Chemistry, 6E John E McMurry Solution Manuals

Chemistry, 6E

By John E. McMurry s

Email: Richard@qwconsultancy.com

Table of Contents Chapter 1

Chemistry: Matter and Measurement

1

Chapter 2

Atoms, Molecules, and Ions

10

Chapter 3

Mass Relationships in Chemical Reactions

22

Chapter 4

Reactions in Aqueous Solution

32

Chapter 5

Periodicity and the Electronic Structure of Atoms

41

Chapter 6

Ionic Bonds and Some Main-Group Chemistry

50

Chapter 7

Covalent Bonds and Molecular Structure

59

Chapter 8

Thermochemistry: Chemical Energy

69

Chapter 9

Gases: Their Properties and Behavior

80

Chapter 10

Liquids, Solids, and Phase Changes

87

Chapter 11

Solutions and Their Properties

98

Chapter 12

Chemical Kinetics

107

Chapter 13

Chemical Equilibrium

120

Chapter 14

Aqueous Equilibria: Acids and Bases

129

Chapter 15

Applications of Aqueous Equilibria

139

Chapter 16

Thermodynamics:

152

Entropy, Free Energy, and Equilibrium Chapter 17

Electrochemistry

160

Chapter 18

Hydrogen, Oxygen, and Water

172

Chapter 19

The Main-Group Elements

183

Chapter 20

Transition Elements and Coordination Chemistry

199

Chapter 21

Metals and Solid-State Materials

212

Chapter 22

Nuclear Chemistry

222

Chapter 23

Organic and Biological Chemistry

229

iii

CHAPTER 1 CHEMISTRY: MATTER AND MEASUREMENT Chapter Learning Goals for Students Section 1.1

Describe the processes involved in the scientific method.

Section 1.2

Give the symbol and name of elements mentioned in this chapter.

Section 1.3

Given the properties of boron and aluminum, show how Mendeleev predicted the properties of eka-aluminum (gallium).

Identify the group number and period to which an element belongs.

Identify the regions of the periodic table.

Section 1.4

Identify properties as extensive or intensive and as chemical or physical.

Identify the regions of the periodic table.

Section 1.5

List the seven basic SI units of measure, and give the numerical equivalent of the common metric prefixes used with these units.

Express numbers in scientific notation (textbook Appendix A).

Section 1.6

Interconvert metric and English masses.

Section 1.7

Interconvert metric and English lengths.

Section 1.8

Interconvert Fahrenheit, Celsius, and Kelvin temperatures.

Section 1.9

Interconvert metric and English volumes.

Section 1.10

Calculate densities from measured masses and volumes.

Use density to interconvert mass and volume.

Section 1.11

Define and perform calculations involving kinetic energy and potential energy.

Relate changes in energy to changes in mass that occur in chemical reactions.

Section 1.12

Distinguish between accuracy and precision.

Determine the number of significant digits in a measured quantity.

Section 1.13

State the result of a calculation involving measured quantities to the correct number of significant digits.

Section 1.14

Use dimensional analysis to solve metric-English conversion problems.

1

Chapter 1—Chemistry: Matter and Measurement

Lecture Outline Introduction 1.1. Approaching Chemistry: Experimentation1 A. Scientific Method – systematic approach to research 1. Experimentation 2. Hypothesis – interpretation explaining results of many experiments 3. Theory – consistent explanation of known observations; logical interpretations of experimental results 1.2. Chemistry and the Elements2 A. Element – fundamental substance that can’t be chemically changed or broken down into anything simpler 1. Chemical symbol –represents a specific element a. Capitalize first letter b. Lower case used for second letter if present B. Periodic table – tabular organization of all 114 elements 1.3. Elements and the Periodic Table3 A. Creation of the periodic table – ideal example of how scientific theory comes into being 1. Random observations 2. Organization of data in ways that make sense 3. Consistent hypothesis emerges a. Explains known facts b. Makes predictions about unknown phenomena B. Mendeleev’s hypothesis about organizing known chemical information – meets criteria for a good hypothesis 1. Listed the known elements by atomic weight 2. Grouped them together according to their chemical reactivity 3. Was able to predict the properties of unknown elements – eka–aluminum, eka–silicon C. Periodic table – grid of the elements arranged in 7 horizontal rows and 18 vertical columns 1. Periods – seven horizontal rows in the periodic table 2. Groups – 18 vertical columns in the periodic table a. Groups numbered 1A → 8A and 1B → 8B (or 1 →18) b. Actually have 32 groups – lanthanides (14 elements after lanthanum) and actinides (14 elements after actinium) not included in the group numbers c. Elements in a given group have similar chemical properties.

D. The periodic table of the elements is the most important organizing principle of chemistry.

1. Regular progression in size of the seven periods – reflects a similar regularity in atomic structure 2. Main Group (or Representative) Elements – Groups 1A-8A; (two larger groups on the left and six larger groups on the right of the table) 3. Transition Metal Elements – Groups 1B-8B; (10 smaller groups in the middle of the table) 4. Inner Transition Metal (or Rare Earth) Elements (14 groups shown separately at the bottom of the table) 1.4. Some Chemical Properties of the Elements4 A. Property – any characteristic used to describe or identify matter 1. Physical properties – characteristics not involving chemical change of the sample 2. Chemical properties – properties that do change chemical makeup of the sample 3. Intensive properties – sample size-independent properties 1

Test Item File Questions: Multiple Choice 1-2; Short Answer 1 Test Item File Questions: Multiple Choice 3-7; Algorithmic 1-5; Short Answer 3-4 3 Test Item File Questions: Multiple Choice 8-19; 116-135; Algorithmic 6-10; Short Answer 5-7 4 Test Item File Questions: Multiple Choice 20-33; Algorithmic 11-16; Short Answer 8-10 2

2

Chapter 1—Chemistry: Matter and Measurement 4. Extensive properties – sample size-dependent properties B. Elements within a group have similar chemical properties.

1. Group 1A – Alkali metals a. Lustrous, silvery metals b. React rapidly with water to form highly alkaline products 2. Group 2A – Alkaline earth metals a. Lustrous, silvery metals b. Less reactive than alkali metals 3. Group 7A – Halogens a. Corrosive, nonmetallic elements b. Salt formers 4. Group 8A – Noble gases: gases with low reactivity C. Three major classes of elements 1. Metals a. Largest category of elements b. Found on the left side of the periodic table (left of the heavy zigzag line) i. Solids (except mercury) ii. Malleable iii. Ductile – can be drawn into thin wires without breaking iv. Conduct heat and electricity 2. Nonmetals a. Relatively small number b. Found on the right side of the periodic table (right of the heavy zigzag line) i. Gases, liquids, or solids ii. Brightly colored iii. Brittle solids iv. Poor conductors of heat and electricity 3. Semimetals (metalloids) a. Elements adjacent to the zigzag boundary between metals and nonmetals b. Properties fall between metals and nonmetals i. Brittle ii. Poor conductors of heat and electricity 1.5. Experimentation and Measurement5 A. International System (SI) of Units – seven base units (and units derived from them) that suffice for all scientific measurements (textbook Table 1.3) B. Common prefixes used to modify SI units 1. giga (G); factor = 109 6 2. mega (M); factor = 10 3 3. kilo (k); factor = 10 –1 4. deci (d); factor = 10 –2 5. centi (c); factor = 10 –3 6. milli (m); factor = 10 –6 7. micro (μ); factor = 10 –9 8. nano (n); factor = 10 C. Scientific notation – exponential format for very large or very small numbers (textbook Appendix A) D. All measurements of physical quantities contain a 1. number.

2. unit label.

1.6. Mass and Its Measurement6 5 6

Test Item File Questions: Multiple Choice 34-46; Algorithmic 17-19; Short Answer 11-12 Test Item File Questions: Multiple Choice 47-54; Algorithmic 20-23; Short Answer 13 3

Chapter 1—Chemistry: Matter and Measurement A. Mass (SI unit = kg) – amount of matter in an object B. Matter – anything physically real C. Weight – pull of gravity on an object D. Mass of an object 1. Measured on a balance 2. A comparison of the weight of the object to the weight of a reference standard of known mass 1.7 Length and Its Measurement7 A. Meter 1. Standard SI unit of length 2. Distance traveled by light in a vacuum in 1/299,792,458th of a second 1.8. Temperature and Its Measurement8 o A. Common unit – Celsius degree ( C).

B. Scientific unit – Kelvin.

C. Celsius and Kelvin scales –100 degrees between the freezing point and boiling point of water Temperature in K = temperature in °C + 273.15 Temperature in °C = temperature in K – 273.15 D. Fahrenheit scale –180° between the freezing point and boiling point of water 1. 180°F encompasses the same range as 100°C. 180 oC 9 1 oC × = × oF 5 100 o F E. Conversion from Celsius to Fahrenheit 1. Size correction 2. Followed by zero-point correction F. Conversion from Fahrenheit to Celsius 1. Zero-point correction 2. Followed by a size correction ⎞ ⎛ 9 oF o F = ⎜ o × oC ⎟ + 32 o F ⎝5 C ⎠ o

C =

5 oC × 9 oF

( F - 32 F) o

o

1.9. Derived Units: Volume and Its Measurement9 A. Derived quantities – quantities expressed in terms of one or more of the seven base units (See textbook Table 1.5.)

B. Volume – amount of space occupied by an object 1. Measured in SI units by the cubic meter (m3) 2. Commonly used measurements: a. Cubic decimeter (dm3) = metric liter (L) b. Cubic centimeter (cm3) = metric milliliter (mL) 1.10. Derived Units: Density and Its Measurement10 A. Density – intensive property relating mass to volume 1. Expressed in units of g/cm3 or g/mL 2. Temperature dependent property 7

Test Item File Questions: Multiple Choice 55-60; Algorithmic 24-27, Short Answer 14 Test Item File Questions: Multiple Choice 61-66; Algorithmic 28-29; Short Answer 15 9 Test Item File Questions: Multiple Choice 67-71; Algorithmic 30-33; Short Answer 16-17 10 Test Item File Questions: Multiple Choice 72-82, 136-138; Algorithmic 34-38, 37-38; Short Answer 18-19 8

4

Chapter 1—Chemistry: Matter and Measurement 3. Useful property – allows conversion of a liquid’s volume to mass 1.11. Derived Units: Energy and Its Measurement11 A. Energy 1. Capacity to do work or supply heat 2. Energy = work + heat B. Kinetic energy 1. Energy of motion 2. EK = 1/2mv2 C. Potential energy – stored energy D. Joule 1. SI unit for energy 2. 1 J = 1(kg⋅m2)/s2 E. Calorie 1. Amount of energy necessary to raise the temperature of 1 g of water by 1°C 2. 1 cal = 4.184 J 3. Nutritional calorie (Calorie) 4. 1 Cal = 1000 cal = 1 kcal = 4.184 kJ 1.12. Accuracy, Precision, and Significant Figures in Measurement12 A. Accuracy – how close a measurement is to the true value B. Precision – how well a number of independent measurements agree with one another 1. To indicate the precision of a measurement use all the digits known with certainty, plus one additional estimated digit.

2. Number of digits in the measurement = number of significant figures C. Determining the number of significant figures – see rules in textbook D. Using scientific notation – helpful in determining number of significant figures E. Exact numbers – infinite number of significant figures 1.13. Rounding Numbers13 A. In carrying out a multiplication or division, the answer can’t have more significant figures than either of the original numbers.

B. In carrying out an addition or subtraction, the answer can’t have more digits to the right of the decimal point than either of the original numbers.

C. Rules for rounding off numbers 1. See text 2. Doing calculations a. Use all figures, significant or not b. Only round off final answer 1.14. Calculations: Converting from One Unit to Another14 A. Dimensional analysis method 1. Quantity described in one unit converted into an equivalent quantity described in a different unit 2. Conversion factor used to express relationship between units Starting quantity × conversion factor = equivalent quantity 3. Units a. Treated like numbers b. Can be multiplied and divided 4. Equation 11

Test Item File Questions: Multiple Choice 83-87; Algorithmic 39; Short Answer 20-21 Test Item File Questions: Multiple Choice 88-96, 139-145; Algorithmic 40-45; Short Answer 22-23 13 Test Item File Questions: Multiple Choice 97-100; Algorithmic 46; Short Answer 24 14 Test Item File Questions: Multiple Choice 101-113; Algorithmic 47-49, Short Answer 25 12

5

Chapter 1—Chemistry: Matter and Measurement a. Set up so that all unwanted units cancel b. Leave only desired units c. Correct answer obtained only if equation set up so unwanted units cancel 5. Method: a. Start with given information, including units b. Identify information needed, including units c. Find relationship between known information and unknown answer d. Plan a strategy to get from known information to unknown answer e. Solve i. Use conversion factor that allows units in given information to cancel ii. Continue conversion factor use to cancel the previous unit until desired unit remains f. Consider whether answer obtained is reasonable (Refer to Ballpark Check in textbook.)

6. Number of significant figures in answer determined from given information Inquiry – What Are the Risks and Benefits of Chemicals?15

15

Test Item File Questions: Multiple Choice 114-115 6

Chapter 1—Chemistry: Matter and Measurement

CHAPTER 1 CHEMISTRY: MATTER AND MEASUREMENT Teaching Tips, Points of Emphasis, and Common Misconceptions Section 1.1

Chemistry is an experimental science. Challenge students to sharpen their skills of observation and measurement, relating laboratory work to concepts discussed in lecture.

Section 1.2

Elements can be broken into simpler substances (protons, neutrons, electrons, etc.), but the energy required is greater than available under ordinary chemical conditions.

Section 1.3

Iodine is a silvery-gray solid that sublimes to a purple gas. The brown solution that was once used as an antiseptic medicine is known as tincture of iodine (iodine dissolved in alcohol).

Section 1.3

Because elements in the same group tend to exhibit similar chemical behavior, a group is also called a family.

Section 1.4

While elements within the same group tend to have similar chemical properties, due to its small size the first member of each main group exhibits a chemistry that differs most from the other members of the group.

Section 1.4

Hydrogen is a unique element. Although listed under group 1A, hydrogen is not a metal under ordinary conditions. In its ionic chemistry, hydrogen sometimes behaves as though it belongs in group 7A. In covalent compounds, hydrogen behaves as though it belongs between boron and carbon.

Section 1.5

All systems of measurement are based on arbitrary standards. The SI system is as arbitrary as the English system, but the relationship of one unit to another is more systematic and easier to remember in the metric system.

Section 1.5

For very large or very small quantities, commonly used prefixes represent amounts that vary by three powers of 10, for example, milli, micro, nano, and pico (10–3, 10–6, 10–9, and 10–12, respectively).

Section 1.9

Remind students to raise both the measured number and unit to the appropriate power when dealing with areas and volumes. For example, (3 cm)3 = 33 cm3 = 27 cm3, not 3 cm3 or 27 cm.

Section 1.10

Density is sometimes confused with mass. A kg of feathers weighs the same as a kg of gold, but gold occupies less space because it has a greater density, the ratio of mass to volume.

Section 1.10

Because students are very familiar with water they tend to use it as a standard when relating the density of a solid to the density of a liquid. Most solids sink in their liquids; water is unusual in that ice floats atop liquid water.

Section 1.12

Significant figures tend to be abstract to students until they realize that significant figures are a shortcut to expressing the limitations of laboratory measurements.

7

Chapter 1—Chemistry: Matter and Measurement

Section 1.12

Only measured quantities may contain a measured error, which limits the number of significant figures. In addition to counted objects (7 days = 1 week), numbers related by definition (1 in = 2.54 cm) are exact numbers with an infinite number of significant figures.

Section 1.13

The answer given to a series of calculations involving measured quantities is limited by the least precise measurement.

Section 1.13

Students can avoid round-off errors by always using one more significant figure for constants or molar masses than required by measured data and rounding off only the final answer and not intermediate ones.

Section 1.13

Based on calculator displays, students often report answers to problems that contain too many or too few significant figures. Remind students that calculators are incapable of distinguishing between measured and exact quantities. One consequence is that calculators do not record a zero at the end of numbers even when the zero may be a significant figure.

Section 1.14

Mathematical problem solving is more difficult for most students than understanding chemistry concepts. Knowing the goal of a series of calculations and where to start is not trivial for many students. Students need to be taught a system, such as writing the units of the desired answer on the right side of the paper, writing a measurement or conversion factor containing the same units on the left, and using appropriate relationships to cancel all but the desired units.

Lecture/Laboratory Demonstration References Section 1.3

James L. Marshall, “A Living Periodic Table,” J. Chem. Educ., Vol. 77, 2000, 979983. Describes a portable and permanent collection of 87 elements.

Section 1.5

Charles W. Earley, “A Simple Demonstration for Introducing the Metric System to Introductory Chemistry Classes,” J. Chem. Educ., Vol. 76, 1999, 1215-1216.

Section 1.10

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Sugar in a Can of Soft Drink: A Density Excercise,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2, (American Chemical Society, Washington, DC, 1988) pp. 126127. Regular and diet soft drinks are identified by differences in their densities.

Section 1.10

David A. Franz and David Speckhard, “Densities and Miscibilities of Liquids and Liquid Mixtures,” J. Chem. Educ., Vol. 68, 1991, 594. Using a graduated cylinder and electronic balance, the densities of water, dichloromethane, and petroleum ether are determined. A heavier mass of water is shown to form a layer on top of the more dense dichloromethane, but below the less dense petroleum ether. Mixtures of the miscible organic solvents are prepared with average densities above or below the density of water.

Literature References Section 1.2

P.G. Nelson, “Important Elements,” J. Chem. Educ., Vol. 68, 1991, 732-737.

8

Chapter 1—Chemistry: Matter and Measurement

Section 1.2

Vivi Ringes, “Origin of the Names of Chemical Elements,” J. Chem. Educ., Vol. 66, 1989, 731-738.

Section 1.3

Milton J. Wieder, “It’s Elementary,” J. Chem. Educ., Vol. 78, 2001, 468-469.

Section 1.5

Mark L. Campbell, “Having Fun with the Metric System,” J. Chem. Educ., Vol. 68, 1991, 1048. A word game.

Section 1.10

Kenneth E. Kolbe and Doris K. Kolbe, “Method for Separating or Identifying Plastics,” J. Chem. Educ., Vol. 68, 1991, 348. A practical use of density.

Section 1.12

Charles J. Guare, “Error, Precision and Uncertainty,” J. Chem. Educ., Vol. 68, 1991, 649-652.

Section 1.12

Richard S. Treptow, “Precision and Accuracy in Measurements: A Tale of Four Graduated Cylinders,” J. Chem. Educ., Vol. 75, 1998, 992-995.

Section 1.12

Arden P. Zipp, “A Simple but Effective Demonstration for Illustrating Significant Figure Rules When Making Measurements and Doing Calculations,” J. Chem.

Educ., Vol. 69, 1992, 291.

Section 1.12

Ben Ruekberg, “A Joke Based on Significant Figures,” J. Chem. Educ., Vol. 71, 1994, 306.

Section 1.12

Kenton B. Abel and William M. Hemmerlin, “Significant Figures,” J. Chem. Educ., Vol. 67, 1990, 213.

Section 1.12

H. Graden Kirksey and Paul Krause, “Significant Figures, A Classroom Demonstration,” J. Chem. Educ., Vol. 69, 1992, 497-498.

Media References Section 1.2

Names of Elements activity 1 from the Instructor Resource Center DVD

Section 1.2

Names of Elements activity 2 from the Instructor Resource Center DVD

Section 1.3

Periodic Property movie from the Instructor Resource Center DVD

Section 1.4

Sodium and Potassium in Water movie from the Instructor Resource Center DVD

Section 1.4

Physical Properties of the Halogens movie from the Instructor Resource Center DVD

Section 1.4

Periodic Table Groups activity from the Instructor Resource Center DVD

Section 1.14

Single-Step Conversion activity from the Instructor Resource Center DVD

Section 1.4

Multiple-Step Conversion activity from the Instructor Resource Center DVD

9

CHAPTER 2 ATOMS, MOLECULES, AND IONS Chapter Learning Goals for Students Section 2.1

Explain the law of a) conservation of mass and b) definite proportions.

Section 2.2

Explain the law of multiple proportions.

For two different compounds comprised of the same two elements, show that the law of multiple proportions is obeyed.

Section 2.3

Explain information about the atom revealed by the experiments of a) Thomson and b) Millikan, c) Rutherford.

Describe the atom in terms of the composition, mass, and volume of the nucleus relative to the mass and volume occupied by the electrons.

Section 2.4

Explain the information about the atom revealed by the experiments of Rutherford.

Describe the atom in terms of the composition, mass, and volume of the nucleus relative to the mass and volume occupied by the electrons.

Perform calculations using atomic size.

Section 2.5

State the difference between atoms and elements.

Write and interpret isotope symbols, determining the number of protons, neutrons, and electrons.

Section 2.6

Given the mass and natural abundance of all isotopes of a given element, calculate the average atomic mass of that element.

For any element, calculate a) the mass in grams of a single atom, the number of moles of atoms, and c) the number of atoms in a given number of grams.

Section 2.7

Define nucleon, nuclide, and nuclear reaction, summarizing the differences between nuclear reactions and chemical reactions and how they are described by their respective nuclear equations and chemical equations.

Section 2.8

Define radioactivity and radioisotope.

Write balanced equations for nuclear reactions, identifying the types of radiation and radioisotopes involved.

Section 2.9

Use the neutron/proton plot for stable isotopes to determine whether a given nuclide is expected to be stable or unstable.

Identify nuclear decay processes and the identity of isotopes involved in nuclear decay processes.

10

Chapter 2—The Structure and Stability of Atoms

Section 2.10

Classify substances as a) pure substances or mixtures and b) elements or chemical compounds.

State the difference between a) atoms and molecules, b) covalent bonds and ionic bonds, c) chemical formulas and structural formulas, and d) ball-and-stick models and space-filling models.

Relate molecular models to structural formulas and chemical formulas.

Section 2.11

Know charges of monatomic and polyatomic anions and cations.

Classify substances as ionic or molecular and bonding as ionic bonding or covalent bonding.

Section 2.12

Give the formulas and names of a) common polyatomic ions, b) ionic compounds, and c) binary molecular compounds.

Lecture Outline 2.1. The Conservation of Mass and the Law of Definite Proportions1 A. Element – a substance that cannot be further broken down B. Law of Mass Conservation – Mass is neither created nor destroyed in chemical reactions.

C. Law of Definite Proportions – Different samples of a pure chemical substance always contain the same proportion of elements by mass; elements do not combine chemically in random proportion.

2.2. The Law of Multiple Proportions and Dalton’s Atomic Theory2 A. Dalton’s Atomic Theory 1. Elements made of tiny particles called atoms 2. Each element characterized by the mass of its atoms a. Atoms of the same element – identical masses b. Atoms of different elements – different masses 3. Atoms combine in small, whole-number ratios – form new substances, called compounds 4. Chemical reactions only rearrange the way that atoms are combined; atoms themselves are not changed.

B. Law of Multiple Proportions – If two elements combine in different ways to form different substances, the mass ratios are small, whole-number multiples of each other.

2.3. Atomic Structure: Electrons3 A. Thomson – found that cathode rays consist of tiny, negatively charged particles called electrons 1. Electrons are emitted from electrodes made of two thin pieces of metal.

2. Many different metals may be used to make electrodes – all contain electrons.

3. Cathode rays can be deflected by bringing either a magnet or an electrically charged plate near the tube. Deflection depends on the a. strength of the deflecting magnetic or electric field. b. size of the negative charge on the electron. c. mass of the electron.

4. Charge-to-mass ratio, e/m of the electron = 1.758 819 × 108 C/g B. Millikan determined that the charge on a drop of oil was always a small, whole-number multiple of e. (See Fig. 2.4 in the textbook.)

1

Test Item File Questions: Multiple Choice 1-7, 141-144; Algorithmic 1-4; Short Answer 1 Test Item File Questions: Multiple Choice 8-14, 145-149; Algorithmic 5; Short Answer 2 3 Test Item File Questions: Multiple Choice 15-16; Short Answer 3 2

11

Chapter 2—Atoms, Molecules, and Ions 1. e = 1.602 177 × 10–19 C 2. Knowing the values for e/m and e for an electron, m can be calculated. a. m = 9.109 390 × 10–28 g 2.4. Atomic Structure: Protons and Neutrons4 A. Rutherford directed a beam of alpha particles at a thin gold foil.

1. Alpha (α) particles a. 7000 times more massive than electrons b. Have a positive charge twice the magnitude of, but opposite in sign to, the charge on an electron 2. Beam of α particles a. Most pass through the thin gold foil b. A few deflected at large angles B. Nuclear model of the atom 1. Nucleus a. A tiny central core in an atom where the mass of the atom is concentrated b. Contains the atom’s positive charges 2. Electrons move in space a relatively large distance away from the nucleus.

C. Nucleus composed of two kinds of particles 1. Protons a. Mass = 1.672 623 × 10–24 g b. Positively (+) charged c. Number of protons = number of electrons in a neutral atom 2. Neutrons a. Neutron mass ≈ proton mass b. Charge = 0 2.5. Atomic Numbers5 A. Elements differ from one another according to the number of protons in their atoms.

B. Atomic number (Z) = the number of protons in an atom = the number of electrons in an atom C. Most nuclei also contain neutrons.

D. Mass number (A) = number of protons + number of neutrons in an atom E. Isotopes – atoms with identical atomic numbers but different mass numbers 1. Mass number written as left superscript 2. Atomic number (Z) written as left subscript (The atomic number is sometimes left off because all atoms of an element always contain the same number of protons.)

3. Number of neutrons in an isotope calculated from A – Z 4. Number of neutrons in an atom has little effect on chemical properties of the atom.

2.6. Atomic Masses and the Mole6 A. Atomic mass unit (amu) 1. Exactly 1/12th the mass of an atom of 126 C 2. 1 amu = 1.660 539 × 10–24 g B. Isotopic mass 1. Mass of an atom in atomic mass units 2. Numerically close to the atom’s mass number C. Atomic mass values – weighted averages for the naturally occurring mixtures of isotopes 1. Atomic mass of an element = Σ(mass of each isotope × the fraction of the isotope) 2. Σ used for the term “the sum of” 3. Use atomic masses to count number of atoms by weighing a sample of the element.

D. One mole of any element is the amount whose mass in grams is numerically equal to its atomic mass.

4

Test Item File Questions: Multiple Choice 17-21; Short Answer 4 Test Item File Questions: Multiple Choice 22-35; Algorithmic 6-12; Short Answer 5-10 6 Test Item File Questions: Multiple Choice 36-44; Short Answer 11-14 5

12

Chapter 2—The Structure and Stability of Atoms 1. Mole (mol) = 6.022 × 1023 of anything a. Avogadro’s number (abbreviated NA) b. 1 mol N atoms = 6.022 × 1023 atoms of N c. Likewise, 1 mol electrons = 6.022 × 1023 electrons 2. Importance of the mole – provides a relationship between numbers of atoms and masses of atoms E. Molar mass of an element 1. One mole of any element has a mass equal to its atomic mass in grams 2. Mass of 6.022 × 1023 atoms of an element 3. Serves as a conversion factor between numbers of atoms and mass 2.7. Nuclear Chemistry: The Change of One Element Into Another7 A. An atom is characterized by its atomic number, Z, and its mass number, A. 1. Z – written as a subscript to the left of the element symbol; gives the number of protons in the nucleus 2. A – written as a superscript to the left of the element symbol, gives the total number of nucleons 3. Nucleon – a general term for both protons (p) and neutrons (n) 4. Uranium-238 238 U 92 a. 92 protons b. 238 – 92 = 146 neutrons c. 238 nucleons B. Isotopes – atoms with identical atomic numbers, but different mass numbers C. Nuclide – nucleus of a specific isotope D. Nuclear reaction – a reaction that changes the nucleus 238 U → 42 He + 23490Th 92 E. Differences between nuclear reactions and chemical reactions 1. Nuclear reactions – a change in an atom’s nucleus a. In nuclear reactions electrons are beta particles 0 e = β-1 b. Superscript 0 – the mass of an electron is essentially zero when compared with that of a proton or neutron c. Subscript (–1) – the charge of an electron is –1 2. Chemical reaction – a change in distribution of the outer shell electrons around an atom 3. Different isotopes of an element a. Have essentially the same behavior in chemical reactions b. Have a different behavior in nuclear reactions 4. Changes in temperature or pressures or the addition of a catalyst do not affect the rate of a nuclear reaction.

5. The nuclear reaction of an atom is essentially the same regardless of whether the atom is in an element or a compound.

6. The energy changes accompanying nuclear reactions are far greater than those accompanying chemical reactions.

2.8. Radioactivity8 A. Radioisotope – an unstable isotope 1. Radioactive nuclei 2. Radioactive – the spontaneous decomposition (decay) of an unstable isotope to form a more stable isotope of the same element or of another element and emission of radiation 7 8

Test Item File Questions: Multiple Choice 45-51; Algorithmic13; Short Answer 15-16 Test Item File Questions: Multiple Choice 52-72, 153-162; Algorithmic 14-15; Short Answer 17-20 13

Chapter 2—Atoms, Molecules, and Ions 3. Nuclear reaction – reaction that changes the nucleus as described by a nuclear equation B. Three common types of radiation with different properties 1. Alpha (α) 2. Beta (β) 3. Gamma (γ) C. Alpha (α) radiation – a stream of particles that consists of two protons and two neutrons 1. Repelled by a positively charged electrode 2. Attracted by a negatively charged electrode 3. A helium ion 4 He 2+ = α 2 4. Emission of an α particle reduces the mass number of the nucleus by four and reduces the atomic number by two.

238 U → 42 He + 23490Th 92 5. Common for heavy radioactive isotopes (radionuclides) D. Beta (β) radiation 1. Attracted by a positively charged electrode 2. Repelled by a negatively charged electrode 3. In nuclear reactions electrons are beta particles 0 e = β-1 4. Superscript 0 – the mass of an electron is essentially zero when compared with that of a proton or neutron 5. Subscript (–1) – the charge of an electron is –1 6. Occurs when a neutron in the nucleus spontaneously decays into a proton plus an electron, which is then ejected.

7. Emission of a beta particle does not affect the mass number but increases the atomic number by one.

14 C → 147 N + -10 e 6 E. Gamma (γ) radiation – electromagnetic radiation of very high energy and short wavelength 1. Not attracted by a positively charged electrode 2. Not repelled by a negatively charged electrode 3. No mass 4. Stream of high-energy photons 5. Almost always accompanies α and β emission – mechanism for release of energy 6. γ emission often not shown in nuclear equations – doesn’t change either the mass number or the atomic number of the product nucleus F. Positron emission and electron capture 1. Positron emission – conversion of a proton in the nucleus into a neutron plus an ejected positron, a particle with the same mass as an electron but opposite charge: 0 e = β+ 1 a. Decrease in atomic number of the product nucleus b. No change in mass number 2. Electron capture – a proton in the nucleus captures an inner-shell electron, which is converted into a neutron a. No change in mass number of the product nucleus b. Atomic number decreases by one G. Balanced nuclear reaction 1. Not balanced in usual chemical sense – nuclei are not the same on both sides of the reaction 2. Sums of nucleons on both sides are equal.

3. Sums of charges on nuclei and any elementary particles on both sides are equal.

14

Chapter 2—The Structure and Stability of Atoms 4. Not concerned with ionic charges on atoms a. Irrelevant to nuclear disintegration b. Disappear during reaction 2.9. Nuclear Stability9 A. Stable radioactive isotope – one that can be prepared and whose half-life can be measured 1. Unstable – those isotopes that can’t be prepared or that decay too rapidly for their halflives to be measured 2. Nonradioactive (stable indefinitely) – isotopes that do not undergo radioactive decay B. Neutron/proton ratio in the nucleus determines if an isotope is radioactive.

C. Plot of number of neutrons (y-axis) versus number of protons (x-axis) – textbook Figure 2.8 1. Stable nuclides – fall in a curved band called the peninsula of nuclear stability 2. Sea of instability a. Area on either side of the peninsula of nuclear stability b. Represents the large number of unstable neutron/proton combinations 3. Island of stability – area predicted to exist for a few super heavy nuclides near 114 protons and 184 neutrons D. Generalizations from plot 1. Every element has at least one radioactive isotope.

2. Hydrogen is the only element whose most abundant stable isotope contains more protons than neutrons.

3. The ratio of neutrons to protons gradually increases for elements heavier than calcium.

4. All isotopes beyond bismuth-209 are radioactive.

5. Nonradioactive isotopes generally have an even number of neutrons.

E. Neutrons function as a kind of nuclear glue that holds nuclei together by overcoming proton– proton repulsions.

F. Magic numbers of protons or neutrons — 2, 8, 20, 28, 50, 82, 126.

1. Give rise to particularly stable nuclei 2. Nucleus with a magic number of either protons or neutrons is unusually stable.

3. Analogous to chemical stability brought about by an octet of electrons G. Trends apparent from the peninsula of nuclear stability 1. Elements with an even atomic number have a larger number of nonradioactive nuclides than do elements with an odd atomic number.

2. Radioactive nuclei on the right side of the peninsula undergo nuclear disintegration by positron emission, electron capture, or alpha emission. a. Lower neutron/proton ratio b. Processes increase the neutron/proton ratio.

3. Nuclei on the left side of the peninsula emit beta particles. a. Higher neutron/proton ratios b. Process decreases the neutron/proton ratio.

H. Some radioisotopes can’t reach a nonradioactive nucleus in a single emission – undergo a decay series of disintegrations 2.10. Mixtures and Chemical Compounds; Molecules and Covalent Bonds10 A. Different kinds of matter on earth classified as either pure substances or mixtures – textbook Figure 2.10 B. Pure substance 1. Elements 2. Compounds

9

Test Item File Questions: Multiple Choice 73-86; Algorithmic 16-18; Short Answer 21 Test Item File Questions: Multiple Choice 87-91, 132-136, 163-164, 166, 168, 171, 178-180, 182184; Algorithmic19-22; Short Answer 22

10

15

Chapter 2—Atoms, Molecules, and Ions C. Chemical compounds – pure substance formed from the combination of atoms of two or more different elements 1. Have constant composition 2. Composition indicated by a chemical formula a. Lists the symbols of individual constituent elements b. Number of each atom is given by subscript 3. Formed when atoms undergo chemical combination in a specific manner 4. Transformation from elements to compound = chemical reaction D. Chemical bonds – connections that join atoms together in a compound 1. Formed by atom’s electrons 2. Classified as: a. Covalent bonds – occur between two nonmetals b. Ionic bonds – occur between a metal and a nonmetal E. Covalent bond – two atoms share electrons 1. Molecule – unit of matter that results when two or more atoms joined by covalent bonds 2. Structural formula a. Shows specific connections between atoms b. Contains more information than the chemical formula 3. Some elements exist as molecules: H2, O2, N2, F2, Cl2, Br2, I2 2.11. Ions and Ionic Bonds11 A. Ionic Bond – complete transfer of one or more electrons from one atom to another 1. Formed between metals and nonmetals a. Metal – gives up electrons b. Nonmetal – accepts electrons 2. Ions – charged particles resulting from loss or gain of electrons a. Cation – positively charged particle resulting from loss of one or more electrons b. Anion – negatively charged particle resulting from gain of one or more electrons c. Polyatomic ions i. Charged, covalently bonded groups of atoms ii. Charged molecules – specific numbers and kinds of atoms joined in a definite way by covalent bonds 3. Ionic solids a. Cations and anions packed together in a regular manner b. Charges cancel 2.12. Naming Chemical Compounds12 A. Binary Ionic Compounds – ionic compounds containing only two elements, a cation and an anion 1. Identify cation first, then anion a. Cation i. Same name as the element ii. Remember metals form cations b. Anion i. First part of its name from the element ii. Adds the ending ide iii. Remember nonmetals form anions 2. Common main group and transition metal ions – textbook Figures 2.14 and 2.15 a. Elements within a group often form similar kinds of ions. b. Main-group metal cations: charge = group number c. Main-group nonmetal anions: charge = group number – 8 11

Test Item File Questions: Multiple Choice 92-114, 137-140, 165, 167, 169-170, 172-177, 181; Algorithmic 23-37; Short Answer 23-25 12 Test Item File Questions: Multiple Choice 115-131; Algorithmic 38-50; Short Answer 26-27 16

Chapter 2—The Structure and Stability of Atoms d. Some metals form more than one kind of cation. i. Charge indicated by a Roman numeral in parentheses ii. Common for transition metal complexes 3. Electrical neutrality – cations and anions combine in such a manner that the overall charge on a compound is equal to zero a. Total positive charge = total negative charge b. Determine number of positive charges on the cation by counting the number of negative charges on the anion (and vice versa) 4. Formulas for ionic compounds always contain the smallest whole number ratio of cation to anion.

B. Binary molecular compounds – molecular compounds (nonmetals) containing only two elements 1. One of the elements is more cation-like – takes the name of the element 2. One of the elements is more anion-like – takes an ide ending 3. Character depends on relative positions of the two elements in the periodic table a. More cationlike – farther left in the periodic table b. More anionlike – farther right in the periodic table 4. To specify the numbers of each element present, use numerical prefixes (textbook Table 2.2) – the mono prefix not used for the atom named first 5. When naming binary molecular compounds that contain hydrogen, it is necessary to indicate whether the molecule is in the gaseous or aqueous (in water) state. a. If molecule a gas i. Use above rules. ii. Indicate gaseous state with (g) after formula b. If molecule in aqueous solution i. Name compound as a binary acid (see below) ii. Indicate aqueous state with (aq) after formula C. Ionic compounds containing polyatomic ions named by following rules for naming binary ionic compounds.

1. Identify cation 2. Identify anion a. Names, formulas, and charge numbers of the most common polyatomic anions found in textbook Table 2.3 b. Most names end with the suffix ite or ate c. Several pairs of ions related by presence or absence of hydrogen atom 3. Oxoanions – an atom of the same element combined with different numbers of oxygen atoms a. Learn the name and formula of the ion whose name ends with ate (including the charge on the anion). b. Add one O; add prefix per. c. Remove one O; change ate to ite. d. Remove two O’s; add prefix hypo and change ate to ite.

