positron decay.
26
Al
Solve
26
13
Al → 10 β + 26
12 Mg
Think about It
We might predict that this process would occur since the neutron-to-proton ratio for 26Al is 1.0 but should be
slightly higher according to the belt of stability shown in Figure 21.4.
21.37. Collect and Organize
For the isotopes listed we can use the belt of stability shown in Figure 21.4 to predict the modes of decay.
Analyze
If the nuclide lies in the belt of stability (green dot on the plot), it is not radioactive and is stable. If it lies
above the belt of stability, then it is neutron rich and tends to undergo β decay to increase the number of
protons and reduce the number of neutrons in its nucleus. If it lies below the belt of stability, it is neutron poor
and it tends to undergo positron emission or electron capture to increase the number of neutrons and reduce
the number of protons in its nucleus.
Solve
(a) 10C has 6 protons and 4 neutrons and is neutron poor; it may undergo electron capture or positron
emission.
(b) 19Ne has 10 protons and 9 neutrons and is neutron poor; it may undergo electron capture or positron
emission.
(c) 50Ti has 22 protons and 28 neutrons and is stable.
Think about It
Notice that all of these isotopes are known, as they appear as orange or green dots in Figure 21.4.
21.39. Collect and Organize
For 56Co and 44Ti we are to predict the mode of decay.
Analyze
If the nuclide lies in the belt of stability (green dot in the plot in Figure 21.4), it is not radioactive and is stable.
If it lies above the belt of stability, then it is neutron rich and tends to undergo β decay to increase the number
of protons and reduce the number of neutrons in its nucleus. If it lies below the belt of stability, it is neutron
poor and it tends to undergo positron emission or electron capture to increase the number of neutrons and
reduce the number of protons in its nucleus.
Solve
56
Co has 27 protons and 29 neutrons and is neutron poor; it may undergo electron capture or positron emission.
44
Ti has 22 protons and 22 neutrons and is neutron poor; it may undergo electron capture or positron emission.
Think about It
Both of these radioisotopes, then, are expected to be positron emitters.
Nuclear Chemistry | 469
21.41. Collect and Organize
In the bombardment of a nucleus we consider why it requires less energy to bombard the nucleus with a
neutron compared to a proton.
Analyze
The mass of the proton and neutron are about the same, but they differ significantly in charge. A proton is
positively charged whereas a neutron is neutral. Any nucleus also has a positive charge, like the proton.
Solve
Neutrons have no charge and are not repelled by the positively charged nucleus like a proton is.
Think about It
Both processes, however, still require high temperatures so that the particles collide with sufficient force to
effect a fusion of the particle with the nucleus.
21.43. Collect and Organize
We are asked to describe how elements from cobalt to bismuth (atomic numbers 27 through 83) are produced.
Analyze
Elements to 26Fe are formed by fusion reactions in stars, and the binding energy per nucleon increases up to
iron. However, after 26Fe, the binding energy per nucleon for the elements decreases. A different process,
then, must form these elements.
Solve
The elements heavier than iron are formed by fusion reactions in supernovae which have the energy necessary
to produce these elements.
Think about It
The elements heavier than iron tend to undergo fission reactions in order to achieve maximum binding energy
per nucleon (Figure 21.3).
21.45. Collect and Organize / Analyze
We are to describe how linear accelerators and cyclotrons are used to prepare radioactive isotopes.
Solve
Both cyclotrons and linear accelerators accelerate small particles to high velocities (high energies) so that the
particles overcome the repulsive forces to fuse atoms or nuclear particles together.
Think about It
The fusing of atoms requires very high energies.
21.47. Collect and Organize
For the absorption of three neutrons by 56Fe, we are to write a balanced nuclear equation and predict the decay
mode for the radionuclide that is formed.
Analyze
The type of decay a nuclide undergoes depends on its neutron-to-proton ratio and where the nuclide lies in the
belt of stability (Figure 21.4).
Solve
1
59
(a) 56
26 Fe + 3 0 n → 26 Fe
(b) The neutron-to-proton ratio in 59Pb is 33/26 = 1.27. This nucleus is neutron rich (lies above the belt of
stability), so this nuclide is likely to undergo β decay.
59
26
59
Fe → –10β + 27
Co
470 | Chapter 21
Think about It
The process of β decay decreases the number of neutrons and increases the number of protons by 1. Notice
that for β decay the mass number does not change.
