the logan notes

Harvard UniversityGeneral ChemistryPractice Problems

“The Logan Notes”2004-2005

PERI

ODI

C TA

BLE

OF

THE

ELEM

ENTS

12

HHe

1.00

84.

003

34

56

78

910

LiBe

BC

NO

FNe

6.94

19.

012

10.8

112

.01

14.0

116

.00

19.0

020

.18

1112

1314

1516

1718

NaM

gAl

SiP

SCl

Ar22

.99

24.3

126

.98

28.0

930

.97

32.0

735

.45

39.9

519

2021

2223

2425

2627

2829

3031

3233

3435

36K

CaSc

TiV

CrM

nFe

CoNi

CuZn

Ga

Ge

AsSe

BrKr

39.1

040

.08

44.9

647

.88

50.9

452

.00

54.9

455

.85

58.9

358

.69

63.5

565

.39

69.7

272

.61

74.9

278

.96

79.9

083

.80

3738

3940

4142

4344

4546

4748

4950

5152

5354

RbSr

YZr

NbM

oTc

RuRh

PdAg

CdIn

SnSb

TeI

Xe85

.47

87.6

288

.91

91.2

292

.91

95.9

4(9

8)10

1.07

102.

9110

6.42

107.

8711

2.41

114.

8211

8.71

121.

7612

7.60

126.

9113

1.29

5556

5772

7374

7576

7778

7980

8182

8384

8586

CsBa

LaHf

TaW

ReO

sIr

PtAu

HgTl

PbBi

PoAt

Rn13

2.91

137.

3313

8.91

178.

4918

0.95

183.

8518

6.21

190.

2019

2.22

195.

0819

6.97

200.

5920

4.38

207.

2020

8.98

(209

)(2

10)

(222

)87

8889

104

105

[106

][1

07]

[108

][1

09]

FrRa

AcRf

Ha(2

23)

226.

0322

7.03

(261

)(2

62)

(263

)(2

62)

(265

)(2

66)

5859

6061

6263

6465

6667

6869

7071

Lant

hani

de s

erie

s

Ce

PrNd

PmSm

EuG

dTb

DyHo

ErTm

YbLu

140.

1214

0.91

144.

24(1

45)

150.

3615

1.96

157.

2515

8.93

162.

5016

4.93

167.

2616

8.93

173.

0417

4.97

9091

9293

9495

9697

9899

100

101

102

103

Actin

ide

serie

s

Th

PaU

NpPu

AmCm

BkCf

EsFm

Md

NoLr

232.

0423

1.04

238.

0323

7.05

(244

)(2

43)

(247

)(2

47)

(251

)(2

52)

(257

)(2

58)

(259

)(2

60)

Harvard University BLB: Ch. 1; PHH: Ch. 1

Significant Figures and Scientific Notation1. Perform the following calculations and express each answer to the proper number

of significant figures:

a) 423.1100.

+ 0.256b) 14.000

6.1c) (6.11)(π)

d) (4/3)π(2.16)3 e) (14.3)(60) f) (6.0+9.7+0.61)(1.113)

g) 6.958×108

5.91×1012h) (1.173×10–3) + 3.6 i) (1.1×106)(2.246×10–10)

1


Dimensional Analysis I: Unit Conversions1. 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 infeet.

2. Given the following conversion factors:

1 atm = 760 torr1 torr = 0.01933 lb/in21 N = 0.225 lb1 in = 2.54 cm1 Pa = 1 N/m2

a) The pressure inside a laboratory vessel is 35 lb/in2. Calculate the equivalentpressure in atmospheres.

b) A certain chemistry experiment takes place in a high-pressure chamber with apressure of 2.7 atmospheres. Calculate the pressure in torr.

c) The atmospheric pressure is measured to be 754 torr. Calculate the pressure inPa.

2


Dimensional Analysis II: Volume and Density1. Here are some old-fashioned units used for measuring volumes:

1 bushel = 1.24 cubic feet (ft3)1 cord = 128 ft31 hogshead = 63 gallons1 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 incords.)

b) How many hogsheads is 3.5 cords?

c) A rectangular tank measures 1.0 × 5.0 × 2.0 meters. Calculate its volume ingallons.

2. A certain crystal of calcium chloride has a volume of 765.3 mm3.

a) Calculate the volume of this crystal in cubic inches (in3).

b) The density of CaCl2 is 2.51 g/cm3. Calculate the mass of this crystal.

3


Atoms, Molecules, and Ions1. 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 andelectrons in a calcium ion (as found in calcium chloride).

3. The chloride ion exists in two common isotopes: 35Cl– and 37Cl–. Determine thenumber 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 onemolecule of NH3. (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 ammoniumion, NH4

+. (Again, assume that the only isotopes present are 14N and 1H.)

4


Oxidation Numbers1. Write the oxidation numbers of each atom in the following species:

BrO3– H3AsO3 AsH3

CrCl3 KClO3 S2O32–

OF2 Na2O2 Fe3O4

2. Identify whether oxidation and reduction is taking place in each of the followingequations. If so, identify which species is being reduced and which is beingoxidized.

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


Naming Compounds1. 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


Chemical Formulas: Percent Composition1. 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 isbelieved that the compound is Sarin. When a 10.0-gram sample of this compoundis completely combusted, 15.6 g CO2 and 6.4 g H2O are produced, along withother combustion products. Using numerical calculations, prove that thisunknown compound can not be Sarin.

2. A pulverized rock sample believed to be pure calcium carbonate is subjected tochemical 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 CaCO3.

7


Empirical and Molecular Formulas1. 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 beenisolated from a tropical plant.

a) When 5.467 grams of compound Z are burned in excess oxygen, 15.02 grams ofCO2 and 2.458 grams of H2O are produced. Determine the empirical formula ofcompound Z.

b) Other experiments suggest that compound Z has a molar mass of approximately250 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 gof H2O and 0.372 g of N2 are recovered. Determine the empirical formula of thiscompound.

b) Other experiments suggest that the molar mass of this compound is about 50g/mol. Determine the molecular formula of this compound.

8


Writing and Balancing Equations1. 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) → P4 (g) + CaSiO3 (l) + CO (g)(Hint: Balance the P first)

2. Photosynthesis (in plants) converts carbon dioxide and water into glucose(C6H12O6) and oxygen. Write and balance the chemical equation forphotosynthesis.

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


Stoichiometry of Reactions1. 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 ofNaOH, how many grams of NaClO can be produced, assuming the reaction goesto completion?

2. For the following unbalanced chemical equation:

NiS + O2 + HCl → NiCl2 + H2SO4

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


Stoichiometry with Limiting Reagents1. Vanadium (V) oxide, V2O5, can be reduced by zinc to form vanadium (II) oxide,

V2O2 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 100.0grams of V2O5 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 P4S5. (Hint:balance the S first.)

P4S3 + Br2 → P4S5 + PBr3

b) What is the maximum quantity of P4S5 that could be prepared from 100.0 g ofP4S3 and 100.0 mL of liquid bromine? (Density of bromine is 3.12 g/mL)

11


Stoichiometry of Mixtures1. Ferrous oxalate, FeC2O4, will decompose on heating:

FeC2O4 → FeO + CO2 + CO

The gaseous products (a mixture of CO2 and CO) are collected. Calculate thepercent by mass of CO2 in this gaseous mixture.

2. A certain mixture of CuO and Cu2O weighs 10.50 grams total. Completereduction of this mixture produces 8.66 grams of pure metallic Cu. Determine theamounts of CuO and Cu2O in the original mixture.

3. A certain mixture of CuS and Cu2S weighs 10.80 grams total. Completereduction of this mixture produces 8.06 grams of pure metallic Cu. Determine theamounts of CuS and Cu2S in the original mixture.

12


Solutions: Molarity1. 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 ofpure (anhydrous) acetic acid.

3. A hydrochloric acid solution is prepared by dissolving 1.97g of hydrogen chloridegas in 27.3 mL of water and then diluting that mixture to a total volume of 250.00mL. Calculate the molarity of the resulting solution.

4. The concentration of NaClO in Clorox is 0.705 M. Calculate the mass of NaClOpresent in 1.0 mL of Clorox.

5. A solution is prepared by dissolving x grams of potassium nitrate in water anddiluting to a total volume of 100.0 mL. Another solution is prepared bydissolving y grams of sodium chloride in water and diluting to a total volume of500.0 mL. Both solutions are then mixed together, giving a final concentration ofKNO3 of 0.073 M and a final concentration of NaCl of 0.128 M.Calculate x and y.

13


Solution Stoichiometry I: Simple Examples1. The toxic compound NCl3 can be formed from the reaction of bleach (NaClO)

with ammonia (NH3):

NaClO (aq) + NH3 (aq) → NCl3 (l) + NaOH (aq)

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? (Cloroxis a 0.705-molar solution of NaClO in water.)

2. A sample of calcium carbonate weighing 6.35 grams is placed in 500.0 mL of0.31M hydrochloric acid and allowed to react to form calcium chloride andcarbon dioxide gas. Calculate the maximum mass of carbon dioxide gas that canbe produced.

14


Solution Stoichiometry II: Acid/Base Neutralizations1. 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 of1.27 M HCl would be required to completely neutralize this solution?

3. Phosphoric acid can be neutralized by sodium hydroxide according to theequation:

H3PO4 + 3 NaOH → Na3PO4 + 3 H2O

What volume of 0.176 M NaOH would be required to neutralize 5.00 mL ofconcentrated (14.8 M) phosphoric acid?

15


Solution Stoichiometry III: Titrations1. 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; completeneutralization 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 waterand 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 1.736grams of HX are titrated with 1.334 M NaOH, the endpoint is reached with 10.92mL of base. Calculate the molar mass of HX.

16


Solution Stoichiometry IV: Precipitation Reactions1. 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 aftercomplete 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 H2SO4; 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 resultingmixture: Ba2+, Cl–, NO3–, NH3, NH4+ (Hint: What is the total volume?)

17


The Ideal Gas Law1. 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 adisinfectant in swimming pools. Humans can detect the odor of chlorine when itis 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 inorder for humans to detect the odor of chlorine?

2. It requires 0.182 mol of O2 gas to exert a pressure of 1.50 atm in a particular tankat 25°C. What mass of O2 would be required to exert a pressure of 17.2 atm inthe same tank at 100°C? Neglect expansion of the tank itself.

18


Reactions Involving Gases: Simple Examples1. 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 oneweek, all of the niobium had been converted to niobium hydride, NbH. Calculatethe final pressure of hydrogen gas in the system at 25°C.

2. Ferrous oxalate, FeC2O4, will decompose on heating:

FeC2O4 (s) → FeO (s) + CO2 (g) + CO (g)

A 1.25-gram sample of FeC2O4 is added to an evacuated 2.00-liter steel vessel.The vessel is heated to 400°C, at which point all the FeC2O4 is decomposed.Calculate the pressure inside the vessel at 400°C.

3. Ethylene, C2H4, will react with hydrogen gas under appropriate conditions toform ethane, C2H6. A 10.0-liter vessel is charged with 1.5 atm of hydrogen and1.0 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


Mixtures of Gases1. 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 inthe 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. Calculatethe 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.

2. A sample of the gas butane (C4H10), of unknown mass, is contained in a vessel ofunknown volume, V, at 24.8°C and a pressure of 560.0 torr. To this vessel8.6787 g of Ne are added in such a way that no butane escapes. The total pressureof the vessel (at the same temperature) is 1420.0 torr. Calculate the volume of thevessel and the mass of the butane.

