Answers to Harvard General Chemistry Practice Problems

1. Math Review: Calculations and Scientific Notation 1. a) 523.4 b) 2.295 c) 19.2 d) 42.2 e) 858 f) 18.15

g) 1.18 × 10–4 h) 3.601 i) 2.47 × 10–4 2. Dimensional Analysis I

1. a) 2.38 atm b) 2.05 × 103 torr or 2050 torr c) 1.00 × 105 Pa 2. 341 mg 3. 3.60 × 106 sheets of paper 3. Dimensional Analysis II

1. a) 4.67 × 10–2 in3 or 0.0467 in3 b) 1.92 g

2. 2.52 × 1010 gallons 4. Atoms, Molecules and Ions 1. a) 18 protons, 22 neutrons, 18 electrons b) 20 protons, 20 neutrons, 18 electrons c) 19 protons, 20 neutrons, 18 electrons d) 19 protons, 20 neutrons, 19 electrons 2. 20 protons, 18 electrons

3. 35Cl–: 17 protons, 18 neutrons, 18 electrons 37Cl–: 17 protons, 20 neutrons, 18 electrons 4. a) 10 protons, 7 neutrons, 10 electrons b) 11 protons, 7 neutrons, 10 electrons 5. Naming Compounds 1. a) Cu3P b) Fe2(SO4)3 c) I2 d) AlCl3

e) Cr2O3 f) NH3 g) Li3N h) Al2(CO3)3

i) Cs3PO4 j) ZnO k) SnCl2 l) N2O5

m) AgF n) H3PO4 o) Ba3(PO4)2 p) (NH4)2SO4

2. a) Carbon tetrachloride b) Copper (II) oxide or cupric oxide (either answer would be acceptable) c) Magnesium nitride 6. Chemical Formulas: Percent Composition 1. ii < iii < i 2. a) C: 34.3% H: 7.2% P: 22.1% O: 22.8% F: 13.6% b) If we combust 10.0 grams of Sarin, we'd get 12.6 g CO2 and 6.5 g H2O. This does not

agree with the results obtained for the unknown compound; thus it can't be Sarin. 7. Empirical and Molecular Formulas 1. C11H17N3O8

2. a) C5H4O b) C15H12O3 ; MW = 240.25 g/mol

3. 64450 g/mol

8. Writing and Balancing Equations 1. a) CO(NH2)2 + 6 HOCl → 2 NCl3 + CO2 + 5 H2O

b) 2 Ca3(PO4)2 + 6 SiO2 + 10 C → P4 + 6 CaSiO3 + 10 CO

2. 6 CO2 + 6 H2O → C6H12O6 + 6 O2

3. 3 Nb + 4 I2 → Nb3I8

4. C8H18 + 25/2 O2 → 8 CO2 + 9 H2O

9. Stoichiometry of Reactions 1. 55.9 g NaClO 2. a) NiS + 2 O2 + 2 HCl → NiCl2 + H2SO4

b) 0.654 g NiCl2

10. Stoichiometry with Limiting Reagents 1. a) V2O5 + 3 Zn → V2O2 + 3 ZnO

b) 68.25 g V2O2

2. a) 5 P4S3 + 12 Br2 → 3 P4S5 + 8 PBr3

b) 77.47 g P4S5

11. Stoichiometry of Mixtures 1. 61.1% CO2 by mass

2. 6.55 g Cu2S, 4.25 g CuS

12. Oxidation Numbers 1. Br: +5 H: +1 As: –3

O: –2 As: +3 H: +1 O: –2

Cr: +3 K: +1 S: +2 Cl: –1 Cl: +5 O: –2 O: –2

O: +2 Na: +1 Fe: +8/3 F: –1 O: –1 O: –2

2. N2 is reduced, H2 is oxidized

FeCl3 is reduced, KI is oxidized

No oxidation or reduction is taking place.