21.49. Collect and Organize
We can write nuclear equations describing the absorption of 3 neutrons by
identify the nuclide produced and assess its stability.
96
Mo followed by β decay to
Analyze
The absorption of 3 neutrons by 96Mo increases the mass number of Mo to 99. The process of β decay does
not change the mass number of 99Mo but does increase the atomic number by 1.
Solve
96
42
99
42
Mo + 3 01 n →
Mo →
0
−1
99
42
99
43
Mo
β + Tc
The neutron-to-proton ratio in 99Tc is 56/43 = 1.3, which lies above the belt of stability. No, this nuclide is not
stable.
Think about It
99
Tc is likely to undergo β emission to become stable.
21.51. Collect and Organize
For the nuclear reactions to prepare some isotopes used in medicine, we are to complete them by supplying
the missing nuclide or nuclear particle.
Analyze
To balance these reactions, the sum of the mass numbers of the reactants must equal that of the products.
Likewise, the sum of the atomic numbers of the reactants must equal that of the products.
Solve
1
(a) 32
16 S + 0 n →
32
15
P + 11 H
(b)
55
25
1
Mn + 11 H → 52
26 Fe + 4 0 n
(c)
75
33
As + 2 11 H →
(d)
124
54
1
0
Xe + n →
77
35
125
54
Br
0
Xe → 125
53 I + 1 β
Think about It
All of these nuclides are produced through the collision of light particles (neutron or hydrogen) with nuclides
of larger mass.
21.53. Collect and Organize
For the nuclear reactions given, we are to complete them by supplying the missing nuclide or nuclear particle.
Analyze
To balance these reactions, the sum of the mass numbers of the reactants must equal that of the products.
Likewise, the sum of the atomic numbers of the reactants must equal that of the products.
Solve
122
(a) 122
53 I → 54 Xe +
4
2
0
−1
β
(b)
10
5
13
7
B + α → N + 01 n
(c)
58
26
Fe + 01 n →
59
26
Fe
Nuclear Chemistry | 471
(d)
68
30
1
Zn + 11 P → 67
31 Ga + 2 0 n
Think about It
Notice that the addition of a neutron to a nucleus as in reaction c does not change the atomic number, only the
mass number.
21.55. Collect and Organize
For 230Th we are to write balanced nuclear equations for all the processes described in the Outside magazine
excerpt to determine the number of alpha and beta emissions and then list the half-lives of the isotopes named
so that we can judge the statement that all these processes take place “within minutes.”
Analyze
The CRC Handbook is an excellent reference for the half-lives of the isotopes. Also there are several sources
available online, such as http://nucleardata.nuclear.lu.se/nucleardata/toi/.
Solve
(a) There are 5 alpha and 2 beta decays.
230
4
90Th → 2 α +
226
88
222
86
218
84
214
82
214
83
214
84
226
88
Ra
t1/2 = 7.54 × 104 years
222
86
Rn
t1/2 = 1600 years
4
2
218
84
Po
t1/2 = 3.82 days
4
2
214
82
Pb
t1/2 = 3.10 min
Ra → 24α +
Rn → α +
Po → α +
Pb →
0
–1
β+
214
83
Bi →
0
–1
β+
214
84
Po → α +
210
82
4
2
Bi
t1/2 = 26.8 min
Po
t1/2 = 19.9 min
Pb
t1/2 = 164.3 µ s
(b) Once 218Po is produced this statement is true, but the rate of the entire process is determined by the ratelimiting step, the α decay of 230Th.
Think about It
Just because a substance is unstable and therefore radioactive does not mean that the rate of decay is fast.
21.57. Collect and Organize
We are asked to write balanced nuclear equations to describe the bombardment of 209Bi to form 211At.
Analyze
Bismuth and astatine differ in atomic number by 2. The mass numbers for the two isotopes for this problem
differ only by 2. Therefore, an appropriate particle with which to bombard 209Bi is the α particle with the
emission of two neutrons.
Solve
209
83
Bi + 24α →
211
85
At + 2 01 n
Think about It
Notice how the emission of neutrons does not change the atomic number, only the mass number.
21.59. Collect and Organize
We are asked to describe how the rate of energy release is controlled in nuclear reactors.
Analyze
Nuclear reactions that power the reactors release neutrons, which promote more nuclear fission processes as
shown in Figure 21.9. The reaction can be controlled by absorbing some of the neutrons.
