1
Answers to Exercises and Problems
for
CHEMISTRY
A Guided Inquiry
Seventh Edition, 2017
Richard S. Moog
Franklin & Marshall College
John J. Farrell
Franklin & Marshall College
Latest Update: December 20, 2017
John Wiley & Sons, Inc.
2
Answers to Exercises and Problems
These answers may be provided to students at the instructor’s discretion.
Do not post on publicly available website.
These answers do not contain full explanations or solutions.
ChemActivity 1
Exercises
1.
Isotope
Mass
Number
A
31
Number of
Electrons
P
Atomic
Number
Z
15
O
8
18
8
39 +
19
39
18
31
18
K
15
58
2.
3.
4.
5.
6.
7.
8.
28
58
26
Ni2+
1.674 × 10–24 g. 1.993 × 10–23 g.
8.67 × 10–17 g.
12.00 g.
2.6621 × 10–23 g.
a) The mass number is the sum of the number of protons and the number of neutrons in the
nucleus. b) The atomic number is the number of protons in the nucleus.
False. 18O has 8 protons and 10 neutrons.
Species
24
Mg
23
Na+
35
Cl
35 Cl
56 3+
Fe
15
N
16 2O
Electrons Protons
12
12
10
11
17
17
18
17
23
26
7
7
10
8
10
13
27
Al3+
Neutrons
12
12
18
18
30
8
8
14
9.
Isotope
59Co2+
14
N
7Li
6Li
58
Zn2+
19 –
F
Atomic
Number
Z
27
7
Mass
Number
A
59
14
Number of
Electrons
3
3
30
7
6
58
3
3
56
9
19
10
25
7
3
10.
All isotopes of an element have the same number of protons in the nucleus. One isotope of
an element is differentiated from another isotope of the same element by the number of
neutrons in the nucleus.
Problems
1.
Using carbon-13 and carbon-12, approx mass of neutron = 13.0034 – 12 = 1.0034 amu.
approx mass of 14C = 13.0034 + 1.0034 = 14.0068 amu.
2.
a) Using 14C and 14C–, mass of electron is approximately 13.0039 amu – 13.0034 amu =
0.0005 amu
b) Using 1H, mass of proton is approximately 1.0078 amu – 0.0005 amu = 1.0073 amu
3.
c) Using 1H and 2H, mass of neutron is approximately 2.0140 amu – 1.0078 amu =
1.0062 amu
(Note that slightly different values for the masses of protons and neutrons will be obtained
if different elements/isotopes are used to calculate these masses.)
The calculated mass of 12C based on the masses of the constituent particles is 12.099 amu;
and the actual mass of 12C is exactly 12 amu.
ChemActivity 2
Exercises
1. a) 1.008 g. b) 39.10 g.
2. a) 45.98 g. b) 57.27 g.
3. Isotopes are versions of atoms of an element that have the same number of protons in the
nucleus but differ in the number of neutrons that they have.
4. 37Cl has two more neutrons in its nucleus.
1
3
5. average mass of a marble = (1 × 5.00 g + 3 × 7.00 g)/4 = 4 × 5.00 g + 4 × 7.00 g =
0.2500 × 5.00 g + 0.7500 × 7.00 g = 6.50 g (this is eqn (2))
6. 10.44 amu
7. 35Cl, 75.76%. 37Cl, 24.24%.
8. a) 4.003 g b) 39.10 g
9. a) 1 He atom ×
×
4.003 g He/mole He atoms
= 6.647 × 10–24 g of He.
b) 1 K atom ×
×
39.098 g K/mole K atoms =
6.493 × 10–23 g. of K
10.
11.
12.
13.
14.
15.
60.06 g
3.613 × 1024 atoms.
2.619 × 1024 atoms.
a) 3.029 × 1025 atoms. b) 1.022 × 1019 atoms. c) 1.878 × 1025 atoms. d) 9.782 × 1027
atoms.
Phosphorus
89.5 g I
4
Problems
1. Assume that mass of 22Ne is 22 amu. Calculated avg mass of Ne is 20.18—close to the
experimental value of 20.179.
2. a) False. 6.941 g per mole of Li atoms. b) False. No H atoms weighs 1.008 amu.
c) True. Na atoms are more massive. d) False. 1H is the lightest atom and has a mass of 1
g/mole. This is (essentially) the mass of one proton. Therefore, there cannot be an atom that
has a lower average atomic mass than 1 g/mole.
3. 17: protons in nucleus and electrons in the neutral atom. 35.453: avg amu of a Cl atom and
grams of one mole of Cl atoms.
187
4.
Re
ChemActivity 3
Exercises
1. 5.47 × 10–18 J.
2. i) IEa = –(2)(–1)/d1 = 2/d1 ii) IEb = –(1)(–1)/2d1 = 1/2d1 IEi > IEii
3. The ionization energy of case i) is larger, 1.20 k/d 1 , than that of case ii), 1.17 k/d 1 .
Problems
1. large and negative
2. V =
ChemActivity 4
Exercises
1. No. The ionization energy of He would be about 4× (twice the charge and half the distance)
the ionization energy of H if this were the case. The data in Table 1 indicate that the
ionization energy of He is only about twice as great.
2. Several possible answers. One possibility: All three electrons at a farther distance (about
three times as far as in H) from the nucleus.
Problem
1. a) V =
b) The IE of He is slightly less than twice the IE of H because the electron-electron
repulsion makes the potential energy more positive. Note that the first two terms in 1a) are
negative and the third term is positive.
ChemActivity 5
Exercises
1. a) 6. b) 5. c) 8.
2. a) +6. b) +5. c) +8.
3. The IE of Br should be less than the IE of Cl. There is about a 0.4 MJ/mole difference
between the IEs of F and Cl. Prediction: Br, 0.8 MJ/mole.
4. The IE of Li+ should be larger than the IE of He because both atoms have 2 electrons in the
1st shell and Li+ has a core charge of +3 whereas He only has a core charge of +2.
5. The IE of F– should be less than the IE of Ne because both atoms have eight electrons in
the 2nd shell and F– has a core charge of +7 whereas Ne has a core charge of +8.
5
6. IE of Kr > IE of Br because they are in the same valence shell and Kr has the higher core
charge (+8 vs. +7). IE of Rb is the lowest because core charge is +1 and its valence shell (n
= 5) is larger than the valence shell (n = 4) of Kr and Br.
7. One of the inner shell electrons is harder to remove because it is closer to the nucleus and
experiences a higher core charge.
Problems
1. a) TRUE. The nuclear charge is 35 and the number of valence electrons is seven. Thus,
there are 28 core electrons and the core charge is 35 - 28 = 7. For neutral atoms, the core
charge is equal to the number of valence electrons.
b) TRUE. H and He are the only elements with a valence shell of n = 1. He has a core
charge of +2 and H has a core charge of +1. Thus, He is expected to have a higher 1st
ionization energy than H because the valence electrons are at roughly the same distance
from the nucleus but the nuclear charge of He is higher. All other atoms have a valence
shell of n = 2 or higher so the IE is lower because the distance from the nucleus is greater.
2. If the fourth electron in Be were added to a third shell, it would be easier to remove
because the core charge would be +1 (as in Li) but the distance to the outermost electron
would be greater than that for Li, resulting in an IE that would be less than the IE of Li.
ChemActivity 6
Exercises
1. Ar: predict r = 150 pm (larger than K+ but smaller than Cl–). N: predict r = 71 pm (larger
than O but smaller than C). F–: predict r = 90 pm (considerably smaller than Cl–, but
probably larger than other 2nd period neutral atoms). Ne: predict r = 50 pm (smaller than
O).
2. a) False. Both have a core charge of +2 and 2 valence electrons but the valence electrons of
Ba are much farther away. b) False. Both have 10 electrons and sodium has more protons.
c) True. Both have 18 electrons and chlorine has fewer protons. d) True. Ar and Ne have
the same core charge and the same number of valence electrons, but Ar has n = 3 and Ne
has n = 2 so Ar has a larger radius. e) False. Ar and Ca2+ are isoelectronic and Ca2+ has
more protons.
3. a) N b) K+ c) Cl d) H e) Mg2+
4. Fe2+
5. a) Pb b) Na c) Ba2+ d) H– e) Rb
f) P3–
Problems
1. Na. The second electron removed would experience a core charge of +8 and has n = 2. The
others have lower core charges and higher values of n (larger radius).
2. Mg2+ is isoelectronic with Ne. Mg2+ is smaller than Ne because it has two more protons
than Ne. S2– is isoelectronic with Ar. S2– is larger than Ar because Ar has two more
protons than S. Ar is larger than Ne because they have the same core charge and number of
valence electrons bu Ar has a larger radius. Therefore, S2– is larger than Mg2+.
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ChemActivity 7
Exercises
1. False. f = c/λ . If c remains constant, then as λ increases, f decreases. Thus, the shorter the
wavelength, λ, the greater the frequency, f.
2.
Energy (J)
Wavelength (m) Frequency (s–1) Region of Spectrum
infrared
9.94 × 10–20
2.00 × 10–6
1.50 × 1014
–19
–6
14
visible
3.97 × 10
0.500 × 10
6.00 × 10
–19
–7
15
ultraviolet
9.94 × 10
2.00 × 10
1.50 × 10
–16
–9
17
X-ray
1.99 × 10
1.00 × 10
3.00 × 10
3. E = hf = hc/λ. The energy is inversely proportional to the wavelength. Thus, the blue
photon with the smaller λ is more energetic.
