1
Answers to Critical Thinking Questions
for
CHEMISTRY
A Guided Inquiry
Fifth Edition, 2011
Richard S. Moog
Franklin & Marshall College
John J. Farrell
Franklin & Marshall College
Latest Update: May 12, 2011
John Wiley & Sons, Inc.
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Answers to Critical Thinking Questions
Please do not give these answers to students.
Quite often, the answers given here are less
detailed than what is expected for a student’s
answer.
ChemActivity 1
1. 6, 6, 6
2. 6, 7, 7
3. 6, 6, 7
4. All carbon atoms and ions have six protons in the nucleus.
5. All hydrogen atoms and ions have one proton in the nucleus.
6. Z is the number of protons in the nucleus of that atom.
7. Twenty-eight protons in the nucleus.
8. a) Because there are six protons and 7 electrons, so there is a net charge of -1. b) In an
ion the number of protons and electrons are not equal. c) The charge on an ion = # of
protons – # of electrons.
9. Because every H atom and ion must have one proton, the number of protons is 1. The
number of neutrons is zero, analogous to 1H and 1H-. The number of electrons is zero
because the charge is 1+ and there has to be one proton.
10. Different isotopes on a particular element have differing numbers of neutrons in the
nucleus (but the same number of protons).
11. The mass number (A) is the sum of the number of protons and the number of neutrons in
the nucleus.
12. The O atom has 8 protons and 8 neutrons in its nucleus; hence, the mass number is 16. The
O atom has 8 protons and 10 electrons; hence the charge on the ion is 2–. The Na atom has
11 protons and 12 neutrons; hence, the mass number is 23. The Na atom has 11 protons
and 10 electrons; hence the charge on the ion is 1+.
13. Most of the mass is in the nucleus—where the protons and neutrons are. Note that the
difference in mass between a 13C and a 13C– atom (which differ by only one electron) is
only 0.0005 amu.
ChemActivity 2
1. 3
2. All isotopes of magnesium have twelve protons in the nucleus. The three isotopes of
magnesium have 12, 13, and 14 neutrons in the nucleus.
3. 12.0000 amu
4. a) 1200.00 amu;
b) 1300.34 amu
5. Slightly more than 1200.00 amu because 1200 amu would be the minimum and there
should be, on average, 1.11 13C atoms among the 100 total atoms.
6. None
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7. You must know the total number of marbles to use the first method. Therefore, the second
method must be used in this case.
8. a) 0.7577  34.9689 amu + 0.2423  36.9659 amu = 35.45 amu. b) Zero.
9. a) 35.45 amu.
b) 5.887 x 10–23 g
10. 6.022 x 1023 atoms x 5.887 x 10-23 g/atom = 35.45 g
11. a) 24.31 amu. b) 4.037 x 1023 g
12. 6.022 x 1023 atoms x 4.037 x 10-23 g/atom = 24.31 g
13. a) It is the same number. b) It is the same number.
14. a) It is the same number. b) It is the same number.
15. 12.001 is a) the average mass in amu of one C atom and b) the mass in grams of 6.022 x
1023 C atoms.
16. Zero %.
17. a) 12. b) They have the same number—12. c) 6.022  1023. d) They have the same
number, 6.022  1023. e) They have the same number, 12. f) They have the same
number, 6.022  1023.
18. a) Two dozen elephants. b) One mole of sodium atoms.
19. There is one mole of H atoms and one mole of argon atoms; they both have the same
number of atoms.
ChemActivity 3
1. The magnitude of V decreases.
2. V = 0.
3. k and d must be positive. q1q2 must be positive.Therefore, V must be positive. That is, V >
0
4. a) q = +1 for a proton. b) q = 0 for a neutron. c) q = +6 for the nucleus of a carbon atom.
5. V is a negative number because k(1)(–1)/d is negative.
6. V is negative because the electron is negative and the proton is positive.
7. I would expect V to be become more negative as d becomes smaller.
8. V (10–18 J) = 0, –.0462, –0.231, –0.462, –1.16, –2.31.
9. V = – IE
10. a) The electron that is closer to the nucleus would have the larger ionization energy because
it is at a lower (more negative) potential energy according to Coulomb’s potential energy
equation.
11. a) The electron that is at d1 from the +2 nucleus would have the larger ionization energy
because it is at a lower (more negative) potential energy according to Coulomb’s potential
energy equation.
12. The ionization energy is larger by a factor of 2.
13. He+ would have a larger ionization energy than H because q = +2 for He and it would have
a lower (more negative) potential energy according to Coulomb’s potential energy
equation.
ChemActivity 4
1. 1.31 MJ/mole
2. The electron that is farthest from the nucleus will have the least negative potential energy
(d is larger) and the lowest ionization energy.
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3. Open-ended question. Students might predict that IE1 increases with atomic number
(higher nuclear charge). Or, they might predict that IE1 decreases if they put additional
electrons farther away from the nucleus.
4. The answer here depends on the students’ answer to CTQ 3.
5. a) The IE1 for He is greater than the IE1 for H because the nuclear charge on a He atom is
6.
7.
8.
9.
+2 whereas the nuclear charge on H is +1. b) The IE1 for Li is less than the IE1 for He
because at least one of the electrons of Li must be farther away from the nucleus than any
electron of He. Otherwise, the +3 nuclear charge of Li would hold the electron more
tightly.
Each electron is held by a +2 charge, rather than a +1 as in H. Therefore, if the electrons is
He and H are at the same distance the Coulombic Potential Energy for He ought twice the
Coulombic Potential for He and IE1 for He ought to be about 2  IE1 for H.
If the 3rd electron was at the same distance as H, the IE1 would be about 3  IE1 for H—
3.93 MJ/mole. However, it would be a bit less because of the repulsion between the
electrons; the best answer is 3.6 MJ/mole.
For this model the IE1 of Li would be about 3.93 MJ/mole (probably a bit less due to the
repulsion between the electrons—see the previous CTQ). The IE1 of Li is 0.52 MJ/mole—
much smaller than 3.93 MJ/mole (see the previous CTQ). This low value is inconsistent
with the least tightly held electron being this close to the nucleus.
If the 3rd electron is farther away from the nucleus than the electron in H the IE1 would be
much lower that the IE1 for H. Because the IE1 of Li is 0.52 MJ/mole and the IE1 of H is
0.52 MJ/mole, this model is consistent with the experimental data.
10. The electrons in the first shell of Li are closer to the nucleus and should be harder to ionize
(have a greater ionization energy) than the electron in the second shell.
ChemActivity 5
1. a) H, one electron b) Li, one electron c) He, two electrons
2. a) H, zero b) Li, two c) He, zero
3. The core charge of Li is +1.
4. Electron “b” experiences more electron-electron repulsion than electron “a” because
electron “a” is farther from the other electrons than electron “b”. Therefore the IE1 of
5.
6.
7.
8.
9.
electron “b” would be less than the IE1 of electron “a”.
a) Be has 4 protons in its nucleus. b) two c) two d) 4 – 2 = 2
e) The core charge is equal to the number of valence electrons for a neutral atom.
Be has a higher core charge (+2) than Li (+1) and a higher IE1 than Li.
Ne: 10–2 = 8
Diagram (a) should have a nuclear charge of +7, two electrons in the first circle (shell), and
five electrons in the second circle (shell). Diagram (b) should have a core charge of +5 and
a circle with five electrons (at about the same distance as the second circle in diagram (a).
a) Diagram (a) should have a nuclear charge of +11, two electrons in the first circle (shell),
and nine electrons in the second circle (shell). Diagram (b) should have a core charge of
+9 and a circle with nine electrons (at about the same distance as the second circle in
diagram (a).
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10.
11.
12.
13.
14.
15.
b) +9
c) Because the core charge of Ne is +8 and the core charge of Na is +9 (and
because the electrons are about the same distance from the nuclei) the IE1 for Na should be
greater than 2.08 MJ/mole.
a)
Diagram (a) should have a nuclear charge of +11, two electrons in the first circle
(shell), eight electrons in the second circle (shell), and one electron in the third circle
(shell). Diagram (b) should have a core charge of +1 and a circle with one electron (at
about the same distance as the third circle in diagram (a).
b) +1
c) Because the core charge of Ne is +8 and the core charge of Na is +1 (and because the
outermost Na electron is farther from the nucleus) the IE1 for Na should be much less than
2.08 MJ/mole.
The experimental IE1 for Na is 0.50 MJ/mole. This is consistent with the model in CTQ 10.
2, 8, 1
Both have a core charge of +1.
The radius of the valence shell of Na is larger than the radius of the valence shell of Li
because they both have a core charge of +1 and Na has the lower IE1.
Na has a core charge of +1 and Ne has a core charge of +8. Na should have a much lower
IE1. The is consistent with the experimental data.
16. IE1 increases as we move from left to right from Na to Ar because the core charge
increases for +1 to +8 and the potential energy of the electron becomes more negative
according to the Coulombic potential energy equation (the distance from the nucleus
remaining fairly constant).
17. a) All three atoms have a core charge of +1 and all three atoms are in the first column
(labeled I). b) The valence shell of H is 1 and it is found in the first row of the table. The
valence shell of Li is 2 and it is found in the second row. The valence shell of Na is 3 and
it is found in the third row. c) For Rb, the valence shell is 5, the core charge is +1, and the
number of valence electrons is 1. d) Rb has the lowest IE1 of these four elements. This is
consistent with all four having one valence electron and a core charge of +1, but the
valence electron of Rb is farther from the nucleus (higher numbered valence shell).
18. Diagram (a) should have a nuclear charge of +9, two electrons in the first shell, and seven
electrons in the second shell. Diagram (b) should have a core charge of +7 and a shell of
seven electrons.
19. a) Both atoms have a core charge of +7 and 7 valence electrons, and both are found in the
column labeled VII. b) The valence shell of F is 2 and it is found in the second row. The
valence shell of Cl is 3 and it is found in the third row. c) The IE 1 of Cl is less than the
IE1 of F because they both have a core charge of +7 but the valence electrons of Cl are
farther from the nucleus than the valence electrons of F.
20. C is found in column IV, second row. Therefore, the core charge is +4 and the four valence
electrons are found in the second shell.
21. The core charge increases as we move from left to right across a period.
22. IE1 increases as we move from left to right across a period because the core charge
increases and the potential energy of the electron becomes more negative according to the
Coulombic potential energy equation (the distance from the nucleus remaining fairly
constant).
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ChemActivity 6
1. The number of the valence shell corresponds to the number of the row of the Periodic
Table.
2. Because the nuclei gain one proton as one moves from left to right across a period but the
number of inner shell electrons remains constant.
3. The radius tends to decrease because the core charge increases and the valence electrons
are in the same shell.
4. The radius tends to increase because the core charge remains the same and the valence
electrons are farther from the nucleus.
5. Nitrogen, 55 pm (less than oxygen because of increased core charge); chlorine, 95 pm (less
than sulfur because of increased core charge); tellurium, 130 pm (greater than selenium
because n = 5 for Te and n = 4 for Se).
6. All have 18 electrons.
7. The diagram should have 19 –1 or 18 electrons—two in the first shell, eight in the second
shell, and eight in the third shell. The third shell is the valence shell. Thus, the core charge
= 19 – 10 = +9.
8. All ions in Table 2 have the same number of electrons (18), but Ca has the most protons
(20) and sulfur the least (16).
9. O2– will be larger than F– because both have 10 electrons and F has more protons.
10. The diagram should have a nuclear charge of +9, two electrons in the first shell and seven
electrons in the second shell. The core charge is 9 – 2 = +7.
11. Experimental observations show that like charges repel each other. Also, according to the
Coulombic potential energy equation, for like charges V > 0. F has 7 electrons in the
second shell and F– has 8 electrons in the second shell. Because 8 electrons will repel each
other more than 7 electrons F– will have a larger radius.
12. N3– has 8 valence electrons and the core charge is +5. The radius difference between O 2–
(core charge +6) and F– (core charge +7) is 7 pm and both have 8 valence electrons.
13.
Therefore, a good prediction for the radius of N3– would be 147 pm.
Core charge; valence shell; charge on species or number of valence electrons.
ChemActivity 7
1. Yes.
2. a) 0.071 sec b) 14 c) Increase. If  > 2.5 cm, it would take longer than 0.71 sec for
one wavelength to pass point X. d) Decrease. The number of wavelengths that would pass
point X is equal to (35 cm/s)/wavelength. If the wavelength is greater than 2.5 cm the
number of wavelengths would decrease.
3. a) The longer the wavelength the smaller the frequency. b) f = c/.
4. The ultraviolet.
5. h = 6.626  10–34 J s
6. c = 2.998  108 m/s
7. E = hf The energy of a photon is proportional to its frequency.
8. E = h c/ The energy of a photon is inversely proportional to its wavelength.
ChemActivity 8
1. a) 6.26 MJ/mole
b) The potential energy of electrons  and  is –6.26 MJ/mole.
The potential energy of electron  is –0.52 MJ/mole. c) The ionization energy of electrons
 and  is is 6.26 MJ/mole. The ionization energy of electron  is 0.52 MJ/mole.
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2. When comparing the energy levels of the two different electrons, the electron with the
lower (more negative) potential energy is the lower energy level. The more negative the
value of the energy level (the lower the energy level), the stronger the electron is held by
the atom.
3. 143.4 MJ/mole - 114.8 MJ/mole = 28.6 MJ/mole.
4. The position of a peak is determined by the ionization energy of the electron ejected.
