Answers to Exercises and Problemsfor

CHEMISTRYA Guided Inquiry

Fifth Edition, 2011

Richard S. MoogFranklin & Marshall College

John J. FarrellFranklin & Marshall College

Latest Update: June 1, 2011

John Wiley & Sons, Inc.

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Answers to Exercises and ProblemsThese answers may be given to students.

ChemActivity 11. A = 31, no. of e– = 15. Z = 8, A = 18. 39K+. Z = 28, no. of e– = 26.2. 1.674 10–24 g. 1.993 10–23 g. 3. 8.67 10–17 g. 4. 12.00 g. 5. 7.305 10–23 g. 6. a) sum of protons and neutrons in the nucleus. b) number of protons in the nucleus.7. False. 18O has 8 protons and 10 neutrons.8. 12, 12, 12. 10, 11, 12. 17, 17, 18. 18, 17, 18. 23, 26, 30. 7, 7, 8. 10, 8, 8. 10, 13, 14.9. 59Co2+ . Z = 7, A = 14, no. of e– = 7. 7Li . Z = 30, A = 58, no. of e– = 28. Z = 9, A =

19, no. of e– = 10.10. 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.

Problems1. 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 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.) 3. 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 21. a) 1.008 g. b) 39.10 g.2. a) 45.98 g. b) 57.27 g.4. 37Cl has two more neutrons in its nucleus.5. average mass of a marble = = 5.00 g + 7.00 g =

0.2500 5.00 g + 0.7500 7.00 g = 6.50 g (this is eqn (2))6. 10.44 amu7. 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.

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b) 1 K atom 39.098 g K/mole K atoms = 6.493 10–23 g. of K

10. 60.06 g11. 3.613 1024 atoms.12. 2.619 1024 atoms.13. a) 3.029 1025 atoms. b) 1.022 1019 atoms. c) 1.878 1025 atoms. d) 9.782 1027

atoms.14. Phosphorus15. 89.5 g I

Problems1. (0.905)(19.9924 amu) + (0.095)(x amu) = 20.179 amu. Solve for x: x = 22.0 amu. Thus,

most likely 22Ne.2. a) False. 6.941 g per mole of Li atoms. b) False. No H atoms weighs 1.008 amu; this is

an average atomic mass. c) True. Na atoms are more massive, so fewer atoms needed for same total mass. D) False. This would imply that an atom has a mass of 0.045 amu which is not possible as it is less than the mass of a proton.

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.

4. Re-187

ChemActivity 31. 5.47 10–18 J.2. a) IEa = –(2)(–1)/d1 = 2/d1 b) IEb = –(1)(–1)/2d1 = 1/2d1 IEa > IEb

3. The ionization energy of case (a) is larger, 1.20 k/d1, than that of case (b), 1.17 k/d1.

Problems1. large and negative

2. V =

ChemActivity 41. 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.2. Open ended. All three electrons at a farther distance (than 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.

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ChemActivity 51. a) 4. b) 6. c) 5. d) 8.2. a) +4. b) +6. c) +5. d) +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. 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.

Problems1. a) True. Br is a group VII element. The number of valence electrons is seven. b) True. The first ionization energy increases as one moves from left to right across a

period and as one moves up a group.2. If the fourth electron in Be were added to a third shell, it would be easier to remove and the

IE would be less than the IE of Li.

ChemActivity 61. 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 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. A whole shell has been added for Ar. 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–

Problems1. Na. The second electron removed would be a core electron.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. Therefore, S2– is larger than Mg2+.

ChemActivity 71. False. The shorter the wavelength, the greater the frequency.2.

Energy (J) Wavelength (m) Frequency (s–1) Region of Spectrum9.94 10–20 2.00 10–6 1.50 1014 infrared

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3.97 10–19 0.500 10–6 6.00 1014 visible9.94 10–19 2.00 10–7 1.50 1015 ultraviolet1.99 10–16 1.00 10–9 3.00 1017 X-ray

3. A blue photon is more energetic. The energy is inversely proportional to the wavelength.

Problem1. 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 81. 140.3 MJ/mole2. a)

3.