472 | Chapter 21
Solve
Control rods made of boron or cadmium are used to absorb the excess neutrons to control the rate of energy
release in a nuclear reactor.
Think about It
When the control rods are removed the reactor core may go critical. This is what occurred in the reactor at
Chernobyl in the Soviet Union on April 26, 1986.
21.61. Collect and Organize
Figure 21.4 shows the belt of stability of the radionuclides. Using this we are to explain why neutrons are byproducts in fission reactions.
Analyze
In a fission reaction a heavier, unstable nucleus splits into two lighter nuclei.
Solve
The neutron-to-proton ratio for heavy nuclei is high and when the nuclide undergoes fission to form smaller
nuclides, it must emit neutrons because the fission products require a lower neutron-to-proton ratio for
stability.
Think about It
In fusing nuclei, more neutrons are needed for the heavier nuclei and so we might expect that β decay often
accompanies those processes.
21.63. Collect and Organize
For the incomplete fission reactions given, we are to determine the missing nuclides.
Analyze
To balance these reactions, the sum of the mass numbers of the reactants must equal that of the products.
Likewise, the sum of the atomic numbers of the reactants must equal that of the products.
Solve
1
(a) 235
92 U + 0 n →
96
40
Zr + 13852Te + 2 01 n
99
41
1
Nb + 133
51Sb + 4 0 n
(b)
235
92
U + 01 n →
(c)
235
92
90
1
U + 01 n → 37
Rb + 143
55 Cs + 3 0 n
Think about It
Notice that all of these fission products (96Zr,
235
U.
138
Te,
99
Nb,
133
Sb,
90
Rb, and
143
Cs) come from the fission of
21.65. Collect and Organize / Analyze
We can use the definitions of the terms to describe the difference between the level of radioactivity and the
dose of radioactivity.
Solve
The level of radioactivity is the amount of radioactive particles present in a given instant of time. The dose is
the accumulation of exposure over a length of time.
Think about It
A person can get a high dose of radioactivity either from a brief exposure to a highly intense radioactive
source or through prolonged exposure to low levels of radiation.
Nuclear Chemistry | 473
21.67. Collect and Organize
We are to describe the dangers of 222Rn.
Analyze
Radon is a colorless, odorless gas that results from the natural decay of uranium in the earth. It is an α emitter
that decays to 218Po, also an α emitter.
Solve
When radon-222 decays to polonium-218 while in the lungs, the 218Po, a reactive solid that is chemically
similar to oxygen, lodges in the lung tissue where it continues to emit α radiation. Alpha radiation is one of
the most damaging kinds of radiation when in contact with biological tissues. The result of exposure to high
levels of radon is an increased risk for lung cancer.
Think about It
Most hardware stores sell radon detection kits for homeowners to learn whether the level of radon-222 in their
homes is unusually high.
21.69. Collect and Organize
We are asked to calculate the grays that are equal to 5 µSv and how much energy 5 µSv corresponds to for a
50 kg person.
Analyze
From Table 21.5, we see that
1 Sv = 1 Gy × RBE
The relative biological effectiveness (RBE) of X-rays is given in the problem as 1, so for dental X-rays
Also from Table 21.5, we see that
1 Sv = 1 Gy × 1 or 1 µSv = 1 µGy × 1
1 Gy = 1 J/kg of tissue mass or 1 µGy = 1 µJ/kg
Solve
5 µSv = 5 µGy
1 µ J/kg
5 µGy ×
× 50 kg = 250 µ J
1 µGy
Think about It
This dosage of 5 µSv is well below the dose that may cause toxic effects (Table 21.6).
21.71. Collect and Organize
For the radioactive isotope 90Sr, we are to write a balanced nuclear equation corresponding to its β decay,
calculate the atoms of 90Sr in 200 mL of milk that has 1.25 Bq/L of 90Sr radioactivity, and give a reason why
90
Sr would be more concentrated in milk rather than other foods.
474 | Chapter 21
Analyze
(a) The process of β decay increases the atomic number, leaving the mass number unchanged.
(b) To calculate the atoms of 90Sr in the milk we can use the equation
mL of milk ×
1.25 Bq 1 disintegration/s
×
= disintegrations/s
1000 mL
1 Bq
This is the rate of the first-order decay of 90Sr that follows the rate law
90
Rate = k[ 90 Sr]
where k = 0.693/t1/2. Given that the t1/2 for Sr is 28.8 yr, we need to convert from years to seconds.