17

Chapter 2—Atoms, Molecules, and Ions

CHAPTER 2 ATOMS, MOLECULES, AND IONS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 2.2

Dalton did not know about isotopes – atoms of the same element that have different masses due to different numbers of neutrons. Challenge students to determine which of Dalton’s postulates are incorrect.

Section 2.4

Demonstrate the relative insignificance of an electron’s mass by calculating its percent contribution to the approximate mass of a simple atom, such as carbon-12.

Section 2.6

The use of carbon-12 to define the amu is another example of a system of measurement based on an arbitrary standard. In earlier times the amu was based on oxygen-16 and hydrogen-1.

Section 2.6

The average atomic mass of an element is a weighted, not a simple, average.

Section 2.6

An effective example showing the difference between simple average and weighted average is the use of a hypothetical student’s test scores of 100, 100, 100, 60. The weighted average = 90; since there are only two different grades, 100 and 60, the simple average is 80.

Section 2.7

Energy changes for nuclear reactions are on the order of a million times greater than energy changes for chemical reactions.

Section 2.8

An alpha (α) particle ( 42 He 2+ ) is a helium nucleus, but the +2 charge is not used in writing nuclear equations.

Section 2.8

0 e or β–) is an electron.

A beta (β) particle ( −1

Section 2.8

For the purpose of balancing nuclear equations, the mass of the electron is taken to be 0 and its “nuclear charge” is taken to be –1.

Section 2.8

Electron capture: proton + core electron → neutron.

Section 2.8

For a nuclear equation to be balanced: sum of atomic numbers of reactants (subscripts) = sum of atomic numbers of products, and sum of nucleons of reactants (superscripts) = sum of nucleons of products. Ionic charges are ignored.

Section 2.9

Transuranium elements, those with atomic numbers higher than uranium, do not occur naturally but are produced by nuclear transmutation reactions.

Section 2.10

The element oxygen is written O2, P4, and S8 are other examples of elements that occur as polyatomic species.

Section 2.11

NaCl is a simplest formula or empirical formula. It does not reveal that the compound is made up of ions (Na+Cl–), nor that it is an extended system (NaCl)x.

Section 2.11

Polyatomic ions are also known as molecular ions because they are molecules with a charge.

18

Chapter 2—The Structure and Stability of Atoms

Section 2.11

Some students think HCl is ionic because it dissolves in water to form H+ and Cl–.

The interaction of a molecule and a solvent can substantially change the properties of the molecule.

Section 2.12

In compounds, elements on the left of the periodic table tend to be relatively positive, and elements on the right tend to be relatively negative.

Lecture/Laboratory Demonstration References Section 2.5

Arthur B. Ellis, Edward A. Adler, and Frederick H. Juergens, “Dramatizing Isotopes: Deuterated Ice Cubes Sink,” J. Chem. Educ., Vol. 67, 1990, 159-160.

Literature References Section 2.1

John J. Fortman, “Analogical Demonstration,” J. Chem. Educ., Vol. 69, 1992, 323324. Discusses the law of conservation of mass and the law of multiple proportions.

Section 2.3

Doris Eckey, “A Millikan Oil Drop Analogy,” J. Chem. Educ., Vol. 73, 1996, 237238.

Section 2.4

Mary V. Lorentz, “Bowling Balls and Beads, A Concrete Analogy to the Rutherford Experiment,” J. Chem. Educ., Vol. 65, 1988, 1082.

Section 2.4

Barrie M. Peake, “The Discovery of the Electron, Proton, and Neutron,” J. Chem.

Educ., Vol. 66, 1989, 738.

Section 2.5

William Spindel and Takanobu Ishida, “Isotope Separation,” J. Chem. Educ., Vol. 68, 1991, 312-318.

Section 2.6

Arthur M. Last and Michael J. Webb, “Using Monetary Analogies to Teach Average Atomic Mass,” J. Chem. Educ., Vol. 70, 1993, 234-235.

Section 2.6

John J. Fortman, “Pictorial Analogies IV: Relative Atomic Weights,” J. Chem.

Educ., Vol. 70, 1993, 235-236.

Section 2.6

Josefina Arce de Sanabia, “Relative Atomic Mass and the Mole: A Concrete Analogy to Help Students Understand These Abstract Concepts,” J. Chem. Educ., Vol. 70, 1993, 233-234.

Section 2.7

Enrique A. Hughes and Anita Zalts, “Radioactivity in the Classroom,” J. Chem.

Educ., Vol. 77, 2000, 613-614.

Section 2.7

“Nuclear Chemistry: State of the Art for Teachers,” a series of articles in the October 1994 issue of the Journal of Chemical Education.

Section 2.7

S. G. Hutchinson and F. I. Hutchinson, “Radioactivity: in Everyday Life,” J. Chem.

Educ., Vol. 74, 1997, 501-505.

19

Chapter 2—Atoms, Molecules, and Ions

Section 2.7

Charles H. Atwood, “Teaching Aids for Nuclear Chemistry,” J. Chem. Educ., Vol. 71, 1994, 845-847.

Section 2.8

C. Ronneau, “Radioactivity: A Natural Phenomenon,” J. Chem. Educ., Vol. 67, 1990, 736-737.

Robert Suder, “Beta Decay Diagram,” J. Chem. Educ., Vol. 66, 1989, 231.

Section 2.8 Section 2.9

Donald J. Olbris and Judith Herzfeld, “Nucleogenesis! A Game with Natural Rules for Teaching Nuclear Synthesis and Decay,” J. Chem. Educ., Vol. 76, 1999, 349352.

Section 2.9

Darleane C. Hoffman and Diana M. Lee, “Chemistry of the Heaviest Elements – One Element at a Time,” J. Chem. Educ., Vol. 76, 1999, 332-347.

Section 2.12

Gerhard Lind, “Teaching Inorganic Nomenclature: A Systematic Approach,” J. Chem. Educ., Vol. 69, 1992, 613-614.

Section 2.12

Steven J. Hawkes, “A Mnemonic for Oxy-Anions,” J. Chem. Educ., Vol. 67, 1990, 149.

Interlude

V. E. Viola, “Teaching Nuclear Science: A Cosmological Approach,” J. Chem.

Educ., Vol. 71, 1994, 840-844.

Media References Section 2.1

Electrolysis of Water movie from the Instructor Resource Center DVD

Section 2.1

Conservation of Mass activity from the Instructor Resource Center DVD

Section 2.2

Multiple Proportions movie from the Instructor Resource Center DVD

Section 2.2

Multiple Proportions activity from the Instructor Resource Center DVD

Section 2.3

Millikan Oil Drop Experiment movie from the Instructor Resource Center DVD

Section 2.3

Separation of Rays activity from the Instructor Resource DVD

Section 2.4

Rutherford Experiment movie from the Instructor Resource Center DVD

Section 2.5

Atomic Number activity from the Instructor Resource Center DVD

Section 2.5

Carbon Isotopes activity from the Instructor Resource Center DVD

Section 2.6

Isotopes of Hydrogen activity from the Instructor Resource Center DVD

Section 2.9

Nuclear Stability activity from the Instructor Resource Center DVD

Section 2.9

Half-Life activity from the Instructor Resource DVD

Section 2.10

Mixtures and Compounds activity from the Instructor Resource Center DVD 20

Chapter 2—The Structure and Stability of Atoms

Section 2.12

Main-Group Ions activity from the Instructor Resource Center DVD

Section 2.12

Nonmetal Anions activity from the Instructor Resource Center DVD

Section 2.12

Transition Metal Ions activity from the Instructor Resource Center DVD

Section 2.12

Naming Polyatomic Ions activity 1 from the Instructor Resource Center DVD

Section 2.12

Naming Polyatomic Ions activity 2 from the Instructor Resource Center DVD

Section 2.12

Naming Polyatomic Ions activity 3 from the Instructor Resource Center DVD

21

CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Chapter Learning Goals for Students Section 3.1

For simple chemical reactions, write and balance chemical equations.

Section 3.2

Differentiate between microscopic and macroscopic interpretations of chemical formulas.

Section 3.3

Determine the formula of a compound from its ball-and-stick model.

Calculate molar mass.

Interconvert grams, moles, and numbers of formula units.

Determine the number of moles and grams of one reactant needed to react with a given number of moles and grams of another reactant and the number of moles and grams of product(s) that result from the reaction.

Section 3.4

Calculate percent yield.

Calculate the number of grams of products produced from a given number of grams of reactants when the theoretical yield is less than 100%.

Section 3.5

Identify the limiting and excess reagents in a reaction mixture.

Determine the number of grams of excess reagent remaining at the end of a reaction and the number of grams of product(s) produced.

Section 3.6

Describe how to prepare a solution of known molarity by dissolving a solid in a solvent.

Calculate the number of moles of solute in a given volume of solution and vice versa.

Section 3.7

Describe how to prepare a solution of known molarity by diluting a more concentrated solution.

Determine the final concentration of solution after dilution.

Section 3.8

Interconvert solution molarity, solution volume, solute moles, and solute grams.

Section 3.9

Determine the volume of one reactant needed to react with a given volume of a second reactant.

Determine the molarity of an acid or base solution by titration.

Section 3.10

Determine the percent composition and empirical formula of a compound.

Section 3.11

Use combustion analysis data to obtain the empirical formula of a compound containing carbon, hydrogen, and one other element.

22

Chapter 3—Mass Relationships in Chemical Reactions From empirical formula and molar mass, determine the molecular formula of a compound.

Section 3.12

Describe how atomic and molecular masses can be determined by mass spectrometry.

23

Chapter 3—Mass Relationships in Chemical Reactions

Lecture Outline 3.1. Balancing Chemical Equations1 A. Balanced equations 1. Numbers and kinds of atoms on both sides of the arrow are equal 2. Chemical equations balanced to obey law of mass conservation B. Formula unit – one unit (an atom, ion, or molecule) corresponding to a given formula C. Balancing process 1. Write unbalanced equation using correct chemical formulas for all reactants and products 2. Use coefficients (numbers placed before each formula) to indicate number of formula units required to balance the equation a. Coefficients can be changed b. Formulas must stay the same 3. Reduce coefficients to smallest whole numbers 4. Check your answer 3.2. Chemical Symbols on Different Levels2 A. Chemical symbols – represent both a microscopic and a macroscopic level B. Microscopic level – chemical symbols represent behavior of individual atoms and molecules C. Macroscopic level – formulas and equations represent large-scale behavior of atoms and molecules giving rise to observable properties 1. Deal with macroscopic behavior in the laboratory 2. Use microscopic models to explain macroscopic behavior 3.3. Chemical Arithmetic: Stoichiometry3 A. Dealing with macroscopic behavior requires that the number ratio of reactants or products from a balanced equation be converted to a mass ratio (In the laboratory we do not count numbers of reactants and products, we weigh grams of reactants and products.)

1. Number ratio determined from coefficients in the balanced equation 2. Mass ratio determined using molecular masses (molecular weights) a. Molecular mass – sum of atomic masses of all atoms in a molecule b. Formula mass – sum of atomic masses of all atoms in one formula unit of any substance B. Mass ratio of molecules (or formula units) = molecular (or formula unit) mass ratio C. One mole of any substance is the amount whose mass is equal to the molecular or formula mass in grams.

1. Mole (mol) = 6.022 × 1023 particles a. Avogadro’s number (abbreviated NA) b. 1 mol HNO3 = 6.022 × 1023 molecules of HNO3 c. 1 mol electrons = 6.022 × 1023 electrons 2. Importance of mole – provides a relationship between numbers of molecules and masses of molecules D. Molar mass of a substance – one mole of any substance has a mass equal to its molecular or formula mass in grams 1. Mass of one mole of a substance 2. Mass of 6.022 × 1023 molecules, ions, atoms, or compounds 3. Molecular mass of substance in grams 4. Serves as a conversion factor between numbers of formula units and mass 5. Formula of the particles is always specified E. Coefficients in a balanced chemical equation indicate the number of moles of each substance needed for the reaction 1 Test Item File Questions: Multiple Choice 1-8, 102-107; Algorithmic 1-3; Short Answer 1-3 2 Test Item File Questions: Multiple Choice 9-11, 98-101 3 Test Item File Questions: Multiple Choice 12-43, 108-109; Algorithmic 4-18; Short Answer 4-10

24

Chapter 3—Mass Relationships in Chemical Reactions F. Stoichiometry – refers to the conversion between moles and grams of reactants and products in a chemical equation G. Converting grams, moles and formula units – use the molar mass of the substance as a conversion factor H. Calculating the grams of reactants needed and products produced, knowing the grams and moles of another reactant 1. Begin with the known grams of reactant, convert to moles 2. Using the balanced chemical equation, convert moles of one substance to moles of another 3. Knowing moles, convert to grams 3.4. Yields of Chemical Reactions4 A. Actual yield 1. Amount of product actually formed 2. Less than the amount predicted by theory B. Theoretical yield – amount of product calculated to form if all of the reactants are converted to products Actual yield × 100% C. Percent yield = Theorectical yield 3.5. Reactions with Limiting Amounts of Reactants5 A. Many reactions carried out using an excess of one reactant B. Excess reactant – reactant present in more than the amount needed according to the stoichiometry 1. Only amount required by stoichiometry reacts 2. Excess reactant acts as a spectator and plays no role in the reaction C. Limiting reactant 1. Reactant present in an insufficient molar amount as required by stoichiometry 2. Determines the extent to which a chemical reaction takes place 3. Reaction stops when limiting reactant depleted D. Determination of whether a limiting amount of one reactant is present and to calculation of the amount of excess reactant consumed: 1. Find moles of each reactant present 2. Look at the coefficients in the balanced equation to see the required mole ratio of the two reactants – compare to number of moles calculated in the first step 3. Subtract number of moles of excess reactant consumed from the number of moles of excess reactant present. a. Convert to grams b. This is grams of nonlimiting reactant in excess 4. Determination of the amount of product formed: a. Base on moles of limiting reactant b. Moles limiting reactant → moles product → grams product 3.6. Concentrations of Reactants in Solution: Molarity6 A. Most chemical reactions carried out in solution – reactants must have considerable mobility for reaction to occur B. Solute – substance dissolved in a solvent to make a solution C. Concentration – relative amount of solute to solvent or solution D. Molarity (M) – number of moles of a solute dissolved in each liter of solution 1. Dissolve solute in enough solution to give a final solution volume of 1.00 L 2. Used as conversion factor 3.7. Diluting Concentrated Solutions7 4 Test Item File Questions: Multiple Choice 44-47; Algorithmic 19-28; Short Answer 11-12 5 Test Item File Questions: Multiple Choice 48-56, 110–116; Algorithmic 29-34; Short Answer 13-16 6 Test Item File Questions: Multiple Choice 57-65; Algorithmic 35-40; Short Answer 17-20 7 Test Item File Questions: Multiple Choice 66-71, 117-118; Algorithmic 41-45; Short Answer 21

25

Chapter 3—Mass Relationships in Chemical Reactions A. Concentrated solution + Solvent → Diluted solution B. Dilution 1. Number of moles of solute remains the same 2. Only volume is changed by adding more solvent 3. A convenient relationship: ⎛ mol ⎞ ⎛ mol ⎞ Moles of solute = M i ⎜ ⎟ × Vi (L)= M f ⎜ ⎟ × V f (L) ⎝ L ⎠ ⎝ L ⎠ 4. Note: This equation should be used only for problems that deal with dilution of a solution!

3.8. Solution Stoichiometry8 A. Knowing the molarity of solutions is important when performing stoichiometry calculations on substances in solution B. Use molarity to calculate either the number of moles (if volume is known) or volume (if the number of moles is known) 3.9. Titration9 A. Procedure for determining the concentration of a solution B. Carefully measured volume of one solution reacts with a second solution of known concentration and volume 3.10. Percent Composition and Empirical Formulas10 A. Composition of a compound – identity and amount of elements present in a compound B. Percent composition 1. Mass percent of each element present in a compound 2. Calculation of chemical formula from percent composition: a. Find relative numbers of moles of each element in the compound. i. Assume 100 g of compound ii. Use molar masses of elements as conversion factors b. Find ratio of the numbers of moles by dividing larger number of moles by smaller number of moles c. Multiply subscripts by small integers in a trial-and-error procedure until whole numbers found

C. Empirical formula – gives only ratios of atoms in a compound D. Molecular formula – gives actual numbers of atoms in a molecule 1. May be the same as the empirical formula 8 Test Item File Questions: Multiple Choice 72-76; Algorithmic 46-50; Short Answer 22 9 Test Item File Questions: Multiple Choice 77-80, 119-120; Algorithmic 51-54 10 Test Item File Questions: Multiple Choice 81-89; Algorithmic 55-57; Short Answer 23-24

26

Chapter 3—Mass Relationships in Chemical Reactions 2. May be a multiple of the empirical formula: Molecular mass Multiple = Empirical formula mass E. Can derive percent composition from chemical formula 3.11. Determining Empirical Formulas: Elemental Analysis11 A. Combustion analysis – compound of unknown composition burned with oxygen to produce volatile combustion products CO2 and H2O 1. CO2 and H2O separated by a gas chromatograph and their masses determined 2. Used to determine empirical formula of a compound B. Determination of empirical formula of a compound from combustion reaction: 1. Perform gram-to-mole conversions to find molar amounts of C and H in CO2 and H2O 2. Perform mole-to-gram conversions to find number of grams of C and H in original sample 3. Subtract masses of C and H from the mass of starting sample to determine number of grams of third element 4. Determine number of moles of third element 5. Find mole ratios by dividing larger number of moles by smaller number of moles C. Determination of molecular formula of a substance: 1. Molecular mass determined from a separate experiment (e.g., using mass spectrometry) 2. Subscripts in empirical formula multiplied by ratio of molecular mass to empirical formula mass 3.12. Determining Molecular Masses: Mass Spectrometry12 A. Most common method for molecular mass determination B. Method 1. Vaporized sample ionized by bombardment with high-energy electrons 2. Some molecular ions fragment into smaller ions 3. Rapidly-moving ions pass through a curved magnetic field a. Pathway followed by ions mass-dependent b. Lighter ions deflected more strongly than heavier ions 4. Ions separated by mass enter a detector assembly a. Signal intensities proportional to the relative number of each ion b. Mass of each ion assigned by comparison to a standard 5. Very precise – molecular masses determined to as many as seven significant figures

11 Test Item File Questions: Multiple Choice 90-95, 121-124; Algorithmic 58-59; Short Answer 25 12 Test Item File Questions: Multiple Choice 96-97

27

Chapter 3—Mass Relationships in Chemical Reactions

CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 3.1

Equations may be balanced using fractions. H2(g) + ½ H2(g) → H2O(l) is a balanced equation.

Section 3.1

When balancing chemical equations involving organic compounds (compounds of carbon), balance the atoms in the order C then H then O.

Section 3.3

Emphasize using chemical formulas with units: 18.0 g H2O/mol H2O instead of 18.0 g/mol.

Just as density provides a link between mass and volume, molar mass links mass and moles, allowing interconversions between these three units.

Coefficients in chemical formulas or balanced chemical equations represent numbers of formula units or moles but never numbers of grams. Show that the mass ratio of C2H4 is not the same as the atomic or mole ratio.

The number of significant figures in the answer to any calculation should be limited only by the number of significant figures in measured quantities. When selecting an atomic mass for use in a calculation, use one more significant figure than is found in the least precise measured quantity involved in the calculation.

Section 3.4

Actual yield is an experimental quantity. It cannot be calculated.

Section 3.5

There are three common methods of determining limiting reactant: (1) comparing stoichiometric ratio to actual ratio of reactants; (2) determining a theoretical yield based on each reactant; and (3) determining the amount of one reactant needed to react with a given quantity of a second reactant.

The excess reactant is not necessarily the reactant present in the greatest amount.

Section 3.6

Molarity is moles of solute per liter of solution, not per liter of solvent. The volume of solvent used is not determined when making a solution of known molarity.

Section 3.7

The relationship M1 × V1 = M2 × V2 is convenient for dilutions only. Students tend to use it for solution stoichiometry problems, which only works if the stoichiometry is 1:1.

Section 3.10

Experimental errors in the data or calculation round-off errors usually result in mole ratios that are not whole numbers.

Mole ratios frequently contain fractions close to 1/4, 1/3, or 1/2 which should be multiplied by 4, 3, and 2, respectively. Mole ratios close to whole numbers should be rounded to that number.

Section 3.11

Combustion analysis can be used to determine the empirical formula of a compound containing carbon, hydrogen, and any third element. Frequently the third element is oxygen.

28

Chapter 3—Mass Relationships in Chemical Reactions When doing combustion analysis problems students frequently try to determine the amount of oxygen in a sample from the amounts of water and carbon dioxide produced. This is not possible because the compound is burned in oxygen which makes up part of the water and carbon dioxide produced.

Molecular weight (molecular mass) cannot be calculated from combustion analysis data. It requires a separate technique, such as mass spectroscopy.

Lecture/Laboratory Demonstration References Section 3.3

Sally Solomon and Chinhyu Hur, “Measuring Avogadro’s Number on the Overhead Projector,” J. Chem. Educ., Vol. 70, 1993, 252-253. A monolayer of stearic acid on water is used to estimate Avogadro's Number.

Section 3.3

William Laurita, “Demonstrations for Nonscience Majors: Using Common Objects to Illustrate Abstract Concepts,” J. Chem. Educ., Vol. 67, 1990, 60-61. A demonstration of the Measurement of Avogadro's Number.

Section 3.10

Bassam Z. Shakhashiri, “Colorful Acid-Base Indicators,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp. 33-40.

Section 3.12

Nancy C. Grim and Jerry L. Saqquis, “Mass Spectrometry Analogy on the Overhead Projector,” J. Chem. Educ., Vol. 72, 1995, 930-931. Mass spectrometry principles are demonstrated on an overhead projector using nickel spheres and magnets.

Literature References Section 3.1

William C. Herndon, “On Balancing Chemical Equations: Past and Present (A Critical Review and Annotated Bibliography) ,” J. Chem. Educ., Vol. 74, 1997, 1359-1362.

Section 3.1

Zoltan Toth, “Balancing Chemical Equations by Inspection,” J. Chem. Educ., Vol. 74, 1997, 1363-1364.

Section 3.3

Dawn M. Wakeley and Hans de Grys, “Developing an Intuitive Approach to Moles,” J. Chem. Educ., Vol. 77, 2000, 1007-1009.

Section 3.3

Henk van Lubeck, “How to Visualize Avogadro’s Number,” J. Chem. Educ., Vol. 66, 1989, 762.

Section 3.3

Miriam Toloudis, “The Size of a Mole,” J. Chem. Educ., Vol. 73, 1996, 348.

Section 3.3

Damon Diemente, “Demonstrations of the Enormity of Avogadro’s Number,” J. Chem. Educ., Vol. 75, 1998, 1565-1566.

Section 3.3

Carmela Merlo and Kathleen E. Turner, “A Mole of M&M’s,” J. Chem. Educ., Vol. 70, 1993, 453.

Section 3.3

Sheryl Dominic, “What’s a Mole For?”, J. Chem. Educ., Vol. 73, 1996, 309.

29

Chapter 3—Mass Relationships in Chemical Reactions

Section 3.3

Shanthi R. Krishnan and Ann C. Howe, “The Mole Concept: Developing an Instrument to Assess Conceptual Understanding,” J. Chem. Educ., Vol. 71, 1994, 635-655.

Section 3.3

Addison Ault, “How to Say How Much: Amounts and Stoichiometry,” J. Chem.

Educ., Vol. 78, 2001, 1347-1348.

Section 3.3

Richard L. Poole, “Teaching Stoichiometry: A Two-Cycle Approach,” J. Chem.

Educ., Vol 66, 1989, 57.

Section 3.3

Karla R. Krieger, “Stoogiometry: A Cognitive Approach to Teaching Stoichiometry,” J. Chem. Educ., Vol. 74, 1997, 306-309.

Section 3.3

John J. Fortman, “Pictorial Analogies XII: Stoichiometric Calculations,” J. Chem.

Educ., Vol. 71, 1994, 571-572.

Section 3.3

John Olmsted III, “Amounts Tables as a Diagnostic Tool for Flawed Stoichiometric Reasoning,” J. Chem. Educ., Vol. 76, 1999, 52-54.

Section 3.5

Ernest F. Silversmith, “Limiting and Excess Reagents, Theoretical Yield,” J. Chem.

Educ., Vol. 62, 1985, 61.

Section 3.5

A. H. Kalantar, “Limiting Reagent Problems Made Simple for Students,” J. Chem.

Educ., Vol. 62, 1985, 106.

Section 3.5

Zoltan Toth, “Limiting Reactant: An Alternative Analogy,” J. Chem. Educ., Vol. 76, 1999, 934.

Section 3.6

Lloyd J. McElroy, “Teaching Dilutions,” J. Chem. Educ., Vol. 73, 1996, 765-766.

Section 3.9

Ara S. Kooser, Judith L. Jenkins, and Lawrence E. Welch, “Acid-Base Indicators: A New Look at an Old Topic,” J. Chem. Educ., Vol. 78, 2001, 1504-1506.

Section 3.10

George L. Gilbert, “Percentage Composition and Empirical Formula – A New View,” J. Chem. Educ., Vol. 75, 1998, 851.

Section 3.10

An easy way to remember the strategy for converting percentage composition to an empirical formula: “Percent to mass, mass to mol, divide by small, multiply ‘til whole”. Joel S. Thompson, “A Simple Rhyme for a Simple Formula," J. Chem.

Educ., Vol. 65, 1988, 704.

Section 3.10

P. K. Thamburaj, “A Known-to-Unknown Approach to Teach About Empirical and Molecular Formulas,” J. Chem. Educ., Vol. 78, 2001, 1050-1052.

Media References Section 3.1

Sodium and Potassium in Water from the Instructor Resource DVD

Section 3.3

Molecular Mass and Mass Percent activity from the Instructor Resource DVD

Section 3.3

Mass-Mole-Mass Conversion activity from the Instructor Resource DVD 30

Chapter 3—Mass Relationships in Chemical Reactions

Section 3.5

Limiting Reagents movie from the Instructor Resource DVD

Section 3.5

Limiting Reagents activity from the Instructor Resource DVD

Section 3.6

Molarity activity from the Instructor Resource DVD

Section 3.6

Solution Formation from a Solid from the Instructor Resource DVD

Section 3.6

Solutions Formation by Dilution from the Instructor Resource DVD

Section 3.9

Acid-Base Titration movie from the Instructor Resource DVD

Section 3.9

Acid-Base Titration activity from the Instructor Resource DVD

Section 3.9

Molarity Titration activity from the Instructor Resource DVD

31

CHAPTER 4 REACTIONS IN AQUEOUS SOLUTION Chapter Learning Goals for Students Section 4.1

Classify reactions as precipitation, acid-base neutralization, or redox.

Section 4.2

Classify substances as strong electrolytes, weak electrolytes, or nonelectrolytes.

Section 4.3

Write molecular, ionic, and net ionic equations for precipitation, acid-base, and redox reactions.

Section 4.4

Use solubility guidelines to determine whether a compound is likely to be soluble in water.

Use solubility guidelines to predict whether a precipitate might form when aqueous salt solutions are mixed.

Use solubility guidelines to predict the product of a precipitation reaction.

Show how a precipitation reaction can be used to prepare a substance.

Section 4.5

Identify the common strong acids and strong bases.

Give the formulas and names of common acids.

Section 4.6

Assign oxidation numbers to each atom in a chemical species.

Section 4.7

In a redox reaction, identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent.

Section 4.8

Use the activity series to predict whether a redox reaction will occur when a metal is placed in contact with a solution containing an ion of a different metal.

Use the activity series to predict the products of a redox reaction.

Section 4.9

Balance redox reactions by the half-reaction method in acid or base.

Section 4.10

Determine the concentration of a solution using data from a redox titration.

Section 4.11

Identify the oxidation and reduction half-reactions in 1) combustion, 2) bleaching, 3) battery, 4) metallurgical, 5) corrosion, and 6) respiration reactions.

32

Chapter 4—Reactions in Aqueous Solution

Lecture Outline 4.1. Some Ways that Chemical Reactions Occur1 A. Reactions – reactants transformed to products by some energetic driving force B. Driving forces of reactions 1. Precipitation Reactions a. Solid precipitate forms and drops out of solution b. Driving force – removal of material from solution 2. Acid-Base Neutralization Reactions a. Acid reacts with base to produce salt and water b. Driving force – formation of water 3. Oxidation–Reduction (Redox) Reactions a. Transfer of electrons between reactants b. Driving force – decrease in electrical potential 4.2. Electrolytes in Aqueous Solution2 A. Electrolytes – dissolve in water to produce ionic solutions 1. Strong Electrolytes – completely dissociate in water a. Soluble ionic compounds b. Strong acids 2. Weak Electrolytes – incompletely dissociate in water; establish dynamic equilibrium between the forward and backward reactions a. Weak acids b. Weak bases B. Nonelectrolytes – do not produce ions in aqueous solution a. Water b. Alcohols c. Sugars d. Other water-soluble organic compounds (compounds of carbon) C. Textbook Table 7.1 4.3. Aqueous Reactions and Net Ionic Equations3 A. Molecular Equations – reactant and product formulas written as if they were molecules B. Ionic Equations – any reactants or products that completely dissociate in water (strong electrolytes) shown as free ions C. Spectator Ions – ions not undergoing change during reaction 1. Spectator ion identity not important 2. Role is to balance charge D. Net Ionic Equation 1. Equation for net change taking place during reaction 2. Spectator ions not included 4.4. Precipitation Reactions and Solubility Guidelines4 A. Solubility – amount of compound dissolved in a given amount of solvent at a given temperature 1. Low solubility – precipitate forms 2. High solubility – no precipitate will form 3. Solubilities predicted using guidelines found in textbook B. Solubility Guidelines 1. Use to predict if a precipitate will form in a reaction 2. Use to prepare and isolate a specific compound by carrying out a precipitation reaction 1

Test Item File Questions: Multiple Choice 1-7; Short Answer 1-3 Test Item File Questions: Multiple Choice 8-13, 103-104; Algorithmic 1-2; Short Answer 4-5 3 Test Item File Questions: Multiple Choice 14-21; Short Answer 6-7 4 Test Item File Questions: Multiple Choice 22-37, 105-110, 119-120; Algorithmic 3-9; Short Answer 8-9 33 2

Chapter 4—Reactions in Aqueous Solution 3. Compounds that are probably soluble usually contain a. group 1A cations, most commonly Na+ or K+. b. ammonium ion, NH4+. c. halides (but not F–), unless the cation is Ag+, Hg22+, or Pb2+. d. nitrate, NO3–. e. perchlorate, ClO4–. f. acetate, CH3CO2–. g. Sulfate, SO42–, unless the cation is Ba2+, Hg22+, or Pb2+.

4. The most soluble compounds contain cations with a +1 charge and anions with a –1 charge. a. Solubility decreases as the charges on the ions increase b. Lattice energy effect 4.5. Acids, Bases, and Neutralization Reactions5 A. Arrhenius Acid – dissociates in water to produce H+ 1. Hydronium Ion a. H+ attached to a water molecule b. Represented as H3O+ 2. Strong acids – strong electrolytes 3. Weak acids – weak electrolytes 4. Polyprotic acids a. Acids with more than one acidic hydrogen b. Dissociate in steps – B. Arrhenius Base – dissociates in water to produce OH 1. Strong bases – strong electrolytes; most metal hydroxides 2. Weak bases – weak electrolytes – NH3(g) + H2O(l) → NH4+(aq) + OH (aq) C. Neutralization Reaction – reaction between acid and base producing salt and water 1. Salt produced from cation of base and anion of acid 2. Net reaction for a strong acid + strong base: H+(aq) + OH– (aq) → H2O(l) 3. Net reaction for a weak acid + strong base: HA(aq) + OH– (aq) → H2O(l) + A– (aq) D. Common acid and bases E. Naming acids 1. Binary acids – contain hydrogen and one other element or a polyatomic anion that is not an oxoanion a. Aqueous solutions b. Use the prefix hydro followed by the first part of the name of the other element, the suffix ic and acid.

1) When writing the formula, indicate aqueous solution by (aq).

2) HCl(aq) is hydrochloric acid but HCl(g) is hydrogen chloride.

2. Oxoacids – contain oxygen in addition to hydrogen and another element a. Yields H+ and a polyatomic oxoanion when dissolved in H2O b. Names related to the names of the corresponding oxoanions 1) Replace suffix ite with ous acid 2) Replace suffix ate with ic acid c. Textbook Table 4.4 4.6. Oxidation-Reduction (Redox) Reactions6 A. Oxidation and Reduction 1. Oxidation a. Loss of one or more electrons b. Increase in oxidation number 5

Test Item File Questions: Multiple Choice 38-55, 111-114, 119-120; Algorithmic 10-18; Short Answer 10-13 6 Test Item File Questions: Multiple Choice 56-61; Algorithmic 19-21; Short Answer 14-15 34

Chapter 4—Reactions in Aqueous Solution 2. Reduction a. Gain of one or more electrons b. Decrease in oxidation number 3. Two processes occur simultaneously B. Redox Reaction 1. Electrons transferred from one substance to another.

2. Number of electrons lost by substance being oxidized = number of electrons gained by substance being reduced C. Oxidation Number 1. Means of determining whether an atom is neutral, electron-rich, or electron-poor 2. Does not necessarily imply ionic charges 3. Rules for assigning oxidation number 4.7. Identifying Redox Reactions7 A. Oxidation and reduction – occur together 1. One atom loses one or more electrons (is oxidized) 2. Another atom must gain those electrons (be reduced) B. Reducing Agent – substance that causes reduction to occur 1. Loses one or more electrons a. Undergoes oxidation b. Oxidation number of atom increases 2. Metals act as reducing agents C. Oxidizing Agent – the substance that causes oxidation to occur 1. Gains one or more electrons a. Undergoes reduction b. Oxidation number of atom decreases 2. Reactive nonmetals act as oxidizing agents 4.8. The Activity Series of the Elements8 A. Simple redox process 1. Reaction of an aqueous cation with a free element to produce a different ion and a different element 2. Process dependent on ease of oxidation of the element and ease of reduction of the cation B. Activity Series – ranks elements in order of reducing ability in aqueous solution (textbook Table 4.5) 1. Any element higher in the activity series will react with the ion of any element lower in the activity series + 2. Position of hydrogen – indicates which metals will react with H (aq) to produce H2(g) 3. Most reactive metals – top of the activity series 4. Least reactive metals – bottom of the activity series 4.9. Balancing Redox Reactions: Half-Reaction Method9 A. Half-reaction method of balancing redox reactions – focuses on transfer of electrons B. Key – overall reaction broken into two parts (half-reactions) 1. Oxidation part of the reaction 2. Reduction part of the reaction C. Steps – textbook Figure 4.4 4.10. Redox Stoichiometry10 A. Redox titration – method for determining concentration of an oxidizing or reducing agent in solution 1. Unknown must react in 100% yield 2. Color change should signal the end of reaction 7

Test Item File Questions: Multiple Choice 62-69; Algorithmic 22; Short Answer 16-17 Test Item File Questions: Multiple Choice 70-81, 121-122; Algorithmic 23; Short Answer 18-21 9 Test Item File Questions: Multiple Choice 82-95; Algorithmic 24; Short Answer 22-25 10 Test Item File Questions: Multiple Choice 96-102, 115-117; Algorithmic 25-29; Short Answer 26 35 8

Chapter 4—Reactions in Aqueous Solution B. Strategy for redox titrations – textbook Figure 4.5: titration of KMnO4 with H2C2O4 1. Measure a known amount of one substance 2. Using the mole ratio of the balanced equation, determine number of moles of a second substrate 3. Volume of solution containing the molar amount of a second substance measured during titration 4. Determine molar concentration of the second substance 4.11. Some Applications of Redox Reactions A. Redox reactions 1. Involve almost every element in the periodic table 2. Occur in a vast number of processes B. Examples of redox reactions 1. Combustion 2. Bleaching 3. Batteries 4. Metallurgy 5. Corrosion 6. Respiration

36

Chapter 4—Reactions in Aqueous Solution

CHAPTER 4 REACTIONS IN AQUEOUS SOLUTIONS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 4.2

Relating properties to structure is an important skill. Later in this text it will be shown that species that are soluble in water are either ionic or polar, like water, as indicated in the rule of thumb “Like dissolves like”. Showing that water, sucrose, methyl alcohol, and ethyl alcohol all contain –OH groups will demonstrate their “likeness”.

Section 4.2

Dissociation is the fragmentation of a neutral species into ionic species.

Section 4.2

Strong electrolytes are usually soluble ionic compounds or strong acids. Weak electrolytes are typically weak acids or weak bases. Common nonelectrolytes are water, alcohols, and sugars.

Section 4.2

Note that HCl, HBr, and HI are strong electrolytes, but HF is a weak electrolyte.

This is an example of how the properties of the first element in a group (fluorine) differ from those of the other members of the group.

Section 4.3

A molecular equation shows complete formulas of reactants and products; an ionic equation shows which of these dissociate into ions; a net ionic equation shows only those ions and molecules that react.

Section 4.3

While not included in the net ionic equation, spectator ions are present in solution and must be included in the computation of the molar mass of the compound dissolved.

Section 4.4

A functional definition of insoluble is that some of the compound remains in the solid state when about 10–3 mole of the compound is placed in contact with one liter of water. It does not mean that none of the compound dissolves.

Section 4.4

Hg22+ is a polyatomic ion with Hg in the +1 oxidation state.

Section 4.5

All compounds that contain H are not acids. NH3 is a base and CH4 is neither an acid nor a base. All compounds that contain OH are not bases. SO2(OH)2 (usually written as H2SO4) is an acid

Section 4.5

H+(aq) and H3O+(aq) are both representations of a proton dissolved in water.

Aquated proton is actually a complex mixture of hydrates, including H9O4+.

Section 4.5

Acidity is a function of both the solute and the solvent. HCl is a strong acid in water, a weak acid in glacial acetic acid, and remains undissociated in benzene.

