Download Harvard University General Chemistry Practice Problems “The

Harvard University
General Chemistry
Practice Problems
“The Logan Notes”
2004-2005
21
22
23
24
25
26
27
28
29
30
8
9
Mg
Na
Ca
K
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
N
O
F
Ne
Si
Al
16
S
15
P
Cl
17
Ar
18
Ga
31
Ge
32
As
33
Se
34
Br
35
Kr
36
26.98 28.09 30.97 32.07 35.45 39.95
14
13
Y
39
Zr
40
41
Nb
Mo
42
Ba
Cs
72
Hf
57
La
Ta
73
W
74
43
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
I
53
Xe
54
Re
75
Os
76
Ir
77
Pt
78
Au
79
Hg
80
Tl
81
Pb
82
Bi
83
Po
84
At
85
Rn
86
(98) 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.76 127.60 126.91 131.29
Tc
Ra
Ac
Rf
Ha
60
Nd
Pm
61
Sm
62
63
Eu
Gd
64
Tb
65
Dy
66
67
Ho
68
Er
69
Tm
70
Yb
Lu
71
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.04 231.04 238.03 237.05 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)
Actinide series Th
140.12 140.91 144.24 (145) 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97
100
101
102
103
90
91
92
93
94
95
96
97
98
99
59
Pr
58
(262) (263) (262) (265) (266)
Lanthanide seriesCe
(223) 226.03 227.03 (261)
Fr
132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.20 192.22 195.08 196.97 200.59 204.38 207.20 208.98 (209) (210) (222)
104
105 [106] [107] [108] [109]
87
88
89
56
55
85.47 87.62 88.91 91.22 92.91 95.94
38
Sr
37
Rb
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.61 74.92 78.96 79.90 83.80
20
19
22.99 24.31
12
11
C
10.81 12.01 14.01 16.00 19.00 20.18
B
Be
7
Li
6.941 9.012
6
4
5
He
4.003
10
H
2
1.008
3
1
PERIODIC TABLE OF THE ELEMENTS
Harvard University
BLB: Ch. 1; PHH: Ch. 1
Significant Figures and Scientific Notation
1.
Perform the following calculations and express each answer to the proper number
of significant figures:
a)
+
423.1
100.
0.256
b)
14.000
6.1
c)
(6.11)(π)
d)
(4/3)π(2.16)3
e)
(14.3)(60)
f) (6.0+9.7+0.61)(1.113)
g)
6.958×10 8
5.91×10 12
h)
(1.173×10 –3) + 3.6
i) (1.1×106 )(2.246×10 –10 )
1
Harvard University
BLB: Ch. 1; PHH: Ch. 1
Dimensional Analysis I: Unit Conversions
1.
Given the following conversion factors:
1 bolt of cloth = 120 ft
1 meter = 3.28 ft
1 meter = 0.55 fathoms
1 hand = 4 inches
1 ft = 12 inches
a)
One bolt of denim cloth costs $65.99. If you paid $37.27 for some denim cloth,
what length (in feet) did you buy?
b)
If a horse stands 15 hands high, what is its height in meters?
c)
A shipping channel is dredged to a depth of 4.5 fathoms. Calculate its depth in
feet.
2.
Given the following conversion factors:
1 atm = 760 torr
1 torr = 0.01933 lb/in2
1 N = 0.225 lb
1 in = 2.54 cm
1 Pa = 1 N/m2
a)
The pressure inside a laboratory vessel is 35 lb/in2. Calculate the equivalent
pressure in atmospheres.
b)
A certain chemistry experiment takes place in a high-pressure chamber with a
pressure of 2.7 atmospheres. Calculate the pressure in torr.
c)
The atmospheric pressure is measured to be 754 torr. Calculate the pressure in
Pa.
2
Harvard University
BLB: Ch. 1; PHH: Ch. 1
Dimensional Analysis II: Volume and Density
1.
Here are some old-fashioned units used for measuring volumes:
1 bushel = 1.24 cubic feet (ft3 )
1 cord = 128 ft3
1 hogshead = 63 gallons
1 gallon = 0.107 bushels
Also recall that 1 ft = 12 inches, and that 1 inch = 2.54 centimeters.
a)
How many bushels is equal to 3.5 cords? (Firewood is usually measured in
cords.)
b)
How many hogsheads is 3.5 cords?
c)
A rectangular tank measures 1.0 × 5.0 × 2.0 meters. Calculate its volume in
gallons.
2. A certain crystal of calcium chloride has a volume of 765.3 mm 3.
a)
Calculate the volume of this crystal in cubic inches (in 3).
b)
The density of CaCl2 is 2.51 g/cm3. Calculate the mass of this crystal.
3
Harvard University
BLB: Ch. 2; PHH: Ch. 2
Atoms, Molecules, and Ions
1.
For each of the following atoms or ions, provide the number of protons, neutrons,
and electrons:
a)
40Ar
b)
40Ca2+
c)
39K +
d)
39K
2.
Calcium chloride is an ionic substance. Determine the number of protons and
electrons in a calcium ion (as found in calcium chloride).
3.
The chloride ion exists in two common isotopes: 35Cl– and 37Cl–. Determine the
number of protons, neutrons, and electrons for both isotopes.
4.
The chemical formuls for ammonia is NH3.
a)
Please calculate the total number of protons, neutrons, and electrons in one
molecule of NH 3. (Assume that the only isotopes present are 14N and 1H.)
b)
In aqueous solution, some of the ammonia is present as the ammonium ion, NH4+ .
Calculate the total number of protons, neutrons, and electrons in one ammonium
ion, NH4+. (Again, assume that the only isotopes present are 14N and 1H.)
4
Harvard University
BLB: Ch. 4; PHH: Ch. 3
Oxidation Numbers
1.
2.
Write the oxidation numbers of each atom in the following species:
BrO3–
H3AsO3
AsH3
CrCl3
KClO3
S2O32–
OF2
Na2O2
Fe3O4
Identify whether oxidation and reduction is taking place in each of the following
equations. If so, identify which species is being reduced and which is being
oxidized.
N2 + 3 H2 → 2 NH3
2 FeCl3 + 3 KI → 2 FeCl2 + KI3 + 2 KCl
2 HCl + Na2CO3 → 2 NaCl + H2O + CO2
3 Cl2 + 6 OH– → 5 Cl– + ClO3– + 3 H2O
5
Harvard University
BLB: Ch. 2; PHH: Ch. 3
Naming Compounds
1.
Write the chemical formula for each of the following species:
a)
Copper (I) phosphide
b)
Iron (III) sulfate
c)
Potassium chlorate
d)
Aluminum chloride
e)
Chromium (VI) oxide
f)
Ammonium iodide
g)
Lithium nitride
h)
Aluminum carbonate
i)
Cesium phosphate
j)
Rhenium (VII) oxide
k)
Tin (IV) chloride
l)
Gallium fluoride
m) Silver (II) Fluoride
n)
Potassium Nitrate
o)
Barium Phosphate
p)
Ammonium Sulfate
2.
Write an acceptable chemical name for each of the following species.
a)
SiCl4
b)
Mn2O7
c)
Mg3N2
6
Harvard University
BLB: Ch. 3; PHH: Ch. 3
Chemical Formulas: Percent Composition
1.
The nerve gas Sarin, which was released in a Tokyo subway station in 1996, has a
molecular formula of C4H10PO2F.
a)
Determine the composition (percent by mass) of each element in Sarin.
b)
An unknown compound is discovered in a raid on a terrorist organization; it is
believed that the compound is Sarin. When a 10.0-gram sample of this compound
is completely combusted, 15.6 g CO2 and 6.4 g H2O are produced, along with
other combustion products. Using numerical calculations, prove that this
unknown compound can not be Sarin.
2.
A pulverized rock sample believed to be pure calcium carbonate is subjected to
chemical analysis and found to contain 51.3% calcium, 7.7% carbon, and 41.0%
oxygen by mass. Demonstrate that this natural sample cannot be pure CaCO 3.
7
Harvard University
BLB: Ch. 3; PHH: Ch. 3
Empirical and Molecular Formulas
1.
A compound containing 79.37% C, 8.88% H, and 11.75% O was found to have a
molecular mass of approximately 270 g. For this compound, determine
a)
the empirical formula
b)
the molecular formula
c)
the exact molar mass
2.
Compound Z, which consists of only carbon, hydrogen, and oxygen, has just been
isolated from a tropical plant.
a)
When 5.467 grams of compound Z are burned in excess oxygen, 15.02 grams of
CO2 and 2.458 grams of H2O are produced. Determine the empirical formula of
compound Z.
b)
Other experiments suggest that compound Z has a molar mass of approximately
250 g/mol. Calculate the true molar mass of compound Z.
3.
When bleach and ammonia are mixed, certain toxic compounds can be formed.
a)
One compound is isolated which contains only nitrogen, hydrogen, and chlorine.
When a 1.376-gram sample of this compound is completely combusted, 0.478 g
of H2O and 0.372 g of N2 are recovered. Determine the empirical formula of this
compound.
b)
Other experiments suggest that the molar mass of this compound is about 50
g/mol. Determine the molecular formula of this compound.
8
Harvard University
BLB: Ch. 3; PHH: Ch. 4
Writing and Balancing Equations
1.
Balance the following equations using the simplest whole-number coefficients.
a)
CO(NH2)2 (aq) + HOCl (aq) → NCl3 (aq) + CO2 (aq) + H2O (l)
(Hint: Balance the N first)
b)
Ca3(PO4)2 (s) + SiO2 (s) + C (s) → P 4 (g) + CaSiO3 (l) + CO (g)
(Hint: Balance the P first)
2.
Photosynthesis (in plants) converts carbon dioxide and water into glucose
(C6 H12O6 ) and oxygen. Write and balance the chemical equation for
photosynthesis.
3.
Niobium metal will react with solid iodine to produce solid triniobium octaiodide.
Write and balance the equation for this process.
4.
Write and balance the chemical equation for the complete combustion of octane,
C8H18 (l).
9
Harvard University
BLB: Ch. 3; PHH: Ch. 4
Stoichiometry of Reactions
1.
Sodium hypochlorite, the active ingredient in Clorox, can be made by the
following reaction:
2 NaOH (aq) + Cl2 (g) → NaCl (aq) + NaClO (aq) + H2O (l)
If chlorine gas is bubbled continuously through a solution containing 60.0 g of
NaOH, how many grams of NaClO can be produced, assuming the reaction goes
to completion?
2.
For the following unbalanced chemical equation:
NiS + O2 + HCl → NiCl2 + H2 SO4
a)
Write the balanced chemical equation for this reaction.
b)
What mass of NiCl2 will be produced if 0.458 g of NiS reacts?
10
Harvard University
BLB: Ch. 3; PHH: Ch. 4
Stoichiometry with Limiting Reagents
1.
Vanadium (V) oxide, V2 O5 , can be reduced by zinc to form vanadium (II) oxide,
V2 O2 and zinc oxide, ZnO.
a)
Write and balance the chemical equation for this process.
b)
What mass of vanadium (II) oxide can be produced from a mixture of 100.0
grams of V2 O5 and 100.0 grams of Zn?
2.
Many binary compounds of phosphorus and sulfur have been prepared.
a)
Balance the following chemical equation for the preparation of P 4S5. (Hint:
balance the S first.)
P4S3 + Br2 → P 4S5 + PBr3
b)
What is the maximum quantity of P4S5 that could be prepared from 100.0 g of
P4S3 and 100.0 mL of liquid bromine? (Density of bromine is 3.12 g/mL)
11
Harvard University
BLB: Ch. 3; PHH: Ch. 4
Stoichiometry of Mixtures
1.
Ferrous oxalate, FeC2 O4 , will decompose on heating:
FeC2 O4 → FeO + CO2 + CO
The gaseous products (a mixture of CO2 and CO) are collected. Calculate the
percent by mass of CO2 in this gaseous mixture.
2.
A certain mixture of CuO and Cu2 O weighs 10.50 grams total. Complete
reduction of this mixture produces 8.66 grams of pure metallic Cu. Determine the
amounts of CuO and Cu2 O in the original mixture.
3.
A certain mixture of CuS and Cu2 S weighs 10.80 grams total. Complete
reduction of this mixture produces 8.06 grams of pure metallic Cu. Determine the
amounts of CuS and Cu2 S in the original mixture.
12
Harvard University
BLB: Ch. 4; PHH: Ch. 4
Solutions: Molarity
1.
A 10.0 mL sample of concentrated sulfuric acid contains 17.7 g of H2SO4.
Determine the molarity of concentrated sulfuric acid.
2.
Pure acetic acid (C2H4O2) has a density of 1.049 g/mL. Calculate the molarity of
pure (anhydrous) acetic acid.
3.
A hydrochloric acid solution is prepared by dissolving 1.97g of hydrogen chloride
gas in 27.3 mL of water and then diluting that mixture to a total volume of 250.00
mL. Calculate the molarity of the resulting solution.
4.
The concentration of NaClO in Clorox is 0.705 M. Calculate the mass of NaClO
present in 1.0 mL of Clorox.
5.
A solution is prepared by dissolving x grams of potassium nitrate in water and
diluting to a total volume of 100.0 mL. Another solution is prepared by
dissolving y grams of sodium chloride in water and diluting to a total volume of
500.0 mL. Both solutions are then mixed together, giving a final concentration of
KNO3 of 0.073 M and a final concentration of NaCl of 0.128 M.
Calculate x and y.
13
Harvard University
BLB: Ch. 4; PHH: Ch. 5
Solution Stoichiometry I: Simple Examples
1.
The toxic compound NCl3 can be formed from the reaction of bleach (NaClO)
with ammonia (NH3):
NaClO (aq) + NH3 (aq) → NCl3 (l) + NaOH (aq)
2.
a)
Please write a complete, balanced, net ionic equation for this process.
b)
You accidentally pour 1.0 mL of Clorox into a large bucket of ammonia solution.
What mass of NCl3 can be produced if the reaction goes to completion? (Clorox
is a 0.705-molar solution of NaClO in water.)
A sample of calcium carbonate weighing 6.35 grams is placed in 500.0 mL of
0.31M hydrochloric acid and allowed to react to form calcium chloride and
carbon dioxide gas. Calculate the maximum mass of carbon dioxide gas that can
be produced.
14
Harvard University
BLB: Ch. 4; PHH: Ch. 5
Solution Stoichiometry II: Acid/Base Neutralizations
1.
What volume of 0.0843 M Ba(OH) 2 would be required to completely neutralize a
1.00-mL sample of 12.0 M acetic acid?
2.
A student has dissolved 87.5 grams of sodium hydroxide in 1.53 liters of water.
This strongly basic solution must be neutralized before disposal. What volume of
1.27 M HCl would be required to completely neutralize this solution?
3.
Phosphoric acid can be neutralized by sodium hydroxide according to the
equation:
H3PO4 + 3 NaOH → Na3PO4 + 3 H2O
What volume of 0.176 M NaOH would be required to neutralize 5.00 mL of
concentrated (14.8 M) phosphoric acid?
15
Harvard University
BLB: Ch. 4; PHH: Ch. 5
Solution Stoichiometry III: Titrations
1.
You are given 2.00 grams of a white solid and told that it is a mixture of NaCl and
Na2CO3. You dissolve the sample in water and titrate with HCl; complete
neutralization requires 23.0 mL of 1.00 M HCl:
Na2CO3 + 2 HCl → H2CO3 + 2 NaCl
Calculate the mass of Na2CO3 in the unknown sample.
2.
A 10.0-gram sample of an unknown solid monoprotic acid is dissolved in water
and titrated with 0.789 M NaOH, requiring 40.6 mL to reach the endpoint.
Determine the molar mass of the unknown acid.
3.
HX is a monoprotic acid; it has only one acidic proton per molecule. When 1.736
grams of HX are titrated with 1.334 M NaOH, the endpoint is reached with 10.92
mL of base. Calculate the molar mass of HX.
16
Harvard University
BLB: Ch. 4; PHH: Ch. 5
Solution Stoichiometry IV: Precipitation Reactions
1.
To 250. mL of a 0.065 M solution of barium nitrate is added 100. mL of a 0.121
M solution of sulfuric acid. Determine the concentration of barium nitrate after
complete precipitation of barium sulfate.
2.
Four individual solutions are prepared and mixed together in the following order:
1. Start with 100. mL of 0.100 M BaCl2
2. Add 50. mL of 0.100 M AgNO3 ; a precipitate of AgCl is formed.
3. Add 50. mL of 0.100 M H2 SO4 ; a precipitate of BaSO4 is formed.
4. Finally, add 250. mL of 0.100 M NH3 to neutralize the acid.
Determine the concentrations of each of the following species in the resulting
mixture: Ba2+ , Cl– , NO3 – , NH3 , NH4 + (Hint: What is the total volume?)
17
Harvard University
BLB: Ch. 10; PHH: Ch. 6
The Ideal Gas Law
1.
2.
Chlorine (Cl2), a toxic, pungent gas, has many uses in manufacturing and
industry. You may be familiar with its odor because it is sometimes used as a
disinfectant in swimming pools. Humans can detect the odor of chlorine when it
is present at a pressure as low as 2.0 × 10–7 atm.
a)
Calculate the volume (in liters) of a classroom which is 20. meters long, 10.
meters wide, and 3.0 meters high. (Recall that 1 mL = 1 cm3).
b)
What mass of Cl2 must be released into the classroom described above at 25°C in
order for humans to detect the odor of chlorine?
It requires 0.182 mol of O2 gas to exert a pressure of 1.50 atm in a particular tank
at 25°C. What mass of O2 would be required to exert a pressure of 17.2 atm in
the same tank at 100°C? Neglect expansion of the tank itself.
18
Harvard University
BLB: Ch. 10; PHH: Ch. 6
Reactions Involving Gases: Simple Examples
1.
In one experiment, 0.780 g Nb (s) was sealed in a 28.0 mL glass tube at 25°C
under 6.33 atm of hydrogen gas, H2. After reacting with the hydrogen for one
week, all of the niobium had been converted to niobium hydride, NbH. Calculate
the final pressure of hydrogen gas in the system at 25°C.