Solve
0
90
(a) 90
38 Sr → –1 β + 39Y
(b) The number of disintegrations in 200 mL of milk is
1.25 Bq 1 disintegration/s
200 mL ×
×
= 0.250 disintegration/s
1000 mL
1 Bq
The rate constant k in reciprocal seconds is
0.693 1 yr
1d
1 hr 1 min
k=
×
×
×
×
= 7.63 ×10–10 s –1
28.8 yr 365 d 24 hr 60 min 60 s
The concentration of 90Sr in the milk from the first-order rate law is
0.250 disintegrations/s
⎡⎣ 90 Sr ⎤⎦ =
= 3.28 ×108 90 Sr atoms
7.63 ×10–10 s –1
(c) Strontium-90 is found in milk and not other foods because it is chemically similar to calcium, and milk is
rich in calcium.
Think about It
In more familiar chemical concentration terms, the concentration of 90Sr in these samples is
3.28 × 108 90Sr atoms ×
1 mol
1
×
= 2.72 × 10–15 M
23
6.022 × 10 atoms 0.200 L
21.73. Collect and Organize
For drinking water we are to calculate the number of decay events per second in 1.0 mL with a radon level of
4.0 pCi/mL. We are then to calculate the number of Rn atoms in 1.0 mL given that t1/2 of 222Rn = 3.8 d.
Analyze
(a) To calculate the number of decay events per second from picocuries, we need the conversions
1 pCi = 1 × 10–12 Ci
1 Ci = 3.70 × 1010 Bq
1 Bq = 1 disintegration/s
(b) The decay of 222Rn follows the first-order rate law
Rate = k[222Rn]
where the rate is the number of decay events per second and k, the rate constant, is
k=
0.693
t1/2
Here t1/2 = 3.8 d, must be converted to seconds.
Solve
(a) 4.0 pCi ×
1 × 10–12 Ci 3.70 × 1010 Bq 1 decay/s
decays
×
×
= 0.15
1 pCi
1 Ci
1 Bq
s
Nuclear Chemistry | 475
(b) [ 222 Rn] =
0.148 decays/s
= 7.0 ×104 atoms
2.11×10–6 s –1
Think about It
Even though this seems like a large number of 222Rn atoms, the percentage of 222Rn atoms in 1.0 mL of water
is very low:
1.0 mL ×
1 g 1 mol 6.022 ×1023 H 2 O molecules
×
×
= 3.35 ×1022 molecules of H 2 O
4
mL 18 g
mole
7.0 ×10 atoms of 222 Rn
% 222 Rn atoms =
×100
3.35 ×1022 molecules of H2 O
= 2.1×10–16 %
21.75. Collect and Organize / Analyze
We are asked to consider how a radioactive isotope for radiotherapy is selected based on its half-life, decay
mode, and properties of its products.
Solve
(a) The half-life should be long enough to effect treatment of the cancerous cells but not so long as to cause
damage to healthy tissues.
(b) Because α radiation does not penetrate far beyond a tumor, the α decay mode is best.
(c) Products should be nonradioactive, if possible, or have short half-lives and be able to be flushed from the
body by normal cellular and biological processes.
Think about It
The investigation and development of new radioisotopes are very active areas of research.
21.77. Collect and Organize
For each of the isotopes given, we can use the belt of stability (Figure 21.4) to predict the mode of decay.
Analyze
If the nuclide has a neutron-to-proton ratio that places it below the belt of stability, either electron capture or
positron emission is likely. If a nuclide lies above the belt of stability, β emission is likely.
Solve
(a) 197
has 80 protons and 117 neutrons. This nuclide lies below the belt of stability and so is likely to
80 Hg
decay by positron emission or electron capture.
75
(b) 34
Se has 34 protons and 41 neutrons. This nuclide lies below the belt of stability and so is likely to decay
by positron emission or electron capture.
(c) 189 F has 9 protons and 9 neutrons. This nuclide lies below the belt of stability and so is likely to decay by
positron emission or electron capture.
Think about It
These imaging agents have relatively short half-lives: 197Hg, 64 hr; 75Se, 120 days; 18F, 110 min.
21.79. Collect and Organize
For a 1.00 mg sample of
half-life of 192Ir.