Cl2 is being oxidized (to ClO3–) and reduced (to Cl–)

13. Solutions: Molarity 1. 18.0 M 2. 17.47 M 3. 0.216 M 4. 0.052 g 5. x = 4.4 g KNO3 y = 4.49 g NaCl

14. Solution Stoichiometry I

1. a) 3 ClO– (aq) + NH3 (aq) → NCl3 (l) + 3 OH– (aq)

b) 0.028 g NCl3

2. 2.79 g CO2

15. Solution Stoichiometry II: Acid/Base Neutralizations 1. 1.72 L 2. 71.2 mL 3. 1.26 L 16. Solution Stoichiometry III: Titrations 1. 1.22 g 2. 312 g/mol 3. 119.2 g/mol 17. Solution Stoichiometry IV: Precipitation Reactions

1. 0.0119 M Ba2+ (aq) 2. CaSO4 = 8.92 g Al2(SO4)3 = 1.08 g

18. Putting It Together: Carminic Acid 1. a) x = 22 y = 20 b) 1 acidic hydrogen (it is a monoprotic acid) 19. The Ideal Gas Law

1. a) 6.0 × 105 L b) 0.348 g Cl2

2. 53.4 g O2

20. Mixtures of Gases 1. a) 0.0863 mol O2

b) 0.0211 torr O2

c) 2.78 × 10–6 torr O2

2. 16.3 g butane volume = 9.286 L 21. Reactions Involving Gases I 1. 0.480 atm 2. 1.5 atm 22. Reactions Involving Gases II

1. χPCl5 = 0.054 χPCl3 = 0.473 χCl2 = 0.473

2. 12.8% O3 by mass

23. Collecting Gases Over Water 1. (ii) 2. 0.109 g H2

3. 3.51 L should be collected

24. Kinetic-Molecular Theory of Gases 1. a) CO2: 752 m/s H2O: 1176 m/s N2: 943 m/s

b) Kinetic energies of all gases are the same at any given temp:

12450 J/mole and 2.07 × 10–20 J/molecule 25. Putting It Together: Gases 1. a) P

Cl2

= 1.63 atm PO2

= 1.07 atm

b) 29.67 g NaAlCl4

c) 628.54 m/s 26. Heat 1. a) 8.2 kJ of heat consumed b) Final temp. = 14.2°C 2. Specific heat = 0.269 J/g°C 27. Calorimetry I 1. ∆H = –27 kJ/mol 2. ∆H = –5152 kJ/mol 3. ∆H = –4219 kJ/mol 28. Calorimetry II 1. 27.08°C 2. 164.3 g of ice 29. Enthalpy, Energy, Heat, and Work 1. a) ∆H = +37.4 kJ/mol b) w = –2800 J/mol c) ∆U = +34.6 kJ/mol 30. Hess’s Law 1. ∆H° = –171.34 kJ/mol 2. ∆H° = –185.1 kJ/mol 31. Standard Enthalpies of Formation 1. ∆H° = –64.9 kJ/mol 2. 268 kJ of heat released 32. Putting It Together: Methanol Fuel 1. a) 3859.8 kJ b) –6.33 × 104 kJ 33. Putting It Together: Butane Heater

1. a) C4H10 (g) + 132

O2 (g) ! 4 CO2 (g) + 5 H2O (l)

b) ΔH°f [C4H10 (g)] = –132.3 kJ/mol

34. Putting It Together: Nitrosyl Chloride 1. q = −3.86 kJ w = +0.124 kJ ΔU = −3.736 kJ

35. Light and Matter 1. 9.31 × 1042 photons from the sun per year

2. 1.74 × 10–35 m/s 36. The Photoelectric Effect

1. a) 1.09 × 1015 s–1

b) KE = 7.46 × 10–20 J Wavelength = 1.80 × 10–9 m 37. Single Electron Atoms 1. n = 7 2. 2.17 × 1010 38. Orbitals and Quantum Numbers

1. Subshell n l total #of e–

1s 1 0 2 3d 3 2 10 5p 5 1 6 2. a) b) c)

2 radial nodes 2 radial nodes 1 radial node 0 angular node 1 angular node in the xz-plane 2 angular nodes in the planes defined by x = y and the z axis

and x = –y and the z axis 39. Electron Configurations of Neutral Atoms

1. C: [He]2s22p2

Mg: [Ne]3s2

Mn: [Ar]4s23d5

Se: [Ar]4s23d104p4

Cu: [Ar]4s13d10 (Remember that Cr and Cu are well-known exceptions to the usual order!)