2.
Ferrous oxalate, FeC2 O4 , will decompose on heating:
FeC2 O4 (s) → FeO (s) + CO2 (g) + CO (g)
A 1.25-gram sample of FeC2 O4 is added to an evacuated 2.00-liter steel vessel.
The vessel is heated to 400°C, at which point all the FeC2 O4 is decomposed.
Calculate the pressure inside the vessel at 400°C.
3.
Ethylene, C2 H4 , will react with hydrogen gas under appropriate conditions to
form ethane, C2 H6 . A 10.0-liter vessel is charged with 1.5 atm of hydrogen and
1.0 atm of ethylene at 25°C. The reaction is allowed to proceed to completion.
Determine the total pressure in the vessel at 25° at the completion of the reaction.
19
Harvard University
BLB: Ch. 10; PHH: Ch. 6
Mixtures of Gases
1.
2.
A 10.0-liter chamber contains a mixture of nitrogen and oxygen at a total pressure
of 760. torr and a constant temperature of 25°C. The mole fraction of oxygen in
the mixture is 0.211.
a)
Calculate the number of moles of oxygen in the chamber.
b)
Gas is pumped out of the chamber until the total pressure is 0.100 torr. Calculate
the new partial pressure of oxygen in the chamber.
c)
Pure nitrogen gas is added to the chamber until the total pressure is again 760.
torr, then gas is pumped out of the chamber until the total pressure is 0.100 torr.
Calculate the new partial pressure of oxygen after this process.
A sample of the gas butane (C4H10), of unknown mass, is contained in a vessel of
unknown volume, V, at 24.8°C and a pressure of 560.0 torr. To this vessel
8.6787 g of Ne are added in such a way that no butane escapes. The total pressure
of the vessel (at the same temperature) is 1420.0 torr. Calculate the volume of the
vessel and the mass of the butane.
20
Harvard University
BLB: Ch. 10; PHH: Ch. 6
Collecting Gases Over Water
1.
Hydrogen gas is generated in the laboratory and collected over water at 25°C and
1.00 atm total pressure. The total volume of gas collected is 1.37 L. Calculate the
mass of hydrogen collected. (The vapor pressure of water at 25°C is 23.8 torr).
2.
Potassium chlorate, KClO3 , decomposes when heated:
KClO3 (s) → KCl (s) + 3 /2 O2 (g)
1.00 gram of potassium chlorate is completely decomposed and the oxygen is
collected over water at 25°C and 0.99 atm. Calculate the total volume of gas
collected. (The vapor pressure of water at 25°C is 23.8 torr).
3.
A 2.00-liter steel cylinder of oxygen gas has a pressure of 17.2 atm at 25°C. You
want to reduce the pressure in the cylinder to 15.5 atm, so you allow oxygen to
escape from the cylinder and collect the escaping gas over water. What volume of
gas should you collect at 25°C and 1.00 atm in order to achieve the desired
pressure in the cylinder? (The vapor pressure of water is 23.76 torr at 25°C.)
21
Harvard University
BLB: Ch. 10; PHH: Ch. 6
Stoichiometry of Gas Mixtures
1.
A 2.42 gram sample of PCl5 was placed into an evacuated 2.00-L flask and
allowed to partially decompose at 250.°C according to the following equation:
PCl5 (g) → PCl3 (g) + Cl2 (g)
The total pressure in the flask after partial decomposition was 359 torr. Calculate
the mole fraction of each gas in the flask.
2.
Ozone (O3) can be prepared in the laboratory by passing an electrical discharge
through a quantity of oxygen gas (O2):
3 O2 (g) → 2 O 3 (g)
An evacuated steel vessel with a volume of 10.00 liters is filled with 32.00 atm of
pure O2 at 25°C. An electric discharge is passed through the vessel, causing some
of the oxygen to be converted into ozone. As a result, the pressure inside the
vessel drops to 30.64 atm at 25°C. Calculate the final percent by mass of ozone
in the vessel.
22
Harvard University
BLB: Ch. 10; PHH: Ch. 6
Reactions Involving Gases: Applications
1.
An automobile air bag is filled with nitrogen gas, which is produced by the rapid
thermal decomposition of sodium azide (NaN 3):
2 NaN3 (s) → 2 Na (s) + 3 N 2 (g)
What mass of NaN3 would be required to produce enough nitrogen gas to fill a
10.0-liter air bag at a pressure of 1.20 atm and a temperature of 25°C?
2.
Consider the reaction used to produce NaAlCl4:
→ 4 NaAlCl4 (l) + 3 O2 (g)
2 Al2O3 (s) + 6 Cl2 (g) + 4 NaCl (l) ←
An evacuated reaction vessel with a volume of 10.0 liters is filled with chlorine
gas at 25°C and a pressure of 1.00 atm. Al2O3 and NaCl are added, and the
mixture is heated to 850°C until the reaction is complete. The final total pressure
in the vessel (at 850°C) is 2.70 atm.
a)
What mass of Cl2 was present in the reactor initially?
b)
Determine the final partial pressures of Cl2 and O2 in the reactor.
c)
Calculate the mass of NaAlCl4 produced in the reaction.
23
Harvard University
BLB: Ch. 10; PHH: Ch. 6
Kinetic-Molecular Theory of Gases
1.
A mixture of ammonium nitrate and butane are placed into a steel cylinder and
detonated. The product gases (CO 2, H2O, and N2) reach a temperature of 725°C.
a)
Calculate the rms velocity of each gas.
b)
Calculate the average kinetic energy per molecule of each gas.
2.
A 2.00-liter steel cylinder of oxygen gas has a pressure of 17.2 atm at 25°C. You
want to reduce the pressure in the cylinder to 15.5 atm, so you drill a tiny pinhole
and allow the oxygen to effuse into a vacuum. You know that, under identical
conditions, this pinhole will allow 0.017 moles of argon gas to effuse every hour.
a)
Calculate the rms velocity of argon gas under these conditions.
b)
How long should you let the oxygen escape from this pinhole in order to reduce
the pressure from 17.2 atm to 15.5 atm?
24
Harvard University
BLB: Ch. 5; PHH: Ch. 7
Heat
1.
2.
50 g of marble chips (heat capacity 0.94 J/g K) are heated from 25°C to 200°C.
a)
How much heat is consumed in this process?
b)
The hot marble chips are placed in 500. g of cold (10°C) water. Calculate the
final temperature of the system.
A coffee-cup calorimeter contains 150. g of water at 24.6°C. A 110-g piece of
molybdenum metal is heated to 100.°C and placed in the water in the calorimeter.
The system reaches equilibrium at a final temperature of 28.0°C. Calculate the
specific heat capacity of molybdenum metal. (The specific heat capacity of water
is 4.184 J/g°C).
25
Harvard University
BLB: Ch. 5; PHH: Ch. 7
Calorimetry I: Simple Examples
1.
Consider the dissolving of sodium carbonate in water:
Na2CO3 (s) → Na2CO3 (aq)
When 2.0 grams of sodium carbonate are dissolved in 100. g of water, the
temperature of the solution increases from 25.0°C to 26.2°C. Calculate ΔH° for
the dissolving of sodium carbonate. (The heat capacity of the solution is 4.2
J/g°C.)
2.
A sample of solid naphthalene, C10H8 , weighing 0.6037 g, is burned by O2 (g)
under pressure to form CO2 (g) and H2 O (l) in a bomb calorimeter. The
temperature rises by 2.270°C. The heat capacity of the calorimeter and its
contents is 10.69 kJ/K. Calculate ΔH for the combustion of naphthalene.
(For this calculation, assume that ΔH ≈ ΔE.)
3.
When a 0.235-gram sample of benzoic acid is burned in a calorimeter, a 1.642°C
rise in temperature is observed. When a 0.265-gram sample of caffeine
(C8 H10O2 N4 ) is burned, a 1.525°C rise in temperature is measured. Using the
value 26.38 kJ/g for the heat of combustion of benzoic acid, calculate the heat of
combustion per mole of caffeine at constant volume. (For this calculation, assume
that ΔH ≈ ΔE.)
26
Harvard University
BLB: Ch. 5; PHH: Ch. 7
Calorimetry II: Mixtures
1.
Ammonium nitrate is a particularly potent explosive when mixed with a
combustible liquid like fuel oil; the Oklahoma City bomb used such a mixture.
The fuel will explode along with the ammonium nitrate.
Consider a mixture of ammonium nitrate and butane, C4H10 (a component of
petroleum fuel). The heat of detonation of ammonium nitrate is –118 kJ/mol,
while the heat of combustion of butane is –2635 kJ/mol.
A mixture of 1.00 g NH4NO3 and 1.00 g C4H10 are burned in a bomb calorimeter
with Ccal = 10.43 kJ/°C. The initial temperature is 22.59°C. Predict the final
temperature of the system. (For this calculation, assume that ΔH ≈ ΔE.)
2.
With all the hot weather this past summer, everyone was drinking a lot of iced tea.
(Iced tea sales increase dramatically when the temperature gets above 70°F.)
Given ΔH° for the melting of ice:
H2O (s) → H2O (l)
ΔH° = +6.01 kJ/mol
Given 400.0 grams of hot tea at 80°C, calculate the mass of ice at 0°C which must
be added to obtain iced tea at 10°C. (The specific heat of the tea is 4.18 J/g°C.)
27
Harvard University
BLB: Ch. 5; PHH: Ch. 7
Hess's Law
1.
Given the following enthalpies of reaction:
N2 O4 (g) → 2 NO2 (g)
NO (g) + 1/2 O2 (g) → NO2 (g)
ΔH° = 57.20 kJ
ΔH° = –57.07 kJ
Calculate the enthalpy of reaction for:
2 NO (g) + O2 (g) → N 2 O4 (g)
2.
Given the following enthalpies of reaction:
NH3 (g) → NH3 (aq)
HNO3 (g) → HNO3 (aq)
NH4NO3 (s) → NH4NO3 (aq)
NH3 (aq) + HNO3 (aq) → NH4NO3 (aq)
Calculate the enthalpy of reaction for:
NH3 (g) + HNO3 (g) → NH4NO3 (s)
28
ΔH = –34.10 kJ/mol
ΔH = –72.3 kJ/mol
ΔH = 26.4 kJ/mol
ΔH = –52.3 kJ/mol
Harvard University
BLB: Ch. 5; PHH: Ch. 7
Standard Enthalpies of Formation
1.
Using the following information:
ΔH°f (CaO (s)) = –635.5 kJ/mol
ΔH°f (H2O (l)) = –285.83 kJ/mol
ΔH°f (Ca(OH)2 (s)) = –986.2 kJ/mol
Calculate ΔH for the following reaction:
CaO (s) + H2 O (l) → Ca(OH)2 (s)
2.
Given the following data for butane (C 4H10):
ΔH°f (C4H10 (g)) = –124.7 kJ/mol
ΔH°f (H2O (g)) = –241.8 kJ/mol
ΔH°f (CO2 (g)) = –393.5 kJ/mol
ΔHvap(C4H10 (l)) = 22.9 kJ/mol
Density of C4H10 (l) = 0.59 g/mL
and the balanced equation for the combustion of liquid butane:
C4H10 (l) + 13/2 O2 (g) → 4 CO 2 (g) + 5 H2O (g)
Calculate the quantity of heat released by the combustion of 10.0 mL of liquid
butane (the amount of butane in a cigarette lighter).
29
Harvard University
BLB: Skip; PHH: Ch. 7
Enthalpy, Energy, Heat, and Work
1.
Consider the evaporation of one mole of methanol, CH3 OH, at its boiling point
(64.4°C) at constant pressure (1.00 atm). The density of liquid methanol is 0.791
g/mL, and you are given the Useful Information table (below).
a)
Calculate ΔΗ for the evaporation of one mole of methanol at its boiling point.
b)
Calculate the work w for the evaporation of one mole of methanol at its boiling
point.
c)
Determine ΔE for the evaporation of one mole of methanol at its boiling point.
Substance:
CH3OH (g)
CH3OH (l)
ΔH°f (kJ/mol)
–201.2
–238.6
30
Harvard University
BLB: Ch. 5; PHH: Ch. 7
Putting It Together: Calorimetry, Hess's Law, and ΔH°f
1.
When a 4.35-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a
coffee-cup calorimeter, the temperature of the solution drops from 22.0°C to
16.9°C.
NH4NO3 (s) → NH4+ (aq) + NO3– (aq)
a)
Calculate ΔH° for the above reaction. (The specific heat capacity of the solution
is 4.18 J/g°C.)
b)
Using the following thermochemical data, and your answer to part (a) above,
determine ΔH°f for NH4NO3 (s).
ΔH°f (NH4+ (aq)) = –132.5 kJ/mol
2.
ΔH°f (NO3– (aq)) = –205.0 kJ/mol
Sodium azide will decompose explosively when heated; this reaction is the source
of the gas used to inflate automobile air bags:
2 NaN3 (s) → 2 Na (s) + 3 N 2 (g)
a)
A 10.0-gram sample of sodium azide is completely decomposed in a bomb
calorimeter. If the heat capacity of the calorimeter and its contents is 2750. J/K,
and the temperature increases from 25.78°C to 27.20°C, calculate ΔH° for the
above reaction. (For this calculation, assume that ΔH ≈ ΔE.)
b)
From this information, calculate ΔH°f for sodium azide.
31
Harvard University
BLB: Ch. 5; PHH: Ch. 7
Putting It Together:
Stoichiometry, Thermochemistry, and Gas Laws
1.
An empty 4.00-liter steel vessel is filled with 1.0 atm CH4 (g) and 4.0 atm O2 (g)
at 300°C. A spark causes the CH4 to completely combust according to the
following equation:
CH4 (g) + 2 O2 (g) → CO 2 (g) + 2 H2O (g)
ΔH° = –802 kJ/mol
a)
Calculate the mass of CO2 (g) which would be produced in this reaction.
b)
Determine the final temperature inside the vessel after combustion; neglect any
loss of heat to the vessel or surroundings. (The mixture of gases has an average
molar heat capacity of 21 J/mol·°C)
c)
Calculate the partial pressure of CO2 (g) in the vessel after combustion.
32
Harvard University
BLB: Ch. 6; PHH: Ch. 9
Energy, Particles, and Waves: A "Chain Problem"
Useful Information:
Speed of light c = 3.00 × 108 m/s
Planck's constant h = 6.63 × 10–34 J·s
Mass of electron m e = 9.11 × 10–31 kg
1. a) A photon produced by an X-ray machine has an energy of 4.70 × 10–16 J. What is
the frequency of this photon?
b)
What is the wavelength of radiation of frequency (a)?
c)
What is the velocity of an electron with a deBroglie wavelength equal to (b)?
d)
What is the kinetic energy of an electron traveling at velocity (c)?
33
Harvard University
BLB: Ch. 6; PHH: Ch. 9
The Photoelectric Effect
The photoelectric binding energy of chromium is 7.21 × 10 –19 J.
1.
a)
Calculate the minimum frequency of light which could produce the photoelectric
effect in chromium.
b)
Light having a wavelength of 2.50 × 10–7 m falls upon a piece of chromium in an
evacuated glass tube. Determine the minimum deBroglie wavelength of the
emitted photoelectrons.
34
Harvard University
BLB: Skip; PHH: Ch. 9
Orbitals and Wavefunctions
1.
Here are three graphs of the radial wavefunction of the 3pz orbital:
(radii are in units of ao )
0.008
0.006
0.004
0.002
5
10
15
5
10
15
5
10
15
0.1
0
-0.1
0.1
0.08
0.06
0.04
0.02
a)
Identify each as representing R, R2 , or r2 R2 .
b)
Which of the following describes a place where the electron will never be found:
c)
2.
(A) θ = 0, r = 2ao
(B) θ = 45°, φ = 90°, r = 6ao
(C) θ = 90°, φ = 45°, r = 12ao
(D) θ = 55°, φ = 0°, r = 10ao
Choose the most likely radius to find an electron:
(A) r = 12ao
(B) r = 2ao
(C) r = 6ao
(D) r = 10ao
Which of the following could be a slice of the 4fz 3 orbital in the (x, z) plane:
A
B
C
35
D
Harvard University
BLB: Ch. 6; PHH: Ch. 9
Orbitals and Quantum Numbers
1.
For each of the following subshells, identify the n and l quantum numbers, and
state the total number of electrons that could be found in each subshell.
Subshell
n
l
total #
of e–
1s
3d
5p
2.
Using the Rydberg equation:
1
1
ΔE = (2.18 × 10–18 J)  2 – 2
n
n
 i
f 
Calculate the energy of the radiation emitted by a hydrogen atom if an electron
undergoes a transition from a 4px orbital to a 2s orbital.
36
Harvard University
BLB: Ch. 6; PHH: Ch. 9
Electron Configurations of Neutral Atoms
1.
Write the electron configurations of the following neutral atoms. You may use
the noble-gas abbreviations (i.e. [Ar] . . . ).
C:
Mg:
Mn:
Se:
Cu:
Xe:
Ba:
Os:
Pb:
37
Harvard University
BLB: Ch. 6; PHH: Ch. 9
Electron Configurations of Ions
1.
Write the complete ground state electron configurations for the following ions.
Do not use the noble-gas abbreviations.
Be+ :
N– :
Al3+ :
H– :
O2– :
2.
Write the ground state electron configurations for the following ions. Use the
noble-gas abbreviations if appropriate.
Zn2+ :
W6+ :
Cu2+ :
Gd3+ :
Se2– :
38
Harvard University
BLB: Ch. 7; PHH: Ch. 10
Periodic Properties I: Overview
1.
Indicate “high” and “low” areas for each property on the mini periodic tables
below. Explain each trend briefly.
a)
Ionization Energy
b)
Atomic Mass
c)
Atomic Radius
d)
Metallic Character
2.
Find the following on the periodic tables below, and indicate their identity:
a)
Most reactive metal
b) Most reactive nonmetal
Identity: __________
Identity: __________
39
Harvard University
BLB: Ch. 7; PHH: Ch. 10
Periodic Properties II: Multiple Choice
1.