5. 28.6 MJ/mole
6. –28.6 MJ/mole.
7. The number of electrons at that energy level.
8. Regardless of the number of electrons at this ionization energy (28.6 MJ/mole), one two,
three,…) the spectrum would look the same.
9. There should be two peaks because there are electrons at two different energy levels.
10. The relative heights of the peaks will be 3 to 2.
11. The peak representing the two electrons will be at a higher ionization energy than the peak
representing the three electrons because the two electrons are at a lower energy level
(harder to remove).
12. a) The single peak indicates that all (one or more than one) electrons are at the same energy
level. Because both electrons in He are at the same energy level, there is only one peak, as
is the case for the single electron in H.
b) Li has electrons in two different energy
levels, so it would have two peaks in the PES spectrum.
13. The unknown atom is He.
14. Electrons that are closer to the nucleus are harder to remove.
15. Because there are 2 electrons in the first shell and 8 electrons in the second shell.
16. The spectrum should have two peaks with a ratio of intensities of 8 (low ionization energy)
to 2 (high ionization energy).
ChemActivity 9
1. a) No. b) There are three peaks. Two peaks have a low IE with a 3:1 ratio. The third
peak, at high IE, has the same intensity as one of the low IE peaks.
2. Only the relative peak height is important. For example, if the relative peak heights of two
peaks are 4 to 1, the number of electrons at the two energy levels must have the same ratio:
4 to 1, or 8 to 2, or 16 to 4, … .
3. The relative peak heights are 3:1:1 (low IE to high IE). Because a Ne atom has ten
electrons, the number of electrons must be 6:2:2 (low IE to high IE).
4. The 84 MJ/mole represents a high IE and the 2 electrons in the first shell of Ne would be
hard to remove.
5. Because they have, relatively, low IEs. The 1s electrons have much higher ionization
energies.
6. Should have 5 peaks. The relative intensities of 3p:3s:2p:2s:1s should be 3:1:3:1:1 and the
peaks will appear with IEs equal to the negative of the energy levels given in the figure.
ChemActivity 10
1. a) N, 1.40 MJ/mole. b) Ar, 1.52 MJ/mole.
2. Use photoelectron spectroscopy.
3. The number of electrons at each energy level.
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4. Table 2 gives the electron configuration of the selected elements but not the ionization
energies for each level.
5. Yes. The electrons with the highest IE will be 1s electrons. The IE of electrons in other
shells will be grouped according to valence shell (the higher the n value, the lower the IE).
Within the same shell, a p electron has a lower IE than an s electron.
6. a) 1s, 340 MJ/mole (higher than 309 of Ar 1s). 2s, 36 MJ/mole (higher than 31.5 of Ar
2s). 2p, 36 MJ/mole (higher than 31.5 of Ar 2p). 3s, 3.2 MJ/mole (higher than 2.8 of Ar
3s). 3p, 1.8 MJ/mole (higher than 1.52 of Ar 3p). Relative intensity, starting with 3p and
working to lower IE— 6:2:6:2:2. b) Closest to 0.42 MJ/mole because the 4s electron in K
should be easier to remove than the 3s electron in Na. c) Closest to 1.4 MJ/mole because
the electron would be slightly easier to remove than a 3p electron in Ar. Note that Be 2s is
0.90 whereas B 2p is 0.80. Also note that Mg 3s is 0.74 whereas Al 3p is 0.58.
ChemActivity 11
1. The experimental spectrum is similar to the predicted spectrum that has the last electron at
the 4s energy level (the IE is closest to 0.42 MJ/mole).
2. The peak at 0.42 MJ/mole is assigned to the 4s energy level in K because it should be
easier to remove than the 3s electron in Na (0.50 MJ/mole).
3. If the electron was in a 4p energy level, it would be easier to remove than a 4s electron.
That is, the peak would be at an IE < 0.63 MJ/mole.
4. All of the elements in a group (column) have the same number of valence electrons and the
same valence-shell electron configuration. Each new period (row) begins a new valence
shell.
5. 10.
6. 1s2 2s2 2p63s2 3p6 4s2 3d10 4p1.
7. All of the elements in a group (column) have the same number of valence electrons and the
same valence-shell electron configuration.
ChemActivity 12
1. An arrow represents an electron— spin "up" (up arrow), spin "down" (down arrow).
2. If there are two electrons in a filled s subshell, one is spin up and the other is spin down.
3. a) Li atoms, split. b) Be atoms, not split. c) B atoms, split. d) C atoms, open ended
question— could split if there are two spin-up electrons— won't split if the 2p level
contains one spin-up electron and one spin-down electron.
4. Apparently, the only "pairs" that can form are one spin-up electron and one spin-down
electron.
5. N > C =O > B > Ne. The greater the magnetic moment, the greater the number of unpaired
electrons.
6. a) The diagram should show two spin-up electrons at the 2p level. Carbon must have more
unpaired electrons than Be (one). b) answer depends on the students' answer to CTQ 3.
7. The diagram should show 3 spin-up electrons at the 2p level. N must have more unpaired
electrons than C.
8. They tend not to pair.
9. a) two. b) zero.
10. 3 pairs.
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ChemActivity 13
1. a) +1.
2. a)
b) +7.
c) +2.
b)
c)
3. Because hydrogen can have only two electrons in its valence shell (n = 1); carbon, nitrogen,
or oxygen can have eight electrons in the valence shell (n = 2).
4. a)
b) three.
c) NH3
5. SH2
6. a) Does the total number of electrons in the Lewis structure equal the total number of
valence electrons in the molecule? b) Are there two electrons "around" each H atom? c)
Are there eight electrons "around" all other atoms?
7. a) 12
b) 10
c) 14
8. a) 12
b) 10
c) 14
9. The total number of electrons used to generate the Lewis structure equals the total number
of valence electrons in the molecule.
10. Yes, if you count the two bonding electrons as being "around" each of the Cl atoms.
11. Yes, if you count all six of the bonding electrons as being "around" each of the N atoms.
12. a) Yes, if you count the eight bonding electrons as being "around" the C atom.
b) Yes, if you count the four bonding electrons in the C–O double bond as being "around"
the O atom.
13. a) Does the total number of electrons in the Lewis structure equal the total number of
valence electrons in the molecule? b) Are there two electrons "around" each H atom?
Note: each bonding electron is counted as being "around" each of the bonding atoms. c)
Are there eight electrons "around" all other atoms?
14. Yes, it is correct. There should be 6  2 + 4 = 16 electrons in the Lewis structure and there
are 16. There are eight electrons around all atoms.
ChemActivity 14
1. Yes, it is correct. There should be 4  2 + 4 = 12 electrons in the Lewis structure and there
are 12. There are eight electrons around all atoms.
2. single: H–H and C–H in H3CCH3 double: C–C in H2CCH2 and C–O in CO2
triple: N–N in N2 and C–C in HCCH.
3. BO=1, single bond. BO=2, double bond. BO=3, triple bond.
4. BO=1, two electrons shared. BO=2, four electrons shared. BO=3, six electrons shared.
5. single < double < triple.
6. a) 1. b) The bond energy decreases in the series from HF to HI. c) The halides increase
in size from F to I; therefore, the H-X bond length should increase in the series.
7. a) 1. b) The bond energy decreases in the series F2 to I2. c) The halides increase in size
from F to I; therefore, the X-X bond length should increase in the series.
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8. The shorter the bond, the stronger the bond.
9. a) single bonds: 151-570 kJ/mole. b) double bonds: 720-804 kJ/mole. c) triple bonds:
945-962 kJ/mole.
10. In general, triple bonds are much stronger than double bonds, and double bonds are much
stronger than single bonds. There were no exceptions in Table 2. Within any particular
bond order, the strongest bonds between two atoms are those with the smallest internuclear
distance.
ChemActivity 15
1. There is a double bond between the two carbon atoms.
2. There is good agreement.
3. The stronger the bond, the shorter the bond length.
4. Halfway between 150 and 133—about 142 pm.
5. For benzene there is not good agreement. The Lewis bond orders are either 1 or 2, whereas
the calculated bond orders are 1.42 (all).
6. The fact that all bond energies are the same, 509 kJ/mole, is consistent with the calculated
bond orders, 1.42 for all C–C bonds, but it is not consistent with the Lewis bond orders,
which are either 1 or 2. If the Lewis bond orders were correct the bond energies would
alternate between single and double bond energies.
7. The fact that all bond lengths are the same, 139 pm, is consistent with the calculated bond
orders, 1.42 for all C–C bonds, but it is not consistent with the Lewis bond orders, which
are either 1 or 2. If the Lewis bond orders were correct the bond lengths would alternate
between 150 and 133 pm (approximately).
8. The alternating single and double bonds in the Lewis structure of benzene are not
consistent with the calculated C–C bond orders.
9. The C1–C2 bond order is 1 in resonance structure I and it is 2 in the resonance structure II.
The average is 1.5.
10. The resonance hybrid gives better agreement with the calculated bond orders than either I
or II.
ChemActivity 16
1. Four from C and six from each O for a total of 16.
2. a) first row: 16 in each column. All remaining rows: 8 in each column
b) Each is a
legitimate Lewis structure because the correct total number of electrons is present, and
there are eight electrons around each atom.
3. Structure I is better because both bonds are identical, experimentally. In structure II, one
C–O bond is a triple bond and one C–O bond is a single bond.
4. a) Structure I is the better choice because all formal charges are zero. In structure II, one
O atom has a formal charge of –1 and the other O atom has a formal charge of +1. b)
Structure I is consistent with the experimental fact that both C–O bonds have the same
bond energy.
5. No, but the sum must be zero—as in CO.
6. a) Structure I is better because there are no formal charges. b) The longer C–O bond
should be the single bond. This is in agreement with Lewis structure I where the C–O with
the attached hydrogen is the single, and longer C–O bond.
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7. a) Does the total number of electrons in the Lewis structure equal the total number of
valence electrons in the molecule? b) Are there two electrons "around" each H atom? c)
Are there eight electrons "around" all second period atoms? d) Does the sum of the formal
charges equal the charge on the molecule (ion)? e) Are the formal charges minimized
(never greater than ±1 on any atom)?
ChemActivity 17
1. a) 5. b) 18. c) 23. d) 24. e) by adding the number of valence electrons for each atom
and by adding one electron because of the negative charge on the molecule.
2. Because all N–O bonds in the molecule are identical.
3. (2 + 1 + 1)/3 = 4/3
4. There are the correct number of electrons: XeO4, 32; PCl5, 40. There are eight electrons
around all O atoms and all Cl atoms. Xe and P have more than eight, but this is OK
because n > 2 in both cases. There are no formal charges.
5. Xe, 16. P, 10.
6. The n = 2 shell can accommodate only 8 electrons.
7. The third period, n = 3, can accommodate 18 electrons.
8. a) Does the total number of electrons in the Lewis structure equal the total number of
valence electrons in the molecule or ion (including any extra or removed electrons to
account for the charge on an ion)? b) Are there two electrons "around" each H atom? c)
Are there eight electrons "around" all other atoms (particularly 2nd shell atoms)? (Atoms in
the 3rd and higher shells can accommodate more than eight electrons.) d) Does the sum of
the formal charges equal the charge on the molecule (ion)? e) Does the structure minimize
the formal charges?
ChemActivity 18
1. The number of bonding domains is equal to the number of bonded atoms (to the given
atom).
2. The number of nonbonding domains is equal to the number of lone pairs on the given atom.
3. 180, 120, 109
4. When the sum of the number of bonding and nonbonding domains is 2, the angle is 180°.
When it is 3, the angle is about 120°, When it is 4, the angle is about 109°.
5. Open ended. Angles should not be 90°.
6. Open ended. For example: tripod with a cap, triangular pyramid, bent
7. a) 180°. b) 120°. c) 109°.
8. a) 4. b) four domains of electrons. c) Yes.
9. a) 4. b) four domains of electrons. c) Yes, approximately.
10. a) 4. b) four domains of electrons. c) Yes, approximately.
11. a) 3. b) three domains of electrons. c) Yes.
12. a) 2. b) two domains of electrons. c) Yes.
13. Bent molecules have exactly two bonding domains around the central atom. If there are two
bonding domains and one non-bonding domain (total of three domains), bond angles close
to 120° are expected. If there are two bonding domains and two non-bonding domains
(total of four domains), angles close to 109° are expected.
14. Determine the total number of electron domains around the atom in question. If there are
two bonding domains (and zero non-bonding domains), the molecule will be linear. If there
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are three domains, the molecule will be trigonal planar (when there are three bonding
domains) or bent (when there are two bonding domains) or linear (when there is only one
bonded atom). If there are four domains, the molecule will be tetrahedral (when there are
four bonding domains) or trigonal pyramidal (when there are three bonding domains) or
bent (when there are two bonding domains) or linear (when there is only one bonded atom).
ChemActivity 19
1. 109.45°. sp3.
2. 109.45° (probably a bit smaller). sp3.
3. 120°. sp2.
ChemActivity 20
1. a) The core charge increases as one goes from left to right across a period with the same
valence shell. b) Ionization energy.
2. a) The distance between the valence electrons and the nucleus increases as one goes down
a group but the core charge remains constant. b) Ionization energy.
3. The core charge on Be is smaller than the core charge on S.
ChemActivity 21
1. 7 is the number of valence electrons on F. 6 is the number of nonbonding electrons on F in
the molecule HF. 2 is the number of bonding electrons in HF.