Xe: [Kr]5s24d105p6

Ba: [Xe]6s2

Os: [Xe]6s24f145d6

Pb: [Xe]6s24f145d106p2

radial nodes

x

y

x

yangular nodes

radial node

40. Electron Configurations of Ions

1. N– : [He] 2s22p4

Al3+: [He]2s22p6 or [Ne]

O2–: [He]2s22p6 or [Ne]

Zn2+: [Ar] 3d10

W6+: [Xe] 4f14

Cu2+: [Ar] 3d9

Gd3+: [Xe] 4f7

Se2–: [Ar] 4s23d104p6 or [Kr]

Fe2+: [Ar] 3d6

Fe3+: [Ar] 3d5 41. Periodic Properties I 1. a) High atomic mass: lower right Low atomic mass: upper left b) Big atomic radius: lower left Small atomic radius: upper right c) High ionization energy: upper right Low ionization energy: lower left d) More negative electron affinity: upper right More positive electron affinity: lower left e) Most metallic: lower left "Least metallic" (most non-metallic): upper right 42. Periodic Properties II 1. a) Radium is the largest of the group; its valence electrons are in the 7s orbital, which is far from the nucleus. b) Indium is the largest of the group; its valence electrons are in the 5p orbital, whereas the others are in

lower-energy orbitals. 2. a) Thallium has the lowest ionization energy of the group; its valence electrons are in a higher energy orbital

than those of Ga or Se, and it has a smaller effective nuclear charge than Po. b) Cesium has the lowest ionization energy of the group; its valence electrons are in a higher energy orbital

than those of Ga or Se, and it has a smaller effective nuclear charge than Bi. 3. Fluorine has the most negative electron affinity because when it gains an electron, it forms the fluoride ion

with a stable, complete octet.

4. Se2– is the largest ion of the group. Of course it will be larger than the 2nd-period ions O2– and F–. Plus, since it is isoelectronic with Rb+, it has the same number of electrons but three fewer protons, meaning that the effective nuclear charge is less and the radius will be larger.

43. Periodic Properties III

1. a) Se: . . . 4s24p4 has 2 electrons in one of its p-orbitals

As: . . . 4s24p3 each p-orbital has one electron Because it is somewhat unfavorable for two electrons to occupy the same orbital, Se is relatively willing to

give up its p-electron to achieve an electron configuration in which each p-electron is in its own orbital. Note that this goes against the usual trend for ionization energy.

b) Adding an electron to Br results in a complete octet (which is particularly stable, and thus quite favorable). c) When Rb and Na react with water, they each lose an electron to form their respective cations. Since Rb has

a lower ionization energy, more total energy can be released in the reaction. d) When an electron is added to fluorine, its octet is completed. e) When fluorine reacts, it almost always accepts an electron, forming fluoride ion and releasing a tremendous

amount of energy (the electron affinity). This large energy released makes fluorine very reactive. f) Xenon can't gain an electron (because it already has a full octet). It also can't readily lose an electron,

because its ionization energy is so high. Thus it is unwilling to participate in chemical reactions. g) Although iodine can't readily lose an electron, it can readily gain an electron (unlike xenon) and form a

stable, complete octet. This is the usual mode of reactivity for iodine. 44. Putting It Together: Bismuth

1. a) wavelength = 1.70 × 10–7 m

b) Bi: [Xe] 6s24f145d106p3

Bi+:[Xe] 6s24f145d106p2 c) n=6, l=1 d) e) Element 113 would have a lower ionization energy than Bi. Element 113 would lie below thallium (Tl) on

the periodic table. Its valence electrons would be in the 7p subshell (more energetic than the 6p valence electrons of Bi). Thus those 7p valence electrons would be easier to ionize.