Circle the element from each set with the largest atomic radius. Explain your
choices.
a)
Ba
Ti
Ra
Li
b)
F
Al
In
As
2.
3.
Circle the element from each set with the smallest ionization energy. Explain
your choices.
a)
Tl
Po
Se
Ga
b)
Cs
Ga
Bi
Se
Circle the element with the most negative electron affinity. Explain your
choice.
Be
4.
N
O
F
Circle the ion with the largest radius. Explain your choice.
Se2–
F–
O2–
40
Rb+
Harvard University
BLB: Ch. 7; PHH: Ch. 10
Periodic Properties III: Explanations
1.
Provide brief explanations for the following observations:
a)
The first ionization energy of Se is less than the first ionization energy of As.
b)
More energy is released upon adding an electron to Br than upon adding one to
Se.
c)
The reaction of Rb with water is much more violent that the reaction of Na with
water.
2.
Two of the most fundamental properties of any element are its ionization energy
and electron affinity.
a)
Fluorine has an electron affinity of –332 kJ/mol—one of the most negative of any
of the elements. Briefly explain why its electron affinity is more negative than
that of oxygen. (One or two sentences should suffice.)
b)
This very negative electron affinity makes fluorine extremely reactive. Explain.
c)
The noble gas xenon has an ionization energy of 1170 kJ/mol. This substantial
ionization energy helps to make xenon very unreactive. Explain.
d)
Xenon’s neighbor on the periodic table, iodine, also has a high ionization energy
(1020 kJ/mol). Iodine is quite reactive, however, unlike xenon. Explain.
41
Harvard University
BLB: Ch. 7; PHH: Ch. 10
Putting It Together:
Quantum Mechanics and Electronic Structure
1.
Consider the electronic structure of the element bismuth (Bi).
a)
The first ionization energy of bismuth is 703 kJ/mol. Calculate the longest
possible wavelength of light which could ionize an atom of bismuth.
b)
Write the electron configurations of neutral Bi and the Bi+ cation. Please use the
noble-gas abbreviations.
Bi :
Bi+ :
c)
What are the n and l quantum numbers of the electron removed when Bi is ionized
to Bi+ ?
n=
d)
l=
Would you expect Element 113 to have an ionization energy which is greater
than, equal to, or less than that of bismuth?
(circle one):
greater than
equal to
less than
Explain briefly. (It is not enough to simply state a trend; you must provide an
explanation for the trend in terms of atomic structure.)
42
Harvard University
BLB: Ch. 8; PHH: Ch. 11
Lewis Structures I: The Octet Rule
1.
For each of the following molecules, draw the best possible Lewis structure.
Include all non-zero formal charges, and indicate resonance if appropriate.
CF4
OF2
HCN
CO
NSF
HNO3
43
Harvard University
BLB: Ch. 8; PHH: Ch. 11
Lewis Structures II: Ions
1.
For each of the following ions, draw the best possible Lewis structure. Include all
non-zero formal charges, and indicate resonance if appropriate.
SF3+
C22–
NO2–
PH2–
CO32–
BH4–
44
Harvard University
BLB: Ch. 8; PHH: Ch. 11
Lewis Structures III: Less Than an Octet
1.
For each of the following molecules or ions, draw the best possible Lewis
structure. Include all non-zero formal charges, and indicate resonance if
appropriate.
BBr3
BeH2
AlF3
ClO2
NO2
NO
45
Harvard University
BLB: Ch. 8; PHH: Ch. 11
Lewis Structures IV: More Than an Octet
1.
For each of the following molecules or ions, draw the best possible Lewis
structure. Include all non-zero formal charges, and indicate resonance if
appropriate.
SeF4
XeF2
AsCl5
AsF6–
XeF3+
BrF2–
46
Harvard University
BLB: Ch. 8; PHH: Ch. 11
Bond Enthalpies
1.
The enthalpy of formation ΔH°f of acetylene (C2 H2 ) is 226.6 kJ/mol.
a)
Draw a correct Lewis structure for acetylene.
b)
Given the following additional information, determine the bond enthalpy of the
C≡C triple bond in acetylene.
ΔHsub (C (s)) = 717 kJ/mol
2.
DH—H = 436 kJ/mol
DC—H = 415 kJ/mol
Cyanogen, NCCN, is a highly toxic gas.
a)
Draw a correct Lewis structure for cyanogen
b)
Given the following average bond enthalpies: (in kJ/mol)
C–N 293
C–O 358
C–C 348
C=N 615
C=O 799
O–O 146
C≡N 891
C≡O 1072
O=O 495
Estimate ΔH for the complete combustion of cyanogen:
NCCN (g) + 2 O2 (g) → N 2 (g) + 2 CO2 (g)
47
N–N 163
N=N 418
N≡N 941
Harvard University
BLB: Ch. 9; PHH: Ch. 11&12
Molecular Geometry I: Neutral Molecules
1.
For each of these molecules, draw the best possible Lewis structure. Using that
Lewis structure, predict the electron-pair geometry, the molecular geometry, the
hybridization of the central atom, and whether the molecule is polar.
CO2
ONF
e- pair geometry:
e- pair geometry:
molecular geometry:
molecular geometry:
hybridization:
hybridization:
polar?
yes
no
polar?
yes
BF3
ICl3
e- pair geometry:
e- pair geometry:
molecular geometry:
molecular geometry:
hybridization:
hybridization:
polar?
yes
no
polar?
48
yes
no
no
Harvard University
BLB: Ch. 9; PHH: Ch. 11&12
Molecular Geometry II: Ions
1.
2.
For each of the following molecules, draw the best possible Lewis structure. On
the basis of that Lewis structure, predict the electron-pair geometry, the molecular
geometry, and the hybridization of the central atom.
IF4–
PCl4+
e- pair geometry:
e- pair geometry:
molecular geometry:
molecular geometry:
hybridization:
hybridization:
SeO32–
I3–
e- pair geometry:
e- pair geometry:
molecular geometry:
molecular geometry:
hybridization:
hybridization:
For extra practice in applying VSEPR theory, go back to the pages of Lewis
structures and predict the geometries of each of those molecules or ions.
49
Harvard University
BLB: Ch. 9; PHH: Ch. 11&12
Molecular Geometry III: Polycentric Molecules and Ions
1.
For each of the following molecules, draw the best possible Lewis structure. On
the basis of that Lewis structure, predict the electron-pair geometry, the molecular
geometry, and the hybridization of every non-terminal atom.
a)
F3S—S—F
b)
CH3—C≡C—CO2–
c)
I5– (shaped like a big “V”)
50
Harvard University
BLB: Ch. 9; PHH: Ch. 11&12
Putting It Together:
Lewis Structures, VSEPR, and Bonding
1.
Consider the phase transition between S8 (g) and S2 (g):
1/
8
S8 (g) → 1/2 S2 (g)
ΔH = 49.13 kJ
a)
The molecule S8 has eight sulfur atoms arranged in a ring, or circle. Draw a
Lewis structure for S8. (Note that each sulfur atom should be identical.)
b)
Specify the hybridization, electron pair geometry, and molecular geometry around
each sulfur atom in S8.
c)
The average S—S single bond energy is 266 kJ/mol. Using the ΔH given at the
top of this page, calculate the S=S double bond energy in S2 (g).
51
Harvard University
BLB: Ch. 9; PHH: Ch. 12
Hybridization and Multiple Bonding: σ and π Bonds
1. a)
Determine the geometry and hybridization of the carbon atom in formaldehyde,
H2C=O.
Explain the bonding in formaldehyde: describe each σ- and π-bond in the
molecule. (e.g. “There are two σ-bonds between H s orbitals and C sp2 hybrid
orbitals”)
c)
Imagine that formaldehyde is positioned in the y,z plane as shown below. On this
framework, draw a perspective sketch of the formaldehyde molecule in this
orientation, showing all σ- and π-bonding orbitals and the lone pairs on the
oxygen atom.
x-axis
b)
H
C
O
z-axis
H
is
x
y-a
2.
Give the hybridization of each carbon atom in the tetrolate ion, CH3–C≡C–CO2–.
Describe each of the σ- and π-bonds in this molecule. (Don’t forget the C–H
bonds, and include resonance!)
52
Harvard University
BLB: Ch. 9; PHH: Ch. 12
Putting It Together: Bonding and Hybridization
1.
Consider the bonding in carbon dioxide, CO2:
a)
Draw the best possible Lewis structure of CO2.
b)
For the carbon atom, determine:
molecular geometry:
hybridization:
c)
What carbon hybrid orbital(s) are participating in σ-bonding in CO2?
d)
What carbon orbital(s) are participating in π-bonding in CO 2?
e)
The molecule allene, H 2C=C=CH2, is similar to CO2 in several respects. Draw a
complete Lewis structure for allene.
f)
For allene, determine the hybridization of:
the central carbon atom:
a terminal carbon atom:
g)
Allene is not planar; the two ends of the molecule are in different perpendicular
planes as shown:
H
H
C
C
C
H
H
Draw π-bonding orbitals on the above diagram, showing clearly why the two ends
of the molecule must be perpendicular to one another.
53
Harvard University
BLB: Skip; PHH: Ch. 12
Covalent Bonding and Orbital Overlap
1.
For each of the following orbital interactions, identify:
• the name of each orbital (1s, 2p, etc.)
• the type of interaction (σ, π*, etc.)
The black and white areas are meant to represent (+) and (–) signs, respectively.
a)
type of interaction:
type of interaction:
b)
c)
type of interaction:
d)
type of interaction:
e)
type of interaction:
54
Harvard University
BLB: Ch. 9; PHH: Ch. 12
Molecular Orbital Theory I: Introduction
1.
For each of the following diatomic molecules or ions:
• Write the Molecular Orbital electron configuration (i.e. (σ2s)2 etc.)
• Determine the bond order
• Identify the species as paramagnetic or diamagnetic
a) B2
b) O2–
c) BN
2.
Consider the bonding in NO from the viewpoint of molecular orbital theory:
a)
Draw an energy-level diagram for the valence molecular orbitals of NO. Be sure
to include the atomic orbitals of N and O as well as the molecular orbitals.
(Note: The molecular orbitals of NO have the same energy order as those for O2.)
b)
The NO molecule can lose or gain an electron to form NO + or NO–, respectively.
Circle the single best choice for each of the following questions.
Has a double bond
NO
NO+
NO–
Is diamagnetic
NO
NO+
NO–
Has the longest bond of the three species
NO
NO+
NO–
Is isoelectronic with CO
NO
NO+
NO–
55
Harvard University
BLB: Ch. 9; PHH: Ch. 12
Molecular Orbital Theory II: Extensions of MO Theory
1.
Here is the molecular orbital correlation diagram (valence orbitals only) for the
triatomic molecule CO2, showing both σ- and π-bonding:
σ2p*
σ2p*
π2p*
2p
2p
nonbonding
π2p
σ2p
σ2p
Note: "nonbonding"
means that the orbital is
not involved in bonding in
any way (neither bonding
nor antibonding).
σ2s*
2s
2s
σ2s
C
CO2
O
O
atomic
orbitals
molecular
orbitals
atomic
orbitals
atomic
orbitals
a)
Fill in the appropriate number of valence electrons in the C, O, and O atomic
orbitals in the diagram above. Use arrows (↑ or ↓) to indicate electron spin.
b)
Fill in the appropriate total number of valence electrons in the CO 2 molecular
orbitals in the diagram above. Use arrows to indicate electron spin.
c)
What is the total number of bonds (total bond order) for CO2?
d)
Is CO2 (circle one) paramagnetic
e)
You add an electron to CO2, forming the CO2– ion.
Would the C–O bond length become (circle one)
shorter
or
56
diamagnetic
or
longer
Harvard University
BLB: Skip; PHH: Ch. 12
Band Theory and Semiconductors
1.
For each of the following pure solids, circle whether it would be an insulator, a
semiconductor, or a conductor. Then briefly explain each answer. (Hint: consider
the number of valence electrons)
a)
SiO2
insulator
semiconductor
conductor
b)
RuO2
insulator
semiconductor
conductor
c)
GaAs
insulator
semiconductor
conductor
d)
TaN
insulator
semiconductor
conductor
e)
Ta3N5
insulator
semiconductor
conductor
2.
The mineral wüstite consists of iron (II) oxide in which some of the Fe 2+ has been
replaced with Fe3+. Wüstite is observed to be a semiconductor. Is it an intrinsic
semiconductor, a n-type semiconductor, or a p-type semiconductor?
(circle one)
intrinsic
n-type
p-type
Explain your answer in terms of the band structure of this mineral.
57
Harvard University
BLB: Ch. 11; PHH: Ch. 13
Intermolecular Forces
1. a) For each of the following substances, circle the one intermolecular force which
would predominate in the solid or liquid phase:
Substance
Intermolecular Forces (circle the predominant one)
Al2O3
London Dispersion
Dipole-Dipole
Hydrogen Bonding
Ionic
F2
London Dispersion
Dipole-Dipole
Hydrogen Bonding
Ionic
H 2O
London Dispersion
Dipole-Dipole
Hydrogen Bonding
Ionic
Br2
London Dispersion
Dipole-Dipole
Hydrogen Bonding
Ionic
ICl
London Dispersion
Dipole-Dipole
Hydrogen Bonding
Ionic
NaCl
London Dispersion
Dipole-Dipole
Hydrogen Bonding
Ionic
b)
Rank the above substances in order of their boiling points
Highest Boiling Point:
Lowest Boiling Point:
58
Harvard University
BLB: Ch. 11; PHH: Ch. 13
Phase Changes
1.
Elemental sulfur exhibits five (5) common phases. There are two solid phases of
sulfur, known as S(I) (s) and S(II) (s), a liquid phase S (l), and two gas phases,
one consisting of molecules of S8 (g) and one consisting of molecules of S 2 (g).
Here are the phases of sulfur at 1.0 atm pressure, with specific heats and the
enthalpy of each phase transition:
below 368 K
•
368 K transition
S(I) (s)
specific heat = 0.73 J/g·K
S(I) (s) → S(II) (s)
ΔH = 0.401 kJ/mol
368 – 388 K
•
388 K transition
S(II) (s)
S(II) (s) → S (l)
specific heat = 0.78 J/g·K
ΔH = 1.722 kJ/mol
388 – 718 K
•
718 K transition
S (l)
S (l) → 1/8 S8 (g)
specific heat = 1.04 J/g·K
ΔH = 45.08 kJ/mol
718 – 882 K
•
882 K transition
1/
above 882 K
S8 (g)
specific heat = 0.73 J/g·K
1
ΔH = 49.13 kJ/mol
8 S 8 (g) → /2 S 2 (g)
S2 (g)
specific heat = 0.59 J/g·K
a)
Identify the normal melting point and normal boiling point of sulfur.
b)
Calculate the total quantity of heat required to heat 1.00 gram of sulfur
from 25°C to 150.°C.
59
Harvard University
BLB: Ch. 11; PHH: Ch. 13
The Clausius-Clapeyron Equation
1.
The vapor pressure of water at 50°C is 92.5 torr. Knowing that the boiling point
of water is 100°C, estimate ΔHvap for water.
2.
Given that the heat of vaporization of water is roughly 42 kJ/mol, and knowing
that water ordinarily boils at 100°C, what would the pressure in a vacuum dome
have to be in order for water to boil at room temperature, 25°C?
60
Harvard University
BLB: Ch. 11; PHH: Ch. 13
Phase Diagrams
1.
Given the following data for iodine:
a)
normal boiling pt.
184°C
normal melting pt.
113°C
triple pt.
68 torr at 106°C
vapor pressure
1.0 torr at 39°C
Sketch a phase diagram for iodine in the space below. Observe the labeled axes
(not drawn to scale).
760
P
(torr)
68
1
39
b)
T (°C)
106
183
Label the following on the above diagram:
solid, liquid, gas, normal boiling point, normal melting point, triple point
c)
A quantity of iodine is introduced into an evacuated flask at 25°C. Two phases
are observed in equilibrium at this temperature. What are they?
d)
At a given temperature, which is more dense: solid or liquid iodine? Explain.
61
Harvard University
BLB: Ch. 11; PHH: Ch. 13
Bonding in Crystalline Solids
1.
In nature, sulfur often appears in sulfide minerals, which contain the S 2– anion.
Lead sulfide, PbS, is one example of such a mineral. Its unit cell (shown below)
is similar to that of NaCl. Given that the edge of the unit cell in PbS is 5.936 Å,
calculate its density in g/cm3.
S
Pb
S
Pb
S
Pb
Pb
S
Pb
S
Pb
S
Pb
S
Pb
Pb
2.
Pb
S
S
S
S
Pb
S
Pb
S
Pb
S
Calcium metal crystallizes in a face-centered cubic unit cell. The density of the
solid is 1.54 g/cm3. Calculate the radius of a calcium atom.
62
Harvard University
BLB: Skip; PHH: Ch. 13
X-Ray Diffraction: The Bragg Equation
1.
Polonium metal is the only element to crystallize in a simple cubic lattice. In a
diffraction experiment using X-rays with a wavelength of 0.7093 Å, reflections
off the parallel faces of the unit cells were observed at θ = 25.09° and θ = 39.49°
(among others).
a)
Assign the Bragg diffraction order (n) to each of these reflections. (There will be
a certain amount of trial and error... but think how you can limit your work. Also
consider the range of reasonable values for d.)
b)
Calculate the length of the unit cell in Å.
c)
Determine the density of polonium.
63
Harvard University
BLB: Ch. 8; PHH: Ch. 13
Lattice Energy: The Born-Haber Cycle
1.
2.
Construct a Born-Haber cycle for FeCl3 and determine its lattice energy.