20


Collecting Gases Over Water1. 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 themass 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 iscollected over water at 25°C and 0.99 atm. Calculate the total volume of gascollected. (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. Youwant to reduce the pressure in the cylinder to 15.5 atm, so you allow oxygen toescape from the cylinder and collect the escaping gas over water. What volume ofgas should you collect at 25°C and 1.00 atm in order to achieve the desiredpressure in the cylinder? (The vapor pressure of water is 23.76 torr at 25°C.)

21


Stoichiometry of Gas Mixtures1. 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. Calculatethe mole fraction of each gas in the flask.

2. Ozone (O3) can be prepared in the laboratory by passing an electrical dischargethrough a quantity of oxygen gas (O2):

3 O2 (g) → 2 O3 (g)

An evacuated steel vessel with a volume of 10.00 liters is filled with 32.00 atm ofpure O2 at 25°C. An electric discharge is passed through the vessel, causing someof the oxygen to be converted into ozone. As a result, the pressure inside thevessel drops to 30.64 atm at 25°C. Calculate the final percent by mass of ozonein the vessel.

22


Reactions Involving Gases: Applications1. An automobile air bag is filled with nitrogen gas, which is produced by the rapid

thermal decomposition of sodium azide (NaN3):

2 NaN3 (s) → 2 Na (s) + 3 N2 (g)

What mass of NaN3 would be required to produce enough nitrogen gas to fill a10.0-liter air bag at a pressure of 1.20 atm and a temperature of 25°C?

2. Consider the reaction used to produce NaAlCl4:

2 Al2O3 (s) + 6 Cl2 (g) + 4 NaCl (l) ←→ 4 NaAlCl4 (l) + 3 O2 (g)

An evacuated reaction vessel with a volume of 10.0 liters is filled with chlorinegas at 25°C and a pressure of 1.00 atm. Al2O3 and NaCl are added, and themixture is heated to 850°C until the reaction is complete. The final total pressurein 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


Kinetic-Molecular Theory of Gases1. A mixture of ammonium nitrate and butane are placed into a steel cylinder and

detonated. The product gases (CO2, 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. Youwant to reduce the pressure in the cylinder to 15.5 atm, so you drill a tiny pinholeand allow the oxygen to effuse into a vacuum. You know that, under identicalconditions, 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 reducethe pressure from 17.2 atm to 15.5 atm?

24


Heat1. 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 thefinal temperature of the system.

2. A coffee-cup calorimeter contains 150. g of water at 24.6°C. A 110-g piece ofmolybdenum 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 thespecific heat capacity of molybdenum metal. (The specific heat capacity of wateris 4.184 J/g°C).

25


Calorimetry I: Simple Examples1. 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, thetemperature of the solution increases from 25.0°C to 26.2°C. Calculate ΔH° forthe dissolving of sodium carbonate. (The heat capacity of the solution is 4.2J/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 H2O (l) in a bomb calorimeter. Thetemperature rises by 2.270°C. The heat capacity of the calorimeter and itscontents 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°Crise in temperature is observed. When a 0.265-gram sample of caffeine(C8H10O2N4) is burned, a 1.525°C rise in temperature is measured. Using thevalue 26.38 kJ/g for the heat of combustion of benzoic acid, calculate the heat ofcombustion per mole of caffeine at constant volume. (For this calculation, assumethat ΔH ≈ ΔE.)

26


Calorimetry II: Mixtures1. 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 ofpetroleum 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 calorimeterwith Ccal = 10.43 kJ/°C. The initial temperature is 22.59°C. Predict the finaltemperature 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 mustbe added to obtain iced tea at 10°C. (The specific heat of the tea is 4.18 J/g°C.)

27


Hess's Law1. Given the following enthalpies of reaction:

N2O4 (g) → 2 NO2 (g) ΔH° = 57.20 kJNO (g) + 1/2 O2 (g) → NO2 (g) ΔH° = –57.07 kJ

Calculate the enthalpy of reaction for:

2 NO (g) + O2 (g) → N2O4 (g)

2. Given the following enthalpies of reaction:

NH3 (g) → NH3 (aq) ΔH = –34.10 kJ/molHNO3 (g) → HNO3 (aq) ΔH = –72.3 kJ/molNH4NO3 (s) → NH4NO3 (aq) ΔH = 26.4 kJ/molNH3 (aq) + HNO3 (aq) → NH4NO3 (aq) ΔH = –52.3 kJ/mol

Calculate the enthalpy of reaction for:

NH3 (g) + HNO3 (g) → NH4NO3 (s)

28


Standard Enthalpies of Formation1. Using the following information:

ΔH°f (CaO (s)) = –635.5 kJ/mol ΔH°f (Ca(OH)2 (s)) = –986.2 kJ/molΔH°f (H2O (l)) = –285.83 kJ/mol

Calculate ΔH for the following reaction:

CaO (s) + H2O (l) → Ca(OH)2 (s)

2. Given the following data for butane (C4H10):

ΔH°f (C4H10 (g)) = –124.7 kJ/mol ΔHvap(C4H10 (l)) = 22.9 kJ/molΔH°f (H2O (g)) = –241.8 kJ/mol Density of C4H10 (l) = 0.59 g/mLΔH°f (CO2 (g)) = –393.5 kJ/mol

and the balanced equation for the combustion of liquid butane:

C4H10 (l) + 13/2 O2 (g) → 4 CO2 (g) + 5 H2O (g)

Calculate the quantity of heat released by the combustion of 10.0 mL of liquidbutane (the amount of butane in a cigarette lighter).

29

Harvard University BLB: Skip; PHH: Ch. 7

Enthalpy, Energy, Heat, and Work1. Consider the evaporation of one mole of methanol, CH3OH, at its boiling point

(64.4°C) at constant pressure (1.00 atm). The density of liquid methanol is 0.791g/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 boilingpoint.

c) Determine ΔE for the evaporation of one mole of methanol at its boiling point.

Substance: ΔH°f (kJ/mol)CH3OH (g) –201.2CH3OH (l) –238.6

30


Putting It Together: Calorimetry, Hess's Law, and ΔH°f1. 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 to16.9°C.

NH4NO3 (s) → NH4+ (aq) + NO3– (aq)

a) Calculate ΔH° for the above reaction. (The specific heat capacity of the solutionis 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 ΔH°f (NO3– (aq)) = –205.0 kJ/mol

2. Sodium azide will decompose explosively when heated; this reaction is the sourceof the gas used to inflate automobile air bags:

2 NaN3 (s) → 2 Na (s) + 3 N2 (g)

a) A 10.0-gram sample of sodium azide is completely decomposed in a bombcalorimeter. 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 theabove reaction. (For this calculation, assume that ΔH ≈ ΔE.)

b) From this information, calculate ΔH°f for sodium azide.

31


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 thefollowing equation:

CH4 (g) + 2 O2 (g) → CO2 (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 anyloss of heat to the vessel or surroundings. (The mixture of gases has an averagemolar heat capacity of 21 J/mol·°C)

c) Calculate the partial pressure of CO2 (g) in the vessel after combustion.

32


Energy, Particles, and Waves: A "Chain Problem"Useful Information:

Speed of light c = 3.00 × 108 m/sPlanck's constant h = 6.63 × 10–34 J·sMass of electron me = 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 isthe 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


The Photoelectric Effect1. The photoelectric binding energy of chromium is 7.21 × 10–19 J.

a) Calculate the minimum frequency of light which could produce the photoelectriceffect in chromium.

b) Light having a wavelength of 2.50 × 10–7 m falls upon a piece of chromium in anevacuated glass tube. Determine the minimum deBroglie wavelength of theemitted photoelectrons.

34


Orbitals and Wavefunctions1. Here are three graphs of the radial wavefunction of the 3pz orbital:

(radii are in units of ao)

5 10 15

0.0020.0040.0060.008

5 10 15

-0.1

0

0.1

5 10 15

0.020.040.060.080.1

a) Identify each as representing R, R2, or r2R2.

b) Which of the following describes a place where the electron will never be found:

(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

c) Choose the most likely radius to find an electron:

(A) r = 12ao (B) r = 2ao

(C) r = 6ao (D) r = 10ao

2. Which of the following could be a slice of the 4fz3 orbital in the (x, z) plane:

A B C D

35


Orbitals and Quantum Numbers1. 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:

ΔE = (2.18 × 10–18 J)

1

ni2 – 1

nf2

Calculate the energy of the radiation emitted by a hydrogen atom if an electronundergoes a transition from a 4px orbital to a 2s orbital.

36


Electron Configurations of Neutral Atoms1. 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


Electron Configurations of Ions1. 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


Periodic Properties I: Overview1. 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


Periodic Properties II: Multiple Choice1. 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. Circle the element from each set with the smallest ionization energy. Explainyour choices.

a) Tl Po Se Ga

b) Cs Ga Bi Se

3. Circle the element with the most negative electron affinity. Explain yourchoice.

Be N O F

4. Circle the ion with the largest radius. Explain your choice.

Se2– F– O2– Rb+

40


Periodic Properties III: Explanations1. 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 toSe.

c) The reaction of Rb with water is much more violent that the reaction of Na withwater.

2. Two of the most fundamental properties of any element are its ionization energyand electron affinity.

a) Fluorine has an electron affinity of –332 kJ/mol—one of the most negative of anyof the elements. Briefly explain why its electron affinity is more negative thanthat 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 substantialionization 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


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 longestpossible wavelength of light which could ionize an atom of bismuth.

b) Write the electron configurations of neutral Bi and the Bi+ cation. Please use thenoble-gas abbreviations.

Bi :Bi+ :

c) What are the n and l quantum numbers of the electron removed when Bi is ionizedto Bi+?

n = l =d) 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 thanExplain briefly. (It is not enough to simply state a trend; you must provide anexplanation for the trend in terms of atomic structure.)

42


Lewis Structures I: The Octet Rule1. 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


Lewis Structures II: Ions1. 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


Lewis Structures III: Less Than an Octet1. For each of the following molecules or ions, draw the best possible Lewis

structure. Include all non-zero formal charges, and indicate resonance ifappropriate.

BBr3 BeH2

AlF3 ClO2

NO2 NO

45


Lewis Structures IV: More Than an Octet1. For each of the following molecules or ions, draw the best possible Lewis

structure. Include all non-zero formal charges, and indicate resonance ifappropriate.

SeF4 XeF2

AsCl5 AsF6–

XeF3+ BrF2–

46


Bond Enthalpies1. The enthalpy of formation ΔH°f of acetylene (C2H2) is 226.6 kJ/mol.

a) Draw a correct Lewis structure for acetylene.

b) Given the following additional information, determine the bond enthalpy of theC≡C triple bond in acetylene.

ΔHsub (C (s)) = 717 kJ/mol DH—H = 436 kJ/molDC—H = 415 kJ/mol

2. 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 N–N 163C=N 615 C=O 799 O–O 146 N=N 418C≡N 891 C≡O 1072 O=O 495 N≡N 941Estimate ΔH for the complete combustion of cyanogen:

NCCN (g) + 2 O2 (g) → N2 (g) + 2 CO2 (g)

47

Harvard University BLB: Ch. 9; PHH: Ch. 11&12

Molecular Geometry I: Neutral Molecules1. For each of these molecules, draw the best possible Lewis structure. Using that

Lewis structure, predict the electron-pair geometry, the molecular geometry, thehybridization of the central atom, and whether the molecule is polar.