45. Lewis Structures I: The Octet Rule

H C NO

NOO

NO

C OC

F

F

F

F

O

F F

N

SF

OOHH

ON

O

O

NO

+

S

F

F FB

H

H H

H

C

O

O O

C C

HP

H

–

2–

2–

(plus two otherresonancestructures)

–

–

–

x

y

radial nodes

46. Lewis Structures II: Less Than an Octet

H Be HB

Br

Br BrAl

F

F F

ONO

ONO

N O

47. Lewis Structures III: More Than an Octet

AsCl

Cl

Cl

Cl

ClF Se F

F FF Xe F As

F

F FF

F F

–

F Xe F

F

+

F Br F–

48. Bond Enthalpies 1. a) :N≡C–C≡N: b) –1017 kJ/mol 2. a) b) –246.5 kJ/mol 49. Lattice Energy: The Born-Haber Cycle 1. Lattice enthalpy = +5400 kJ/mol 2. Lattice enthalpy = +2021 kJ/mol

50. Molecular Geometry I: Neutral Molecules

O

NF

Cl I Cl

ClO C O

B

F

F F

e- pair geo:

mol. geo:

hybridization:

polar?

linear

linear

sp

no

trig. planar

bent

sp2

yes

trig. planar

trig. planar

sp2

no

trig. bipyramidal

T-shaped

sp3d

yes

C O Cl ClC

O

Cl Cl

!H°sub (Ba(s))

DCl-Cl

Ba (s) + Ba(ClO3)2 (s)

!H°f

Ba (g) Ba2+ (g)IE1

2EA (ClO3)

2Cl (g)

"LE

Cl2 (g)

Ba+ (g)IE2

+ 3 O2 (g)

2ClO3– (g)

6O (g)

3DO=O

2ClO3 (g)

"6DCl-O

51. Molecular Geometry II: Ions

2–

Se

O

O OI

F

F

F

FI I I

Cl P

Cl

Cl

Cl

–

+

–

e- pair geo:

mol. geo:

hybridization:

octahedral

sq. planar

sp3d

2

tetrahedral

tetrahedral

sp3

tetrahedral

trig. pyramidal

sp3

trig. bipyramidal

linear

sp3d

52. Molecular Geometry III: Polycentric Molecules and Ions 1.

53. Sigma and Pi Bonding I

1. a) Trigonal planar; sp2 hybridization

b) 2 σ-bonds between H s orbitals and C sp2 hybrid orbitals

1 σ-bond between O orbital and C sp2 hybrid orbital 1 π-bond between O p-orbital and C p-orbital c)

2. a) Linear geometry; sp-hybridization b) sp hybrid orbitals c) p orbitals

O C O

C

O

HH

C OH

H!

sp2

!

!

sp2

sp2

s orbital

"

p orbitals

s orbital

H3C CH

CH

C

O

O

tetrahedral

tetrahedral

sp3

trig. planar

trig. planar

sp2

trig. planar

trig. planar

sp2

trig. planar

trig. planar

sp2 I

I

I

I

I

–

trig. bipy.

linear

sp3d

tetrahedral

bent

sp3

trig. bipy.

linear

sp3d

S

F

F

F

SF

trig. bipy.

see-saw

sp3d

tetrahedral

bent

sp3

54. Sigma and Pi Bonding II 1. a) Central carbon atom: sp hybridization

Terminal carbon atom: sp2 hybridization

b) or 2. a) b) From left to right: • 3 σ-bonds between C and H

• 1 σ-bond between C sp3 hybrid and C sp hybrid • 1 σ-bond between C sp hybrid and C sp hybrid

2 π-bonds between C p orbitals and C p orbitals (these three bonds together make up the C≡C triple bond)

• 1 σ-bond between C sp hybrid and C sp2 hybrid

• 2 σ-bonds between C sp2 hybrids and O orbitals 1 π-bond which is shared (by resonance) between the C p orbital and the two O p orbitals (these three bonds together make up the CO2– resonance portion of the molecule)

55. Molecular Orbital Theory: Orbital Overlap 1. a) σ*1s

b) σ2p

c) π2p

d) π∗2p

C C

H

H

C

H

H

C

H

H

C CH

H

!!