Arrhenius theory pertains to aqueous solutions.

Section 4.5

Acids with one, two, and three ionizable protons are called monoprotic, diprotic, and triprotic, respectively.

Section 4.5

Ammonium hydroxide and aqueous ammonia are the same substance: NH3(g) + H2O(l) ⇌ NH4+(aq) + OH–( aq). There is no evidence for the undissociated species NH4OH. 37

Chapter 4—Reactions in Aqueous Solution

Section 4.5

Due to economic considerations NaOH and KOH are the only two common strong bases with group 1A metal ions. The others (LiOH, RbOH, CsOH) are equally strong bases.

Section 4.6

Two commonly used redox reaction memorization aids are: 1) LEO the lion goes GER (Lose Electrons Oxidation, Gain Electrons Reduction); and OIL RIG (Oxidation Involves Loss of electrons, Reduction Involves Gain of electrons).

Section 4.6

Oxidation numbers of atoms in a compound are the charges the atoms would have if the compound were ionic.

Section 4.6

Oxygen has an oxidation number of –2 except when it is bonded to itself. The oxidation number of oxygen is 0 in O2, 1/2 on O2– (superoxide), and –1 in O22– (peroxide).

Section 4.6

A useful rule of thumb: For any main group element, maximum oxidation number = group number; minimum oxidation number = group number – 8.

Section 4.7

Species to the left and bottom of the periodic table tend to lose electrons and are good reducing agents; species at the top and right of the periodic table tend to gain electrons and are good oxidizing agents.

Section 4.8

The activity series can be roughly divided into three groups based on periodic table location: 1) active metals from Groups 1A, 2A, and Al; 2) lighter transition metals (Cr, Mn, Fe, Co, Ni, Zn), Sn and Pb; 3) Cu and heavier transition metals (Ag, Pt, Au, Hg).

Section 4.8

A memorization aid: the elements below hydrogen in the periodic table form an upside-down T on the periodic table: Cu Ag Pt Au Hg

Section 4.9

The most common error that students make when balancing redox reactions by the half-reaction method is to miscount charges, which results in an incorrect number of electrons used to balance the half-reaction.

Section 4.9

Balancing a redox equation by the half-reaction method will reveal that a halfreaction is an oxidation when electrons appear on the right and a reduction when electrons appear on the left of the arrow.

Section 4.9

The number of electrons used to balance the half-reaction should agree with the change in oxidation number of the element undergoing oxidation or reduction.

Lecture/Laboratory Demonstration References Section 4.2

Bassam Z. Shakhashiri, “Conductivity and Extent of Dissociation of Acids in Aqueous Solution,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.140-145.

Universal indicator and a conductivity probe are used to explore the relative acidity and conductivity of a series of aqueous acids.

38

Chapter 4—Reactions in Aqueous Solution

Section 4.4

Lee R. Summerlin, Christie. L. Borgford, and Julie. B. Ealy, “Solubility of Some Silver Compounds,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 83-85. The solubility of a series of silver salts and complexes is explored in this demonstration.

Section 4.4

L. R. Summerlin, C. L. Borgford, and J. B. Ealy, “Name That Precipitate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 121-123. Blue, brown, red, white, and yellow precipitates are formed when solutions containing +2 cations of mercury , nickel, lead, and barium are mixed with solutions containing carbonate, chloride, iodide, and sulfate anions. Students are asked to name and determine the formula of each precipitate and to write an ionic equation for each reaction.

Section 4.6

C. E. Ophardt, “Redox Demonstrations and Descriptive Chemistry: Part I. Metals,” J. Chem. Educ., Vol. 64, 1987, 716. Redox reactions of iron(III) with thiosulfate, iron(II) with permanganate, and tin(II) with mercury(I) are used to show how an abbreviated table of standard reduction potentials is used to predict the products of these reactions from the relative positions of the oxidizing agents and reducing agents in the table.

Section 4.6

Lee R. Summerlin and James. L. Ealy, Jr., “Oxidation States of Manganese: Mn7+, Mn6+, Mn4+, and Mn2+,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 133-134.

Section 4.8

Bassam Z. Shakhashiri, “Reactions of Metals and Hydrochloric Acid,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983) pp. 25-26. Reactions of iron, zinc, and magnesium with hydrochloric acid to produce aqueous solutions of the +2-metal ions and hydrogen gas are shown to be exothermic. Observed temperature changes are related to enthalpies of reaction.

Section 4.8

Bassam Z. Shakhashiri, “An Activity Series: Zinc, Copper, and Silver Half Cells,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992) pp.101-106.

Section 4.8

Lee R. Summerlin and James L. Ealy, Jr., “Activity Series for Some Metals,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 150. An overhead projector demonstration showing hydrogen gas generated by a redox reaction.

Section 4.8

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Producing Hydrogen Gas from Calcium Metal,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 51-52.

Literature References Section 4.2

John J. Fortman, “Pictorial Analogies X: Solutions of Electrolytes,” J. Chem. Educ., Vol. 71, 1994, 27-28.

Section 4.2

Albert Kowalak, “When Is a Strong Electrolyte Strong?” J. Chem. Educ., Vol. 65, 1988, 607.

39

Chapter 4—Reactions in Aqueous Solution

Section 4.3

Betty J. Wruck, “Reinforcing Net Ionic Equation Writing,” J. Chem. Educ., Vol. 73, 1996, 149-150.

Section 4.4

Richard A. Kjonas, “An Analogy for Solubility: Marbles and Magnets,” J. Chem.

Educ., Vol. 61, 1984, 765.

Section 4.5

H. van Lubeck, “Significance, Concentration Calculations, Weak and Strong Acids,” J. Chem. Educ., Vol. 60, 1983, 189.

Section 4.6

Gion Calzaferri, “Oxidation Numbers,” J. Chem. Educ., Vol. 76, 1999, 362-363.

Section 4.9

Marten J. ten Hoor, “Redox Balancing Without Puzzling,” J. Chem. Educ., Vol. 74, 1997, 1368-1376.

Media References Section 4.2

Electrolytes and Nonelectrolytes movie from the Instructor Resource DVD

Section 4.2

Strong and Weak Electrolytes movie from the Instructor Resource DVD

Section 4.2

Dissolution of NaCl in Water movie from the Instructor Resource DVD

Section 4.2

Dissolution of KMnO4 in Water movie from the Instructor Resource DVD

Section 4.3

Net Ionic Equation activity from the Instructor Resource DVD

Section 4.4

Precipitation Reactions movie from the Instructor Resource DVD

Section 4.5

Introduction to Aqueous Acids movie from the Instructor Resource DVD

Section 4.5

Introduction to Aqueous Bases movie from the Instructor Resource DVD

Section 4.6

Oxidation-Reduction Reactions – Part 1 movie from the Instructor Resource DVD

Section 4.7

Oxidation-Reduction Reactions – Part 2 movie from the Instructor Resource DVD

Section 4.7

Redox Chemistry of Tin and Zinc movie from the Instructor Resource DVD

Section 4.8

Formation of Silver Crystals movie from the Instructor Resource DVD

Section 4.8

Precipitation, Redox, and Neutralization Reactions activity from the Instructor Resource DVD

Section 4.9

Stepwise Balancing of Redox Equations – Acidic activity from the Instructor Resource DVD

40

CHAPTER 5 PERIODICITY AND THE ELECTRONIC STRUCTURE OF ATOMS Chapter Learning Goals for Students Section 5.1

Interconvert wavelength, frequency, and energy of electromagnetic radiation.

Section 5.2

Using the Balmer-Rydberg equation, calculate the wavelength and energy of a photon absorbed or released when an electron changes orbitals.

Section 5.3

Interconvert the amount of energy associated with a quantum of radiant energy, frequency, and wavelength.

Section 5.4

Using the de Broglie equation, calculate the wavelength of a moving object knowing its mass and velocity.

Section 5.5

Determine the uncertainty in the position of an object given the uncertainty in its momentum and vice versa.

Section 5.6

Use quantum numbers to identify an orbital.

Assign quantum numbers to an orbital.

Section 5.7

Sketch and name each of the s, p, and d orbitals.

Determine numbers of nodal planes present in an orbital.

Section 5.8

Calculate the energy difference between two orbitals given the wavelength of absorption or emission of radiation required for an electron to undergo a transition from one orbital to another.

Calculate the energy required to remove an electron from hydrogen.

Section 5.9

State the Pauli Exclusion Principle, Hund’s Rule and the Aufbau Principle.

Section 5.10

Explain what is meant by effective nuclear charge, Zeff.

Section 5.11

Predict ground-state electron configurations for elements; use orbital-filling diagrams to determine the number of unpaired electrons in these species.

Section 5.12

Predict which elements might be expected to have anomalous electron configurations.

Predict the anomalous electron configurations of elements.

Section 5.13

Write the general valence-shell electron configuration for each group of the periodic table, and identify the blocks in which the elements are located.

Section 5.14

Given a set of atoms, determine which atom is expected to have the largest radius.

41

Chapter 5—Periodicity and the Electronic Structure of Atoms

Lecture Outline 5.1. Light and the Electromagnetic Spectrum1 A. Electromagnetic radiation – forms of radiant energy (light in all its varied forms) B. Electromagnetic spectrum – a continuous range of wavelengths and frequencies of all forms of electromagnetic radiation C. Radiant energy – has wavelike properties 1. Frequency (ν) – the number of peaks (maxima) that pass by a fixed point per unit time (s–1 or Hz) 2. Wavelength (λ) – the length from one wave maximum to the next 3. Amplitude – the height measured from the middle point between peak and trough (maximum and minimum) 2 4. Intensity of radiant energy is proportional to amplitude D. Speed of light (c) – rate of travel of all electromagnetic radiation in a vacuum 8 1. c = 3.00 × 10 m/s 2. Wavelength × Frequency = Speed λ(m) × ν (s–1) = c (m/s) 3. Frequency and wavelength inversely related: c c λ = or ν = ν λ a. long λ; low ν b. short λ; high ν 5.2. Electromagnetic Radiation and Atomic Line Spectra2 A. Individual energetically-excited atoms emit light 1. Provides clue to atomic makeup 2. Consists of only a few λ 3. Line spectrum – series of discrete lines (or wavelengths) separated by blank areas 4. Each element has its own unique line spectrum B. Balmer – pattern in atomic line spectra for the hydrogen atom 1. All four lines in the hydrogen spectrum expressed by ⎛1 1 1⎞ = R⎜ 2 − 2 ⎟ λ n ⎠ ⎝2 2. R (Rydberg constant) = 1.097 ×10–2 nm–1 3. Rydberg – every line in the entire hydrogen spectrum fits the generalized Balmer-Rydberg equation ⎛1 1⎞ 1 = R ⎜ 2 − 2 ⎟ where n > m λ ⎝2 n ⎠ 5.3. Particlelike Properties of Electromagnetic Energy3 A. Blackbody radiation – the visible glow emitted by heated solid objects 1. Intensity does not rise indefinitely as λ decreases 2. Planck – energy radiated by a heated object is quantized a. Radiant energy emitted in discrete units or quanta hc b. E = hν ; E = ; h = 6.626 x 10–34 J⋅s (Planck’s constant)

λ

c. High-energy radiation – higher ν, shorter λ d. Low-energy radiation – lower ν, longer λ 5.4. Wavelike Properties of Matter4 1

Test Item File Questions: Multiple Choice 1-7, 77-81; Short Answer 1-6 Test Item File Questions: Multiple Choice 8-11; Short Answer 7-8 3 Test Item File Questions: Multiple Choice 12-15; Short Answer 9-11 42 2

Chapter 3—Periodicity and the Electronic Structure of Atoms A. Einstein – relationship between mass and λ E hc / λ h m= 2 = = λc c c2 B. de Broglie – matter can behave in some respects like light 1. Both light and matter wavelike as well as particlelike 2. Relationship between λ of an electron or of any other particle or object of mass m moving at velocity v h h m= ; λ= mv λv C. Dual wave/particle description of light and matter – mathematical model that accounts for atomic properties and behavior 5.5. Quantum Mechanics and the Heisenberg Uncertainty Principle5 A. Bohr 1. Hydrogen atom contains an electron circling the nucleus 2. Specific orbits of the electron correspond to specific energy levels B. Schrödinger 1. Quantum mechanical model of the atom 2. Abandoned idea of an electron as a small particle moving around the nucleus in a defined path 3. Concentrated on the electron’s wavelike properties C. Heisenberg Uncertainty Principle – both position (Δx) and momentum (Δmv) of an electron cannot be known beyond a certain level of precision h ( Δx )( Δmv) ≥ 4π 1. Cannot know both the position and the momentum of an electron with a high degree of certainty 2. If momentum is known with a high degree of certainty, a. Δmv is small b. Δ x (position of the electron) is large 3. If the exact position of the electron is known, a. Δx is small b. Δmv is large 5.6. Wave Functions and Quantum Numbers6 A. Quantum mechanical model of atomic structure 1. Mathematical form is a wave equation 2. Wave function (ψ) or orbital – solution to wave equation a. Has a specific energy b. Contains information about an electron’s position in 3-D space c. ψ2 – gives the probability of finding an electron within a given region in space d. Defines a volume of space around the nucleus where there is a high probability of finding an electron e. Says nothing about the electron’s path or movement B. Wave function – contains a set of three variables, quantum numbers 1. Describes the energy level of an orbital 2. Defines the shape and orientation of the region in space where the electron is most likely to be found C. Principal quantum number (n) – describes the size and energy level of the orbital 1. A positive integer (n = 1,2,3,4,…)

4

Test Item File Questions: Multiple Choice 16-20; Short Answer 12 Test Item File Questions: Multiple Choice 21-24 6 Test Item File Questions: Multiple Choice 25-28, 86-89; Algorithmic 1-4; Short Answer 13-17 43 5

Chapter 5—Periodicity and the Electronic Structure of Atoms 2. As the value of n increases, a. number of allowed orbitals increases b. size of the orbitals increases c. energy of the electron in the orbital increases 3. Shell – grouping of orbitals according to the principal quantum number D. Angular-momentum quantum number (l) – defines the 3-D shape of the orbital 1. Integral value from 0 to n – 1 2. Within each shell, n different shapes for orbitals 3. Subshells – grouping of orbitals according to the angular-momentum quantum number 4. Referred to by letter rather than by number a. quantum number l: 0 1 2 3 4 ... b. subshell notation: s p d f g ...

E. Magnetic quantum number (ml) – defines the spatial orientation of the orbital along a standard set of coordinate axes 1. Integral values from – l to +l 2. Within each subshell (same n and same l), 2 l + 1 different spatial orientations F. Energy level of various orbitals – textbook Figure 5.8 1. Hydrogen – energy levels depend only on n 2. Multielectron atoms – energy levels depend on both n and l 5.7. The Shapes of Orbitals7 A. s orbitals – textbook Figure 5.9 1. Spherical 2. Probability of finding an electron – depends only on electron from the nucleus 3. Differences among s orbitals in different shells a. Size increases in successively higher shells b. Electron distribution in outer s orbitals – several different regions of maximum probability separated by a node 4. Node – a surface of zero probability a. Intrinsic property of a wave – zero amplitude at node (textbook Figure 3.10) B. p orbitals – textbook Figure 5.11 1. Dumbbell shaped 2. Electron distribution concentrated in identical lobes on either side of the nucleus 3. Nodal plane cuts through the nucleus 4. Probability of finding a p electron near the nucleus is zero 5. Have different phases (mathematical signs) a. Crucial for bonding b. Only lobes with same phase interact to form covalent bonds 6. Three p orbitals oriented along the x-, y-, and z-axes. (px, py, pz) C. d and f orbitals 1. d orbitals – textbook Figure 5.12 a. Four (dxy , dxz, dyz, dx2– y2) – cloverleaf shaped i. Four lobes of maximum electron probability separated by two nodal planes through the nucleus b. dz2 – similar in shape to a pz orbital with additional donut-shaped region of electron probability in the xy plane c. Alternating lobes have different phases 2. f orbitals – eight lobes of maximum electron probability a. Three nodal planes through the nucleus 5.8. Quantum Mechanics and Atomic Line Spectra8 A. Electron in an atom 1. Occupies an orbital 7 8

Test Item File Questions: Multiple Choice 39-44, 81-85, Short Answer 18-20 Test Item File Questions: Multiple Choice 45-52; Algorithmic 5-6; Short Answer 21 44

Chapter 3—Periodicity and the Electronic Structure of Atoms 2. Each orbital has a specific energy 3. Electron energies quantized – have only specific energy values associated with the orbital B. Addition of energy to an atom causes an electron to jump from a lower-energy orbital to a higher-energy orbital.

1. Atom excited 2. Excited atom unstable a. Electron returns to a lower-energy level by emission of amount of energy equal to the energy difference between the higher and lower-energy orbitals b. Because orbital energies quantized, the energy of emitted light also quantized 3. Calculate energy differences between orbitals by measuring the frequencies emitted C. Balmer-Rydberg equation – variables m and n correspond to the principal quantum numbers of the two orbitals involved in the transition 1. n – principal quantum number of the outer shell (orbital the transition is from) 2. m – principal quantum number of the inner shell (orbital the transition is to) 5.9. Electron Spin and the Pauli Exclusion Principle9 A. Fourth quantum number – ms 1. Related to property called electron spin 2. Electron behaves as if spinning around an axis a. Gives rise to a tiny magnetic field 3. Has two values: +1/2 ( ↑ ) or –1/2 ( ↓ ) 4. Independent of other quantum numbers B. Pauli Exclusion Principle 1. No two electrons in an atom can have same four quantum numbers 2. Only two electrons with opposite spins per orbital 5.10. Orbital Energy Levels in Multielectron Atoms10 A. Energy level of an orbital in multielectron – depends on both n and l.

1. Energy difference due to electron-electron repulsions 2. Outer-shell electrons a. Pushed farther from nucleus b. Held less tightly 3. Partially cancels electron-nucleus attractions 4. Electrons shielded from nucleus by the other electrons.

B. Effective nuclear charge, Zeff – net nuclear charge actually felt by an electron 1. Zeff = Z – electron shielding 2. Lower than the actual nuclear charge 3. For the same shell a. Lower value of l corresponds to higher value of Zeff b. Corresponds to lower energy for the orbital 4. Useful for explaining various chemical phenomena 5.11. Electron Configurations of Multielectron Atoms11 A. Electron configuration – describes the orbitals occupied by electrons in an atom B. Aufbau principle 1. Fill lowest-energy orbitals first 2. Only two electrons with opposite spin per orbital 3. Follow Hund’s rule: if two or more orbitals with same energy available, one electron with parallel spin in each until all are half full C. Ground-state configuration – lowest-energy electron configuration D. Degenerate orbitals – orbitals with the same energy level E. Orbital-filling diagrams – electrons are represented by arrows 5.12. Some Anomalous Electron Configurations12 9

Test Item File Questions: Multiple Choice 53, 102 Test Item File Questions: Multiple Choice 54-57; Algorithmic 7 11 Test Item File Questions: Multiple Choice 58-59, 103-105 45 10

Chapter 5—Periodicity and the Electronic Structure of Atoms A. Unusual stability when subshells half-filled or filled – leads to anomalies in electron configurations B. Anomalies 1. Occur when energy differences between subshells small 2. Transfer of an electron from one subshell to another a. Lowers total energy of the atom b. Due to decrease in electron-electron repulsions c. More than 40 anomalies 5.13. Electron Configurations and the Periodic Table13 A. Valence-shell electrons – outermost shell of electrons 1. Elements in each group of the periodic table have similar valence-shell electron configurations.

2. Most loosely held 3. Determine an element’s properties B. Similar electron configurations explain why the elements in a given group have similar chemical behavior.

C. Blocks of elements in the periodic table – depends upon the valence orbitals being filled 1. s-block elements – Groups 1A and 2A (filling of an s orbital) 2. p-block elements – Groups 3A through 8A (filling of p orbitals; ns orbitals are filled) 3. d-block elements – transition metals (filling of (n – 1)d orbitals) 4. f-block elements – lanthanide and actinide elements (filling of (n – 2) f orbitals) 5.14. Electron Configurations and Periodic Properties: Atomic Radii14 A. Radius of an atom – half the distance between nuclei of two identical atoms when covalently bonded together B. Atomic radius 1. Increases down a group 2. Successively larger valence-shell orbitals occupied C. Atomic radius 1. Decreases across a period 2. Due to increase in Zeff for valence-shell electrons across a period a. Value of Zeff – dependent upon the amount of shielding felt by an electron b. Amount of shielding – depends on both the shell and subshell of other electrons c. Valence-shell electrons i. Strongly shielded by electrons in inner shells ii. Less strongly shielded by electrons in same shell – order: s > p > d > f iii. Weakly shielded by electrons in same subshell

12

Test Item File Questions: Multiple Choice 60-61, 106; Short Answer 22 Test Item File Questions: Multiple Choice 62-73, 90-101, 107-110; Algorithmic 8-12; Short Answer 23-24 14 Test Item File Questions: Multiple Choice 74-76, 111-118; Algorithmic 13-15; Short Answer 25 46 13

Chapter 3—Periodicity and the Electronic Structure of Atoms

CHAPTER 5 PERIODICITY AND THE ELECTRONIC STRUCTURE OF ATOMS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 5.1

Radiation refers to any form of energy that travels in all directions from a single source, such as light from the sun.

Section 5.1

Many students associate radiation only with nuclear decay processes and their negative connotations. This form of radiation represents only a small segment of the electromagnetic radiation spectrum.

Section 5.1

Wavelength can be related to colors of the spectrum ranging from 400 nm (violet) to 700 nm (red) using the acronym ROY G BIV (Red, Orange, Yellow, Green, Blue, Indigo, Violet).

Section 5.1

Students best relate some wave properties to radio waves: the frequency of a favorite radio station and loudness (amplitude).

Section 5.3

Wave-particle duality does not mean that energy is sometimes a wave and other times a particle; it means we must use both wave and particle theories to describe energy.

Section 5.3

A photon is a quantum of light energy.

Section 5.6

Quantum number n is related to the size and energy of an orbital. The larger the value of n, the larger the orbital and the higher the energy of an electron occupying the orbital.

Section 5.6

Quantum number l defines orbital shape. Some students believe that the shape of an orbital shows how the electron moves about the atom. An orbital defines the region in which the electron moves with an indeterminable motion.

Section 5.6

Quantum number ml defines the different orientations an orbital of a given shape may assume. This gives the number of orbitals of a given type: s orbital (one), p orbital (three), d orbital (five), f orbital (seven).

Section 5.6

For neutral atoms 4s < 3d but for cations 3d < 4s in energy.

Section 5.7

An electron’s orbital is a wave theory concept. Since a wave exists simultaneously on both sides of a node, an electron does not “move” from one lobe of an orbital to another according to this mathematical model.

Section 5.8

The closer an electron is brought to the nucleus of an atom, the greater the energy of the photon emitted and the lower the energy of the electron.

Section 5.8

The Planck equation is for a single photon or single electron energy. On a macroscopic scale it is useful to convert to energy per mole of photons or electrons by multiplying by Avogadro’s number.

Section 5.9

If a beam of atoms having one unpaired electron (such as atoms of silver) pass through a magnetic field, half are attracted to one pole and half to the other pole of the magnetic field.

47

Chapter 5—Periodicity and the Electronic Structure of Atoms

Section 5.9

Electrons are said to be paired on an atom or ion if there are the same number of electrons with ms = + 1/2 as with ms = – 1/2; any electron with ms = + 1/2 without a corresponding electron with ms = – 1/2 is said to be unpaired. Experimentally, atoms and ions that have only paired electrons are repelled from a magnetic field and are called diamagnetic; atoms and ions with one or more unpaired electrons are attracted to a magnetic field and are called paramagnetic.

Section 5.11

Electron orbital-filling diagrams explicitly show the number of unpaired electrons; electron configurations do not. Students who ignore the use of orbital-filling diagrams are prone to say that all electrons are paired on an atom with an electron configuration such as 1s22s22p2.

Section 5.14

Core electron configuration does have some effect on the chemistry of an element.

Due to filling of the d orbitals , p-block elements in period 4 have an additional 10 core electrons relative to elements in period 3. This influences some properties, such as atomic radius. A similar and more dramatic effect, called the lanthanide contraction, caused by f orbital filling, occurs for period 6 d-block and p-block elements.

Lecture/Laboratory Demonstration References Section 5.2

N. C. Thomas and R. Brown, “A Spectacular Demonstration of Flame Tests,” J. Chem. Educ., Vol. 69, 1992, 326. Brightly-colored flashes of light resulting from the ignition of cellulose nitrate impregnated with salts of sodium, potassium, strontium, and barium ions demonstrate light emission from these energetically excited ions.

Literature References Section 5.3

Robert R. Perkins, “Put Body to Them,” J. Chem. Educ., Vol. 72, 1995, 151-152.

Analogies for quantized states.

Section 5.5

Bianca L. Haendler, “Presenting the Bohr Atom,” J. Chem. Educ., Vol. 59, 1982, 372-376.

Section 5.5

Oliver G. Ludwig, “On a Relation Between the Heisenberg and de Broglie Principles,” J. Chem. Educ., Vol. 70, 1993, 28.

Section 5.5

Pedro L. Muino, “Introducing the Uncertainty Principle Using Diffraction of Light Waves,” J. Chem. Educ., Vol. 77, 2000, 1025-1027.

Section 5.5

Lawrence S. Bartell, “Perspectives on the Uncertainty Principle and Quantum Reality,” J. Chem. Educ., Vol. 62, 1985, 192-196.

Section 5.6

Ngai Ling Ma, “Quantum Analogies on Campus,” J. Chem. Educ., Vol. 73, 1996, 1016-1017.

Section 5.6

Mali Yin and Raymond S. Ochs, “The Mole, The Periodic Table, and Quantum Numbers: An Introductory Trio,” J. Chem. Educ., Vol. 78, 2001, 1345-1347.

48

Chapter 3—Periodicity and the Electronic Structure of Atoms

Section 5.6

Geoff Raynner-Canham, “A Student's Travels, Close Dancing, Bathtubs, and the Shopping Mall: More Analogies in Teaching Introductory Chemistry,” J. Chem.

Educ., Vol. 71, 1994, 943-944.

Section 5.6

John J. Fortman, “Pictorial Analogies VII: Quantum Numbers and Orbitals,” J. Chem. Educ., Vol. 7o, 1993, 649-650.

Section 5.8

Elvin Hughes, Jr., and Arnold George, “Suitable Light Sources and Spectroscopes for Student Observation of Emission Spectra in Lecture Halls,” J. Chem. Educ., Vol. 61, 1984, 908-909.

Section 5.11

Robert D. Freeman, “New Schemes for Applying the Aufbau Principle,” J. Chem.

Educ., Vol. 67, 1990, 576.

Section 5.11

James R. Hanley, III, and James R. Hanley, Jr., “A Low-Cost Classroom Demonstration of the Aufbau Principle,” J. Chem. Educ., Vol. 56, 1979, 747.

Section 5.11

Ronald J. Gillespie, James N. Spencer, and Richard S. Moog, “Demystifying Introductory Chemistry; Part 1. Electron Configurations from Experiment,” J. Chem. Educ., Vol. 73, 1996, 617-622.

Section 5.11

Anthony Garofalo, “Housing Electrons: Relating Quantum Numbers, Energy Levels, and Electron Configurations,” J. Chem. Educ., Vol. 74, 1997, 709-719.

Section 5.11

M. Bonneau, “The Quantum Shoe Store and Electron Structure,” J. Chem. Educ., Vol. 68, 1991, 837.

Section 5.12

Peter Cann, “Ionization Energies, Parallel Spins, and the Stability of Half-Filled Shells,” J. Chem. Educ., Vol. 77, 2000, 1056-1061.

Section 5.13

Judith A.Strong, “The Periodic Table and Electron Configurations,” J. Chem. Educ., Vol. 63, 1986, 834.

Media References Section 5.2

Flame Tests for Metals movie from the Instructor Resource DVD

Section 5.3

The Photoelectric Effect movie from the Instructor Resource DVD

Section 5.7

Shapes of Orbitals activity from the Instructor Resource DVD

Section 5.7

Radial Electron Distribution movie from the Instructor Resource DVD

Section 5.10

Line Spectrum of Sodium activity from the Instructor Resource DVD

Section 5.10

Effective Nuclear Charge movie from the Instructor Resource DVD

Section 5.11

Electron Configuration movie from the Instructor Resource DVD

Section 5.14

Periodic Trends: Atomic Radii movie from the Instructor Resource DVD

49

CHAPTER 6 IONIC BONDS AND SOME MAIN-GROUP CHEMISTRY Chapter Learning Goals for Students Section 6.1

Predict the ground-state electron configuration for ions.

Section 6.2

Given a set of atoms and ions, determine which atom or ion is expected to have the largest radius.

Section 6.3

For any two elements predict which has the higher first ionization energy.

Section 6.4

For any two elements, predict which has the higher second, third, fourth, etc. ionization energy.

Section 6.5

For any two elements, predict which has the more favorable (more negative) electron affinity.

Section 6.6

Identify the energies involved in a Born-Haber calculation of lattice energy. Know whether these energies are positive or negative, large or small.

Given all other energies, use the Born-Haber cycle to calculate the remaining energy in the cycle.

Section 6.7

On the basis of ionic charges and ionic radii, predict which of two ionic compounds should have the greater lattice energy.

Section 6.8

Use the octet rule to generalize the chemistry of each family of elements studied in this chapter.

Know when to expect the octet rule to be valid and when it can fail.

Give the noble gas configuration of cations and anions in ionic compounds.

Use the octet rule to determine the number of electrons gained or lost by a given element in a redox reaction.

Section 6.9

Know which alkali metals form 1) oxides, 2) peroxides, and 3) superoxides upon reaction with oxygen. Assign oxidation numbers to the oxygen atoms in these compounds.

Give the formulas of products formed when alkali metals react with halogens, hydrogen, nitrogen, oxygen, water, or ammonia. Balance the equations.

Section 6.10

Know which alkaline earth metals form 1) oxides, 2) peroxides, and 3) superoxides upon reaction with oxygen. Assign oxidation numbers to the oxygen atoms in these compounds.

Give the formulas of products formed when alkaline earth metals react with halogens, hydrogen, oxygen, or water. Balance the equations.

Section 6.11

Give the formulas of products formed when aluminum reacts with halogens, nitrogen, oxygen, acids, or bases. Balance the equations.

50

Chapter 6 — Ionic Bonds and Some Main–Group Chemistry

Section 6.12

Give the formulas of products formed when halogens react with metals, hydrogen, or other halogens.

Section 6.13

Give the formulas of products formed when noble gases react with metals, hydrogen, or other halogens.

Assign oxidation numbers to elements in noble gas compounds.

Balance redox equations representing reactions studied throughout this chapter. Identify which species are oxidized and which are reduced and which elements have undergone a change in oxidation state.

51

Chapter 6—Ionic Bonds and Some Main–Group Chemistry

Lecture Outline 6.1. Electron Configurations of Ions1 A. Main-group elements 1. Electrons lost by a metal come from the highest-energy occupied orbital 2. Electrons gained by a nonmetal go into lowest-energy unoccupied orbital B. Transition metals – lose valence-shell s electrons first, then d electrons 6.2. Ionic Radii2 A. Cations have smaller radii than neutral atoms 1. Electrons removed from larger valence-shell orbitals 2. Increase in Zeff when electrons removed B. Anions have larger radii than neutral atoms 1. Decrease in Zeff when electrons added 2. Increase in electron-electron repulsions 6.3. Ionization Energy3 A. Ionization energy (Ei) – amount of energy required to remove the outermost electron from an isolated neutral atom in the gaseous state B. Periodic trends in ionization energy 1. Minimum Ei – group 1A alkali metals 2. Maximum Ei – group 8A noble gases 3. Ei increases across a period 4. Ei decreases down a group C. Periodicity due to electron configurations 1. Single s electron in valence-shell of alkali metals feels low Zeff – a. Single valence electron well shielded by the core (inner shell) electrons b. Electron loosely held. c. Low energy needed to remove electron 2. Electrons in filled valence-subshell of noble gas elements feel high Zeff a. Electrons tightly held b. Radius of atom shrinks c. High energy needed to remove electron 3. Increase in atomic number down a group a. Average distance of the electron from the nucleus increases as the value of n increases b. Valence-shell electrons less tightly held than core electrons c. Ei decreases D. Minor irregularities occur from left to right across a row of the periodic table 1. Group 2A elements – Ei (Be) > Ei (B) (due to electron configuration) a. 2p electron of boron shielded somewhat by 2s electrons b. Feels a smaller Zeff c. More easily removed 2. Group 6 A elements – Ei (N) > Ei (O) (due to electron configuration) a. 2p electron removed from nitrogen removed from a half-filled orbital (stable configuration) b. 2p electron removed from oxygen removed from a filled orbital i. Electrons in filled orbitals forced together and have slightly higher energy ii. Easier to remove electrons from filled orbitals 6.4. Higher Ionization Energies4 A. Energy involved with the removal of two, three, or even more electrons sequentially from an atom B. Successively larger amounts of energy required for each ionization step – more difficult to remove a negatively charged electron from a positively charged ion 1

Test Item File Questions: Multiple Choice 1-12, 90-93; Algorithmic 1-2; Short Answer 1-12 Test Item File Questions: Multiple Choice 1-18, 94-109; Short Answer 13 3 Test Item File Questions: Multiple Choice 19-26, 110-111; Algorithmic 3-7; Short Answer 14 4 Test Item File Questions: Multiple Choice 27-34, 112-113; Algorithmic 8-11; Short Answer 15-16 52 2

Chapter 6 — Ionic Bonds and Some Main–Group Chemistry C. Large jumps in successive ionization energies of the elements – textbook Table 6.2 1. Valence-shell electrons easily lost during ionization 2. Core electrons relatively difficult to remove D. Valence-shell electron configuration of an atom controls its chemistry.

6.5. Electron Affinity5 A. Electron affinity (Eea) – energy change when electron added to an isolated gaseous atom B. Eea usually has a negative value 1. Energy usually released when an atom adds an electron 2. More negative the value, a. greater the tendency of the atom to accept an electron b. more stable the anion C. Eea > 0 for atoms that form an unstable anion on addition of an electron D. No experimental measurements can be made.