ChemActivity 9

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1. a) Two. b) Lower energy peak (1s) is 2 x the intensity of the higher energy peak (2s). c) The nuclear charge for H, He, and Li is 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.

Problems1. 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. b) True. Both have 17 electrons and 17 protons.

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 101. 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 electron in its nucleus.

ChemActivity 111. Kr2. C<Ne<Zn<Ba<Gd<Pt

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4. P: [Ne] 3s2 3p3 P3–: [Ar] Ba: [Xe] 6s2 Ba2+: [Xe] S: [Ne] 3s2 3p4

S2–: [Ar] Ni: [Ar] 4s2 3d8 Zn: [Ar] 4s2 3d10 5. three

Problems1. 5d2. Pd [Kr] 5s2 4d8. Pd2+ [Kr] 4d8. Gd either [Xe] 6s2 5d1 4f7 or [Xe] 6s2 4f8 is

correct (depending on how you use the periodic table to determine electron configurations; experimentally, we find the configuration to be [Xe] 6s2 5d1 4f7). Gd3+ [Xe] 4f7

(regardless of your answer for Gd!).

ChemActivity 121. 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 should have the same valence shell electron configuration as C, which has two unpaired electrons. c) True. Sulfur has four 2p electrons; two are paired and two are unpaired. 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.

5. One unpaired electron. Predicted magnetic moment, 1.7 magnetons (same as H).6. F is smaller than Ar. F– has no unpaired electrons. F has one unpaired electron and F+ has

two.

Problems1. 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 13

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9. 14, 24, 18, 12, 34, 32, 20,

Problems1. 722. 83. 14

ChemActivity 141. 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 (have greater bond energy) than double bonds, which share four electrons between two atoms. Single bonds, which share two electrons, are weaker (have lower bond energy) than double bonds.

5. True. 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.

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. C–F has the shortest bond length. b) 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. b) The N–N triple bond is stronger than the O–O double bond. c) O–H. O is smaller than P. d) O–H. O is smaller than S. e) S–H. S is smaller than Se. f) N–H. N is smaller than P. 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. 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.

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c) True. Draw the Lewis structures. The C-N bond in H3CNH2 is a single bond and the C-N bond in HCN is a triple bond.

Problem1. 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 151. 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

H2CO (a C–O double bond) than in CH3OH (a C–O single bond).5. a)

b) 1.5

c)

Problems1. N2 (BO = 3) < HNNH (BO = 2) < H2NNH2 (BO = 1) 2. N2 (a triple bond) < O2 (a double bond) < F2 (a single bond)3. 143 pm The C–O bond should be shorter because an oxygen atom is smaller than a

nitrogen atom.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 161. NH4

+: N(+1), H(0). CS2: All formal charges are zero. N2O5: the three O atoms with 2 bonds are 0; two O atoms with single bonds are -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

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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 171. 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.

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.

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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).

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Problems1. 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 Xe–O bond order is two.

e)

The C–O bond order is 4/3.

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2. The atoms in the molecule must have 34 valence electrons. Therefore, S or Se or Te are good candidates. 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

ChemActivity 18

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5. H—CC—H The central C atoms have two electron domains. The angle must be 180.

6. The central N atoms have three electrons domains. The angle must be close to 120.

7. N(109); C(120); O(109). 8. O(109); both Cs (120); N(109)

Problems1. 109°; 120°2. a) 180°, linear b) 180°, linear c) 120°, trigonal planar d) 109°, bent 3. 109°, trigonal pyramidal

ChemActivity 19

1. a) b) four c) tetrahedral d) about 109 e) bent f) sp3

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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, spPH3, see Ex 4 in CA 17, tetrahedral, sp3 H2CO, see Ex 4 in CA 17, trigonal planar, sp2

3. glycine: N(sp3); C(sp2); O(sp3). PABA: O(sp3); both Cs (sp2); N(sp3)

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 201. B, 1.17 MJ/mole. F, 2.31 MJ/mole.2. a) Mg<P<Cl. b) Se<S<O<F. c) K<P<O.3. Metals: Ca, Co, K, Cu. Nonmetals: Br, S. Metalloids: Si.