IEFe (1,2,3) = 758, 1558, and 2952 kJ/mol
ΔHsub (Fe) = 415.5 kJ/mol
EACl = –349 kJ/mol
ΔH°f (FeCl3 ) = –401 kJ/mol
DCl–Cl = 243.4 kJ/mol
Given the following thermodynamic information:
Na(g)
Na (s)
Na (s) + C (s) + 1/2 N2 (g)
HCN (g)
1/2 H 2 (g) + C (s) + 1/2 N 2 (g)
H2 (g)
CN (g) + e–
→
→
→
→
→
→
→
Na+ (g) + e–
Na (g)
NaCN (s)
H (g) + CN (g)
HCN (g)
2 H (g)
CN– (g)
IE = 496 kJ/mol
ΔHsub = 108 kJ/mol
ΔH°f = –87 kJ/mol
DC–H = 502 kJ/mol
ΔH°f = 135 kJ/mol
DH–H = 436 kJ/mol
EA = –382 kJ/mol
Construct a Born-Haber cycle to calculate the lattice energy of NaCN (s).
(Hint: You will need to use all of the above information in your answer!)
64
Harvard University
BLB: Ch. 13; PHH: Ch. 14
Expressing Solution Concentrations
1.
A 1.457-molar solution of maltose (C12H22O11) has a density of 1.188 g/mL. For
this solution, calculate the following:
a)
the percent by mass of maltose
b)
the molality of maltose
c)
the mole fraction of maltose
d)
the mole fraction of water
e)
the molarity of water
65
Harvard University
BLB: Ch. 13; PHH: Ch. 14
Colligative Properties I: Non-Dissociating Solutes
1.
A 1.06 M solution of sucrose (C12H22O11) has a density of 1.14 g/mL at 25°C.
a)
Predict the boiling point of this solution.
b)
Predict the vapor pressure of water over this solution at 25.0°C. (The vapor
pressure of pure water at 25°C is 23.76 torr.)
2.
A solution containing 0.674 g of one type of interferon in 157 mL of aqueous
solution has an osmotic pressure of 4.15 torr at 25°C. Calculate the molar mass of
this type of interferon.
3.
A solution of Br2 in CCl4 has a total vapor pressure of 129 torr at 25°C. Calculate
the mole fraction of each component in the solution.
(At 25°C, P°(Br2 ) = 209 torr; P°(CCl4 ) = 109 torr)
66
Harvard University
BLB: Ch. 13; PHH: Ch. 14
Colligative Properties II: Dissociating Solutes
1.
2.
A 2.02 M solution of NaBr has a density of 1.15 g/mL at 25°C.
a)
Predict the boiling point of this solution.
b)
Predict the vapor pressure of water over this solution at 25°C. (Hint: Consider
whether you have a dissociating or non-dissociating solute, and be sure to
calculate the mole fraction appropriately.) The vapor pressure of pure water is
23.76 torr.
Sodium aluminum chloride, NaAlCl4, is soluble in water. One chemist suggests
that, on dissolving, it completely dissociates into six individual ions:
NaAlCl4 (s) → Na+ (aq) + Al3+ (aq) + 4 Cl– (aq)
(proposal A)
Another chemist believes that it dissociates into only two ions: a sodium cation
and the tetrachloroaluminate anion:
NaAlCl4 (s) → Na+ (aq) + AlCl4– (aq)
(proposal B)
When 6.00 grams of NaAlCl4 are dissolved in 100. grams of pure water, the
freezing point of the resulting solution is –3.5°C. Do the results of this
experiment support proposal A, proposal B, or neither? Provide calculations to
support your choice.
67
Harvard University
BLB: Ch. 13; PHH: Ch. 14
Colligative Properties III: Multiple Solutes
1.
When excess chlorine gas is bubbled into water, the resulting solution has the
following composition:
Cl2 (aq)
HOCl (aq)
HCl (aq)
0.062 M
0.030 M
0.030 M
Predict the freezing point of this solution. (Its density is 1.04 g/mL.)
Note:
2.
• Hypochlorous acid (HOCl) is a weak acid.
• HCl is a strong acid.
• Be sure to consider the total concentration of dissolved species.
Consider the following solution: 0.500 moles of the weak acid HN3 plus
0.750 moles of the ionic compound NaN3 in a total of 1.00 liter of solution. The
density of this solution is 1.02 g/mL.
Calculate the vapor pressure of water over this solution at 25°C. (The vapor
pressure of pure water at 25°C is 23.76 torr). (Hint: Be sure to consider ALL the
moles of solute particles in your calculation; assume ideal behavior.)
68
Harvard University
BLB: Ch. 13; PHH: Ch. 14
Colligative Properties IV: Non-Ideal Solutions
1.
2.
A 1.00-molar solution of magnesium sulfate, MgSO4, has a density of 1.11 g/mL.
The freezing point of this solution is –1.66 °C.
a)
Determine the van’t Hoff factor i for a 1.00-M solution of MgSO 4.
b)
What is the expected (ideal) van’t Hoff factor for MgSO4?
c)
Briefly discuss any discrepancy between the observed and expected van’t Hoff
factors.
In aqueous solution, the metasilicate ion (SiO32–) forms aggregates such as the
following:
n = 2:
n = 3:
n = 4:
2 SiO32– (aq) → Si2O64– (aq)
3 SiO32– (aq) → Si3O96– (aq)
4 SiO32– (aq) → Si4O128– (aq)
and so on . . . .
A chemist dissolves 0.0100 moles of Na2SiO3 in enough water to make 1.00 liter
of solution. Using a semipermeable membrane that allows Na+ ions to pass freely
but prevents the passage of the silicate ions, the osmotic pressure of this solution
is found to be 0.011 atm at 25°C.
Determine the average aggregation state (n) for the silicate ions under these
conditions. (Hint: the osmotic pressure is due to the concentration of the silicate
ions only.)
69
Harvard University
BLB: Ch. 14; PHH: Ch. 15
Rate Laws
1.
Using the following data, write a rate law for the reaction 2A + B + C → P.
[A], M
0.01
0.02
0.01
0.01
2.
[B], M
0.01
0.01
0.03
0.01
initial rate, M·s–1
0.3
1.2
0.9
0.3
[C], M
0.01
0.01
0.01
0.04
Nitrogen (II) oxide, NO, also called nitric oxide, reacts with chlorine according to
the following reaction:
2 NO (g) + Cl2 (g) → 2 NOCl (g)
The following initial rates of reaction have been obtained:
Experiment
1
2
3
[NO], M
0.50
1.00
1.00
[Cl2 ], M
0.50
0.50
1.00
rate, M/hr
1.14
4.56
9.12
a)
Write the rate law for this reaction. What is the overall order? the order with
respect to each reactant?
b)
Calculate the rate constant k. Be sure to include the correct units.
c) What rate would you expect if the initial concentration of NO is 2.0 mol/L and the
initial concentration of Cl2 is 1.50 mol/L?
70
Harvard University
BLB: Ch. 14; PHH: Ch. 15
First-Order Kinetics
1.
The decomposition of SO2Cl2 is first-order with a half-life of 245 minutes at
600K.
SO2Cl2 (g) → SO2 (g) + Cl2 (g)
An evacuated flask is filled with SO2Cl2 at a pressure of 9.00 atm at 600 K.
Calculate the partial pressure of each gas after 6.0 hours have elapsed.
2.
Colorless N2O4 will decompose to form brown NO2:
N2O4 (g) → 2 NO2 (g)
The half-life for this rapid first-order decomposition is 1.3 × 10–5 seconds.
An evacuated flask is charged with 17.0 torr of N2O4. After how many seconds
will the pressure of NO2 reach 1.3 torr?
71
Harvard University
BLB: Ch. 14; PHH: Ch. 15
Higher-Order Kinetics
1.
2.
Some reactions are so rapid that they are “diffusion-controlled”; that is, the
reactants will react as quickly as they can collide. The neutralization of H3 O+ by
OH– is one such example; this reaction has a second-order rate constant of
1.3 × 10 11 M–1·s–1 at room temperature.
a)
If equal volumes of 2.0 M HCl and 2.0 M NaOH are mixed instantaneously
(affording a solution 1.0 M in each), calculate the time required for 99.999% of
the acid to be neutralized.
b)
Under normal laboratory conditions, would you expect the rate of acid-base
neutralization to be limited by the rate of the reaction or the speed of mixing?
The reaction:
2 NO (g) + O2 (g) → 2 NO2 (g)
has the third-order rate law:
rate = k [NO]2[O2]
with k = 25 M–2s–1.
Under the condition that [NO] = 2[O2], the rate law integrates to yield:
1
1
= 8kt +
[O2]2
([O2]o)2
An experiment begins with [NO] = 0.020 M and [O2] = 0.010 M. Determine the
concentrations of NO, O2, and NO2 after 100. seconds.
72
Harvard University
BLB: Ch. 14; PHH: Ch. 15
Temperature and Rate
1.
Tree crickets (Oecanthurs) chirp twice as often at 26°C as they do at 18°C.
Calculate the activation energy for the act of chirping.
2.
A certain industrial chemical process has an activation energy of 65 kJ/mol. If
this process can produce 1.0 kg per hour at 100.°C, what temperature would be
required in order to produce 2.0 kg per hour?
3.
You have seen several factors that can affect reaction rates. Consider the firstorder decomposition of SO2Cl2 at 600 K:
SO2Cl2 (g) → SO2 (g) + Cl2 (g)
Predict the change in the rate of decomposition of SO 2Cl2:
If you:
i) add more SO2Cl2
Rate of decomposition of SO2Cl2 will:(circle)
speed up
slow down
remain same
ii) decrease the volume of the flask speed up
slow down
remain same
iii) lower the temperature
speed up
slow down
remain same
iv) add a catalyst
speed up
slow down
remain same
v) add some Cl2
speed up
slow down
remain same
73
Harvard University
BLB: Ch. 14; PHH: Ch. 15
Putting It Together: Macroscopic Kinetics
1.
Given the following data for the reaction: N2 O4 (g) → 2 NO2 (g):
Experiment No.
1
2
3
4
Temp., °C
25
25
40
40
[N2 O4 ], M
0.10
0.20
0.10
0.30
Initial rate, M·s–1
5.5 × 10 3
1.1 × 10 4
1.5 × 10 4
4.5 × 10 4
a)
Write the rate law for this reaction.
b)
Calculate the specific rate constant, k, at 25°C. Include units.
c)
In Experiment No. 1, after how many seconds will [N2 O4 ] reach 0.002 M?
d)
Calculate the half-life of this reaction at 40°C.
e)
Calculate the activation energy for this reaction.
74
Harvard University
BLB: Ch. 14; PHH: Ch. 15
Introduction to Mechanisms
1.
Write a rate law for each of the following elementary reactions:
Cl2 + H → HCl + Cl
C3H6N2 → C3H6 + N2
2.
For the reaction A + C → D, the steps of the mechanism are as follows:
k1
k1 = rate const. of forward rxn
k–1
k–1 = rate const. of reverse rxn
← B
A →
k2
B + C → D
k2 = rate const. of forward rxn
Write the rate law for the reaction under each of the following assumptions:
a)
The first step is rate-determining.
b)
The second step is rate-determining and the first step is in equilibrium.
c)
The intermediate B is in steady-state.
75
Harvard University
BLB: Ch. 14; PHH: Ch. 15
Advanced Mechanisms
1.
2.
For the gas-phase reaction: CO + Cl2 → COCl2
the following mechanism has been proposed:
→ 2 Cl
Cl2 ←
fwd. rate const. k1 , reverse rate const. k–1
Cl + CO →
← COCl
fwd. rate const. k2 , reverse rate const. k–2
COCl + Cl2 → COCl2 + Cl
fwd. rate const. k3
a)
Assume that the first step is in equilibrium, [COCl] is steady-state, and the final
step is rate-determining, and derive the rate of formation of product (COCl2).
b)
Show that if k3 is very small, the rate law simplifies to the form k[CO][Cl2 ]3/2 .
Given the following mechanism for the reaction: A + 2B + D → F
→
A+B ←
C
fwd. rate const. k1 , reverse rate const. k–1
C+D →
E
fwd. rate const. k2
E+B →
F
fwd. rate const. k3
Assuming that the first step is in rapid equilibrium, and that the intermediate E is
in steady state, derive the rate law for the formation of the product F.
76
Harvard University
BLB: Ch. 15; PHH: Ch. 16
Writing Equilibrium Constants
1.
Write the equilibrium constant expression (Kc) for these chemical equations:
a)
→ PCl3 (g) + Cl2 (g)
PCl5 (g) ←
b)
NH4Cl (s) →
← NH3 (g) + HCl (g)
c)
→ H2O (g)
H2O (l) ←
2.
3.
Write the equilibrium constant expression (Kc) for these chemical equations. The
numerical value of the equilibrium constant is given in each case:
a)
H2 (g) + CO2 (g) →
← CO (g) + H2O (g)
Kc = 0.771
b)
SnO2 (s) + 2 H2 (g) →
← Sn (s) + 2 H2O (g)
Kc = 8.12
c)
SnO2 (s) + 2 CO (g) →
← Sn (s) + 2 CO2 (g)
Kc = ????
d)
Given the data from parts (a) and (b), above, calculate the numerical value of the
equilibrium constant for the reaction in part (c).
The equation:
N2O4 (g) →
← 2 NO2 (g)
has an equilibrium constant Kc of 6.1 × 10 –3 at 25°C. Calculate the equilibrium
constant for the related reaction:
NO2 (g) →
← 1/2 N2O4 (g)
77
Harvard University
BLB: Ch. 15; PHH: Ch. 16
Kp, Kc, and Q
1.
Nitrogen oxides are partly responsible for smog. Nitrogen monoxide, NO, is
released from the burning of fossil fuels, and is oxidized by oxygen in the air to
form brown nitrogen dioxide, NO2:
Kp = 2.4 × 1012 at 25°C
2 NO (g) + O2 (g) →
← 2 NO2 (g)
a)
Calculate the value of Kc for this reaction at 25°C.
b)
On a certain day, the partial pressures of NO, O2, and NO2 were as follows:
PNO = 1.1 × 10–5 torr
PO2 = 152 torr
PNO2 = 3.6 × 10–5 torr
Calculate the value of Q , and determine if NO 2 will be produced or consumed
under these conditions.
2.
The decomposition of NOBr has an equilibrium constant Kp = 0.028 at 350 K:
NOBr (g) →
← NO (g) + 1/2 Br2 (g)
a)
Calculate the value of Kc for this reaction at 350 K.
b)
A 2.00-liter flask contains 0.50 mol NOBr, 0.40 mol NO, and 0.20 mol Br2
at 350 K. Calculate the reaction quotient Q, and determine if NOBr will be
produced or consumed under these conditions.
78
Harvard University
BLB: Ch. 15; PHH: Ch. 16
Le Châtelier’s Principle
1.
Consider the decomposition of SO2Cl2 at 600 K.
SO2Cl2 (g) →
← SO2 (g) + Cl2 (g)
ΔH° = +67 kJ/mol
Predict the change in SO2Cl2 concentration at equilibrium if:
2.
If you:
Concentration of SO2Cl2 will: (circle)
i) add more SO2
increase
decrease
remain same
ii) decrease the volume of the flask increase
decrease
remain same
iii) raise the temperature
increase
decrease
remain same
iv) add some inert Argon gas
increase
decrease
remain same
v) add a catalyst
increase
decrease
remain same
Consider the following exothermic reaction, used to obtain lead from its ore:
2 PbS (s) + 3 O2 (g) + 2 CO (g) →
← 2 Pb (l) + 2 SO2 (g) + 2 CO2 (g)
Assume that this reaction is at equilibrium. Given each of the following changes
to the system, will the quantity of lead increase, decrease, or remain the same:
3.
If you:
Total quantity of lead will: (circle)
i) add more lead sulfide
increase
decrease
remain same
ii) add more carbon monoxide
increase
decrease
remain same
iii) remove some oxygen
increase
decrease
remain same
iv) compress the entire system
increase
decrease
remain same
v) add argon (an inert gas)
increase
decrease
remain same
vi) increase the temperature
increase
decrease
remain same
What conditions (high vs. low temperature, high vs. low pressure) would you use
to increase the yield of products in the following reactions:
a)
→ 2 NO (g) + Cl2 (g)
2 NOCl (g) ←
endothermic
b)
2 SO2 (g) + O2 (g) →
← 2 SO3 (g)
exothermic
79
Harvard University
BLB: Ch. 15; PHH: Ch. 16
Calculating Equilibrium Constants
1.
CO2 + H2 →
← CO + H2O
In an investigation of the reaction:
a chemist put 0.50 mol CO2 and 0.50 mol H2 into a 2.00 L flask and allowed them
to reach equilibrium at 690 K. At equilibrium 0.38 mol CO2 remained. What is
the equilibrium constant for this reaction at 690 K?
2.
For the reaction:
2 NOCl →
← 2 NO + Cl2
2.00 moles of NOCl (g) were heated at 225°C in a 400.-liter steel reaction vessel.
After reaching equilibrium, the total pressure in the vessel was 0.246 atm.
Calculate Kp for this reaction at this temperature.
3.
At 500°C, fluorine gas (F2) is stable and does not dissociate. However, at 840°C,
some dissociation to fluorine atoms occurs:
→ 2 F (g)
F2 (g) ←
A flask is filled with 0.600 atm F2 at 500°C. The temperature is raised to 840°C,
and the pressure is measured to be 0.984 atm (at 840°C with dissociation).
Calculate the equilibrium constant Kp for the dissociation of fluorine gas at
840°C.
80
Harvard University
BLB: Ch. 15; PHH: Ch. 16
Determining Final Equilibrium Concentrations
1.
Consider the equilibrium decomposition of SO2Cl2 at 600 K:
SO2Cl2 (g) →
← SO2 (g) + Cl2 (g)
Kc = 6.1 at 600 K
A 1.00-liter evacuated flask is filled with 0.183 moles of SO2Cl2 at 600 K.
Calculate the moles of each gas at equilibrium, and the total pressure.
2.
Given the following equilibrium:
Br2 (g) + F2 (g) →
← 2 BrF (g)
K = 54.7 at 300. K
An evacuated flask is charged with 2.2 atm of Br2 and 2.2 atm of F2. Calculate
the final pressure of BrF once equilibrium is reached at 300. K.