C

H

H

C CH

H

C C CC

H

H

HO

O

sp3 sp sp2sp

56. MO Diagram: Homonuclear Diatomic Molecules 1. a)

b) O2 (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)2

O22– (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)4

c) Yes d) O22– will have a weaker bond because there are two more electrons in the antibonding orbital.

e)

!2p* "2p*

HOMO LUMO

!2p*

2s

2p

2s

!2s

!2s*

!2p*

!2p

"2p

molecularorbitals

atomicorbitals

atomicorbitals

OO O2

"2p*

2p

57. MO Diagram: Heteronuclear Diatomic Molecules 1. a)

b) Double bond: NO– Diamagnetic: NO+ Longest bond: NO– Isoelectronic w/CO: NO+

58. Chemistry of Coordination Compounds (There are no problems on this page. A separate packet of problems for coordination chemistry will be

distributed, and answers will be provided with that packet.) 59. Intermolecular forces 1. a) Al2O3 Ionic

F2 London Dispersion

H2O Hydrogen Bonding

Br2 London Dispersion

ICl Dipole-Dipole NaCl Ionic b) In order from highest BP to lowest BP: Al2O3, NaCl, H2O, ICl, Br2, F2

2. Yes. 60. Phase Changes 1. a) MP: 115°C (388 K) BP: 445°C (718 K) b) total of 169.4 joules required 61. The Clausius-Clapeyron Equation 1. +42.2 kJ/mol 2. 25.1 torr

2p

2s

2p

2s

!2s

!2s*

!2p*

!2p

!2p*

!2p

N NO O

molecularorbitals

atomicorbitals

atomicorbitals

62. Phase Diagrams 1. a) and b)

18439

760

100

P

T (°C)

(torr)

110

1

solid

liquid

gas

triple pt.

melting pt. boiling pt.•

•

•

c) Solid and gas phases will be observed at equilibrium at 25°C. d) Solid iodine is more dense than liquid; this is why the s/l line slopes to the right on the diagram. 63. Bonding in Crystalline Solids

1. 7.60 g/cm3 2. radius = 1.97 Å 64. Differential Rate Laws 1. a) The reaction is first order with respect to A and second order with respect to B. It is third order overall. b) Rate = k[A][B]2, where k is 0.0090 M–2·s–1

c) The new reaction rate will be 18 times greater in magnitude. 2. Rate = k[CO][Cl2]1.5 k = 11.3 M–1.5 s–1 65. Integrated Rate Laws 1. P(SO2Cl2) = 3.25 atm

P(SO2) = 5.75 atm

P (Cl2) = 5.75 atm

2. 94.7% 66. Temperature and Rate 1. 24.13 kJ/mol 2. a) 43.4 kJ/mol b) 20.5 hours 3. i) speed up ii) speed up iii) slow down iv) speed up v) remain same

67. Putting It Together: Degradation of Rotenone 1. a) 27.4 days b) 39.7 kJ/mol 68. Introduction to Reaction Mechanisms 1. a) rate = k1 [A]

b) rate = k1k2k–1 [A][C]

c) rate = k1k2[A][C]k–1 + k2[C]

69. Advanced Mechanisms

1. a) rate = k11/2 k2k3 [CO][Cl2]3/2

k–11/2(k–2 + k3[Cl2])

b) if k3[Cl2] <<< k–2

70. Putting It Together: Enzyme-Catalyzed Production of I2

1. a) Rate =

!

d I2[ ]

dt =

k1k2E[ ] H2O2[ ] I!"# $%

k!1 + k2 I!"# $%

b) Rate = k1E[ ] H2O2[ ]

71. Putting It Together: Cisplatin 1. a) 63 kJ/mol

b) k1k2k!1

cpt ! Cl2"# $% DNA!"# $%

72. Nuclear Reactions 1. a) positron

b) 15C

c) 90Zr

d) 226Ra

e) 247Es

f) 143Xe

g) 4He (α particle) h) 7 alpha particles 4 beta particles 73. Radioactive Dating 1. a) 41.5 years b) 83 years 74. Energy Changes in Nuclear Reactions

1. 1.34 × 10–13 J

2. 7.8 × 10–14 J

75. Putting It Together: Carbon-14

1. a) 14C → 14N + β–

b) 2.51 × 10–14 J

c) 2.35 × 108 m/s d) 92 % 76. Putting It Together: Uranium-235 Fission

1. 2.23 × 106 g 77. Writing Equilibrium Constant Expression

1. a) Kc = [PCl3][Cl2]

[PCl5] Kp =

!