2. The atom with the greater AVEE will have the partial negative charge.
3. The sum of the charge on a neutral molecule must be zero. Or, any electrons gained by F
must be lost from H.
4. 0.
5. Cl has a greater AVEE than H. Therefore, Cl will have a negative partial charge in HCl.
6. F has a greater AVEE than Cl. Therefore, F will have a more negative partial charge in HF
than Cl has in HCl. Therefore, H will have a more positive partial charge in HF.
7. F has a higher AVEE than C. Therefore, the F atoms will have a negative partial charge in
CF4. C has a higher AVEE than H. Therefore, the H atoms will have a positive partial
charge in CH4. In both cases the sum of the partial charge must equal zero.
8. N has a greater AVEE than H.
9. a) O has a greater AVEE than H. Therefore, O will have a negative charge in H2O.
b) S has a greater AVEE than H. Therefore, S will have a negative charge in H2S.
c) O and S both have greater AVEE than H. O has a greater AVEE than S. So, the
difference in AVEE between O and H is greater than the difference in AVEE between S
and H. So, the O atom in H2O will have a greater negative charge than the S atom in H2S.
10. a) zero. b) zero.
c) +1
d) –2
11. The sum of the partial charges on a molecule or ion must equal the charge on the molecule
or ion.
ChemActivity 22
1. F is more electronegative than Br. Both are more electronegative than H. Therefore, the
partial charge on F in HF should be more negative than the partial charge on Br in HBr.
Thus, the partial charge on H in HF should be more positive.
13
2. Electronegativities tend to increase as one moves from left to right across a period of the
periodic table.
3. Electronegativities tend to decrease as one moves down a group of the periodic table.
4. The trend in ionization energy is the same as the trend in electronegativity. This is
reasonable because they both are a measure of how tightly an atom holds its electrons.
5. The electronegativities of the two atoms are identical; they will share electrons equally.
6. a) Atoms with low electronegativities (metals) will tend to lose electrons. b) Atoms with
high electronegativities (nonmetals) will tend to gain electrons.
ChemActivity 23
1. Both H atoms have the same partial charge (+). The center of positive charge is midway
between the two H atoms.
2. a) The center of positive charge is at the center of the C nucleus. b) The center of the
negative charge is also at the center of the C nucleus. c) The distance between the center
of positive charge and the center of negative charge is zero.
3. O is more electronegative than S. Therefore the center of negative charge will not be at the
nucleus of the C atom. There will be a distance between the center of positive charge and
the center of negative charge,
4. a) Yes. b) The center of positive is located at the nucleus of the C atom. c) The center of
negative is located at the nucleus of the C atom. d) The distance between the center of
positive charge and the center of negative charge is zero.
5. a) The one with the larger d because u = q  d and the qs are the same in both cases. b)
The one with the larger q because u = q  d and the ds are the same in both cases..
6. a) Cl is bigger than F (more shells) b) HCl has the longer bond. c) F has the greater
partial charge. d) The partial charge appears to be a more important factor: the partial
charge can go from zero to one, whereas there is not as large of a variation (% or fraction)
in the distance.
7. The electronegativity difference.
ChemActivity 24
1. The first electron is relatively easy to remove. The second electron is much harder to
remove because it is being removed from what normally would be considered the core.
That is, the core charge for Na is +1. The core charge for Na+ is +9.
2. The first two electrons are relatively easy to remove. The third electron is much harder to
remove because it is being removed from what normally would be considered the core.
That is, the core charge for Mg is +2. The core charge for Mg2+ is +10.
3. The halogens can add one electron to the valence shell. An additional electron would be
added to the shell beyond the valence shell—much farther from the nucleus, with a “core
charge” that would be a negative number.
4. Na+ and Cl–. The alkali metals tend to lose one electron; the halogens tend to gain one
electron.
5. Na+ is the small ion and Cl– is the larger ion. Reasoning: Na+ is smaller than Ne; Cl– is
larger than Ar; Ar is larger than Ne.
6. a) +2. b) –2. c) MgO
7. a) Cl– b) NaF because the smaller the d value, the greater the force between the ions
(according to the Coulombic potential energy equation).
14
9. a) The Coulombic force of attraction is greater in NaF. b) We predict the melting point of
NaF must be higher than the melting point of NaCl because of the greater attraction in
NaF—993°C.
10. a) The Coulombic force of attraction is greater in MgO. b) We predict the melting point
of MgS must be lower than the melting point of MgO because of the smaller attraction in
MgS—about 2000°C.
11. The charge appears to be more important: the ±1 ions have melting points around 1000°C,
whereas the ±2 ions have melting points around 2000°C.
ChemActivity 25
1. Covalent.
2. Neither. Metals have different properties than nonmetals, so the bonding is expected to be
different, so it isn’t covalent. The bonding can’t be ionic because there is only one type of
atom present, so there is no basis for the formation of ions.
3. If the EN is high (>2) the element will tend to form a covalent compound. If the
electronegativity is low (<2) the element will tend to form a metallic compound.
4. Be has an electronegativity of 1.58 and B has an electronegativity of 2.05. Therefore, the
dividing line in the 2nd period will be between groups IIA and IIIA. Sb has an
electronegativity of 1.98 and Te has an electronegativity of 2.16. Therefore, the dividing
line in the 5th period will be between groups VA and VIA.
5. Yes because the ENs are low.
6. Metals have low EN values whereas nonmetals have high EN values. That is metals hold
their electrons loosely and nonmetals hold their electrons strongly.
ChemActivity 26
1. metallic bonding: ∆ENs are low and ENs are low.
2. ionic bonding: ∆ENs are high and ENs are high.
3. covalent bonding: ∆ENs are low__
and ENs are high.
4. a) ∆EN = (2.36 – 2.30)
__ = 0.06 EN = (2.36 + 2.30)/2 = 2.33
 covalent
b) ∆EN is low and EN is high:
bonding.
__
5. a) ∆EN = (2.21 – 1.76) =
0.45 EN = (2.21 + 1.76)/2 = 1.99
 as the examples of either covalent or metallic bonding and
b)
__∆EN is low but not as low
EN is lower than the examples of covalent bonds and greater than the examples of metallic
bonding. It appears to be perhaps somewhat intermediate between metallic and covalent, so
can’t really classify.
6. Points are OK.
__
7. The NaCl point falls
__ in section A. The Na point (EN = 0.87; ∆EN = 0) falls
__ in section B.
The CH4 point (EN = 2.42; ∆EN = 0.06) falls in section C. The Si point (EN = 1.92; ∆EN
= 0) falls in section SM.
8. A: ionic. B: metallic. C: covalent.__
9. ∆EN = (3.61 – 1.92) = 1.69
EN = (3.61 + 1.92)/2 = 2.78
The SiO2 point falls in
section C, covalent.
ChemActivity 27
1. No.
2. H2O(l) because the molecules are much closer in the liquid than in the gas.
3. Alkane: nonpolar. Ketone: polar. Alcohol: polar.
15
4. a) The boiling point increases as the molecular weight of the alkane increases.
b) The boiling point increases as the molecular weight of the ketone increases.
c) The boiling point increases as the molecular weight of the alcohol increases.
5. The intermolecular forces increase as the molecular weight increases.
6. Butane, acetone, and 1-propanol all have MW = 60 ±2 g/mole. The boiling points increase
as follows: butane < acetone < 1-propanol.
7. a) Pentane, 2-butanone, and 1-butanol all have MW = 72±2 g/mole. The boiling points
increase as follows: alkane < ketone < alcohol. Hexane, 2-pentanone, and 1-pentanol all
have MW = 87±2 g/mole. The boiling points increase as follows: alkane < ketone <
alcohol. b) For roughly equal MWs the boiling points increase as follows: alkane < ketone
< alcohol.
8. For roughly equal MWs the intermolecular forces increase as follows: alkane < ketone <
alcohol.
9. A dipole moment tends to increase the strength of the intermolecular forces; alkanes are
nonpolar and have the lowest boiling points (for a given MW).
10. No. The bond strength within the molecule can be very strong, but the intermolecular
forces depend on the interactions between molecules.
11. Intramolecular bonds refer to the bond within a molecule. Intermolecular forces refer to
the bonds or forces between molecules.
12. hydrogen bonding > dipole-dipole > induced dipole-induced dipole
13. a) dispersion. b) dispersion
14. a) dispersion, dipole-dipole. b) dipole-dipole
15. a) dispersion, dipole-dipole, hydrogen bonding. b) hydrogen bonding
16. The boiling point increases as the MW increases because the molecule has more electrons
and the dispersion forces increase.
17. The three compounds have about the same MW and the dispersion forces are about the
same. The ketone and the alcohol have dipole-dipole forces. Only the alcohol has
hydrogen bonding—which increases the intermolecular forces substantially.
ChemActivity 28
1. 12.
2. 48.
3. 12.
4. 48.
5. 6.022  1023
6. 2.409  1024
7. 6.022  1023
8. 2.409  1024
9. 16.04 amu
10. 16.04 g
11. The numerical value for the mass in amu of one molecule of a compound is equal to the
numerical value in grams for one mole of that compound.
ChemActivity 29
1. (1) Reactants: I2, H2. Products: HI. (2) Reactants: CO, O2. Products: CO2.
16
2. The arrow represents the transformation of the material on the back-side of the arrow to the
material on the front-side of the arrow.
3. a) H, two.
I, two.
b) H, two.
I, two
4. a) C, two.
O, four.
b) C, two.
O, four.
5. The number of atoms of each element must be identical on the reactant and product sides of
a balanced chemical equation.
6. (3)
Ag, one.
Cl, one.
(4)
Zn, one.
Cu, one.
(5)
Cl, three.
O, three.
(6)
Cr, two.
Mg, one.
7. a) (3) zero for each side.
(4) +2 for each side.
(5) –3 for each side.
(6) +6 for each side.
b) i) statement is incorrect; see eqns. (4), (5), (6).
ii) statement is incorrect; see eqns. (3) and (5).
iii) statement is incorrect; see eqns. (3), (4) and (6).
8. The sum of the charges must be identical on the reactant and product sides of a balanced
chemical equation.
9. Two.
10. a) Two.
b) One.
c) Four.
d) Ten.
11. Six.
12. For reaction (1): two reactant molecules; two product molecules.
For reaction (2): three reactant molecules; two product molecules.
13. a) No. b) The number of moles of gas decreases when reaction (2) occurs.
14. A given atom can be a component of many different molecules. It is the atoms that are
conserved, not the molecules.
15. No. 100 g of I2 must produce more than 100 g of HI because the 100 g of iodine will be
found in the HI and, in addition, hydrogen will be found in the HI as well.
16. (1) Calculate the moles of O2 that are consumed by dividing the mass X grams of O2 by
the MW of O2: 32.00 g/mole. (2) The moles of CO2 produced is equal to two times the
moles of O2 consumed. (3) The grams of CO2 produced is the moles of CO2 produced
multiplied by the MW of CO2.
ChemActivity 30
1. 100 chocolate bars; 200 marshmallows.
2. a) You can make 500 S'mores. b) You will have 500 graham crackers and 300 chocolate
bars left.
3. The limiting reagent is marshmallows.
4. a) A chocolate bar weighs the most. A gross of chocolate bars weighs more than a gross
of graham crackers or a gross of marshmallows. b) A marshmallow weighs the least.
5. You have 4 gross of graham crackers, 1 gross of chocolate bars, and 12 gross of
marshmallows. a) Most: marshmallows. b) Least: chocolate bars.
17
6. a) You run out of chocolate bars first. b) 1 gross. c) 3 gross of graham crackers and 10
gross of marshmallows. d) 27 pounds of graham crackers and 30 pounds of marshmallows.
e) 51 pounds.
7. G + Ch + 2 M  GChM2
8. We will run out of chocolate bars and we will not be able to use the remaining graham
crackers and marshmallows to make additional S'mores.
9. G + Ch + 2M 
2
wt of 1 gross GChM2 = [9.0 + 36.0 + 2(3.0)]lb = 51.0 lb
416 lb Ch 
1 gross Ch 1 gross GChM 2 51.0 lb GChM 2
x
x
 589 lb GChM 2
36.0 lb Ch
1 gross Ch
1 gross GChM 2
142 lb G 
1 gross G 1 gross GChM 2 51.0 lb GChM 2
x
x
 805 lb GChM 2
9.0 lb G
1 gross G
1 gross GChM 2
58.2 lb M 
1 gross M 1 gross GChM 2 51.0 lb GChM 2
x
x
 495 lb GChM 2
3.0 lb M
2 gross M
1 gross GChM 2
A maximum of 495 lb of S'mores can be made from the given starting materials. This
assumes none is eaten while being prepared.
ChemActivity 31
1. Easy to show.
2. Table : C6H6, 92.26% C, 7.74% H; C4H8, 85.63% C, 14.37% H; C8H16, 85.63% C,
14.37% H
3. Yes. % H = 100 % – 85.63%
4. No. Ethyne and benzene have identical percent compositions but different molecular
formulas.
5. They all have the same molecular ratios. For example, the H to C molar ratio is 1:1 in
C2H2 and C6H6.
6. The empirical formula.
7. ethyne and benzene: CH cyclobutane, 2-butene, and 1-octene: CH2
ChemActivity 32
1. The ionic compound in Figure 1 is NaCl because the charge of the anion is –1 (Cl–) not –2
(SO42–) andd there are equal numbers of positive and negative ions.
2. a) The cation is Na+. b) The anion is Cl–.
3. Oxygen, because the oxygen atoms have a negative partial charge and the cations are
positive.
4. Hydrogen, because the hydrogen atoms have a positive partial charge and the anions are
negative.
5. a) two b) one
18
1 mole Na2SO4
6. a) 10.0 g Na2SO4  142.05 g Na SO
2
4
b) 7.04 
10–2
= 7.04  10–2 mole Na2SO4
1 mole SO42–
mole Na2SO4  1 mole Na SO
2
4
= 7.04  10–2 mole SO42–
2 mole Na+
c) 7.04  10–2 mole Na2SO4 1 mole Na SO = 1.41  10–2 mole Na+
2
4
7. a) 0.282 M
b) 0.282 M c) 0.562 M
+
8. [Na ] in the NaCl solution = 1.71 M. [Na+] in the Na2SO4 solution = 1.66 M.
9. The 0.30 M Na2SO4 has a sodium ion concentration of 0.60 M, and it is more concentrated
with respect to sodium ions than 0.50 M NaCl.
10. CTQ 9 was easier to answer because the molarity is directly proportional to number of
sodium ions in solution. If grams are given, the molarity must be calculated from the
grams, MW, stoichiometry, and volume of solution.
1 mole glucose
11. a) 5.26 g glucose 180.16 g glucose = 2.92 – mole glucose
b) 2.92  10–2 mole glucose 
= 1.76