3.
At 1800 K, oxygen dissociates slightly into individual atoms:
→ 2 O (g)
O2 (g) ←
Kp = 1.2 × 10–10 at 1800 K
You place 1.0 mol of O2 in a 10.0-liter flask and heat it to 1800 K. How many
single atoms of oxygen (O) will be present in the flask at equilibrium?
81
Harvard University
BLB: Ch. 16; PHH: Ch. 17
Introduction to Acids, Bases, and pH
1.
For the following Bronsted-Lowry acid-base reactions, identify the reactant which
is acting as an acid and the reactant acting as a base:
a)
HN3 + NH2OH → NH 3OH+ + N3–
b)
CN– + B(OH)3 → HCN + BO(OH)2–
Si(OH)4 + C6H5O– → SiO(OH)3– + C6H6O
c)
2.
Calculate the pH of each of the following solutions.
a)
0.027 M KOH
b)
0.020 M HBr
c)
1.0 × 10 –3 M Ba(OH)2
d)
1.0 × 10 –9 M KOH
3.
A solution is prepared by mixing 500. mL of 0.10 M NaOH with 500. mL of
0.0400 M H2SO4.
a)
Write the equation for the chemical reaction which will take place.
b)
Calculate the pH of the resultant solution.
82
Harvard University
BLB: Ch. 16; PHH: Ch. 17
Weak Acids
1.
Calculate the pH of a 0.100 M solution of HF (Ka = 6.8 × 10–4)
2.
Calculate the pH of a 0.0500 M solution of HIO3. (Ka = 1.69 × 10–1)
83
Harvard University
BLB: Ch. 16; PHH: Ch. 17
Weak Bases
1.
Calculate the pH of a 0.117 M solution of CH3NH2. (Kb = 4.3 × 10–4)
2.
Calculate the pH of a 0.075 M solution of sodium benzoate (C7H5O2Na).
(Ka of C7H5O2H = 6.46 × 10–5)
3.
Many home pools are disinfected by adding calcium hypochlorite, Ca(OCl) 2. The
hydrolysis of the hypochlorite ion yields hypochlorous acid. Given that Ka for
HOCl = 3.0 × 10–8, calculate the pH of an 0.100 M solution of calcium
hypochlorite, Ca(OCl)2.
84
Harvard University
BLB: Ch. 16; PHH: Ch. 17
Polyprotic Acids and Bases
1.
Calculate the concentration of each ionic species in a 0.100 M solution of H3PO4.
For H3PO4, Ka1 = 7.5 × 10–3 Ka2 = 6.2 × 10–8
Ka3 = 4.2 × 10–13
2.
Calculate the pH of a 0.050 M H2SO4 solution. For H2SO4, Ka2 = 0.012
3.
Calculate the pH of a 0.100 M solution of Na 2C2O4.
For the conjugate acid H2C2O4,
Ka1 = 5.9 × 10–2
85
Ka2 = 6.4 × 10–5
Harvard University
BLB: Ch. 16; PHH: Ch. 17
Acidic Behavior of Metal Cations (Hydrolysis)
1.
Calculate the pH of a 0.100 M solution of Al(NO3)3.
Ka (Al(H2O)63+ ) = 1.4 × 10–5
2.
A solution is prepared by dissolving 0.0030 moles of Cu(NO3)2 in 500. mL of
distilled water. Calculate the pH of this solution.
Ka (Cu(H2O)42+ ) = 1.0 × 10–8
86
Harvard University
BLB: Ch. 16; PHH: Ch. 17
Putting It Together: Aqueous Acid Equilibria
1.
When NO2 is bubbled into water, it completely decomposes into HNO3 and
HNO2:
2 NO2 (g) + H2O (l) → HNO3 (aq) + HNO2 (aq)
A solution is prepared by dissolving 0.0500 moles of NO2 in 1.00 L of water.
Determine the pH of this solution. (Ka for HNO2 = 4.5 × 10–4)
2.
Sulfur dioxide is quite soluble in water. It dissolves according to the equation:
SO2 (g) + H2O (l) →
← H2SO3 (aq)
[H SO ]
K = P2 3 = 1.33
SO2
The H2SO3 thus produced is a weak, diprotic acid:
Ka1 (H2SO3) = 1.54 × 10–2
Ka2 (H2SO3) = 1.02 × 10–7
A sulfur dioxide solution is prepared by continuously bubbling SO 2 at a pressure
of 1.00 atm into pure water. Calculate the pH of this solution, and the
concentrations of H2SO3, HSO3–, and SO32–.
87
Harvard University
BLB: Ch. 17; PHH: Ch. 18
An Introduction to Buffers
1.
Calculate the pH of a solution which is 0.050 M in HF and 0.112 M in NaF.
Ka for HF = 6.8 × 10–4
2.
You are given two burettes. One is filled with 0.10 M acetic acid. The other is
filled with 0.15 M sodium acetate. How much of each would you mix together to
produce a total of 20.0 mL of solution with a pH of 4.70? You may not add any
water to the solution.
3.
A buffer is prepared by dissolving 0.500 moles of HN3 and 0.750 moles of NaN3
in enough water to make 1.00 liter of solution. (K a for HN3 = 1.9 × 10–5)
a)
Calculate the pH of this buffer solution.
b)
You prepare 1.00 L of 0.100-molar HN3. How will the following changes affect
the quantity of HN3 present at equilibrium:
If you:
Quantity of HN3 at equilibrium will: (circle)
add 1 gram of solid NaN3
increase
decrease
remain same
dilute the solution with water
increase
decrease
remain same
add 1 gram of solid NaOH
increase
decrease
remain same
add 1 gram solid NaCl
increase
decrease
remain same
lower the pH with HCl
increase
decrease
remain same
88
Harvard University
BLB: Ch. 17; PHH: Ch. 18
Calculating Concentrations in Buffered Solutions
1.
Calculate the concentration of NH 4+ required for an ammonia / ammonium
chloride buffer in which [NH 3] = 0.200 M and the pH is 9.00.
(Kb (NH3) = 1.8 × 10–5)
2.
4.00 mL of 0.100 M NH4Cl are mixed with 1.00 mL of 0.100 M Na2CO3.
(The total volume of solution is 5.00 mL.)
a)
The resulting solution has [NH3] = 0.0200 M and [NH4+ ] = 0.0600 M. Calculate
the pH of this buffered solution. (Kb (NH3) = 1.8 × 10–5)
b)
Calculate the concentrations of H 2CO3, HCO3–, and CO32– at this pH once
equilibrium is reached. Note that the carbonate species must add up, that is:
[H2CO3] + [HCO3–] + [CO32–] = 0.0200 M.
(Hint: Apply the Henderson-Hasselbalch equation to each acid-base pair.)
Ka1 (H2CO3) = 4.3 × 10–7
Ka2 (H2CO3) = 5.6 × 10–11
89
Harvard University
BLB: Ch. 17; PHH: Ch. 18
Titrations
1.
What volume of 1.00-molar ammonium chloride must be added to 1.00-liter of
0.100-molar LiOH in order to achieve a pH of 9.4? (Kb (NH3) = 1.8 × 10–5)
2.
Calculate the pH that would result if 60.0 mL of 1.00-molar H3PO4 is added to
1.00 liter of 0.100-molar LiOH.
For H3PO4, Ka1 = 7.5 × 10–3 Ka2 = 6.2 × 10–8
90
Ka3 = 4.2 × 10–13
Harvard University
BLB: Ch. 17; PHH: Ch. 19
Solubility and Ksp
1.
Experiments show that in a saturated solution of cadmium iodate, Cd(IO3)2, at
25°C, [Cd2+ ] = 1.79 × 10–3 M.
a)
What is the concentration of IO 3–?
b)
Calculate the solubility product constant (Ksp) of cadmium iodate.
The Ksp for silver sulfate (Ag2SO4) is 1.2 × 10 –5.
2.
3.
a)
Calculate the solubility of Ag2SO4 in pure water.
b)
Calculate the solubility of Ag2SO4 in 0.100 M AgNO3 solution.
The solubility product constant of calcium fluoride is 3.9 × 10–11. How many
moles of solid NaF must be added to 1.00 L of 0.100 M CaCl2 in order to
precipitate CaF2?
91
Harvard University
BLB: Ch. 17; PHH: Ch. 19
Solubility of Metal Hydroxides
1.
Magnesium hydroxide, also known as Milk of Magnesia, is practically insoluble.
Ksp (Mg(OH)2) = 1.2 × 10–11.
a)
Determine the pH of a saturated solution of Mg(OH)2.
b)
Calculate the molar solubility of Mg(OH)2 in a solution buffered at pH = 10.0.
92
Harvard University
BLB: Ch. 17; PHH: Ch. 19
Other Aqueous Equilibria: Dealing With Large K’s
1.
Copper (II) forms a complex ion with ammonia as follows:
→ Cu(NH3)42+
Cu2+ (aq) + 4 NH3 (aq) ←
Kf = 5 × 1012
Calculate the concentrations of Cu2+ , NH3, and Cu(NH3)22+ , if 0.020 moles of
Cu(NO3)2 are dissolved in 1.0 liter of 1.00 M NH3. (Ignore the basicity of NH 3.)
2.
In a test tube, 1.00 mL of 0.100 M Mn(NO3)2 is mixed with 3.00 mL of
0.100 M Na2CO3 and 1.00 mL of 0.882 M H2O2. Dark brown MnO2 (s)
precipitates according to the following equilibrium:
–
Mn2+ (aq) + 2 CO32– (aq) + H2O2 (aq) →
← MnO2 (s) + 2 HCO3 (aq)
[HCO3–]2
= 6 × 1038
[Mn2+ ][CO32–]2[H2O2]
Calculate the concentration of Mn2+ at equilibrium. (Neglect the effect of the
hydrolysis of the ions in water.)
K=
93
Harvard University
BLB: Ch. 17; PHH: Ch. 19
Solubility and Complex Ion Formation
1.
In a watch glass, 2.00 mL of 0.100 M NH4Cl are mixed with 1.00 mL of
0.100 M AgNO3; a precipitate of AgCl is formed. To this solution and precipitate
is added 2.00 mL of 0.100 M Na2S2O3. The AgCl precipitate dissolves due to the
formation of a complex ion in the following reaction:
3–
–
AgCl (s) + 2 S2O32– (aq) →
← Ag(S2O 3)2 (aq) + Cl (aq)
Calculate the concentration of Ag+ at equilibrium. (Note total volume = 5.00 mL)
Ksp (AgCl) = 1.8 × 10–10
2.
Kf (Ag(S2O3)23–) = 2.9 × 1013
Chromium (III) hydroxide, Cr(OH)3, is practically insoluble (K sp = 6.3 × 10–31).
However, chromium hydroxide will dissolve in excess base due to the formation
of a complex ion:
Cr3+ (aq) + 4 OH– (aq) →
← Cr(OH)4– (aq)
Kf = 8.0 × 1029
Calculate the solubility of Cr(OH)3 in a 0.010 M solution of NaOH.
94
Harvard University
BLB: Ch. 17; PHH: Ch. 19
Putting It Together: Solubility and Acid/Base Behavior
Aluminum phosphate is quite insoluble: Ksp (AlPO4) = 9.8 × 10–21
1.
a)
Determine the solubility of AlPO4 in pure water. Assume that neither of the ions
undergoes hydrolysis.
b)
Aluminum phosphate is actually more soluble than your answer to part (a) might
indicate, due to hydrolysis of the ions involved. The actual dissolving of AlPO4
in water can be represented:
AlPO4 (s) + 6 H2O (l) →
← Al(H2O)5(OH)2+ (aq) + HPO42– (aq)
Given the following information:
Ksp (AlPO4) = 9.8 × 10–21
Ka (Al(H2O)63+ ) = 1.1 × 10–5
Ka (H3PO4) = 7.5 × 10–3
Ka (H2PO4–) = 6.2 × 10–8
Ka (HPO42–) = 2.2 × 10–13
Calculate the solubility of AlPO4 in pure water, including the effect of hydrolysis
as represented above. Recall that Al 3+ (aq) is actually Al(H2O)63+ (aq).
95
Harvard University
BLB: Ch. 19; PHH: Ch. 20
An Introduction to Thermodynamics
1.
Consider the evaporation of methanol, CH3 OH. Using the Useful Information
(below):
a)
Calculate the heat of vaporization of methanol.
b)
Calculate the entropy of vaporization of methanol.
c)
Calculate the boiling point of methanol.
Substance:
CH3OH (g)
CH3OH (l)
ΔH°f (kJ/mol)
–201.2
–238.6
S° (J/mol·K)
237.6
126.8
96
Harvard University
BLB: Ch. 19; PHH: Ch. 20
Molecular Interpretation of Entropy
1.
Acetic acid, CH3 COOH, can form a dimer in the gas phase which is held together
by hydrogen bonds:
2 CH3COOH (g) →
← [CH3COOH–HOOCCH3]
2.
ΔH° = –66.5 kJ/mol
a)
Would you expect the entropy change to be positive or negative? Briefly explain
your reasoning.
b)
At 25°C, ΔG° for this reaction is +16.5 kJ/mol. Calculate ΔS° for the reaction, and
compare it with your prediction.
When HF (aq) dissociates in water, one might expect an increase in entropy when
the neutral HF molecule dissociates into ions. In fact, however, ΔS for this
dissociation is negative (–88 J/mol K). Consider carefully the issues of order and
disorder in the solution, and speculate briefly why this must be so.
HF (aq) + H2 O (l) →
← H3 O+ (aq) + F– (aq) ΔS° = –88 J/mol K
97
Harvard University
BLB: Ch. 19; PHH: Ch. 20
Using Standard Free Energies and Enthalpies
1.
Given the following reaction and the Useful Information (below):
2 RbCl (s) + 3 O2 (g) → 2 RbClO3 (s)
a)
Calculate ΔG° at 25°C.
b)
Calculate ΔH° at 25°C.
c)
Calculate ΔS° at 25°C.
d)
Calculate ΔG° at 60°C.
e)
Calculate the standard molar entropy S° for O2 (g) at 25°C.
Substance:
RbCl
RbClO3
ΔH°f (kJ/mol)
–430.5
–392.4
ΔG°f (kJ/mol)
–412.0
–292.0
98
S° (J/mol·K)
92
152
Harvard University
BLB: Ch. 19; PHH: Ch. 20
ΔG° and Keq
Gadolinium (III) is used in medical imaging (MRI) applications. Because Gd3+ is
itself toxic, gadolinium (III) is usually administered as a complex of DTPA5–:
(DTPA5– is a chelating, or complexing, agent)
1.
2–
Gd3+ + DTPA5– →
← Gd(DTPA)
Kf = 2.9 × 1022 at 25°C
a)
Determine ΔG° for the above reaction at 25°C.
b)
Calculate Kf for the above reaction at 37°C (normal body temperature), given that
its ΔH° is –32.6 kJ/mol.
2.
Consider the following equilibrium, and the Useful Information (below):
→ 2 NO2 (g)
N2 O4 (g) ←
a)
At what temperature will an equilibrium mixture of 1.00 atm total pressure
contain twice as much NO2 as N2 O4 ?
b)
At what temperature will an equilibrium mixture of 1.00 atm total pressure
contain equal amounts of the two gases?
Substance:
NO2 (g)
N2O4 (g)
ΔH°f (kJ/mol)
33.84
9.66
ΔG°f (kJ/mol)
51.84
98.28
99
S° (J/mol·K)
240.45
304.3
Harvard University
BLB: Ch. 19; PHH: Ch. 20
ΔG at Nonstandard Conditions
1.
The vapor pressure of water at 27°C is 27 torr.
a)
Calculate the standard free energy of vaporization ΔG°vap for water at 27°C.
b)
Calculate the nonstandard ΔG for the evaporation of water at 27°C if the partial
pressure of water vapor is 11 torr.
2.
Methanol, CH3 OH, can be synthesized by reacting CO and H 2 directly:
CO (g) + 2 H2 (g) →
← CH3 OH (l)
a)
Using the Useful Information (below), calculate ΔG° and K for this reaction at
25°C.
b)
Calculate the nonstandard ΔG if the pressure of H2 is 3.0 atm and the pressure of
CO is 5.0 atm.
Substance:
CO (g)
CH3OH (l)
ΔG°f (kJ/mol)
–137.2
–166.23
100
Harvard University
BLB: Ch. 19; PHH: Ch. 20
Calculating K at Different Temperatures
1.
For the synthesis of NH3 by the Haber process:
N2 (g) + 3 H2 (g) →
← 2 NH3 (g)
a)
Using the Useful Information (below), determine ΔH°, ΔS°, and ΔG° for this
reaction at 25°C.
b)
Determine the equilibrium constant K at 25°C.
c)
Determine ΔG° and K at 200°C.
Substance:
NH3 (g)
ΔH°f (kJ/mol)
–46.19
ΔG°f (kJ/mol)
–16.66
101
S° (J/mol·K)
192.5
Harvard University
BLB: Ch. 19; PHH: Ch. 20
Applications: Determining ΔG°, ΔH°, and ΔS° from Keq
1.
Consider the Haber ammonia synthesis:
at 300°C: Kp = 4.34 × 10–3
at 500°C: Kp = 1.45 × 10–5
N2 (g) + 3 H2 (g) →
← 2 NH3 (g)
a)
Calculate ΔH° for this reaction over this temperature range.
b)
Calculate ΔS° for this reaction over this temperature range.
Consider the reaction between xylenol orange (H4Q) and aluminum ion (Al3+ ),
which you studied in the laboratory:
2.
→ AlQ– (aq) + 4 H+ (aq)
Al3+ (aq) + H4Q (aq) ←
at 55°C: K = 7.6 × 10–5
at 90°C: K = 6.5 × 10–4
a)
Calculate ΔG° for this reaction at 55°C.
b)
Calculate ΔH° for this reaction over this temperature range.
c)
Calculate ΔS° for this reaction over this temperature range.