Problem
1. No. The energy required to ionize a sodium atom is 8.30 × 10–19 J. A photon with a
wavelength of 500 nm has only 3.96 × 10–19 J.
ChemActivity 8
Exercises
1. 140.3 MJ/mole
2. a)
7
3.
ChemActivity 9
Exercises
1. a) Two. b) Peak with higher ionization energy (1s) is twice the intensity of the peak with
lower ionization energy (2s). c) The nuclear charges for H, He, and Li are 1, 2, and 3,
respectively. Therefore, the electrons in the first shell will be held most tightly by Li and
least tightly by H. d) H and Li have the same core charge; the electron is farther away in
Li. Therefore, Li will hold its valence electron less tightly than H.
2. Be. Two peaks. Both peaks have the same intensity. C. Three peaks. All three peaks
have the same intensity.
4. Mg. Two electrons in the 1s. Two electrons in the 2s. Six electrons in the 2p. Two
electrons in the 3s.
Problems
1. a) False. Both have 10 electrons. The number of peaks and the relative intensities will be
the same, but the IEs of Mg2+ will be greater than the equivalent IEs of Ne because Mg2+
has 2 more protons. b) True. Both have 17 electrons and 17 protons. The number of
neutrons is not relevant.
8
2. 273 MJ/mole. The energy required to remove an electron from the 1s of Cl must be much
higher than the energy required to remove an electron from the 1s of F because Cl has 17
protons in its nucleus and F only has 9 protons in its nucleus.
ChemActivity 10
Exercises
1. Na has 11 protons in its nucleus and Ne only has 10 protons. Therefore, the 1s electrons
will be held more tightly in Na.
2. It would require more than 0.50 MJ/mole because Mg+1 and Na are isoelectronic and Mg
has an additional proton in its nucleus.
3. It would require less than 1.52 MJ/mole because Cl– and Ar are isoelectronic and Cl– has
one fewer protons in its nucleus.
4. Kr
5. C<Ne<Zn<Ba<Gd<Pt
6.
P3–: [Ar] Ba: [Xe] 6s2 Ba2+: [Xe]
7. P: [Ne] 3s2 3p3
S2–: [Ar] Ni: [Ar] 4s2 3d8 Zn: [Ar] 4s2 3d10
8. three
Problems
1. 5d
2. Pd: [Kr] 5s2 4d8. Pd2+ [Kr] 4d8
S: [Ne] 3s2 3p4
ChemActivity 11
Exercises
1. 13C has a small nucleus consisting of 6 protons (positively charged) and 7 neutrons (no
charge). 13C has six electrons in shells around the nucleus. Electrons (negatively charged)
1 and 2 are paired (one spin up, one spin down) in the first shell, the 1s orbital, which is the
shell closest to the nucleus. There are four electrons in the second shell (farther from the
nucleus than the 1s electrons). Electrons 3 and 4 are paired in the 2s orbital. Electron 5 is
found in one of the three 2p orbitals in the 2p subshell. Electron 6 is also found in one of
the three 2p orbitals, but not the same 2p orbital as electron 5. Electrons 5 and 6 are
unpaired (both have spins in the same direction).
2. a)True. Both have eight electrons. b) False. Si has 2 electrons in the 3p subshell (similar to
C) and so it has two unpaired electrons. c) True. Sulfur has four 2p electrons; two are
paired and two are unpaired. Si also has two unpaired electrons (as described in part b
above). d) False. Carbon, for example, has 6 electrons and two are not paired.
3. There are three 2p orbitals in the 2p subshell. Experimental evidence indicates that N has
three unpaired electrons.
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5. One unpaired electron. Predicted magnetic moment, 1.7 magnetons (same as H).
6. The element is F. F is smaller than Ar. F– has no unpaired electrons. F has one unpaired
electron and F+ has two. All other halogens (Cl, Br, I) would be possible but they are all
larger than Ar. H would work except that H+ has no electrons.
Problems
1. Ti (2 unpaired electrons), Na (1 unpaired electron) , Sm (6 unpaired electrons), Sm3+(5
unpaired electrons) Cl (1 unpaired electron).
2. Both have four unpaired electrons.
ChemActivity 12
Exercises
9. 14, 24, 18, 12, 34, 32, 20,
Problems
1. 72
2. 8
3. 14
ChemActivity 13
Exercises
1. The C–C double bond is harder to break. Double bonds are stronger than single bonds.
2. The C–C triple bond is harder to break. Triple bonds are stronger than double bonds.
3. The C–N triple bond is harder to break. Triple bonds are stronger than double bonds.
4. The bond energy (in MJ/mole) is the energy required to break one mole of the specified
bonds. Triple bonds share six electrons between two atoms and are stronger than double
bonds, which share four electrons between two atoms. Single bonds, which share two
electrons, are weaker than double bonds. Bond orders indicate the number of pairs of
electrons in a bond. Single bonds have bond order = 1, etc.
5. When Lewis structures are drawn, the bond between C and O in formaldehyde is a double
bond and the bond between C and O in methanol is a single bond. Thus, it is harder to
break the C–O bond in formaldehyde, which is a double bond, than to break the C–O bond
in methanol, which is a single bond.
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6. Predictions: BE of double bonds~600 kJ/mole; BE of triple bonds~900 kJ/mole.
The single bonds in the table vary from 243-552 kJ/mole. The double bonds vary from
532-782 kJ/mole. The triple bonds vary from 945-962 kJ/mole. For this table, the rule of
thumb is a bit low.
7. a) C–I has the longest bond length because the atomic radii increase in the series F, Cl, Br,
I. b) C–F has the shortest bond length. C–F has the strongest bond because the shorter the
bond, the stronger the bond (all are single bonds).
8. a) C–H. C is smaller than Si so the bond is shorter and stronger. b) The N–N triple bond
is stronger than the O–O double bond. c) O–H. O is smaller than P so the bond is shorter
and stronger. d) O–H. O is smaller than S so the bond is shorter and stronger. e) S–H. S
is smaller than Se so the bond is shorter and stronger. f) N–H. N is smaller than P so the
bond is shorter and stronger. g) The O–O double bond is stronger than the F–F single
bond.
9. All are triple bonds. The bonds increase in the series: As2< P2 < N2 because N2 has the
shortest bond.
10. a) True. H is smaller than F. b) True. Cl is smaller than Br and the C–Cl bond is stronger.
c) True. Draw the Lewis structures. The C-N bond in H 3 CNH 2 is a single bond and the CN bond in HCN is a triple bond.
Problem
1. The Lewis structure for all five molecules (ions) indicates a single bond between the two
atoms. The shortest bond is HF. Therefore, HF has the strongest bond.
ChemActivity 14
Exercises
1. The C–C double bond is shorter.
2. The C–C triple bond is shorter.
3. The C–N double bond is longer.
4. False. Write Lewis structures for both molecules. The C–O bond length is shorter in
H 2 CO (a C–O double bond) than in CH 3 OH (a C–O single bond).
5. a)
b) 1.5
c)
Problems
1. N 2 (BO = 3) < HNNH (BO = 2) < H 2 NNH 2 (BO = 1)
2. N 2 (a triple bond) < O 2 (a double bond) < F 2 (a single bond)
3. 143 pm The C–O bond should be shorter because an oxygen atom is smaller than a
nitrogen atom.
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4. Yes. In the model, the C–C single bond energy is 376 kJ/mole and the C–C double bond
energy is 720 kJ/mole. The average of these two values is 548 kJ/mole—reasonably close
to 509 kJ/mole.
ChemActivity 15
Exercises
1. NH4+: N(+1), H(0). CS2: All formal charges are zero. N2O5: Oxygens with double bond
and central oxygen have no formal charge. Oxygens on the ends have formal charge of –1.
N(+1). HN3: H(0), nitrogen atoms from left to right in Lewis structure: (0), (+1), (–1).
2. There are only six electrons around the C atom.
3. a) 3/2. b) Yes. The bond order in ozone is 1.5; the bond length should be intermediate
between the bond length of a single bond and the bond length of a double bond.
4. True. Write Lewis structures. The C–N bond in H3CNH2 is a single bond; the C–N bond
in HCN is a triple bond.
Problem
1.
The carbon-oxygen bond order is 1.5 and we would expect the bond length to be between
that of a single and a double bond, say 127 pm.
ChemActivity 16
Exercises
1. 24; 26; 32; 32; 8; 30; 106; 105; 36; 32; 32.
The N–O bond order is 3/2.
3. NO2– has the shorter bond length because the bond order is 3/2 in NO2– and only 4/3 in
NO3–.
4.
5.
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6.
Sulfate ion on left: S(+2), each O(–1). Sulfate ion on right: S(0), two O(–1), two O(0).
The Lewis structure on the right has the lower formal charges and is better. Perchlorate ion
on left: Cl(0), three O(0), one O(–1). Perchlorate ion on right: Cl(+3), each O(–1). The
Lewis structure on the left has the lower formal charges and is better.