(PPCl3)(PCl2

)

(PPCl5)

b) Kc = [NH3][HCl] Kp = (PNH3)(PHCl)

c) Kc = [H2O(g)] Kp = (PH2O(g))

2. Kc = [CO2]2

[CO]2 = 13.7

3. Kc = 13

78. Reaction Quotient

1. a) Kc = 5.9 × 1013

b) Qp = 5.4 × 101, so reaction goes to right, and NO2 will be produced under these conditions.

2. a) Kc = 5.2 × 10–3

b) Qc = 0.25, so reaction goes to left, and NOBr will be produced under these conditions.

79. LeChâtelier’s Principle 1. i) remain same ii) increase iii) decrease iv) increase v) remain same vi) decrease 2. a) high temperature, low pressure b) low temperature, high pressure 80. Calculating Equilibrium Constants 1. Kp = 0.0196

2. Kp = 0.077

81. Determining Final Equilibrium Concentrations 1. Pressure of BrF: 3.5 atm

2. 1.7 × 1018 oxygen (O) atoms in the flask

82. Putting It Together: Decomposition of Sulfuryl Chloride 1. a) SO2Cl2: 0.0052 moles

SO2: 0.178 moles

Cl2: 0.178 moles

Total pressure: 17.8 atm b) Add more SO2(g) Decrease the volume of the flask

83. Putting It Together: Production of Sulfuric Acid

1. a) K = 2.64 ×1079

b)

!

PSO

3= 0.0265 atm

c) i) increase ii) decrease 84. Introduction to Acids, Bases, and pH 1. a) HN3 is the acid, NH2OH is the base

b) B(OH)3 is the acid, CN– is the base

c) Si(OH)4 is the acid, C6H5O– is the base

2. a) 12.43 b) 1.70 c) 11.30 d) 7.00 85. Weak Acids 1. 5.10 2. [H3PO4] = 0.0761 M

[H2PO4–] = 0.0239 M

[HPO42–] = 6.2 × 10–8 M

[PO43–] = 1.1 × 10–18 M

[H+] = 0.0239 M

[OH–] = 4.2 × 10–13 M 3. 1.84 86. Weak Bases 1. 11.84 2. 10.41 3. 8.60 87. Acidic Behavior of Metal Cations (Hydrolysis) 1. 2.93 2. 5.11

88. Putting It Together: Aqueous Acid Equilibria 1. pH = 0.84 [H2SO3] = 1.33 M

[HSO3–] = 0.143 M

[SO32–] = 1.02 × 10–7 M

89. An Introduction to Buffers 1. 3.52 2. pH = 4.90 3. (in order from top to bottom) increase, decrease, decrease, remain same, increase 90. Creating Buffers

1. [NH4+] = 0.360 M

2. m-chlorobenzoic acid is the best choice. When choosing a buffer the ideal situation is for pKa and pH to be equal or as close to equal as possible. The

reason for this is that the ratio of a weak acid to its conjugate base (or vice versa) will be more resistant to change when they are present in equal or nearly equal amounts.