CO2

e- pair geometry:

molecular geometry:

hybridization:

polar? yes no

ONF

e- pair geometry:

molecular geometry:

hybridization:

polar? yes no

BF3

e- pair geometry:

molecular geometry:

hybridization:

polar? yes no

ICl3

e- pair geometry:

molecular geometry:

hybridization:

polar? yes no

48


Molecular Geometry II: Ions1. 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 moleculargeometry, and the hybridization of the central atom.

IF4–

e- pair geometry:

molecular geometry:

hybridization:

PCl4+

e- pair geometry:

molecular geometry:

hybridization:

SeO32–

e- pair geometry:

molecular geometry:

hybridization:

I3–

e- pair geometry:

molecular geometry:

hybridization:

2. For extra practice in applying VSEPR theory, go back to the pages of Lewisstructures and predict the geometries of each of those molecules or ions.

49


Molecular Geometry III: Polycentric Molecules and Ions1. 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 moleculargeometry, 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


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 aLewis structure for S8. (Note that each sulfur atom should be identical.)

b) Specify the hybridization, electron pair geometry, and molecular geometry aroundeach sulfur atom in S8.

c) The average S—S single bond energy is 266 kJ/mol. Using the ΔH given at thetop of this page, calculate the S=S double bond energy in S2 (g).

51


Hybridization and Multiple Bonding: σ and π Bonds1. a) Determine the geometry and hybridization of the carbon atom in formaldehyde,

H2C=O.

b) Explain the bonding in formaldehyde: describe each σ- and π-bond in themolecule. (e.g. “There are two σ-bonds between H s orbitals and C sp2 hybridorbitals”)

c) Imagine that formaldehyde is positioned in the y,z plane as shown below. On thisframework, draw a perspective sketch of the formaldehyde molecule in thisorientation, showing all σ- and π-bonding orbitals and the lone pairs on theoxygen atom.

z-axisC OH

H

y-axis

x-ax

is

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–Hbonds, and include resonance!)

52


Putting It Together: Bonding and Hybridization1. 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 CO2?

e) The molecule allene, H2C=C=CH2, is similar to CO2 in several respects. Draw acomplete 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 perpendicularplanes as shown:

C C

H

HH

HC

Draw π-bonding orbitals on the above diagram, showing clearly why the two endsof the molecule must be perpendicular to one another.

53


Covalent Bonding and Orbital Overlap1. 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:

b)type of interaction:

c)

type of interaction:

d)


e)


54


Molecular Orbital Theory I: Introduction1. 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 sureto 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


Molecular Orbital Theory II: Extensions of MO Theory1. Here is the molecular orbital correlation diagram (valence orbitals only) for the

triatomic molecule CO2, showing both σ- and π-bonding:

2p

2s

2p

2s

σ2s

σ2s*

σ2p*

σ2p*

σ2p

σ2p

π2p*

nonbonding

π2p

C CO2 O Oatomicorbitals

molecularorbitals

atomicorbitals

atomicorbitals

Note: "nonbonding"means that the orbital isnot involved in bonding inany way (neither bondingnor antibonding).

a) Fill in the appropriate number of valence electrons in the C, O, and O atomicorbitals in the diagram above. Use arrows (↑ or ↓) to indicate electron spin.

b) Fill in the appropriate total number of valence electrons in the CO2 molecularorbitals 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 or diamagnetic

e) You add an electron to CO2, forming the CO2– ion.Would the C–O bond length become (circle one) shorter or longer

56


Band Theory and Semiconductors1. 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: considerthe 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 Fe2+ has beenreplaced with Fe3+. Wüstite is observed to be a semiconductor. Is it an intrinsicsemiconductor, 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


Intermolecular Forces1. 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

H2O 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


Phase Changes1. 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 S2 (g).

Here are the phases of sulfur at 1.0 atm pressure, with specific heats and theenthalpy of each phase transition:

below 368 K S(I) (s) specific heat = 0.73 J/g·K• 368 K transition S(I) (s) → S(II) (s) ΔH = 0.401 kJ/mol368 – 388 K S(II) (s) specific heat = 0.78 J/g·K• 388 K transition S(II) (s) → S (l) ΔH = 1.722 kJ/mol388 – 718 K S (l) specific heat = 1.04 J/g·K• 718 K transition S (l) → 1/8 S8 (g) ΔH = 45.08 kJ/mol718 – 882 K S8 (g) specific heat = 0.73 J/g·K• 882 K transition 1/8 S8 (g) → 1/2 S2 (g) ΔH = 49.13 kJ/molabove 882 K 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 sulfurfrom 25°C to 150.°C.

59


The Clausius-Clapeyron Equation1. 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 knowingthat water ordinarily boils at 100°C, what would the pressure in a vacuum domehave to be in order for water to boil at room temperature, 25°C?

60


Phase Diagrams1. Given the following data for iodine:

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

a) Sketch a phase diagram for iodine in the space below. Observe the labeled axes(not drawn to scale).

18339

760

68

P

T (°C)

(torr)

106

1

b) 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 phasesare observed in equilibrium at this temperature. What are they?

d) At a given temperature, which is more dense: solid or liquid iodine? Explain.

61


Bonding in Crystalline Solids1. In nature, sulfur often appears in sulfide minerals, which contain the S2– 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

Pb

SPb

S Pb

S

S

Pb

S Pb

S

Pb

S

Pb

S

Pb

S

Pb

S

S

Pb

Pb

S Pb S

2. Calcium metal crystallizes in a face-centered cubic unit cell. The density of thesolid is 1.54 g/cm3. Calculate the radius of a calcium atom.

62


X-Ray Diffraction: The Bragg Equation1. 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, reflectionsoff 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 bea certain amount of trial and error... but think how you can limit your work. Alsoconsider the range of reasonable values for d.)

b) Calculate the length of the unit cell in Å.

c) Determine the density of polonium.

63


Lattice Energy: The Born-Haber Cycle1. 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 DCl–Cl = 243.4 kJ/mol ΔH°f (FeCl3) = –401 kJ/mol

2. Given the following thermodynamic information:

Na(g) → Na+ (g) + e– IE = 496 kJ/molNa (s) → Na (g) ΔHsub = 108 kJ/mol

Na (s) + C (s) + 1/2 N2 (g) → NaCN (s) ΔH°f = –87 kJ/molHCN (g) → H (g) + CN (g) DC–H = 502 kJ/mol

1/2 H2 (g) + C (s) + 1/2 N2 (g) → HCN (g) ΔH°f = 135 kJ/molH2 (g) → 2 H (g) DH–H = 436 kJ/mol

CN (g) + e– → CN– (g) 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


Expressing Solution Concentrations1. 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


Colligative Properties I: Non-Dissociating Solutes1. 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 vaporpressure 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 aqueoussolution has an osmotic pressure of 4.15 torr at 25°C. Calculate the molar mass ofthis type of interferon.

3. A solution of Br2 in CCl4 has a total vapor pressure of 129 torr at 25°C. Calculatethe mole fraction of each component in the solution.(At 25°C, P°(Br2) = 209 torr; P°(CCl4) = 109 torr)

66


Colligative Properties II: Dissociating Solutes1. 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: Considerwhether you have a dissociating or non-dissociating solute, and be sure tocalculate the mole fraction appropriately.) The vapor pressure of pure water is23.76 torr.

2. Sodium aluminum chloride, NaAlCl4, is soluble in water. One chemist suggeststhat, 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 cationand 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, thefreezing point of the resulting solution is –3.5°C. Do the results of thisexperiment support proposal A, proposal B, or neither? Provide calculations tosupport your choice.

67


Colligative Properties III: Multiple Solutes1. When excess chlorine gas is bubbled into water, the resulting solution has the

following composition:

Cl2 (aq) 0.062 MHOCl (aq) 0.030 MHCl (aq) 0.030 M

Predict the freezing point of this solution. (Its density is 1.04 g/mL.)

Note: • Hypochlorous acid (HOCl) is a weak acid.• HCl is a strong acid.• Be sure to consider the total concentration of dissolved species.

2. Consider the following solution: 0.500 moles of the weak acid HN3 plus0.750 moles of the ionic compound NaN3 in a total of 1.00 liter of solution. Thedensity of this solution is 1.02 g/mL.

Calculate the vapor pressure of water over this solution at 25°C. (The vaporpressure of pure water at 25°C is 23.76 torr). (Hint: Be sure to consider ALL themoles of solute particles in your calculation; assume ideal behavior.)

68


Colligative Properties IV: Non-Ideal Solutions1. 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 MgSO4.

b) What is the expected (ideal) van’t Hoff factor for MgSO4?

c) Briefly discuss any discrepancy between the observed and expected van’t Hofffactors.

2. In aqueous solution, the metasilicate ion (SiO32–) forms aggregates such as thefollowing:

n = 2: 2 SiO32– (aq) → Si2O64– (aq)n = 3: 3 SiO32– (aq) → Si3O96– (aq)n = 4: 4 SiO32– (aq) → Si4O128– (aq)

and so on . . . .

A chemist dissolves 0.0100 moles of Na2SiO3 in enough water to make 1.00 literof solution. Using a semipermeable membrane that allows Na+ ions to pass freelybut prevents the passage of the silicate ions, the osmotic pressure of this solutionis found to be 0.011 atm at 25°C.

Determine the average aggregation state (n) for the silicate ions under theseconditions. (Hint: the osmotic pressure is due to the concentration of the silicateions only.)

69


Rate Laws1. Using the following data, write a rate law for the reaction 2A + B + C → P.

[A], M [B], M [C], M initial rate, M·s–1

0.01 0.01 0.01 0.30.02 0.01 0.01 1.20.01 0.03 0.01 0.90.01 0.01 0.04 0.3

2. Nitrogen (II) oxide, NO, also called nitric oxide, reacts with chlorine according tothe following reaction:

2 NO (g) + Cl2 (g) → 2 NOCl (g)

The following initial rates of reaction have been obtained:

Experiment [NO], M [Cl2], M rate, M/hr1 0.50 0.50 1.142 1.00 0.50 4.563 1.00 1.00 9.12

a) Write the rate law for this reaction. What is the overall order? the order withrespect 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 theinitial concentration of Cl2 is 1.50 mol/L?

70


First-Order Kinetics1. 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 secondswill the pressure of NO2 reach 1.3 torr?

71


Higher-Order Kinetics1. 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 H3O+ byOH– is one such example; this reaction has a second-order rate constant of1.3 × 1011 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% ofthe acid to be neutralized.

b) Under normal laboratory conditions, would you expect the rate of acid-baseneutralization to be limited by the rate of the reaction or the speed of mixing?

2. 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[O2]2 = 8kt + 1

([O2]o)2

An experiment begins with [NO] = 0.020 M and [O2] = 0.010 M. Determine theconcentrations of NO, O2, and NO2 after 100. seconds.

72


Temperature and Rate1. 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. Ifthis process can produce 1.0 kg per hour at 100.°C, what temperature would berequired in order to produce 2.0 kg per hour?

3. You have seen several factors that can affect reaction rates. Consider the first-order decomposition of SO2Cl2 at 600 K:

SO2Cl2 (g) → SO2 (g) + Cl2 (g)

Predict the change in the rate of decomposition of SO2Cl2:

If you: Rate of decomposition of SO2Cl2 will:(circle)

i) add more SO2Cl2 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


Putting It Together: Macroscopic Kinetics1. Given the following data for the reaction: N2O4 (g) → 2 NO2 (g):

Experiment No. Temp., °C [N2O4], M Initial rate, M·s–1

1 25 0.10 5.5 × 103

2 25 0.20 1.1 × 104

3 40 0.10 1.5 × 104

4 40 0.30 4.5 × 104

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 [N2O4] reach 0.002 M?

d) Calculate the half-life of this reaction at 40°C.

e) Calculate the activation energy for this reaction.