102
Harvard University
BLB: Ch. 19; PHH: Ch. 20
Applications: Phase Changes
1.
Using the Useful Information (below), calculate:
a)
The boiling point of TiCl4 .
b)
The vapor pressure of ethanol (C2 H5 OH) at 25°C.
2.
Using the Useful Information (below):
a)
Calculate ΔG° for the evaporation of mercury at 25°C.
b)
Calculate ΔG for the evaporation of mercury at 25°C at a pressure of 0.100 torr.
c)
Calculate the standard entropy S° for Hg (l) at 25°C.
Substance:
TiCl4 (g)
TiCl4 (l)
C2H5OH (g)
C2H5OH (l)
Hg (g)
ΔH°f (kJ/mol)
–763.2
–804.2
–235.1
–277.7
60.83
ΔG°f (kJ/mol)
–726.8
–728.1
–168.5
–174.76
31.76
103
S° (J/mol·K)
354.9
221.9
282.7
160.7
174.89
Harvard University
BLB: Ch. 19; PHH: Ch. 20
Applications: Acid/Base Equlilbria
1.
An 0.100-molar solution of butyric acid (call it HBu) has a pH that varies with
temperature:
at 10°C: pH = 2.90
at 50°C: pH = 2.95
2.
a)
Calculate the concentrations of HBu, H+ , and Bu– at 10°C.
b)
Calculate the enthalpy of ionization (ΔH°) for butyric acid between 10°C and
50°C.
Consider the first step of the reaction between phosphoric acid and hydroxide ion:
H3PO4 (aq) + OH– (aq) →
← H2PO4– (aq) + H2O (l)
Given that, at 25°C, Ka1 for phosphoric acid is 7.5 × 10–3, and Kw = 1.0 × 10–14,
calculate ΔG° for the above reaction at 25°C.
104
Harvard University
BLB: Ch. 19; PHH: Ch. 20
Applications: Buffers
1.
A buffer is prepared by dissolving 0.500 moles of HN3 and 0.750 moles of NaN3
in enough water to make 1.00 liter of solution.
→ H3O+ (aq) + N3– (aq)
HN3 (aq) + H2O (l) ←
2.
Ka = 1.9 × 10–5 at 25°C
a)
Calculate ΔG° for the above reaction at 25°C.
b)
Calculate ΔG (nonstandard) for the above reaction at 25°C under the given buffer
conditions (above), but at pH 7.0.
Consider the equilibrium dissociation of HN3:
HN3 (aq) + H2O (l) →
← H3O+ (aq) + N3– (aq)
Ka = 1.9 × 10–5 at 25°C
ΔH° = +14.7 kJ/mol
For the same buffer system used above (0.500 M HN 3 plus 0.750 M NaN3),
calculate the pH at 50°C.
105
Harvard University
BLB: Ch. 20; PHH: Ch. 5
Balancing Redox Reactions: Acidic Solution
1.
Balance each of the following redox reactions in acidic solution.
a)
S4O62– (aq) + Al (s) → H2S (aq) + Al3+ (aq)
b)
S2 O3 2– (aq) + Cr2 O7 2– (aq) → S 4 O6 2– (aq) + Cr3+ (aq)
c)
ClO3 – (aq) + As2 S3 (s) → Cl– (aq) + H2 AsO4 – (aq) + SO4 2– (aq)
d)
IO3– (aq) + Re (s) → ReO 4– (aq) + I– (aq)
e)
HSO4– (aq) + As4 (s) + Pb3O4 (s) → PbSO 4 (s) + H3AsO4 (aq)
f)
HNO2 (aq) → NO3– (aq) + NO (g)
106
Harvard University
BLB: Ch. 20; PHH: Ch. 5
Balancing Redox Reactions: Basic Solution
1.
Balance each of the following redox reactions in basic solution
a)
C4H4O62– (aq) + ClO3– (aq) → CO 32– (aq) + Cl– (aq)
b)
Al (s) + BiONO3 (s) → Bi (s) + NH 3 (aq) + AlO2– (aq)
c)
H2 O2 (aq) + Cl2 O7 (aq) → ClO2 – (aq) + O2 (g)
d)
Tl2O3 (s) + NH2OH (aq) → TlOH (s) + N2 (g)
e)
Cu(NH3 )4 2+ (aq) + S2 O4 2– (aq) → SO3 2– (aq) + Cu (s) + NH 3 (aq)
f)
Mn(OH)2 (s) + MnO4– (aq) → MnO2 (s)
107
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Standard Reduction Potentials at 25°C
from Brown, LeMay, and Bursten, Chemistry: The Central Science, 8th ed.
108
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Voltaic Cells: Calculating Cell Potential
For the unbalanced reaction: Fe3+ (aq) + I– (aq) → Fe2+ (aq) + I2 (s)
1.
a)
Balance the reaction. Label the oxidation and reduction half-reactions.
b)
Identify the anode and cathode.
c)
What is the oxidizing agent? the reducing agent?
d)
Using the table of Standard Reduction Potentials, calculate E° for this cell.
2.
Calculate the standard cell potential for the following cell:
Mn (s) | Mn2+ || Co2+ | Co (s)
An electrochemical cell is set up by connecting a standard Cu/Cu2+ half-cell with
a standard Ag/Ag+ half-cell.
3.
a)
A voltmeter is connected to this system in such a way as to read a positive
voltage. What voltage would it display?
b)
While keeping the copper electrode connected to the voltmeter, the Ag/Ag + halfcell is replaced with a Zn/Zn2+ half-cell. What would the voltmeter read now?
(Be sure to consider the sign.)
109
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Cell Potential and Free Energy
1.
A standard electrochemical cell is set up as follows:
Sn (s) | Sn2+ (1.0 M) || Cu2+ (1.0 M) | Cu (s)
Calculate the standard free energy change ΔG° for the overall cell reaction.
2.
For the following unbalanced half-reaction:
SF6 (g) + e– → H 2SO3 (aq) + HF (aq)
a)
Balance this half-reaction in acidic solution.
b)
Using the data below, determine E° for this reduction half-reaction.
Substance:
SF6 (g)
H2SO3 (aq)
HF (aq)
H2O (l)
ΔG°f (kJ/mol)
–1105.4
–537.81
–296.82
–237.13
110
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Reduction Potentials of Half-Reactions
1.
Given the following standard reduction potentials:
S4O62– + 2 e– → 2 S 2O32–
E° = 0.090 V
2 SO32– + 3 H2O + 4 e– → S 2O32– + 6 OH–
E° = –0.58 V
Determine the standard reduction potential for the following half-reaction:
S4O62– + 12 OH– → 4 SO32– + 6 H2O + 6 e–
2.
For the unbalanced reduction half-reaction:
E° = ???
HSO4– (aq) + e– → H 2S (g)
a)
Balance this half-reaction in acidic solution.
b)
Using the table of Standard Reduction Potentials, determine E° for this halfreaction.
111
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Nonstandard Conditions: The Nernst Equation
1.
The reaction between hydrogen and oxygen can be carried out in a fuel cell, which
uses the energy released from this redox reaction to generate electricity:
→ H2O (l)
H2 (g) + 12 O2 (g) ←
A certain fuel cell is operated at 25°C with an oxygen pressure of 10. torr and a
hydrogen pressure of 40. torr. The observed potential under these conditions is
1.16 V. Calculate the standard potential for this electrochemical cell.
Using the table of Standard Reduction Potentials, show that Fe2+ can be
spontaneously oxidized to Fe3+ by O2 (g) at 25°C assuming the following
(reasonable) environmental conditions:
2.
[Fe2+ ] = [Fe3+ ] = 1 × 10–7 M
pH = 7
PO2 = 160 torr
a)
Write a complete, balanced equation for the oxidation of Fe 2+ by O2.
b)
Calculate E for the reaction under these conditions.
112
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Concentration Cells
1.
A cell is constructed as follows: On the left is a solution of 0.010 M AgNO3 with
a silver electrode. On the right is a solution of 1.0 M AgNO3 with another silver
electrode. The cells are separated by a salt bridge and the electrodes are
connected through a voltmeter.
a)
Identify the anode and cathode.
b)
Calculate the potential E for this cell.
c)
Water is added to the left half-cell, reducing the Ag+ concentration. The new cell
potential is 0.210 volts. Calculate the new Ag+ concentration in the left half-cell.
113
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Putting It Together: Electrochemistry, ΔG°, and Keq
1.
Consider a cell constructed from the following half-cells:
Pb2+ + 2 e– → Pb (s)
–0.13 V
Sn2+ + 2 e– → Sn (s)
–0.14 V
a)
Write the anode and cathode reactions and the overall cell reaction. Label each
clearly.
b)
Calculate E° for this cell.
c)
Determine the free energy change ΔG° for this reaction.
d)
Calculate the equilibrium constant K for this reaction.
e)
Calculate E if the actual concentrations are [Pb 2+ ] = 1.5 M and [Sn2+ ] = 0.10 M.
114
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Electrolysis: Faraday’s Law
1.
If an object is copper-plated in a solution of CuSO4 for 1.0 minute at a current of
15.0 amps, what mass of copper will be deposited on the object?
2.
A fuel cell is powered by hydrogen and oxygen:
→ H2O (l)
H2 (g) + 12 O2 (g) ←
A certain such fuel cell, given excess oxygen, consumes at most 0.370 grams of
pure hydrogen gas per hour at 25°C. Calculate the maximum current which can
be produced by this fuel cell.
3.
Aluminum is produced by the electrolytic reduction of alumina, Al2O3.
a)
Given a current of 1.0 × 105 amps, how many hours would it take to produce
1000. kg of aluminum?
b)
The anode in the above reaction is graphite (carbon), which is oxidized to CO2 (g)
during the reaction. What mass of graphite must be consumed in order to produce
1000. kg of aluminum?
115
Harvard University
BLB: Ch. 20; PHH: Ch. 21
Putting It Together: Electrochemistry and Solubility
1.
A cell is constructed as follows: A silver electrode is immersed in a 1.0 M
AgNO3 solution. Another electrode, also made of silver, is immersed in 1.0 M
KCl solution with some solid AgCl. The voltmeter reads 0.578 V. Determine the
Ksp for AgCl, and compare with the value given in the textbook.
2.
For the following unbalanced half-reaction:
BaCrO4 (s) + e– → Ba2+ (aq) + Cr3+ (aq)
a)
Balance this half-reaction in acidic solution.
b)
Using the following data, and the table of Standard Reduction Potentials,
determine E° for this reduction half-reaction.
Ksp (BaCrO4) = 2.1 × 10–10
Substance:
CrO42– (aq)
Cr2O72– (aq)
H+ (aq)
H2O (l)
ΔG°f (kJ/mol)
–727.75
–1301.1
0
–237.18
116
Harvard University
BLB: Ch. 21; PHH: Ch. 26
Nuclear Reactions: Introduction
1.
Fill in the blanks with a correct word, number, or chemical symbol:
a)
b)
c)
d)
Carbon-11 decays by
emission to yield boron-11.
decays by beta emission to yield nitrogen-15.
Strontium-90 emits two beta particles and finally yields
.
decays by alpha emission to yield 222
86Rn.
e)
14
Bombarding 238
92U with 7 N affords
f)
90
Neutron-induced fission of 235
92U can give 3 neutrons, 38Sr, and
.
g)
Two atoms of 126C can react inside a star to produce 20
10Ne and
.
h)
207
The decay of 235
92U to 82Pb yields a total of
along with 5 neutrons.
117
beta particles.
Harvard University
BLB: Ch. 21; PHH: Ch. 26
Radioactive Dating
1.
The age of wine can be determined by measuring its radioactive tritium content.
Tritium (3H, an isotope of hydrogen) is formed in the upper atmosphere and
comes down to the Earth in rainwater. Once the wine is bottled, the tritium
content decays by a first-order process with a half-life of 12.5 years.
a)
You are offered a valuable vintage wine which has only 20% of the tritium
activity of a freshly bottled sample. Calculate the age of this wine.
b)
Once the tritium activity has decayed to 1% of its original activity, this dating
technique is no longer reliable. What is the maximum age of wine that could be
dated using this technique?
118
Harvard University
BLB: Ch. 21; PHH: Ch. 26
Energy Changes in Nuclear Reactions
1.
Cobalt-57 decays by electron capture to yield the stable nuclide 57Fe. Using the
following data, calculate the total energy released from the decay of one atom of
57Co.
Masses of Selected Neutral Atoms
4He
4.002603 amu
7Li
7.016004 amu
10B
10.012937 amu
57Fe
56.935398 amu
57Co
56.936296 amu
Masses of Subatomic Particles
Electron
0.000549 amu
Proton
1.007276 amu
Neutron
1.008665 amu
2.
A cancer treatment called boron neutron-capture therapy works as follows: A
drug containing boron-10 is injected into the patient; this drug selectively binds to
cancer cells. Irradiating the affected area with neutrons induces the following
reaction:
10B
+ 1n → 4He + 7Li + γ
The alpha radiation kills the cancer cells, leaving the surrounding tissue
unharmed. In this nuclear reaction, the alpha particle and 7Li together have a total
kinetic energy of 2.31 MeV. The reactants (the boron atom and the neutron) have
essentially no kinetic energy. Calculate the energy of the gamma photon released
in this process. (Masses of relevant nuclides are given above.)
(1 MeV = 1.60 × 10–13 J)
119
Harvard University
BLB: Ch. 21; PHH: Ch. 26
Radiation Dose
1.
Potassium-40 decays by beta emission with a half-life of 1.28 × 10 9 years; the
emitted beta particles have an average kinetic energy of 0.55 MeV. A person
weighing 70.0-kg contains about 0.020 grams of 40K. Determine the total dose of
radiation absorbed per year from this decay of 40K. (1 MeV = 1.60 × 10–13 J)
2.
Cobalt-57 decays with a half-life of 271 days. An individual with a mass of
70.0 kg ingests 8.0 × 10–10 grams of 57Co.
a)
Calculate the mass of 57Co remaining in this person after one year.
b)
Each atom of 57Co releases 0.122 MeV of gamma radiation when it decays.
Calculate the total dose of gamma radiation (in rad) absorbed by this person
during the first year after ingestion. (Hint: how many atoms of 57Co decayed in
that time?)
120
Harvard University
BLB: Ch. 21; PHH: Ch. 26
Putting It Together: Nuclear Chemistry
1.
Radioactive dating based on carbon-14 is often used to determine the age of
archeological discoveries.
a)
When 14C decays, it emits an electron, or beta particle ( –10 β ). Write the complete
nuclear equation for this decay process. What isotope is produced when carbon14 decays?
b)
Carbon-14 has a mass of 14.003242 amu. The isotope produced when 14C decays
has a mass of 14.003074 amu., and a beta particle a mass of 0.000549 amu.
Calulate the energy released when one atom of carbon-14 decays by the above
process.
c)
Virtually all the energy from the above decay goes into the kinetic energy of the
ejected electron. What would be the velocity of the electron ejected from a
decaying atom of carbon-14?
d)
The Shroud of Turin, long believed to be Christ’s burial shroud, was probably
made around 1320 A.D. Carbon-14 dating played an important role in the
determination of its age. What fraction of the 14C present in 1320 is still present
today? The half-life of 14C is 5730 years.
121
Harvard University
BLB: Ch. 24; PHH: Ch. 25
Coordination Complexes: Introduction
1.
Provide the transition metal oxidation state, number of d electrons, and the correct
name for each of the following:
a)
[Pt(NH3)4](ClO4)2
b) [ReH9]2– (think about the ox. state of H!)
c)
Na2[Fe(CO)4]
d) [IrF6]3–
e)
K3[Fe(CN)6]
f) [PtCl2(NH3)2]
122
Harvard University
BLB: Ch. 24; PHH: Ch. 25
Geometry and Isomerism I: Monodentate Ligands
1.
Each of the following complexes contains a metal atom in an octahedral
geometry. For each complex, draw all the possible stereoisomers. Indicate if any
of the isomers are chiral.
a)
[Co(NH3)3(OH2)3]3+
b)
[Pt(NH3)2Cl2Br2]
2.
When the octahedral complex [Co(NO2)3(NH3)3] is treated with HCl, one can
isolate a complex [CoCl2(NH3)3(OH2)]1+ in which the two chloride ligands are
trans to one another.
a)
Draw the two possible stereoisomers of the starting material [Co(NO2)3(NH3)3].
(The NO2– ligands are all bound through the nitrogen atom.)
b)
Assuming that the NH3 groups remain in place, which of the two starting isomers
could give rise to the observed product?
123
Harvard University
BLB: Ch. 24; PHH: Ch. 25
Geometry and Isomerism II: Chelates
1.
Consider the octahedral complex [Co(Cl) 2(OH2)2(en)]+ ,
where en = ethylenediammine (H2NCH2CH2NH2).
Using the following octahedral templates, draw all the isomers of this complex.
Circle the appropriate term to indicate whether each isomer is chiral. (Note: there
are more templates than you actually need.... cross out any which you don’t
need).
Co
Co
Co
chiral
chiral
chiral
achiral
achiral
achiral
Co
Co
Co
chiral
chiral
chiral
achiral
achiral
achiral
124
Harvard University
BLB: Ch. 24; PHH: Ch. 25
Electronic Structure of Coordination Compounds
1.
For each of the following , give the electronic configuration of the d-orbitals (e.g.
t2g3eg1), indicate whether the complex is high-spin or low-spin, and give the
number of unpaired electrons:
a)
[Fe(OH2)6]2+
b) [Ni(OH2)6]2+
c)
[Fe(CN)6]4–
d) W(CO)6
2.
The three complexes [Co(NH3)6]3+ , [Co(NH3)5(OH2)]3+ , and [CoF6]3– are
colored red, yellow, and blue (but not in that order). Match each complex with its
color, and explain your reasoning.
3.
Explain why [FeF6]3– is basically colorless whereas [CoF6]3– is colored.
125
Harvard University
BLB: Ch. 24; PHH: Ch. 25
Putting It Together: Coordination Compounds
1.