192
Ir, 0.756 mg remains after 30 days. From this information we are to calculate the
Analyze
We need to rearrange Equation 21.18 to solve this problem:
t
N
t = – 1/2 ln t
0.693
N0
476 | Chapter 21
t1/2 = –
0.693t
ln ( Nt N 0 )
Solve
t1/2 = –
0.693 × 30 d
= 74.3 d
ln ( 0.756 mg 1.00 mg )
Think about It
After 30 days, then, this isotope has decayed through not even one half-life (Equation 21.3):
0.756
= 0.5n
1.00
0.756 = 0.5n
n = 0.404 half-lives
21.81. Collect and Organize
Using the information given about the half-life of 131I (t1/2 = 8.1 d) and the initial and residual activity of the
isotope after 30 days (108 and 4.1 counts/min, respectively), we are to determine whether the brain cells took
up any of the 131I.
Analyze
We need to calculate the expected activity of the 131I if it were not taken up by the cells. This is found through
the use of Equation 21.18 to solve for Nt.
t
N
t = – 1/2 ln t
0.693
N0
If the activity of the sample after 30 days is less than the calculated activity, then the cell did take up 131I.
Solve
N
8.1 d
ln t
0.693
108
Nt
−2.567 = ln
108
Nt
0.07679 =
108
Nt = 8.29 counts/min
Because this is more than the counts/min in the sample, yes, the brain cells must have incorporated some of
the 131I.
30 days = –
Think about It
The sample was taken after the 131I had decayed through 3.7 half-lives:
8.29
= 0.5n
108
0.0768 = 0.5n
n = 3.7 half-lives
21.83. Collect and Organize
Given that the half-life of 11C is 20.4 min, we are to calculate the time it takes for 99% of the injected 11C to
decay.
Analyze
We can use Equation 21.18 directly to solve this problem:
Nuclear Chemistry | 477
t1/2
N
ln t
0.693
N0
where Nt = 1 and N0 = 100 for the 99% decay of the isotope.
t=–
Solve
t=–
20.4 min
1
ln
= 136min
0.693
100
Think about It
This time is the equivalent of over 6 half lives:
1
= 0.0100 = 0.5n
100
n = 6.64 half-lives
21.85. Collect and Organize
For the decay of 11B formed during boron neutron-capture therapy (BNCT) for cancers and which decays to
7
Li, we are to write balanced nuclear equations for the neutron absorption and α decay, then calculate the
energy released in the process using Einstein’s equation, and finally consider why this process where
an α emitter is produced by neutron capture is an effective cancer treatment.
Analyze
(a) In writing the balanced equation we need to balance the sum of the mass numbers of the products with that
of the reactants. Likewise, the sum of the atomic numbers of the products must equal that of the reactants.
(b) In the Einstein equation
ΔE = Δmc 2
c is the speed of light and ∆m is the mass defect. The mass defect is the mass lost in the reaction and must be
expressed in kilograms. Since the masses in this problem are expressed in amu we need to use the conversion
factor
1 amu = 1.6605402 × 10–27 kg
Solve
(a) 105 B + 01 n → 115 B
11
5
B → 73 Li + 24α
Overall process: 105 B + 01 n → 73 Li + 24α
(b) Δm = (7.01600 amu + 4.00260) – (10.0129 amu + 1.008665 amu) = –0.002965 amu
0.002965 amu ×
1.6605402 × 10–27 kg
= 4.9235 × 10–30 kg
1 amu
(
)
2
ΔE = 4.9235 × 10–30 kg × 3.00 × 108 m/s = 4.43 × 10 –13 J
(c) Alpha particles have a high RBE, and they do not penetrate into healthy tissue if the radionuclide is placed
inside a tumor.
Think about It
BNCT is especially useful in the treatment of brain tumors. However, the only source of neutrons is from
nuclear reactors and there are only a few sites worldwide that have this treatment available.
21.87. Collect and Organize
We are to explain why radiocarbon dating is reliable only for artifacts younger than 50,000 years.
478 | Chapter 21
Analyze
The half life of
8.726 half-lives.
14
C is 5730 yr. The number of half-lives that 50,000 years represents is 50,000/5730 =
Solve
After 8.726 half-lives, the ratio of 14C present to that originally in an artifact is
Nt
= 0.508726 = 0.00236 or 0.236%
N0
This is too little to detect.
Think about It
Longer-lived isotopes would be better suited for dating old artifacts. For example,
useful for objects older than 300,000 years!