74


Introduction to Mechanisms1. 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:

Ak1←→k–1

Bk1 = rate const. of forward rxn

k–1 = rate const. of reverse rxn

B + Ck2→

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


Advanced Mechanisms1. For the gas-phase reaction: CO + Cl2 → COCl2

the following mechanism has been proposed:Cl2 ←→ 2 Cl 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 finalstep 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.

2. Given the following mechanism for the reaction: A + 2B + D → FA + 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 isin steady state, derive the rate law for the formation of the product F.

76


Writing Equilibrium Constants1. Write the equilibrium constant expression (Kc) for these chemical equations:

a) PCl5 (g) ←→ PCl3 (g) + Cl2 (g)

b) NH4Cl (s) ←→ NH3 (g) + HCl (g)

c) H2O (l) ←→ H2O (g)

2. Write the equilibrium constant expression (Kc) for these chemical equations. Thenumerical 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 theequilibrium constant for the reaction in part (c).

3. The equation: N2O4 (g) ←→ 2 NO2 (g)

has an equilibrium constant Kc of 6.1 × 10–3 at 25°C. Calculate the equilibriumconstant for the related reaction:

NO2 (g) ←→ 1/2 N2O4 (g)

77


Kp, Kc, and Q1. 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 toform brown nitrogen dioxide, NO2:

2 NO (g) + O2 (g) →← 2 NO2 (g) Kp = 2.4 × 1012 at 25°C

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:

PO2 = 152 torr PNO = 1.1 × 10–5 torrPNO2 = 3.6 × 10–5 torr

Calculate the value of Q , and determine if NO2 will be produced or consumedunder 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 Br2at 350 K. Calculate the reaction quotient Q, and determine if NOBr will beproduced or consumed under these conditions.

78


Le Châtelier’s Principle1. 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:

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

2. 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 changesto the system, will the quantity of lead increase, decrease, or remain the same:

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

3. What conditions (high vs. low temperature, high vs. low pressure) would you useto increase the yield of products in the following reactions:

a) 2 NOCl (g) ←→ 2 NO (g) + Cl2 (g) endothermic

b) 2 SO2 (g) + O2 (g) ←→ 2 SO3 (g) exothermic

79


Calculating Equilibrium Constants1. In an investigation of the reaction: CO2 + H2 ←→ CO + H2O

a chemist put 0.50 mol CO2 and 0.50 mol H2 into a 2.00 L flask and allowed themto reach equilibrium at 690 K. At equilibrium 0.38 mol CO2 remained. What isthe 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:

F2 (g) ←→ 2 F (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 at840°C.

80


Determining Final Equilibrium Concentrations1. 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. Calculatethe final pressure of BrF once equilibrium is reached at 300. K.

3. At 1800 K, oxygen dissociates slightly into individual atoms:

O2 (g) ←→ 2 O (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 manysingle atoms of oxygen (O) will be present in the flask at equilibrium?

81


Introduction to Acids, Bases, and pH1. 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 → NH3OH+ + N3–

b) CN– + B(OH)3 → HCN + BO(OH)2–

c) Si(OH)4 + C6H5O– → SiO(OH)3– + C6H6O

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 of0.0400 M H2SO4.

a) Write the equation for the chemical reaction which will take place.

b) Calculate the pH of the resultant solution.

82


Weak Acids1. 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


Weak Bases1. 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. Thehydrolysis of the hypochlorite ion yields hypochlorous acid. Given that Ka forHOCl = 3.0 × 10–8, calculate the pH of an 0.100 M solution of calciumhypochlorite, Ca(OCl)2.

84


Polyprotic Acids and Bases1. 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 Na2C2O4.For the conjugate acid H2C2O4, Ka1 = 5.9 × 10–2 Ka2 = 6.4 × 10–5

85


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 ofdistilled water. Calculate the pH of this solution.

Ka (Cu(H2O)42+) = 1.0 × 10–8

86


Putting It Together: Aqueous Acid Equilibria1. 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) K = [H2SO3]PSO2

= 1.33

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 SO2 at a pressureof 1.00 atm into pure water. Calculate the pH of this solution, and theconcentrations of H2SO3, HSO3–, and SO32–.

87


An Introduction to Buffers1. 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 isfilled with 0.15 M sodium acetate. How much of each would you mix together toproduce a total of 20.0 mL of solution with a pH of 4.70? You may not add anywater to the solution.

3. A buffer is prepared by dissolving 0.500 moles of HN3 and 0.750 moles of NaN3in enough water to make 1.00 liter of solution. (Ka 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 affectthe 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


Calculating Concentrations in Buffered Solutions1. Calculate the concentration of NH4+ required for an ammonia / ammonium

chloride buffer in which [NH3] = 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 H2CO3, HCO3–, and CO3

2– at this pH onceequilibrium is reached. Note that the carbonate species must add up, that is:

[H2CO3] + [HCO3–] + [CO3

2–] = 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


Titrations1. 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 to1.00 liter of 0.100-molar LiOH.

For H3PO4, Ka1 = 7.5 × 10–3 Ka2 = 6.2 × 10–8 Ka3 = 4.2 × 10–13

90


Solubility and Ksp1. 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 IO3–?

b) Calculate the solubility product constant (Ksp) of cadmium iodate.

2. The Ksp for silver sulfate (Ag2SO4) is 1.2 × 10–5.

a) Calculate the solubility of Ag2SO4 in pure water.

b) Calculate the solubility of Ag2SO4 in 0.100 M AgNO3 solution.

3. The solubility product constant of calcium fluoride is 3.9 × 10–11. How manymoles of solid NaF must be added to 1.00 L of 0.100 M CaCl2 in order toprecipitate CaF2?

91


Solubility of Metal Hydroxides1. 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


Other Aqueous Equilibria: Dealing With Large K’s1. Copper (II) forms a complex ion with ammonia as follows:

Cu2+ (aq) + 4 NH3 (aq) ←→ Cu(NH3)42+ Kf = 5 × 1012

Calculate the concentrations of Cu2+, NH3, and Cu(NH3)22+, if 0.020 moles ofCu(NO3)2 are dissolved in 1.0 liter of 1.00 M NH3. (Ignore the basicity of NH3.)

2. In a test tube, 1.00 mL of 0.100 M Mn(NO3)2 is mixed with 3.00 mL of0.100 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)

K = [HCO3–]2

[Mn2+][CO32–]2[H2O2] = 6 × 1038

Calculate the concentration of Mn2+ at equilibrium. (Neglect the effect of thehydrolysis of the ions in water.)

93


Solubility and Complex Ion Formation1. 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 precipitateis added 2.00 mL of 0.100 M Na2S2O3. The AgCl precipitate dissolves due to theformation of a complex ion in the following reaction:

AgCl (s) + 2 S2O32– (aq) →← Ag(S2O3)2

3– (aq) + Cl– (aq)

Calculate the concentration of Ag+ at equilibrium. (Note total volume = 5.00 mL)

Ksp (AgCl) = 1.8 × 10–10 Kf (Ag(S2O3)23–) = 2.9 × 1013

2. Chromium (III) hydroxide, Cr(OH)3, is practically insoluble (Ksp = 6.3 × 10–31).However, chromium hydroxide will dissolve in excess base due to the formationof 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


Putting It Together: Solubility and Acid/Base Behavior1. Aluminum phosphate is quite insoluble: Ksp (AlPO4) = 9.8 × 10–21

a) Determine the solubility of AlPO4 in pure water. Assume that neither of the ionsundergoes hydrolysis.

b) Aluminum phosphate is actually more soluble than your answer to part (a) mightindicate, due to hydrolysis of the ions involved. The actual dissolving of AlPO4in 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 (H3PO4) = 7.5 × 10–3

Ka (Al(H2O)63+) = 1.1 × 10–5 Ka (H2PO4–) = 6.2 × 10–8

Ka (HPO42–) = 2.2 × 10–13

Calculate the solubility of AlPO4 in pure water, including the effect of hydrolysisas represented above. Recall that Al3+ (aq) is actually Al(H2O)63+ (aq).

95


An Introduction to Thermodynamics1. Consider the evaporation of methanol, CH3OH. 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: ΔH°f (kJ/mol) S° (J/mol·K)CH3OH (g) –201.2 237.6CH3OH (l) –238.6 126.8

96


Molecular Interpretation of Entropy1. Acetic acid, CH3COOH, can form a dimer in the gas phase which is held together

by hydrogen bonds:

2 CH3COOH (g) ←→ [CH3COOH–HOOCCH3] ΔH° = –66.5 kJ/mol

a) Would you expect the entropy change to be positive or negative? Briefly explainyour reasoning.

b) At 25°C, ΔG° for this reaction is +16.5 kJ/mol. Calculate ΔS° for the reaction, andcompare it with your prediction.

2. When HF (aq) dissociates in water, one might expect an increase in entropy whenthe neutral HF molecule dissociates into ions. In fact, however, ΔS for thisdissociation is negative (–88 J/mol K). Consider carefully the issues of order anddisorder in the solution, and speculate briefly why this must be so.

HF (aq) + H2O (l) ←→ H3O+ (aq) + F– (aq) ΔS° = –88 J/mol K

97


Using Standard Free Energies and Enthalpies1. 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: ΔH°f (kJ/mol) ΔG°f (kJ/mol) S° (J/mol·K)RbCl –430.5 –412.0 92RbClO3 –392.4 –292.0 152

98


ΔG° and Keq

1. Gadolinium (III) is used in medical imaging (MRI) applications. Because Gd3+ isitself toxic, gadolinium (III) is usually administered as a complex of DTPA5–:(DTPA5– is a chelating, or complexing, agent)

Gd3+ + DTPA5– →← Gd(DTPA)2– 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 thatits ΔH° is –32.6 kJ/mol.

2. Consider the following equilibrium, and the Useful Information (below):

N2O4 (g) ←→ 2 NO2 (g)

a) At what temperature will an equilibrium mixture of 1.00 atm total pressurecontain twice as much NO2 as N2O4?

b) At what temperature will an equilibrium mixture of 1.00 atm total pressurecontain equal amounts of the two gases?

Substance: ΔH°f (kJ/mol) ΔG°f (kJ/mol) S° (J/mol·K)NO2 (g) 33.84 51.84 240.45N2O4 (g) 9.66 98.28 304.3

99


ΔG at Nonstandard Conditions1. 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 partialpressure of water vapor is 11 torr.

2. Methanol, CH3OH, can be synthesized by reacting CO and H2 directly:

CO (g) + 2 H2(g) ←→ CH3OH (l)

a) Using the Useful Information (below), calculate ΔG° and K for this reaction at25°C.

b) Calculate the nonstandard ΔG if the pressure of H2 is 3.0 atm and the pressure ofCO is 5.0 atm.

Substance: ΔG°f (kJ/mol)CO (g) –137.2CH3OH (l) –166.23

100


Calculating K at Different Temperatures1. 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 thisreaction at 25°C.

b) Determine the equilibrium constant K at 25°C.

c) Determine ΔG° and K at 200°C.