The drug Nipride, Na2[Fe(CN)5NO], is an inorganic complex which is used as a
source of NO to lower blood pressure during surgery.
a)
Provide a systematic name for the compound Na 2[Fe(CN)5NO]. Note: The NO
ligand in this complex is neutral and is named “nitrosyl.”
b)
For the complex [Fe(CN)5NO]2–, give the electronic configuration of the dorbitals (e.g. t2g3eg1), indicate whether the complex is high-spin or low-spin, and
give the number of unpaired electrons.
c)
The [Fe(CN)5NO]2– complex is red-violet in color (“ruby red”), while the similar
complex [Fe(CN)6]3– is red-orange in color (“bright red”).
The CN– ligand is a strong-field ligand. By comparison, the NO ligand is:
(circle one)
much
weaker
slightly
weaker
slightly
stronger
much
stronger
Briefly explain your choice using the language of crystal field theory.
d)
Would you expect NO to dissociate rapidly from the complex [Fe(CN) 5NO]2–?
Briefly explain your answer.
126
Harvard University
BLB: Ch. 25; PHH: Ch. 27
Organic Chemistry: Isomers and Functional Groups
1.
Draw all the possible isomers with molecular formula C3H6O. Identify the
functional groups found in each case. (For now, don't worry about the question of
which isomers are stable or not!)
2.
Ethanol (an alcohol with the molecular formula C2H6O) is metabolized in the
liver through a two-step oxidation process. The initial product has the formula
C2H4O, and is responsible for the red flushed appearance which often
accompanies alcohol consumption. This is then oxidized again to the final
product, which has the molecular formula C 2H4O2.
a)
Draw the structure of ethanol:
b)
Draw the structure of each product, and name the functional groups.
Structure
Initial product (C2H4O):
Final product (C2H4O2):
127
Functional Group
Harvard University
BLB: Ch. 25; PHH: Ch. 27
Organic Chemistry: Polymers
1.
For each of the following polymers, identify the monomer unit which could be
used to synthesize the polymer. Also circle the type of polymerization which
would be required.
Monomer:
Poly(tetrafluoroethylene), Teflon®
F
F
F
F
F
F
C
C
C
C
C
C
F
F
F
F
F
F
Type of Polymerization: (circle)
n
addition
Monomer:
Polyalanine (a polypeptide)
2.
H
O
N
C
C
H
CH3
H
O
N
C
C
H
CH3
condensation
H
O
N
C
C
H
CH3
Type of Polymerization: (circle)
n
addition
condensation
The molecules below can combine to form a polymer (a type of polyester).
O
O
C
HO
C
C
H2
OH
HO
OH
A
C
H2
B
a)
What is the functional group in molecule A?
b)
What is the functional group in molecule B?
c)
Draw a portion of the polymer chain which would be formed. The polyester is
formed by condensation polymerization. Include at least 4 monomers: A-B-A-B.
128
Answers to Harvard General Chemistry Practice Problems
1.
Significant Figures and Scientific Notation
1.
2.
2.
4.
c) 19.2
d) 42.2
e) 860 or 900
f) 18.2
g) 1.18 × 10–4
h) 3.6
i) 2.5 × 10 –4
a) 67.77 ft
b) 1.5 m
c) 27 ft
a) 2.4 atm
b) 2.1 × 103 torr
c) 1.00 × 105 Pa
Dimensional Analysis II: Volume and Density
1.
a) 3.6 × 102 bushels b) 54 hogsheads
2.
a) 4.670 × 10–2 in3
c) 2.7 × 103 gallons
b) 1.92 g
Atoms, Molecules and Ions
1.
5.
b) 2.3
Dimensional Analysis I: Unit Conversions
1.
3.
a) 523
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
Oxidation Numbers
1.
2.
Br: +5
O: –2
H: +1
As: +3
O: –2
As: –3
H: +1
Cr: +3
Cl: –1
K: +1
Cl: +5
O: –2
S: +2
O: –2
O: +2
F: –1
Na: +1
O: –1
Fe: +8/3
O: –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–)
6.
Naming Compounds
1.
2.
a) Cu3P
e) CrO3
b) Fe 2(SO4) 3
f) NH4I
c) KClO3
d) AlCl3
g) Li3N
h) Al 2(CO3) 3
i) Cs3PO4
j) Re2O7
k) SnCl4
l) GaF3
m) AgF2
n) KNO3
o) Ba3(PO4) 2
p) (NH4) 2SO4
a) Silicon tetrachloride
b) Manganese (VII) oxide
c) Magnesium nitride
129
7.
Chemical Formulas: Percent Composition
1.
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.
2.
8.
Pure CaCO3 would have 40% calcium, 12% carbon, and 48% oxygen. This does not agree
with the analysis of the rock sample.
Empirical and Molecular Formulas
1.
a) C9H12O
b) C18H24O2
2.
a) C5H4O
b) C15H12O3 ; MW = 240.25 g/mol
b) NH 2Cl
3.
9.
c) 272.39 g/mol
a) NH2Cl
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.
3.
6 CO 2 + 6 H2O → C6H12O6 + 6 O2
3 Nb + 4 I2 → Nb3I 8
C8H18 + 25/2 O2 → 8 CO2 + 9 H2O
10. Stoichiometry of Reactions
4.
1.
55.9 g NaClO
2.
a) NiS + 2 O 2 + 2 HCl → NiCl2 + H2SO4
b) 0.654 g NiCl2
11. 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 PBr 3
b) 77.47 g P 4S5
12. Stoichiometry of Mixtures
1.
61.1% CO2 by mass
2.
7.46 g CuO, 3.04 g Cu2O
3.
6.55 g Cu2S, 4.25 g CuS
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: Simple Examples
1.
a) 3 ClO – (aq) + NH3 (aq) → NCl3 (l) + 3 OH– (aq)
b) 0.028 g NCl3
2.
2.79 g CO 2
130
15. Solution Stoichiometry II: Acid/Base Neutralizations
1.
71.2 mL
2.
1.72 L
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 Ba(NO3) 2
2.
Ba2+ : 0.011 M
Cl–: 0.033 M
NH3: 0.033 M
NH4+ : 0.022 M
NO3–: 0.011 M
18. The Ideal Gas Law
1.
a) 6.0 × 105 L
b) 0.348 g Cl2
2.
53.4 g O2
19. Reactions Involving Gases: Simple Examples
1.
2.66 atm
2.
0.480 atm
3.
total pressure = 1.5 atm
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
21. Collecting Gases Over Water
1.
0.109 g H2
2.
0.312 L collected
3.
3.51 L should be collected
22. Stoichiometry of Gas Mixtures
1.
χ PCl5 = 0.054
2.
12.8% O3 by mass
χ PCl3 = 0.473
χ Cl2 = 0.473
23. Reactions Involving Gases: Applications
1.
21.3 g NaN3
2.
a) 29.0 g Cl2
b) 1.07 atm O2
1.63 atm Cl 2
c) 29.7 g NaAlCl 4 produced
131
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: 2.07 × 10–20 J/molecule
2.
a) 431 m/s
b) 7.3 hours
25. Heat
1.
a) 8.2 kJ of heat consumed
b) Final temp. = 14.2°C
2.
Specific heat = 0.269 J/g°C
26. Calorimetry I: Simple Examples
1.
ΔH° = –27 kJ/mol
2.
ΔH° = –5152 kJ/mol
3.
ΔH° = –4219 kJ/mol
27. Calorimetry II: Mixtures
1.
Final temp. = 27.08 °C
2.
311 g of ice
28. Hess’s Law
1.
ΔH° = –171.34 kJ/mol
2.
ΔH° = –185.1 kJ/mol
29. Standard Enthalpies of Formation
1.
ΔH° = –64.9 kJ/mol
2.
268 kJ of heat released
30. Enthalpy, Energy, Heat, and Work
1.
a) +37.4 kJ/mol
b) –2800 J/mol
c) +34.6 kJ/mol
31. Putting It Together: Calorimetry, Hess’s Law, and ΔH°f
1.
a) ΔH° = 25.3 kJ/mol
2.
b) ΔH°f (NH4NO3(s)) = –362.8 kJ/mol
a) ΔH° = –50.8 kJ/mol for the reaction as written (i.e. consuming 2 mol NaN3)
b) ΔH°f (NaN3) = 25.4 kJ/mol
32. Putting It Together: Stoichiometry, Thermochemistry, and Gas Laws
1.
a) 3.74 g CO 2
b) 7938°C
c) 14.3 atm CO 2
132
33. Energy, Particles, and Waves: A “Chain Problem”
1.
a) 7.09 × 1017 s–1
b) 4.23 × 10–10 m
c) 1.72 × 106 m/s
d) 1.35 × 10–18 J
34. The Photoelectric Effect
1.
a) 1.09 × 1015 s–1
b) 1.80 × 10–9 m
35. Orbitals and Wavefunctions
1.
a) The first graph is R2, the second one is R, and the third is r 2 R2 .
b) B and C.
c) A.
2. C.
36. Orbitals and Quantum Numbers
1.
2.
Subshell
n
l
total #
of e–
1s
1
0
2
3d 3
2
10
5p 5
1
6
4.09 × 10–19 J emitted
37. 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
133
38. Electron Configurations of Ions
1.
Be+ : 1s 22s1
N– : 1s22s22p4
Al3+ :1s22s22p6
H– : 1s2
O2–: 1s 22s22p6
2.
Zn2+ : [Ar] 3d10
W6+ : [Xe] 4f 14
Cu2+ : [Ar] 3d9
Gd3+ : [Xe] 4f7
Se2–: [Ar] 4s23d104p6
39. Periodic Properties I: Overview
1.
a)
High ionization energy: upper right
Low ionization energy: lower left
b)
High atomic mass: lower right
Low atomic mass: upper left
c)
Big atomic radius: lower left
Small atomic radius: upper right
d)
Most metallic: lower left
"Least metallic" (most non-metallic): upper right
2.
a) Most reactive metal: Francium (Fr)
b) Most reactive non-metal: Fluorine (F)
40. Periodic Properties II: Multiple Choice
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.
134
41. Periodic Properties III: Explanations
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.
2.
a) When an electron is added to fluorine, its octet is completed.
b) 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 release make fluorine very
reactive.
c) 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.
d) 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.
42. Putting It Together: Quantum Mechanics and Electronic Structure
a) wavelength = 1.70 × 10–7 m
1.
b) Bi: [Xe] 6s24f145d106p3
Bi+:[Xe] 6s24f145d106p2
c) n=6, l=1
d) 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.
43. Lewis Structures I: The Octet Rule
F
F
C
F
O
F
F
F
H
C
O
N
N
C
S
O
F
O
N
H
H
O
O
O
O
2–
N
O
44. Lewis Structures II: Ions
F
S
O
+
F
N
–
O
2–
F
C
C
O
N
–
H
O
135
P
–
H
O
C
O
(plus two other
resonance
structures)
–
H
H
B
H
H
45. Lewis Structures III: Less Than an Octet
O
Br
Br
B
F
H Be H
Br
F
Al
Cl
O
O
O
O
N
O
N O
F
O
Cl
N
O
46. Lewis Structures IV: More Than an Octet
F Se F
F F
F
Cl
Xe F
Cl
Cl
As
Cl
F
Cl
F
F
As
F
–
F
F Xe F
F
F
+
F
–
Br F
47. Bond Enthalpies
1.
a) H–C≡C–H
2.
b) D C≡C = 813 kJ/mol
a) :N≡C–C≡N:
b) –1017 kJ/mol
48. Molecular Geometry I: Neutral Molecules
F
O
e- pair geo:
mol. geo:
hybridization:
polar?
C O
O
linear
linear
sp
no
N
F
F
trig. planar
bent
sp2
yes
B
Cl
F
I
Cl
Cl
trig. planar
trig. planar
sp2
no
trig. bipyramidal
T-shaped
sp3d
yes
49. Molecular Geometry II: Ions
F
F
e- pair geo:
mol. geo:
hybridization:
I
F
F
octahedral
sq. planar
sp3d2
–
+
Cl
Cl
P
O
Cl
Cl
tetrahedral
tetrahedral
sp3
136
O
2–
Se O
tetrahedral
trig. pyramidal
sp3
I
I
I
–
trig. bipyramidal
linear
sp3d
50. Polycentric Molecules and Ions
1.
trig. bipy.
see-saw
sp3d
F
S
F
S
tetrahedral
bent
sp3
F
F
tetrahedral
tetrahedral
sp3
O
H 3C
C
C
trig. planar
trig. planar
sp2
C
O
linear
linear
sp
linear
linear
sp
–
I
I
I
I
I
trig. bipy.
linear
sp3d
tetrahedral
bent
sp3
trig. bipy.
linear
sp3d
51. Putting It Together: Lewis Structures, VSEPR, and Bonding
1.
a)
S
S
S
S
S
S
S
S
b) Hybridization: sp3
Electron pair geometry: tetrahedral
Molecular geometry: bent
c) DS=S = 434 kJ/mol
137
52. Hybridization and Multiple Bonding: σ and π Bonds
1.
a) Trigonal planar; sp2 hybridization
b) 2 σ-bonds between H s orbitals and C sp 2 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) There’s a good drawing in your textbook, section 9.6, p. 327 (in the 8th edition).
2.
From left to right: sp3, sp, sp, sp2 (Refer to page 59, above, for further explanation)
•
3 σ-bonds between C and H
•
1 σ-bond between C sp 3 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)
53. Putting It Together: Bonding and Hybridization
O
1.
C
O
a)
b) Linear geometry; sp-hybridization
c) sp hybrid orbitals
d) p orbitals
H
H
C
e) H
C
C
H
Terminal carbon atom: sp2 hybridization
f) Central carbon atom: sp hybridization
g)
H
H
C
C
C
H
H
54. Covalent Bonding and Orbital Overlap
1.
a) 2p and 3s making σ bonding orbital
b) 2p and 2p making σ bonding orbital
c) 3d and 2p making π* antibonding orbital
d) 3p and 1s making σ* antibonding orbital
e) 3d and 3d making π bonding orbital
138
Each p-orbital must
be perpendicular to
the sp2–hybrid plane,
and the two π-bonds
must thus be perpend.
to each other.
55. Molecular Orbital Theory I: Introduction
1.
2.
a) (σ2s) 2(σ*2s) 2(π2p) 2
Bond order = 1 Paramagnetic
b) (σ2s) 2(σ*2s) 2(σ2p) 2(π2p) 4(π*2p) 3
c) (σ2s) 2(σ*2s) 2(π2p) 4
Bond order = 1.5
Paramagnetic
Bond order = 2 Diamagnetic
a)
σ2p*
π2p*
2p
2p
π2p
σ2p
σ2s*
2s
2s
σ2s
N
NO
atomic
orbitals
b)
Double bond:
Diamagnetic:
Longest bond:
Isoelectronic w/CO:
molecular
orbitals
NO–
NO+
NO–
NO+
139
O
atomic
orbitals
56. Molecular Orbital Theory II: Extensions of MO Theory
1.
a) and b)
σ2p*
σ2p*
π2p*
2p
2p
nonbonding
π2p
σ2p
σ2p
Note: "nonbonding"
means that the orbital is
not involved in bonding in
any way (neither bonding
nor antibonding).
σ2s*
2s
2s
σ2s
C
CO2
O
O
atomic
orbitals
molecular
orbitals
atomic
orbitals
atomic
orbitals
c) 4
d) diamagnetic
e) longer
57. Band Theory and Semiconductors
1.
a) insulator. Si4+ and O2– are the oxidation states, so no extra valence electrons.
b) conductor. Ru4+ and O2–, but Ru4+ does have extra valence electrons, so it can conduct.
c) semiconductor. No extra valence electrons, so not a conductor. But these large atoms have weak
bonding and will have a smallish band gap.
d) conductor. Ta3+ and N3–. The Ta3+ has extra valence electrons, so it can conduct.
e) insulator. This time it's Ta5+ , which has no extra valence electrons.
2.
p-type. The Fe3+ has one fewer valence electron, so it provides a "hole" which can allow electricity to
flow.
140
58. 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, F 2
59. Phase Changes
1.
a) MP: 115°C (388 K)
BP: 445°C (718 K)
b) total of 169.4 joules required
60. The Clausius-Clapeyron Equation
1.
+42.2 kJ/mol
2.
25.1 torr
61. Phase Diagrams
1.
a) and b)
• boiling pt.
• melting pt.
760
P
(torr)
liquid
solid
•triple pt.
100
gas
1
39
T (°C)
110
184
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.
62. Bonding in Crystalline Solids
1.
7.60 g/cm3
2.
radius = 1.97 Å
63. X-Ray Diffraction: The Bragg Equation
1.
a) They must be n1 = 4 and n2 = 6. (These are the only combinations which give a reasonable
value for the radius of the Po atom!)
b) 3.34 Å
c) 9.28 g/cm3
141
64. Lattice Energy: The Born-Haber Cycle
1.
Lattice enthalpy = +5400 kJ/mol
2.
Lattice enthalpy = +728 kJ/mol
65. Expressing Solution Concentrations
1.
a) 41.9% maltose
b) 2.11 m
c) χ maltose = 0.037
d) χ H2O = 0.963
e) 38.3 M water
66. Colligative Properties I: Non-Dissociating Solutes
1.
a) 100.71°C
b) 23.2 torr
2.
19200 g/mol
3.
χ CCl4 = 0.80 χ Br2 = 0.20
67. 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.
68. 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).
69. 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.
n = 22.2 (Silicate ions tend to form long chains and rings in solution.)
70. Rate Laws
1.
rate = k [A]2 [B]
2.
a) rate = k [NO]2 [Cl2]
b) k = 9.12 M–2 hr–1
c) 54.7 M hr–1
142
71. First-Order Kinetics
1.
P(SO2Cl2) = 3.25 atm
P(SO2) = 5.75 atm
P (Cl2) = 5.75 atm
2.
t = 7.31 × 10 –7 sec
72. Higher-Order Kinetics
1.
a) 7.7 × 10–7 sec
b) This reaction is so fast that the speed of mixing would actually be the limiting factor.