E. Periodic trends of Eea related to electron configurations 1. Sign and magnitude of Eea due to offsetting factors a. Negative Eea due to attraction between the additional electron and the atomic nucleus b. Positive Eea due to repulsions between the additional electron and the electrons in the atom 2. Halogens have large, negative Eea a. High Zeff b. Room in valence-shell for an additional electron c. High attraction between additional electron and atomic nucleus 3. Noble-gas elements have positive Eea a. Filled s and p sublevels b. Additional electron goes into the next higher shell – feels a low Zeff c. Small attraction between the additional electron and the atomic nucleus d. High electron-electron repulsions 4. Alkaline earth metals have Eea ≈ 0 a. Filled s subshell b. Added electron goes into p subshell of higher energy c. Relatively low Zeff 6.6. The Octet Rule6 A. Octet rule – main-group elements tend to undergo reactions that leave them with eight valence electrons 1. Useful for making predictions; provides insights about chemical bonds 2. "Magic number" = 8 a. Filled octet – s and p subshells both filled b. Can’t take electrons away – electrons tightly held by high Zeff c. No low-energy orbitals available to accept electrons 3. Failure of rule a. Elements on right side of periodic table in third or lower period b. More than eight electrons sometimes accommodated by larger atoms B. Factors that determine cation or anion formation 1. Cations a. Electrons highly shielded from nucleus by core electrons i. Feel low Zeff ii. Easily lost b. Charge based on electronic configuration – energy required unavailable to lose another electron once next-lower noble gas configuration achieved 2. Anions a. Electrons poorly shielded from nucleus by core electrons i. Feel high Zeff 5 6

Test Item File Questions: Multiple Choice 35-42, 114; Algorithmic 12; Short Answer 17-18 Test Item File Questions: Multiple Choice 43-49, 131; Algorithmic 13-14; Short Answer 19 53

Chapter 6—Ionic Bonds and Some Main–Group Chemistry ii. Can gain electrons because of Zeff b. Charge based on electronic configuration – low-energy orbitals required for additional electrons unavailable once next-higher noble gas configuration achieved 6.7. Ionic Bonds and the Formation of Ionic Solids7 A. Element with low Ei can transfer an electron to an element with negative Eea 1. Produces a cation and an anion 2. Cation and anion mutually attracted by electrostatic forces 3. Ionic bond created 4. Three-dimensional network of ions (ionic solid) formed a. Each cation surrounded by and attracted to many anions b. Each anion surrounded by and attracted to many cations B. Ionic bond formation leads to large gain in stability 1. Overcomes unfavorable energy change of electron transfer C. Born-Haber cycle – series of hypothetical steps each of which contributes to the overall energy change during the formation of an ionic compound 1. Net process is the sum of individual steps 2. Typical individual steps a. Sublimation of a metal atom to a gaseous atom b. Dissociation of nonmetal molecules c. Ionization of isolated metal atoms d. Formation of nonmetal anions e. Formation of solid ionic compound 6.8. Lattice Energies in Ionic Solids8 A. Lattice energy (U) – sum of electrostatic interaction energies between ions in a solid 1. Refers to the breakup of a crystal into individual ions a. Positive value b. When using the lattice energy in a Born-Haber cycle must change the sign because last step is the reverse of the breakup of the crystal 2. Lattice energy mostly coulombic zz Coulomb's law: F = k × 1 22 d 3. Effect of ion charge – compounds of ions with higher charges have greater lattice energies than compounds of ions with lower charges 4. Effect of ion size – compounds of ions with smaller sizes have greater lattice energies than compounds of ions with larger sizes 2. Anions a. Electrons poorly shielded from nucleus by core electrons i. Feel high Zeff ii. Can gain electrons because of Zeff b. Charge based on electronic configuration – low-energy orbitals required for additional electrons unavailable once next-higher noble gas configuration achieved 6.9. Some Chemistry of the Alkali Metals9 A. General properties: 1. Valence-shell electron configuration – ns1 a. Common ion M+ dominates chemistry b. Lowest ionization energies of all elements c. Most powerful reducing agents in the periodic table 2. Low densities and melting points 3. Metals 7

Test Item File Questions: Multiple Choice 50-58, 115-122; Short Answer 20 Test Item File Questions: Multiple Choice 59-64, 123-124; Algorithmic 15-17 9 Test Item File Questions: Multiple Choice 65-71, 128; Short Answer 21-22 54 8

Chapter 6 — Ionic Bonds and Some Main–Group Chemistry a. Bright, silvery solids b. Malleable c. Good conductors of electricity d. Soft enough to cut with a knife 4. Very reactive a. Must be stored under oil to prevent instantaneous reaction with oxygen and water 5. Occur in nature as salts B. Produced by reaction of their chloride salts 1. Lithium and sodium produced by electrolysis – electric current is passed through molten salt 2. Potassium, rubidium, and cesium produced by chemical reduction C. Reactions of alkali metals 1. Halogens a. General reaction: 2 M(s) + X2 (g) → 2 MX(s) b. Products are halides, crystalline ionic salts c. Lower the ionization energy, greater the reactivity 2. Hydrogen a. General reaction: 2 M(s) + H2 (g) → 2 MH(s) b. Products are hydrides (oxidation number of H = –1) c. Sluggish reaction at room temperature 3. Nitrogen – only lithium reacts: 6 Li(s) + N2(g) → 2 Li3N(s) 4. Oxygen a. All react rapidly b. Form different kinds of products i. Li → oxide (oxidation number of O = –2) ii. Na → peroxide (oxidation number of O = –1 iii. K, Rb, Cs → superoxide (oxidation number O = –1/2) 5. Water a. General reaction: 2 M(s) + 2 H2O(l) → 2 M+(aq) + 2 OH–(aq) + H2(g) b. Reactivity increases down the group c. Produces an alkaline solution d. Redox process 6. Ammonia a. General reaction: 2 M(s) + 2 NH3(l) → 2 M+(soln) + 2 NH2–(soln) + H2(g) b. Analogous to reaction between metal and water c. Metal amide, MNH2, solutions have powerful reducing properties 6.10. Some Chemistry of the Alkaline Earth Metals10 A. General properties 1. Similar to alkali metals 2. Valence-shell electron configuration – ns2 a. Common ion – M2+ b. Powerful reducing agents 3. Soft, silvery metals 4. Higher melting points and densities than alkali metals 5. Less reactive toward oxygen and water than alkali metals 6. Occur in nature as salts B. Produced by reduction of their salts C. Reactions of alkaline earth metals 1. Same kinds of redox reactions as alkali metals 2. Less reactive than alkali metals – Ei (alkaline earth) > Ei (alkali) 3. Halogens → ionic halide salts, MX2 4. Oxygen → oxides, MO; Sr, Ba → peroxides, MO2 10

Test Item File Questions: Multiple Choice 72-78, 125-127; Short Answer 23 55

Chapter 6—Ionic Bonds and Some Main–Group Chemistry 5. Water → metal hydroxides, M(OH)2 6.11. Some Chemistry of the Halogens11 A. General properties 1. Exist as diatomic molecules 2. Valence–shell electron configuration – ns2np5 a. Gain electrons b. Powerful oxidizing agents c. Large negative Eea 3. Occur in nature as salts and minerals B. Produced by oxidation of their anions C. Reactions 1. Most reactive elements in the periodic table 2. Less reactive going down the periodic table 3. Metals → metal halides, MXn 4. Hydrogen → hydrogen halides, HX – behave as acids when dissolved in water 5. Other halogens → XY a. Interhalogen compounds b. Redox process – the lighter, more reactive element is the oxidizing agent and the heavier, less reactive element is the reducing agent c. Properties intermediate between parent elements d. Act as strong oxidizing agents in redox reactions 6.12. Some Chemistry of the Noble Gases12 A. General properties 1. Colorless, odorless, unreactive gases 2. Valence-shell electron configuration – ns2np6 a. Stable configuration b. Do not form cations or anions c. Normally do not undergo redox reactions d. Commercial use – applications that require inert atmospheres B. Reactions of noble gases 1. Kr and Xe react only with fluorine a. Xenon fluorides – powerful oxidizing agents 2. Lack of reactivity due to valence-shell configuration a. Large Ei b. Small Eea

11 12

Test Item File Questions: Multiple Choice 79-87, 130 Test Item File Questions: Multiple Choice 88-89, 129; Short Answer 24-25 56

Chapter 6 — Ionic Bonds and Some Main–Group Chemistry

CHAPTER 6 IONIC BONDS AND SOME MAIN-GROUP CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 6.1

Electrons are always lost from orbitals having the highest value of the principal quantum number, n, before they are lost from orbitals with lower values of n.

Section 6.2

The larger the proton-to-electron ratio, the greater the attraction of the nucleus for the valence electrons and the smaller the size of the ion.

Section 6.3

Ionization energy is defined for isolated, gaseous-state atoms. Other factors, such as lattice energy or solvation energy must be considered when forming ions in a condensed state.

Section 6.3

Core electrons shield valence electrons effectively; electrons in the same shell shield one another ineffectively.

Section 6.3

The closer an electron is to the nucleus, the higher its ionization energy. Hence, small atoms have high ionization energies. Due to its small size, the first element in very group tends to differ significantly from the other elements in the same group.

Section 6.4

A large change in two successive ionization energies signifies that a core electron is being ionized. In ordinary chemical reactions there is not enough energy available to ionize core electrons. Hence, only valence electrons are involved in chemical reactions.

Section 6.5

Electron affinity is defined for isolated, gaseous-state atoms. Other factors, such as lattice energy or solvation energy must be considered when forming ions in a condensed state.

Section 6.5

Just as there are successive ionization energies: Ei1, Ei2, Ei3,..., there are successive electron affinities: Eea1, Eea2, Eea3,....While energy is generally released on addition of one electron, energy is always required to add two or more electrons to a neutral gaseous atom.

Section 6.6

An ionic compound will form when an element from the far left-hand side combines with an element from the far right-hand side of the periodic table (a notable exception being BeCl2). Compounds encountered in a general chemistry course are usually ionic if formed from reaction of a metal with a nonmetal.

Section 6.6

Formulas for ionic compounds are empirical formulas giving the relative number of cations and anions. Although the formula for sodium chloride is NaCl, molecules of NaCl do not exist under ordinary laboratory conditions.

Section 6.6

Because the sum of ionization energy and electron affinity is always positive, obtaining noble gas electron configurations upon formation of ions is not the driving force behind the formation of an ionic compound. The ions must be stabilized by strong forces arising from electrostatic attraction.

Section 6.6

The Born-Haber cycle is an example of Hess's law, introduced in Chapter 6. It is used to calculate properties that are impossible to measure directly.

57

Chapter 6—Ionic Bonds and Some Main–Group Chemistry

Section 6.7

When an actual lattice energy is more negative than that calculated on the basis of an ionic model, some covalent bonding between ions is occurring. This typically happens when the electronegativity of the metal is high (approximately > 1.5).

Lecture/Laboratory Demonstration References Section 6.5

Lee R. Summerlin, Christie. L. Borgford, and Julie. B. Ealy, "Halogens Compete for Electrons," Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 60-61. The relative tendency of halogens to gain or lose electrons is demonstrated.

Section 6.7

Kristin A. Johnson, Rodney Schreiner, and Jon Loring, "A Dramatic Flame Test Demonstration," J. Chem. Educ., Vol. 78, 2001, 640-641.

Section 6.7

M.D. Alexander, "Reactions of the Alkali Metals with Water: A Novel Demonstration," J. Chem. Educ., Vol. 69, 1992, 418. The reaction of sodium metal with water to produce an aqueous solution of sodium hydroxide and hydrogen gas is performed at the interface between paint thinner and the more dense water. Periodically, bubbles of hydrogen gas carry the sodium metal into the organic layer, temporarily stopping the reaction. The presence of the aqueous layer is shown by a phenolphthalein indicator color change.

Section 6.8

Lee R. Summerlin, Christie L. Borgford, and Julie. B. Ealy, "Producing Hydrogen Gas from Calcium Metal," Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 51-52.

Literature References Section 6.

None

Media References Section 6.1

Ion Electron Configuration activity from the Instructor Resource DVD

Section 6.2

Gain and Loss of Electrons movie from the Instructor Resource DVD

Section 6.3

Effective Nuclear Charge movie from the Instructor Resource DVD

Section 6.3

Periodic Trends: Ionization Energies movie from the Instructor Resource DVD

Section 6.5

Electron Affinity movie from the Instructor Resource DVD

Section 6.5

Periodic Trends: Electron Affinity movie from the Instructor Resource DVD

Section 6.9

Sodium and Potassium in Water movie from the Instructor Resource DVD

Section 6.9

Formation of Aluminum Bromide movie from the Instructor Resource DVD

Section 6.11

Physical Properties of Halogens movie from the Instructor Resource DVD

58

CHAPTER 7 COVALENT BONDS AND MOLECULAR STRUCTURE Chapter Learning Goals for Students Section 7.1

Determine the H–H bond length from a plot of potential energy versus internuclear distance for the hydrogen mlecule.

Use atomic radii to estimate bond length.

Section 7.2

Define bond dissociation energy, D. Give a range of bond dissociation energies for common bonds.

Section 7.3

From a list of compounds, predict which are ionic and which are molecular.

Section 7.4

Using only the periodic table, predict which of two elements is more electronegative.

Using only the periodic table, predict whether a given bond is ionic, polar covalent, or nonpolar covalent.

Using a table of electronegativities, predict which of two bonds is expected to be more polar.

Section 7.5

Write Lewis symbols for atoms, and tell how many electrons must be shared to enable the atom to achieve a completed valence shell. Give the symbol of the noble gas with the same number of valence electrons.

Draw electron-dot structures for simple molecules.

Section 7.6

Draw electron-dot structures for polyatomic molecules and ions, recognizing when multiple bonding is needed.

For each atom in an electron-dot structure, give the number of bonded electron pairs and the number of nonbonded electron pairs.

For a given electron-dot structure, give the number of single bonds, double bonds, and triple bonds. Give the bond order of each bond.

Section 7.7

Draw electron-dot structures for polyatomic molecules and ions, recognizing when resonance structures are needed.

Give the bond order of each bond in an electron-dot structure that requires resonance structures.

Section 7.8

Calculate the formal charge on each atom in a molecule.

Determine the formal charge on each atom in a resonance structure, and use the formal charges to select the best resonance structure.

Section 7.9

Use the VSEPR model to predict the geometries of molecules and polyatomic ions, including those with more than one central atom.

59

Chapter 7—Covalent Bonds and Molecular Structure

Section 7.10

For molecules and polyatomic ions, sketch and identify the orbitals used by each atom to form bonds. Show which orbital overlaps result in σ bonds and which result in π bonds.

Section 7.11

Describe the formation of sp3 hybrid orbitals, and give the spatial orientation of the four sp3 hybrid orbitals.

Describe the bonding in polyatomic molecules and tell which atoms use of sp3 hybrid orbitals to form bonds.

Section 7.12

Describe the formation of sp and sp2 hybrid orbitals, and give the spatial orientation of the two sp hybrid orbitals and the three of sp2 hybrid orbitals, and identify the orbitals involved in bonding.

Describe the bonding in polyatomic molecules, and tell what kinds of orbitals on each atom overlap to form bonds.

Section 7.13

Identify bonding and antibonding molecular orbitals.

Sketch a molecular orbital diagram for a diatomic molecule. Use the molecular orbital diagram to determine the number of unpaired electrons and to calculate the bond order of the molecule described.

60

Chapter 7—Covalent Bonds and Molecular Structure

Lecture Outline 7.1. Covalent Bonding in Molecules1 A. Formation of a covalent bond 1. Two atoms come close together, and electrostatic interactions develop a. Nuclei repel each other; electrons repel each other b. Each nucleus attracts the same electrons; electrons attract both nuclei 2. If attractive forces > repulsive forces; then covalent bond formed B. Distance between the two atoms affects the magnitude of attractive and repulsive forces.

1. Bond length – optimum point where net attractive forces maximized a. Each covalent bond has a characteristic length that leads to maximum stability. b. Can predict from atomic radii 7.2. Strengths of Covalent Bonds2 A. Formation of covalent bonds leads to lower energy.

B. Bond dissociation energy, D 1. Amount of energy necessary to break a chemical bond in an isolated gaseous-state molecule 2. Only positive values possible 3. When a bond forms, ΔH = – D a. Bond breaking requires energy (D) b. Bond formation releases energy (– D) C. Bonds between the same pairs of atoms usually have similar bond dissociation energies.

7.3. A Comparison of Ionic and Covalent Compounds3 A. Ionic compounds 1. High-melting solids 2. Must overcome every ionic attraction in the entire crystal B. Covalent compounds 1. Low-melting solids, liquids, or even gases 2. Bonds within individual molecule may be very strong 3. Attractive forces between different molecules (intermolecular forces) are relatively weak.

7.4. Polar Covalent Bonds: Electronegativity4 A. Ionic and covalent bonds – extremes of a continuous spectrum of possibilities B. Polar covalent bonds – bonding electrons attracted somewhat more strongly by one atom in a bond 1. Electrons not completely transferred 2. More electronegative atom: δ– (δ represents partial charge formed.)

3. Less electronegative atom: δ+ C. Electronegativity (EN) – ability of an atom in a molecule to attract shared electrons in a bond 1. Metallic elements – low electronegativities 2. Halogens and other elements in upper right-hand corner of periodic table – high electronegativities D. Predicting bond polarity 1. Atoms with similar electronegativities (ΔEN ≤ 0.4) – form nonpolar bonds 2. Atoms whose electronegativities differ by more than two (ΔEN > 2 – form ionic bonds 3. Atoms whose electronegativities differ by less than two (ΔEN < 2) – form polar covalent bonds 7.5. Electron-Dot Structures5 1

Test Item File Questions: Multiple Choice 1-3 Test Item File Questions: Multiple Choice 4-8; Short Answer 1 3 Test Item File Questions: Multiple Choice 9-12; Short Answer 2 4 Test Item File Questions: Multiple Choice 13-26, 174-176; Algorithmic 1-6; Short Answer 3-8 5 Test Item File Questions: Multiple Choice 27-29; Short Answer 9 61 2

Chapter 7—Covalent Bonds and Molecular Structure

A. Electron-dot structure – atom’s valence-electron distribution in a molecule B. Maximum stability when s and p subshells filled for each atom in a molecule C. Electron-dot structures for elements 1. Electrons represented by dots around element symbol a. One to each side until all four sides occupied b. Then pair dots until all valence-shell electrons used c. Note: pairing of dots does not correspond to pairing of electrons in electron configuration D. Simple molecules of second-period main-group elements – textbook Table 7.3 1. Octet rule used to determine number of covalent bonds formed by an element 2. Unpaired electrons form bonds 3. Electrons shared until each atom surrounded by four pairs of electrons (an octet) 4. Lone pairs (nonbonded pairs) – electron pairs not used in bonding 5. Bonding pairs – electrons shared in bond formation E. Multiple covalent bonds – sometimes more than one pair of electrons shared 1. Shorter and stronger than single bonds 2. Double bond – two pairs of electrons shared 3. Triple bond – three pairs of electrons shared 4. Bond order – number of electron pairs shared/2 a. Bond order = 1 i. Single bond ii. Two electron pairs shared b. Bond order = 2 i. Double bond ii. Four electron pairs shared c. Bond order = 3 i. Triple bond ii. Three electron pairs shared F. Coordinate covalent bonds 1. Two electrons (a lone pair) donated by one atom to another atom in bond formation 2. Second atom has vacant valence orbital 2. Often formed by N, O, P, and S 7.6. Electron-Dot Structures of Polyatomic Molecules6 A. Compounds of second-row elements: C, H, N, O 1. Octet rule almost always applies 2. Easy to predict number of bonds formed by each element – similar to above 3. Number of bonds formed equals number of unpaired dots in the Lewis formula 4. Indicate covalent single bond by a single line a. Double bond – two lines b.Triple bond – three lines B. Compounds with elements beyond the second row 1. Third period and lower elements larger than second-period elements 2. Larger size allows bonds to more than four atoms 3. “Expanded octet” C. General method for drawing electron-dot structures – textbook pages 222 ff 7.7. Electron-Dot Structures and Resonance7 A. Sometimes more than one electron-dot structure needed for molecular description B. Resonance hybrid – average of various possible electron-dot structures for a molecule 1. Resonance indicated by straight double-headed arrows (↔) joining resonance structures 2. Resonance structures differ only in valence electron placement 7.8. Formal Charges8 6 7

Test Item File Questions: Multiple Choice 30-41, 156-158; Algorithmic 7-11; Short Answer 10-11 Test Item File Questions: Multiple Choice 42-47; Short Answer 12 62

Chapter 7—Covalent Bonds and Molecular Structure A. Result of electron bookkeeping 1. Formal charge = (number of valence electrons in free atom) – 1/2(number of bonding electrons) – (number of nonbonding electrons) 2. Sum of formal charges on all atoms = overall charge on the molecule B. Use to evaluate relative importance of different resonance structures 1. Structures with smallest formal charges more stable 2. Structures with more negative formal charge on more electronegative atom more stable 7.9. Molecular Shapes: The VSEPR Model9 A. Molecular shape – determined by numbers of valence electrons around atoms B. Valence-shell electron-pair repulsion model – predicts approximate molecular shape 1. Bonded and lone electrons occupy charge clouds 2. Count the number of charge clouds surrounding atom 3. Geometry based on number of charge clouds a. Charge clouds – stay as far away as possible from each other b. Textbook Table 7.4 7.10. Valence Bond Theory10 A. Easily visualized orbital picture of how electron pairs are shared in a covalent bond B. Orbital overlap – bond formation 1. Greater the overlap, stronger the bond 2. Singly-occupied valence orbital on one atom overlaps a singly occupied valence orbital on another atom C. For orbitals other than s orbitals, orbitals used in bond formation have directionality D. Each bonded atom maintains its own orbital but shares electrons 7.11. Hybridization and sp3 Hybrid Orbitals11 A. Hybrid orbitals – combination of wave functions for atomic orbitals that forms a new set of equivalent wave functions 3 B. Combine one s with three p – form four equivalent new orbitals, each named sp 1. Each hybrid a. Two lobes b. One larger than the other 2. Four large lobes point toward corners of a tetrahedron 3. Bonds formed with sp3 orbitals are very strong C. A tetrahedral arrangement of charge clouds always implies sp3 hybridization.

7.12. Other Kinds of Hybrid Orbitals12 A. Geometries related to hybrid orbitals used 1. Textbook Table 7.5 2. Each electron-cloud geometry has a corresponding set of hybrid orbitals B. Sigma (σ) bond 1. Bond formed by head-on overlap 2. Has shared electrons centered about the axis between the two nuclei C. Pi (π) bond 1. Bond formed by side-by-side overlap 2. Has shared electrons occuping a region above and below a line connecting the two nuclei 7.13. Molecular Orbital Theory: The Hydrogen Molecule13 8

Test Item File Questions: Multiple Choice 48-57, 170-173; Algorithmic 12; Short Answer 13-14 Test Item File Questions: Multiple Choice 58-70, 102-149, 165-169; Algorithmic 13-21; Short Answer 15-18 10 Test Item File Questions: Multiple Choice 71 11 Test Item File Questions: Multiple Choice 72-75, 152, 155, 159-161, 163-164; Short Answer 19 12 Test Item File Questions: Multiple Choice 76-84, 150-151, 153-154, 162; Algorithmic 22; Short Answer 20-21 13 Test Item File Questions: Multiple Choice 85-86, 177-178, 181 63 9

Chapter 7—Covalent Bonds and Molecular Structure A. Molecular orbital model 1. More complex bonding model than valence bond 2. Considers molecule as a whole rather than concentrating on individual atoms 3. Sometimes bonding description offers better agreement with experimental observations B. Atomic orbital – wave function whose square gives probability of finding the electron within a given region of space in an atom C. Molecular orbital – a wave function whose square gives probability of finding the electron within a given region of space in a molecule 1. Specific energy level 2. Specific shape 3. Occupied by a maximum of two electrons with opposite spins D. Orbital interactions in molecular orbital theory 1. Additive interaction a. For H2 – formation of egg-shaped molecular orbital b. Denoted σ c. Lower in energy than the two isolated 1s orbitals d. Bonding molecular orbital (MO) i. Electrons in bonding MO spend most time in region between the two nuclei ii. Bonds atoms together 2. Subtractive interaction a. For H2 – formation of molecular orbital that has a node between the two atoms b. Denoted σ* c. Higher in energy than the two isolated 1s orbitals d. Antibonding molecular orbital i. Electrons in antibonding MO do not occupy central region between the nuclei ii. Don't contribute to bonding E. Bond order = 1/2(number of bonding electrons – number of antibonding electrons) F. Number of molecular orbitals formed = the number of atomic orbitals combined G. Key ideas of molecular orbital theory – see textbook 7.14. Molecular Orbital Theory of Other Diatomic Molecules14 A. Paramagnetic – substances with unpaired electrons are attracted by magnetic fields B. Diamagnetic – substances whose electrons are all spin-paired C. Molecular orbital of O2 1. Two 2s orbitals interact to form σ2s and σ*2s molecular orbitals 2. Two 2p orbitals on the internuclear axis a. Interact head-on b. Form σ2p and σ*2p molecular orbitals 3. Remaining 2p orbitals perpendicular to the internuclear axis a. Interact in sideways manner b. Form π2p and π*2pmolecular orbitals 4. Energy levels of molecular orbitals – textbook Figure 7.18 D. Molecular orbital diagrams require some experience to generate.

7.15. Combining Valence Bond Theory and Molecular Orbital Theory15 A. Molecules with resonance structures 1. σ-bond framework a. Localized b. Describe with valence bond theory 2. π-bond framework a. Delocalized 14 15

Test Item File Questions: Multiple Choice 87-98, 179-180, 182-185, Short Answer 22-24 Test Item File Questions: Multiple Choice 99-101, Short Answer 25 64

Chapter 7—Covalent Bonds and Molecular Structure b. Describe with molecular orbital theory

65

Chapter 7—Covalent Bonds and Molecular Structure

CHAPTER 7 COVALENT BONDS AND MOLECULAR STRUCTURE Teaching Tips, Points of Emphasis, and Common Misconceptions Section 7.2

The units of bond dissociation energy (bond energy) are kJ/mol. As with ionization energy and electron affinity, bond energies are given for isolated, gaseous-state molecules.

Section 7.5

The Lewis structure is a helpful tool for understanding the configuration and chemistry of molecules of main-group elements. In general, Lewis structures cannot be drawn for transition-metal compounds because of the complex way that partially filled d orbitals influence the configuration and chemistry of these compounds.

Section 7.5

Once formed, there is no difference between an ordinary covalent bond and a coordinate covalent bond.

Section 7.5

Students sometimes inquire about the origin of a positive or negative charge on a molecule. Showing how a charge on an atom occurs (H → H+ + e–) and how that simple ion can combine with a neutral molecule to form a polyatomic ion may be helpful (H+ + H2O → H3O+).

Section 7.6

This general method for writing electron-dot structures works equally well for the simple molecules discussed in the previous section.

Section 7.6

Warn students that it is crucial to follow the steps for writing electron-dot structures exactly in the order given. Otherwise, an incorrect elelectron-dot structure may result.

Section 7.6

Except for hydrogen, terminal atoms always have complete octets.

Section 7.6

Multiple bonds are never introduced into the electron-dot structure before step 5.

Section 7.8

The best resonance structure has (a) formal charges as close to zero as possible and (b) a negative formal charge on the most electronegative element and a positive formal charge on the least electronegative element.

Section 7.9

A good way to introduce the tetrahedron is to draw a cube with atoms on alternating corners. Each atom will be on a face diagonal to each other atom, and the symmetry will be more evident. Likewise, an octahedron can be shown as a cube with an atom on each face.

Section 7.9

Molecular geometries are described by the relative positions of the nuclei, not the location of electron clouds. NH3 is trigonal pyramidal, not tetrahedral.

Section 7.9

Electron-dot structures do not imply geometry. It makes no difference whether the two pairs of nonbonded electrons on H2O are placed 90o or 180o to one another in the Lewis structure.

Section 7.9

Use models to show that a tetrahedron is sometimes viewed down its three-fold symmetry axis (for example, CHCl3) and sometimes viewed down its two-fold symmetry axis (for example, CH2Cl2).

66

Chapter 7—Covalent Bonds and Molecular Structure

Section 7.9

Nonbonded electrons are always placed in the trigonal plane of a trigonal bipyramid.

Section 7.10

When VSEPR is used to describe the shape of an organic molecule, it is applied to each individual carbon, nitrogen, and oxygen atom.

Section 7.10

Electron-dot structures describe a bond as a sharing of a pair of electrons. Valence bond theory explains how electrons become shared–by the overlap of atomic orbitals.

Section 7.12

Hybrid orbital sets have the same geometries as electron cloud sets predicted by VSEPR. Therefore, each VSEPR geometry has a corresponding set of hybrid orbitals used by the central atom.

Section 7.13

Valence bond theory assumes electrons are in atomic orbitals; molecular orbital theory places electrons in orbitals belonging to the entire molecule.

Lecture/Laboratory Demonstration References Section 7.

None

Literature References Section 7.1

Georgios R. Tsaparlis, "The Chemical Bond as an Atomic Tug-of-War," J. Chem.

Educ., Vol. 61, 1984, 677.

Section 7.1

Roger L. DeKock, "The Chemical Bond," J. Chem. Educ., Vol. 64, 1987, 934-941.

Section 7.2

Raymond F. Peterson and David F. Treagust, "Grade-12 Students’ Misconceptions of Covalent Bonding and Structure," J. Chem. Educ., Vol. 66, 1989, 459-460.

Section 7.4

Gordon Sproul, "Electronegativity and Bond Type: Predicting Bond Type," J. Chem.

Educ., Vol. 78, 2001, 387-390.

Section 7.4

James N. Spencer, Richard S. Moog, and Ronald J. Gillespie, "Demystifying Introductory Chemistry: Part 3. Ionization Energies, Electronegativity, Polar Bonds and Partial Charges," J. Chem. Educ., Vol. 73, 1996, 627-631.

Section 7.4

William B. Jensen, "Electronegativity from Avogadro to Pauling, Part I: Origins of the Electronegativity Concept," J. Chem. Educ., Vol. 73, 1996, 11-20.

Section 7.5

Natalie Foote Tiernan, "Gilbert Newton Lewis and the Amazing Electron Dots," J. Chem. Educ., Vol. 62, 1985, 569-570.

Section 7.5

Anthony N. Stranges, "Reflections on the Electron Theory of the Chemical Bond: 1900-1925," J. Chem. Educ., Vol. 61, 1984, 185-190.

Section 7.5

William B. Jensen, "Abegg, Lewis,Langmuir, and the Octet Rule," J. Chem. Educ., Vol. 61, 1984, 191-200.

67

Chapter 7—Covalent Bonds and Molecular Structure

Section 7.5

Linus Pauling, "G. N. Lewis and the Chemical Bond," J. Chem. Educ., Vol. 61, 1984, 201-203.

Section 7.5

S. R.Logan, "The Role of Lewis Structures in Teaching Covalent Bonding," J. Chem. Educ., Vol 78, 2001, 1457-1458.

Section 7.5

Wan-Yaacob Ahmad and Mat B. Zakaria, "Drawing Lewis Structures from Lewis Symbols: A Direct Electron Pairing Approach" J. Chem. Educ., Vol. 76, 1999, 329331.

Section 7.6

Gordon H. Purser, "Lewis Structures are Models for Predicting Molecular Structure, Not Electronic Structure" J. Chem. Educ., Vol. 76, 1999, 1013-1017.

Section 7.6

Juan Quilex Pardo, "Teaching a Model for Writing Lewis Structures," J. Chem.

Educ., Vol. 66, 1989, 456-458.

Section 7.6

James Allen Carroll, "Drawing Lewis Structures Without Anticipating Octets," J. Chem. Educ., Vol. 63, 1986, 28-31.

Section 7.6

Melvin E. Zandler and Erach R. Talty, "The '6N + 2 Rule' for Writing Lewis Octet Structures," J. Chem. Educ., Vol. 61, 1984, 124-127.

Section 7.6

Thomas J. Clark, "Another Procedure for Writing Lewis Structures," J. Chem.

Educ., Vol. 61, 1984, 100.

Section 7.7

Todd P. Silverstein, "The 'Big Dog-Puppy Dog' Analogy for Resonance," J. Chem.

Educ., Vol. 76, 1999, 206-207.

Section 7.7

Ronald Starkey, "Resonance Analogy Using Cartoon Characters," J. Chem. Educ., Vol. 72, 1995, 542.

Section 7.7

Kenton B. Abel and William M. Hemmerlin, "Explaining Resonance–A Colorful Approach," J. Chem. Educ., Vol. 68, 1991, 834.

Section 7.7

Francis Delvigne, "A Visual Aid for Teaching the Resonance Concept," J. Chem.

Educ., Vol. 66, 1989, 461-462.

Section 7.8

David G. Dewit, "Using Formal Charges in Teaching Descriptive Inorganic Chemistry," J. Chem. Educ., Vol. 71, 1994, 750-755.

Section 7.8

Packer and Sheila D. Woodgate, "Lewis Structures, Formal Charge, and Oxidation Numbers: A More User-Friendly Approach," J. Chem. Educ., Vol. 68, 1991, 456458.

Section 7.9

James P. Birk and Soraya Abbassian, "Teaching VSEPR: The Plastic Egg Model," J. Chem. Educ., Vol. 73, 1996, 636-637.

Section 7.9

Ronald J. Gillespie, James N. Spencer, and Richard S. Moog, "Demystifying Introductory Chemistry: Part 2. Bonding and Molecular Geometry WIthout Orbitals–The Electron Domain Model," J. Chem. Educ., Vol. 73, 1996, 622-627.

68

Chapter 7—Covalent Bonds and Molecular Structure

Section 7.9

Brian W. Pfennig and Richard L. Frock, "The Use of Molecular Modeling and VSEPR Theory in the Undergraduate Curriculum to Predict the Three-Dimensional Structure of Molecules," J. Chem. Educ., Vol. 76, 1999, 1018-1022.

Section 7.9

Ronald J. Gillespie, "Multiple Bonds and the VSEPR Model," J. Chem. Educ., Vol. 69, 1992, 116-121.

Section 7.11

D. W. Emerson, "A Colorful Demonstration to Simulate Orbital Hybridization," J. Chem. Educ., Vol. 65, 1988, 454.

Section 7.11

John Barbaro, "Orbital Bartending," J. Chem. Educ., Vol. 71, 1994, 1012. An analogy for orbital hybridization.

Section 7.13

A. B. Sannigrahi and Tapas Kar, "Molecular Orbital Theory of Bond Order and Valency," J. Chem. Educ., Vol. 65, 1988, 674-676.

Section 7.14

Albert Haim, "The Relative Energies of Molecular Orbitals for Second-Row Homonuclear Diatomic Molecules; The Effect of s-p Mixing," J. Chem. Educ., Vol. 68, 1991, 737-738.

Media References Section 7.1

H2 Bond Formation movie from the Instructor Resource DVD

Section 7.4

Periodic Trends: Electronegativity movie from the Instructor Resource DVD

Section 7.5

Electron Dot Structures I activity from the Instructor Resource DVD

Section 7.6

Electron Dot Structures II activity from the Instructor Resource DVD

Section 7.7

Resonance activity from the Instructor Resource DVD

Section 7.8

Formal Charges movie from the Instructor Resource DVD

Section 7.11

Promotion and Hybridization activity from the Instructor Resource DVD

Section 7.11

Hybridization movie from the Instructor Resource DVD

Section 7.13

Molecular Orbital Theory activity from the Instructor Resource DVD

69

CHAPTER 8 THERMOCHEMISTRY: CHEMICAL ENERGY Chapter Learning Goals for Students Section 8.1

Define energy.

Define potential and kinetic energy.

State the First Law of Thermodynamics, and understand its application to thermochemistry.

Differentiate between the concepts of heat and temperature.

Section 8.2

Identify a state function.

Section 8.3

Define and calculate PV work. Know whether work is being done by the system or on the system.

Section 8.4

Differentiate between energy and enthalpy, and perform calculations interconverting the two.

From ΔH or ΔE, tell whether energy is being lost from or gained by the system.

Section 8.5

Define thermodynamic standard state.

Calculate ΔE for a reaction.

Section 8.6

Given a balanced chemical equation and enthalpy change for a chemical reaction, calculate the enthalpy change per mole or per gram of each reactant and product.

Section 8.7

Perform calculations involving specific heat (or molar heat capacity), heat flow, and temperature change.

Calculate ΔH or ΔE in a calorimetry experiment.

Section 8.8

Use Hess’s law to calculate a standard heat of reaction.

Section 8.9

Use standard heats of formation to calculate a standard heat of reaction.

Section 8.10

Use bond dissociation energies to approximate a standard heat of reaction.

Section 8.11

Write balanced equations for combustion reactions.

Use Hess’s Law to calculate enthalpies of combustion.

Section 8.12

Predict whether entropy increases or decreases for a chemical reaction or physical change.

Section 8.13

Predict the signs of ΔH, ΔS and ΔG for a reaction.

Use the equation ΔG = ΔH – TΔS to determine whether the forward reaction or the reverse reaction is favored.

69

Chapter 8—Thermochemistry: Chemical Energy

Use ΔH and ΔS to determine the temperature at which a reversible system is at equilibrium.

70

Chapter 8—Thermochemistry: Chemical Energy

Lecture Outline 8.1. Energy and Its Conservation1 A. Energy 1. Capacity to do work or supply heat 2. Energy = work + heat B. Kinetic energy 1. Energy of motion 2. EK = 1/2mv2 C. Potential energy – stored energy D. Joule 1. SI unit for energy 2. 1 J = 1(kg⋅m2)/s2 E. Calorie 1. Amount of energy necessary to raise the temperature of 1 g of water by 1°C 2. 1 cal = 4.184 J F. Law of Conservation of Energy 1. Energy can be neither created nor destroyed 2. It can only be converted from one form into another G. Many forms of energy 1. Thermal energy a. Kinetic energy of molecular motion b. Temperature – measure of the kinetic energy of molecular motion 2. Heat a. Energy transferred from one object to another b. Result of a temperature difference between them 3. Chemical energy a. A type of potential energy b. Chemical bonds of molecules act as the storage medium H. First Law of Thermodynamics: The energy of the universe is constant.

8.2. Internal Energy and State Functions2 A. System – everything we focus on in an experiment B. Surroundings – everything other than the system C. Internal energy – energy of the system D. System isolated from the surroundings 1. No energy transfer to surroundings 2. ΔE = 0 3. First Law of Thermodynamics: The total internal energy of an isolated system is constant.

E. System is not isolated from surroundings 1. Energy flows to or from surroundings 2. ΔE = Efinal – Einitial F. Energy changes measured from the point of view of the system 1. Energy flows out of system to surroundings – negative value 2. Energy flows into system from surroundings – positive value G. State Function – function or property whose value depends only on the present state (condition) of the system, not on path used to arrive at that condition 1. Reversible 2. Overall change is zero if system returns to its original condition 8.3 Expansion Work3 1

Test Item File Questions: Multiple Choice 1-5; Algorithmic 1; Short Answer 1-7 Test Item File Questions: Multiple Choice 6-8; Short Answer 8-12 3 Test Item File Questions: Multiple Choice 9-12; Algorithmic 2 71 2

Chapter 8—Thermochemistry: Chemical Energy A. Work 1. Force (f) that produces movement of an object times distance moved (d) 2. w = F × d B. Expansion work (PV work) 1. Work done as result of volume change in the system 2. w = – (P × ΔV) 3. Expansion of the system. a. System does work on surroundings b. ΔE is negative c. w = – PΔV i. Work is negative ii. ΔV is positive 4. Contraction of the system a. Surroundings do work on the system b. ΔE is positive c. w = – PΔV i. Work is positive ii. ΔV is negative 8.4. Energy and Enthalpy4 A. Total energy change of a system 1. ΔE = q + w (q = heat) 2. ΔE = q + (– PΔV) B. Amount of heat transferred: q = ΔE + PΔV C. Reactions carried out with constant volume 1. ΔV = 0 2. No PV work done 3. qv = ΔE D. Reactions carried out at constant pressure 1. ΔV ≠ 0 2. Energy change due to both heat transfer and PV work 3. qp = ΔE + PΔV E. Heat of reaction, ΔH 1. Enthalpy change of a system 2. ΔH = ΔE + PΔV 3. State function – value depends only on the current state of the system 4. ΔH = Hproducts – Hreactants 5. Amount of heat released in a specific reaction depends on actual amounts of reactants.

8.5. The Thermodynamic Standard State5 A. Value of enthalpy change, ΔH, reported for a reaction represents amount of heat released when reactants converted to products in molar amounts represented by coefficients of the balanced equation.