1022
glucose molecules
12. 5.84  10–2 M
ChemActivity 33
1. The volume increases.
2. The volume decreases.
3. The volume increases.
nRT
1
1
4. a) V= P . When n and T are constant, V= (nRT ) P = kB P
nRT
nR
b) V= P . When n and P are constant, V= P T = kC T
nRT
RT
c) V= P . When T and P are constant, V= P n = kA n
ChemActivity 34
1. Because 100 pm is the distance between the centers of the two nuclei.
2. Because each nucleus is attracted to the electron; there are two Coulombic terms of equal
magnitude.
3. The two nuclei repel each other (V >0). An electron and a proton are attracted to each other
V < 0).
19
4. Energy is required, 6.93  10–18 J, to separate the nuclei because they are attracted to the
electron.
5. Endothermic.
6. Positive.
7. H + H  H2
8. C(g) is the reference state for C. Therefore, no energy is released or required to form C(g)
from C(g); nothing happens.
9. Energy is always released in bond formation.
10. a) N2 has a triple bond and O2 has a double bond. b) Triple bonds are stronger than
double bonds, and more energy is released upon triple bond formation than upon double
bond formation.
11. Zero.
12. a) 3,324.18 kJ is released upon formation of two moles of methane from two moles of C(g)
and eight moles of H(g). b) H = 1.5 moles x -1662.09 kJ/mole = -2493.14 kJ or -2.5 x
103 kJ.
ChemActivity 35
1. 1,932.93 kJ is required.
2. Because the bonds in one mole of N2O4 are being broken to form 2 moles of N(g) and 4
moles of O(g), so energy must be provided.
3. Because the bonds in 2 moles of NO2 are being formed from 2 moles of N(g) and 4 moles
of O(g), 1875.72 kJ are released.
4. Energy is released.
5. a) Energy is required. b) Endothermic.
6. +57.21 kJ/mole
8. Obtain the energy required to break the bonds of the reacting molecules: multiply the
enthalpy of atom combination for H2(g) by –3 (3 moles and the negative sign because
bonds are being broken) and add –1 times the enthalpy of atom combination for N2(g).
Obtain the energy released when the bonds of the product molecules are formed: multiply
the enthalpy of atom combination for NH3(g) by 2. The ∆H is the sum of these two
energies.
9. Positive. More energy is required to break all of the bonds of the reactants than is released
when the product is formed, so the reaction is endothermic. (The length of the “up” arrow
will be greater than the length of the “down” arrow.)
20
10. a) HBr < HCl < HF because HBr has the longest bond length and HF has the shortest. b)
o
o
HF has the most negative H ac
and HBr has the least negative H ac
in this series. This is
consistent with part a). The stronger the bond, the more energy released upon bond
formation.
11. a) C–H should have a stronger bond than C–Cl because H is smaller than Cl. b) The