Problem1. (F-Cl) = (2.30–1.59) = 0.71 and (O–S) = 0.54. So, (N–P) ~ 0.37. The AVEE of N is

1.82, so we predict an AVEE for P of 1.45. The actual value is fairly close, 1.42.

ChemActivity 211. Br = –0.78; H = +0.078. These values are close to, but slightly less than, the values in

Table 2.2. Br = 0. Yes, each Br has the same electron pulling power so neither Br gains negative

charge.3. Zero.4. B = –0.13.5. All of the atoms have the same partial charge. There are three resonance structures for the

carbonate ion and the average charge on each O atom is –2/3 or –0.67. According to the CAChe calculations the charge is slightly more negative, –0.80. The oxygen atoms are negative because O has a greater AVEE than C.

6. +0.213.7. –1.388. –0.049. Calculations: 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)

ChemActivity 22

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1. O2, nonpolar. NaF, ionic. I2, nonpolar. KCl, ionic. CO, polar. NO, polar. CuO, ionic. CN–, polar. ICl, polar.

2. All are polar except Na–F, K–Cl, and Al–O which are ionic. The C-H bond is sometimes considered to be nonpolar because the electronegativity difference is so small.

3. Largest F; smallest Cs (excluding Fr).4. a) S b) P c) P5. H2O

Problems1. Open ended.2. a) CF4 b) SeF2

ChemActivity 23

1.2. 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.

3. O2, nonpolar; I2, nonpolar; CO, polar; NO, polar; H2O, polar; CO32–, nonpolar; SO42–, nonpolar; SO2, polar. chlorobenzene, polar.

4. a) N2 b) CH4 c) SO3 d) NO3– e) SO42- f) CO2 5. False. CCl4 has a dipole moment of zero, but NH3 does not.6. a) CH3Cl b) CH3Cl c) NF3, the electronegativity difference is greater. d) H2O7. 1.95 (must be somewhat greater than 1.89).

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Problems

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1. The dipole moment of I = 2(2d1) = 4d1. The dipole moment of II = 1(3d1) = 3d1. The dipole moment of I is larger.

2. a) CH3Cl b) SO23. The nitrate ion (trigonal planar) has a dipole moment of zero; the other two molecules have

a non-zero dipole moment.

ChemActivity 241. Al3+. Al is a metal and tends to lose electrons. Al has three valence electrons.2. S2–. Cl–. Cs+. Br–. O2–. Be2+. N3–.3. a) NaBr. b) Li2O. c) AlN. d) MgBr2. e) CaO.4. A possible metal is Pb. The metal must sustain a +4 charge. Pb is a metal and has four

valence electrons.5. a) NaCl. b) NaCl. c) MgO. d) CaO. e) MgS. f) NaCl. g) LiF. 6. a) NaCl. b) NaCl. c) MgO. d) CaO. e) MgS. f) NaCl. g) LiF.

Problems1. KCl (+1,–1)<K2S (+1,–2)<CaS (+2,–2)<CaO (+2, –2 and O2– smaller than S2–)

2. a) LiF b) CaO c) CaSO4 d) Al2O3 e) CaSO4

ChemActivity 251. Co(s) ; Pb (s)2. Open ended.

ChemActivity 261.Compound

EN ∆ENType of Bonding

firstatom

secondatom

CO2 2.54 3.61 3.08 1.07 covalentNH3 3.07 2.30 2.69 0.77 covalentBaO 0.88 3.61 2.24 2.73 ionicSO2 2.59 3.61 3.10 1.02 covalentAlSb 1.61 1.98 1.80 0.37 semimetalGaAs 1.76 2.21 1.98 0.45 semimetalCdLi 1.52 0.91 1.22 0.61 metallicBaBr2 0.88 2.69 1.78 1.81 ionicZnO 1.59 3.61 2.60 2.02 ionic/covalentNaH 0.87 2.30 1.58 1.43 ionic2. a) NaK b) NaCl c) CO2 d) CuZn e) NaK f) GaAs3. a) Metallic b) covalent

Problems1. 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,

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that ∆EN would be as large as indicated by point A (point A might represent some compound such as SrI2) It is harder to predict whichE N value will be higher (AgCl or SnI4). Therefore, "B" is the most likely point for AgCl.