1. Actual amount of heat involved in a reaction depends on actual amounts of reactants.

2. Physical states of reactants and products must be specified.

3. Temperature and pressure also must be reported.

B. Thermodynamic Standard State = 298.15 K (25°C), 1 atm pressure of each gas, 1 M concentration (for solutions) 1. Allows different reactions to be compared 2. Indicated by addition of a superscript o, ΔH° C. Standard enthalpy of reaction – enthalpy change measured under standard conditions 4 5

Test Item File Questions: Multiple Choice 13-22, 95-98; Algorithmic 3; Short Answer 13-17 Test Item File Questions: Multiple Choice 23-28; Algorithmic 4-5 72

Chapter 8—Thermochemistry: Chemical Energy

8.6. Enthalpies of Physical and Chemical Change6 A. Enthalpies of physical change 1. Heat of fusion – amount of heat required for melting 2. Heat of vaporization – amount of heat required for evaporation 3. Sublimation a. Direct conversion of solid to vapor without going through a liquid state b. Heat of sublimation = Heat of fusion + Heat of vaporization B. Enthalpies of chemical change 1. Heats of reaction – enthalpies of chemical change 2. Endothermic reactions a. Hproducts > Hreactants b. Heat flows into system from surroundings c. ΔH is positive 3. Exothermic reactions a. Hproducts < Hreactants b. Heat flows to surroundings from system c. ΔH is negative 4. ΔH° values for a given equation a. Equation balanced for number of moles of reactants and products b. All substances are in standard states c. Physical state of each substance specified d. Refers to reaction going in direction written e. Reversing direction of reaction changes sign of ΔH° 8.7. Calorimetry and Heat Capacity7 A. Calorimetry – experimental technique that allows energy change associated with a chemical or physical process to be determined 1. Temperature change observed when system gains or loses energy in the form of heat 2. Performed in a calorimeter 3. For an exothermic reaction: amount of heat released by reaction = amount of heat gained by calorimeter + amount of heat gained by solution B. Heat capacity (C) q C= ΔT 1. Amount of heat required to raise the temperature of an object or substance a given amount 2. Extensive property C. Specific heat 1. Amount of heat necessary to raise the temperature of exactly 1 g of a substance by exactly 1°C 2. q = (specific heat) × (mass of substance) × (ΔT) D. Molar heat capacity (Cm) 1. Amount of heat necessary to raise the temperature of 1 mole of a substance by 1°C 2. qm = (Cm) × (moles of substance) × (ΔT) 8.8. Hess’s Law8 A. Hess’s law 1. Overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction 2. Reactants and products in the individual steps added and subtracted like algebraic quantities in determining the overall equation 6

Test Item File Questions: Multiple Choice 29-40; Algorithmic 6-12 Test Item File Questions: Multiple Choice 41-49; Algorithmic 13-18; Short Answer 18-20 8 Test Item File Questions: Multiple Choice 50-55, 99-104 73 7

Chapter 8—Thermochemistry: Chemical Energy B. Combine individual reactions so that their sums will be the desired reaction – textbook example 8.6 8.9. Standard Heats of Formation9 A. Standard heat of formation o 1. Enthalpy change ΔH f for the hypothetical formation of 1 mole of a substance in its standard state from the most stable forms of it constituent elements in their standard states o 2. The most stable forms of all elements in their standard state have ΔH f = 0.

B. Standard enthalpy change found by subtracting the sum of the heats of formation of reactants from the sum of the heats of formation of products: o o ΔH o = Σ ΔH products − Σ ΔH reactants C. Textbook Appendix B 8.10. Bond Dissociation Enthalpies10 A. Bond dissociation enthalpies 1. Enthalpy changes, ΔH°, for corresponding bond–breaking reactions 2. ΔH° = D = bond dissociation energy 3. Always positive 4. Always need energy to break a bond B. ΔH° = D(bonds broken) – D( bonds formed) C. Textbook Table 8.1 8.11. Fossil Fuels, Fuel Efficiency, and Heats of Combustion11 o A. Heat of combustion (ΔH c) – amount of energy released on burning a substance B. Fuel efficiency o 1. Calculate ΔH c in kJ/g or kJ/mL 2. Can compare efficiency for different fuels C. Fossil fuels 1. Decayed remains of organisms from previous geological eras 2. Coal, natural gas and petroleum a. Coal and petroleum – complex mixture of compounds b. Coal – vegetable origin, compounds structurally similar to graphite c. Petroleum – viscous liquid mixture of hydrocarbons, primarily marine origin 8.12. An Introduction to Entropy12 A. Spontaneous process 1. Process that proceeds on its own without continuous external influence 2. Need either a release of energy or an increase in disorder of the system B. Entropy (S) 1. Amount of molecular disorder or randomness in a system 2. Larger the value of S, greater the molecular randomness 3. ΔS = Sfinal – Sinitial 4. Sfinal > Sinitial a. ΔS is positive b. System has become more random 5. Sfinal < Sinitial a. ΔS is negative b. System has become less random C. Spontaneous process 1. Favored by decrease in H (negative ΔH) 2. Favored by increase in S (positive ΔS) 9

Test Item File Questions: Multiple Choice 56-63 Test Item File Questions: Multiple Choice 64-68 11 Test Item File Questions: Multiple Choice 69-71 12 Test Item File Questions: Multiple Choice 72-80; Algorithmic 19-21; Short Answer 22 74 10

Chapter 8—Thermochemistry: Chemical Energy 8.13. An Introduction to Free Energy13 A. Gibbs free-energy change (ΔG); ΔG = ΔH – TΔS. B. Sign of ΔG used as a criterion for determining spontaneity of a process a. ΔG negative – spontaneous b. ΔG positive ≠ nonspontaneous C. Temperature dependence (TΔS) term for ΔG. 1. Spontaneity of some processes depends on temperature 2. Low temperatures – ΔH dominates and controls spontaneity 3. High temperatures – TΔS dominates and controls spontaneity D. ΔG = 0 1. Process at equilibrium 2. Balanced between spontaneous and nonspontaneous 3. Temperature at which a nonspontaneous reaction becomes spontaneous ΔH T= ΔS

13

Test Item File Questions: Multiple Choice 81-94, 105-117; Short Answer 23-25 75

Chapter 8—Thermochemistry: Chemical Energy

CHAPTER 8 THERMOCHEMISTRY: CHEMICAL ENERGY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 8.2

Students need to relate direction of heat flow with observed temperature change.

Demonstrate that for a chemical reaction (system) performed in a beaker (surroundings), if heat energy is released by the system, the beaker absorbs the heat and becomes hotter; if heat energy is absorbed by the system, the beaker contributes the heat and becomes colder.

Section 8.2

It is important for students to understand that the amount of energy released or gained in a reaction is proportional to the amounts of reactants and products involved in the reaction.

Section 8.3

Gases do work when the molecules push against the walls of their container and the volume of the container increases. If there is no volume change, no work is done.

Section 8.4

Relate energy change to physical work: When a person does work, the person’s energy diminishes, so when a system does work, w is negative and ΔE is negative.

Section 8.4

If the volume of the system does not change, the heat transferred to or from a system is the energy change of that system.

Section 8.4

If the pressure of the system does not change, the heat transferred to or from a system is the enthalpy change of that system.

Section 8.4

When reactions are performed in an open container, such as a coffee-cup calorimeter, the pressure (= atmospheric pressure) does not change and qP = ΔH; in a closed container, such as a bomb calorimeter, the volume (= volume of the bomb) does not change and q qx = ΔE. Even in an open container, unless a gas is involved in the reaction, volume changes are usually negligible and ΔH ≅ ΔE.

Section 8.5

The states of reactants and products are important in determining an energy change or enthalpy change for a reaction. The energy change will be different for a reaction producing a product as a liquid (with lower energy) versus producing the same product as a gas (with higher energy).

Section 8.5

Standard conditions as defined for thermodynamics (P = 1 atm; T = 298.15 K) are different from standard conditions as defined for gas laws (P = 1 atm; T = 273.15 K).

Section 8.6

Temperature does not change during a change of state; hence, ΔHfusion for water refers to H2O(s, 0oC) → H2O(l, 0oC) and ΔHvap for water refers to H2O(l, 100oC) → H2O(g, 100oC).

Section 8.6

Endothermic means heat flows into a system and ΔH > 0. Exothermic means heat flows out of a system and ΔH < 0.

Section 8.7

Just as density provides a link between mass and volume, heat capacity links heat flow and temperature change, allowing the three variables to be interconverted.

Section 8.7

The amount of heat transferred and, therefore, the temperature change depend on the amount of substance involved. Specific heat is heat capacity per gram of substance 76

Chapter 8—Thermochemistry: Chemical Energy and has units of J/(g⋅oC); molar heat capacity is heat capacity per mole of substance and has units J/(mol⋅oC).

Section 8.7

Specific heat and molar heat capacity are intensive properties that depend on the state of the substance.

Section 8.8

Students have already employed Hess’s law when performing Born-Haber cycle calculations.

Section 8.8

When an equation is reversed, ΔH must be multiplied by –1. When an equation is multiplied by a coefficient, ΔH must be multiplied by the same coefficient.

Section 8.9

Remind students to multiply the heat of formation of each substance by its coefficient in the chemical equation and also to multiply the standard heat of formation of each product by –1.

Section 8.10

Bond dissociation energies are defined for isolated, gaseous-state species. Other factors must be considered when dealing with condensed-state systems.

Section 8.10

Reactant bonds are broken and product bonds are formed: ΔHo = D(reactant bonds broken) – D(product bonds formed). Students should not confuse this equation for that used for heats of formation: ΔHo = ΔHof(products) – ΔHof(reactants), in which products and reactants appear on the opposite side of the minus sign.

Section 8.10

Electron-dot structures should be drawn for each reactant and product to see the number of each type of bond being broken and formed. These numbers must also be multiplied by the coefficients in the balanced chemical equation.

Section 8.12

Another way to introduce entropy is to use a spontaneous process for which ΔH = 0, such as the mixing of two ideal gases.

Section 8.13

ΔH has units of kJ, ΔS has much smaller units of J/K. Thus, reaction spontaneity as measured by ΔG = ΔH –TΔS is usually dominated by ΔH except at high temperatures or when the reaction results in an increase in the number of moles of gaseous products.

Lecture/Laboratory Demonstration References Section 8.6

Bassam Z. Shakhashiri, “Endothermic Reactions of Hydrated Barium Hydroxide and Ammonium Salts,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983) pp. 10-12.

Section 8.7

Bassam Z. Shakhashiri, “Boiling Water in a Paper Cup: Heat Capacity of Water,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp. 239-241.

Section 8.7

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “A Chemical Hand Warmer,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 101-102.

77

Chapter 8—Thermochemistry: Chemical Energy

Section 8.7

Bassam Z. Shakhashiri, “Chemical Cold Pack,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983) pp. 8-9.

Section 8.7

Bassam Z. Shakhashiri, “Heat of Neutralization,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983) pp. 15-16.

Section 8.7

Lee R. Summerlin and and James L. Ealy, Jr., “Endothermic Reaction: Ammonium Nitrate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 65.

Section 8.11

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Making Canned Heat,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 111-112.

Literature References Section 8.7

John J. Fortman, “Analogical Demonstrations,” J. Chem. Educ., Vol. 69, 1992, 323324.

Section 8.7

Charles M. Wynn, Sr., “Heat Flow vs. Cash Flow: A Banking Analogy,” J. Chem.

Educ., Vol. 74, 1997, 397-398.

Section 8.7

Brother Thomas McCullogh, “A Specific Heat Analogy,” J. Chem. Educ., Vol. 57, 1980, 896.

Section 8.7

Doris R. Kimbrough, “Heat Capacity, Body Temperature and Hypothermia,” J. Chem. Educ., Vol. 75, 1998, 48-49.

Section 8.11

Donald J. Wink, “The Conversion of Chemical Energy. Part 1. Technological Examples,” J. Chem. Educ., Vol. 69, 1992, 108-111.

Section 8.11

Gustav P. Dinga, “Hydrogen: The Ultimate Fuel and Energy Carrier,” J. Chem.

Educ., Vol. 65, 1988, 688-691.

Section 8.11

Israel Dostrovsky, “Chemical Fuels from the Sun,” Scientific American., Vol. 265, 1991, pp. 102-107.

Section 8.11

Harold H. Schobert, “The Geochemistry of Coal. Part II: The Components of Coal,” J. Chem. Educ., Vol. 66, 1989, 290-293.

Section 8.12

John J. Fortman, “Pictorial Analogies III: Heat Flow, Thermodynamics, and Entropy,” J. Chem. Educ., Vol. 70, 1993, 102-103.

Media References Section 8.3

Expansion Work activity from the Instructor Resource DVD

Section 8.3

Work of Gas Expansion movie from the Instructor Resource DVD 78

Chapter 8—Thermochemistry: Chemical Energy

Section 8.4

Signs for q and w activity from the Instructor Resource DVD

Section 8.4

Heat Transfer activity from the Instructor Resource DVD

Section 8.6

Enthalpies of Chemical Change activity from the Instructor Resource DVD

Section 8.6

Thermite movie from the Instructor Resource DVD

Section 8.6

Changes of State movie from the Instructor Resource DVD

Section 8.9

Formation of Aluminum Bromide movie from the Instructor Resource DVD

Section 8.10

Bond Dissociation activity from the Instructor Resource DVD

Section 8.12

Airbags movie from the Instructor Resource DVD

Section 8.12

Spontaneous Process activity from the Instructor Resource DVD

Section 8.13

Formation of Water movie from the Instructor Resource DVD

Section 8.13

Formation of Water activity from the Instructor Resource DVD

79

CHAPTER 9 GASES: THEIR PROPERTIES AND BEHAVIOR Chapter Learning Goals for Students Section 9.1

Perform calculations converting between different units of pressure.

Show how the height of a liquid in a barometer depends on the density of the liquid.

Determine the pressure of a gas using an open-end manometer.

Section 9.2

Understand visual representations of the gas laws.

Section 9.3

Use the ideal gas law to calculate pressure, volume, moles of gas, or temperature, given the other three variables.

Use the ideal gas law to calculate final pressure, volume, moles of gas, or temperature from initial pressure, volume, moles of gas, and temperature.

Section 9.4

Perform stoichiometric calculations relating the mass of a reactant to the mass, moles, and volume or pressure of a gaseous product.

Use the ideal gas law to calculate the density of a gas.

Use the ideal gas law to calculate the molar mass of a gas.

Section 9.5

Use Dalton’s law to calculate the partial pressure or mole fraction of a gas in a mixture.

Section 9.6

Use the kinetic-molecular theory of gases to explain each of the gas laws.

Section 9.7

Use Graham’s law to calculate the relative rates of effusion of two different gases.

Section 9.8

State the conditions under which a gas is expected to behave ideally or nonideally.

Use the van der Waals equation to calculate the pressure of a real gas.

Section 9.9

Describe the four regions of the earth’s atmosphere.

Explain the origins of air pollution, acid rain, global warming, and the ozone hole.

80

Chapter 9—Gases: Their Properties and Behavior

Lecture Outline 9.1 Gases and Gas Pressure1 A. Gases 1. Little attraction between constituent atoms or molecules 2. Free to move about in available volume B. Some properties of gases 1. Mixtures always homogeneous a. Very weak attraction between gas molecules b. Identity of neighbor is irrelevant 2. Compressible a. Volume contracts when pressure applied b. Only 0.10% of volume of gas occupied by molecules 3. Exert a measurable pressure on the walls of their container C. Pressure 1. Force exerted per unit area, P = F/A 2. SI unit a. Pascal (Pa) b. 1 Pa = 1 N/m2 (1 N = 1 (kg⋅m)/s2) 3. Alternative units a. Millimeters of mercury (mm Hg) b. Atmosphere (atm) 4. 1 atm = 760 mm Hg = 101,325 Pa D. Atmospheric pressure 1. Pressure created from the mass of atmosphere pressing down on the earth’s surface 2. Standard atmospheric pressure at sea level = 760 mm Hg = Patm E. Measuring pressure 1. Barometer a. Long thin mercury-filled tube sealed at one end and inverted into a dish of mercury b. Downward pressure of Hg in column equals outside atmospheric pressure 2. Manometer a. U-tube filled with mercury i. One end connected to the gas-filled container ii. Other end open to the atmosphere b. Pgas = Patm: liquid level in both arms is equal c. P gas > Patm i. Liquid level in the arm connected to the gas-filled cylinder will be lower ii. P gas = Patm + PHg (PHg = the difference in the heights of the two mercury columns) d. P gas < Patm i. Liquid level in the arm connected to the gas-filled cylinder will be higher ii. P gas + PHg = Patm 9.2. The Gas Laws2 A. Different gases show similar physical behavior B. Defined by four variables – pressure, temperature, volume, and number of moles a. Relationships of variables – gas laws b. Ideal gas – behavior follows the gas laws exactly C. Boyle’s law 1. Relationship between volume and pressure at constant temperature 2. V ∝ 1/P 1 Test Item File Questions: Multiple Choice 1-14, 98-107, 119; Algorithmic 1-3; Short Answer 1-3 2 Test Item File Questions: Multiple Choice 15-26, 108-112; Algorithmic 4-7; Short Answer 4-6

81

Chapter 9—Gases: Their Properties and Behavior D. Charles’ law 1. Relationship between volume and temperature at constant pressure 2. V ∝ T (temperature expressed in Kelvins) E. Avogadro’s law 1. Relationship between volume and amount of gas at constant pressure and temperature 2. V ∝ n (where n = number of moles of gas) 3. 1 mol of gas at 273.15 K and 1.00 atm = 22.4 L of gas 9.3 The Ideal Gas Law3 A. Describes how volume of gas affected by changes in pressure, temperature, and amount L⋅ atm 1. PV = nRT; R = gas constant = 0.08206 K⋅ mol 2. Can be rearranged in different ways to take the form of Boyle’s law, Charles’ law, and Avogadro’s law B. Standard temperature and pressure (STP) 1. T = 273.15 K 2. P = 1 atm 9.4. Stoichiometric Relationships with Gases4 A. Stoichiometric calculations involving the application of the ideal gas law B. Three of four variables, P, V, T, and n known, the fourth calculated C. Includes calculations involving gas density and molar mass 9.5. Partial Pressure and Dalton’s Law5 A. Gas laws applied to mixtures of gases B. Dalton’s law of partial pressures 1. Ptotal = P1 + P2 + P3 + .... at constant V, T 2. P1, P2, .... refer to the pressures of individual gases in the mixture C. Partial pressure 1. Refers to the pressure an individual gas would exert if alone in the container (P1, P2, .....)

2. Total pressure – depends on total molar amount of gas present 3. Mole fraction (X) a. Number of moles of one component divided by total number of moles in the mixture moles of component b. Mole fraction ( X ) = total moles in mixture 4. P1 = X1⋅Ptotal 9.6. The Kinetic–Molecular Theory of Gases6 A. Model that explains gas behavior B. Assumptions 1. Gas particles in constant random motion 2. Most gas volume is empty space 3. Negligible attractive forces between gas molecules 4. Constant total kinetic energy of gas particles at constant T 5. Average EK ∝ T C. Above assumptions used to explain gas laws 9.7. Graham’s Law: Diffusion and Effusion of Gases7 A. Consequence of constant motion and high velocities of gas particles – gases mix rapidly 1. Diffusion – mixing of different gases a. Random molecular motion 3 Test Item File Questions: Multiple Choice 27-45; Algorithmic 8-21; Short Answer 7-9 4 Test Item File Questions: Multiple Choice 46-53; Algorithmic 22-25; Short Answer 10-12 5 Test Item File Questions: Multiple Choice 54-62, 113-118; Algorithmic 26-30; Short Answer 13-14 6 Test Item File Questions: Multiple Choice 63-77; Algorithmic 31-34; Short Answer 15-18 7 Test Item File Questions: Multiple Choice 78-85; Algorithmic 35; Short Answer 19-20

82

Chapter 9—Gases: Their Properties and Behavior b. Frequent collisions 2. Effusion a. Gas molecules escape through a tiny hole in a membrane into a vacuum b. No collisions B. Graham’s law 1. Rate of gas effusion inversely proportional to the square root of its molar mass 1 Rate ∝ M 2. Two gases at same temperature and pressure: Rate1 = Rate 2

M2 M1

3. Different rates of effusion used to separate gas mixture into separate components 9.8. The Behavior of Real Gases8 A. Ideal gas 1. No attractive forces between molecules – true at low pressures 2. Most of gas is empty space – true at low pressures B. Real gas 1. Attractive forces more important at higher pressures – decreases volume of gas from that predicted by the ideal gas law 2. Molecular volume – actual gas volume at high pressure larger than predicted by the ideal gas law C. Two effects – molecular volume and intermolecular forces cancel out at intermediate pressures D. van der Waals equation: ⎛ an2 ⎞ ⎜ P + 2 ⎟ V − nb = nRT V ⎠ ⎝ 1. a = correction for intermolecular attractions 2. b = correction for molecular volume 9.9. The Earth’s Atmosphere9 A. Troposphere – region nearest earth’s surface 1. Has greatest effect on the earth’s surface 2. Air pollution a. Release of unburned hydrocarbon molecules b. Production of NO – produces photochemical smog 3. Acid rain a. Results from production of SO2 b. 2 SO2(g) + O2(g) → 2 SO3(g); SO3(g) + H2O(l) → H2SO4(aq) 4. Greenhouse effect and global warming a. CO2 absorbs radiation from the sun b. Recent increase in concentration of atmospheric carbon dioxide c. May lead to widespread global warming B. Ozone layer – an atmospheric band stretching from about 20 to 40 km above the earth’s surface 1. Absorbs intense ultraviolet radiation from the sun 2. Shields high-energy solar radiation from reaching the earth’s surface 3. Ozone depletion – due to the presence of chlorofluorocarbons (CFCs) in the stratosphere 4. Reaction sequence – chain reaction in which generation of few chlorine atoms leads to destruction of many ozone molecules

(

)

8 Test Item File Questions: Multiple Choice 86-87, 120-122; Short Answer 21 9 Test Item File Questions: Multiple Choice 88-97; Short Answer 22-25

83

Chapter 9—Gases: Their Properties and Behavior

CHAPTER 9 GASES: THEIR PROPERTIES AND BEHAVIOR Teaching Tips, Points of Emphasis, and Common Misconceptions Section 9.1

Atmospheric pressure is measured with a barometer; the pressure of any gas is measured with a manometer. If gas pressure > atmospheric pressure, Pgas = Patm + Δhmercury; if gas pressure < atmospheric pressure, Pgas = Patm – Δhmercury

Section 9.2

Students tend to remember the standard molar volume while forgetting that this value holds true only under standard conditions. One mole of an ideal gas occupies 22.4 L only at 1.00 atm and 273 K.

Section 9.2

Students sometimes think that some substances are ideal gases while other substances are nonideal. All gases are ideal under certain conditions and nonideal under others.

Lecture/Laboratory Demonstration References Section 9.1

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Boiling at Reduced Pressure,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 24-25.

Section 9.2

Bassam Z. Shakhashiri, “Boyle’s Law,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985) pp. 14-19.

Section 9.2

Bassam Z. Shakhashiri, “Effect of Pressure on the Size of a Balloon,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985) pp. 12-13.

Section 9.2

Bassam Z. Shakhashiri, “Boyle’s Law and the Mass of a Textbook,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985) pp. 20-23.

Section 9.2

Rick Broniec, “Boyle’s Law and the Monster Marshmallow,” J. Chem. Educ., Vol. 59, 1982, 974.

Section 9.2

N. Szczepanski, “A Simple Demonstration of Charles’s Law,” J. Chem. Educ. 71, 1994, 433.

Section 9.2

Bassam Z. Shakhashiri, “Thermal Expansion of Gases,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985) pp. 24-27.

Section 9.2

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Charles’ Law: The Relationship Between Volume and Temperature of a Gas,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 23.

Section 9.4

Bassam Z. Shakhashiri, “Determination of the Molecular Mass of a Volatile Liquid,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The 84

Chapter 9—Gases: Their Properties and Behavior University of Wisconsin Press, Madison, 1983) pp. 51-54. A boiling-water bath is used to heat a volatile liquid until it is vaporized, completely filling an Erlenmeyer flask covered by aluminum foil with a pinhole orifice. From atmospheric pressure, bath temperature, volume of the flask, and mass of recondensed vapor, the molar mass of the volatile liquid is determined.

Section 9.4

Lee R. Summerlin, and James L. Ealy, Jr., “Determining the Molecular Weight of a Gas,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 19-20. The molar mass of butane is determined by the vapor density method.

Section 9.7

Bassam Z. Shakhashiri, “Ratio of Diffusion Coefficients: The Ammonium Chloride Ring,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983) pp. 59-62. Ammonia vapor and hydrogen chloride gas are simultaneously introduced into opposite ends of dry glass tubing. A white ring of solid ammonium chloride forms inside the tube closer to the end at which the hydrogen chloride was introduced, showing an inverse relationship between the rate of diffusion and the molar mass of each reactant. No distinction is made between diffusion and effusion, and Graham's law is erroneously used in explaining the data.

Section 9.7

Lee R. Summerlin, and James L. Ealy,Jr., “Diffusion of Gases,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 14-15. Ammonia and hydrogen chloride diffusion rates are determined.

Section 9.7

Dianne N. Epp, “Overhead Projection of Graham’s Law of Gaseous Diffusion,” J. Chem. Educ. 67, 1990, 1061.

Section 9.7

Bassam Z. Shakhashiri, “Relative Velocity of Sound Propagation: Musical Molecular Weights,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985) pp. 88-89.

Literature References Section 9.1

Joseph S. Schmuckler, “Gases and Their Behavior,” J. Chem. Educ., Vol. 57, 1980, 85.

Section 9.2

John T. Petty, “Charles’ Law of Gases: A Simple Demonstration,” J. Chem. Educ.

72, 1995, 257.

Section 9.2

R. P. Graham, “Guy-Lussac: Chemist Extraordinary,” J. Chem. Educ. 58, 1981, 789.

Section 9.2

Harold Goldwhite, “Guy-Lussac after 200 Years,” J. Chem. Educ. 55, 1978, 366368.

Section 9.4

Andreas Madlung, “The Chemistry Behind the Air Bag,” J. Chem. Educ. 73, 1996, 347-348.

Section 9.4

William L. Bell, “Chemistry of Air Bags,” J. Chem. Educ. 67, 1990, 61.

85

Chapter 9—Gases: Their Properties and Behavior

Section 9.6

Dacio R. Hartwig and Romeu C. Rocha Filho, “Cinema, Flirts, Snakes, and Gases,” J. Chem. Educ. 59, 1982, 295. A kinetic molecular theory of gases analogy is presented.

Section 9.6

Reggie L. Hudson, “Toy Flying Saucers and Molecular Speeds,” J. Chem. Educ. 59, 1982, 1025-1026.

Media References Section 9.1

Gas Pressure in Manometer activity from the Instructor Resource DVD

Section 9.2

Gas Laws activity from the Instructor Resource DVD

Section 9.2

P-V Relationships movie from the Instructor Resource DVD

Section 9.4

Airbags movie from the Instructor Resource DVD

Section 9.5

Kinetic Energies in a Gas movie from the Instructor Resource DVD

Section 9.5

Kinetic-Molecular Theory of Gases activity from the Instructor Resource DVD

Section 9.6

Diffusion of Bromine Vapor movie from the Instructor Resource DVD

Section 9.7

Diffusion and Effusion activity from the Instructor Resource DVD

Section 9.8

Real Gases activity from the Instructor Resource DVD

Section 9.9

Carbon Dioxide Behaves as an Acid in Water movie from the Instructor Resource DVD

Section 9.9

CFCs and Stratospheric Ozone movie from the Instructor Resource DVD

86

Chapter 10—Liquids, Solids, and Phase Changes

CHAPTER 10 LIQUIDS, SOLIDS, AND PHASE CHANGES Chapter Learning Goals for Students Section 10.1

Calculate percent ionic character from a dipole moment.

Using only VSEPR geometries and electronegativity trends, determine whether a molecule is expected to be polar and have a dipole moment.

Give the direction of a dipole moment.

Section 10.2

Identify the major type of intermolecular forces present in substances, and determine which of two substances exhibits the stronger intermolecular force.

Relate relative strengths of intermolecular forces to physical properties of molecules.

Section 10.3

Describe viscosity and surface tension, and relate these properties to intermolecular forces.

Section 10.4

Describe fusion, freezing, vaporization, condensation, sublimation, and deposition.

For a phase change, determine whether enthalpy is increasing or decreasing and whether entropy is increasing or decreasing.

Use the ΔG = ΔH – TΔS equation to calculate the entropy change for a phase change or the temperature (boiling point, melting point, sublimation point) at which the phase change occurs.

Section 10.5

Use the Clausius-Clapeyron equation to calculate vapor pressure or heat of vaporization.

Section 10.6

Classify solids as ionic, molecular, covalent network, or metallic, and give the major forces of attraction of each.

Section 10.7

Use the Bragg equation to calculate the distance between layers of atoms in a crystal.

Section 10.8

Identify unit cells.

For metals crystallizing in one of the three cubic unit cells, determine the number of atoms, mass, volume, density, atomic radius, and packing efficiency.

Section 10.9

From unit cell information for ionic solids, determine the types and numbers of ions, the oxidation state and geometry around each ion, and the formula of the ionic compound.

Section 10.10

Describe structure of the allotropes of carbon and SiO2.

Section 10.11

Sketch a phase diagram, labeling the axes and each of the regions, and locate the triple point, critical point, normal melting point, and the normal boiling point.

Use phase diagrams to describe physical changes.

87

Chapter 10—Liquids, Solids, and Phase Changes

Lecture Outline 10.1 Polar Covalent Bonds and Dipole Moments1 A. Bond dipole 1. A bond with partial positive (δ+) and partial negative (δ–) ends 2. Due to in electronegativity difference of the atoms in the bond 3. Represented by +→; indicates direction of electron displacement a. Point of arrow represents δ– end of dipole b. Crossed end represents δ+ end of dipole B. Polar molecules – due to the net sum of individual bond polarities and lone-pair contributions in the molecule 1. Molecular dipoles a. Center of mass of positive charge (nuclei) doesn’t coincide with the center of mass of negative charge (electrons). b. Results in a net molecular polarity 2. Lone pairs of electrons make substantial contributions to net molecular polarity.

3. Individual bond polarities cancel in symmetrical molecules.

C. Dipole moment (μ) 1. Magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges 2. μ = Q × r 10.2. Intermolecular Forces2 A. Intermolecular forces 1. Attractive forces between molecules that hold them together at certain temperatures 2. Often called van der Waals forces 3. Divided into categories a. Ion-dipole b. Dipole-dipole c. London dispersion forces d. Hydrogen bonding 4. Electrical in nature B. Ion-dipole forces 1. Result of electrical interactions between an ion and partial charges on a polar molecule 2. Favored orientation a. δ+ end of molecule near the anion b. δ– end of molecule near the cation 3. Important in aqueous solutions of ionic substances, where dipolar water molecules surround the ions C. Dipole-dipole force 1. Result from electrical interactions among dipoles on neighboring molecules 2. Generally weak – strength depends on sizes of the dipole moments involved 3. Significant only when molecules in close contact 4. Correlation between dipole moment and boiling point a. High dipole moment corresponds to strong intermolecular forces b. Substance must overcome intermolecular forces to boil c. Stronger intermolecular forces require higher boiling points D. London dispersion forces 1. Result from motion of electrons around atoms a. Due to instantaneous unsymmetrical electron distribution b. Creates a short-lived dipole moment (instantaneous dipole) 1 Test Item File Questions: Multiple Choice 1-11, 89-98; Algorithmic 1-3; Short Answer 1-6 2 Test Item File Questions: Multiple Choice 12-18, 99-101; Algorithmic 4-10; Short Answer 7-9

88

Chapter 10—Liquids, Solids, and Phase Changes 2. Instantaneous dipole induces a temporary dipole on neighboring atoms 3. Weak attractive forces 4. Polarizability – ease of molecule’s electron cloud distorted by a nearby electric field a. Smaller molecules and lighter atoms with fewer electrons i. Relatively nonpolarizable ii. Smaller dispersion forces b. Larger molecules and heavier atoms with more electrons i. More polarizable ii. Larger dispersion forces c. Molecules with more spread-out shapes i. Maximize molecular surface area ii. Greater contact between molecules iii. Have higher dispersion forces E. Hydrogen bonds – attractive interaction between a hydrogen atom bonded to an electronegative O, N, or F atom and an unshared electron pair on another nearby electronegative atom 1. Quite strong 2. Responsible for water’s remarkable properties 3. Results in higher boiling points than might be expected 4. Contributing factors a. Extraordinary electronegativity and small sizes of F, O, and N: H–F, H–O, H–N bonds are highly polar b. Small size of hydrogen 10.3. Some Properties of Liquids3 A. Viscosity 1. Measure of a liquid’s resistance to flow 2. SI unit: N⋅s/m2 3. Ease with which molecules move around in the liquid 4. Related to intermolecular forces – stronger the forces, higher the viscosity B. Surface tension 1. Resistance of a liquid to spreading out and increasing its surface area 2. Due to difference in intermolecular forces felt by liquid surface molecules and liquid interior molecules 3. Related to intermolecular forces – stronger the forces, greater the surface tension C. Properties temperature dependent 1. Increase in temperature corresponds to increase in kinetic energy 2. Molecules with high kinetic energies more easily overcome intermolecular forces 10.4. Phase Changes4 A. Phase changes (changes of state) 1. Physical form but not chemical identity of a substance changes 2. Matter in any one state can change into either of the other two 3. Sublimation – solid changes directly into a gas B. Phase change associated with free-energy change, ΔG 1. ΔG = ΔH – TΔS 2. Enthalpy part – energy change associated with making or breaking intermolecular attractions holding liquids and solids together 3. Entropy part – associated with change in disorder between various states 4. Solid → liquid, solid → gas, liquid → gas: ΔH and ΔS both positive 5. Gas → liquid, gas → solid, liquid → solid: ΔH and ΔS both negative 6. Knowing ΔH and ΔS for a phase transition can calculate temperature at which two phases are in equilibrium.

3 Test Item File Questions: Multiple Choice 19-23; Short Answer 10-11 4 Test Item File Questions: Multiple Choice 24-34; Algorithmic 11-12; Short Answer 12

89

Chapter 10—Liquids, Solids, and Phase Changes a. At equilibrium, ΔG = 0 b. T = ΔH/ΔS C. Heating curve – graphical display of results of adding heat to a sample (See textbook Figure 10.10.)

1. Melting point a. Temperature at which solid and liquid coexist in equilibrium as molecules break free from their position in the crystal and enter the liquid phase b. Heat of fusion (ΔHfusion) – amount of energy required to overcome enough intermolecular forces to convert solid into liquid 2. Boiling point a. Temperature at which liquid and vapor coexist in equilibrium as molecules break free from the surface of the liquid and enter the gas phase b. Heat of vaporization (ΔHvap) – amount of energy necessary to convert liquid into gas 3. ΔHvap >> ΔHfusion a. For vaporization, must overcome all intermolecular forces in compound b. For fusion, must overcome fewer intermolecular forces 10.5. Evaporation, Vapor Pressure, and Boiling Point5 A. Evaporation – escape of molecules from a liquid surface B. Vapor pressure – pressure exerted by vapor molecules over a liquid in a closed container 1. Dynamic equilibrium 2. Number of molecules escaping liquid = number of molecules returning to liquid 3. Constant total number of molecules in both liquid and vapor phases 4. Individual molecules constantly passing back and forth from one phase to another C. Explained by kinetic molecular theory – increase in temperature results in higher fraction of molecules with sufficient kinetic energy to overcome surface tension and escape into the vapor D. Value of vapor pressure 1. Related to intermolecular forces and temperature 2. Smaller the intermolecular forces, higher the vapor pressure – molecules loosely held in liquid and easily escape 3. Higher the temperature, higher the kinetic energy – molecules have sufficient kinetic energy to escape the liquid E. Clausius-Clapeyron equation 1. Relates vapor pressure of a liquid to inverse of its temperature ⎛ - ΔH ⎞ vap ⎟ + C, where C = a constant characteristic of each substance ln Pvap = ⎜⎜ ⎟ R T ⎝ ⎠

( )

2. Plot ln (Pvap) versus 1/T a. Straight line b. Slope = –ΔHvap/R 3. Knowing vapor pressure at several temperatures, can calculate ΔHvap of a liquid from ΔH vap =

(ln P - ln P )R 2

1

⎛1 1⎞ ⎜⎜ - ⎟⎟ ⎝ T1 T2 ⎠

F. Boiling point 1. Temperature at which vapor pressure of a liquid equals the external pressure pushing on the surface, and all of the liquid can change into the vapor phase 2. Normal boiling point – external pressure = 1 atm 3. External pressure < 1 atm – liquid boils at a lower temperature 4. External pressure > 1 atm – liquid boils at a higher temperature 5 Test Item File Questions: Multiple Choice 35-47, 102-111; Short Answer 13

90

Chapter 10—Liquids, Solids, and Phase Changes 10.6. Kinds of Solids6 A. Crystalline solids 1. Solids with atoms, ions, or molecules in an ordered, long-range arrangement 2. Macroscopically – observe flat faces and sharp angles 3. Types of crystalline solids – textbook Table 10.9 a. Ionic solids i. Constituent particles are ions ii. Example: NaCl iii. Ordered into a 3-D arrangement held together by ionic bonds b. Molecular solids i. Constituent particles are molecules held together by intermolecular forces ii. Examples: sucrose and ice c. Covalent network solids i. Atoms linked together by covalent bonds into a giant 3-D array ii. Examples: diamond and quartz d. Metallic solids i. Comparable to network solids but consist of metal atoms ii. Examples: Ag and Fe iii. Have metallic properties B. Amorphous solids 1. Constituent particles randomly arranged 2. No ordered long-range structure 10.7. Probing the Structure of Solids: X–Ray Crystallography7 A. Diffraction 1. Beam of electromagnetic radiation scattered by an object containing regularly spaced lines or points (e.g., atoms in a crystal) 2. Spacing must be comparable to the λ of the radiation 3. Interference between two λ’s passing through the same region of space at the same time a. Constructive interference i. Waves in-phase (peak-to-peak and trough-to-trough) ii. Increases intensity of wave b. Destructive interference i. Waves out-of-phase ii. Waves cancel B. Bragg analysis 1. X-rays diffracted by different layers of atoms in the crystal undergo constructive and destructive interference nλ 2. Bragg equation: nλ = 2d × sin θ; d = 2 sin θ a. λ = known b. Sin θ = angle at which incoming rays are reflected (can be measured) c. n = an integer (usually 1) 10.8. Unit Cells and the Packing of Spheres in Crystalline Solids8 A. Particles pack together in crystals so that they can be as close together as possible to maximize intermolecular attractions.