40
K (Problem 21.89) is
21.89. Collect and Organize
For this question we consider why 40K is a useful isotope only for dating objects older than 300,000 years.
Analyze
The half-life of 40K is 1.28 × 109 yr. The number of half-lives that 300,000 years represents is
300,000/1.28 × 109 = 0.00023 half-lives.
Solve
After 0.00023 half-lives the ratio of 40K present to that originally in a sample is
Nt
= 0.500.00023 = 0.9998 or 99.98%
N0
This level is just when we can detect the difference in amounts of 40K.
Think about It
For objects younger than 300,000 years, an isotope with a shorter half-life must be used.
21.91. Collect and Organize
For a piece of charcoal that is 8700 years old, we are to calculate the fraction of 14C remaining.
Analyze
We can use Equation 21.18 to solve for Nt/N0:
t=–
t1/2
N
ln t
0.693
N0
Solve
8700 yr = –
N
5730 yr
ln t
0.693
N0
Nt
= 0.35 or 35%
N0
Think about It
Alternatively, this problem can be solved by first calculating n, the number of half-lives:
n=
8700
= 1.518 half-lives
5730
Nt
= 0.51.518 = 0.35 or 35%
N0
Nuclear Chemistry | 479
21.93. Collect and Organize
For a wood sample from a giant sequoia tree that in 1891 was 1342 years old, we are to compare the fraction
of 14C in the innermost ring with that in the outermost ring.
Analyze
We can use Equation 21.18 to solve for Nt/N0:
t=–
t1/2
N
ln t
0.693
N0
Solve
1342 yr = –
N
5730 yr
ln t
0.693
N0
Nt
= 0.850 or 85.0%
N0
Think about It
Alternatively, this problem can be solved by first calculating n, the number of half-lives:
n=
1342
= 0.2342 half-lives
5730
Nt
= 0.50.2342 = 0.850 or 85.0%
N0
21.95. Collect and Organize
Given that the 14C to 12C ratio is only 1.19% for the ancient mammoth tusk compared to the ratio in elephants
today, we are to calculate the age of the mammoth tusk.
Analyze
We can use Equation 21.18 to solve for t:
t=–
t1/2
N
ln t
0.693
N0
Solve
t=–
5730 yr
1.19
ln
= 36,640 yr
0.693
100
Think about It
The number of half-lives this represents for 14C is
n=
36,640
= 6.39 half-lives
5730
21.97. Collect and Organize
We are asked to explain why antihydrogen would have been a suitable fuel for the starship Enterprise.
Analyze
In Problem 21.15 we calculated that a large amount of energy (1.813 × 1014 J) is released when one mole of
hydrogen collides with one mole of antihydrogen.
Solve
Besides releasing a large amount of energy to power the starship Enterprise, hydrogen is an abundant fuel in
the universe and therefore could easily react with any antihydrogen produced.
Think about It
Antihelium, although much more difficult to produce, might also have been a good choice.
480 | Chapter 21
21.99. Collect and Organize
We consider whether dating by determining the activity of tritium would be able to determine if a wine were
made from grapes grown in 1969.
Analyze
In order to know if the grapes were grown in 1969 we need to be able to discriminate using the tritium
activity between grapes grown in 1969 versus those in 1968 or 1970. We can use the following equation to
determine the expected activity of 1969 grapes versus 1968 or 1970 grapes:
Nt
= 0.5n
N0
age of wine
n=
t1/ 2 of tritium
Solve
For grapes grown in 1968
40 yr
n=
= 3.33
12 yr
For grapes grown in 1969
39 yr
n=
= 3.25
12 yr
For grapes grown in 1970
38 yr
n=
= 3.17
12 yr
Nt
Nt
Nt
= 0.53.33 = 0.0994
= 0.53.25 = 0.105
= 0.53.17 = 0.111
N0
N0
N0
These ratios are all very close, so it is unlikely that we would be able to confirm that the grapes were grown
in 1969.
Think about It
We might be able to discriminate a difference of 5 years, however.
21.101. Collect and Organize
We need to calculate the energy released using Einstein’s equation for the fusion of four hydrogen atoms to
compare that to the energy released by the fission of 235U. Because this fusion reaction produces positrons,
which are immediately annihilated, we include their energy in the calculation.
Analyze
To determine the energy released in the fusion reaction, we need to first calculate the mass defect for the
reactions and then solve Einstein’s equation.