2. CO32– covalent ; BaCO3 ionic, but covalent within the carbonate ions; CaSO4 ionic, but

covalent within the sulfate ions ; NaClO4 ionic, but covalent within the perchlorate ions.

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ChemActivity 271. 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 100C.

a) heptane. 100C. b) ethanol. 77C. c) 2-octanone. 165C.

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.

Problems1. HOCH2CH2OH

2. a) CH3CH2CH2NH2 All of these compounds have about the same MW. This compound has hydrogen bonding. b) NaCl Ionic compounds have much higher melting points. c) CaO CaO and LiF are ionic. CaO has the higher charges (±2).

3.4. The H–F bond in hydrogen fluoride.

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ChemActivity 281. 63.55 g2. 41.99 g3. 0.0167 moles of C. 1.00 1022 C atoms.4. 6.02 1023 H2 molecules. 12.0 1022 H atoms. 2.02 g5. 46.07 g/mole. 7.650 10–23 g/molecule.6. 25.90 g.7. a) 2.17 mole. b) C, 4.34 moles; H, 13.0 moles; O, 2.17 moles. c) C, 52.2 g; H, 13.1 g;

O, 34.7 g.8. 5.77 moles of CO2 molecules.9. 11.5 moles of O atoms.

10. 8.02 1021 carbon atoms11. 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, 1.1 mole of C, and 13 g of C. c) False, one mole of N2 has two moles of N; whereas, one mole of NH3 has only one mole of N. d) False, there are 100 g of Cu in 100 g of Cu, but there is less than 100 g of Cu in 100 g of CuO. 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.

12. The number of moles of H is 3z. The number of hydrogen atoms is Avogadro's number times 3z.

ChemActivity 291. a) 16 Cr(s) + 3 S8(s) 8 Cr2S3(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. 2C(s) + O2(g) + N2(g) + H2(g) H2NCH2COOH(s)4. 2O3(g) 3O2(g)5. a) is not atom balanced.

c) is not charged balanced.6. 19.2 g Cr2S3 7. 0.0163 g H2S8. 82.5 g FeCl3 9. 7.25 g Al and 21.4 g Fe2O3

10. N2 + 3 H2 2 NH3 176 g NH3 11. True. 2 CO(g) + O2(g) 2 CO2(g)

Problems 1. 35.9 mass % Ni

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2. “Q” is oxygen. 2Fe(s) + O2(g) Fe2O3(s)

ChemActivity 301. 46 g NO2 2. a) Zn(s) + I2(s) ZnI2(s) b) I2 is totally consumed Limiting reagent is I2 62.9 g

ZnI2 37.2 g Zn left3. a) HCCH(g) + 2 H2(g) C2H6(g) b) 31.2 g of C2H6 can be produced4. a) TiCl4 + 2 Mg Ti + 2 MgCl2 b) 6.43 103 g of Ti5. 9.16 g of P4O10(g) can be produced.6. 18.0 g of N2 (28.01 g/mole) can be obtained.

ChemActivity 311. 14.37% H2. C, 40.00%. H, 6.71%. O, 53.29%.3. a) 36.8 g N 63.2 g O b) 2.63 moles N 3.95 moles O c) 1.50 moles O/mole N

d) N2O3

4. C2H3Cl3

Problems 1. False. Ethyne, HCCH, and benzene, C6H6, are not isomers of each other.2. COH3

3. Zn3P2O8 or Zn3(PO4)2

ChemActivity 321. [Al3+] = 0.125 M [Cl–] = 0.375 M2. [Cr3+] = 0.0850 M [SO42–] = 0.128 M3. 8.64 g Na. 3.95 10–3 g Pb. 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 M6. 3.12 10–3 moles of AgCl