B. Simple cubic packing – spheres in one layer sit directly on top of those in the previous layer 1. All layers identical 2. Primitive-cubic unit cell

6 Test Item File Questions: Multiple Choice 48-56; Algorithmic 13-15; Short Answer 14-16 7 Test Item File Questions: Multiple Choice 57-58; Short Answer 17-18 8 Test Item File Questions: Multiple Choice 59-72, 112-117; Algorithmic 16-20; Short Answer 19-24

91

Chapter 10—Liquids, Solids, and Phase Changes 3. Coordination number = 6: sphere touches four neighbors in the same layer, one above and one below 4. Uses only 52% of available volume C. Body–centered cubic packing – spheres in alternate layers in an a-b-a-b arrangement where spheres in b layers fit into small depressions between spheres in neighboring a layers 1. Body-centered cubic cell 2. Coordination number = 8; four neighbors above and four neighbors below 3. Occupies 68% of available volume D. Hexagonal closest-packed – noncubic unit cell with two alternating layers (a-b-a-b) 1. Hexagonal arrangement of touching spheres 2. Spheres in b layer fit into small triangular depressions between spheres in a layer.

3. Coordination number = 12: six neighbors in the same layer, three above and three below E. Cubic closest-packed – face-centered cubic unit cell with three alternating layers, a-b-c-a-b-c 1. a-b layers identical to hexagonal closest-packed 2. Third layer offset from both a and b 3. Coordination number = 12 F. Unit cells – small repeating units found in crystals 1. Symmetrical geometries 2. Stack to minimize space 3. Fourteen different geometries a. Parallelepipeds (six-sided geometric solids whose faces are parallelograms) b. Differ in lengths of cell edges and angles between edges G. Cubic cells – all edges equal in length, and all angles 90° 1. Primitive-cubic unit cell for metals a. An atom at each of the eight corners b. Each atom shared with seven other neighboring cubes that come together at the same point c. One-eighth of each corner atom belongs to a single cube 2. Body-centered cubic unit cell – additional atom in the center of the cube 3. Face-centered cubic unit cell – additional atom on each of the six cube faces a. Shared with one other neighboring cube b. One-half of each face atom belongs to a single cube 10.9. Structure of Some Ionic Solids9 A. Spheres not all the same size – anions larger than cations B. Adopt a variety of different unit cells depending on ion sizes and charges 1. Face-centered cubic – NaCl, KCl. 2. Other common ionic unit cells – textbook Figure 10.25 10.10. Structure of Some Covalent Network Solids10 A. Carbon 1. Allotropes a. Different structural forms of the same element b. Differ in physical and chemical properties 2. More than 40 amorphous forms of carbon 3. Diamond a. Each carbon atom sp3 hybridized b. Covalently bonded with tetrahedral geometries 4. Graphite a. Two-dimensional sheets of fused six-membered rings b. Each carbon sp2 hybridized 5. Fullerene 9 Test Item File Questions: Multiple Choice 73-78, 88, 118-124; Algorithmic 21-22; Short Answer 25 10 Test Item File Questions: Multiple Choice 79-83

92

Chapter 10—Liquids, Solids, and Phase Changes a. Spherical C60 molecule b. Shape of a soccer ball B. Silica (SiO2) 1. Four single bonds between silicon and four oxygens in a covalent network structure 2. Quartz glass – result of heating silica above 1600°C then cooling the viscous liquid a. Si–O bonds reform in a random arrangement b. Amorphous solid c. Mix in additives – prepare a wide variety of glass i. Window glass – add CaCO3 and Na2CO3 ii. Colored glass – add transition metal ions iii. Borosilicate glass (Pyrex) – add B2O3; resistant to thermal shock because it doesn’t expand much on heating 10.11. Phase Diagrams11 A. Change any one state of matter spontaneously into either of the other two, depending on temperature and pressure B. Phase diagram – graphical method of illustrating pressure and temperature dependencies of a pure substance in a closed system 1. Boundary line – points on this line represent pressure/temperature combinations at which two phases are in equilibrium 2. Triple point – a unique combination of pressure and temperature at which all three phases coexist in equilibrium 3. Critical temperature – temperature beyond which a gas cannot be liquefied 4. Critical pressure – pressure required to liquefy a gas at the critical point 5. Critical point – point defined by critical temperature and critical pressure 6. Supercritical fluid – substance that is neither liquid nor gas a. Pressure of a gas at the critical point is so high and the molecules are so close together that it is difficult to distinguish between gas and liquid. b. Temperature of a liquid at the critical point is so high that it is hard to distinguish between liquid and gas.

7. Effect of pressure on the slope of solid/liquid boundary line depends on the relative densities of the solid and liquid phases

11 Test Item File Questions: Multiple Choice 84-87, 126-139

93

Chapter 10—Liquids, Solids, and Phase Changes

CHAPTER 10 LIQUIDS, SOLIDS, AND PHASE CHANGES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 10.1

A bond is polar if the atoms joined have different electronegativities. A molecule is polar if the bond polarities are not symmetrically arranged, which results in an uneven distribution of charges about the molecule and a net dipole moment.

Section 10.2

Intramolecular forces are the relatively strong bonding forces that hold atoms together in the molecular unit. Intermolecular forces are the relatively weak forces that hold one molecular unit to other molecular units.

Section 10.4

A phase change usually refers to what occurs when a solid, liquid, or gas is converted to one of the other two states. Sometimes more than one solid state exists for a substance, such as the metallic and nonmetallic forms of solid tin, and a phase change can occur between the two.

Section 10.4

On a heating curve for a pure substance the temperature remains constant during a phase change. Within a given state, temperature rises with a slope that depends on the heat capacity of the substance in that state.

Section 10.7

The variable n in the Bragg equation is the order of diffraction. After a rainfall, double or triple rainbows are sometimes observed with each successive rainbow fainter than the previous one. The most intense spectrum is due to a first-order diffraction. The fainter spectra are due to higher orders of diffraction.

Section 10.8

Students tend to consider the center atom of a body-centered cubic cell as different from the corner atoms. All atoms in the unit cell are identical. The unit cell can be reconfigured so that the atom originally in the center is positioned on a corner and vice versa.

Section 10.8

Primitive cubic is also known as simple cubic.

Section 10.8

Atoms that belong to more than one unit cell are classified as corner, edge, or face atoms, which contribute, respectively, 1/8, 1/4, and 1/2 atom per unit cell.

Section 10.8

Visualizing three-dimensional unit cells is difficult for many students. It is helpful to introduce the unit cell using a two-dimensional analog, such as a square array and a closest packed array of pennies laid flat on a table. The number of pennies per unit cell, the coordination number of each penny, and the percent empty space can be determined.

Section 10.8

When determining the density of a metal, students sometimes use the volume of the atoms instead of the volume of the entire unit cell. The density of a substance includes the empty space between atoms.

Section 10.9

The ratio of cations to anions in a unit cell must be consistent with the empirical formula of the ionic compound, as predicted in Chapter 2.

94

Chapter 10—Liquids, Solids, and Phase Changes

Lecture/Laboratory Demonstration References Section 10.1

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Bending a Stream of Water,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 91. This demonstration compares the polarities of water and hexane.

Section 10.2

William M. Hemmerlin and Kenton Abel, “Viscosity Races,” J. Chem. Educ., Vol. 68, 1991, 417.

Section 10.2

Andrew J. Rosenthal, “Demonstration of Surface Tension,” J. Chem. Educ., Vol. 78, 2001, 332-333.

Section 10.2

Todd P. Silverstein, “Polarity, Miscibility, and Surface Tension of Liquids,” J. Chem. Educ., Vol. 70, 1993, 253.

Section 10.2

Paul G. Jasian, Glenn Barnett, and David Speckhard, “Lowering the Surface Tension of Water: An Illustration of the Scientific Method,” J. Chem. Educ., Vol. 70, 1993, 251-252.

Section 10.2

Lee R. Summerlin and James L. Ealy, Jr., “Surface Tension of Water: The Magic Touch,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 45.

Section 10.2

Bassam Z. Shakhashiri, “At the Water's Edge: Surface Spreading and Surface Tension,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 301-304.

Section 10.5

Bassam Z. Shakhashiri, “Evaporation as an Endothermic Process,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 249-251.

Section 10.5

J. A. Sears and C. J. Grieve, “A Vapor Pressure Demonstration,” J. Chem. Educ., Vol. 67, 1990, 427. Hexane is introduced into a sealed Erlenmeyer flask containing hot water and fitted with a delivery tube. The hexane is vaporized by the hot water, forcing water from the flask into a graduated cylinder. As the flask cools, water is drawn back into the flask, showing that the vapor pressure of the hexane decreases with a decrease in temperature.

Section 10.5

Lee R. Summerlin and James L. Ealy, Jr., “The Effect of Pressure on Boiling Point,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 21.

Section 10.11

Stephen J. Hawkes, “A Simple Experiment for the Demonstration of the Phase Diagram of Carbon Dioxide,” J. Chem. Educ., Vol. 73, 1996, 837-838.

Literature References Section 10.1

R. J. Tykodi, “Identifying Polar and Nonpolar Molecules,” J. Chem. Educ., Vol. 66, 1989, 1007-1011.

95

Chapter 10—Liquids, Solids, and Phase Changes

Section 10.1

Thomas H. Bindel and Timothy C. Smiley, “The Ropes: A Molecular Polarity Activity,” J. Chem. Educ., Vol. 71, 1994, 945.

Section 10.1

Carlos Furio and Ma. Luisa Calatayud, “Difficulties with the Geometry and Polarity of Molecules: Beyond Misconceptions,” J. Chem. Educ., Vol. 73, 1996, 36-41.

Section 10.1

Gordon H. Purser, “The Significance of the Bond Angle in Sulfur Dioxide,” J. Chem. Educ., Vol. 66, 1989, 710-713.

Section 10.2

Robert R. Perkins, “Put the Body to Rest,” J. Chem. Educ., Vol. 72, 1995, 151-152.

An analogical demonstration of molecular polarity and chemical and physical changes involving intramolecular and intermolecular forces.

Section 10.2

John J. Fortman, “Pictorial Analogies I: States of Matter,” J. Chem. Educ., Vol. 70, 1993, 56-57.

Section 10.2

Ponnadural Ramasami, “Students as Solids, Liquids, and Gases,” J. Chem. Educ., Vol. 76, 1999, 485.

Section 10.2

Doris R. Kimbrough and Ronald DeLorenzo, “Solving the Mystery of Fading Fingerprints with London Dispersion Forces,” J. Chem. Educ., Vol. 75, 1998, 13001301.

Section 10.2

C. Jayne Wilcox, “London Dispersion Forces and the Wave,” J. Chem. Educ., Vol. 75, 1998, 1300-1301.

Section 10.2

John W. Hill, “A People-and Velcro Model for Hydrogen Bonding,” J. Chem. Educ., Vol. 67, 1990, 223.

Section 10.3

Todd P. Silverstein, “Why Do Alcoholic Beverages Have Legs?” J. Chem. Educ., Vol. 75, 1998, 723-724.

Section 10.6

John J. Fortman, “Pictorial Analogies II: Types of Solids,” J. Chem. Educ., Vol. 70, 1993, 57-58.

Section 10.8

Keenan E. Dungey, George Lisensky, and S. Michael Condren, “Kixium Monolayers: A Simple Alternative to the Bubble Raft Model for Close-Packed Spheres,” J. Chem. Educ., Vol. 76, 1999, 618-619.

Section 10.8

Daryl L. Ostercamp, “Close Packing of Identical Spheres,” J. Chem. Educ., Vol. 69, 1992, 162.

Section 10.8

Frank Galasso, “The Importance of Understanding Structure,” J. Chem. Educ., Vol. 70,1993, 287-290. A discussion of the relationship between unit cells and the structure of solids.

Section 10.9

James P. Birk, “A Model to Illustrate the Brittleness of Ionic and Metallic Crystals,” J. Chem. Educ., Vol. 62, 1985, 667.

Section 10.10

Robert F. Curl and Richard E. Smalley, “Fullerenes,” Scientific American, October 1991, 54-63.

96

Chapter 10—Liquids, Solids, and Phase Changes

Section 10.11

Nicholas K. Kildahl, “Journey Around a Phase Diagram,” J. Chem. Educ., Vol. 71, 1994, 1052-1054.

Section 10.11

Stephen J. Hawkes, “There Is No Perceptible Inflection at the Triple Point,” J. Chem. Educ., Vol. 76, 1999, 226.

Media References Section 10.4

Heating Curves activity from the Instructor Resource DVD

Section 10.4

Changes of State movie from the Instructor Resource DVD

Section 10.5

Equilibrium Vapor Pressure activity from the Instructor Resource DVD

Section 10.8

Close Packing activity from the Instructor Resource DVD

Section 10.11

Phase Diagram of Water movie from the Instructor Resource DVD

97

CHAPTER 11 SOLUTIONS AND THEIR PROPERTIES Chapter Learning Goals for Students Section 11.1

Classify homogeneous mixtures as solutions or colloids.

Identify the solvent and the solute in a solution.

Section 11.2

Explain the rule of thumb, “like dissolves like” by analyzing the solution process in terms of forces overcome in the solute and solvent and forces formed between solute and solvent particles.

Predict which solute is more soluble in a given solvent by correlating solubility with chemical structure.

Section 11.3

Define solution density, molarity, mole fraction, mass percent, parts per million, parts per billion, and molality, and perform calculations using these quantities.

Section 11.4

Predict the effect of changing pressure and temperature on the solubility of a given solute.

Perform calculations using Henry’s law.

Section 11.5

Define and give examples of colligative properties.

Section 11.6

Use Raoult’s law to calculate the vapor pressure over a solution containing a nonvolatile solute and a solution containing two volatile liquids.

Interpret vapor pressure curves.

Section 11.7

Perform calculations involving freezing–point depression and boiling–point elevation, and determine the molality of a solution and the molar mass of the solute.

Section 11.8

Perform calculations involving the osmotic pressure equation, and determine the molarity of a solution.

Section 11.9

Use osmotic pressure to calculate the molecular mass of a solute.

Section 11.10

Use temperature versus composition diagrams to explain fractional distillation.

98

Chapter 11 — Solutions and Their Properties

Lecture Outline 11.1. Solutions1 A. Heterogeneous mixtures – mixtures in which the mixing of components is visually nonuniform B. Homogeneous mixtures (solutions) – mixtures in which the mixing of components is visually uniform 1. Classified according to the size of the constituent particles a. Suspensions – contain particles greater than 1000 nm in diameter and visible with a low-power microscope b. Colloids – contain particles with diameters 2–1000 nm c. Solutions – contain particles the size of a typical ion or covalent molecule C. Any one state of matter can form a solution with any other state.

1. Seven different kinds of solutions – textbook Table 11.1 2. Gas or solid dissolved in a liquid a. Solute – the dissolved substance b. Solvent – the liquid 3. Liquid dissolved in a liquid a. Solute – the minor component b. Solvent – the major component 11.2. Energy Changes and the Solution Process2 A. Most solutions involve condensed phases.

B. Intermolecular forces important for explaining the properties of solutions C. Three types of interactions among particles to take into account 1. Solvent-solvent interactions 2. Solute-solute interactions 3. Solvent-solute interactions D. “Like dissolves like” 1. Solutions form when the three types of interactions are similar in kind and in magnitude.

2. Ionic solids dissolve in polar solvents.

3. Nonpolar organic substances dissolve in nonpolar organic solvents.

E. Solvated 1. Ions surrounded and stabilized by a shell of solvent molecules 2. Hydrated – water is the solvent F. Free-energy change for solution formation 1. ΔG is negative – spontaneous process and the substance dissolves 2. ΔG is positive – nonspontaneous process and the substance does not dissolve 3. ΔHsoln – enthalpy of solution a. Difficult to predict b. Can be either negative (exothermic) or positive (endothermic) 4. ΔSsoln – entropy of solution a. Usually positive b. Increase in molecular randomness during dissolution F. Variations in heats of solution due to interplay of three kinds of interactions 1. Solvent-solvent interactions – positive ΔH. a. need to overcome intermolecular forces between solvent molecules b. separate molecules to make room for solute molecules 2. Solute-solute interactions – positive ΔH a. Need to overcome intermolecular forces between solute molecules b. Lattice energy for ionic solids 3. Solvent-solute interactions – negative ΔH 1 Test Item File Questions: Multiple Choice 1-6; Short Answer 1 2 Test Item File Questions: Multiple Choice 7-28, 108-119; Algorithmic 1; Short Answer 2-3

99

Chapter 11 – Solutions and Their Properties a. Solvent molecules cluster around solute particles and solvate them b. Ionic substances in water i. Increase in hydration energy with decrease in cation size ii. Increase in hydration energy with an increase in ionic charge 4. Sum of three interactions determines whether ΔHsoln is endothermic or exothermic textbook Figure 11.4 11.3. Units of Concentration3 A. Concentration of a solution – exact amount of solute dissolved in a given amount of solvent B. Molarity Moles of solute Molarity (M) = Liters of solution 1. Advantages a. Simplifies stoichiometry calculations – uses moles instead of mass b. Simplifies volumetric titrations – amounts of solution measured by volume rather than by mass 2. Disadvantages a. Exact concentration depends on temperature – volume changes as temperature changes b. Need to know the density of the solution to determine amount of solvent present C. Mole fraction Moles of component Mole Fraction (X ) = Total moles in the solution 1. Advantages a. Independent of temperature b. Useful for calculations involving gas mixtures 2. Disadvantages – not convenient for liquid solutions D. Mass percent Mass of component Mass percent (mass%) = ×100 Total mass of solution 1. For very dilute solutions: Mass of component a. Parts per million (ppm) = × 10 6 Total mass of solution

b. Parts per billion (ppb) =

Mass of component × 10 9 Total mass of solution

2. For dilute aqueous solutions at room temperature, 1 kg solution = 1 L solution 1 mg solute a. 1 ppm = 1 L soln 1 μg solute b. 1 ppb = 1 L soln 3. Advantage a. Values independent of temperature b. Masses don't change when temperature changes 4. Disadvantages a. Difficult to measure mass of a liquid solution b. Need to know the density of a solution to convert to molarity E. Molality

3 Test Item File Questions: Multiple Choice 29-66; Algorithmic 2-17; Short Answer 4-10

100

Chapter 11 — Solutions and Their Properties

Molality (m) =

Moles of solute Mass of solvent (kg)

1. Advantages a. Temperature-independent b. Well suited for calculating certain properties of solutions 2. Disadvantages a. Difficult to measure mass of a liquid solution b. Need to know the density of a solution to convert to molarity 3. When dissolving ionic compounds, moles of solute = total moles of ions 11.4. Some Factors Affecting Solubility4 A. Saturated solution – solution in which number of ions leaving a crystal to go into solution = number of ions returning from solution to the crystal 1. Solute + Solvent = Solution 2. At equilibrium with undissolved solid B. Supersaturated solution – contains a greater than equilibrium amount of solute C. Solubility – amount of solute per unit of solvent needed to form a saturated solution 1. Physical property, characteristic of a particular substance 2. Temperature-dependent D. Effect of temperature on solubility 1. Gases less soluble as temperature increased 2. More difficult to predict effect on solubility of solids E. Effect of pressure on solubility 1. None on solids and liquids 2. Profound effect on solubility of gases a. Henry’s law b. Solubility = k × P i. k = Henry's law constant ii. P = partial pressure of gas 4. Increase in pressure leads to increase in solubility a. Due to change in position of equilibrium between dissolved and undissolved gas b. More gas particles forced into solution 11.5. Physical Behavior of Solutions: Colligative Properties5 A. Colligative properties – properties that depend on amount of dissolved solute but not on chemical identity of the solute 1. Boiling-point elevation 2. Freezing-point depression 3. Vapor pressure of a solution 4. Osmosis – migration of solvent and other small molecules through a semipermeable membrane 11.6. Vapor-Pressure Lowering of Solutions: Raoult’s Law6 A. Solutions with a nonvolatile solute — lower vapor pressure than pure solvent.

1. Raoult’s law: Psoln= Psolv × Xsolv 2. For ionic substances, calculate mole fractions based on total number of solute particles rather than on number of formula units.

B. Raoult’s law applies only to ideal solutions.

1. Solute concentrations are low.

2. Solute and solvent particles have similar intermolecular forces.

C. Solute-solvent intermolecular forces < solvent intermolecular forces; vapor pressure higher than predicted by Raoult’s law 4 Test Item File Questions: Multiple Choice 67-74, 120-121; Algorithmic 18-19; Short Answer 11-13 5 Test Item File Questions: Multiple Choice 75; Short Answer 14 6 Test Item File Questions: Multiple Choice 76-81, 122-126; Algorithmic 20-22; Short Answer 15-19

101

Chapter 11 – Solutions and Their Properties D. Solute-solvent intermolecular forces > solvent intermolecular forces; vapor pressure less than predicted by Raoult’s law E. Solutions with a volatile solute – vapor pressure of the solution always intermediate between the vapor pressures of the two pure substances 1. Ptotal = PA + PB 2. PA and PB calculated from Raoult's law PA = X A (PAo ) PB = X B (PBo ) Ptotal = X A (PAo ) + X B (PBo )

F. Free-energy change accompanies vapor pressure lowering – more negative ΔG, easier the vaporization process G. Liquid → gas: 1. ΔH is positive a. Need energy to overcome intermolecular forces b. Similar value for both solvent and solution 2. ΔS is positive a. Molecular disorder increases b. ΔS for liquid → vapor is smaller for solution than for pure solvent 3. Larger ΔG for solution (subtracting a smaller TΔS from ΔH) a. Vaporization more difficult b. Lower vapor pressure 11.7. Boiling-Point Elevation and Freezing-Point Depression of Solutions7 A. Boiling-point elevation and freezing-point depression of a solution relative to that of a pure solvent depends on number of solute particles.

1. ΔTb = Kb⋅ m 2. ΔTf = Kf ⋅ m 3. Use molality as the concentration unit because numbers of solute and solvent particles are independent of temperature.

B. Due to an entropy difference between pure solvent and solvent in a solution 1. At equilibrium ΔH vap Tb = ΔS vap 2. ΔSvap smaller for a solution, Tb larger 3. At equilibrium ΔH fus Tf = ΔSfus 4. ΔSfusion larger for a solution, Tf smaller.

11.8. Osmosis and Osmotic Pressure8 A. Semipermeable membrane – membranes that allow water or other small molecules to pass through, but block the passage of large solute molecules or ions B. Osmosis – migration of solvent and other small molecules through a semipermeable membrane 1. Passage from pure solvent side to solution side a. More favored b. Decrease in amount of liquid on solvent side c. Increase in amount of liquid on solution side d. Decrease in concentration of solution

7 Test Item File Questions: Multiple Choice 82-95; Algorithmic 23-27; Short Answer 20-21 8 Test Item File Questions: Multiple Choice 96-101, 127-128; Short Answer 22-24

102

Chapter 11 — Solutions and Their Properties C. Osmotic pressure of a solution ( Π ) – pressure needed to prevent osmotic flow of solvent through a semipermeable membrane 1. Π = MRT 2. Can use molarity because measurements made at the temperature specified in the equation.

3. Due to an increase in entropy when pure solvent passes through the membrane and mixes with the solution 11.9 Some Uses of Colligative Properties9 A. Common uses 1. Freezing-point depression a. Sprinkling of salt to melt snow b. Antifreeze in automobiles c. De-icing of airplane wings B. Desalination of seawater 1. Reverse osmosis – increase osmotic pressure to force molecules from solution side to solvent side C. Molar mass determinations 1. Can use any of the four colligative properties 2. Most accurate is osmotic pressure – magnitude of osmosis effect is so great 11.10. Fractional Distillation of Liquid Mixtures10 A. Fractional distillation – separation of a mixture of volatile liquids into fractions by boiling and condensing the vapors 1. Vapor enriched in the more volatile component 2. Condensed vapor also enriched in more volatile component 3. Repeat boil/condense cycle with condensed vapor phase many times. a. Can have complete purification of the more volatile liquid component b. Occurs naturally in a distillation column – textbook Figure 11.19 B. Represent fractional distillation with a liquid/vapor phase diagram 1. Plot T vs. composition – textbook Figure 11.18 2. Lower region represents liquid phase.

3. Upper region represents vapor phase.

4. Equilibrium region – between lower and upper regions a. Liquid and vapor coexist b. Lower curve – represents composition of liquid phase at a particular temperature c. Upper curve – represents composition of vapor phase at a particular temperature d. Tie line – connects points on upper and lower curve at a particular temperature 5. Fractional distillation – walk across tie lines to achieve desired purity

9 Test Item File Questions: Multiple Choice 102-106 10 Test Item File Questions: Multiple Choice 107, 129-140; Short Answer 25

103

Chapter 11 – Solutions and Their Properties

CHAPTER 11 SOLUTIONS AND THEIR PROPERTIES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 11.2

Hydration involves ion-dipole forces. Because opposite charges attract, water molecules differ in their orientation around an anion and around a cation.

Section 11.2

ΔSsoln is usually positive; ΔHsoln can be either negative or positive.

Section 11.2

The strength of an ion-dipole interaction increases and the hydration energy becomes more negative as the charge-to-radius ratio of the ion increases.

Section 11.3

Molarity contains a volume term in the denominator. As temperature increases, volume increases and molarity decreases. Mole fraction, weight percent, and molality contain no volume terms and are independent of temperature.

Section 11.3

In environmental applications parts per million is often taken to mean milligrams solute/liter of aqueous solution, which seems to contradict the use of ppm as a temperature-independent unit. However, the mass of solute is negligible relative to the mass of water, so 1 L of water ≈ 1kg of water ≈ 1 kg of solution.

Section 11.3

Students tend to be careless in their use of units. Stress that concentration units refer to solute, solvent, or solution, and that it is helpful to include these terms as part of the unit. For example, molality (mol/kg) means mol solute/kg solvent.

Section 11.4

Solubility of solids in liquids is usually given in units of grams solute/100 mL of solvent. Because volume changes with temperature, the temperature is also specified (usually 20oC ≈ room temperature in the United States).

Section 11.6

For a solution of a nonvolatile solid in a liquid, the vapor pressure of a solution is always lower than the vapor pressure of the pure solvent. For a solution of two volatile liquids, the vapor pressure is lower than the vapor pressure of the more volatile liquid and higher than the vapor pressure of the less volatile liquid.

Lecture/Laboratory Demonstration References Section 11.1

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Nonadditivity of Volumes,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 14. The volume of a solution of two miscible liquids is shown to not equal the sum of the volumes of the individual liquids before mixing.

Section 11.2

W. Bergquist, “Do ‘Likes Dissolve Likes’? An Illustration of Polar and Nonpolar Solvents,” J. Chem. Educ., Vol. 69, 1992, 158. Iodine is shown to dissolve in ethyl alcohol and an alcohol/water mixture, but not in water alone. The relationship between solubility and the molecular polarities of solvent and solute are noted.

Section 11.2

Walter H. Corkern and Linda L. Munchausen, “Solubility of Alcohols,” J. Chem.

Educ., Vol. 69, 1992, 928. An overhead projector demonstration.

104

Chapter 11 — Solutions and Their Properties

Section 11.2

Katia Pravia and David F. Maynard, “Why Don’t Oil and Water Mix?” J. Chem.

Educ., Vol. 73, 1996, 497. An overhead projector demonstration.

Section 11.4

Lee R. Summerlin, Christie L. Borgford, and Julie. B. Ealy, “Supersaturation,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 121-122. Excess solute suddenly precipitates from a supersaturated solution of sodium acetate trihydrate in this demonstration.

Section 11.4

C. A. Snyder and D. C. Snyder, “Simple Soda Bottle Solubility and Equilibria,” J. Chem. Educ., Vol. 69, 1992, 573. Bromocresol green indicator, added to a glass bottle of seltzer water, changes color from yellow to green to blue as carbon dioxide concentration is decreased. Observations are related to temperature and pressure effects on gas solubility, and LeChâtelier's principle is applied to the equilibria involved.

Section 11.4

Bassam Z. Shakhashiri, “Effect of Temperature and Pressure on the Solubility of Gases in Liquids,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp. 280-282.

Section 11.8

Joseph G. Morse and Ed Vitz, “A Simple Demonstration Model of Osmosis,” J. Chem. Educ., Vol. 76, 1999, 64-65.

Section 11.8

Bassam Z. Shakhashiri, “Osmotic Pressure of a Sugar Solution,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp. 286-289.

Section 11.8

Bassam Z. Shakhashiri, “Osmosis Through the Membrane of an Egg,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp. 283-285.

Section 11.8

Lee R. Summerlin, Christie L. Borgford, and Julie. B. Ealy, “Osmosis and the Egg Membrane,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 136-137.

Literature References Section 11.2

Todd P. Silverstein, “Polarity, Miscibility, and Surface Tension,” J. Chem. Educ., Vol. 70, 1993, 253.

Section 11.2

Barry K. Thornton, “An Analogy to Illustrate Miscibility of Liquids,” J. Chem.

Educ., Vol. 71, 1994, 156.

Section 11.2

Michael J. Sanger and Steven M. Badger III, “Using Computer-Based Visualization Strategies to Improve Students’ Understanding of Molecular Polarity and Miscibility,” J. Chem. Educ., Vol. 78, 2001, 1412-1416.

Section 11.3

Alvin D. White, “An Alternative Introduction to Mole Fraction,” J. Chem. Educ., Vol. 59, 1982, 153.

Section 11.3

Donald DeLorenzo, “Mole Fraction Analogies,” J. Chem. Educ., Vol. 57, 1980, 733.

105

Chapter 11 – Solutions and Their Properties

Section 11.3

Clifton E. Meloan, Mindy L. Meloan, and John M. Meloan “Candy Sprinkles to Illustrate One Part per Million,” J. Chem. Educ., Vol. 71, 1994, 658.

Section 11.4

Jamil Ahmad, “Crystallization from a Supersaturated Solution of Sodium Acetate,” J. Chem. Educ., Vol. 77, 2000, 1446.

Section 11.4

James H. Cragin, “Soft Drink Bubbles,” J. Chem. Educ., Vol. 60, 1983, 71.

Section 11.4

Doris R. Kimbrough, “Henry’s Law and Noisy Knuckles,” J. Chem. Educ., Vol. 76, 1999, 1509-1510.

Section 11.8

Sarah F. McDuffie and Catherine E. Matthews, “Antifreeze Solutions: The Colligative Properties of Antifreeze,” The Science Teacher, Vol. 63, 1996, 41-43.

Section 11.8

Donald DeLorenzo, “Freeze-Proof Bugs,” J. Chem. Educ., Vol. 58, 1981, 788.

Section 11.9

Gerald Parkinson, Charlene Crabb, and Takeshi Kamiya, “Seawater Gets Fresh,” Chem. Eng., Vol. 106(3), 1999, 32-35. Desalination by reverse osmosis.

Media References Section 11.2

Dissolution of NaCl in Water movie from the Instructor Resource DVD

Section 11.3

Solution Formation from a Solid movie from the Instructor Resource DVD

Section 11.3

Expressions of Concentration activity from the Instructor Resource DVD

Section 11.4

Henry’s Law movie from the Instructor Resource DVD

Section 11.4

Henry’s Law activity from the Instructor Resource DVD

Section 11.5

Colligative Properties activity from the Instructor Resource DVD

Section 11.6

Vapor Pressure activity from the Instructor Resource DVD

Section 11.7

Boiling-Point Elevation and Freezing-Point Depression activity from the Instructor Resource DVD

Section 11.9

Determination of Molar Mass from the Instructor Resource DVD

106

CHAPTER 12 CHEMICAL KINETICS Chapter Learning Goals for Students Section 12.1

Use a table of concentration versus time data to calculate an average rate of reaction over a period of time.

From the coefficients of a balanced chemical equation, express the relative rates of consumption of reactants and formation of products.

Section 12.2

Find the reaction order with respect to each reactant and the overall reaction order from a rate law.

Section 12.3

From a table of initial concentrations of reactants and initial rates, determine the order of reaction with respect to each reactant, the overall order of reaction, the rate law, the rate constant, and the initial rate for any other set of initial concentrations.

Section 12.4

Use the integrated first-order rate law to find the value of one variable, given values of the other variables.

Use concentration-versus-time, ln concentration-versus-time, and 1/concentrationversus-time plots to verify a first-order reaction.

Section 12.5

From a plot of concentration versus time, estimate the half–life of a first–order reaction.

Use the expression for half-life of a first-order reaction to determine t½ from k, or vice versa.

Section 12.6

Show how radiocarbon dating is used to determine the age of an object.

Perform first-order decay calculations involving medical uses of radioactivity.

Section 12.7

Use the integrated second-order rate law to find the value of one variable, given values of the other variables.

Use concentration-versus-time, ln concentration-versus-time, and 1/concentrationversus-time plots to verify a second-order reaction.

Use the expression for half-life of a second-order reaction to determine t½ from k, or vice versa.

Section 12.8

Use the integrated zeroth-order rate law to find the value of one variable, given values of the other variables.

Use concentration-versus-time, ln concentration-versus-time, and 1/concentrationversus-time plots to verify a zeroth-order reaction.

Section 12.9

Given a reaction mechanism, write the chemical equation for the overall reaction, identify the reaction intermediates, determine the molecularity of each elementary reaction and the molecularity of the overall reaction.

107

Chapter 12—Chemical Kinetics

Section 12.10

Write rate laws for elementary reactions.

Section 12.11

Given an experimental rate law and a reaction mechanism with an initial slow step, determine if the mechanism is consistent with the experimental rate law.

Given an experimental rate law and a reaction mechanism with an initial fast step, write a balanced equation for the overall reaction, determine if the mechanism is consistent with the experimental rate law, and relate the rate constant k to the rate constants for the elementary reactions.

Section 12.12

Sketch a potential energy profile, showing the activation energies for the forward and reverse reactions, determine whether the reaction is exothermic or endothermic, and suggest a plausible structure for the transition state.

Section 12.13

Prepare an Arrhenius plot, and determine the activation energy from the slope of the line.

Solve the Arrhenius equation for any variable, given the others.

Section 12.14

Sketch a potential energy profile, showing the activation energies for the forward and reverse reactions and how they are affected by the addition of a catalyst.

Given the mechanism for a reaction, identify the catalyst.

Section 12.15

Classify catalysts as homogeneous or heterogeneous.

108

Chapter 12—Chemical Kinetics

Lecture Outline 12.1. Reaction Rates1 A. Rate of a reaction – how fast the concentration of a reactant or a product changes per unit time Δ concentration Rate = Δ time 1. Increase in the concentration of a product per unit time 2. Decrease in the concentration of a reactant per unit time 3. Units: M/s or mol/(L⋅ s) a. Allows rate to be independent of the scale of the reaction b. Use a minus sign in calculating rate of disappearance of reactant B. Relative rates of product formation and reactant consumption 1. Depend on coefficients in the balanced equation 2. Must specify the reactant or product when quoting a rate.

C. Rate changes as the reaction proceeds.

1. Must specify the time.

2. Reaction rates decrease as the reaction mixture runs out of reactants.

D. Plot concentration (y-axis) versus time (x-axis).

1. Δ[concentration] and Δtime represent vertical and horizontal sides of a right triangle.

2. Slope of the triangle hypotenuse is the average rate during that time period.

E. Instantaneous rate at time t – the slope of the tangent to a concentration-time curve at time t F. Initial rate – the instantaneous rate at the beginning of a reaction (t = 0) 12.2. Rate Laws and Reaction Order2 A. Rate law – states the dependence of the reaction rate on concentration 1. Equation shows how the rate depends on the concentration of each reactant.

2. For the reaction a A + b B → products a. Rate law: Δ A m n Rate = =k A B Δt b. k = proportionality constant called the rate constant B. Reaction order – determined by the values of the exponents 1. Values of exponents in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction.

2. Values of m and n indicate reaction order with respect to A and B. a. Exponent = 1: first order b. Exponent = 2: second order c. Exponent = 3: third order 3. Overall reaction order = m + n 4. Indicates how change in concentration affects rate 5. Unrelated to coefficients in the balanced equation 6. Usually small positive integers, but can be negative, zero, or even fractions a. Exponent = 1: rate depends linearly on the concentration of the corresponding reactant b. Exponent = 0: rate is independent of the concentration of the corresponding reactant c. Exponent < 1: rate decreases as concentration of the corresponding reactant increases 12.3. Experimental Determination of a Rate Law3 A. Determination of reaction order (values of the exponents in a rate law) – measure initial rate of a reaction as a function of different sets of initial concentrations.

1 Test Item File Questions: Multiple Choice 1-14, 112-116; Algorithmic 1-4; Short Answer 1-3 2 Test Item File Questions: Multiple Choice 15-30; Algorithmic 5-7; Short Answer 4-10 3 Test Item File Questions: Multiple Choice 31-39, 117-129, 134; Short Answer 11

109

Chapter 12—Chemical Kinetics 1. Design pairs of experiments to investigate the effect of initial concentration of a single reactant on initial rate of change.

2. If, by doubling the concentration of a reactant, the rate also doubles, then the reaction is first order with respect to that reactant.

3. If, by doubling the concentration of a reactant, the rate increases by a factor of 22 = 4, then the reaction is second order with respect to that reactant.

4. If, by doubling the concentration of a reactant, the rate of the reaction increases by a factor of 23 = 8, then the reaction is third order with respect to that reactant.

5. Use initial rates to avoid complications from the reverse reaction. a. Initial rates measure only the rate of the forward reaction. b. Only reactants and catalysts appear in the rate law.

B. Can determine the value of k from the rate law.

1. Characteristic of a reaction 2. Depends on temperature 3. Does not depend on concentration 4. Units depend on the number of concentration terms in the rate law and values of the exponents.

12.4. Integrated Rate Law for a First–Order Reaction4 A. Integrated rate law 1. A concentration-time equation that allows us to calculate the concentration of a reactant at any time t or the fraction of a reactant that remains at any time t.

2. Can be used to calculate the time required for the initial concentration of a reactant to drop to any particular value or to any particular fraction of its initial concentration.