7.
S—O bond order = 1.5.
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8.
9. nitrate ion: 10 electrons around the N atom. HCN: 6 electrons around the C atom.
ethene: too many electrons in the diagram (14). N2: too many electrons in the diagram (12).
HCCH: too few electrons in the diagram (8). too many electrons in the diagram (26).
10. A single bond from the Xe to each O requires a formal charge of –1 on each O atom. A
better structure—with no formal charges is shown below:
14
Problems
1. a)
The P–O bond order is 4/3.
b)
The Cl–O bond order is 5/3
c)
The Te–Cl bond order is one.
d)
.
The C–O bond order is 4/3.
2. The atoms in the molecule must have 34 valence electrons, 8 around each Br atom plus two
for the lone pair. Br has 7 valence electrons, so X must have 6 valence electrons to yield a
total of 34 valence electrons. Therefore, S or Se, or Te are good candidates because they
have 6 valence electrons and are capable of accommodating more than eight electrons.
Oxygen is not a possible candidate because it can only accommodate eight electrons.
3. Write the Lewis structures. The C–O bond order in methanol is one. The C–O bond order
in formaldehyde is two. The C–O bond order in the carbonate ion is 4/3. Therefore, the
C–O bond length in the carbonate ion should be closer to 143 pm than to 116 pm. We can
calculate a bond length by assuming that bond length vs. bond order is linear between bond
orders one and two.
bond length =
= 134 pm
15
ChemActivity 17
Exercises
16
HF
single bond between H and F; 3 lone pairs on the F; linear; 180°
5. H—C≡C—H The central C atoms have two electron domains. The angle must be 180°.
6.
The central N atoms have three electrons domains and one lone pair. The
angle must be close to 120° but slightly smaller because of the presence of the lone pair.
7. N(109° or slightly smaller); C(120°); O(109° or slightly smaller).
8. O(109° or slightly smaller); both Cs (120°); N(109° or slightly smaller)
Problems
1. a) 109° b) 120°
2. a) 180°, linear b) 180°, linear
c) 120°, trigonal planar d) 107° (slightly smaller
than 109°) bent
ChemActivity 18
Exercises
1. a)
b) four c) tetrahedral d) about 107° (slightly less than 109°) e) bent f) sp3
2. carbonate, see Ex 4 in CA 17, trigonal planar, sp2
CCl4, see Ex 4 in CA 17, tetrahedral, sp3
SO2, see Ex 3 in CA 17, trigonal planar, sp2
CO2, see see Ex 4 in CA 17, linear, sp
PH3, see Ex 4 in CA 17, tetrahedral, sp3
H2CO, see Ex 4 in CA 17, trigonal planar, sp2
3. glycine: N(109°); C(120°); O(109°). PABA: O(109°); both Cs (120°); N(109°)
17
Problems
1. Write the Lewis structure.
H–O–O bond angle,109°; hybridization, sp3
2. a) All six carbon atoms have sp3 hybridization. b) All six C-C bonds have BO = 1.5. All
six C–H bonds have BO = 1. c) All HCC bonds have an angle of 109°. All CCC bonds
have an angle of 109°; all twelve atoms lie in a single plane.
ChemActivity 19
Exercises
1. B, 1.17 MJ/mole. F, 2.31 MJ/mole. These results agree with the values in Table 2.
2. a) Mg<P<Cl. b) Se<S<O<F. c) K<P<O.
3. a) F
b) Rb
4. a) S
b) As
c) P
Problems
1. ∆(F-Cl) = (2.30–1.59) = 0.71 and ∆ (O–S) = 0.54. So, ∆ (N–P) ~ 0.37. Given that the
AVEE of N is 1.82, we predict an AVEE for P of 1.45. The actual value is fairly close,
1.42.
2. Electronegativity is a measure of how strongly an atom holds on to its valence electrons.
The electronegativity of the elements in any group decrease, from the top to bottom of a
group, because the atoms get larger and the valence electrons are farther from the nucleus.
ChemActivity 20
Exercises
1. Using equation 2, p Br = 0.539. The partial charge on Br = δBr = 7 – 6 – (0.539)(2) = –
0.078.
2. δBr = 0. Yes, each Br has the same electron pulling power so neither Br gains negative
charge.
3. Zero.
4. δB = –0.13.
5. a)
6.
7.
8.
9.
b) The partial charge on the oxygen atoms is negative because the oxygen atoms are more
electronegative than the carbon atom.
c) The average charge is the same on all three of the oxygen atoms. There are three Lewis
structures and three oxygen atoms. From one point of view one could consider a total of –2
charge spread over three oxygen atoms or a –2/3 on each atom.
+0.213
–1.38
–0.04
Estimates: a) A(0), A(0) b) C(0), C(0) c) A(-0.4), B(+0.4) d) A(–0.1), C(–0.1)
Calculation: a) A(0), A(0)
b) C(0), C(0)
Assume that A, B, and C have seven
valence electrons and that there is a single bond in all molecules. c) A(-0.50), B(+0.50)
d) A(–0.10), C(0.10)
18
Problem
1. F (EN = 4.19) is more electronegative than O (EN = 3.61). Therefore, the partial charge on
the O should be positive. The ∆EN for H2O is approximately 1.3 and the ∆EN for F2O is
approximately 0.6—not as much. My prediction for the partial charge on O is +0.32.
ChemActivity 21
Exercises
1. The bonds in all of the molecules are polar. However, C and H have very similar
electronegativities and often C-H bonds are considered nonpolar.
2. The H2O molecule has more polar bonds than NH3 because O is more electronegative than
N and therefore the ∆EN in H2O is greater than the ∆EN in NH3.
3.
4. See Model 1 of CA 16. The dipole moment is zero. The center of positive charge is at the
nucleus of the N atom. All three oxygen atoms are negatively charged (equally). All
∠ONO are equal to 120° and the center of negative charge must be at the nucleus of the N
atom.
5. O2, nonpolar; I2, nonpolar; CO, polar; H2O, polar; CO32–, nonpolar;. chlorobenzene,
polar.
6. a) N2 b) CH4 c) SO3 d) NO3– e) SO42- f) CO2
7. False. CCl4 has a dipole moment of zero.
b) CH3Cl
c) H2O
8. a) CH3Cl
9. 1.95 (must be somewhat greater than 1.89).
10.
19
20
Problems
b) SeF 2
1. a) CF 4
2. For I, dipole moment is 2 x (2d) or 4d. For II, diploe moment = 1 x (3d) or 3d. The dipole
moment of I is larger.
3. a) CH3Cl. Cl is more electronegative than Br and the other have no dipole moment.
b) SO2 . The others have no dipole moment.
4. Both species have a nonzero dipole moment.
ChemActivity 22
Exercises
1. O2, nonpolar. NaF, ionic. I2, nonpolar. KCl, ionic. CO, polar. NO, polar. CuO, ionic.
CN–, polar. ICl, polar.
2. Al3+. Al is a metal and tends to lose electrons. Al has three valence electrons.
3. S2–. Cl–. Cs+. Br–. O2–. Be2+. N3–.
4. a) NaBr. b) Li2O. c) AlN. d) MgBr2. e) CaO.
5. A possible metal is Pb. The metal must sustain a +4 charge. Pb is a metal and has four
valence electrons.
6. a) NaCl. b) NaCl. c) MgO. d) CaO. e) MgS. f) NaCl. g) LiF.
7. a) NaCl. b) NaCl. c) MgO. d) CaO. e) MgS. f) NaCl. g) LiF.
Problems
1. KCl (+1,–1)<K2S (+1,–2)<CaS (+2,–2)<CaO (+2, –2 and O2– smaller than S2–)
2. a) LiF b) CaO
c) CaSO 4
d) Al 2 O 3
e) CaSO 4
3. In covalent bonding the valence electrons of two atoms are shared, more or less, equally
between the bonding atoms. In ionic bonding the two charged atoms, one a positive ion
and the other a negative ion, are held together by Coulombic forces.
ChemActivity 23
Exercises
1. Co(s); Pb(s)
2. MgF 2 is an ionic compound comprised of Mg2+ ions and F– ions. The bonding is a result of
the Coulombic attraction between the positive and negative ions. Zn is a metal and its
valence electrons are free to move throughout the entire piece of metal. The bonding occurs
from the attraction of the positive Zn2+ iond and the sea of free electrons.
3. Ca is a metal; Br is a nonmetal; S is a nonmetal; Si is a metalloid; Co is a metal; K is a
metal; Cu is a metal.
ChemActivity 24
Exercises
1.
Compound
EN
first
atom
2.54
CO2
3.07
NH3
second
atom
3.61
2.30
3.08
2.69
∆EN
Type of Bonding
1.07
0.77
covalent
covalent
21
BaO
0.88
3.61
2.24
2.73 ionic
2.59
3.61
3.10
1.02 covalent
SO2
AlSb
1.61
1.98
1.80
0.37 semimetal
GaAs
1.76
2.21
1.98
0.45 semimetal
CdLi
1.52
0.91
1.22
0.61 metallic
0.88
2.69
1.78
1.81 ionic
BaBr2
ZnO
1.59
3.61
2.60
2.02 ionic/covalent
NaH
0.87
2.30
1.58
1.43 ionic
2. a) CuNi b) NaCl c) CO2 d) CuZn e) NaK f) GaAs (these are examples)
3. a) metallic b) covalent
Problems
1. Ag is likely to be somewhat less electronegative than Sn; Cl is more electronegative than I.
Therefore, ∆EN for AgCl will be larger than ∆EN for SnI4. We would not expect,however,
that ∆EN would be as large as indicated by point A (point A might represent some
compound such as SrI 2 ) It is harder to predict which
value will be higher (AgCl or
SnI4). Therefore, "B" is the most likely point for AgCl.