Problems1. Cl2 is limiting. 2.24 g of M. 0.43 M/l x 0.1 L = 0.043 moles of M. 2.24 g of M/0.043

moles of M = 51 g/mole. M = vanadium.2. a) [Cl–] = 3.42 M b) [Cl–] = 2.70 M c) [Cl–] = 3.03 M so a) is highest.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 mole Cl–/L

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ChemActivity 331. 44.8 L2. 520 L3. 1.19 104 L4. 6.0 1011 g 5. 252 K6. d = (MW)P/RT 7. 0.621 g/L8. PO2 = 48.0 atm PHe = 32.5 atm P = 80.5 atm9. 29.02 g/mole

Problems1. a) PHe = 50 torr PNe = 1.0 102 torr PAr = 1.2 102 torr b) P = 2.7 102 torr2. 4.22 L3. a) C2H3 b) 54.2 g/mole c) C4H6 d)

ChemActivity 341. a) endothermic b) exothermic c) exothermic2. 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 than diamond.5. True. 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 SiCl4(g) is –1,599.3 kJ/mole whereas the enthalpy of atom combination of SnCl4(g) is –1,260.3 kJ/mole.)

6. a) O-H bond energy in H2O is 463 kJ/mole; S–H bond energy in H2S is 367 kJ/mole.b) O–H has the stronger bond c) O is smaller than S.

ChemActivity 351. a) –37.01 kJ b) –696.56 kJ c) –67.18 kJ 2. a) Cl2 has the strongest bond (all are single bonds) and should have the most negative

. b) N2 has the strongest bond (all are triple bonds, Lewis structures) and should

have the most negative .3. C–H (smaller bond distance than C–Cl; stronger bond)). C–H bond energy = 1662/4 = 416

kJ/mole. C–Cl bond energy = 1306/4 = 326 kJ/mole. Yes.4. O is smaller than C. Therefore, the O–H bond should be stronger than the C–H bond.

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Problem1. 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 SO3 transformations).

ChemActivity 361. a) False. Depends on the stoichiometric coefficients. 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/s3. rate of ozone consumption = 9.00 10–5 M/s rate of reaction = 4.50 10–5 M/s4. rate of I– consumption = 7.5 10–2 M/s rate of IO2– consumption = 2.5 10–2 M/s

rate of H+ consumption = 10 10–2 M/s rate of I2 production = 5.0 10–2 M/s rate of reaction = 2.5 10–2 M/s

ChemActivity 371. At equilibrium, A molecules are being converted to B molecules and B molecules are being

converted to A molecules, but these conversions are occurring at the same rate. Thus, 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 381. 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

Problem1. a) kinetic region, < about 400 seconds; equilibrium region, > about 500 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 391. 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 = 0.33 No.4. a) Kc = b) Kc = c) Kc =

d) Kc = e) Kc = f) Kc = 5. Kc for reaction c) = [Kc for reaction b)]1/2

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6. Kc = 23 7. Kc = 6.18. Kc = 6.8 10–3 9. [NO2] = 2.7 10–3

10. 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

11. Kc = 6.04 10–2 12. 0.010013. a) Kc for the second reaction = (0.78)1/2 = 0.88 b) 1.114. 9.0 108

Problems1. False. Kfirst reax = 1/Ksecond reax . If Kfirst reax is small, Ksecond reax must be large.2. a) Kc = 10 b) The forward rate is larger; the moles of B are still increasing and the moles

of A are still decreasing. c) The forward rate is larger; the moles of B are still increasing and the moles of A are still decreasing. d) They are equal; the reaction is at equilibrium.

ChemActivity 402.

CO2 H2 CO H2Oinitial moles 1.00 2.00 0 0change in moles –x –x x xequilibrium moles 1.00–x 2.00–x x xequilibrium conc (1.00–x)/5 (2.00–x)/5 x/5 x/5equilibrium conc value (no "x")

0.11 0.31 0.086 0.086

3. x moles of O2 react. 2x moles of NO are formed.4.