3. 12.5 mL of 0.10 M acetic acid 7.5 mL of 0.15 M sodium acetate 91. Putting It Together: Trona 1. a) 10.25 b) 0.046 moles HCl 92. Titrations of Weak Acid and Strong Base 1. a) 3.57 b) 8.36 c) 12 93. Titrations of Weak Base and Strong Acid 1. a) 11.62 b) 5.76 94. Titration Curves 1. a) 2.9 b) 38.7 mL 95. Solubility and Ksp

1. a) 3.58 × 10–3 M

b) Ksp = 2.29 × 10–8

2. a) 0.0144 M b) 0.0012 M

3. must add at least 2.0 × 10–5 moles of NaF 96. Solubility of Metal Hydroxides 1. a) 10.46

b) 1.2 × 10–3 M

97. Complex Ion Formation

1. [Cu2+] = 5.6 × 10–15 [NH3] = 0.92

[Cu(NH3)4+] = 0.020

2. [Mn2+] = 4 × 10–38 98. Solubility and Complex Ion Formation 1. 0.0033 M

2. [Ag+] = 4.5 × 10–9 99. Entropy and Physical Changes 1. a) negative b) positive c) uncertain 2. 327.3 J/K 3. –543.68 J/mol·K 100. Gibbs Free Energy 1. a) +240 kJ/mol b) +76.2 kJ/mol c) –550 J/mol·K d) +259 kJ/mol e) +223 J/mol·K 101. Free Energy and Equilibrium 1. 64.4°C 2. 60°C 102. ∆G at Nonstandard Conditions 1. –38.4 kJ/mol 2. a) Since ∆G = 0, the reactants and the products have the same free energies at the given partial pressures. The

system is already at equilibrium, and no shift occurs. b) The partial pressures given are all 1.00 atm, indicating that the system is in the standard state. The negative

value for ∆G° (–7.336 kJ/mol) means that in their standard states the products have a lower free energy than the reactants. Therefore the system moves to the right to reach equilibrium.

103. The Temperature Dependence of K

1. 1.74 × 1022

2. a) K = 6.9 × 105 b) K = 0.71 ∆G° = +1.35 kJ/mol 104. Putting It Together: Xylenol Orange-Aluminum Complex 1. a) +25.9 kJ/mol b) +60.7 kJ/mol c) +106.1 J/mol·K

105. Putting It Together: Phase Change of Mercury 1. a) +31.76 kJ/mol b) +9.63 kJ/mol c) +77.34 J/mol·K 106. Putting It Together: Incandescent Light Bulb 1. a) 5.00 b) 69.7 kJ/mol non-spontaneous (the reaction is spontaneous in the reverse direction. Therefore, the

tungsten is continually recycled from the walls of the bulb back to the filament, allowing the bulb to last longer.)

107. Putting It Together: Buffers 1. a) +26.9 kJ/mol b) –12.0 kJ/mol c) 4.70 108. Balancing Redox Reactions: Acidic Solution

1. a) 20 H+ (aq) + S4O62– (aq) + 6 Al (s) → 4 H2S (aq) + 6 Al3+ (aq) + 6 H2O (l)

b) 14 H+ (aq) + 6 S2O32– (aq) + Cr2O72– (aq) → 3 S4O62– (aq) + 2 Cr3+ (aq) + 7 H2O (l)

c) 18 H2O + 14 ClO3– (aq) + 3 As2S3 (s) → 14 Cl– (aq) + 6 H2AsO4– (aq) + 9 SO42– (aq) + 24 H+ (aq)

d) 3 H2O + 7 IO3– (aq) + 6 Re (s) → 6 ReO4– (aq) + 7 I– (aq) + 6 H+ (aq)

e) 3 HNO2 (aq) → NO3– (aq) + 2 NO (g) + H2O + H+ (aq)

109. Balancing Redox Reactions: Basic Solution

1. a) 3 C4H4O62– (aq) + 5 ClO3– (aq) + 18 OH– (aq) → 12 CO32– (aq) + 5 Cl– (aq) + 15 H2O (l)

b) 11 Al (s) + 3 BiONO3 (s) + 11 OH– (aq) → 3 Bi (s) + 3 NH3 (aq) + 11 AlO2– (aq) + H2O (l)

c) 4 H2O2 (aq) + Cl2O7 (aq) + 2 OH– (aq) → 2 ClO2– (aq) + 4 O2 (g) + 5 H2O (l)

d) Tl2O3 (s) + 4 NH2OH (aq) → 2 TlOH (s) + 2 N2 (g) + 5 H2O (l)

e) 3 Mn(OH)2 (s) + 2 MnO4– (aq) → 5 MnO2 (s) + 2 OH– (aq) + 2 H2O (l)