B. For the reaction A → products, the integrated rate law is [A]t = - kt ln [A]o 1. Can rearrange the equation to give ln[A]t = - kt + ln[A]o 2. Plot of ln[A]t versus time gives a straight line if the reaction is first order in A. 3. k = slope 12.5. Half-Life of a First-Order Reaction5 A. Half-life = t1/2 B. Time required for reactant concentration to drop to one-half its initial value 0.693 t1/2 = k C. For a first-order reaction half-life is constant.

D. Depends only on the rate constant 12.6. Radioactive Decay Rates6 A. Decay of radioactive nuclei – a first-order process 1. Integrated rate law – analogous to that given in section 12.4.

Nt = - kt ln N0 a. N0 is the number of radioactive nuclei initially present. b. Nt is the number of radioactive nuclei present at time t. c. k is the decay constant.

4 Test Item File Questions: Multiple Choice 40-46, 130-131, 136-139; Algorithmic 8-12; Short

Answer 12 5 Test Item File Questions: Multiple Choice 47-51, 132-133, 135; Algorithmic 13-16; Short Answer 13 6 Test Item File Questions: Multiple Choice 52-61, 140-147; Algorithmic 16-17; Short Answer 14 110

Chapter 12—Chemical Kinetics 2. Decay rate – characterized by half-life 0.693 t1/2 = k 3. Calculation of the ratio of remaining and initial amounts of radioactive sample Nt/N0 at any time t ⎛ Nt ⎞ ⎛ t ⎞ ln ⎜ ⎟ = - 0.693 ⎜ ⎟ ⎝ N0⎠ ⎝ t1/2 ⎠ 4. Relationship between decay rate at time t, initial decay rate, half-life, and sample age Decay rate at time t kNt Nt = = Initial decay rate kN 0 N 0 Therefore ⎛ Decay rate at time t ⎞ ⎛ t ⎞ ln ⎜ ⎟ = - 0.693 ⎜ ⎟ ⎝ Initial decay rate ⎠ ⎝ t 1/2 ⎠ B. Radiocarbon dating 1. Depends on the slow and constant production of radioactive carbon-14 in the upper atmosphere by neutron bombardment of nitrogen atoms 14 N + 01 n → 146 C + 11 H 7 2. Carbon-14 reacts to form CO2, which mixes with 12CO2 and is taken up by plants during photosynthesis.

14 C + O 2 → 14 CO 2 3. Ratio of 14C to 12C in a living organism is the same as that in the atmosphere.

4. 14C /12C decreases when the organism dies because 14C undergoes radioactive decay.

14 C → 147 N + −10 e 6 5. For 14C, t1/2 = 5715 y 6. Can determine how long ago the organism died by measuring amount of 14C remaining C. Use other radionuclides for dating rocks.

1. 238U a. For 238U, t1/2 = 4.46 × 109 y b. 238U decays through a series to give 206Pb c. Measurement of 238U /206Pb ratio can determine the age of uranium-containing rock.

40 2. K a. For 40K, t1/2 = 1.28 × 109 y b. Potassium-40 decays through electron capture and positron emission 40 Ar K + −01e - → 40 19 18 40 19

K → 4108 Ar + 01e -

c. Compare amount of 40Ar to 40K remaining in sample 12.7. Second-Order Reactions7 A. Integrated rate law for a second-order reaction 1 1 = kt + [A]t [A]o B. Plot of 1/[A]t versus time gives a straight line if the reaction is second order.

1. Slope of line = k 2. Intercept = 1/[A]o C. Half-life for a second-order reaction – depends on both the rate constant and the initial concentration.

7 Test Item File Questions: Multiple Choice 62-69, 130-131; Algorithmic 18-21; Short Answer 15

111

Chapter 12—Chemical Kinetics

t1/2 =

1 k[A]o

12.8. Zeroth–Order Reactions8 A. Integrated rate law for a zeroth-order reaction [A] = - kt + [A]o B. Rate is independent of the concentration of the reactant.

C. Plot of [A] versus time gives a straight line if zeroth–order. a. Slope = – k b. Intercept = [A]o D. Relatively uncommon; occurs under special circumstances.

12.9. Reaction Mechanisms9 A. Reaction mechanism – the sequence of molecular events, or reaction steps, that defines the pathway from reactants to products 1. Reaction steps – involve breaking of chemical bonds and/or making of new bonds 2. Knowing reaction mechanisms allows for better control of known reactions and predictions of new reactions.

B. Elementary step – a single step in a reaction mechanism 1. Description of an individual molecular event (collisions of individual molecules) 2. Classified on the basis of molecularity a. Molecularity – the number of molecules on the reactant side of the chemical equation b. Unimolecular reaction – elementary reaction that involves a single reactant molecule c. Bimolecular reaction – elementary reaction that results from energetic collisions between two reactant molecules d. Termolecular reaction – involves three atoms or molecules; rare 3. Reaction intermediate – a species that is formed in one step of a reaction mechanism and consumed in a subsequent step a. Does not appear in the net equation for the overall reaction b. Presence noticed only in elementary steps C. Balanced equation for an overall reaction 1. Provides no information about how the reaction occurs 2. Merely describes reaction stoichiometry 12.10. Rate Laws for Elementary Reactions10 A. Rate law for overall reaction – determined by experimentation B. Rate law for an elementary reaction – determined from its molecularity 1. Elementary reaction is an individual molecular event.

2. Contains the concentration of each reactant raised to an exponent equal to its coefficient in the chemical equation for the elementary reaction 3. Applies only to elementary reactions, not overall reactions 4. Rate of a unimolecular reaction is first order in the concentration of the reactant molecule.

5. Overall reaction order for an elementary reaction is equal to its molecularity – textbook Table 12.5 12.11. Rate Laws for Overall Reactions11 A. Experimentally observed rate law for an overall reaction depends on the reaction mechanism.

1. If the overall reaction occurs in a single elementary step: experimental rate law = rate law for the elementary step 2. If the overall reaction occurs in two or more steps: a. Rate-determining step = the slowest step in a reaction mechanism 8 Test Item File Questions: Multiple Choice 70-73; Short Answer 16 9 Test Item File Questions: Multiple Choice 74-79, 148-150; Short Answer 17-18 10 Test Item File Questions: Multiple Choice 80-83 11 Test Item File Questions: Multiple Choice 84-86; Short Answer 19

112

Chapter 12—Chemical Kinetics b. Limits the rate at which reactants can be converted to products c. Overall reaction can occur no faster than the speed of the rate-determining step. d. A rate law derived for a reaction for which the slowest step is not the first step of the mechanism will require combining information from two or more steps in order for it to take the form of the experimentally-determined rate law in which only initial reactants appear.

B. Two criteria for an acceptable reaction mechanism 1. Elementary steps must sum to give the overall reaction.

2. Mechanism must be consistent with the observed rate law for the overall reaction.

C. Procedure used for establishing a reaction mechanism 1. Determine the overall rate law experimentally.

2. Devise a series of elementary steps.

3. Predict the rate law based on the reaction mechanism.

4. If observed and predicted rate laws agree, the proposed mechanism is a plausible pathway for the reaction.

5. Easy to disprove a mechanism; impossible to “prove” a mechanism.

12.12. Reaction Rates and Temperature: The Arrhenius Equation12 A. Reaction rates tend to double when the temperature is increased by 10°C. B. Collision theory model: a bimolecular reaction occurs when two properly oriented reactant molecules come together in a sufficiently energetic collision.

C. For the reaction: A + BC → AB + C 1. For a single-step reaction, a new bond, A–B, develops at the same time as the old bond, B–C, breaks.

2. Nuclei pass through a configuration in which all three atoms are weakly linked together. a. A + B–C → A- -B- -C → A–B + C 3. Need energy to overcome repulsions a. Comes from kinetic energy of the colliding particles b. Stored as potential energy in A- -B- -C. c. A- -B- -C has more potential energy than either the reactants or products.

4. A potential energy barrier exists that must be surmounted before reactants can be converted to products. a. Potential energy profile – plot of potential energy versus reaction progress b. Activation energy, Ea, – the height of the barrier c. Transition state (or activated complex) – configuration of atoms at the maximum in the potential energy profile 5. All the energy needed to climb the potential energy barrier must come from the kinetic energy of the colliding molecules.

D. Comparison of collision rates and reaction rates leads to experimental evidence for the idea of an activation energy barrier.

1. Only a small fraction of collisions lead to reaction. a. Very few collisions occur with a kinetic energy as large as the activation energy. b. Area under the curve to the right of Ea represents the fraction of the collision with an energy equal to or greater than the activation energy. c. Fraction of molecules having sufficient energy for reaction - E / RT

f =e a 2. Distribution of collision energies broadens and shifts to higher energies with an increase in temperature.

3. Exponential increase in the fraction of collisions that lead to products 4. Accounts for exponential dependence of reaction rates on temperature 5. Explains why reaction rates are so much lower than collision rates.

E. Orientation requirement also reduces the fraction of collisions that lead to products.

12 Test Item File Questions: Multiple Choice 87-95, 151-159

113

Chapter 12—Chemical Kinetics 1. Reactants won’t react unless the orientation of the reaction partners is correct for formation of the transition state.

2. Steric factor, p – fraction of collisions having proper orientation F. For a bimolecular collision between A and B. 1. Collision rate = Z [A][B]; Z is a constant related to collision frequency.

2. Reaction rate < collision rate by a factor of p × f.

3. Reaction rate = p f Z [A][B] 4. Reaction rate = k[A][B] 5. Rate constant, k

k = pfZ = pZe G. Arrhenius equation k = Ae

- Ea / RT

- Ea / RT

where A = frequency factor = pZ

12.13. Using the Arrhenius Equation13 A. Can determine the activation energy if values of the rate constant are known at different temperatures.

B. Rearrange the equation to get ⎛ −E ⎞ ⎛ 1 ⎞ ln k = ⎜⎜ a ⎟⎟ ⎜ ⎟ + ln A ⎝ R ⎠⎝T ⎠ 1. Plot of ln k versus 1/T gives a straight line.

2. slope = –Ea/R C. Can estimate the activation energy from rate constants at just two temperatures, using another form of the equation. ⎛ k ⎞ ⎛ −E ⎞ ⎛ 1 1 ⎞ ln ⎜⎜ 2 ⎟⎟ = ⎜⎜ a ⎟⎟ ⎜⎜ − ⎟⎟ ⎝ k ⎠ ⎝ R ⎠ ⎝ T2 T1 ⎠ 12.14. Catalysis14 A. Catalyst – a substance that increases the rate of a reaction without being consumed in the reaction 1. Important in chemical industry and in living organisms 2. Chemical industry – favors formation of specific products and lower reaction temperatures 3. Living organisms – enzymes (catalysts) facilitate specific reactions of crucial biological importance.

B. Accelerates the rate of a reaction by making a new and more efficient pathway available for the conversion of reactants to products 1. Speeds up reaction in two ways a. Increases the frequency factor A b. Decreases the activation energy 2. Rate constant is more sensitive to the Ea, and a catalyst usually functions by lowering the activation energy.

12.15. Homogeneous and Heterogeneous Catalysts15 A. Homogeneous catalyst – one that exists in the same phase as the reactants B. Heterogeneous catalyst – exists in a different phase from the reactants 1. Mechanism complex and not well understood 2. Important steps a. Adsorption of reactants onto the surface of the catalyst b. Conversion of reactants to products on the surface c. Desorption of products from the surface 13 Test Item File Questions: Multiple Choice 96-102; Algorithmic 22; Short Answer 20-22 14 Test Item File Questions: Multiple Choice 103-108; Short Answer 23 15 Test Item File Questions: Multiple Choice 109-111; Short Answer 24-25

114

Chapter 12—Chemical Kinetics 3. Adsorption steps a. Chemical bonding of reactants to the highly reactive metal atoms on the surface b. Breaking or weakening of bonds in the reactants 4. Industrial chemical processes use mostly heterogeneous catalysts due to the ease of separation of the catalyst from the reaction products.

5. Used in automobile catalytic converters

115

Chapter 12—Chemical Kinetics

CHAPTER 12 CHEMICAL KINETICS Teaching Tips, Points of Emphasis, and Common Misconceptions None

Lecture/Laboratory Demonstration References Section 12.1

Bassam Z. Shakhashiri, “Color Variations of the Landolt Reaction,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp. 26-28. Three pairs of colorless solutions are mixed. After 10 seconds the three mixtures turn red, yellow, and blue, respectively.

Each clock reaction is the reduction of iodate by bisulfite. The observed colors are related to the amount of triiodide produced, which is dependent upon the relative amounts of iodate and bisulfite reacted, and whether or not starch is present.

Section 12.1

Lee R. Summerlin and James L. Ealy, “The Starch-Iodine Clock Reaction,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 107-108. The iodine clock experiment.

Section 12.1

Bassam Z. Shakhashiri, “Hydrogen Peroxide Iodine Clock: Oxidation of Potassium Iodide by Hydrogen Peroxide,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp. 37-43. The iodine clock experiment.

Section 12.1

Xavier Creary and Karen M. Morris, “A New Twist on the Iodine Clock Reaction: Determining the Order of a Reaction,” J. Chem. Educ., Vol. 76, 1999, 530-531.

Section 12.2

Steven W. Wright, “Tick Tock, a Vitamin C Clock,” J. Chem. Educ., Vol. 79, 2002, 40A-40B. Presents an activity that uses supermarket chemicals to perform a clock reaction in which the endpoint is signaled by an abrupt change in the appearance from colorless to blue-black. This activity can be used to explore reaction kinetics and the effect of reactant concentrations on the apparent rate of reaction.

Section 12.3

Bassam Z. Shakhashiri, “Hydrolysis of 2-chloro-2-methylpropane,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp. 56-64. The time required for the color change in this reaction is shown to increase with increasing sodium hydroxide concentration, decrease with increasing 2-chloro-2-methylpropane concentration, and increase as the ratio of water/acetone increases. These observations are related to the mechanism of 2-chloro-2-methylpropane hydrolysis.

Section 12.11

Jose E. Cortes-Figueroa and Deborah A. Moore, “Using CBL Technology and a Graphing Calculator toTeach the Kinetics of Consecutive First-Order Reactions,” J. Chem. Educ., Vol. 76, 1999, 635-638.

Section 12.12

Bassam Z. Shakhashiri, “Lightsticks,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 146-152. Cyalume® lightsticks are used to demonstrate the effect of temperature on reaction rates.

116

Chapter 12—Chemical Kinetics

Section 12.12

Bassam Z. Shakhashiri, Lloyd G. Williams, Glen E. Dirreen, and Ann Francis, “Cool-Light Chemiluminescence,” J. Chem. Educ., Vol. 58, 1981, 70-72. Cyalume® lightstick chemiluminescence reactions.

Section 12.15

A. Sae, “Hydrogen Peroxide Demo Curing the Brown-Bottle-with-Cap Syndrome,” J. Chem. Educ., Vol. 71, 1994, 433. Hydrogen peroxide solutions from three different consumer products are decomposed catalytically with iodide producing oxygen gas and water in the presence of bubble bath. The rate of oxygen production is proportional to the amount of bubbling observed. Observations are related to the reaction mechanism.

Section 12.15

Lee R. Summerlin and James L. Ealy, “Catalytic Decomposition of Hydrogen Peroxide: Foam Production,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 101102. Oxygen gas is catalytically produced in the presence of detergent resulting in the formation of a large quantity of foam.

Section 12.15

Christine L Copper and Edward Koubeck, “An Experiment To Demonstrate How a Catalyst Affects the Rate of a Reaction,” J. Chem. Educ., Vol. 76, 1999, 1714-1715.

Describes a chemistry experiment that allows students to calculate rates of reaction, orders of reaction, and activation energies. The activity demonstrates that to increase a reaction’s rate, a catalyst need only provide any additional pathway for the reaction, not necessarily a pathway having lower activation energy.

Literature References Section 12.2

Lionello Pobliani and Mario N. Berberan-Santos, “Inflation Rates, Car Devaluation, and Chemical Kinetics,” J. Chem. Educ., Vol. 73, 1996, 950-952.

Section 12.2

Charles J. Marzzacco, “An Analogy to Help Students Understand Reaction Orders,” J. Chem. Educ., Vol. 75, 1998, 482.

Section 12.4

Edward Todd Urbansky, “Don’t Be Tricked by Your Integrated Rate Plot,” J. Chem.

Educ., Vol. 78, 2001, 921-923.

Section 12.6

Kent J. Crippin and Robert D. Cutright, “Modeling Nuclear Decay: A Point of Integration Between Chemistry and Mathematics,” J. Chem. Educ., Vol. 75, 1998, 434-436.

Section 12.6

M. W. Rowe, “Radioactive Dating: A Method for Geochronology,” J. Chem. Educ., Vol. 62, 1985, 580-584.

Section 12.6

M. W. Rowe, “Archeological Dating,” J. Chem. Educ., Vol. 63, 1986, 16-20.

Section 12.6

Ron DeLorenzo, “California Earthquakes: Predicting the Next Big One Using Radiocarbon Dating,” J. Chem. Educ., Vol. 57, 1980, 601.

Section 12.8

Francisco J. Arnaiz, “Mice in the Box for Zero-Order Kinetics,” J. Chem. Educ., Vol. 76, 1999, 1458.

Section 12.9

John J. Fortman, “Pictorial Analogies XIII: Kinetics and Mechanisms,” J. Chem.

Educ., Vol. 71, 1994, 848-849.

117

Chapter 12—Chemical Kinetics

Section 12.9

Arthur M. Last, “Doing the Dishes: An Analogy for Use in Teaching Reaction Kinetics,” J. Chem. Educ., Vol. 62, 1985, 1015-1016.

Section 12.11

Richard A. Potts, “Auto Analogies,” J. Chem. Educ., Vol. 62, 1985, 579.

Section 12.11

David W. Ball, “Another Auto Analogy: Rate-Determining Steps,” J. Chem. Educ., Vol. 64, 1987, 486-487.

Section 12.12

Ilya A. Leenson, “The Arrhenius Law and Storage of Food,” J. Chem. Educ., Vol. 76, 1999, 504-505.

Section 12.12

Keith L. Laidler, “Just What Is a Transition State?” J. Chem. Educ., Vol. 65, 1988, 540-542.

Section 12.12

Kent W. Piepgrass, “Audience-Appropriate Analogies: Collision Theory,” J. Chem.

Educ., Vol. 75, 1998, 72.

Section 12.12

Lee A. Krug, “The Collision Theory and an American Tradition,” J. Chem. Educ., Vol. 64, 1987, 1000.

Section 12.14

Doris Kolb, “Catalysis,” J. Chem. Educ., Vol. 56, 1979, 743-747.

Section 12.14

Albert Haim, “Catalysis: New Reaction Pathways, Not Just a Lowering of the Activation Energy,” J. Chem. Educ., Vol. 66, 1989, 935-937.

Section 12.15

Cynthia M. Friend, “Catalysis on Surfaces,” Scientific American, April 1993, 74-79.

Section 12.15

Sir John Meurig Thomas, “Solid Acid Catalysts,” Scientific American, April 1992, 112-118.

Section 12.15

Mitch Jacoby, “Getting Auto Exhausts to Pristine,” Chem. Eng. News, January 25, 1999, 36-44.

Section 12.15

Robert J. Farrauto, Ronald M. Heck, and Barry K. Speronello, “Environmental Catalysts,” Chem. Eng. News., September 7, 1992, 34-44.

Section 12.15

Reed A. Howald, “The Fizz Keeper, a Case Study in Chemical Education, Equilibrium, and Kinetics,” J. Chem. Educ., Vol. 76, 1999, 208-209.

Media References Section 12.1

Rate of Decomposition of N2O5 activity from the Instructor Resource DVD

Section 12.5

First-Order Process movie from the Instructor Resource DVD

Section 12.7

Integrated Rate Law, Second-Order Reaction activity

Section 12.9

Bimolecular Reaction movie from the Instructor Resource DVD

Section 12.13

Arrhenius Equation activity from the Instructor Resource DVD

Section 12.14

Catalysis movie from the Instructor Resource DVD 118

Chapter 12—Chemical Kinetics

Section 12.15

Surface Reaction – Hydrogenation movie from the Instructor Resource DVD

119

CHAPTER 13 CHEMICAL EQUILIBRIUM Chapter Learning Goals for Students Section 13.1

Describe the equilibrium state in terms of the rate of the forward reaction, the rate of the reverse reaction, and concentrations of reactants and products.

Section 13.2

Given any balanced chemical equation representing a homogenous equilibrium, write the equilibrium equation.

From the equilibrium concentrations of products and reactants, calculate the equilibrium constant Kc. Judge whether a mixture is at equilibrium by comparing the equilibrium constant expression to the value of the equilibrium constant Kc.

Section 13.3

Given the equilibrium partial pressures of reactants and products, calculate the equilibrium constant Kp. From a value of Kc and a balanced equation, calculate Kp. From a value of Kp and a balanced equation, calculate Kc.

Section 13.4

Given any balanced chemical equation representing a heterogeneous equilibrium, write the equilibrium equation.

Section 13.5

From the value of Kc or Kp, determine whether mainly products or mainly reactants exist at equilibrium.

For a given mixture of reactants and products, determine whether a system is at equilibrium. If it is not, determine the direction in which the reaction must go to achieve equilibrium.

Given Kc and initial concentrations of reactants and/or products, calculate the equilibrium concentrations of reactants and/or products.

Given Kp and initial partial pressures of reactants and/or products, calculate the equilibrium partial pressures of reactants and/or products.

Section 13.6

State the three factors that can be employed to alter the composition of an equilibrium mixture.

Section 13.7

Apply Le Châtelier’s principle to determine the reaction direction when a system at equilibrium reacts to a stress applied to the system in the form of a concentration change.

Section 13.8

Apply Le Châtelier’s principle to determine the reaction direction when a system at equilibrium reacts to a stress applied to the system in the form of a pressure or volume change.

Section 13.9

Apply Le Châtelier’s principle to determine the reaction direction when a system at equilibrium reacts to a stress applied to the system in the form of a temperature change.

120

Chapter 13—Chemical Equilibrium

Section 13.10

Describe the effect of adding a catalyst to a system at equilibrium

Section 13.11

Describe the relationship between the equilibrium constant and the ratio of the rate constants for the forward and reverse reactions. Solve problems involving this relationship.

121

Chapter 13—Chemical Equilibrium

Lecture Outline 13.1. The Equilibrium State1 A. Reversible reaction – reaction in which product molecules recombine to form reactant molecules 1. Reactants – substances on the left side of the equation 2. Products – substances on the right side of the equation 3. Use a double arrow to indicate a reaction at equilibrium.

Reactants ⇌ Products B. With time, concentrations of reactants decrease and concentrations of products increase until all concentrations level off at constant equilibrium values.

1. Rates of the forward and reverse reactions are equal.

2. State of chemical equilibrium a. Equilibrium mixture – the mixture of reactants and products in the equilibrium state b. Dynamic state – forward and reverse reactions continue at equal rates c. No net conversion of reactants to products 13.2. The Equilibrium Constant Kc 2 A. The equilibrium mixture obeys an equilibrium equation (or law of mass action).

B. For the reaction: a A + b B ⇌ c C + d D [C]c [D]d Kc = [A]a [B]b C. Kc = equilibrium constant 1. Units depend on the particular equilibrium expression.

2. Units usually omitted when quoting values of equilibrium constants.

D. Equilibrium constant expression – expression on right side of equilibrium equation E. For the reverse reaction 1. Equilibrium expression = reciprocal of the original expression 2. Equilibrium constant for reverse reaction 1 K c' = Kc 13.3. The Equilibrium Constant Kp3 A. For gas-phase reactions, use partial pressures rather than molar concentrations in equilibrium equations.

B. Kp – equilibrium constant defined using partial pressures C. Relationship between Kc and Kp 1. Kp = Kc(RT)Δn 2. Δn = the sum of the coefficients of gaseous products minus the sum of the coefficients of gaseous reactants 13.4. Heterogeneous Equilibria4 A. Homogeneous equilibria – reactants and products in a single phase B. Heterogeneous equilibria – reactants and products present in more than one phase 1. Solids – constant molar concentrations a. Calculated from the densities and molar masses b. Independent of its amount 2. Concentrations of pure solids or pure liquids not included when writing the equilibrium equation for any heterogeneous equilibrium 1 Test Item File Questions: Multiple Choice 1, 98; Short Answer 1-2 2 Test Item File Questions: Multiple Choice 2-15, 99-115; Algorithmic 1; Short Answer 3-4 3 Test Item File Questions: Multiple Choice 16-30; Algorithmic 2; Short Answer 5-10 4 Test Item File Questions: Multiple Choice 31-41, 127-128; Short Answer 11

122

Chapter 13—Chemical Equilibrium 3. The concentrations are constant and included in value of equilibrium constant.

13.5. Using the Equilibrium Constant5 A. Judging extent of reaction (how much reactant is converted to product) 3 1. Large value of Kc ( > 10 ) a. Products predominate over reactants. b. Reaction proceeds almost to completion. c. Position of reaction is to the right.

–3 2. Small value of Kc ( < 10 ) a. Reactants predominate over products. b. Reaction hardly proceeds. c. Position of the reaction is to the left.

–3 3 3. Intermediate value of Kc (10 < Kc < 10 ) a. Appreciable amounts of both reactants and products present. b. Both forward and reverse reactions are important.

B. Predicting the direction of reaction 1. Reaction quotient, Qc, same as the equilibrium constant expression except that concentrations are not necessarily equilibrium values.

2. Predict direction of reaction by comparing the value of Qc to Kc. a. Qc < Kc; reaction goes form left to right. b. Qc > Kc; reaction goes from right to left. c. Qc = Kc; reaction is at equilibrium.

13.6. Factors That Alter the Composition of an Equilibrium Mixture: LeChâtelier’s Principle6 A. Factors that alter the composition of an equilibrium mixture 1. Concentrations of reactants or products 2. Pressure and volume 3. Temperature B. Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, a net reaction occurs in the direction that relieves the stress.

1. Predicts changes in equilibrium mixture composition if one of the above changes occurs.

2. Reaction moves in a direction that relieves the stress.

13.7. Altering an Equilibrium Mixture: Changes in Concentration7 A. Disturb an equilibrium by adding or removing reactant or product.

1. Stress of an added reactant or product relieved by reaction in the direction that consumes the added substance. a. Add reactant – reaction shifts right (toward product) b. Add product – reaction shifts left (toward reactant) 2. Stress of removing reactant or product is relieved by reaction in the direction that replenishes the removed substance. a. Remove reactant – reaction shifts left b. Remove product – reaction shifts right B. Changes occur due to change in value of Qc. 1. Add reactant – denominator in Qc expression becomes larger. a. Qc < Kc b. To return to equilibrium, Qc must increase. i. Numerator of Qc expression must increase and the denominator must decrease. ii. Implies net conversion of reactants to products. iii. Reaction shifts right.

2. Remove reactant – denominator in Qc expression becomes smaller. a. Qc > Kc 5 Test Item File Questions: Multiple Choice 42-69, 116-118; Algorithmic 3-14; Short Answer 12-17 6 Test Item File Questions: Multiple Choice 70-72 7 Test Item File Questions: Multiple Choice 73-77, 124; Short Answer 18-19

123

Chapter 13—Chemical Equilibrium b. To return to equilibrium, Qc must decrease. i. Numerator of Qc expression must decrease and the denominator must increase. ii. Implies net conversion of products to reactants. iii. Reaction shifts left.

13.8. Altering an Equilibrium Mixture: Changes in Pressure and Volume8 A. If the number of moles of gaseous reactants in the balanced equation is different from the number of moles of gaseous products, a change in pressure (due to changing volume) changes the equilibrium mixture composition.

B. Increase in pressure (due to decrease in volume) results in reaction in the direction of fewer number of moles of gas.

C. Decrease in pressure (due to increase in volume) results in reaction in the direction of greater number of moles of gas.

D. Changes occur due to change in value of Qc. 1. Decrease volume – molarity (= n/V) increases.

2. If reactant side has more moles of gas a. Increase in denominator is greater than increase in numerator. b. Qc < Kc c. To return to equilibrium, Qc must increase. i. Numerator of Qc expression must increase and the denominator must decrease. ii. Implies net conversion of reactants to products (shifts toward fewer moles of gas).

3. If product side has more moles of gas a. Increase in numerator is greater than increase in denominator. b. Qc > Kc c. To return to equilibrium, Qc must decrease. i. Denominator of Qc expression must decrease and the numerator must increase. ii. Implies net conversion of products to reactants (shifts toward fewer moles of gas).

E. If reaction involves no change in the number of moles of gas, there is no effect on equilibrium mixture composition.

F. For a heterogeneous equilibrium 1. Volume is nearly independent of pressure.

2. Effect of pressure changes on solids and liquids can be ignored.

G. Change in pressure due to addition of an inert gas.

1. No change in the molar concentrations of reactants or products 2. No effect on composition of equilibrium mixture 13.9. Altering an Equilibrium Mixture: Changes in Temperature9 A. Disturb an equilibrium by a change in concentration, pressure, or volume.

1. Qc ≠ Kc 2. Value of Kc remains constant with constant T. B. Change in temperature always changes the equilibrium constant.

1. Depends on sign of ΔH° for the reaction.

2. Exothermic reaction (ΔH° = –) – equilibrium constant decreases as the temperature increases.

3. Endothermic reaction (ΔH°= +) – equilibrium constant increases as the temperature increases.

C. Le Châtelier’ principle – add heat to an equilibrium mixture, net reaction occurs in the direction that relieves the stress of the added heat.

1. Endothermic reaction – heat is absorbed by reaction in the forward direction (heat is written on the reactant side of the reaction). a. Equilibrium mixture contains more product than reactant. b. Kc increases with increasing temperature.

8 Test Item File Questions: Multiple Choice 78-80, 119-120, 123; Short Answer 20-22 9 Test Item File Questions: Multiple Choice 81-86, 121-122, 125; Short Answer 23-24

124

Chapter 13—Chemical Equilibrium 2. Exothermic reaction – heat is absorbed by reaction in the reverse direction (heat is written on the product side of the reaction). a. Equilibrium mixture contains more reactant than product. b. Kc decreases with decreasing temperature.

13.10. The Effect of a Catalyst on Equilibrium10 A. Catalyst increases the rate of a chemical reaction.

1. Provides a new, lower energy pathway 2. Forward and reverse reactions pass through the same transition state.

3. Rates for forward and reverse reactions increase by the same factor.

4. Does not affect the composition of the equilibrium mixture a. Does not alter the equilibrium constant or equilibrium concentrations b. Does not appear in the balanced chemical equation 5. Can influence choice of optimum reaction conditions 13.11. The Link Between Chemical Equilibrium and Chemical Kinetics11 A. For the reaction: A + B ⇌ C + D 1. Rate of forward reaction = kf [A][B] 2. Rate of reverse reaction = kr [C][D] 3. At equilibrium: kf[A][B] = kr[C][D] kf [C][D] = = Kc kr [A][B] B. Relative values of kf and kr determine the composition of the equilibrium mixture.

1. kf >> kr a. Kc is very large b. Reaction goes to completion c. Irreversible reaction d. Reverse reaction too slow to be detected 2. kf ≈ kr a. Kc ≈ 1 b. Reactants and products are present at equilibrium c. Reversible reaction

10 Test Item File Questions: Multiple Choice 87-91, 126; Short Answer 25 11 Test Item File Questions: Multiple Choice 92-97

125

Chapter 13—Chemical Equilibrium

CHAPTER 13 CHEMICAL EQUILIBRIUM Teaching Tips, Points of Emphasis, and Common Misconceptions Section 13.2

The subscript c in Kc means that the amounts of reactants and products are expressed as molar concentrations.

Section 13.2

Both the equilibrium equation and the value of the equilibrium constant depend on how the chemical equation is written and balanced.

Section 13.5

For an equilibrium system the concentration of any reactant or product can never be zero. No matter how small or how large the equilibrium constant, there must always be some products and some reactants in order for an equilibrium to exist.

Section 13.5

The equilibrium constant expression = Kc only at equilibrium. The equilibrium constant expression = Qc, the reaction quotient, whether or not the system is at equilibrium.

Section 13.5

Students should apply these criteria results not by rote memorization but by analysis of what must occur when Qc ≠ Kc

Section 13.5

Strongly encourage using tables. Although not necessary, it is less confusing for many students to make the coefficient used with the unknown, x, the same as the coefficient in the balanced chemical equation.

Section 13.5

Mathematical tricks are not apparent to many students. Point out the value of setting up the problem to obtain a perfect square.

Section 13.5

Mathematics sometimes yields results that have no physical meaning. Students should learn to analyze answers to problems to determine whether they make sense.

Section 13.7

When an equilibrium is upset by a change in concentration, pressure and volume, or temperature, equilibrium will be restored, but the concentrations of reactants and products will be different from those in the original equilibrium mixture.

Section 13.7

If a reactant or product is added to a system at equilibrium, the reaction will occur in a direction away from that reactant or product. If a reactant or product is removed from a system at equilibrium, the reaction will occur in a direction toward that reactant or product.

Section 13.10

The addition of a catalyst equally affects the rate of the forward and of the reverse reaction; hence, a catalyst does not change the equilibrium mixture or the value of the equilibrium constant, K.

Lecture/Laboratory Demonstration References Section 13.1

Lee R. Summerlin and James L. Ealy, Jr., “Equilibrium in the Gas Phase,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 60-61. Reddish-brown nitrogen dioxide is produced and placed in two test tubes where it exists in equilibrium with colorless nitrogen dioxide. One test tube is placed in a boiling water bath, causing an increase in the 126

Chapter 13—Chemical Equilibrium intensity of the color due to an increase in the intensity in the amount of NO2. The second test tube is placed in an ice-water bath, which causes the gaseous mixture to become almost colorless due to a decrease in the amount of NO2. When returned to room temperature, the gaseous mixtures in both test tubes exhibit the same brown color. Pressure and temperature effects on the equilibrium are related to the stoichiometry and enthalpy of the reaction.

Section 13.7

Lee R. Summerlin and James L. Ealy, Jr., “Equilibrium and Le Châtelier’s Principle,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 77-78.

Section 13.8

Lee R. Summerlin and James L. Ealy, Jr., “Equilibrium in the Gas Phase,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 85-86.

Section 13.9

Lee R. Summerlin and James L. Ealy, Jr., “Effect of Temperature Change on Equilibrium: Cobalt Complex,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 60-61.

Concentrated hydrochloric acid is added to pink [Co(H2O)6]2+ until blue [CoCl4]2– is formed. When heated the solution turns darker blue; when cooled the solution turns pink, indicating that the reaction is endothermic. Students are asked to examine the equilibrium reaction and predict how the system will shift upon the addition of water.

Literature References Section 13.1

Vladimir E. Fainzilberg and Stewart Karp, “Chemical Equilibrium in the General Chemistry Curriculum,” J. Chem. Educ., Vol. 71, 1994, 769-770.

Section 13.1

Carl W. David, “An Elementary Discussion of Chemical Equilibrium,” J. Chem.

Educ., Vol. 65, 1988, 407-409.

Section 13.1

Jan H. Van Driel, Wobbe de Vos, and Nico Verloop, “Introducing Dynamic Equilibrium as an Explanatory Model,” J. Chem. Educ., Vol. 76, 1999, 559-561.

Louise Tyson, David F. Treagust, and Robert B. Bucat, “The Complexity of Teaching and Learning Chemical Equilibrium,” J. Chem. Educ., Vol. 76, 1999, 554558.

Section 13.1

Section 13.1

Audrey H. Wilson, “Equilibrium: A Teaching/Learning Activity,” J. Chem. Educ., Vol 75, 1998, 1176-1177.

Section 13.1

Penelope A. Huddle, Margie W. White, and Fiona Rogers, “Simulations for Teaching Chemical Equilibrium,” J. Chem. Educ., Vol 77, 2000, 920-926.

Section 13.5

John Olmsted III, “Amounts Tables as a Diagnostic Tool for Flawed Stoichiometric Reasoning,” J. Chem. Educ., Vol. 76, 1999, 52-54.

Section 13.5

William R. Smith and Ronald W. Munson, “Chemical Equilibrium and Polynomial Equations: Beware of Roots,” J. Chem. Educ., Vol. 66, 1989, 489-490.

Section 13.5

E. Weltin, “Calculating Equilibrium Concentration by Iteration: Recycle Your Approximations,” J. Chem. Educ., Vol. 72, 1995, 36-38.

127

Chapter 13—Chemical Equilibrium

Section 13.6

Martin R. Feldman and Monica L. Tarver, “Fritz Haber,” J. Chem. Educ., Vol. 60, 1983, 463-464.

Section 13.8

Ron DeLorenzo, “From Chicken Breath to the Killer Lakes of Cameroon: Uniting Seven Interesting Phenomena with a Single Chemical Underpinning,” J. Chem.

Educ., Vol. 78, 2001, 191-194.

Section 13.8

Reed A, Howard, “The Fizz Keeper, a Case Study in Chemical Education,” J. Chem.

Educ., Vol. 76, 1999, 208-209.

Media References Section 13.1

N2O4/NO2 Equilibrium Experiment activity from the Instructor Resource DVD

Section 13.1

NO2 – N2O4 Equilibrium movie from the Instructor Resource DVD

Section 13.1

Chemical Equilibrium activity from the Instructor Resource DVD

Section 13.2

Equilibrium Constant activity from the Instructor Resource DVD

Section 13.4

Equilibrium Expressions activity from the Instructor Resource DVD

Section 13.5

Predicting the Direction of Reaction activity from the Instructor Resource DVD

Section 13.5

Figure 13.6 Examples activity from the Instructor Resource DVD

Section 13.6

Le Châtelier’s Principle movie from the Instructor Resource DVD

Section 13.7

Altering an Equilibrium Mixture activity from the Instructor Resource DVD

128

CHAPTER 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES Learning Goals for Students Section 14.1

Define an acid and a base according to the Arrhenius and Brønsted-Lowry theories.

Write balanced equations for the dissociation of Brønsted-Lowry acids.

From electron-dot structures, determine which chemical species can act as a Brønsted-Lowry acid, a Brønsted-Lowry base, or both.