N2 O2 NOinitial moles 5.00 10.00 0change in moles –x –x 2xequilibrium moles 5.00 – x 10.00–x 2x

5.N2 H2 N2H4

initial moles 1.00 1.50 0change in moles –x –2x xequilibrium moles 1.00–x 1.50–2x xequilibrium conc expression

(1.00–x)10 (1.50–2x)/10 x/10

[N2H4] = 1.1 10–5 M [N2] = 0.10 M [H2] = 0.15 Mc) is correct

26

27

6.CO2 H2 CO H2O

initial moles 1.00 2.00 1.00 2.00change in moles –x –x x xequilibrium moles 1.00–x 2.00–x 1.00+x 2.00+xequilibrium conc 1.00–x 2.00–x 1.00+x 2.00+xequilibrium conc value

0.84 1.84 1.16 2.16

7.NH3 N2 H2

initial moles 2.65 0 0change in moles –0.84 0.42 1.26equilibrium moles 1.81 0.42 1.26equilibrium conc value

0.90 0.21 0.63

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.

Problems1. 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 it is not generally true.c) False. The reaction must be going to the left; the rate of the reverse reaction must be greater than rate of the forward reaction.

2. a) K = 1.09 103 b) After the addition of the water the initial concentrations are: (FeSCN2+) = 0.163 M; (Fe3+) = ..00963; (SCN–) = 0.0116. Q = 1.46 103. Because Q > K, the reaction must go to the left.

3. Q = 0.81, the reaction must go to the left. The following equation must be solved for x:

0.26 = 3 where x is the number of moles of CH4 produced. The concentration of H2 is

equal to . 4. a) Q = 17.8 therefore, the reaction is going to the left

b) 16 =

c) The reaction is almost at equilibrium, Q = 17.8 and K = 16. Therefore, "x" must be small, say 0.1 moles; it will take only a small change to get Q = 16.

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ChemActivity 41

1. a) K =

b) K = c) Ksp = [Ba2+][SO4

2–]d) K = [NH3(g)][H2S(g)]e) K = [CO2(g)]f) K = [NH3(g)][HCl(g)]g) Ksp = [Ag+]2[SO4

2–]

h) K = = 2. a) False. K = [CO2(g)]. The concentration of CO2 is a constant (at a given temperature)

as long as some CaCO3(s) is present.b) False. At equilibrium the forward rate is equal to the reverse rate.

3. 1.63 10–3 4. 9.2 10–3 g5. 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 M7. a) Yes, Q = 6.2 10–7 b) No, Q = 6.2 10–9 Ksp for MgF2 = 6.5 x 10–9.8. No, Q = 7.8 10–9

9. = 40.5 mole/L

Problems1. 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 Au3. a) The NaCl solution had 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 PbCl2 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 421. 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.

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4. a) NH4+ b) NH2– c) NH3(l) + NH3(l) = NH4+(am) + NH2–(am) 5.

Acid BaseH2S HS–

HS– S2–

HNO2 NO2–

H3PO4 H2PO4–

HOCN OCN–

H3O+ H2O

OH– O2–

HF F–

H2PO4– HPO42–

HOCl OCl–

ChemActivity 431. 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 M6. 3.0 10–3 M7. 2.9 10–10 M; 1.4 10–14 M; 2.2 10–5 M; 4.8 10–8 M8. 2.9 10–10 M; 1.4 10–14 M; 1.8 10–3 M; 9.1 10–8 M9. 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, this occurs when [H3O+] > 10–7.

ChemActivity 441. 1.1 10–6

2. 1.0 10–9

3. [HONO] = 0.80 M; HONO + H2O = H3O+ + NO2– ; Ka = [H3O+] [NO2–]/[HNO2] [H3O+] = 0.020 M Note that if you use [HONO] = 0.80, [H3O+] is still 0.020 M.