2. CO 3 2– covalent ; BaCO 3 ionic, but covalent within the carbonate ions; CaSO 4 ionic, but
covalent within the sulfate ions ; NaClO 4 ionic, but covalent within the perchlorate ions.
ChemActivity 25
Exercises
1. a) heptane. A plot of bp versus MW for the alkanes in Table 1 is shown below. The data fit
the equation: bp = - 188.03 + 3.6709*MW - 7.9015 × 10–3 *(MW)2 . The MW of heptane
is 100.2 g/mole. The predicted bp is 100°C.
One can do the same sorts of plots for alcohols ketones to obtain b) ethanol, 77°C and
c) 2-octanone, 165°C. However, it is possible to eyeball the data and make very less
reliable predictions.
22
2. The cis compound should have the higher boiling point because it has a dipole moment
and, therefore, a dipole-dipole interaction that is not present in the trans compound.
3. a) He (dispersion only, low MW) < CH4 (dispersion only, slightly higher MW) < CH3F
(dispersion and dipole-dipole) < NH3 (dispersion, dipole-dipole, hydrogen bonding)
b) Ne (dispersion only, low MW) < CH3CN (dispersion and dipole-dipole) < CH3Br
(dispersion with higher MW than CH3CN and dipole-dipole) < CH3OH (dispersion,
dipole-dipole, hydrogen bonding) c) CH4 < SiH4< GeH4< SnH4 (all have dispersion
only; trend goes according to MW)
4. The O–H bond is a very strong covalent (single) bond because both atoms are small. The
hydrogen bonding that exists between two water molecules is quite strong in comparison to
other intermolecular forces, but it is only about 7–10% of a normal covalent (single) bond.
5. CH3F and CH3OH have about the same molecular weight, but CH3OH has hydrogen
bonding. CH3OH should, and does, have the higher boiling point.
Problems
1. a) CH 3 CH 2 CH 2 NH 2 All of these compounds have about the same MW. This compound
has hydrogen bonding and should have the highest boiling point. b) NaCl Ionic compounds
have much higher melting points and therefore much higher boiling points also.
c) CaO CaO and LiF are ionic. CaO has the higher charges (±2).
2.
3. This compound has an abnormally high boiling point for its MW. It must be that hydrogen
bonding is involved. A possible compound is HOCH 2 CH 2 OH, which has two site of
hydrogen bonding.
4. Hydrogen bonding is very weak compared to a normal covalent single bond. All of the rest
are single covalent bonds. Of these, the H–F bond length is the shortest and strongest.
ChemActivity 26
Exercises
1. 63.5 g
2. 42.0 g
3. 0.0167 moles of C.
1.00 × 1022 C atoms.
23
4. a) 6.02 × 10 H2 molecules. b) 12.0 × 1022 H atoms
c) 2.02 g
5. 46.07 g/mole. 7.650 × 10–23 g/molecule.
6. 25.90 g.
7. a) 2.171 moles. b) C, 4.341 moles; H, 13.02 moles; O, 2.171 moles. c) C, 52.14 g; H,
13.12 g; O, 34.74 g.
8. 5.77 moles of CO2 molecules.
9. 11.5 moles of O atoms.
10. 8.02 × 1021 carbon atoms
11. a) False. One mole of NH3 has a mass of 17 g and one mole of H2O has a mass of 18 g.
b) True, 48 g of CO2 is 1.1 mole of CO2, which contains 1.1 mole of C. 12 grams of
diamond is 1.0 mole of C, which contains fewer atoms than 1.1 mole of C.
23
c) False, one mole of N2 has two moles of N; one mole of NH3 has only one mole of N.
d) False. All of the atoms in 100 grams of Cu are Cu atoms; in 100 grams of CuO, some of
the atoms are O atoms and thus there can’t be as many Cu atoms to make 100 grams.
e) True, there are 100 moles of Ni atoms in both 100 moles of Ni and 100 moles of NiCl2.
f) False, there are six moles of hydrogen atoms in two moles of NH3, whereas there are
eight moles of hydrogen in two moles of CH4.
ChemActivity 27
Exercises
1. a) Fe 2 O3(s) + 2 Al(s) → 2 Fe(l) + Al 2 O3 (s)
b) 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
c) Fe2S3(s) + 6 HCl(g) → 2 FeCl3(s) + 3 H2S(g)
d) CS2(l) + 2 NH3(g) → H2S(g) + NH4SCN(s)
2. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
3.
2 C(s) + O 2 (g) +
N 2 (g) +
H 2 (g) → H 2 NCH 2 COOH(s)
4. 2 O 3 (g) → 3 O 2 (g)
5. a) is not atom balanced.
c) is not charged balanced.
6. 15.6 g FeCl3
7. 0.0163 g H2S
8. 82 g FeCl3
9. 7.25 g Al and 21.4 g Fe2O3 used. 13.7 grams of Al2O3 produced
10. N2 + 3 H2 → 2 NH3 176 g NH3
11. True. 2 CO(g) + O2(g) → 2 CO2(g). Thus, every time that 2 CO2 molecules are produced,
3 molecules have reacted, decreasing the total number of gas molecules present.
Problems
1. 35.9 mass % Ni
2. “Q” is oxygen. 2Fe(s) +
O 2 (g) → Fe 2 O 3 (s)
ChemActivity 28
Exercises
1. 46.0 g NO2
2. a) HCCH(g) + 2 H2(g) → C2H6(g) b) 31.2 g of C2H6 can be produced
3. a) Zn(s) + I2(s) → ZnI2(s) b) I2 is totally consumed Limiting reagent is I2
ZnI2
37.2 g Zn left
4. a) TiCl4 + 2 Mg = Ti + 2 MgCl2 b) 6.43 × 103 g of Ti
5. 9.16 g of P4O10(g) can be produced.
6. 18.0 g of N2 (28.01 g/mole) can be obtained.
ChemActivity 29
Exercises
62.9 g
24
1. 14.37% H
2. C, 40.00%. H, 6.71%. O, 53.29%.
3. a) 36.8 g N 63.2 g O
b) 2.63 moles N
d) N2O3
3.95 moles O
c) 1.50 moles O/mole N
4. C 2 H 3 Cl 3
Problems
1. False. For example, ethyne, HCCH, and benzene, C6H6, have the same empirical formula
but are not isomers of each other.
2. COH3
3. Zn 3 P 2 O 8 [or Zn 3 (PO 4 )2]
ChemActivity 30
Exercises
1. [Al3+] = 0.125 M [Cl–] = 0.375 M
2. [Cr3+] = 0.0850 M [SO42–]= 0.128 M
3. 8.64 g Na+. 3.95 × 10–3 g Pb2+. 151 g Na+.
4. False. There are three times as many solute particles in Na2SO4 because each molecular
unit dissociates into two sodium ions and one sulfate ion.
5. [Fe3+] = 9.066 × 10–4 M
6. 3.12 × 10–3 moles of AgCl
Problems
1. Cl2 is limiting. 2.2 g of M. 0.43 M/l x 0.1 L = 0.043 moles of M. 2.2 g of M/0.043 moles
of M = 51 g/mole. M = vanadium.
2. i) [Cl–] = 3.42 M ii) [Cl–] = 2.70 M iii) [Cl–] = 3.03 M so i) has highest concentration.
3. Weigh 10.65 g Na2SO4 into the 500-mL volumetric flask. Dilute to the neck. Swirl to
mix. Dilute to the 500 mL mark. Mix.
4. 8.67 × 10–2 M Cl–
ChemActivity 31
Exercises
1. 44.8 L
2. 5.20 × 102 L
3. 1.19 × 104 L
4. 6.0 × 1011 g
5. 252 K
6. d = (MW)P/RT
7. 0.621 g/L
8. a) PO2 = 48.0 atm b) PHe = 32.5 atm c) P = 80.5 atm
9. a) 29.02 g/mole b) 25.5% O2 and 74.5% N2
Problems
1. a) PHe = 50 torr PNe = 1.0 × 102 torr PAr = 1.2 × 102 torr b) P = 2.7 × 102 torr
2. 4.22 L
25
3. a) C 2 H 3
b) 54.2 g/mole c) C 4 H 6
d)
ChemActivity 32
Exercises
1. a) endothermic b) exothermic c) exothermic
2. a) positive b) negative c) negative
3. The value for I2(s) is obviously incorrect. The standard enthalpy of atom combination for
any molecule must be negative.
4. Graphite, it has a more negative ∆H° ac than diamond. Thus, more energy is required to
break all of the bonds in graphite than in diamond.