o
o
H ac
of CH4(g) is –1662.09 kJ/mole and the H ac
of CH3Cl(g) is –1572.15 kJ/mole (less
o
o
negative). c) Prediction: H ac
of CH3F(g) = –1600 kJ/mole (more negative than H ac

o
o
of CH3Cl(g) and less negative than H ac
of CH4(g)); H ac
CH3Br(g) = –1540 kJ/mole

o
(less negative than H ac
of CH3Cl(g)).


ChemActivity 36


1. M/s or moles/(liter s)

2. ∆ is change or {state or condition at t2} – {state or condition at t1}. ( ) is concentration in
M.
3. a) There is less reactant at t2 than at t1, so ∆(reactant) is negative. b) The negative sign
will make the rate of consumption a positive quantity.
4. rate of consumption of product = ∆(product)/∆time
5. 0.40 moles Cl– and 0.20 moles ClO3–
6. a) 3.0  10–3 mole ClO–/Ls b) 2.0  10–3 mole Cl–/Ls c) 1.0  10–3 mole ClO3–/Ls
7. 1.0  10–3 M/s
8. a) rate of reaction = rate of production of ClO3–
1
b) rate of reaction =
rate of consumption of ClO–
3
1
9. rate of reaction =
x (rate of production of product)
stoich
coef
P

10. The rate of reaction is equal to the rate of consumption of a reactant divided by the
stoichiometric coefficient of the reactant.

ChemActivity 37
1. You must break at least one of the double bonds.
2. Molecules will collide but no reaction will take place.
3. Some of the cis-2-butene will be converted to the trans-2-butene.
4. a) 2500 b) 2500 c) A molecules become B molecules
5. a) 705 b) 705 c) B molecules become A molecules
6. a) 0.25 b) 0.25 c) 0.25 d) 0.25
7. 3686(0.25) = 921
8. 714. After 21 seconds 714 A molecules react in the next second.
9. all are 0.10
10. 6314(0.10) = 631
11. 714. After 21 seconds 714 B molecules react in the next second.
12. a) 21 s.
b) Yes.
c) Yes.
13. kB = 0.10/s
21
14. a) rate of conversion of A to B = kA  number of A molecules b) kA = 0.25/s
ChemActivity 38
3686 A molecules mole
1
1.
= 0.0369 M
23
6.022 x10 molecules 1.661x10 19 L
2. a) (0.25/s)(0.10) = 2.5  10–2 M/s c) b) (0.25/s)(0.0286) = 7.1  10–3 M/s
3. (0.25/s)(0.0369) = 9.2  10–3 M/s
4. a) (0.10/s)(0) = 0
b) (0.10/s)(0.0714) = 7.1  10–3 M/s