Solve
The mass defect for the fusion reaction is
Δm = ⎡⎣(2 × 5.485799 ×10 – 4 amu) + 4.00260 amu ⎤⎦ − ⎡⎣(4 ×1.00728 amu ) + (2 × 5.485799 ×10 – 4 amu )⎤⎦
1.6605402 × 10–27 kg
= −0.02652 amu
= 4.404 × 10–29 kg
amu
The energy released in the fusion reaction is
ΔE = 4.404 × 10 –29 kg × (3.00 × 108 ) 2 = 3.9633 × 10 –12 J/atom 4 He
0.02652 amu ×
3.9633 × 10–12 J
= 9.91 × 10–13 J/nucleon
4
The energy released in the fission reaction is
ΔE per nucleon =
∆E = 3.2 × 10–11 J/atom × 1 atom/235 nucleons = 1.4 × 10–13 J/nucleon
On a per nucleon basis, the fusion reaction generates more energy.
Think about It
On a per gram basis, the fusion reaction also provides more energy.
Nuclear Chemistry | 481
21.103. Collect and Organize
For the absorption of a neutron by 11B and the subsequent α or β decay of the product 12B, we are to write
balanced equations for these processes and then decide if any of the final nuclear reaction products are stable.
Analyze
We can use the usual method to balance the nuclear reactions. For each we must make sure that the sum of
the atomic numbers for the reactants equals the sum of the atomic numbers of the products. Likewise, the
sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products.
Solve
(a) 115 B + 01 n → 125 B
12
5
B → 126 C + –10β
12
5
B → 83 Li + 24α
(b) 12C is stable.
Think about It
We expect that 8Li would decay by β decay because it is neutron rich and lies above the belt of stability
(Figure 21.4).
21.105. Collect and Organize
For the incomplete nuclear reactions to synthesize the superheavy elements, we are to fill in the missing
nuclide.
Analyze
To balance these reactions, the sum of the mass numbers of the reactants must equal that of the products.
Likewise, the sum of the atomic numbers of the reactants must equal that of the products.
Solve
58
(a) 26
Fe +
209
83
Bi →
266
109
Mt + 01 n
(b)
64
28
Ni + 209
83 Bi →
272
111
Rg + 01 n
(c)
62
28
Ni + 208
82 Pb →
269
110
Ds + 01 n
(d)
22
10
Ne +
(e)
58
26
Fe + 208
82 Pb →
249
97
Bk →
267
107
265
108
Bh + 4 01 n
Hs + 01 n
Think about It
These superheavy elements all belong to the transition metal series.
21.107. Collect and Organize
For the synthesis of 294Uuo, we are to describe its decay reactions with balanced equations and, by looking at
its position in the periodic table, select another element that has similar properties to Uuo.
Analyze
To balance these reactions, the sum of the mass numbers of the reactants must equal that of the products.
Likewise, the sum of the atomic numbers of the reactants must equal that of the products. By balancing the
equations we can identify the nuclides in parts b, c, and d of this problem.
Solve
(a) 249
98 Cf +
(b)
294
118
48
20
Ca →
Uuo → 24α +
294
118
290
116
Uuo + 3 01 n
Uuh
482 | Chapter 21
(c)
290
116
Uuh → 24α +
286
114
(d)
286
114
Uuq → 24α + 282
112 Cn
Uuq
(e) Because 294Uuo is a member of the noble gas family, it has chemical and physical properties similar to
naturally occurring radon.
Think about It
Even though these superheavy elements are short-lived, their half-lives are sufficiently long (milliseconds) to
allow for some of their chemical and physical properties to be experimentally determined.
21.109. Collect and Organize
By examining the half-lives for nuclides for a decay series for
present in the highest amount in a sample after 1 year.
214
Bi, we can identify which element will be
Analyze
The decay process can only go as fast as the slowest step. This means that there will be a buildup of the
isotope with the longest half-life.
Solve
Because
1 year.
210
Pb has the longest half-life in the decay series, it will be the most abundant of the isotopes after
Think about It
In the language of kinetics, we can think of the decay of 210Pb as the rate-limiting step.
21.111. Collect and Organize
For the synthesis of Pt by two fusion reactions, we can determine which isotopes of Pt are formed by writing
the balanced nuclear reactions for their formation.
Analyze
To balance these reactions, the sum of the mass numbers of the reactants must equal that of the products.
Likewise, the sum of the atomic numbers of the reactants must equal that of the products.