5. Si is smaller than Sn so we would expect that Si–Cl bonds would be stronger than Sn–Cl
bonds. This is verified by the fact that the enthalpy of atom combination of SiCl 4 (g) is –
1,599.3 kJ/mole whereas the enthalpy of atom conbination of SnCl 4 (g) is –1,260.3
kJ/mole.
6. a) O-H bond energy in H2O is 463.15 kJ/mole; S–H bond energy in H2S is 367.37 kJ/mole.
b) O–H has the stronger bond
c) O is smaller than S, so the O-H bond is shorter than
the S-H bond, making the O-H bond stronger.
ChemActivity 33
Exercises
1. a) –37.01 kJ b) –696.56 kJ
c) –67.18 kJ
2. a) Because Cl is the smallest atom, Cl2 has the shortest and strongest bond (all are single
bonds), it should have the most negative ∆H° ac . b) Because N is the smallest atom, N2
has the shortest and strongest bond (all are triple bonds, Lewis structures) and should have
the most negative ∆H° ac .
3. a) C–H (smaller bond distance than C–Cl; stronger bond)). b) C–H bond energy = 1662/4
= 416 kJ/mole. c) C–Cl bond energy = 1306/4 = 326 kJ/mole. d) Yes.
4. a) O is smaller than C. Therefore, the O–H bond should be shorter and stronger than the
C–H bond given that both are single bonds.
b) The average O–H bond energy is 464 kJ/mole and the C–H bond energy from Exercise
3b (above) is 416 kJ/mole. So, the prediction is consistent with the data.
Problem
1. The ∆H° for all four transformations are positive; 38.00, 44.01, 13.46, and 58.78 kJ/mole.
Intermolecular forces are strongest in the solid phase; it requires energy to convert the solid
to the liquid. It also requires energy to convert a substance from the liquid phase to the gas
phase. Therefore, the intermolecular forces are weakest in the gas phase (which is
demonstrated by the two SO 3 transformations).
26
ChemActivity 34
Exercises
1. a) False. the relationship of the rate of reaction to the rate of production of a product
depends on the stoichiometric coefficients for the products. If the stoichiometric coefficents
are not 1, then the rates are not equal. b) False. The rate of consumption of PCl5 is equal
to the rate of production of Cl2.
2. rate of ammonia production = 2.33 × 10–4 M/s rate of reaction = 1.17 × 10–4 M/s
3. rate of ozone consumption = 9.00 × 10–5 M/s rate of reaction = 4.50 × 10–5 M/s
4. rate of I– consumption = 7.5 × 10–2 M/s rate of IO2– consumption = 2.5 × 10–2 M/s
rate of H+ consumption = 0.10 M/s
rate of I2 production = 5.0 × 10–2 M/s
rate of reaction = 2.5 × 10–2 M/s
ChemActivity 35
Exercise
1. At equilibrium, A molecules are being converted to B molecules and B molecules are being
converted to A molecules, but the number of A molecules and the number of B molecules
at any time does not change.
2. a) 172 B molecules
b) 172 B molecules
c) 344 A molecules
ChemActivity 36
Exercises
1. t = 1, 0.33
t = 4, 1.42 t = 15, 2.48 t = 20, 2.50 t = 40, 2.50 . The values of (A)
and (B) are not as accurate when read from the graph.
2. (B)eq = 0.0286 M (A)eq = 0.0714 M
Problem
1. a) kinetic region, < 400 seconds; equilibrium region, > 400 seconds.
b) (A)o = 1.3 M, (B)o = 6.0 M, (C)o = 0 M c) [A] = 0.3 M, [B] = 3.0 M, [C] = 2.0 M.
d) A + 3 B = 2 C
ChemActivity 37
Exercises
1. If kY > kZ, there is a tendency for Y molecules to be converted to Z molecules faster than Z
molecules are converted to Y molecules. The forward rate and the reverse rate must be
equal at equilibrium. This can be accomplished if there is more Z than Y at equilibrium.
2. At equilibrium, Y molecules are being converted to Z molecules and Z molecules are being
converted to Y molecules; the ratio [Z]/[Y], however, is constant. .
3. Set M: K = 0.49
Set N: K = 3.0 Set O: K = 1.0 Set P: K = 5.7 Set Q: K = 1.3
Set R: K = 0.72 Set S: K = 0.59. No.
[H2] [I2]
[NH3]2
[NH3]
4. a) K =
b)
K
=
c)
K
=
[HI]2
[H2]3/2 [N2]1/2
[H2]3 [N2]
[CO]2
d) K = [NH3] [H2S] e) K = [CO ]
f) K = [HCl] [NH3]
2
5. K c for reaction c) = [K c for reaction b)]1/2
27
6.
7.
8.
9.
10.
11.
12.
13.
14.
Kc = 23
Kc = 6.1
Kc = 6.8 × 10–3
[NO 2 ] = 2.7 × 10–3
a) 1.8 moles A; 3.4 moles B; 0.2 moles C. b) [A] = 1.8 M; [B] = 3.4 M; [C] = 0.2 M [D]
= 0.4 M. c) Kc = 4.5 × 10–4
Kc = 6.04 × 10–2
0.0100
a) Kc for the second reaction = (0.78)1/2 = 0.88 b) 1.1
9.0 × 108
ChemActivity 38
Exercises
2.
initial moles
change in moles
equilibrium moles
equilibrium conc
equilibrium conc
value (no "x")
3. a) x moles of O2 react.
4.
CO2
1.00
–x
1.00–x
(1.00–x)/5
0.11
initial moles
change in moles
equilibrium moles
H2
2.00
–x
2.00–x
(2.00–x)/5
0.31
CO
0
x
x
x/5
0.086
H2O
0
x
x
x/5
0.086
b) 2x moles of NO are formed.
NO
0
2x
2x
N2
5.00
–x
5.00 – x
O2
10.00
–x
10.00–x
N2
1.00
–x
1.00–x
(1.00–x)/10
H2
N2H4
1.50
0
–2x
x
1.50–2x
x
(1.50–2x)/10 x/10
CO2
1.00
–x
1.00–x
1.00–x
0.84
H2
2.00
–x
2.00–x
2.00–x
1.84
5.
initial moles
change in moles
equilibrium moles
equilibrium conc
expression
a) [N2H4] = 1.1 × 10–5 M b) [N2] = 0.10 M [H2] = 0.15 M
6.
initial moles
change in moles
equilibrium moles
equilibrium conc
equilibrium conc
value
CO
1.00
x
1.00+x
1.00+x
1.16
H2O
2.00
x
2.00+x
2.00+x
2.16
28
7.
NH3
2.65
–0.84
1.81
0.90
N2
0
0.42
0.42
0.21
H2
0
1.26
1.26
0.63
initial moles
change in moles
equilibrium moles
equilibrium conc
value
Kc = 6.5 × 10–2
8. a) At equilibrium. b) Not at equilibrium; reaction must go to left. c) Not at equilibrium;
reaction must go to right.
Problems
1. a) False. The rate of production of O2 is one-half of the rate of consumption of SO3.
b) False. This statement is true if the initial concentrations of SO2 and O2 are zero but there
is no general relationship between [SO2] and [O2].
c) False. The reaction must produce more reactants to reach equilibrium; therefore, the rate
of the reverse reaction must be greater than rate of the forward reaction.
2. a) K = 1.1 × 103 b) After the addition of the water the initial concentrations are:
(FeSCN2+) = 0.163 M; (Fe3+) = 0.00963; (SCN–) = 0.0116. Q = 1.46 × 103. Because Q >
K, the reaction must go to the left and the number of moles of FeSCN2+ decreases.
3. Q = 0.81, the reaction must go to the left. The following equation must be solved for x:
1.8–x  1.8–3x 3 1.8+x 1.8+x
0.26 =  2   2  ÷  2   2  where x is the number of moles of CH4






1.8–3x
produced. The concentration of H2 is equal to
.
2
4. a) Q = 18, proceed to the left
b) 16 = [(4.0 – 2x)/(2.0)]2 / [(1.0 + 2x)/(2.0)]2 [(1.8 + 2x)/(2.0)] where x is the number of
moles of O2 produced to reach equilibrium
c) Because Q is relatively close to K, the value of x is likely to be pretty small, probably
about 0.05 moles.
ChemActivity 39
Exercises
1.
a) K =
b) K =
c)
d)
e)
f)
g)
K sp = [Ba2+][SO 4 2–]
K = [NH 3 (g)][H 2 S(g)]
K = [CO 2 (g)]
K = [NH 3 (g)][HCl(g)]
K sp = [Ag+]2[SO 4 2–]
29
h) K =
=
2. a) False. K = [CO 2 (g)]. The concentration of CO 2 is a constant (at a given temperature)
as long as some CaCO 3 (s) is present.
b) False. At equilibrium the forward rate is equal to the reverse rate.
3. 2.0 × 10–3 g
3. 2.04 × 10–3 g
4. 9.2 × 10–3 g
5. K = [Ag+] [Cl–] = 1.6 × 10–9
6. a) x moles of Pb2+. 2x moles of Cl–. b) [Pb2+] = 1.6 × 10–2 M [Cl–] = 3.2 × 10–2 M
7. a) Yes, Q = 6.2 × 10–7 b) No, Q = 6.2 × 10–9
8. No, Q = 7.8 × 10–9
9.