30

4. [CH3NH2] = 1.5 M; CH3NH2 + H2O = CH3NH3+ + OH– ; Kb = [CH3NH3+] [OH–]/[CH3NH2] [OH–] = 2.7 10–2

Problems1. a) 2.4 10–12 b) c)

d) Ka = 1.8 10–4

ChemActivity 451. a) pH = 6.49, pOH = 7.51; acidic b) pH = 3.70, pOH = 10.30; acidic

c) pH = 4.49, pOH = 9.51; acidic d) pH = 7.40, pOH = 6.60; basic2. 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 neutrald) [H3O+] = 5 10–10 [OH–] = 2 10–5 basic

3. x molar HCl (strong acid) < x molar acetic acid (weak acid) < pure water (neutral) < x molar NaOH (strong base)

4. pure water (neutral) < 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 ammonia9. pH = 0.90 pOH = 13.10

10. pH = 0.90 pOH = 13.1011. [H3O+] = 0.800 M [OH–] = 1.25 10–14 M 12. [OH–] = 0.800 M [H3O+] = 1.25 10–14 M13. 2.5914. 8.9415. a) 13.65 b) 0.35 c) 2.55 d) 12.17

Problem1. [Mg2+] = 0.025 M; [OH–] = 2.7 10–5; Ksp = 1.8 x 10–11

ChemActivity 461. 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)2SO2. The H–Q bond strengths are the same; S is more electronegative than Se; there will be a greater positive charge on

31

the acidic hydrogen atom in (HO)2SO2. 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).

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.

Problems1. (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 471. 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 .3a. 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.v) The equilibrium constant is Ka = = = 5.6 10–10

3b. i) CsI is an ionic compound. Cs+ and I– ions exist in solution.

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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.

3c. i) CH3COONa is an ionic compound. Na+ and CH3COO– ions exist in solution.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.

v) The equilibrium constant is Kb = = = 5.6 10–10

3d. 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.

3e. 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.

v) The equilibrium constant is Kb = = = 5.6 10–10

4. a) pH = 4.63 b) pH = 7.00 c) pH = 9.37 d) pH = 7.00 e) pH = 4.48 5. a) neutral b) neutral c) neutral d) basic e) acidic f) acidic g) basic

h) neutral i) basic j) basic k) neutral6. pH = 11.19

Problem1. (a) acidic (b) CH3(CH2)8NH3

+ + H2O = CH3(CH2)8NH2 + H3O+

ChemActivity 481. 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 agente) S2O82– = Ox agent Zn = Red agent f) Au3+ = Ox agent Fe = Red agent

2. a) 2 moles b) 2 moles c) 2 moles d) 2 moles e) 2 moles f) 3 moles3. True. According to the data in Model 2, Cu2+ does oxidize Co, but Cr3+ does not.

ChemActivity 49

33

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).

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(+7/2), O(–2); P(+3), O(–2); P(+4), O(–2); P(+9/2), O(–2).9. a) Yes. b) No. c) Yes. d) Yes. e) Yes. f) Yes. g) No.

10. Yes.11. Yes.12. Yes.

Problem1. HOBrO2 (+5); HOBr (+1); HOBrO (+3); the higher the ox number, the stronger the acid.

ChemActivity 501. 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)

34

4.

b) Al is the anode. Pt in SHE 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. Reduction takes place at the SHE, so it has a stronger pull on the electrons, which means a greater (more positive) standard reduction potential. SHE’s standard reduction potential (0.0 V) is 1.66 V greater than that of Al(s)/ Al3+, so the standard reduction potential for Al(s)/ Al3+ must be -1.66 V.

ChemActivity 511. 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– 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. 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) +0.80 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.

Problems1. a) T b) F c) F d) T e) T2. 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– ERed = 1.24 V4. No reaction will occur because the cell voltage is negative, – 0.09 volts for:

Sn + Ni2+ = Ni + Sn2+

5. 0.467 M

35

ChemActivity 521. 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. ∆S is expected to be positive because there are three moles of gas produced for every two

moles of gas consumed.

ChemActivity 531. Entropy only. The reaction is endothermic, not energy favored. The increase in entropy

makes this a naturally occurring process.2. Both enthalpy and entropy. The process is exothermic (favorable) and the entropy

increases (favorable).

ChemActivity 541. a) prediction: ∆S< 0. actual: –198.73 J/K b) prediction: ∆S < 0. actual: –626.91 J/K

c) prediction: ∆S 0. actual: –20.07 J/K d) prediction: ∆S > 0. actual: 156.27 J/Ke) 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. a) 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.