110. Standard Reduction Potentials at 25°C (No problems on this page)

111. Voltaic Cells 1. a)

Anode Cathode

Pt Pt

V

Cr2+ NO3- NO3

-Fe3+

KNO3

Cr3+ Fe2+

e

b) Anode: Cr2+ (aq) → Cr3+ (aq) + e–

Cathode: Fe3+ (aq) + e– → Fe2+ (aq)

Overall: Cr2+ (aq) + Fe3+ (aq) → Cr3+ (aq) + Fe2+ (aq)

Pt (s) | Cr2+ (aq), Cr3+ (aq) || Fe3+ (aq), Fe2+ (aq) | Pt (s) c) 1.181 V

d) If [Fe2+] = 1 M, [Fe3+] = 0.299 M If [Fe3+] = 1 M, [Fe2+] = 3.34 M 112. Reduction Potentials of Half-Reactions 1. 0.803 V

2. a) 9 H+ (aq) + HSO4– (aq) + 8 e– → H2S (g) + 4 H2O (l)

b) 0.303 V 113. The Nernst Equation 1. a) 1.47 V Spontaneous b) The reduction potential of Fe is –0.44 V, Ni is –0.28 V, Zn is –0.76 V, and Na is –2.71 V. Since zinc and

sodium are, relative to iron, harder to reduce they are therefore easier to oxidize. Based on this standard both will be oxidized before iron. However, sodium reacts explosively with water, so zinc is the best choice.

114. Concentration Cells

1. a) Anode: cell on the left (0.010 M Ag+) Cathode: cell on the right (1.0 M Ag+) b) 0.118 V

c) New [Ag+] = 0.00028 M 115. Faraday's Law 1. 9.85 amps 2. a) 30 hours b) 334 kg graphite

116. Putting It Together: A Voltaic Cell 1. a)

b) 2Al (s) + 3 Cu2+ (aq) → 2 Al3+ (aq) + 3 Cu (s) c) 1.994 V d) 1.981 V 117. Putting It Together: Intensive Property of E

1. a) 8 H+ (aq) + BaCrO4 (s) + 3 e– → Ba2+ (aq) + Cr3+ (aq) + 4 H2O (l)

b) 1.28 V 118. Putting It Together: Ethanol Fuel Cells 1. a) oxidation (top) reduction (bottom) b) 1.14 V

c) 6.2 × 10–2 moles 119. Expressing Solution Concentrations 1. a) 41.9% maltose b) 2.11 m

c) χmaltose = 0.037

d) χH2O = 0.963

e) 38.3 M water 120. Colligative Properties I: Non-Dissociating Solutes 1. a) 100.71°C b) 23.2 torr 2. 19200 g/mol

3. χCCl4 = 0.80 χBr2 = 0.20

121. Colligative Properties II: Dissociating Solutes 1. a) 102.23°C b) 22.06 torr 2. Assuming proposal A, van’t Hoff factor i = 6, and predict FP = –3.49°C.

Assuming proposal B, van’t Hoff factor i = 2, and predict FP = –1.16°C. Since observed FP = –3.5°C, we expect that proposal A is correct.

Anode Cathode

Al Cu

V

NO3-

NO3-

KNO3

Al3+Cu2+

e

122. Colligative Properties III: Multiple Solutes 1. –0.27°C

2. 22.9 torr Note: NaN3 dissociates into two ions: Na+ and N3– (the azide ion).

123. Colligative Properties IV: Non-Ideal Solutions 1. a) i = 0.88 b) Expected i = 2 c) In concentrated solutions, charged ions will tend to associate into ion pairs or other aggregates. In this

case, since i is less than one, there must be net aggregation of Mg2+ and SO42– ions into groups of 4 ions (or perhaps even larger).

2. a) i = 2.70 b) The more concentrated the solution, the greater the ion-pairing and the smaller the measured value of i.