5.
(0.10/s)(0.0631) = 0.63  10–2 M/s
6. At equilibrium, the forward rate is equal to the reverse rate; both rates are 7.1  10–3 M/s.
7. a) kinetic region, t < 15 sec; equilibrium region, t > 15 sec
b) In the kinetic region the concentration of the reactant decreases as a function of time.
8. t = 1, 0.33
t = 4, 1.1 t = 15, 2.5 t = 20, 2.6 t = 40, 2.6
9. Equilibrium region.
ChemActivity 39
1. (Y)o is the initial conc of Y in M. [Y] is the equil conc of Y.
2. forward rate = 0.20(67) = 13 M/s reverse rate = 0.40(33) = 13 M/s They are equal.
3. forward rate = 0.20(75) = 15 M/s reverse rate = 0.60(25) = 15 M/s They are equal.
4. At equilibrium, the forward rate is equal to the reverse rate.
5. a) If kY < kZ, then [Y] > [Z] (at equilibrium); see Sets M and P. b) If kY > kZ, then [Z] >
[Y] (at equilibrium); see Sets N and O.
6. a) kY [Y] = kZ [Z] , [Z]/[Y] = kY / kZ = a constant.
b) [Z]/[Y] = kY / kZ
7.
8.
9.
10.
Note that when kY < kZ, then [Y] > [Z], Also,
when kY > kZ, then [Z] > [Y].
H2O(g). Kc is a very large number, 1083, so the water concentration (the product in the
numerator) must be much larger than [H2]2 [O2] (the reactants in the denominator).
a) K = [Z]/[Y]
b) K = 33/67 = 0.49
c) [Y] + [Z] = 200
[Z] = 0.49 [Y]
1.49 [Y] = 200
[Y] = 134 [Z] = 66
[PCl5]
[PCl3] [Cl2]
a) Kc = [PCl ] [Cl ]
b) Kc =
c) Reaction (2) is the reverse of
[PCl5]
3
2
reaction (1).
d) The equilibrium expression of reaction (2) is the inverse of the
equilibrium expression of reaction (1).
1.00 x 10–3
ChemActivity 40
1. [PCl3] = 0.0500 mole/5.00 L = 1.00  10–2 M
2. (0.200/5)/{(0.0200/5)(0.0500/5)} = 1.00  103
3. a) 0.1100 mole PCl3
b) 2.2  10–2 M c) No. One of the concentrations has changed
so the ratio cannot be the same, 1.00  10–3
22
[PCl5]
3
[PCl3] [Cl2] is < 1.00  10 . PCl3 and Cl2 must react to reduce the
denominator and increase the numerator.
(0.200/5)/{(0.0200/5)(0.110/5)} = 4.55  102
PCl3 and Cl2 must react to reduce the denominator and increase the numerator; the ratio
must be 1.00  103 at equilibrium, So, b) describes what will happen.
a) QC must get smaller.
b) The concentrations of the products must decrease and the
concentrations of the reactants must increase.
The reaction quotient (1) indicates if the reaction is at equilibrium or not, and (2) indicates
which direction the reaction must go to reach equilibrium.
a) 0.0200 b) 0.200
a) x moles of Cl2 must react. b) x moles of PCl5 are formed.
4. The ratio
5.
6.
7.
8.
9.
10.
10. c) through 13.
initial moles
change in moles
equilibrium moles
equilibrium conc
equilibrium conc
value
PCl3
Cl2
PCl5
0.1100
0.0200
0.200
–x
–x
x
0.1100–x
0.0200–x
0.200+x
(0.1100–x)/5 (0.0200–x)/5 (0.200+x)/5
0.0201
2.1010–3
4.1910–2
14. (4.1910–2)/{(0.0201)(2.1010–3)} = 993 which is about 1.00 x 103.
15. Qc is too small; PCl3 and Cl2 will react to form PCl5.
16. Qc is too large; PCl5 will react to form Cl2 and PCl3.
ChemActivity 41
1. 0.00963 g Mg(OH)2 
mole Mg(OH) 2
= 1.65  10–4 mole Mg(OH)2
58.32g Mg(OH) 2
2. a) There are 8.26  10–4 moles of Mg2+ in 10.0 L;
8.26 x10 4 moles Mg 2
2+
[Mg ] =
= 8.26  10-5 M
10.0 L

3.
4.
5.
6.
7.
b) There are 2  8.26 x 10–4 moles of OH– in 10.0 L.
0.19260 g – 0.09630 g = 0.09630 g dissolve
0.09700 g –0.00070 g = 0.09630 g dissolve
1.65  10–3 mole of Mg(OH)2
a) 1.65  10–4 mole
b) 0.009630 g
–4
–4 2
(1.65  10 ) (3.30  10 ) = 1.80 x 10–11
8.
Mg2+
9.
OH–
23
10.
1.65 x10 4 moles
For Mg2 :
= 2.90 x 10-5 moles
6
3.30 x10 4 moles
–
For OH :
= 2.90 x 10-5 moles
12
+

11. Yes. According to Table 1 no more than 1.65  10-3 moles of Mg(OH)2 can dissolve in
10.0
L of water.
12. The solubility of Mg(OH)2 is 1.65  10 –3 M; this is greater than 1  10 –3 M but less than
0.1 M. Therefore, Mg(OH)2 is moderately soluble.
13.
14.
Ksp = (1.65  10 –3 ) (2 x 1.65  10 –3 )2 = 1.80 x 10–11
a) 0.050 moles of Mg2+ b) 0.060 moles OH– c) 1000 mL or 1.000 L
d) 0.050 M
2+
–2
e) 0.060 M f) Q = (Mg )(OH ) = 1.8  10–4
g) No. Q is too large; the reaction
will shift toward more solid Mg(OH)2. h) Yes.
15. a) Ksp = [Ag+] [Cl–]
b) Ksp = [Cu2+]3 [PO43–]2
ChemActivity 42
1. a) (1) HCl (2) none (3) HCN
b) (1) none (2) NH3 (3) none
c) (1) HCl (2) H2O (3) HCN
d) (1) H2O (2) NH3 (3) H2O
2. H2O reacts as an acid in (2); H2O reacts as a base in (1). I used the Bronsted-Lowry
definition.
3. HCl, Kc is much larger—the reaction proceeds further to the right.
4. a) Cl– b) As a base. c) H3O+ d) As an acid. e) In (2), NH3 is a base and NH4+
is an acid. In (3), HCN is an acid and CN– is a base. When an acid loses a proton, the
species that remains is a base.
5. When neutral H2S loses a proton the remaining portion will be charged –1, HS–.
H2S(aq) + H2O = H3O+(aq) + HS–(aq)
6. CO32– (aq) + H2O =
7. a) NH4+
b) OH–
OH–(aq) + HCO3– (aq)
ChemActivity 43
1. Kc = Ka/55 = 1.75  10–5/55 = 3.2  10–7
2. Ka = (0.26) (0.26)/(1.33) = 5.2  10–4
3. a) HI because it has the largest value for Ka.
b) H2S because it has the smallest value for Ka.
[H 3O] [NO3 – ]
4. Ka =
HNO3
5. a) 1.00 - 0.967 = 0.033
b) Ka = (0.967)2/0.033 = 28
6. HI > HBr > HOClO3 > HCl > H2SO4 > HNO3
7. a) [H3O+] = [NO2– ] = 1.00 x 0.023 = 0.023 M
24
[HNO2] = 1.00 - 0.023 = 0.98 M
b)
The larger the value of the % dissociation the larger the value of Ka.
8. When Ka > 1 the % dissociation values are near or equal to 100%. When Ka < 1 the %
dissociation values less than 10%—some are very close to zero.
9. Acid 1 is a strong acid because only A— ions are found in the solution (no HA molecules).
10. Acid 2 is a weak acid because there are many more HA molecules than A— ions.
11. In Table 2, the six acids with Ka > 1 are strong acids. All of the other acids are weak acids.
12. a) HI
b) HNO3
13. a) H2S b) H3PO4
14. 0 M HOCl (pure water) is neutral. The other three solutions are acidic.
15. HOCl is an acid because in its solutions the [H3O+] > [OH–].
16. [H3O+]  [OH–] is a constant; 1  10–14 .
17. Water is neither acidic or basic because [H3O+] = [OH–]. Pure water is neutral.
[H3O+] [OH–]
[H3O+] [OH–]
18. a) Kc = [H O][H O]
Ka =
[H2O]
2
2
b) [H3O+]  [OH–] = 1.8  10–16  55 = 9.9  10–15 c) good agreement
19. The comparison is good.
20. [H3O+] = [OH–] = 1.0  10–7
ChemActivity 44
1. a) HCl is a strong acid—100% dissociation.
b) HF is a weak acid—less than 10% dissociation.
2. a) Solution A represents HCl and solution B represents HF.
b) H3O+, Cl–, HF, F–
3. The hydroxide ion is not shown because its concentration would be extremely small in
these acidic solutions.
4. HOCl(aq) + H2O(l) = H3O+(aq) + OCl–(aq) Ka = [H3O+][OCl–]/[HOCl]
5,6.
H3O+
OCl–
HOCl
initial moles
0.30
0
0
change in moles
–x
x
x
equilibrium moles
0.30–x
x
x
equilibrium conc
(0.30–x)/1
x/1
x/1
–5
equilibrium conc value
0.30
9.3  10
9.3  10–5
Ka = (9.3  10–5)2/0.30 = 2.9  10–8
7. [HOCl] = 0.30 M [OCl–] = 9.3  10–5 M
8. Very little HOCl reacts. To two significant figures, the concentrations before and after
reaction are the same.
9. One molecule of H3O+ and one molecule of OCl– are produced for every molecule of
HOCl that reacts.
25
10. B + H2O = BH+ + OH–
11. Kw  1  10–14 Yes, this value agrees with the previously calculated value.
12. C5H5N(aq) + H2O = C5H5NH+(aq) + OH–(aq) Kb = [C5H5NH+] [OH–]/[C5H5N]
13,14.
15.
16.
17.
18.
19.
20.
21.
22.
C5H5N
OH–
C5H5NH+
initial moles
0.30
0
0
change in moles
–x
x
x
equilibrium moles
0.30–x
x
x
equilibrium conc
(0.30–x)/1
x/1
x/1
–5
equilibrium conc value
0.30
2.3  10
2.3  10–5
Kb = (2.3  10–5)2/0.30 = 1.8  10–9
[C5H5N] = 0.30 M [C5H5NH+] = 2.3  10–5 M
Very little C5H5N reacts. To two significant figures, the concentrations before and after
reaction are the same.
One molecule of C5H5NH+ and one molecule of OH– are produced for every molecule of
C5H5N that reacts.
HA(aq) + H2O(l) = H3O+(aq) + A–(aq) One molecule of H3O+ and one molecule of A–
are produced for every molecule of HA that reacts.
HA is a weak acid. Very little HA reacts.
(CH3COOH)o = 0.50 M [CH3COOH] = 0.50 M [H3O+] = 3.0  10–3 M Ka = 1.8  10–5
Ka = 1.9  10–5 using equation (3) and using equation (4).
[BH+] [OH–]
x2
–
[OH–]  [BH+] [B]  (B)o Kb =
=
[B]
(B)o–x where x = [OH ]
x2
Kb  (B)
o
ChemActivity 45
1. –log(5.0  10–4) = 3.30
2. a) HCl is a strong acid (100% dissociation). Therefore, the chloride ion concentration will
equal the hydronium ion concentration. Add eight chloride ions to the Model.
b) 5.0 x 10-5 moles
c) [OH–] = 2.0 x 10-10 M so there are so few OH– ions present that
only a tiny fraction of a symbol would be needed.
3. a) 7.0 b) 7.0
4. a) acidic solution, pH < 7.0 b) basic solution, pH > 7.0
5. pKw = 14.00
6. pKw = pOH + pH
[A–]
7. a) Ka = [H3O+][A–]/[HA]
b) –log(Ka) = –log(H3O+) – log [HA]
[A–]
[A–]
[HA]
pKa = pH – log [HA]
pH = pKa + log[HA]
pH = pKa – log –
[A ]
26
ChemActivity 46
1. S–H
2. The H–S bond is weaker than the O–H bond because the H–S bond is longer than the O–H
bond (S is bigger than O).
3. Because the H–X bond length increases in the series HF, HCl, HBr, HI.
4. H2S because the S–H bond is weaker; it ought to be easier to remove a proton.
5. The P–H bond is weaker (longer) than the N–H bond (shorter). Therefore, PH4+ should be
a stronger acid than NH4+.
6. The second statement best describes relative acid strengths.
7. Zero.
8. a) The H atom attached to the O atom (it has the most positive partial charge). b) The H
atom attached to the O atom (it has the most positive partial charge). c) In each case the
hydrogen atom with the most positive partial charge is the most acidic.
9. CCl3COOH is most likely the stronger acid because its acidic hydrogen atom is more
positively charged than the acidic hydrogen atom in CH3COOH.
10. The acidic hydrogen atom in HCOOH is the H atom bonded to the O atom.
11. The acidic hydrogen atom in CH3CH2OH is the H atom bonded to the O atom.
12. In both CH3COOH and CCl3COOH the hydrogen atom with the most positive partial
charge (the most acidic hydrogen atom) is the hydrogen atom bonded to the oxygen atom.
13. Because CCl3COOH has three very electronegative Cl atoms which are drawing electrons
from the O–H bond.
14. The electronegativities decrease down a group: Cl > Br > I. The greater the
electronegativity of X, the more the electrons will be pulled toward X and, therefore, the H
atoms will have a more positive partial charge.
15. The results in Table 2 are consistent with the first statement. For example: the partial
charge on the H atom in HOCl is larger than the partial charge on the H atom in HOI and
HOCl has a larger Ka than HOI.
16. Prediction: Ka of CF3COOH = 10. F is more electronegative than Cl. The Ka of
CF3COOH should be larger than the Ka of CCl3COOH (0.22)
17. a) In Table 1, the H atom is always attached to a different atom. b) In Table 2, the H atom
is always attached to the same atom, O. c) In Table 1 the H–Q bond strengths are similar.
In Table 2 the H–O bond strengths are essentially the same.
18. a) In order of increasing partial charge on the H atom i) H2S < H2O ii) PH4+ < NH4+
iii) HI < HBr < HCl < HF. b) If partial positive charge were the most important factor,
H2O would be a stronger acid than H2S, NH4+ would be a stronger acid than PH4+, and HF
would be a stronger acid than HI. This is not consistent with the known acid strengths
given in Table 1. c) Bond strength is more important. Note that the partial charge on the
H atom in H2O is greater than the partial charge on the H atom in H2S, but H2S is the
stronger acid.
ChemActivity 47
1.
27
Acid
Ka
HF
hydrofluoric acid
HONO
nitrous acid
[ H 3O + ] [ONO – ]
[ HONO]
+
NH4
+
[H 3O ] [NH 3 ]
+
[NH 4 ]
Conjugate Base
F–
fluoride ion
ONO–
NH3
ammonia
Ka  Kb
Kb
[ H 3O + ] [OH– ]
[OH – ] [HONO ] [ H 3O + ] OH– 
[ONO– ]
–