Substance: ΔH°f (kJ/mol) ΔG°f (kJ/mol) S° (J/mol·K)NH3 (g) –46.19 –16.66 192.5

101


Applications: Determining ΔG°, ΔH°, and ΔS° from Keq

1. Consider the Haber ammonia synthesis: N2 (g) + 3 H2 (g) ←→ 2 NH3 (g)at 300°C: Kp = 4.34 × 10–3

at 500°C: Kp = 1.45 × 10–5

a) Calculate ΔH° for this reaction over this temperature range.

b) Calculate ΔS° for this reaction over this temperature range.

2. Consider the reaction between xylenol orange (H4Q) and aluminum ion (Al3+),which you studied in the laboratory:

Al3+ (aq) + H4Q (aq) ←→ AlQ– (aq) + 4 H+ (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


Applications: Phase Changes1. Using the Useful Information (below), calculate:

a) The boiling point of TiCl4.

b) The vapor pressure of ethanol (C2H5OH) 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: ΔH°f (kJ/mol) ΔG°f (kJ/mol) S° (J/mol·K)TiCl4 (g) –763.2 –726.8 354.9TiCl4 (l) –804.2 –728.1 221.9C2H5OH (g) –235.1 –168.5 282.7C2H5OH (l) –277.7 –174.76 160.7Hg (g) 60.83 31.76 174.89

103


Applications: Acid/Base Equlilbria1. 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

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 and50°C.

2. 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


Applications: Buffers1. 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.

HN3 (aq) + H2O (l) ←→ H3O+ (aq) + N3– (aq) 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 bufferconditions (above), but at pH 7.0.

2. 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 HN3 plus 0.750 M NaN3),calculate the pH at 50°C.

105


Balancing Redox Reactions: Acidic Solution1. Balance each of the following redox reactions in acidic solution.

a) S4O62– (aq) + Al (s) → H2S (aq) + Al3+ (aq)

b) S2O32– (aq) + Cr2O72– (aq) → S4O62– (aq) + Cr3+ (aq)

c) ClO3– (aq) + As2S3 (s) → Cl– (aq) + H2AsO4– (aq) + SO42– (aq)

d) IO3– (aq) + Re (s) → ReO4– (aq) + I– (aq)

e) HSO4– (aq) + As4 (s) + Pb3O4 (s) → PbSO4 (s) + H3AsO4 (aq)

f) HNO2 (aq) → NO3– (aq) + NO (g)

106


Balancing Redox Reactions: Basic Solution1. Balance each of the following redox reactions in basic solution

a) C4H4O62– (aq) + ClO3– (aq) → CO32– (aq) + Cl– (aq)

b) Al (s) + BiONO3 (s) → Bi (s) + NH3 (aq) + AlO2– (aq)

c) H2O2 (aq) + Cl2O7 (aq) → ClO2– (aq) + O2 (g)

d) Tl2O3 (s) + NH2OH (aq) → TlOH (s) + N2 (g)

e) Cu(NH3)42+ (aq) + S2O42– (aq) → SO32– (aq) + Cu (s) + NH3 (aq)

f) Mn(OH)2 (s) + MnO4– (aq) → MnO2 (s)

107


Standard Reduction Potentials at 25°Cfrom Brown, LeMay, and Bursten, Chemistry: The Central Science, 8th ed.

108


Voltaic Cells: Calculating Cell Potential1. For the unbalanced reaction: Fe3+ (aq) + I– (aq) → Fe2+ (aq) + I2 (s)

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)

3. An electrochemical cell is set up by connecting a standard Cu/Cu2+ half-cell witha standard Ag/Ag+ half-cell.

a) A voltmeter is connected to this system in such a way as to read a positivevoltage. What voltage would it display?

b) While keeping the copper electrode connected to the voltmeter, the Ag/Ag+ half-cell is replaced with a Zn/Zn2+ half-cell. What would the voltmeter read now?(Be sure to consider the sign.)

109


Cell Potential and Free Energy1. 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– → H2SO3 (aq) + HF (aq)

a) Balance this half-reaction in acidic solution.

b) Using the data below, determine E° for this reduction half-reaction.

Substance: ΔG°f (kJ/mol)SF6 (g) –1105.4H2SO3 (aq) –537.81HF (aq) –296.82H2O (l) –237.13

110


Reduction Potentials of Half-Reactions1. Given the following standard reduction potentials:

S4O62– + 2 e– → 2 S2O32– E° = 0.090 V

2 SO32– + 3 H2O + 4 e– → S2O32– + 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– E° = ???

2. For the unbalanced reduction half-reaction: HSO4– (aq) + e– → H2S (g)


b) Using the table of Standard Reduction Potentials, determine E° for this half-reaction.

111


Nonstandard Conditions: The Nernst Equation1. 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:

H2 (g) + 12 O2 (g) ←→ H2O (l)

A certain fuel cell is operated at 25°C with an oxygen pressure of 10. torr and ahydrogen pressure of 40. torr. The observed potential under these conditions is1.16 V. Calculate the standard potential for this electrochemical cell.

2. Using the table of Standard Reduction Potentials, show that Fe2+ can bespontaneously oxidized to Fe3+ by O2 (g) at 25°C assuming the following(reasonable) environmental conditions:

[Fe2+] = [Fe3+] = 1 × 10–7 M pH = 7 PO2 = 160 torr

a) Write a complete, balanced equation for the oxidation of Fe2+ by O2.

b) Calculate E for the reaction under these conditions.

112


Concentration Cells1. 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 silverelectrode. The cells are separated by a salt bridge and the electrodes areconnected 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 cellpotential is 0.210 volts. Calculate the new Ag+ concentration in the left half-cell.

113


Putting It Together: Electrochemistry, ΔG°, and Keq1. 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 eachclearly.

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 [Pb2+] = 1.5 M and [Sn2+] = 0.10 M.

114


Electrolysis: Faraday’s Law1. 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:

H2 (g) + 12 O2 (g) ←→ H2O (l)

A certain such fuel cell, given excess oxygen, consumes at most 0.370 grams ofpure hydrogen gas per hour at 25°C. Calculate the maximum current which canbe 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 produce1000. 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 produce1000. kg of aluminum?

115


Putting It Together: Electrochemistry and Solubility1. 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 MKCl solution with some solid AgCl. The voltmeter reads 0.578 V. Determine theKsp 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)


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: ΔG°f (kJ/mol)CrO42– (aq) –727.75Cr2O72– (aq) –1301.1H+ (aq) 0H2O (l) –237.18

116


Nuclear Reactions: Introduction1. Fill in the blanks with a correct word, number, or chemical symbol:

a) Carbon-11 decays by emission to yield boron-11.

b) decays by beta emission to yield nitrogen-15.

c) Strontium-90 emits two beta particles and finally yields .

d) decays by alpha emission to yield 22286Rn.

e) Bombarding 23892U with 14

7N affords along with 5 neutrons.

f) Neutron-induced fission of 23592U can give 3 neutrons, 90

38Sr, and .

g) Two atoms of 126C can react inside a star to produce 20

10Ne and .

h) The decay of 23592U to 207

82Pb yields a total of beta particles.

117


Radioactive Dating1. 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 andcomes down to the Earth in rainwater. Once the wine is bottled, the tritiumcontent 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 tritiumactivity 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 datingtechnique is no longer reliable. What is the maximum age of wine that could bedated using this technique?

118


Energy Changes in Nuclear Reactions1. 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 of57Co.

Masses of Selected Neutral Atoms4He 4.002603 amu7Li 7.016004 amu10B 10.012937 amu57Fe 56.935398 amu57Co 56.936296 amu

Masses of Subatomic ParticlesElectron 0.000549 amuProton 1.007276 amuNeutron 1.008665 amu

2. A cancer treatment called boron neutron-capture therapy works as follows: Adrug containing boron-10 is injected into the patient; this drug selectively binds tocancer cells. Irradiating the affected area with neutrons induces the followingreaction:

10B + 1n → 4He + 7Li + γ

The alpha radiation kills the cancer cells, leaving the surrounding tissueunharmed. In this nuclear reaction, the alpha particle and 7Li together have a totalkinetic energy of 2.31 MeV. The reactants (the boron atom and the neutron) haveessentially no kinetic energy. Calculate the energy of the gamma photon releasedin this process. (Masses of relevant nuclides are given above.)(1 MeV = 1.60 × 10–13 J)

119


Radiation Dose1. Potassium-40 decays by beta emission with a half-life of 1.28 × 109 years; the

emitted beta particles have an average kinetic energy of 0.55 MeV. A personweighing 70.0-kg contains about 0.020 grams of 40K. Determine the total dose ofradiation 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 of70.0 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 personduring the first year after ingestion. (Hint: how many atoms of 57Co decayed inthat time?)

120


Putting It Together: Nuclear Chemistry1. 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 ( 0–1β ). Write the complete

nuclear equation for this decay process. What isotope is produced when carbon-14 decays?

b) Carbon-14 has a mass of 14.003242 amu. The isotope produced when 14C decayshas 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 aboveprocess.

c) Virtually all the energy from the above decay goes into the kinetic energy of theejected electron. What would be the velocity of the electron ejected from adecaying atom of carbon-14?

d) The Shroud of Turin, long believed to be Christ’s burial shroud, was probablymade around 1320 A.D. Carbon-14 dating played an important role in thedetermination of its age. What fraction of the 14C present in 1320 is still presenttoday? The half-life of 14C is 5730 years.

121


Coordination Complexes: Introduction1. 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


Geometry and Isomerism I: Monodentate Ligands1. Each of the following complexes contains a metal atom in an octahedral

geometry. For each complex, draw all the possible stereoisomers. Indicate if anyof 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 canisolate a complex [CoCl2(NH3)3(OH2)]1+ in which the two chloride ligands aretrans 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 isomerscould give rise to the observed product?

123


Geometry and Isomerism II: Chelates1. 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: thereare more templates than you actually need.... cross out any which you don’tneed).

Co Co Co

chiral

achiral

chiral

achiral

chiral

achiral

Co Co Co

chiral

achiral

chiral

achiral

chiral

achiral

124


Electronic Structure of Coordination Compounds1. 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 thenumber 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– arecolored red, yellow, and blue (but not in that order). Match each complex with itscolor, and explain your reasoning.

3. Explain why [FeF6]3– is basically colorless whereas [CoF6]3– is colored.

125


Putting It Together: Coordination Compounds1. 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 Na2[Fe(CN)5NO]. Note: The NOligand in this complex is neutral and is named “nitrosyl.”

b) For the complex [Fe(CN)5NO]2–, give the electronic configuration of the d-orbitals (e.g. t2g3eg1), indicate whether the complex is high-spin or low-spin, andgive the number of unpaired electrons.

c) The [Fe(CN)5NO]2– complex is red-violet in color (“ruby red”), while the similarcomplex [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) muchweaker

slightlyweaker

slightlystronger

muchstronger

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


Organic Chemistry: Isomers and Functional Groups1. 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 ofwhich isomers are stable or not!)

2. Ethanol (an alcohol with the molecular formula C2H6O) is metabolized in theliver through a two-step oxidation process. The initial product has the formulaC2H4O, and is responsible for the red flushed appearance which oftenaccompanies alcohol consumption. This is then oxidized again to the finalproduct, which has the molecular formula C2H4O2.

a) Draw the structure of ethanol: b) Draw the structure of each product, and name the functional groups.

Structure Functional Group

Initial product (C2H4O): Final product (C2H4O2):

127


Organic Chemistry: Polymers1. For each of the following polymers, identify the monomer unit which could be

used to synthesize the polymer. Also circle the type of polymerization whichwould be required.