2.
[NO] = 0.0115 M
[O 2] = 0.00577 M
[NO 2] = 0.00845 M
73. Temperature and Rate
1.
Ea = 62.7 kJ/mol
2.
386 K
3.
i) speed up
ii) speed up
iii) slow down
iv) speed up
v) remain same
74. Putting It Together: Macroscopic Kinetics
1.
a) rate = k [N2O4]
b) k = 5.5 × 104 sec–1
c) t = 7.11 × 10 –5 sec
d) half-life = 4.62 × 10 –6 sec
e) Ea = 51.9 kJ/mol
75. Introduction to Mechanisms
1.
rate = k [Cl2][H]
rate = k [C3H6N2]
2.
a) rate = k1 [A]
k k
b) rate = k1 2 [A][C]
–1
k 1k 2[A][C]
c) rate = k + k [C]
–1
2
76. Advanced Mechanisms
1.
2.
k 1/2 k k [CO][Cl2] 3/2
a) rate = 1 1/22 3
k –1 (k–2 + k3[Cl2])
k 1/2 k k
b) The term in the denominator with k3[Cl2] would vanish, leaving k = 1 1/2 2 3
k –1 k–2
k 1k 2[A][B][D]
rate =
k –1
143
77. Writing Equilibrium Constants
[PCl3][Cl2]
1. a) Kc = [PCl
5]
b) Kc = [NH3][HCl]
2.
3.
c) Kc = [H2O(g)]
[CO][H2O]
a) Kc = [H ][CO
2
2]
b) Kc =
[H 2O] 2
[H 2] 2
c) Kc =
[CO2] 2
[CO] 2
d) 13.7
Kc = 13
78. Kp, Kc, and Q
1.
a) Kc = 5.9 × 1013
b) Q p = 5.4 × 101, so reaction goes to right, and NO2 will be produced under these conditions.
2.
a) Kc = 5.2 × 10–3
b) Q c = 0.25, so reaction goes to left, and NOBr will be produced under these conditions.
79. LeChâtelier’s Principle
1.
i) increase
ii) increase
iii) decrease
iv) remain same
v) remain same
2.
i) remain same
ii) increase
iii) decrease
iv) increase
v) remain same
vi) decrease
3.
a) high temperature, low pressure
b) low temperature, high pressure
80. Calculating Equilibrium Constants
1. Kc = 0.10
2.
Kp = 0.0196
3.
Kp = 0.077
144
81. Determining Final Equilibrium Concentrations
1.
SO2Cl2: 0.0052 moles
SO2: 0.178 moles
Cl2: 0.178 moles
Total pressure: 17.8 atm
2.
Pressure of BrF: 3.5 atm
3.
1.7 × 1018 individual oxygen atoms in the flask
82. 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
3.
a) H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
b) 12
83. Weak Acids
1.
2.10
2.
1.39
84. Weak Bases
1.
11.84
2.
8.53
3.
10.41
85. Polyprotic Acids and Bases
1.
[H 3PO4] = 0.0761 M
[H 2PO4–] = 0.0239 M
[HPO 42–] = 6.2 × 10–8 M
[PO43–] = 1.1 × 10–18 M
[H + ] = 0.0239 M
[OH –] = 4.2 × 10–13 M
2.
1.23
3.
8.60
86. Acidic Behavior of Metal Cations (Hydrolysis)
1.
2.93
2.
5.11
145
87. Putting It Together: Aqueous Acid Equilibria
1.
1.59
2.
pH = 0.84
[H 2SO3] = 1.33 M
[HSO 3–] = 0.143 M
[SO32–] = 1.02 × 10 –7 M
88. An Introduction to Buffers
1.
3.52
2.
12.5 mL of 0.10 M acetic acid
7.5 mL of 0.15 M sodium acetate
3.
a) pH = 4.90
b) (in order from top to bottom) increase, decrease, decrease, remain same, increase
89. Calculating Concentrations in Buffered Solutions
1.
[NH 4+ ] = 0.360 M
2.
a)
pH = 8.78
b)
[H 2CO3] = 7.44 × 10 –5 M
[HCO3–] = 0.0193 M
[CO32–] = 6.53 × 10 –4 M
90. Titrations
1.
172 mL
2. 7.51
91. 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
92. Solubility of Metal Hydroxides
1.
a) 10.46
b) 1.2 × 10–3 M
93. Other Aqueous Equilibria: Dealing With Large K’s
1.
[Cu 2+ ] = 5.6 × 10–15
[NH 3] = 0.92
[Cu(NH3) 4+ ] = 0.020
2.
[Mn2+ ] = 4 × 10–38
94. Solubility and Complex Ion Formation
1.
[Ag+ ] = 4.5 × 10–9
2.
0.0033 M
146
95. Putting It Together: Solubility and Acid/Base Behavior
1.
9.9 × 10–11 M
2.
7.0 × 10–7 M
96. An Introduction to Thermodynamics
1.
a) +37.4 kJ/mol
b) +110.8 J/mol·K
c) 64.4°C
97. Molecular Interpretation of Entropy
1.
a) ΔS should be negative; the system is becoming more ordered.
b) ΔS° = –279 J/mol·K
2.
Ions in water are surrounded by waters of hydration; these water molecules are highly ordered.
Thus the two ions produced each have a lot of well-ordered water molecules surrounding them and the
total order is increased.
98. Using Standard Free Energies and Enthalpies
1.
a) +240 kJ/mol
b) +76.2 kJ/mol
c) –550 J/mol·K
d) +259 kJ/mol
e) +223 J/mol·K
99. ΔG° and Keq
1.
a) –128 kJ/mol
b) 1.74 × 1022
2.
a) 60°C
b) 45°C
100. ΔG at Nonstandard Conditions
1.
a) +8.32 kJ/mol
b) –2.24 kJ/mol
2.
a) ΔG° = –29.0 kJ/mol
K = 1.2 × 105
b) –38.4 kJ/mol
101. Calculating K at Different Temperatures
1.
a) ΔH° = –92.38 kJ/mol
ΔS° = –198.2 J/mol·K
b) K = 6.9 × 105
c) K = 0.71
147
ΔG° = –33.32 kJ/mol
102. Applications: Determining ΔG°, ΔH°, and ΔS° from Keq
1.
a) –104.9 kJ/mol
b) –228 J/mol·K
2.
a) +25.9 kJ/mol
b) +60.7 kJ/mol
c) +106.1 J/mol·K
103. Applications: Phase Changes
1.
a) 35.3°C
b) 60.7 torr
2.
a) +31.76 kJ/mol
b) +9.63 kJ/mol
c) +77.34 J/mol·K
104. Applications: Acid/Base Equilibria
1.
a) [H + ] = 0.00126 M [Bu –] = 0.00126 M
[HBu] = 0.0987 M
b) –4.4 kJ/mol
2.
–67.7 kJ/mol
105. Applications: Buffers
1.
a) +26.9 kJ/mol
b) –12.0 kJ/mol
2.
pH = 4.70
106. 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 S 2 O3 2– (aq) + Cr2 O7 2– (aq) → 3 S4 O6 2– (aq) + 2 Cr3+ (aq) + 7 H2O (l)
c) 18 H2O + 14 ClO3 – (aq) + 3 As2 S3 (s) → 14 Cl– (aq) + 6 H2 AsO4 – (aq) + 9 SO4 2– (aq)
+ 24 H + (aq)
d) 3 H2O + 7 IO3– (aq) + 6 Re (s) → 6 ReO4– (aq) + 7 I– (aq) + 6 H+ (aq)
e) 30 H+ (aq) + 30 HSO 4– (aq) + As4 (s) + 10 Pb3O4 (s) → 30 PbSO4 (s) + 4 H3AsO4 (aq)
+ 24 H 2O
f) 3 HNO2 (aq) → NO3– (aq) + 2 NO (g) + H2O + H+ (aq)
107. Balancing Redox Reactions: Basic Solution
1.
a) 3 C 4H4O62– (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 H2 O2 (aq) + Cl2 O7 (aq) + 2 OH – (aq) → 2 ClO 2 – (aq) + 4 O2 (g) + 5 H2O (l)
d) Tl2O3 (s) + 4 NH2OH (aq) → 2 TlOH (s) + 2 N2 (g) + 5 H2O (l)
e) Cu(NH3 ) 4 2+ (aq) + S2 O4 2– (aq) + 4 OH– (aq) → 2 SO3 2– (aq) + Cu (s) + 4 NH3 (aq)
+ 2 H2O (l)
f) 3 Mn(OH) 2 (s) + 2 MnO 4– (aq) → 5 MnO2 (s) + 2 OH– (aq) + 2 H2O (l)
108. Standard Reduction Potentials at 25°C
148
109. Voltaic Cells: Calculating Cell Potential
1.
a) Overall: 2 Fe3+ (aq) + 2 I– (aq) → 2 Fe2+ (aq) + I2 (s)
Reduction: Fe3+ → Fe2+
Oxidation: 2 I– → I 2
b) Anode: I– / I2
Cathode: Fe3+ / Fe 2+
c) Oxidizing agent: Fe3+
Reducing agent: I–
d) E° = 0.235 V
2.
0.903 V
3.
a) 0.462 V
b) –1.100 V
110. Cell Potential and Free Energy
1.
–91.3 kJ/mol
2.
a) 2 H+ (aq) + 3 H2O (l) + SF 6 (g) + 2 e– → H2SO3 (aq) + 6 HF (aq)
b) 2.60 V
111. 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
112. Nonstandard Conditions: The Nernst Equation
1.
1.23 V
2.
a) 4 H+ (aq) + 4 Fe 2+ (aq) + O2 (g) → 4 Fe3+ (aq) + 2 H2O (l)
b) 0.036 V
113. 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
114. Putting It Together: Electrochemistry, ΔG°, and Keq
1.
a) Anode: Sn → Sn2+ + 2 e –
Cathode: Pb2+ + 2 e– → Pb
b) 0.010 V
c) –1.9 kJ/mol
d) K = 2.2
e) 0.045 V
115. Electrolysis: Faraday's Law
1.
0.296 g
2.
9.85 amps
3.
a) 30 hours
b) 334 kg graphite
149
116. Putting It Together: Electrochemistry and Solubility
1.
1.7 × 10–10
2.
a) 8 H+ (aq) + BaCrO4 (s) + 3 e– → Ba2+ (aq) + Cr3+ (aq) + 4 H2O (l)
b) 1.28 V
117. Nuclear Reactions: Introduction
1.
a) positron
b) 15C
c) 90Zr
d) 226Ra
e) 247Es
f) 143Xe
g) 4He (α particle)
h) 4 beta particles
118. Radioactive Dating
1.
a) 29 years
b) 83 years
119. Energy Changes in Nuclear Reactions
1.
1.34 × 10–13 J
2.
7.8 × 10–14 J
120. Radiation Dose
1.
0.0205 rad
2.
a) 3.14 × 10–10 g
b) 0.143 rad
121. Putting It Together: Nuclear Chemistry
1.
a) 14C → 14N + β–
b) 2.51 × 10–14 J
c) 2.35 × 108 m/s
d) 92 %
122. Coordination Complexes: Introduction
1.
a) Pt(II), d8, tetraammineplatinum(II) perchlorate
b) Re(VII), d0, nonahydridorhenate(VII)
c) Fe(-2), d 10, sodium tetra(carbonmonoxide)ferrate(–2) or sodium tetracarbonylferrate(–2)
d) Ir(III), d6, hexafluoroiridate(III)
e) Fe(III), d5, potassium hexacyanoferrate(III)
f) Pt(II), d 8, diamminedichloroplatinum(II)
150
123. Geometry and Isomerism I: Monodentate Ligands
1.
a)
NH3
H2
O
OH2
NH3
H2
O
Co
H 2O
NH3
Co
N
H3
NH3
H 2O
both achiral
N
H3
OH2
b)
NH3
NH3
Br
Cl
Cl
Pt
Cl
Cl
Br
Br
Pt
Br
Cl
Br
NH3
all trans
achiral
NH3
Cl
NH3 trans
achiral
NH3
Cl
Pt
Br
NH3
Br
NH3
Br
Pt
N
H3
Cl
Br
Cl trans
achiral
NH3
H3
N
NH3
Pt
N
H3
Br
Pt
Cl
Br trans
achiral
Br
Cl
Br
Cl
Cl
all cis
chiral
all cis
chiral
2.
NH3
O2
N
a)
NH3
O2
N
NO2
Co
O2N
N
H3
NH3
NH3
Co
O2N
N
H3
NO2
b) Since the Cl groups are replacing the NO2– ligands, and the Cl's end up trans in the product, there
must be a trans pair of NO2– ligands in the starting material. Thus the isomer on the left must have
been the starting material.
124. Geometry and Isomerism II: Chelates
1.
Cl
OH2
N
OH2
N
Co
N
Cl
O
H2
Cl trans
achiral
OH2
Cl
N
Co
N
OH2
H2
O
OH2
Co
Cl
N
OH2
H2O trans
achiral
Cl
Cl
N
Cl
Cl
all cis
chiral
all cis
chiral
Note: ethylenediammine (en) is represented as:
151
Co
N
N
N
125. Electronic Structure of Coordination Compounds
1.
a) d 6 high-spin; 4 "downstairs" and 2 "upstairs"; 4 unpaired electrons.
b) d8 (high/low doesn't matter); 6 "downstairs" and 2 "upstairs"; 2 unpaired electrons
c) d 6 low-spin; 6 "downstairs"; no unpaired electrons.
d) d6 low-spin; 6 "downstairs"; no unpaired electrons.
2.
3.
[Co(NH3) 6] 3+
[Co(NH3) 5(OH 2)]3+
[CoF 6] 3–
strongest ligands
intermediate
weakest ligands
largest Δo
intermediate Δo
smallest Δ o
absorbs violet
absorbs green
absorbs orange
appears yellow
appears red
appears blue
FeF63– is d5 high-spin electron configuration with 5 unpaired electrons, all spin "up". In order to
promote an electron from the lower to the higher energy level, its spin would have to "flip," which is
forbidden. Thus the d5 high-spin configuration is usually colorless. (This isn't a problem for d6
Co(III)).
126. Putting It Together: Coordination Compounds
1.
a) Sodium pentacyanonitrosylferrate (III)
b) d5 low spin; 5 "downstairs"; one unpaired electron.
c) NO must be slightly weaker than CN. The complex with the NO ligand absorbs light of slightly
lower energy (absorbs yellow-green) in comparison with the all-CN complex (absorbs green-blue).
Thus the splitting Δ must be less when there is an NO ligand, and we conclude that the NO ligand is
slightly weaker than the CN ligand. (If it were much weaker we would have seen a much more
dramatic change of color.)
d) Since the complex has 5 electrons in the more stable “downstairs” (t2g) orbitals, and no electrons in
the “upstairs” (eg) orbitals, it will be very stable, kinetically inert, and thus the ligands will dissociate
slowly.
127. Organic Chemistry: Isomers and Functional Groups
1.
H3C
HO
O
O
C
H2
C
H
C
H
C
H
C
H2
H2
C
CH2
H3 C
CH3
C
H
C
H
OH
CH
CH3
H2
C
H2 C
C
CH3
O
OH
HO
CH
C
H2
O
CH2
CH3
C
H2C
CH3
(cis+trans isomers)
C
H2
152
H2
C
O
2.
a) H3C
C
H2
OH
b) Initial product: H3C
C
H
O , which is an aldehyde.
O
c) Final product: H3C
, which is a carboxylic acid.
C
OH
128. Organic Chemistry: Polymers
1.
Poly(tetrafluoroethylene), Teflon®
Monomer:
F
F
F
F
F
F
F
C
C
C
C
C
C
F
F
F
F
F
F
C
2.
N
C
C
H
CH3
H
O
N
C
C
H
CH3
addition
H
O
N
C
C
H
CH3
H
c)
O
O
C
O
C
O
O
C
H2
O
N
C
C
H
CH3
addition
b) alcohol
C
H
OH
Type of Polymerization: (circle)
a) carboxylic acid
O
condensation
Monomer:
n
O
F
Type of Polymerization: (circle)
Polyalanine (a polypeptide)
O
C
F
n
H
F
C
O
C
H2
O
C
H2
153
C
H2
condensation
154
21
22
23
24
25
26
27
28
29
30
8
9
Mg
Na
Ca
K
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
N
O
F
Ne
Si
Al
16
S
15
P
Cl
17
Ar
18
Ga
31
Ge
32
As
33
Se
34
Br
35
Kr
36
26.98 28.09 30.97 32.07 35.45 39.95
14
13
Y
39
Zr
40
41
Nb
Mo
42
Ba
Cs
72
Hf
57
La
Ta
73
W
74
43
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
I
53
Xe
54
Re
75
Os
76
Ir
77
Pt
78
Au
79
Hg
80
Tl
81
Pb
82
Bi
83
Po
84
At
85
Rn
86
(98) 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.76 127.60 126.91 131.29
Tc
Ra
Ac
Rf
Ha
60
Nd
Pm
61
Sm
62
63
Eu
Gd
64
Tb
65
Dy
66
67
Ho
68
Er
69
Tm
70
Yb
Lu
71
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.04 231.04 238.03 237.05 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)
Actinide series Th
140.12 140.91 144.24 (145) 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97
100
101
102
103
90
91
92
93
94
95
96
97
98
99
59
Pr
58
(262) (263) (262) (265) (266)
Lanthanide seriesCe
(223) 226.03 227.03 (261)
Fr
132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.20 192.22 195.08 196.97 200.59 204.38 207.20 208.98 (209) (210) (222)
104
105 [106] [107] [108] [109]
87
88
89
56
55
85.47 87.62 88.91 91.22 92.91 95.94
38
Sr
37
Rb
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.61 74.92 78.96 79.90 83.80
20
19
22.99 24.31
12
11
C
10.81 12.01 14.01 16.00 19.00 20.18
B
Be
7
Li
6.941 9.012
6
4
5
He
4.003
10
H
2
1.008
3
1
PERIODIC TABLE OF THE ELEMENTS