= 40.5 mole/L
Problems
1. No, Q = 5.3 × 10–6
2. a) AuCl3(s) = Au3+(aq) + 3 Cl–(aq) b) Ksp = [Au3+] [Cl–]3 c) 2.1 x 10–4 g Au
3. a) The NaCl solution has the higher concentration because the original solutions had the
same volume and there are 13 Cl– in the diagram and only seven Pb2+.
b)
If the solution were prepared from PbCl 2 there would be twice as many Cl– as Pb2+.
c)
Ksp = [Pb2+] [Cl–]2 = (2.5 x 10–2) (2.5 x 10–2) 2 = 1.6 x 10–5
ChemActivity 40
Exercises
1. SO42– ; CO32– ; OH– ; O2– ; H2O ; NH3 ; CH3NH2 ; F– ; CH3COO– .
2. HSO4– ; HCO3– ; H3O+; H2O; OH– ; NH4+; CH3NH3+ ; HCN ; CH3COOH ; HF ;
H2CO3; NH3 .
3. a) H2SO4 , HSO4– , H2O , HCN , H2S
b) H2O , H2O , H2O , CO32– , NH3
c) H2SO4 and HSO4– ; H3O+ and H2O. HSO4– and SO42– ; H3O+ and H2O.
H3O+ and H2O; H2O. and OH– . HCN and CN– ; HCO3– and CO32– .
H2S and HS– ; NH4+ and NH3.
4. a) NH4+
5.
b) NH2–
Acid
H2S
HS–
HNO2
H3PO4
HOCN
H3O+
c) NH3(l) + NH3(l) = NH4+(am) + NH2–(am)
Base
HS–
S2–
NO2–
H2PO4–
OCN–
H2O
30
OH–
HF
H2PO4–
HOCl
O2–
F–
HPO42–
OCl–
ChemActivity 41
Exercises
1. I–, iodide; Br–, bromide; Cl–, chloride; ClO4–, perchlorate; HSO4–, hydrogen sulfate; NO3,
nitrate; CH3COO–, acetate; HCO3–, hydrogen carbonate; F–, fluoride; HS–, hydrogen
sulfide; NO2–, nitrite; H2PO4–, dihydrogen phosphate.
2. phosphoric acid > hydrofluoric acid > nitrous acid > acetic acid > carbonic acid >
hydrosulfuric acid
3. HF + H2O = H3O+ + F– Ka = [H3O+] [F–]/[HF]
HCl + H2O = H3O+ + Cl– Ka = [H3O+] [Cl–]/[HCl]
H2S + H2O = H3O+ + HS– Ka = [H3O+] [HS–]/[H2S]
H2CO3 + H2O = H3O+ + HCO3– Ka = [H3O+] [HCO3–]/[H2CO3]
4. False, nitrous acid is a stronger acid (greater Ka) than acetic acid.
5. 1.6 × 10–12 M
6. 3.0 × 10–3 M
7. 2.9 × 10–10 M; 1.4 × 10–14 M; 2.2 × 10–5 M; 4.8 × 10–8 M
8. 2.9 × 10–10 M; 1.4 × 10–14 M; 1.8 × 10–3 M; 9.1 × 10–8 M
9. All are acidic except [H3O+] = 4.5 × 10–10 M, which is basic.
10. All are basic except [OH–] = 7.1 × 10–10 M and 5.7 × 10–12 M which are acidic.
11. a) True. If [OH–] < 10–7 then [H3O+] > 10–7. b) False. A solution is considered acidic
when [H3O+] > [OH–]. For water at 25 °C, this occurs when [H3O+] > 10–7.
ChemActivity 42
Exercises
1. 1.1 × 10–6
2. 1.0 × 10–9
3. a) [HONO] = 0.80 M
b) HONO + H2O = H3O+ + NO2–
c) Ka = [H3O+] [NO2–]/[HNO2]
d) [H3O+] = 0.020 M
4. a) [CH3NH2] = 1.5 M b) CH3NH2 + H2O = CH3NH3+ + OH–
d) [OH–] = 2.7 × 10–2 M
c) Kb = [CH3NH3+] [OH–]/[CH3NH2]
Problem
1. a) 2.4 × 10–12
b)
c)
plus another resonance structure
d) Ka = 1.8 × 10–4
31
ChemActivity 43
Exercises
1. a) pH = 6.49, pOH = 7.51 b) pH = 3.70, pOH = 10.30 c) pH = 4.49, pOH = 9.51
d) pH = 7.40, pOH = 6.60
2. a) [H3O+] = 1 × 10–2 [OH–] = 8 × 10–13 acidic b) [H3O+] = 6 × 10–5 [OH–]
= 2 × 10–10 acidic c) [H3O+] = 1 × 10–7 [OH–] = 1 × 10–7 neutral
d) [H3O+] = 5 × 10–10 [OH–] = 2 × 10–5 basic
3. x molar HCl (strong acid) < x molar acetic acid (weak acid) < pure water
< x molar
NaOH (strong base)
4. pure water < x molar NH3 (weak base; smaller Kb than C5H5N) < x molar C5H5N < x
molar NaOH (strong base)
5. 2.4 × 10–9
6. 1.8 × 10–4
7. 1.5 × 10–6
8. 0.48 moles of ammonia
9. pH = 0.90 pOH = 13.10
10. pH = 0.90 pOH = 13.10
11. [H3O+] = 0.800 M [OH–] = 1.25 × 10–14 M
12. [OH–] = 0.800 M [H3O+] = 1.25 × 10–14 M
13. 2.59
14. 8.94
15. a) 13.65 b) 0.35 c) 2.55 d) 12.17
Problem
1. [Mg2+] = 0.025 M; [OH–] = 2.7 × 10–5; Ksp = 1.8 x 10–11
ChemActivity 44
Exercises
1. a) H2Se. H–Q bond strengths are different; Se–H has the weaker bond. b) HONO. The
H–Q bond strengths are the same; N is more electronegative than P; there will be a greater
positive charge on the acidic hydrogen atom in HONO. c) Cl3NH+. The H–Q bond
strengths are the same; Cl is more electronegative than H; there will be a greater positive
charge on the acidic hydrogen atom in Cl3NH+. d) (HO) 2 SO 2 . The H–Q bond strengths
are the same; S is more electronegative than Se; there will be a greater positive charge on
the acidic hydrogen atom in (HO) 2 SO 2 . e) H2Te. H–Q bond strengths are different; Te–
H has the weaker bond. f) HONO2. The H–Q bond strengths are the same; two oxygen
atoms attached to the N is more electronegative than one oxygen atom attached to the N;
there will be a greater positive charge on the acidic hydrogen atom in HONO2.
2. HBr (strong acid) < CF3COOH (weak acid, but stronger than CH3COOH because of the
more electronegative fluorine atoms) < CH3COOH (weak acid) < KBr (neutral) < NH3
(weak base).
32
3. HF is a weak acid. H2O is neutral. The H–F bond should be stronger than the H–O bond
(this tends to make H2O the stronger acid). The H atom in H–F should be more positively
charged because F is more electronegative than O (this tends to make HF the stronger
acid). Partial charge seems to be the more important factor.
Problems
1.
(a)
(b) pH = 2.17
2. HSIO3 will have the larger value for Ka because the S–H bond is weaker than the O–H bond.
ChemActivity 45
Exercises
1. CH3COO–, Kb = 5.6 × 10–10 ; HCO3–, Kb = 2.2 × 10–8 ; HS–, Kb = 1.0 × 10–7 ;
NO2–, Kb = 2.0 × 10–11 ; NH3, Kb = 1.8 × 10–5 .
2. NH4+, Ka = 5.6 × 10–10 ; CH3COOH, Ka = 1.8 × 10–5 ; C6H5NH3+, Ka =2.5 × 10–5 ;
HNO2, Ka = 5.0 × 10–4 ; H2NNH3+, Ka = 8.3 × 10–9
3. pH = 11.19
4. a) neutral b) neutral c) neutral d) basic e) acidic f) acidic g) basic
h) neutral i) basic j) basic k) neutral
5a. i) NH4NO3 is an ionic compound. NH4+ and NO3– ions exist in solution.
ii) The NH4+ is a weak acid (the conjugate acid of the weak base NH3).
iii) The NO3– ion is not a weak base (the conjugate base of the strong acid HNO3). The
solution will be acidic.
iv) The predominant reaction will be NH4+(aq) + H2O = NH3(aq) + H3O+(aq) .
This is the chemical reaction that makes the solution acidic.
Kw
1.0 ∞ 10–14
=
= 5.6 × 10–10
v) The equilibrium constant is Ka = K
b
1.8 ∞ 10–5
5b.
5c.
i) CsI is an ionic compound. Cs+ and I– ions exist in solution.
ii) The Cs+ ion is not an acid (Group I cation).
iii) The I– is not a base (the conjugate base of the strong acid HI). The solution will be
neutral.
i) CH3COONa is an ionic compound. Na+ and CH3COO– ions exist in solution.
33
ii) The Na+ is not an acid (Group I cation).
iii) The CH3COO– ion is a weak base (the conjugate base of the weak acid CH3COOH).
The solution will be basic.
iv) The predominant reaction will be CH3COO–(aq) + H2O = CH3COOH(aq) + OH(aq).