N C C N C C N C C

H

CH3

O

H

H

CH3

O

H

H

CH3

O

Hn

C C C C C C

F

F

F

F

F

F

F

F

F

F

F

Fn

Monomer:

Type of Polymerization: (circle)

addition condensation

Poly(tetrafluoroethylene), Teflon®

Monomer:



Polyalanine (a polypeptide)

2. The molecules below can combine to form a polymer (a type of polyester).

C C

HO

O O

OHCH2

CH2

HO OH

A 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 Problems1. Significant Figures and Scientific Notation

1. a) 523 b) 2.3 c) 19.2d) 42.2 e) 860 or 900 f) 18.2g) 1.18 × 10–4 h) 3.6 i) 2.5 × 10–4

2. Dimensional Analysis I: Unit Conversions1. a) 67.77 ft b) 1.5 m c) 27 ft2. a) 2.4 atm b) 2.1 × 103 torr c) 1.00 × 105 Pa

3. Dimensional Analysis II: Volume and Density1. a) 3.6 × 102 bushels b) 54 hogsheads c) 2.7 × 103 gallons2. a) 4.670 × 10–2 in3 b) 1.92 g

4. Atoms, Molecules and Ions1. 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 electrons2. 20 protons, 18 electrons3. 35Cl–: 17 protons, 18 neutrons, 18 electrons 37Cl–: 17 protons, 20 neutrons, 18 electrons4. a) 10 protons, 7 neutrons, 10 electrons b) 11 protons, 7 neutrons, 10 electrons

5. Oxidation Numbers1. Br: +5 H: +1 As: –3

O: –2 As: +3 H: +1O: –2

Cr: +3 K: +1 S: +2Cl: –1 Cl: +5 O: –2

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

2. N2 is reduced, H2 is oxidizedFeCl3 is reduced, KI is oxidizedNo oxidation or reduction is taking place.Cl2 is being oxidized (to ClO3–) and reduced (to Cl–)

6. Naming Compounds1. a) Cu3P b) Fe2(SO4)3 c) KClO3 d) AlCl3

e) CrO3 f) NH4I g) Li3N h) Al2(CO3)3i) Cs3PO4 j) Re2O7 k) SnCl4 l) GaF3m) AgF2 n) KNO3 o) Ba3(PO4)2 p) (NH4)2SO4

2. a) Silicon tetrachlorideb) Manganese (VII) oxidec) Magnesium nitride

129

7. Chemical Formulas: Percent Composition1. 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 notagree with the results obtained for the unknown compound; thus it can't be Sarin.

2. Pure CaCO3 would have 40% calcium, 12% carbon, and 48% oxygen. This does not agreewith the analysis of the rock sample.

8. Empirical and Molecular Formulas1. a) C9H12O b) C18H24O2 c) 272.39 g/mol2. a) C5H4O b) C15H12O3 ; MW = 240.25 g/mol3. a) NH2Cl b) NH2Cl

9. Writing and Balancing Equations1. a) CO(NH2)2 + 6 HOCl → 2 NCl3 + CO2 + 5 H2O

b) 2 Ca3(PO4)2 + 6 SiO2 + 10 C → P4 + 6 CaSiO3 + 10 CO2. 6 CO2 + 6 H2O → C6H12O6 + 6 O23. 3 Nb + 4 I2 → Nb3I84. C8H18 + 25/2 O2 → 8 CO2 + 9 H2O

10. Stoichiometry of Reactions1. 55.9 g NaClO2. a) NiS + 2 O2 + 2 HCl → NiCl2 + H2SO4

b) 0.654 g NiCl211. Stoichiometry with Limiting Reagents

1. a) V2O5 + 3 Zn → V2O2 + 3 ZnOb) 68.25 g V2O2

2. a) 5 P4S3 + 12 Br2 → 3 P4S5 + 8 PBr3b) 77.47 g P4S5

12. Stoichiometry of Mixtures1. 61.1% CO2 by mass2. 7.46 g CuO, 3.04 g Cu2O3. 6.55 g Cu2S, 4.25 g CuS

13. Solutions: Molarity1. 18.0 M2. 17.47 M3. 0.216 M4. 0.052 g5. x = 4.4 g KNO3 y = 4.49 g NaCl

14. Solution Stoichiometry I: Simple Examples1. a) 3 ClO– (aq) + NH3 (aq) → NCl3 (l) + 3 OH– (aq)

b) 0.028 g NCl32. 2.79 g CO2

130

15. Solution Stoichiometry II: Acid/Base Neutralizations1. 71.2 mL2. 1.72 L3. 1.26 L

16. Solution Stoichiometry III: Titrations1. 1.22 g2. 312 g/mol3. 119.2 g/mol

17. Solution Stoichiometry IV: Precipitation Reactions1. 0.0119 M Ba(NO3)22. Ba2+: 0.011 M Cl–: 0.033 M NO3–: 0.011 M

NH3: 0.033 M NH4+: 0.022 M18. The Ideal Gas Law

1. a) 6.0 × 105 Lb) 0.348 g Cl2

2. 53.4 g O219. Reactions Involving Gases: Simple Examples

1. 2.66 atm2. 0.480 atm3. total pressure = 1.5 atm

20. Mixtures of Gases1. a) 0.0863 mol O2

b) 0.0211 torr O2c) 2.78 × 10–6 torr O2

2. 16.3 g butane21. Collecting Gases Over Water

1. 0.109 g H22. 0.312 L collected3. 3.51 L should be collected

22. Stoichiometry of Gas Mixtures1. χPCl5 = 0.054 χPCl3 = 0.473 χCl2 = 0.4732. 12.8% O3 by mass

23. Reactions Involving Gases: Applications1. 21.3 g NaN32. a) 29.0 g Cl2

b) 1.07 atm O2 1.63 atm Cl2c) 29.7 g NaAlCl4 produced

131

24. Kinetic-Molecular Theory of Gases1. 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/molecule2. a) 431 m/s

b) 7.3 hours25. Heat

1. a) 8.2 kJ of heat consumedb) Final temp. = 14.2°C

2. Specific heat = 0.269 J/g°C26. Calorimetry I: Simple Examples

1. ΔH° = –27 kJ/mol2. ΔH° = –5152 kJ/mol3. ΔH° = –4219 kJ/mol

27. Calorimetry II: Mixtures1. Final temp. = 27.08 °C2. 311 g of ice

28. Hess’s Law1. ΔH° = –171.34 kJ/mol2. ΔH° = –185.1 kJ/mol

29. Standard Enthalpies of Formation1. ΔH° = –64.9 kJ/mol2. 268 kJ of heat released

30. Enthalpy, Energy, Heat, and Work1. a) +37.4 kJ/mol

b) –2800 J/molc) +34.6 kJ/mol

31. Putting It Together: Calorimetry, Hess’s Law, and ΔH°f1. a) ΔH° = 25.3 kJ/mol

b) ΔH°f (NH4NO3(s)) = –362.8 kJ/mol2. a) ΔH° = –50.8 kJ/mol for the reaction as written (i.e. consuming 2 mol NaN3)

b) ΔH°f (NaN3) = 25.4 kJ/mol32. Putting It Together: Stoichiometry, Thermochemistry, and Gas Laws

1. a) 3.74 g CO2b) 7938°Cc) 14.3 atm CO2

132

33. Energy, Particles, and Waves: A “Chain Problem”1. a) 7.09 × 1017 s–1

b) 4.23 × 10–10 mc) 1.72 × 106 m/sd) 1.35 × 10–18 J

34. The Photoelectric Effect1. a) 1.09 × 1015 s–1

b) 1.80 × 10–9 m35. Orbitals and Wavefunctions

1. a) The first graph is R2, the second one is R, and the third is r2R2.b) B and C.c) A.

2. C.36. Orbitals and Quantum Numbers

1. Subshell n l total #of e–

1s 1 0 2 3d 3 2 10 5p 5 1 6

2. 4.09 × 10–19 J emitted37. 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 Ions1. Be+: 1s22s1

N– : 1s22s22p4

Al3+:1s22s22p6

H– : 1s2

O2–: 1s22s22p6

2. Zn2+: [Ar] 3d10

W6+: [Xe] 4f14

Cu2+: [Ar] 3d9

Gd3+: [Xe] 4f7

Se2–: [Ar] 4s23d104p6

39. Periodic Properties I: Overview1. a) High ionization energy: upper right

Low ionization energy: lower leftb) High atomic mass: lower right

Low atomic mass: upper leftc) Big atomic radius: lower left

Small atomic radius: upper rightd) Most metallic: lower left

"Least metallic" (most non-metallic): upper right2. 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 thenucleus.b) Indium is the largest of the group; its valence electrons are in the 5p orbital, whereas the others are inlower-energy orbitals.

2. a) Thallium has the lowest ionization energy of the group; its valence electrons are in a higher energyorbital 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 energyorbital 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 fluorideion with a stable, complete octet.

4. Se2– is the largest ion of the group. Of course it will be larger than the 2nd-period ionsO2– and F–. Plus, since it is isoelectronic with Rb+, it has the same number of electrons but three fewerprotons, meaning that the effective nuclear charge is less and the radius will be larger.

134

41. Periodic Properties III: Explanations1. a) Se: . . . 4s24p4 has 2 electrons in one of its p-orbitals

As: . . . 4s24p3 each p-orbital has one electronBecause it is somewhat unfavorable for two electrons to occupy the same orbital, Se is relatively willingto give up its p-electron to achieve an electron configuration in which each p-electron is in its ownorbital. 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 quitefavorable).c) When Rb and Na react with water, they each lose an electron to form their respective cations. SinceRb 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 atremendous amount of energy (the electron affinity). This large energy release make fluorine veryreactive.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 forma stable, complete octet. This is the usual mode of reactivity for iodine.

42. Putting It Together: Quantum Mechanics and Electronic Structure1. a) wavelength = 1.70 × 10–7 m

b) Bi: [Xe] 6s24f145d106p3

Bi+:[Xe] 6s24f145d106p2

c) n=6, l=1d) 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 6pvalence electrons of Bi). Thus those 7p valence electrons would be easier to ionize.