Problem1. (a) would have the most positive change in entropy. There are two moles of gas on the left-

hand-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.

ChemActivity 551. a) ∆H° = –296.81 ∆S° = 39.11 K > 1

{assumes S(s) is S8(s)}b) ∆H° = +52.26 ∆S° = –53.28 K < 1

(assumes C(graphite))c) ∆H° = –128.5 ∆S° = –70.25 K can't be deducedd) ∆H° = –92.21 ∆S° = –198.57 K can't be deducede) ∆H° = +44.01 ∆S° = +118.34 K can't be deduced

2. a) ∆H° = –12.55 ∆S° = –102.5 ∆H° – T∆S° = 18.01 b) ∆H° = –92.21 ∆S° = –198.77 ∆H° – T∆S° = –32.95 c) ∆H° = 23.42 ∆S° = –12.50 ∆H° – T∆S° = 27.15 Largest K: b) Smallest K: c)

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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.

4. ∆H° = +24.9 ; ∆S° = +307.6 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 561. ∆G° = 18.03 = –RT lnK

lnK =–18,030 J/[(8.3144 J/K )(298.15 K)] = -7.27Ka = 7.0 x 10–4

2. ∆H° = 41.16 ; ∆S° = 41.50 ; ∆G° = 28.79 ; K = 9.0 10–6

K will increase as T increases.3. ∆H° > 0 ∆S° > 0

As T increases, K increases, causing the concentration of NH3(g) to increase.4. ∆H° = –92.21 . Since ∆H° < 0, K will decrease as T increases.5. a) Liquid boils when ∆G° = 0, so that ∆H° = T∆S°. Thus, the boiling point is T = . 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.b) ∆H° = 38.00

∆S° = 113.0 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

Problem1. (a) E° = 0.44 V (b) I3

– (aq) (c) K = 7 1014

ChemActivity 571. 5.40 10–7 M sec–1

2. a) first order in bothb) rate = k (Fe2+) (Ce4+)c) 1.0 x 103 M–1 sec–1

d) 1.8 x 10–7 M 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.

Problems1. 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

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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

3. (a) rate = k (UO2+)2 (H+) (b) k =

4. rate = k (NO)2 (O2)

ChemActivity 581. a) first order; k = 6.93 10–3 sec–1 b) 0.093 M2. Calculate ln(BrO–) and at each time. Prepare two plots: ln(BrO–) vs. time (a first-order

plot) and vs. time (a second-order plot). The plot with the best straight line determines the order.

3. 6.6 10–4 sec–1; 2.1 103 sec4. 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. IV6. 1.1 103 sec; 2.1 103 sec7. (d) 1/88. 87%

19. t = 7.35 103 yr

Problems1. 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.

ChemActivity 591. forward reaction has the larger activation energy2. 210.5 kJ/mole3. c) reactions not energetic enough to break bonds; molecules not oriented properly when

collision occurs

Problems1. The activation energy is small because:

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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 N2 bond and the H2 bond must be broken to form NH3. The Lewis structure for N2

shows that the N2 bond is a triple bond—very strong. The H2 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 601. bimolecular; rate = k' (ONBr)2

unimolecular; rate = k' (N2O2) bimolecular; rate = k' (NO)2 bimolecular; rate = k' (I) (H2)

2. b; the sum and the stoichiometry are the same3. No4. Yes – very little product will form from the reactants because K is very small.5. exothermic6. a) II

b) IVc) Id) II

7. a) rate = k (NO2) (O3) ; 5.0 x 104 M–1 sec–1

b) –150.7 kJ/mol Kc) 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. However, if one considers the first and second step to actually represent an equilibrium process, then the mechanism could be considered consistent.

Problems1. 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 611. Activation energy is lowered.2. Diagram looks similar to figure in Exercise 1.

Problem1. (a) Diagram not given here. (b) Ea (reverse) = 172 kJ (c) H° = 9.5 kJ/mol

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ChemActivity 621. exp = 1.972. exp = 2.12, so roughly correct!3. False. The higher the activation energy, the slower a reaction occurs at a given

temperature.

Problems1. 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.

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