This is the chemical reaction that makes the solution basic.
Kw
1.0 ∞ 10–14
= 5.6 × 10–10
v) The equilibrium constant is Kb = K =
a
1.8 ∞ 10–5
5d. i) KClO4 is an ionic compound. K+ and ClO4– ions exist in solution.
ii) The K+ ion is not an acid (Group I cation).
iii) The ClO4– is not a base (the conjugate base of the strong acid HClO4). The solution
will be neutral.
5e. i) (CH3COO)2Mg is an ionic compound. Mg2+ and CH3COO– ions exist in solution.
ii) The Mg2+ is not an acid (Group II cation).
iii) The CH3COO– ion is a weak base (the conjugate base of the weak acid CH3COOH).
The solution will be basic.
iv) The predominant reaction will be CH3COO–(aq) + H2O = CH3COOH(aq) + OH–(aq)
.
This is the chemical reaction that makes the solution basic.
Kw
v) The equilibrium constant is Kb = K =
= 5.6 × 10–10
a
6.
a) pH = 4.63 b) pH = 7.00 c) pH = 9.37 d) pH = 7.00 e) pH = 4.48
Problem
1. (a) acidic (b) CH3(CH2)8NH3+ + H2O = CH3(CH2)8NH2 + H3O+
ChemActivity 46
Exercises
1. a) Br2 = Ox agent Hg = Red agent b) Co3+ = Ox agent Br– = Red agent
c) Cl2 = Ox agent Br– = Red agent d) H+ = Ox agent Zn = Red agent
e) S2O82– = Ox agent Zn = Red agent f) Au3+ = Ox agent Fe = Red agent
2. a) 2 b) 2 c) 2 d) 2 e) 2 f) 3
3. True. See Table 1. Cu2+ does oxidize Zn, but K+ does not.
ChemActivity 47
Exercises
1. Br(0); Na(+1), Cl(–1); Cu(+2), Cl(–1); C(–4), H(+1); Si(+4), Cl(–1); C(+4), Cl(–1); S(+2),
Cl(–1); Br(+1), O(–2), Br(+1) .
2. Ni(2+); N(+5), O(–2); C(+4), O(–2); S(+6), O(–2); N(–3), H(+1); Cl(+7),O(–2); Mn(+7),
O(–2); C(+2), N(–3); I(+5), F(–1); P(+3), O(–2).
3. Ni(+2), Cl(–1); H(+1), N(+5), O(–2); Na(+1), C(+4), O(–2); Al(+3), S(+6), O(–2); N(–3),
H(+1), Cl(–1); K(+1), Mn(+7), O(–2); K(+1), C(+2), N(–3); H(+1), Cl(+7), O(–2).
4. H(+1), C(+4), O(–2); H(+1), S(+6), O(–2); H(+1), P(+5), O(–2); N(–3), H(+1); Cr(+6),
O(–2).
34
5. C(–2), H(+1), O(–2); C(–2), H(+1), O(–2); C(–2), H(+1); C(–2), H(+1), Cl(–1); C(+4),
Cl(–1).
6. N(–3). Cu(+2).
7. O(–2), H(+1). Al(+3).
8. P(+14/4), O(–2); P(+3), O(–2); P(+4), O(–2); P(+18/4), O(–2).
9. a) Yes. b) No. c) Yes. d) Yes. e) Yes. f) Yes. g) No.
10. Yes.
11. Yes.
12. Yes.
Problem
1. HOBrO2 (+5); HOBr (+1); HOBrO (+3); the higher the oxidation number on the Br atom,
the stronger the acid.
ChemActivity 48
Exercises
1. Cu2+
2. 1.10 V
Cu2+(1 M) + 2e– = Cu(s)
H2(g; 1 atm) = 2 H+(1 M) + 2e–
Cu2+(1 M) + H2(g; 1 atm) = Cu(s) + 2 H+(1 M)
4.
35
b) Al is the anode. Pt is the cathode. c) Al is negative. Pt is positive.
d)
Al(s) = Al3+(aq) + 3e–
2 H+(1 M) + 2e– = H2(g; 1 atm)
2 Al(s) + 6 H+(1 M) = 2 Al3+(aq) + 3 H2(g; 1 atm)
e) The standard reduction potential for Al3+(aq) + 3e– = Al(s) is –1.66 V. According to
the above, the standard reduction potential of the SHE is 1.66 V greater than the standard
reduction potential for the Al3+/Al half-cell. Because the standard reduction of the SHE is
zero, then the standard reduction potential for Al3+/Al is -1.66 V.
ChemActivity 49
Exercises
1. a) Cathode is Zn, anode is Al. b) Electrons flow from the Al to the Zn.
c) Al is negative, Zn is positive. d) Cathode: Zn2+ + 2e– = Zn Anode: Al = Al3+ + 3e–
e) 3 Zn2+(aq) + 2 Al(s) = 3 Zn(s) + 2 Al3+(aq)
f) The bar of zinc will become heavier. The bar of aluminum will become lighter.
2. a) False. The half-cell with the most positive standard reduction potential is always the
cathode in a galvanic cell because is has a stronger pull on the electrons.
b) True. If an atom or ion loses an electron it must lose it to some other species.
3. (a) +0.78 V, the reaction will proceed as written. (b) +).80 V, the reaction will proceed
as written. (c) –1.97 V, the reaction will proceed from right to left. (d) 0.88 V, the
reaction will proceed as written. (e) –0.69 V, the reaction will proceed from right to left.
4. (a), (b), and (d) will occur. (c) and (e) will not occur.
Problems
1. a) T b) F c) F d) T e) T
2. The reaction: Zn + 2 H+ = H2 + Zn2+ has a standard cell voltage of +0.76 V. The
reaction: Cu + 2 H+ = H2 + Cu2+ has a standard cell voltage of –0.34 V. The Zn
reaction should occur and the Cu reaction should not occur.
3. Ozone. O3 + H2O + 2e– = O2 + 2 OH– E°Red = 1.24 V
4. No reaction will occur because the cell voltage is negative, – 0.09 volts for:
Sn + Ni2+
= Ni + Sn2+
5. 0.467 M
ChemActivity 50
Exercises
1. a) negative b) negative c) positive d) positive e) negative
2. ∆S for the second reaction would be more negative than ∆S for the first reaction. In the
second reaction, 3 moles of particles become one mole of particles (more order).
3. False. ∆S is expected to be positive because there are three moles of gas produced for every
two moles of gas consumed.
ChemActivity 51
Exercises
1. Entropy only. The reaction is endothermic, not energy favored. The increase in entropy
makes this a naturally occurring process.
36
2. Both enthalpy and entropy. The process is exothermic (favorable) and the entropy
increases (favorable).
Problems
1. a) ∆S is positive because the arrangement in Figure B is more disordered than the
arrangement in Figure A. b) ∆H is negative. The H atom of each molecule will have a
partial negative charge and the F atom of each molecule will have a partial negative charge.
In Figure A the molecules are arranged such that these partial charges will repel. In Figure
B these partial charge will attract. Thus, Figure B is a lower energy arrangement. c) When
∆S is positive and ∆H is negative the process is favored by both energy and entropy and
will be naturally occurring at high and low temperatures (see the table in Model 2).
2. a)
plus resonance
b) There are three bonds on the left=hand-side of the equation, O 3 , and only two bonds on
the right-hand-side, O 2 and O. Therefore, the left-hand-side is the lower enthalpy state. c)
∆H is > 0 (going from a lower enthalpy state to a higher enthalpy state. d) The right–handside is more disordered than the left-hand-side. Therefore, ∆S > 0. e) When a process is
entropy favored but not favored by enthalpy it will tend to be naturally occurring at high
temperatures (see the table in Model 2).
ChemActivity 52
Exercises
1. a) prediction: ∆S° < 0. actual: –198.73 J/K b) prediction: ∆S° < 0. actual: –626.91 J/K
c) prediction: ∆S° probably close to 0. actual: –20.07 J/K
d) prediction: ∆S° > 0. actual: 156.27 J/K e) prediction: ∆S° < 0. actual: –137.57 J/K
f) prediction: ∆S° < 0. actual: –274.88 J/K
2. a) False. The standard entropy of atom combination for any diatomic molecule is negative
because the molecule is more ordered than the separate atoms. b) True. The reaction to
form CH4 has ∆ngas = –4, whereas the reaction to form NH3 has ∆ngas = –3.
3. i) is incorrect because the entropy of atom combination of any liquid must be negative (the
liquid is more ordered than the monatomic gas).
4. a) ∆S° = –332.3 J/mol K b) CO (C=+2; O = –2); H2 (H = 0); CH3OH (C = –2, H = +1; O
= –2). Yes, this is a redox reaction.
5. a) ∆S° = +173.3 J/K and ∆H° = –851.5 kJ/mole. b) Favorable c) Favorable
d) Yes, it is an oxidation-reduction process. The aluminum, for example, changes oxidation
number from zero to +3.
Problem
1. (i) would have the most positive change in entropy. There are two moles of gas on the lefthand-side and two moles of gas on the right-hand-side—so standard entropy change should
be near zero. Each of the other four chemical reactions should have a negative entropy
change because there are more moles of gas on the left-hand-side.