43. Lewis Structures I: The Octet Rule

H C NO

NOO

NOC OC

FF

FF

OF F

NS

FOO HH

44. Lewis Structures II: Ions

ON

O

ON

O

+

SF

F F BH

H HH

CO

O OC C

HP

H

–2–

2–

(plus two otherresonance

structures)

–

–

–

135

45. Lewis Structures III: Less Than an Octet

H Be HBBr

Br BrAl

F

F F

OCl

O

OCl

O

ON

O

ON

ON O

46. Lewis Structures IV: More Than an Octet

AsClCl

ClCl

ClF Se FF F F Xe F As

F

F FF

F F–

F Xe FF

+F Br F –

47. Bond Enthalpies1. a) H–C≡C–H

b) DC≡C = 813 kJ/mol2. a) :N≡C–C≡N:

b) –1017 kJ/mol48. Molecular Geometry I: Neutral Molecules

ON

FCl I Cl

ClO C OBF

F F

e- pair geo:mol. geo:hybridization:polar?

linearlinearsp

no

trig. planarbentsp2

yes

trig. planartrig. planarsp2

no

trig. bipyramidalT-shapedsp3dyes

49. Molecular Geometry II: Ions

2–

SeO

O OIFF

FF I I I

Cl PCl

ClCl

–+

–

e- pair geo:mol. geo:hybridization:

octahedralsq. planarsp3d2

tetrahedraltetrahedralsp3

tetrahedraltrig. pyramidalsp3

trig. bipyramidallinearsp3d

136

50. Polycentric Molecules and Ions1.

SF

FF

SF

trig. bipy.see-sawsp3d

tetrahedralbentsp3

H3C C C CO

O

tetrahedraltetrahedralsp3

linearlinearsp

linearlinearsp

trig. planartrig. planarsp2

II

II

I–

trig. bipy.linearsp3d

tetrahedralbentsp3

trig. bipy.linearsp3d

51. Putting It Together: Lewis Structures, VSEPR, and Bonding1. a)

SS S S

SSSS

b) Hybridization: sp3

Electron pair geometry: tetrahedral Molecular geometry: bentc) DS=S = 434 kJ/mol

137

52. Hybridization and Multiple Bonding: σ and π Bonds1. a) Trigonal planar; sp2 hybridization

b) 2 σ-bonds between H s orbitals and C sp2 hybrid orbitals 1 σ-bond between O orbital and C sp2 hybrid orbital 1 π-bond between O p-orbital and C p-orbitalc) 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 sp3 hybrid and C sp hybrid• 1 σ-bond between C sp hybrid and C sp hybrid

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

• 1 σ-bond between C sp hybrid and C sp2 hybrid• 2 σ-bonds between C sp2 hybrids and O orbitals

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

53. Putting It Together: Bonding and Hybridization

1. a) O C O

b) Linear geometry; sp-hybridizationc) sp hybrid orbitalsd) p orbitals

e)

C CH

HC

H

H

f) Central carbon atom: sp hybridization Terminal carbon atom: sp2 hybridizationg)

C C

H

H

H

HC

Each p-orbital mustbe perpendicular tothe sp2–hybrid plane,and the two π-bondsmust thus be perpend.to each other.

54. Covalent Bonding and Orbital Overlap1. a) 2p and 3s making σ bonding orbital

b) 2p and 2p making σ bonding orbitalc) 3d and 2p making π* antibonding orbitald) 3p and 1s making σ* antibonding orbitale) 3d and 3d making π bonding orbital

138

55. Molecular Orbital Theory I: Introduction1. 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 Bond order = 1.5 Paramagneticc) (σ2s)2(σ*2s)2(π2p)4 Bond order = 2 Diamagnetic

2. a)

2p

2s

2p

2s

σ2s

σ2s*

σ2p*

σ2p

π2p*

π2p

N NO Omolecularorbitals

atomicorbitals

atomicorbitals

b) Double bond: NO–

Diamagnetic: NO+

Longest bond: NO–

Isoelectronic w/CO: NO+

139

56. Molecular Orbital Theory II: Extensions of MO Theory1. a) and b)

2p

2s

2p

2s

σ2s

σ2s*

σ2p*

σ2p*

σ2p

σ2p

π2p*

nonbonding

π2p

C CO2 O Oatomicorbitals

molecularorbitals

atomicorbitals

atomicorbitals

Note: "nonbonding"means that the orbital isnot involved in bonding inany way (neither bondingnor antibonding).

c) 4d) diamagnetice) longer

57. Band Theory and Semiconductors1. 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 weakbonding 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 toflow.

140

58. Intermolecular forces1. a) Al2O3 Ionic

F2 London DispersionH2O Hydrogen BondingBr2 London DispersionICl Dipole-DipoleNaCl Ionic

b) In order from highest BP to lowest BP: Al2O3, NaCl, H2O, ICl, Br2, F259. 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 Equation1. +42.2 kJ/mol2. 25.1 torr

61. Phase Diagrams1. a) and b)

18439

760

100

P

T (°C)

(torr)

110

1

solid

liquid

gas

triple pt.

melting pt. boiling pt.•

•

•

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

62. Bonding in Crystalline Solids1. 7.60 g/cm3

2. radius = 1.97 Å63. X-Ray Diffraction: The Bragg Equation

1. a) They must be n1 = 4 and n2 = 6. (These are the only combinations which give a reasonablevalue for the radius of the Po atom!)

b) 3.34 Åc) 9.28 g/cm3

141

64. Lattice Energy: The Born-Haber Cycle1. Lattice enthalpy = +5400 kJ/mol2. Lattice enthalpy = +728 kJ/mol

65. Expressing Solution Concentrations1. a) 41.9% maltose

b) 2.11 mc) χmaltose = 0.037d) χH2O = 0.963e) 38.3 M water

66. Colligative Properties I: Non-Dissociating Solutes1. a) 100.71°C

b) 23.2 torr2. 19200 g/mol3. χCCl4 = 0.80 χBr2 = 0.20

67. Colligative Properties II: Dissociating Solutes1. a) 102.23°C

b) 22.06 torr2. 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 Solutes1. –0.27°C2. 22.9 torr Note: NaN3 dissociates into two ions: Na+ and N3– (the azide ion).

69. Colligative Properties IV: Non-Ideal Solutions1. a) i = 0.88

b) Expected i = 2c) In concentrated solutions, charged ions will tend to associate into ion pairs or other aggregates. Inthis case, since i is less than one, there must be net aggregation of Mg2+ and SO42– ions into groups of4 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 Kinetics1. P(SO2Cl2) = 3.25 atm

P(SO2) = 5.75 atmP (Cl2) = 5.75 atm

2. t = 7.31 × 10–7 sec72. Higher-Order Kinetics

1. a) 7.7 × 10–7 secb) This reaction is so fast that the speed of mixing would actually be the limiting factor.

2. [NO] = 0.0115 M[O2] = 0.00577 M[NO2] = 0.00845 M

73. Temperature and Rate1. Ea = 62.7 kJ/mol2. 386 K3. i) speed up

ii) speed upiii) slow downiv) speed upv) remain same

74. Putting It Together: Macroscopic Kinetics1. a) rate = k [N2O4]

b) k = 5.5 × 104 sec–1

c) t = 7.11 × 10–5 secd) half-life = 4.62 × 10–6 sece) Ea = 51.9 kJ/mol

75. Introduction to Mechanisms1. rate = k [Cl2][H]

rate = k [C3H6N2]2. a) rate = k1 [A]

b) rate = k1k2k–1

[A][C]

c) rate = k1k2[A][C]k–1 + k2[C]

76. Advanced Mechanisms

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

k–11/2(k–2 + k3[Cl2])

b) The term in the denominator with k3[Cl2] would vanish, leaving k = k11/2 k2k3k–11/2 k–2

2. rate = k1k2[A][B][D]k–1

143

77. Writing Equilibrium Constants

1. a) Kc = [PCl3][Cl2][PCl5]

b) Kc = [NH3][HCl]c) Kc = [H2O(g)]

2. a) Kc = [CO][H2O][H2][CO2]

b) Kc = [H2O]2

[H2]2

c) Kc = [CO2]2

[CO]2

d) 13.73. Kc = 13

78. Kp, Kc, and Q

1. a) Kc = 5.9 × 1013

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

2. a) Kc = 5.2 × 10–3

b) Qc = 0.25, so reaction goes to left, and NOBr will be produced under these conditions.79. LeChâtelier’s Principle

1. i) increaseii) increaseiii) decreaseiv) remain samev) remain same

2. i) remain sameii) increaseiii) decreaseiv) increasev) remain samevi) decrease

3. a) high temperature, low pressureb) low temperature, high pressure

80. Calculating Equilibrium Constants1. Kc = 0.102. Kp = 0.01963. Kp = 0.077

144

81. Determining Final Equilibrium Concentrations1. SO2Cl2: 0.0052 moles

SO2: 0.178 molesCl2: 0.178 molesTotal pressure: 17.8 atm

2. Pressure of BrF: 3.5 atm3. 1.7 × 1018 individual oxygen atoms in the flask

82. Introduction to Acids, Bases, and pH1. a) HN3 is the acid, NH2OH is the base

b) B(OH)3 is the acid, CN– is the basec) Si(OH)4 is the acid, C6H5O– is the base

2. a) 12.43b) 1.70c) 11.30d) 7.00

3. a) H2SO4 + 2 NaOH → Na2SO4 + 2 H2Ob) 12

83. Weak Acids1. 2.102. 1.39

84. Weak Bases1. 11.842. 8.533. 10.41

85. Polyprotic Acids and Bases1. [H3PO4] = 0.0761 M

[H2PO4–] = 0.0239 M[HPO42–] = 6.2 × 10–8 M[PO43–] = 1.1 × 10–18 M[H+] = 0.0239 M[OH–] = 4.2 × 10–13 M

2. 1.233. 8.60

86. Acidic Behavior of Metal Cations (Hydrolysis)1. 2.932. 5.11

145

87. Putting It Together: Aqueous Acid Equilibria1. 1.592. pH = 0.84

[H2SO3] = 1.33 M[HSO3–] = 0.143 M[SO32–] = 1.02 × 10–7 M

88. An Introduction to Buffers1. 3.522. 12.5 mL of 0.10 M acetic acid

7.5 mL of 0.15 M sodium acetate3. a) pH = 4.90

b) (in order from top to bottom) increase, decrease, decrease, remain same, increase89. Calculating Concentrations in Buffered Solutions

1. [NH4+] = 0.360 M2. a) pH = 8.78

b) [H2CO3] = 7.44 × 10–5 M[HCO3–] = 0.0193 M[CO32–] = 6.53 × 10–4 M

90. Titrations1. 172 mL2. 7.51

91. Solubility and Ksp1. a) 3.58 × 10–3 M

b) Ksp = 2.29 × 10–8

2. a) 0.0144 Mb) 0.0012 M

3. must add at least 2.0 × 10–5 moles of NaF92. Solubility of Metal Hydroxides

1. a) 10.46b) 1.2 × 10–3 M

93. Other Aqueous Equilibria: Dealing With Large K’s1. [Cu2+] = 5.6 × 10–15

[NH3] = 0.92[Cu(NH3)4+] = 0.020

2. [Mn2+] = 4 × 10–38

94. Solubility and Complex Ion Formation1. [Ag+] = 4.5 × 10–9

2. 0.0033 M

146

95. Putting It Together: Solubility and Acid/Base Behavior1. 9.9 × 10–11 M2. 7.0 × 10–7 M

96. An Introduction to Thermodynamics1. a) +37.4 kJ/mol

b) +110.8 J/mol·Kc) 64.4°C

97. Molecular Interpretation of Entropy1. a) ΔS should be negative; the system is becoming more ordered.

b) ΔS° = –279 J/mol·K2. 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 thetotal order is increased.