[OH ] [NH 4 ]
[NH 3 ]
2. HF + H2O = H3O+ + F–
Ka = [H3O+][F–]/[HF]
–]/[HONO]
HONO + H2O
= H3O+ + NO2– Ka = [H3O+][NO2
NH4+ + H2O = H3O+ + NH3
Ka = [H3O+][NH3]/[NH4+]
NO2– + H2O = HNO2 + OH–
Kb = [OH–][HNO2]/[NO2–]
NH3 + H2O = NH4+ + OH–
Kb = [OH–][NH4+]/[NH3]
3. a) All expressions for Ka have [H3O+]  [base]/[acid]. b) All expressions for Kb have
[OH–]  [acid]/[base]. c) Ka  Kb = [OH–]  [H3O+].
4. Ka  Kb = Kw
5. Kb = Kw /Ka
6. a) HA b) X– c) The stronger the acid, the weaker its conjugate base.
7. potential acids: Al3+, CH3NH3+, NH4+. potential bases: F–- both: HPO42-, H2O
8.



–
[H3O+][OH ]
9. HCl is a strong acid, its conjugate base is so weak that it does not affect the pH of the
solution.
10. pH = 7.00. The Na+ ion is not an acid; the Br– is not a base (conjugate base of the strong
acid HBr).
11. The conjugate bases of strong acids are extremely weak bases—so weak that they do not
affect the pH of a solution.
ChemActivity 48
1. a) Zn(s) loses electrons. b) Cu 2+(aq) gains electrons
2. a) two electrons
b) two moles of electrons
3. a) K = [Zn2+]/[Cu2+]. b) Essentially all of the Cu2+ ions have been plated out, [Cu2+]
0 . The [Zn]  1.0 M.
c) K is a very large number, K>>1.
4. a) Zn is oxidized.
b) Cu2+ is reduced.
5. a) Cu2+ is the oxidizing agent. b) Zn is the reducing agent.
28
6. Zn + Cu2+ = Cu + Zn2+
Ox agent = Cu2+ Red agent = Zn
+
2+
Zn + 2 K = 2 K + Zn
Ox agent = K+ Red agent = Zn
Co + Ni2+ = Ni + Co2+
Ox agent = Ni2+ Red agent = Co
Co + Cu2+ = Cu + Co2+
Ox agent = Cu2+ Red agent = Co
3 Co + 2 Cr3+ = 2 Cr + 3 Co2+
Ox agent = Ni2+ Red agent = Co
7. K>1
K<1 K>1 K>1 K<1
2+
8. Cu , because Cu2+ oxidizes Zn and K+ does not oxidize Zn.
9. Cu2+ > Ni2+ > Cr3+
10. Cu2+ > K+ . In order to rank Ni2+ and Cr3+, it would be necessary to test Zn/Ni2+ and
Zn/Cr3+ or Co/K+.
ChemActivity 49
1. The partial charge. Formal charge and oxidation state are merely bookkeeping methods.
Partial charge is a real attempt to accurately describe the charge on an atom in a molecule.
2. MgO; NaF; NiCl2. The ionic compounds.
3. Left-hand side: Cr(+3), Mn(+7), O(-2) H(+1) O(-2). Right-hand side: Cr(+6), O(–2),
Mn(+2), H(+1). Yes, Cr increases its oxidation number from +3 to +6.
4. If the oxidation number is increased, that species is oxidized. It is the reducing agent.
5. Cr3+ is oxidized. MnO4– is reduced.
6. Cu+ is oxidized. Cu+ is reduced.
7. Determine the oxidation numbers for all atoms in the chemical equation. If there is any
atom that undergoes a change in oxidation number, the reaction is an oxidation-reduction
reaction.
ChemActivity 50
1. Cu2+ + 2 e– = Cu (more Cu is produced)
2. Zn (anode), Cu (cathode)
3. The electrons flow to the Cu electrode from the Zn electrode.
4. Zn is the negative electrode.
5. a) Zn is oxidized. b) Cu2+ is reduced.
6. The electrons cannot flow and the chemical reaction stops.
7. A flow of electrons can be made to do work: light a bulb, heat a room, run a TV, etc.
8. The voltaic cell can (1) produce a flow of electrons that can do work and (2) can be stopped
and started by throwing a switch.
9. Cu2+ + 2 e– = Cu
10. Zn = Zn2+ + 2 e–
11. Cu2+(aq)
12. Co = Co2+ + 2 e–
13. Cu2+ + 2 e– = Cu
14. Cu2+(aq)
15. Co2+ has a stronger pulling power than Zn2+ because the voltage of the Cu2+ /Zn2+ cell is
larger than the voltage of the Cu2+ /Co2+ cell and the voltage is a measure of the difference
in electron-pulling power. Both Co2+ and Zn2+ lose out to Cu2+ but the Zn2+ loses by
more.
29
16.
17.
18.
19.
Cu2+ is the species that gets reduced so it is the stronger oxidizing agent.
0.34 V
0.34 V
Au+, it has the greater standard electron potential.
ChemActivity 51
1. a) Cu/Cu2+ b) Cu2+ + 2 e– = Cu c) Zn = Zn2+ + 2 e–
d) Eo = 0.34 V – (– 0.76 V) = 1.10 V
2. Cl2 (1.36 V); Br2 (1.09 V); Ag+ (0.80 V); Cu2+ (0.34 V); H+ (0.00V); Zn2+ (–0.76V); K+
(–2.92 V).
3. a) Strongest oxidizing agent, Cl2 (most favorable reduction potential). b) Weakest
oxidizing agent, K+ (least favorable reduction potential).
4. a) K is the strongest reducing agent. b) Cl– is the weakest reducing agent.
ChemActivity 52
1. a) The ball at the bottom of the hill is the lower energy state.
b) The change in the potential energy is negative.
c) There is tendency for processes to go from the higher energy state to the lower energy
state. Going up the hill would require an input of energy from somewhere!
2. a) 1 mole NaCl. b) ∆H is negative. c) NaCl is the lower energy state. Breaking the
ionic compound apart into its constituent atoms would require an input of energy from
somewhere!
3. H2O(s) = H2O(l) ∆H > 0 Occurs at T > 0C
4. H2O(l) = H2O(s) ∆H < 0 Occurs at T < 0C
5. No, you must know the temperature as well.
6. The temperature must be known.
7. (c)
8. (c)
9. Sc > Sa. ∆S is positive for this naturally occurring process.
10. Sc > Sr. ∆S is negative for this naturally occurring process.
11. ∆S is positive because the number of particles increases and the amount of disorder
increases.
12. ∆S is positive because the volume in which the particles can move increases and the
amount of disorder increases.
13. ∆S is positive because the temperature increases (and the volume in which the particles can
move increases) and the amount of disorder increases.
14. H2O(s) < H2O(l) < H2O(g)
ChemActivity 53
1. No.
2. ∆S is positive for the melting of ice. Water has more disorder than ice.
3. Yes.
4. ∆S is negative for the freezing of water. Water has more disorder than ice.
5. a) When ice melts, H > 0 and S > 0 (row 4).
b) When water freezes, H < 0 and S < 0 (row 1).
30
c) There is agreement with these observations with the data in Model 2: ice melts at high
temperatures (higher than 0°C) and water freezes at low temperatures (lower than 0°C).
6. In row 1, the process is exothermic and the process will occur naturally at low
temperatures. In row 2, the process is exothermic and the process occurs naturally at any
temperature.
7. a) Chemical reactions generally occur naturally when S > 0 (S is positive).
b) When S > 0 the process is going from a lower entropy state to a higher entropy state.
8. Positive ∆S, high temperature (according to Model 2).
ChemActivity 54
1. ∆S is negative, fewer gaseous particles on the right.
2. The reference state for elements is gaseous, monatomic atoms.
3. The number of particles decreases.
4. The number of particles decreases. In addition, the products could be liquids or solids –
which have lower entropies than gases.
5. As the number of atoms in the molecule increases, the decrease in the number of particles
becomes greater.
6. Liquids are more ordered than gases.
7. ∆S = +646.53 kJ (The negative of the entropy of formation of N2O4(g).
8. a) –235.35 kJ
b) –470.70 kJ
9. Positive, more gaseous particles on the right.
10. a) S° is a positive number because 2 moles of N(g) and 4 moles of O(g) is more
disordered than 1 mole of N2O4(g). b) The magnitude is equal to the magnitude of the
entropy of formation of N2O4(g).
11. a) S° is a negative number because 2 moles of N(g) and 4 moles of O(g) is more
disordered than 2 moles of NO2(g).
b) The magnitude is equal to two times the
magnitude of the entropy of formations of NO2(g).
12. ∆S = +175.83 J/K
31
14. The ∆S for the reaction is equal to the sum of the following: two times the entropy of atom
combination of XB(g), the entropy of atom combination of B2(g), the negative of the
entropy of atom combination of A2X2(g), the negative of the entropy of atom combination
of B2(g).
ChemActivity 55
1. a) N, N, Y, N, Y
b) Y, Y, N, Y, N
2. No.
3. No.
4. The first 3; the last 2.
5. ∆H° – T∆S°
6. Negative.
7. Positive.
8. The more negative the value of ∆H° – T∆S°, the larger the magnitude of K.
9. Negative.
10. Positive.
11. ∆G° = 0; K = 1
ChemActivity 56
1. d.
2. 2.48 kJ/mole
J
kJ
3. 8.314 mole K  298.15 K  1000J = 2.479 kJ/mole
∆H°
∆S°
4. a) ln K = – RT + R
b) K increases.
c) K decreases.
32
5. The reaction involves separating molecules, so it is endothermic and ∆H° > 0. Thus, as T
increases, K increases.
6. Cu2+(aq) + Zn(s)
Cu(s) + Zn2+(aq)
Cu2+(aq) + H2(g)
Cu(s) + 2H+(aq)
Br2(aq) + Zn(s)
2Br–(aq) + Zn2+(aq)
Zn2+(aq) + 2K(s)
Zn(s) + 2K+(aq)
Cl2(aq) + 2Ag(s)
2Cl–(aq) + 2Ag+(aq)
7. d.
8. 1.28  10–2 V = 1.28  10–2 J/C
J 