Solve
124
188
(a) 64
28 Ni + 50 Sn → 78 Pt
(b)
64
28
196
Ni + 132
50 Sn →78 Pt
Think about It
Both of these isotopes of Pt lie well above the belt of stability (Figure 21.4), so we would expect that they
might decay by β or α emission.
21.113. Collect and Organize
For the synthesis of one atom of
bombarded with 62Ni.
269
Ds, we are to write the balanced nuclear reaction when
208
Pb is
Analyze
To balance this reaction, the sum of the mass numbers of the reactants must equal that of the products.
Likewise, the sum of the atomic numbers of the reactants must equal that of the products.
Solve
208
82
Pb + 62
28 Ni →
269
110
Ds + 01 n
Think about It
Notice that this fusion reaction also produces a mole of neutrons for every 1 mol of 269Ds produced.
Nuclear Chemistry | 483
21.115. Collect and Organize
For the dating of ancient human skulls in Ethiopia based on the amount of 40Ar, we are to propose a decay
mechanism for this nuclide to form from 40K and explain why researchers used 40Ar to date the skulls and not
14
C.
Analyze
(a) The mass number in the decay of 40K to 40Ar does not change, which indicates either a β decay or
positron emission.
(b) To explain why researchers used 40Ar instead of 14C dating, we must compare the half-lives of the parent
isotopes: 14C, 5730 yr; 40K, 1.28 × 109 yr.
Solve
40
0
(a) 40
19 K → 18 Ar + 1 β
(b) Because the half-life of 40K is so much longer than that of 14C, 40Ar can be used to date much, much older
objects.
Think about It
The amount of 14C remaining after 154,000 yr would be too small to measure
Nt
= 0.5154,000 / 5730 = 8.1 × 10–9
N0
21.117. Collect and Organize
We are to determine the ratio of 14C present in a bone sample that is 15,000 years old to another sample that
is 25,000 years old.
Analyze
We can determine the ratio using
Nt /N 0 (15, 000 years old) 0.5n
=
Nt /N 0 (25, 000 years old) 0.5n
where n for 15,000 years ago is 15,000 yr/5730 yr and n for 25,000 years ago is 25,000 yr/5730 yr.
Solve
Nt /N 0 (15, 000 years ago) 0.515,000 5730
=
= 3.35
Nt / N 0 (25, 000 years ago) 0.525,000 5730
Think about It
The number of half-lives of
4.4 half-lives.
14
C that have passed for the 25,000-year-old sample is 25,000 yr/5730 yr =
21.119. Collect and Organize
Given the counts per minute over time for the decay of 208Tl we can use a first-order plot to find the value of
the rate constant k and then compute the half-life of the decay.
Analyze
Radioactive decay follows first-order kinetics. A plot of ln(counts/min) versus time gives a straight line with
slope = –k. The half-life of the first-order decay is equal to 0.693/k.
Solve
The first-order plot gives a straight line with slope = –5.87 × 10–3, so k = 5.87 × 10–3 s–1. The half-life of the
decay is
t1 2 =
0.693
= 118 s
5.87 × 10–3
484 | Chapter 21
Think about It
If a radioactive decay is fast enough, experiments such as those described in Chapter 15 can be performed to
measure with confidence the half-life of the decay.
21.121. Collect and Organize
For a radon detector that collects 214Pb and 214Bi, we are to explain how these isotopes are related to
We are also to explain why it is lead and bismuth that are collected, not radon.
222
Rn.
Analyze
Both bismuth and lead are solids. Radon is a gas.
Solve
(a) Both 214Pb and 214Bi are formed in the nuclear decay of 222Rn. 214Pb is produced after two α decays and
214
Bi is produced when 214Pb undergoes β decay.
(b) Radon is an inert gas and does not stick to the charcoal like solid Pb and Bi will.
Think about It
Radon is the second leading cause of lung cancer after smoking.
21.123. Collect and Organize
We are to explain why radioactive radon is more harmful than radioactive uranium from which it is
produced.
Analyze
Uranium is a heavy metal that is a solid. Radon is a gas.
Solve
Because radon is a gas, it seeps up from the ground to fill enclosed spaces such as basements. Uranium does
not spread in that way; rather, it stays locked in the uranium ore.
Think about It
County by county radon risk levels in the United States are shown at the EPA’s website
(http://epa.gov/radon/zonemap.html).