98. Using Standard Free Energies and Enthalpies1. a) +240 kJ/mol

b) +76.2 kJ/molc) –550 J/mol·Kd) +259 kJ/mole) +223 J/mol·K

99. ΔG° and Keq1. a) –128 kJ/mol

b) 1.74 × 1022

2. a) 60°Cb) 45°C

100. ΔG at Nonstandard Conditions1. a) +8.32 kJ/mol

b) –2.24 kJ/mol2. a) ΔG° = –29.0 kJ/mol K = 1.2 × 105

b) –38.4 kJ/mol101. Calculating K at Different Temperatures

1. a) ΔH° = –92.38 kJ/mol ΔS° = –198.2 J/mol·K ΔG° = –33.32 kJ/molb) K = 6.9 × 105

c) K = 0.71

147

102. Applications: Determining ΔG°, ΔH°, and ΔS° from Keq1. a) –104.9 kJ/mol

b) –228 J/mol·K2. a) +25.9 kJ/mol

b) +60.7 kJ/molc) +106.1 J/mol·K

103. Applications: Phase Changes1. a) 35.3°C

b) 60.7 torr2. a) +31.76 kJ/mol

b) +9.63 kJ/molc) +77.34 J/mol·K

104. Applications: Acid/Base Equilibria1. a) [H+] = 0.00126 M [Bu–] = 0.00126 M [HBu] = 0.0987 M

b) –4.4 kJ/mol2. –67.7 kJ/mol

105. Applications: Buffers1. a) +26.9 kJ/mol

b) –12.0 kJ/mol2. pH = 4.70

106. Balancing Redox Reactions: Acidic Solution1. a) 20 H+ (aq) + S4O62– (aq) + 6 Al (s) → 4 H2S (aq) + 6 Al3+ (aq) + 6 H2O (l)

b) 14 H+ (aq) + 6 S2O32– (aq) + Cr2O72– (aq) → 3 S4O62– (aq) + 2 Cr3+ (aq) + 7 H2O (l)c) 18 H2O + 14 ClO3– (aq) + 3 As2S3 (s) → 14 Cl– (aq) + 6 H2AsO4– (aq) + 9 SO42– (aq)

+ 24 H+ (aq)d) 3 H2O + 7 IO3– (aq) + 6 Re (s) → 6 ReO4– (aq) + 7 I– (aq) + 6 H+ (aq)e) 30 H+ (aq) + 30 HSO4– (aq) + As4 (s) + 10 Pb3O4 (s) → 30 PbSO4 (s) + 4 H3AsO4 (aq)

+ 24 H2Of) 3 HNO2 (aq) → NO3– (aq) + 2 NO (g) + H2O + H+ (aq)

107. Balancing Redox Reactions: Basic Solution1. a) 3 C4H4O62– (aq) + 5 ClO3– (aq) + 18 OH– (aq) → 12 CO32– (aq) + 5 Cl– (aq) + 15 H2O (l)

b) 11 Al (s) + 3 BiONO3 (s) + 11 OH– (aq) → 3 Bi (s) + 3 NH3 (aq) + 11 AlO2– (aq) + H2O (l)c) 4 H2O2 (aq) + Cl2O7 (aq) + 2 OH– (aq) → 2 ClO2– (aq) + 4 O2 (g) + 5 H2O (l)d) Tl2O3 (s) + 4 NH2OH (aq) → 2 TlOH (s) + 2 N2 (g) + 5 H2O (l)e) Cu(NH3)42+ (aq) + S2O42– (aq) + 4 OH– (aq) → 2 SO32– (aq) + Cu (s) + 4 NH3 (aq)

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

108. Standard Reduction Potentials at 25°C

148

109. Voltaic Cells: Calculating Cell Potential1. a) Overall: 2 Fe3+ (aq) + 2 I– (aq) → 2 Fe2+ (aq) + I2 (s)

Reduction: Fe3+ → Fe2+

Oxidation: 2 I– → I2b) Anode: I– / I2 Cathode: Fe3+ / Fe2+

c) Oxidizing agent: Fe3+ Reducing agent: I–

d) E° = 0.235 V2. 0.903 V3. a) 0.462 V

b) –1.100 V110. Cell Potential and Free Energy

1. –91.3 kJ/mol2. a) 2 H+ (aq) + 3 H2O (l) + SF6 (g) + 2 e– → H2SO3 (aq) + 6 HF (aq)

b) 2.60 V111. Reduction Potentials of Half-Reactions

1. 0.803 V2. a) 9 H+ (aq) + HSO4– (aq) + 8 e– → H2S (g) + 4 H2O (l)

b) 0.303 V112. Nonstandard Conditions: The Nernst Equation

1. 1.23 V2. a) 4 H+ (aq) + 4 Fe2+ (aq) + O2 (g) → 4 Fe3+ (aq) + 2 H2O (l)

b) 0.036 V113. 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 Vc) New [Ag+] = 0.00028 M

114. Putting It Together: Electrochemistry, ΔG°, and Keq1. a) Anode: Sn → Sn2+ + 2 e– Cathode: Pb2+ + 2 e– → Pb

b) 0.010 Vc) –1.9 kJ/mold) K = 2.2e) 0.045 V

115. Electrolysis: Faraday's Law1. 0.296 g2. 9.85 amps3. a) 30 hours

b) 334 kg graphite

149

116. Putting It Together: Electrochemistry and Solubility1. 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: Introduction1. a) positron

b) 15Cc) 90Zrd) 226Rae) 247Esf) 143Xeg) 4He (α particle)h) 4 beta particles

118. Radioactive Dating1. a) 29 years

b) 83 years119. Energy Changes in Nuclear Reactions

1. 1.34 × 10–13 J2. 7.8 × 10–14 J

120. Radiation Dose1. 0.0205 rad2. a) 3.14 × 10–10 g

b) 0.143 rad121. Putting It Together: Nuclear Chemistry

1. a) 14C → 14N + β–

b) 2.51 × 10–14 Jc) 2.35 × 108 m/sd) 92 %

122. Coordination Complexes: Introduction1. a) Pt(II), d8, tetraammineplatinum(II) perchlorate

b) Re(VII), d0, nonahydridorhenate(VII)c) Fe(-2), d10, 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), d8, diamminedichloroplatinum(II)

150

123. Geometry and Isomerism I: Monodentate Ligands1. a)

H2O

H2O NH3

OH2

Co

NH3

NH3

H2O

H2O NH3

NH3

Co

NH3

OH2

both achiral

b)

Br

Cl Br

ClPt

NH3

NH3

Cl

Cl Br

BrPt

NH3

NH3

Br

Br NH3

NH3

Pt

Cl

Cl

Cl

Cl NH3

NH3

Pt

Br

Br

Br

Br Cl

NH3

Pt

NH3

Cl

H3N

Cl Br

BrPt

NH3

Cl

all transachiral

NH3 transachiral

Cl transachiral

Br transachiral

all cischiral

all cischiral

2.

O2N

O2N NH3

NO2

Co

NH3

NH3

O2N

O2N NH3

NH3

Co

NH3

NO2

a)

b) Since the Cl groups are replacing the NO2– ligands, and the Cl's end up trans in the product, theremust be a trans pair of NO2– ligands in the starting material. Thus the isomer on the left must havebeen the starting material.

124. Geometry and Isomerism II: Chelates1.

N

N OH2

OH2

Co

Cl

Cl

Cl transachiral

N

N Cl

ClCo

OH2

OH2

H2O transachiral

N

N Cl

OH2

Co

OH2

Cl

all cischiral

H2O

Cl N

NCo

OH2

Cl

all cischiral

Note: ethylenediammine (en) is represented as: N N

151

125. Electronic Structure of Coordination Compounds1. a) d6 high-spin; 4 "downstairs" and 2 "upstairs"; 4 unpaired electrons.

b) d8 (high/low doesn't matter); 6 "downstairs" and 2 "upstairs"; 2 unpaired electronsc) d6 low-spin; 6 "downstairs"; no unpaired electrons.d) d6 low-spin; 6 "downstairs"; no unpaired electrons.

2. [Co(NH3)6]3+ [Co(NH3)5(OH2)]3+ [CoF6]3–

strongest ligands intermediate weakest ligandslargest Δo intermediate Δo smallest Δoabsorbs violet absorbs green absorbs orangeappears yellow appears red appears blue

3. FeF63– is d5 high-spin electron configuration with 5 unpaired electrons, all spin "up". In order topromote an electron from the lower to the higher energy level, its spin would have to "flip," which isforbidden. Thus the d5 high-spin configuration is usually colorless. (This isn't a problem for d6Co(III)).

126. Putting It Together: Coordination Compounds1. 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 slightlylower 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 isslightly weaker than the CN ligand. (If it were much weaker we would have seen a much moredramatic change of color.)d) Since the complex has 5 electrons in the more stable “downstairs” (t2g) orbitals, and no electrons inthe “upstairs” (eg) orbitals, it will be very stable, kinetically inert, and thus the ligands will dissociateslowly.

127. Organic Chemistry: Isomers and Functional Groups1.

H3C O CH

CH2

HO CH2

CH

CH2

HO CH

CH

CH3

O CH

CH2

CH3

H2C

CH2

CH

H2C

OCH

H2C

CH2

O

H2C

OH

CH3

H3C C CH3

O

H2C C CH3

OH

(cis+trans isomers)

152

2. a) H3C CH2

OH

b) Initial product: H3C CH

O , which is an aldehyde.

c) Final product: H3C C

O

OH

, which is a carboxylic acid.

128. Organic Chemistry: Polymers1.

C C

F

F F

F

N C C N C C N C C

H

CH3

O

H

H

CH3

O

H

H

CH3

O

Hn

C C C C C C

F

F

F

F

F

F

F

F

F

F

F

Fn

Monomer:



Poly(tetrafluoroethylene), Teflon®

Monomer:



Polyalanine (a polypeptide)

N C C

H

H

CH3

O

H OH

2. a) carboxylic acidb) alcoholc)

C C

O

O O

CH2

CH2

O

C C

O

O O

CH2

CH2

O O

153

PERI

ODI

C TA

BLE

OF

THE

ELEM

ENTS

12

HHe

1.00

84.

003

34

56

78

910

LiBe

BC

NO

FNe

6.94

19.

012

10.8

112

.01

14.0

116

.00

19.0

020

.18

1112

1314

1516

1718

NaM

gAl

SiP

SCl

Ar22

.99

24.3

126

.98

28.0

930

.97

32.0

735

.45

39.9

519

2021

2223

2425

2627

2829

3031

3233

3435

36K

CaSc

TiV

CrM

nFe

CoNi

CuZn

Ga

Ge

AsSe

BrKr

39.1

040

.08

44.9

647

.88

50.9

452

.00

54.9

455

.85

58.9

358

.69

63.5

565

.39

69.7

272

.61

74.9

278

.96

79.9

083

.80

3738

3940

4142

4344

4546

4748

4950

5152

5354

RbSr

YZr

NbM

oTc

RuRh

PdAg

CdIn

SnSb

TeI

Xe85

.47

87.6

288

.91

91.2

292

.91

95.9

4(9

8)10

1.07

102.

9110

6.42

107.

8711

2.41

114.

8211

8.71

121.

7612

7.60

126.

9113

1.29

5556

5772

7374

7576

7778

7980

8182

8384

8586

CsBa

LaHf

TaW

ReO

sIr

PtAu

HgTl

PbBi

PoAt

Rn13

2.91

137.

3313

8.91

178.

4918

0.95

183.

8518

6.21

190.

2019

2.22

195.

0819

6.97

200.

5920

4.38

207.

2020

8.98

(209

)(2

10)

(222

)87

8889

104

105

[106

][1

07]

[108

][1

09]

FrRa

AcRf

Ha(2

23)

226.

0322

7.03

(261

)(2

62)

(263

)(2

62)

(265

)(2

66)

5859

6061

6263

6465

6667

6869

7071

Lant

hani

de s

erie

s

Ce

PrNd

PmSm

EuG

dTb

DyHo

ErTm

YbLu

140.

1214

0.91

144.

24(1

45)

150.

3615

1.96

157.

2515

8.93

162.

5016

4.93

167.

2616

8.93

173.

0417

4.97

9091

9293

9495

9697

9899

100

101

102

103

Actin

ide

serie

s

Th

PaU

NpPu

AmCm

BkCf

EsFm

Md

NoLr

232.

0423

1.04

238.

0323

7.05

(244

)(2

43)

(247

)(2

47)

(251

)(2

52)

(257

)(2

58)

(259

)(2

60)

154

the logan notes

Documents