37
ChemActivity 53
Exercises
kJ
1. a) ∆H° = –296.83 mol
J
∆S° = 39.11 mol K K > 1
kJ
b) ∆H° = +52.26 mol
J
∆S° = –53.28 mol K K < 1
kJ
c) ∆H° = –128.5 mol
J
∆S° = –70.25 mol K K can't be deduced
kJ
d) ∆H° = –92.21 mol
J
∆S° = –198.57 mol K K can't be deduced
kJ
e) ∆H° = +44.01 mol
J
∆S° = +118.34 mol K K can't be deduced
kJ
2. i) ∆H° = –12.55 mol
J
∆S° = –102.5 mol K
kJ
∆H° – T∆S° = 18.01 mol
kJ
J
kJ
ii) ∆H° = –92.21 mol
∆S° = –198.77 mol K
∆H° – T∆S° = –32.95 mol
kJ
J
kJ
iii) ∆H° = 23.42 mol ∆S° = –12.50 mol K
∆H° – T∆S° = 27.15 mol
a) K > 1 is ii
b) Largest K is ii
c) Smallest K: iii
3. ∆H° < 0 because reaction is exothermic.
∆S° > 0 because number of particles increases.
Since both enthalpy and entropy factors are favorable, the equilibrium constant is expected
to be > 1.
kJ
J
4. ∆H° = +24.9 mol ; ∆S° = +307.6 mol K
Although ∆H° is positive, the term ∆H°– T∆S° will become negative at high T. As T
increases, K will become larger and flammable hydrogen gas will be released.
5. K<1 for ∆H° > 0 and ∆S° < 0. K>1 for ∆H° < 0 and ∆S° > 0.
ChemActivity 54
Exercises
kJ
1. ∆G° = 18.03 mol = –RT lnK
lnK =–18,030 J/[(8.3144 J/K )(298.15 K)] = -7.27
K a = 7.0 x 10–4
kJ
J
kJ
× –6
2. ∆H° = 41.16 mole ; ∆S° = 41.50
;
∆G°
=
28.79
mol ; K = 9.0 10
mol.K
K will increase as T increases.
3. ∆H° > 0
∆S° > 0
As T increases, K increases, causing the concentration of NH3(g) to increase.
kJ
4. ∆H° = –92.21 mole . Since ∆H° < 0, K will decrease as T increases.
38
∆H°
5. a) Liquid boils when ∆G° = 0, so that ∆H° = T∆S°. Thus, the boiling point is T = ∆S° .
When salt is added to water, ∆H° does not change, but ∆S° decreases, so that ∆H°/∆S°
increases. Thus, the boiling point T is increased.
kJ
b) ∆H° = 38.00 mole
J
∆S° = 113.0 mole K
Boiling pt = 336.3 K = 63.1 °C.
6. a) ∆G° = 2.74 × 104 J b) ∆S° must be > 0 (ions in solutions are more disordered than the
solid); because ∆G° = ∆H° – T∆S°, ∆H° must be positive.
7. a) 6.2 × 1025 b) 7.9 × 1063 c) 3.2 × 10–12
Problem
1. (a) E° = 0.44 V (b) I3– (aq) (c) K = 7 × 1014
ChemActivity 55
Exercises
1. 5.40 × 10–7 M sec–1
2. a) first order in both
b) rate = k (Fe2+) (Ce4+)
c) 1.0 x 103 M–1 sec–1
d) 1.8 x 10–7 M–1 sec-1
3. rate = k (NO)2 (Br)1; k = 0.65 M–2 min–1
4. rate = k (I-) (S2O82-); 1.9 × 10–6 M sec–1
5. False. The rate law must be determined experimentally.
Problem
1. Consider expt's 1 and 2. Init conc of C2H4 is the same; init conc of O3 triples; rate triples.
Therefore, the rate is first order with respect to O3: rate = k (O3) (C2H4)x
Now consider expt's 2 and 3. The init conc of O3 falls by 2/3; if the init conc of C2H4
stayed the same, 1.0 x 10–8 M, the rate should decrease by 2/3 to 2.0 x 10–12 M/sec. The
actual rate is 2 x 2.0 x 10–12 or 4.0 x 10–12, and the actual initial conc of C2H4 was
doubled. Therefore, rate is first order with respect to C2H4: rate = k (O3) (C2H4).
2. a) 1st order in HgCl2, 2nd order in C2O42–.
b) rate = k (HgCl2)1(C2O42-)2
c) k = 1.3 × 10–4 M–2 sec–1
130
3. (a) rate = k (UO2+)2 (H+) (b) k = 2
M sec
4. rate = k (NO)2 (O 2 )
ChemActivity 56
Exercises
1. a) first order; k = 6.93 × 10–3 sec–1
b) 0.093 M
39
2. Calculate ln(BrO–) and
1
at each time. Prepare two plots: ln(BrO–) vs. time (a first(BrO–)
1
vs. time (a second-order plot). The plot with the best straight line
(BrO–)
determines the order.
3. 6.6 × 10–4 sec–1; 2.1 × 103 sec
4. False. As the reaction proceeds, (A) decreases. The rate of reaction is proportional to (A)
because this is a first order reaction. Thus, as the reaction proceeds and (A) decreases, the
rate of reaction also decreases.
5. IV
6. 1.1 × 103 sec; 2.1 × 103 sec
7. iv) 1/8
8. 87%
19. t = 7.35 × 103 yr
Problems
1. Plot 1/(A2B2) vs time, the rate constant is the slope of the line.
2. If the time for (A) to reach 50% of its original value is the same as the time for (A) to
decrease from 50% to 25% of its original value, then the reaction is first order. This can be
concluded because in this case the half-life would be seen to be independent of the starting
concentration, and for first order reactions this is the case. For a second order reaction, the
half-life depends on original concentration, and so the time to decrease from 50% to 25%
of the original concentration will be twice as long as the time to reach 50%.
3. The half life is 50 seconds and does not depend on concentration. This is not true for 2nd
order reactions, but it is true for 1st order reactions.
order plot) and
ChemActivity 57
Exercises
1. forward reaction has the larger activation energy
2. 210.5 kJ/mole
3. c) reactions not energetic enough to break bonds; molecules not oriented properly when
collision occurs
Problems
1. The activation energy is small because:
a) No bonds are broken in either reactant; b) The bare proton, H+, can simply bond to one
of the lone pairs on the oxygen atom; c) The proton is positively charged and the lone pair
is negatively charged; these two species are attracted to each other and there is no
repulsion.
2. Both the N 2 bond and the H 2 bond must be broken to form NH 3 . The Lewis structure for
N 2 shows that the N 2 bond is a triple bond—very strong. The H 2 bond is a single bond,
but it is a very strong single bond because of the short bond length. Thus, the activation
energy for this reaction is large.
ChemActivity 58
Exercises
1. a) bimolecular
b) unimolecular
rate = k' (NOBr)2
rate = k' (N2O2)
40
c) bimolecular
rate = k' (NO)2
d) bimolecular
rate = k' (I) (H2)
2. a) ii
b) the sum and the stoichiometry are the same
3. No, the forward reaction may or may not be under kinetic control.
4. Yes, there will be very little reaction regardless of whether equilibrium is reached or not
because the value of K is very small so very little reactant will be present at equilibrium.
5. exothermic
6. a) II
b) IV
c) I
d) II
7. a) rate = k (NO2) (O3) ; 5.0 x 104 M–1 sec–1
b) –150.7 kJ/mol K
c) The rate of reaction is fast, indicating that the activation energy for the forward reaction
is relatively small. The reaction must be an exothermic reaction, so the enthalpy of the
products are shown to be lower than the enthalpy of the reactants.
d) Not consistent because the sum of the steps in the mechanism does not yield the correct
stoichiometry.
Problems
1. a) rate = k (NO2)2 (CO)0 = k (NO2)2 . k = 24/Ms b) Yes. rate = k (NO2)2
2. (a) From the slow step, rate = k (O) (O3). From the fast step, equilibrium, K = (O)
(O2)/O3). Combining the two equations above: rate = kK(O3)/(O2). (b) The concentration
of O2 is on the denominator of the rate law.
3.
2 Fe2+(aq) + I2(aq) = 2 Fe3+(aq) + 2 I2–(aq)
ChemActivity 59
Exercise
1. Activation energy is lowered.
Problem
1. (a) Diagram not given here. (b) Ea (reverse) = 173.5 kJ
(c) ∆H° = 9.5 kJ/mol
ChemActivity 60
Exercises
50000 J/mol
1
1
1. exp 8.315 J/mol K (298 K – 308K) = 1.97
50000 J/mol
1
1
2. exp 8.315 J/mol K (233 K – 240K) = 2.12 The Martian chemist is correct!
3. False. The higher the activation energy, the slower a reaction occurs at a given
temperature.
Problems
1. 9 × 10–4 M–1 sec–1
2. The reaction needs a large, negative ∆H° (assuming that ∆S° is small because ∆ngas for the
reaction is zero) to make ∆G° very negative and cause a large equilibrium constant. The
activation energy (without catalyst) must be large so that the rate will be slow. The
activation with catalyst must be much smaller and the rate will be much faster.
41