8.314 mole °K (298.15 K)


9.
= 1.285 10–2 J/C
C 

(2) 96485 mole


10.
–nF° = G°
ChemActivity 57
1. 1.35 10–7 M/sec
2. 1.80  10–8 M/sec
3. Rate decreases as (NO2–) decreases.
4. Much less than 1.80  10–8 M/S; close to zero. Rate of reaction continues to get much
slower as (NO2–) approaches zero.
5. a) Exp. 3.
b) Exp. 1.
c) Because the initial concentrations of reactants are different.
6. a) Yes.
0.010
b) No. 0.0050 = 2
2.70 x10 7
c) No.
= 2
1.35 x10 7
d) No. Because the initial concentrations are identical.
e) Yes. Because they have different initial concentrations.
f) Yes. 1

0.200
7. a) No. 0.100 = 2
b) Yes.
5.40 x10 7
c) No.
= 2
2.70 x10 7
d) Yes. Because they have different initial concentrations.
e) No. Because the initial concentrations are identical.
f) Yes. 1
8. First order.
9. First order.
10. a) All three experiments yield k = 2.7  10–4 M–1 sec–1 .
33
b) The rate constant is a characteristic for a given reaction at a given temperature. It
should have the same value for all experiments for this reaction at 25°C.
11. No.
12. a) is not appropriate, as shown by Table 2.
b) is the only appropriate method.
ChemActivity 58
1. As t increases, kt increases. For the first order rate law, this results in a larger quantity
being subtracted from ln (R)o, so that ln (R) becomes smaller, so (R) decreases. For the
1
1
second order rate law, this results in (R) + kt increasing, so that (R) increases. Thus, (R)
o
decreases.
2. a) slope = –k b) intercept = ln (R)o
1
3. a) slope = k b) intercept =
(R)o
4. a) 0.010 M
b) 0.005 M; 1/2
c) 0.0025 M; 1/2
d) 0.00125 M; 1/2
5. 0.008 M, 0.004 M; Yes, because after 130 seconds, concentration is reduced by 1/2.
6. No. Regardless of the concentration, it always takes the same amount of time to reach half
of that concentration, so t1/2 is constant.
1
7. a) (R) = 2 (R) o
(R)o
ln 2 – ln (R)o
ln (R)o – ln (2) – ln (R)o
ln 2
b) t1/2 =
=
= k
–k
–k
1
8. a) (R) = 2 (R) o
1
1
2
1
–
–
(R) (R)o
(R)o (R)o
1
b) t1/2 =
=
=
k
k
k(R)o
ChemActivity 59
1. ii.
2. a) Negative.
b) Favorable.
3. ii.
4. a) Positive.
b) Favorable.
5. a) Negative.
b) Favorable.
c) Yes.
6. The balls cannot jump over the barrier on their own. Either the barrier must be removed
temporarily, or someone must reach in and move the balls.
34
7. A violent collision is more likely to break the bonds because there is more kinetic energy
available to use.
8. High temperature.
9. No, because the two Br atoms are too far apart to form a bond and become a Br2 molecule.
10. ONBr BrNO
11. The collision may not be strong enough, and the molecules may not be oriented properly.
12. a) The vertical line should extend between the two parallel lines: one line of the parallel
lines is the horizontal line representing H[ONBr(g) + ONBr(g)] and the other line of the
parallel lines is the horizontal line representing H [NO(g) + NO(g) + Br2(g)]. {You must
extend one of the lines above to create the parallel lines.}
b) H is negative because the products are at a lower enthalpy than the reactants: Hp – Hr
< 0.
c) The reaction is exothermic because the products are at a lower enthalpy than the
reactants.
13. ON--Br--Br--NO
14. Ea for the forward reaction can be indicated by drawing a vertical line between H[ONBr(g)
+ ONBr(g)] and the horizontal line representing H[ON–Br–Br–NO(g)].
15. Ea for the reverse reaction can be indicated by drawing a vertical line between H[NO(g) +
NO(g) + Br2(g)] and the horizontal line representing H[ON–Br–Br–NO(g)].
16. Ea (forward) – Ea (reverse) = H
ChemActivity 60
1. When the concentration of hydroxide ions was doubled, the rate of the reaction did not
change.
2. Unimolecular: 1
Bimolecular: 2, 3
3. The rate of a forward step is proportional to the concentration of each species involved in
that step.
4. Because the rate law cannot be determined from the stoichiometry of the reaction.
6. Because step 1 is the slowest step in the proposed mechanism.
7. Yes.
8. Unimolecular: none
Bimolecular: 1, 2
9. Yes.
10. Because this is the slowest step of the proposed mechanism.
11. Yes.
13. Thermodynamic control.
14. Kinetic control.
15. c.
ChemActivity 61
1. No effect on stoichiometry. The catalyst only increases the rate of reaction but is not
produced or consumed.
2. No effect on ∆H°. There is no change in the enthalpy of reactants or products.
3. The rate-limiting step in a mechanism involving a catalyst is faster than the rate-limiting
step in the mechanism without the catalyst. This must be true because the overall rate of
35
4.
5.
6.
7.
8.
9.
reaction with the catalyst is faster, and the overall rate is controlled by the rate-limiting
step.
The catalyst decreases the activation energy.
On each side there are: 2 N atoms; 4 O atoms; 4 C atoms; 10 H atoms.
O-H is attached to C atom; H is attached to N atom.
a) This reaction is not a redox reaction—no atom undergoes a change in oxidation state.
b) The reaction is not an acid/base reaction—a proton is not donated from one species to
another species.
The uncatalyzed reaction is subject to kinetic control because the reaction is
thermodynamically favorable yet the reaction is extremely slow.
Co2+ is a positively charged species, so it attracts electrons.
ChemActivity 62
Ea
1. slope = – RT ; intercept = ln A
2. Reaction W will have the larger rate constant because it has a smaller activation energy.
Ea
According to equation (1), if RT is smaller, then ln k will be larger, so that k will be
larger.
Ea
3. The rate constant increases because RT gets smaller as T gets larger, and so a smaller
value is subtracted from ln A.
4. As temperature increases, molecules move faster so that collisions are more frequent and
more violent, so effective collisions occur more frequently.
5. 2.