From a chemical equation for a proton-transfer reaction, identify the conjugate acidbase pairs.

Section 14.2

Given the extent of dissociation of an acid in water, determine whether the acid is a stronger or weaker acid than water and whether the conjugate base of the acid is a stronger or weaker base than water.

Given a chemical equation representing a proton-transfer reaction and the relative strengths of each acid and base involved in the reaction, determine whether the reaction is favored to the right or to the left.

Section 14.3

Draw electron-dot structures of H3O+ and H9O4+ and explain how both represent a hydrated proton.

Section 14.4

Calculate H3O+ concentration from OH concentration and Kw and vice versa. From these concentrations, determine whether the solution is acidic, neutral, or basic.

Section 14.5

Interconvert pH and [H3O+]. Classify the solution as acidic, neutral, or basic.

Section 14.6

Determine pH from the color of indicator solutions.

Section 14.7

Given the molar concentration of a strong acid or a strong base, determine the pH of the solution.

Section 14.8

Given the pH of a weak acid solution, determine the Ka and pKa of the acid.

Section 14.9

Given the Ka value and the initial concentration of a weak monoprotic acid, calculate the concentrations of all species at equilibrium, and the pH of the solution.

Section 14.10

From molarity and Ka calculate the percent dissociation of a weak acid.

Section 14.11

Given the Ka values and the initial concentration of a weak diprotic acid, calculate the concentrations of all species at equilibrium and the pH of the solution.

Section 14.12

Given the Kb value and the initial concentration of a weak base, calculate the concentrations of all species at equilibrium and the pH of the solution.

Section 14.13

Interconvert Ka and Kb, pKa, and pKb

Section 14.14

Classify salt solutions as acidic, neutral, or basic. Calculate the pH of these solutions.

–

129

Chapter 14—Aqueous Equilibria: Acids and Bases

Section 14.15

Identify which of two binary acids or two oxoacids is more acidic.

Section 14.16

Identify the Lewis acid and the Lewis base in a chemical reaction.

130

Chapter 14—Aqueous Equilibria: Acids and Bases

Lecture Outline 14.1. Acid–Base Concepts: The Brønsted–Lowry Theory1 A. Arrhenius theory of acids and bases 1. Acids – substances that dissociate in water to produce hydrogen ions + – HA (aq) ⇌ H (aq) + A (aq) 2. Bases – substances that dissociate in water to produce hydroxide ions + – MOH (aq) ⇌ M (aq) + OH (aq) 3. Limitations to theory a. Restricted to aqueous solutions b. Doesn’t account for the basicity of substances that don’t contain OH groups B. Brønsted-Lowry theory of acids and bases 1. Acid – any substance that can transfer a proton to another substance (proton donor) 2. Base – any substance that can accept a proton (proton acceptor) 3. Acid-base reaction – proton-transfer reactions C. Conjugate acid-base pair – chemical species whose formulas differ only by one proton – 1. Conjugate base has one less proton than its acid: A is the conjugate base of HA. + 2. Conjugate acid has one more proton than its base: BH is the conjugate acid of B. D. Acid-dissociation equilibrium 1. Acid transfers a proton to the solvent, which acts as a base. + – 2. HA(aq) + H2O(l) ⇌ H3O (aq) + A (aq) E. Base-dissociation equilibrium 1. Base accepts a proton from the solvent, which acts as an acid.

– + 2. NH3(aq) + H2O(l) ⇌ OH (aq) + NH4 (aq) F. All Brønsted-Lowry bases have one or more lone pairs of electrons: unshared pair of electrons are used for bonding to the proton.

14.2. Acid Strength and Base Strength2 – A. In an acid-dissociation equilibrium, the two bases, H2O and A , are competing for protons.

B. Direction of reaction to reach equilibrium is proton-transfer from the stronger acid to the stronger base.

C. Strong acid – completely dissociated in aqueous solution 1. Has weak conjugate base – 2. H2O is a stronger base than A D. Weak acid – partially dissociates in aqueous solution 1. Has strong conjugate base – 2. A is a stronger base than H2O E. Inverse relationship between the strength of an acid and the strength of its conjugate base textbook Table 14.1 14.3. Hydrated Protons and Hydronium Ions3 + A. Hydronium ion, H3O – ion in which H+ is bonded to the oxygen atom of a solvent water molecule + B. Simplest hydrate of the proton – [H(H2O)] C. Produces higher hydrates through hydrogen bonding with other water molecules + D. General formula – [H(H2O)n] + + E. Symbols H (aq) and H3O (aq) used to represent a proton hydrated by an unspecified number of water molecules 14.4. Dissociation of Water4 1 Test Item File Questions: Multiple Choice 1-16, 131-136; Algorithmic 1; Short Answer 1 2 Test Item File Questions: Multiple Choice 17-30, 137-138, 144; Algorithmic 2; Short Answer 2 3 Test Item File Questions: Multiple Choice 31-32; Short Answer 3-4 4 Test Item File Questions: Multiple Choice 33-41; Algorithmic 3-6; Short Answer 5-8

131

Chapter 14—Aqueous Equilibria: Acids and Bases A. Water can act both as an acid and as a base.

B. Dissociation of water – one water molecule donates a proton to another water molecule. + – 1. H2O(l) + H2O(l) ⇌ H3O (aq) + OH (aq) 2. Ion-product constant for water, Kw + – Kw = [H3O ] [OH ] 3. Forward and reverse reactions are rapid.

4. Position of the equilibrium lies far to the left. + – –14 5. In an aqueous solution, [H3O ][OH ] = 1.0 × 10 . + – C. Relative values of H3O and OH concentrations can be used to determine whether an aqueous solution is neutral, acidic, or basic.

1.0 ×10 −14 1. [H3 O+ ] = − [OH ] 2. [OH − ] =

1.0 × 10−14

[H3 O+ ] – + –7 3. [H3O ] = [OH ], neutral solution and [H3O ] = 1.0 × 10 M + – + –7 4. [H3O ] > [OH ], acid solution and [H3O ] > 1.0 × 10 M + – + –7 5. [H3O ] < [OH ], basic solution and [H3O ] < 1.0 × 10 M 14.5. The pH Scale5 A. pH scale – logarithmic scale used to express the hydronium ion concentration + 1. pH = –log[H3O ] + –pH 2 [H3O ] = antilog(–pH) = 10 3. Significant figures in a logarithm are the digits to the right of the decimal point.

4. Number to the left of the decimal point is an exact number related to the integral power of + 10 in the exponential expression for [H3O ]. + B. pH decreases as [H3O ] increases.

1. Acidic solutions, pH < 7 2. Basic solutions, pH > 7 3. Neutral solutions, pH = 7 14.6. Measuring pH6 A. Acid-base indicators (HIn) – substances that change color in a specific pH range 1. Weak acids 2. Have different colors in their acid (HIn) and conjugate base (In–) forms.

3. Change color over a range of ≈ 2 pH units 4. Can determine the pH of a solution to within ≈ ±1 pH unit B. pH meter 1. Electronic instrument that measures a pH-dependent electrical potential of the test solution 2. Determines pH values more accurately than acid-base indicators 14.7. The pH of Strong Acids and Strong Bases7 A. Strong monoprotic acids – 100% dissociated in aqueous solution (HClO4, HCl, HNO3) 1. Contain a single dissociable proton + 2. pH = –log[H3O ] + – 3. [H3O ] = [A ] = initial concentration of the acid 4. [undissociated HA] = 0 B. Strong bases 1. Alkali metal hydroxides, MOH a. Water-soluble ionic solids b. Exist in aqueous solution as alkali metal cations and hydroxide anions +

5 Test Item File Questions: Multiple Choice 42-51, 141, 147; Algorithmic 7-12; Short Answer 9-10 6 Test Item File Questions: Multiple Choice 52-54; Short Answer 11-12 7 Test Item File Questions: Multiple Choice 55-63; Algorithmic 13-19; Short Answer 13

132

Chapter 14—Aqueous Equilibria: Acids and Bases c. Calculate pH from [OH–] 2. Alkaline earth metal hydroxides, M(OH)2 (M = Mg, Ca, Sr, Ba) a. Less soluble than alkali hydroxides, therefore lower [OH–] b. Most important strong base – lime, CaO i. Produces Ca(OH)2 when mixed with water ii. Used in steelmaking, water purification, and chemical manufacture 14.8. Equilibria in Solutions of Weak Acids8 A. Acids – partially dissociated B. Can write an acid–dissociation equilibrium equation.

– HA(aq) + H2O(l) ⇌ H3O+(aq) + A (aq) C. Position of acid-dissociation equilibrium – characterized by the acid-dissociation constant, Ka

[H 3 O + ][A - ] [HA] D. The larger the value of Ka, the stronger the acid E. Ka values can be determined from pH measurements.

14.9. Calculating Equilibrium Concentrations in Solutions of Weak Acids9 A. Equilibrium concentrations and solution pH of a weak acid can be calculated from the value of Ka. B. To solve acid-base equilibrium problems, think about the chemistry, and use the following steps: Ka =

14.10. Percent Dissociation in Solutions of Weak Acids10

8 Test Item File Questions: Multiple Choice 64-71, 139, 145; Short Answer 14-17 9 Test Item File Questions: Multiple Choice 72-78; Algorithmic 20-22 10 Test Item File Questions: Multiple Choice 79-81, 140, 146; Short Answer 18-20

133

Chapter 14—Aqueous Equilibria: Acids and Bases

[HA] dissociated × 100 [HA] undissociated B. Useful measure of acid strength C. Value depends on the acid D. Increases with increasing value of Ka E. Increases with increasing dilution.

14.11. Polyprotic Acids11 A. Polyprotic acid – acids that contain more than one dissociable proton.

B. Dissociate in a stepwise manner: each dissociation step has its own Ka. C. Stepwise dissociation constants decrease in the order Ka1 > Ka2 > Ka3.

D. More difficult to remove a positively charged proton from a negative ion.

– E. Diprotic acid solution contains a mixture of acids: H2A, HA , and H2O. 1. Strongest acid – H2A 2. Principal reaction – dissociation of H2A + 3. All of H3O comes from first dissociation step.

14.12. Equilibria in Solutions of Weak Bases12 A. Weak bases – partially dissociated.

1. Can write a base-dissociation equilibrium equation. + – 2. B(aq) + H2O(l) ⇌ BH (aq) + OH (aq) B. Position of base-dissociation equilibrium – characterized by the base-dissociation constant, Kb [BH + ][OH - ] Kb = [B] C. Amines – organic compounds that are weak bases 1. Derivatives of ammonia 2. One or more hydrogen atoms replaced by another group 3. Basicity due to lone pair of electrons on nitrogen atom: can be used for bonding to a proton D. To solve for equilibrium concentrations and solution pH, use the same procedure as for solving weak acid problems.

14.13. Relation Between Ka and Kb 13 A. For a conjugate acid-base pair, can calculate either Ka or Kb from the other B. Sum of an acid-dissociation reaction and a base-dissociation reaction is the dissociation of water.

1. Equilibrium constant for the net reaction (dissociation of water) = the product of the equilibrium constants for the individual reactions.

2. Ka × Kb = Kw 3. For Ka × Kb to remain constant, the strength of a conjugate base must decrease as the strength of the acid increases.

14.14. Acid–Base Properties of Salts14 A. pH of a salt solution is determined by acid-base properties of the constituent cations and anions.

1. In an acid-base reaction, the influence of the stronger partner is dominant.

2. Strong acid + Strong base → Neutral solution 3. Strong acid + Weak base → Acidic solution 4. Weak acid + Strong base → Basic solution B. Salts that yield neutral solutions 1. Derived from a strong base and a strong acid

A. Percent dissociation =

11 Test Item File Questions: Multiple Choice 82-88, 142-143; Algorithmic 23-27 12 Test Item File Questions: Multiple Choice 89-94, 148-150; Algorithmic 28-31; Short Answer 21 13 Test Item File Questions: Multiple Choice 95-99 14 Test Item File Questions: Multiple Choice 100-112 151-153; Algorithmic 32-39; Short Answer 22

134

Chapter 14—Aqueous Equilibria: Acids and Bases +

–

2. Neither the cation nor the anion reacts with water to produce H3O or OH ions.

3. Cations that do not react with H2O a. Group 1A b. Group 2A 4. Anions that do not react with H2O: anions from strong, monoprotic acids C. Salts that yield acidic solutions 1. Derived from a weak base and a strong acid 2. Anion is neither an acid nor a base.

3. Cation is a weak acid + a. Reacts with water (hydrolysis reaction) to produce H3O b. Small, highly charged cations are also acidic.

D. Salts that yield basic solutions 1. Derived from a strong base and a weak acid 2. Cation is neither an acid nor a base.

– 3. Anion is a weak base: reacts with water to produce OH .

E. Salts that contain acidic cations and basic anions 1. Derived from a weak acid and a weak base 2. Both the cation and the anion can undergo proton-transfer reactions.

3. pH depends on the relative acid strength of the cation and the base strength of the anion. a. Ka (for the cation) > Kb (for the anion); acidic solution b. Kb (for the anion) > Ka (for the cation); basic solution c. Ka (for the cation) ≈ Kb (for the anion); neutral solution 14.15. Factors That Affect Acid Strength15 A. The extent of dissociation of an acid, HA, is often determined by the strength and polarity of the H–A bond.

1. Weaker the H–A bond, stronger the acid 2. More polar the H–A bond, stronger the acid B. For binary acids 1. H–A bond strength decreases down a group in the periodic table. a. Increase in size of A down a group produces poorer orbital overlap and a weaker bond. b. H–A bond strength is the most important factor for determining acid strength for binary acids of elements in the same column. c. Acidity increases down a group.

2. H–A bond polarity is the most important factor for determining acid strength for binary acids of elements in the same row. a. Increase in H–A bond polarity as electronegativity of A increases b. In general, electronegativities increase from left to right across the periodic table. c. Acidity increases across a period.

C. Oxoacids, HnYOm (Y = a nonmetallic atom, n and m are integers) 1. Y is always bonded to one or more hydroxyl (OH) groups (can also be bonded to one or more oxygen atoms).

2. Dissociation of oxoacid involves breaking an O–H bond.

3. Strength of the acid increases as the O–H bond is weakened or becomes more polar. a. For oxoacids that contain the same number of OH groups and the same number of O atoms, acid strength increases with increasing electronegativity of Y. b. For oxoacids that contain the same atom Y but different numbers of oxygen atoms, acid strength increases with increasing oxidation number of Y, which increases, in turn, with an increasing number of oxygen atoms.

14.16. Lewis Acids and Bases16 15 Test Item File Questions: Multiple Choice 113-122, 154-157; Short Answer 23 16 Test Item File Questions: Multiple Choice 123-130, 158-160; Short Answer 24-25

135

Chapter 14—Aqueous Equilibria: Acids and Bases A. Lewis base – electron-pair donor 1. All Lewis bases are Brønsted-Lowry bases.

2. All Brønsted-Lowry bases are Lewis bases.

B. Lewis acid – electron-pair acceptor 1. More general definition than Brønsted-Lowry 2. Includes cations and neutral molecules having vacant valence orbitals that can accept a share in a pair of electrons from a Lewis base 3. Cationic Lewis acids – metal ions 4. Neutral Lewis acids – halides of group 3A elements and oxides of nonmetals

136

Chapter 14—Aqueous Equilibria: Acids and Bases

CHAPTER 14 AQUEOUS EQUILIBRIA: ACIDS AND BASES Teaching Tips, Points of Emphasis and Common Misconceptions Section 14.5

Typically pH measurements fall within the range 0–14; however, the scale extends beyond those values in both directions and includes negative pHs.

Section 14.11

A polyprotic acid has two or more dissociable protons. Acetic acid, CH3CO2H, contains four protons, only one of which is dissociable. Hence, acetic acid is not a polyprotic acid.

Section 14.13

Caution students that the equilibrium constants are multiplied, and they should not confuse this principle with Hess’s Law for which enthalpy changes are added.

Lecture/Laboratory Demonstration References Section 14.6

J. J. Fortman and K. M. Stubbs, “Demonstrations with Red-Cabbage Indicator,” J. Chem. Educ., Vol. 69, 1992, 66 Anthocyanin pigment extract from red cabbage is used as a pH indicator exhibiting colors that range from green to aqua to violet to red in the presence of various concentrations of household ammonia and white vinegar.

Section 14.14

Bassam Z. Shakhashiri, “Hydrolysis: Acidic and Basic Properties of Salts,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.103-108.

Section 14.14

Edward Koubek, “Acid-Base Chemistry of the Aluminum Ion in Aqueous Solution,” J. Chem. Educ., Vol. 75, 1998, 60.

Section 14.15

K. E. Kolb and D. Kolb, “Illustrating the Inductive Effect on Acid Strength of Carboxylic Acids,” J. Chem. Educ., Vol. 66, 1989, 1034. Substituent effects on the acidity of carboxylic acids are demonstrated from the colors formed when methyl violet indicator is added to aqueous carboxylic acid solutions.

Section 14.15

Bassam Z. Shakhashiri, “Effect of Molecular Structure on the Strength of Organic Acids in Aqueous Solutions,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.158161.

Literature References Section 14.1

Doris Kolb, “Acids and Bases,” J. Chem. Educ., Vol. 55, 1978, 459-464.

Section 14.1

George B. Kauffman, “The Brønsted-Lowry Acid-Base Concept,” J. Chem. Educ., Vol. 65, 1988, 28-31.

Section 14.1

Jamie L. Adcock, “Teaching Brønsted-Lowry Acid-Base Theory in a Direct Comprehensive Way,” J. Chem. Educ., Vol. 78, 2001, 1495-1496.

Section 14.5

Doris Kolb, “The pH Concept,” J. Chem. Educ., Vol. 56, 1979, 49-53.

137

Chapter 14—Aqueous Equilibria: Acids and Bases

Section 14.5

Stephen J. Hawkes, “Teaching the Truth about pH,” J. Chem. Educ., Vol. 71, 1994, 747-749.

Section 14.5

Addison Ault, “Do pH in Your Head,” J. Chem. Educ., Vol. 76, 1999, 936-938.

Section 14.8

Todd P. Silverstein, “Weak vs Strong Acids and Bases,” J. Chem. Educ., Vol. 77, 2000, 849-850.

Section 14.8

John J. Fortman, “Pictoral Analogies XI: Concentrations and Acidity of Solutions,” J. Chem. Educ., Vol. 71, 1994, 430-432.

Section 14.15

Manus Monroe and Karl Abrams, “The Relative Strength of Oxyacids and Its Application,” J. Chem. Educ., Vol. 62, 1985, 41-43.

Section 14.16

Glenn T. Seaborg, “The Research Style of Gilbert N. Lewis: Acids and Bases,” J. Chem. Educ., Vol. 61, 1984, 93-100.

Media References Section 14.2

Introduction to Aqueous Acids movie from the Instructor Resource DVD

Section 14.2

Introduction to Aqueous Bases movie from the Instructor Resource DVD

Section 14.5

pH Scale activity from the Instructor Resource DVD

Section 14.6

Natural Indicators movie from the Instructor Resource DVD

Section 14.9

Equilibrium Constants for Acids activity from the Instructor Resource DVD

Section 14.13

Relationship between Ka and Kb for a Conjugate…activity from the Instructor Resource DVD

Section 14.13

Determining Ka and Kb for a Conjugate activity from the Instructor Resource DVD

Section 14.14

Acid-Base Properties of Salts activity from the Instructor Resource DVD

Section 14.16

Lewis Acid-Base Theory movie from the Instructor Resource DVD

138

CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA Chapter Learning Goals for Students Section 15.1

Write balanced net ionic equations for the four types of neutralization reactions.

From the relative strengths of the acid and base in a neutralization reaction, predict whether the pH will be equal to, greater than, or less than 7.00 at the equivalence point.

Section 15.2

Describe the effect on pH when the conjugate base of a weak acid is added to a solution of the weak acid and the conjugate acid of a weak base is added to a solution of the weak base. Calculate the concentrations of all species present at equilibrium and percent dissociation.

Section 15.3

Differentiate between buffer and non-buffer solutions.

From a table of weak acids and their Ka values, select the weak acid/conjugate base pair that would make the best buffer at a given pH. Given the initial concentration of weak acid (or weak base) and its conjugate base (or conjugate acid), determine the equilibrium concentrations of all species and the pH of a buffer solution.

Calculate the pH of a buffer after the addition of OH– or H3O+.

Section 15.4

Use the Henderson-Hasselbalch equation to calculate the pH of a buffer.

Section 15.5

Identify the equivalence point on a titration curve.

Given a titration curve, select which indicator(s) could be used to detect the equivalence point.

Section 15.6

Recognize a strong acid-strong base titration curve.

Calculate pH values for a strong acid–strong base titration before titrant is added, before the equivalence point, at the equivalence point, and after the equivalence point.

Section 15.7

Recognize a weak acid–strong base titration curve.

Calculate pH values for a weak acid–strong base titration before titrant is added, before the equivalence point, at the half-equivalence point, at the equivalence point, and after the equivalence point.

Section 15.8

Recognize a weak base–strong acid titration curve.

Calculate pH values for a weak base-strong acid titration before titrant is added, before the equivalence point, at the half-equivalence point, at the equivalence point, and after the equivalence point.

Section 15.9

Recognize a polyprotic acid–strong base titration curve.

139

Chapter 15—Applications of Aqueous Equilibria

Calculate pH values for a polyprotic acid–strong base titration before titrant is added, halfway to the first equivalence point, at the first equivalence point, halfway between the first and second equivalence points, and at the second equivalence point.

Section 15.10

Write the solubility product expression for a given ionic compound.

Section 15.11

Given the Ksp of an ionic compound, calculate its solubility and vice versa.

Section 15.12

Show qualitatively how the addition of a common ion affects solubility.

Given the Ksp of an ionic compound, calculate its solubility in the presence of a common ion.

Identify ionic compounds that have an enhanced solubility at low pH. Write chemical equations showing why the solubility increases as [H3O+] increases.

Identify ionic compounds that have an enhanced solubility in the presence of a complexing agent. Write chemical equations showing why the solubility increases in the presence of the complexing agent.

Given Kf, calculate the concentrations of the species present in a complex-ion equilibrium.

Given Ksp and Kf, calculate the solubility of a slightly soluble ionic compound in an excess of the complexing agent.

Write chemical equations showing how a given oxide or hydroxide exhibits amphoteric behavior.

Section 15.13

From Ksp values and ion concentrations, determine whether a precipitate will form on mixing solutions of ionic compounds.

Section 15.14

Determine which metal sulfides will precipitate from a solution of metal ions on addition of H2S at a specified pH. Determine whether two ions can be separated by selective precipitation.

Section 15.15

Identify ions using a qualitative analysis scheme.

140

Chapter 15—Applications of Aqueous Equilibria

Lecture Outline 15.1. Neutralization Reactions1 A. Strong Acid–Strong Base 1. Net ionic equation for the neutralization reaction: H3O+(aq) + OH–(aq) ⇌ 2 H2O(l) 2. When equal numbers of moles of acid and base are mixed together, –7 [H3O+] = [OH–] = 1 × 10 M. 3. Reaction proceeds far to the right.

4. Equilibrium constant, Kn, for the reaction is the reciprocal of the ion-product constant for water.

1 1 = Kn = + K [H 3O ][OH ] w 4. pH = 7.00 B. Weak Acid–Strong Base 1. Net ionic equation for the neutralization reaction involves proton transfer from HA to the strong – base, OH .

HA(aq) + OH–(aq) ⇌ H2O(l) + A–(aq) 2. Equilibrium constant, Kn, is obtained from multiplying the equilibrium constants for the reactions that add to give the net ionic equation: Kn = Ka(1/Kw).

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) Ka – + H3O (aq) + OH (aq) ⇌ 2 H2O(l) 1/Kw – 3. Neutralization of any weak acid by a strong base goes 100% to completion because OH has a great affinity for protons.

4. pH > 7.00 due to the hydrolysis of A–, which behaves as a weak base.

C. Strong Acid–Weak Base 1. Net ionic equation for the neutralization reaction involves proton transfer from the strong acid, H3O+ to the weak base, B. H3O+(aq) + B(aq) ⇌ H2O(l) + BH+(aq) 2. Equilibrium constant, Kn, is obtained from multiplying the equilibrium constants for the reactions that add to give the net ionic equation: Kn = Kb(1/Kw).

– Kb B(aq) + H2O(l) ⇌ BH+(aq) + OH (aq) – 1/Kw H3O+(aq) + OH (aq) ⇌ 2 H2O(l) 3. Neutralization of any weak base with a strong acid goes 100% to completion because H3O+ is a powerful proton donor 4. pH < 7.00 due to the hydrolysis of BH+, which behaves as a weak acid.

D. Weak Acid–Weak Base 1. Neutralization reaction involves proton transfer from the weak acid to the weak base.

HA(aq) + B(aq) ⇌ BH+(aq) + A–(aq) 2. Obtain Kn from Ka for the weak acid, Kb for the weak base, and 1/Kw: Kn = KaKb(1/Kw).

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) Ka – Kb B(aq) + H2O(l) ⇌ BH+(aq) + OH (aq) – 1/Kw H3O+(aq) + OH (aq) ⇌ 2 H2O(l) 3. Less tendency to proceed to completion than neutralizations involving strong acids or strong bases 15.2. The Common-Ion Effect2 A. Common-ion effect – the shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium B. Example of Le Châtelier’s principle C. Determination the pH of a solution prepared from a weak acid and a salt of its conjugate base 1 Test Item File Questions: Multiple Choice 1-14; Short Answer 1-4 2 Test Item File Questions: Multiple Choice 15-21; Algorithmic 1-4; Short Answer 5-7

141

Chapter 15—Applications of Aqueous Equilibria 1. Identify the acid-base properties of the various species in solution.

2. Consider the possible proton-transfer reaction these species can undergo.

3. Principal reaction – dissociation of the weak acid 4. Set up a table of concentrations of species involved in the principal reaction.

– 5. A comes from 2 sources.

– a. Salt of the conjugate base, which provides the initial concentration of A – b. Dissociation of weak acid, which determines the change in the concentration of A 15.3. Buffer Solutions3 A. Buffer solutions 1. Resist drastic changes in pH 2. Contain a weak acid and its conjugate base or a weak base and its conjugate acid B. Add a small amount of base to a buffer solution – acid component of the solution neutralizes the added base.

C. Add a small amount of acid to a buffer solution – base component of the solution neutralizes the added acid.

D. Important in biological systems – pH of human blood (pH = 7.4) controlled by conjugate acid-base – pairs (H2CO3/HCO3 ).

E. For a buffer prepared from a weak acid, HA, and its conjugate base, B. [HA] [H 3O + ] = K a [B] 1. Can use initial concentrations in the calculations 2. Ka is small for commonly used buffer solutions.

3. Initial concentrations are relatively large.

4. Change in concentration is generally negligible.

F. Addition of acid or base to a buffer solution 1. Must take neutralization into account before calculating [H3O+] – a. Addition of OH – – HA(aq) + OH (aq) ⇌ H2O(l) + A (aq) + b. Addition of H3O – A (aq) + H3O+(aq) ⇌ H2O(l) + HA(aq) 2. Neutralization alters the number of moles of acid and conjugate base.

3. Concentration ratio, [HA]/[B], changes by only a small amount.

4. [H3O+] changes by only a small amount.

G. Buffer Capacity – a measure of the amount of acid or base that a buffer solution can absorb without a significant change in pH 1. Depends on how much weak acid and conjugate base is present.

2. For equal volumes of solutions, the more concentrated the solution, the greater the buffer capacity.

3. For solutions with the same concentration, increasing the volume increases the buffer capacity.

15.4. The Henderson-Hasselbalch Equation4 A. Explains how pH affects the percent dissociation of a weak acid. [base] pH = pK a + log [acid] [base] = pH - pK a [acid] B. Useful for determining how to prepare a buffer solution 1. Select a weak acid whose pKa is close to the desired pH of the buffer solution.

2. pKa of the weak acid should be within ±1 pH units of the desired pH. log

3 Test Item File Questions: Multiple Choice 22-35; Algorithmic 5-6 4 Test Item File Questions: Multiple Choice 36-44, 94-117; Short Answer 8-14

142

Chapter 15—Applications of Aqueous Equilibria 3. Adjust the [base]/[acid] ratio.

C. The pH of a buffer solution does not depend on the volume of the solution; depends only on pKa and the relative molar amounts of weak acid and conjugate base.

15.5. pH Titration Curves5 A. Acid-base titration – a titrant, a solution containing a known concentration of base (or acid), is slowly added from a buret to a solution containing an unknown concentration of acid (or base).

B. Equivalence point – the point at which stoichiometrically equivalent quantities of acid and base have been mixed together.

C. pH titration curve – a plot of the pH of the solution versus the volume of added titrant.

1. Use to identify the equivalence point in a titration 2. Useful in selecting a suitable indicator to signal the equivalence point 3. Calculate pH titration curves from the principles of aqueous solution equilibria 15.6. Strong Acid–Strong Base Titrations6 A. Before addition of any strong base – initial [H3O+] = concentration of strong acid B. Addition of strong base before the equivalence point + 1. Decrease in [H3O+] – added base will neutralize some of the H3O present.

2. mmol H3O+ after neutralization = mmol H3O+ initial – mmol OH– added. [mmol H 3O + after neutralization] [H 3O + ] after neutralization = [total volume of acid and base] C. Addition of strong base at the equivalence point.

1. Add just enough base to neutralize all the acid initially present 2. pH = 7.00 3. Solution contains water and a salt derived from a strong base and a strong acid.

D. Addition of strong base after the equivalence point – 1. Excess of OH present 2. mmol of excess OH– = mmol of OH– added – mmol of acid initially present [mmol of excess OH - ] [OH - ] after neutralization = [total volume of acid and base] [H 3O + ] =

[ Kw ]

[OH - ] E. pH titration curve 1. Plot of pH versus milliliters of strong base added 2. Sharp increase in pH in region near the equivalence point a. Characteristic of the titration curve for any strong acid-strong base titration b. Can use to identify the equivalence point when the concentration of the acid is unknown 15.7. Weak Acid–Strong Base Titrations7 A. Before addition of strong base – pH calculated in the same manner as for a solution of a weak acid B. Addition of strong base before the equivalence point 1. mmol of base, A–, after neutralization = mL of strong base × [strong base] mmol of A − 2. [A − ] = total mL of acid and base 3. mmol of acid, HA, after neutralization = mmol of HA initially – mmol of A– mmol HA after neutralization [HA] after neutralization = total volume of acid and base

5 Test Item File Questions: None 6 Test Item File Questions: Multiple Choice 44-45, 118-119, 121; Algorithmic 7-14; Short Answer 15-17 7 Test Item File Questions: Multiple Choice 46-54, 106-109, 120, 122-125; Short Answer 18

143

Chapter 15—Applications of Aqueous Equilibria

[A - ] [HA] C. Addition of strong base at the equivalence point 1. Basic salt solution.

– 2 mmol A after neutralization = mL of strong base × [strong base] mmol of A − [A − ] = total mL of acid and base 3. Calculate pH from the method outlined in Section 15.14 in your textbook.

4. pH > 7.00 due to hydrolysis of the anion, which behaves as a weak base.

D. Addition of strong base after the equivalence point.

– 1. mmol OH added = mL of strong base × [strong base] – – 2. mmol A present = mmol A at the equivalence point – – – 3. mmol OH present = mmol OH added – mmol A present mmol OH - present [OH - ] present = total mL of acid and base pH = pK a + log

[H 3O + ] =

[ Kw ]

[OH - ] E. Can use pH range at the equivalence point to select a suitable indicator.

15.8. Weak Base–Strong Acid Titrations8 A. Before addition of strong acid – calculate pH in the same manner as a weak base.

B. Addition of strong acid before the equivalence point 1. Have a B/BH+ buffer solution 2. Observe a leveling of the titration curve in the buffer region between the start of the titration and the equivalence point 3. pH calculated from the Henderson Hasselbalch equation C. Addition of strong acid at the equivalence point 1. Acidic salt solution of BH+ 2. pH calculated using the method outlined in textbook Section 15.14 3. pH < 7.00 D. Addition of strong acid after the equivalence point 1. Excess H3O+ present.

2. pH calculated from excess [H3O+] 15.9. Polyprotic Acid–Strong Base Titrations9 A. Diprotic acid – has two dissociable protons and reacts with two molar amounts of OH– B. Amino acids – both acidic and basic and can be protonated by strong acids 1. H2A+– protonated form; neutralized to HA H2A+(aq) + H2O(l) ⇌ H3O+(aq) + HA(aq) Ka1 2. HA – neutralized to A– HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) K a2 C. Before addition of strong base 1. Calculate the pH of a diprotic acid.

2. Principal reaction – dissociation of H2A+ D. Addition of strong base before the first equivalence point 1. H2A+ is converted to HA. 2. H2A+/HA buffer solution.

3. Calculate [H2A+] and [HA] of the amino acid in the same way [HA] and [A– ] are calculated in a weak acid–strong base titration.

8 Test Item File Questions: Multiple Choice 55-59; Algorithmic 15-17; Short Answer 19 9 Test Item File Questions: Multiple Choice 60-64, 126-134; Short Answer 20

144

Chapter 15—Applications of Aqueous Equilibria 4. Use the Henderson-Hasselbalch equation to calculate pH. E. Addition of strong base at the first equivalence point 1. All of the H2A+ is converted to HA. 2. Principal reaction – proton transfer between HA molecules 2 HA(aq) ⇌ H2A+(aq) + A–(aq) K = Ka2/Ka1 3. pH at the first equivalence point = average of pKa1 and pKa2 pK a1 + pK a2 pH = 2 a. Isoelectric point – the [HA] is at a maximum and the [H2A+] and [A–] are very small and equal b. Useful for separating mixtures of amino acids F. Addition of strong base between the first and second equivalence points 1. HA converted to A– 2. HA/ A– buffer solution 3. Calculate [HA] and [A–] of the amino acid in the same way [HA] and [A–] are calculated in a weak acid–strong base titration.

4. Use the Henderson-Hasselbalch equation to calculate pH. G. Addition of strong base at the second equivalence point 1. All of the HA is converted A– 2. [A–] = [H2A+] initially 3. Solution of a basic salt a. Principal reaction Kb = Kw/Ka2 A–(aq) + H2O(l) ⇌ HA(aq) + OH–(aq) b. Calculate [OH–] from equilibrium equation; calculate [H3O+] and pH from [OH–] H. Addition of strong base beyond the second equivalence point – pH calculated from the excess [OH–] 15.10. Solubility Equilibria10 A. Dissolution or precipitation of a sparingly soluble ionic compound B. Principles of solubility equilibria – examine quantitative aspects of solubility and precipitation phenomena C. Solubility product constant = Ksp b+ a– 1. MaXb(s) ⇌ a M (aq) + b X (aq) b+ a a– b 2. Ksp = [M ] [X ] 15.11. Measuring Ksp and Calculating Solubility from Ksp11 A. Ksp – measured by experiment B. Temperature dependent C. Use to calculate solubility of a compound D. Can be calculated from the molar solubility of a compound 15.12. Factors That Affect Solubility12 A. Common-ion effect – solubility of a slightly soluble ionic compound is decreased by the presence of a common ion in the solution B. pH of the solution – solubility of an ionic compound increases with decreasing pH of solution if the compound contains a basic anion C. Formation of complex ions 1. Solubility of an ionic compound increases dramatically if the solution contains a Lewis base that can form a coordinate covalent bond to the metal cation.

2. Complex – ion that contains a metal cation bonded to one or more small molecules or ions.

3. Formation constant, Kf a. Equilibrium constant for complex ion formation from the hydrated metal cation b. Measures complex ion stability 10 Test Item File Questions: Multiple Choice 65-66 11 Test Item File Questions: Multiple Choice 67-72, 135-145; Algorithmic 18-19; Short Answer 21-23 12 Test Item File Questions: Multiple Choice 73-82; Algorithmic 20-24; Short Answer 24

145

Chapter 15—Applications of Aqueous Equilibria D. Amphoterism 1. Amphoteric oxides – soluble both in strongly acidic and in strongly basic solutions 15.13. Precipitation of Ionic Compounds13 A. Ion product (IP) – determines if a precipitate will form when solutions containing constituent ions are mixed 1. For the salt, MaXb; IP = [Mb+]a[Xa–]b 2. Defined in the same way as Ksp 3. Concentrations are initial concentrations, not equilibrium concentrations.

B. IP > Ksp 1. Supersaturated solution 2. Precipitation occurs C. IP = Ksp 1. Saturated solution 2. Equilibrium exists D. IP < Ksp 1. Unsaturated solution 2. Precipitation will not occur 15.14. Separation of Ions by Selective Precipitation14 A. To separate a mixture of ions in solution, add a reagent that will precipitate some ions but not others.

B. Can separate insoluble metal sulfides from the more soluble metal sulfides 1. Carry out separation in acidic solution.

2. MS(s) + 2 H3O+(aq) ⇌ M+(aq) + H2S(aq) + 2 H3O(l) 3. Kspa, solubility product in acid [M 2+ ][H 2S] K spa = [H 3O + ] 4. Qc > Kspa; metal sulfide precipitates out.

5. Adjust [H3O+] so that Qc > Kspa for the more insoluble metal sulfide.

6. The more insoluble metal sulfide will precipitate out from the more soluble metal sulfide.

15.15. Qualitative Analysis A. Procedure for identifying ions present in an unknown solution B. Traditional scheme of analysis for metal cations involves separation of 20 cations into five groups by selective precipitation – textbook Figure 15.17 C. Determine the presence or absence of the ions in each group with further separations and tests D. Excellent vehicle for developing laboratory skills and learning about acid-base, solubility, and complex-ion equilibria

13 Test Item File Questions: Multiple Choice 83-91; Algorithmic 25; Short Answer 25 14 Test Item File Questions: Multiple Choice 92-93

146

Chapter 15—Applications of Aqueous Equilibria

CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA Teaching Tips, Points of Emphasis, and Common Misconceptions Section 15.2

The value of x is negligible compared with the initial concentration and can be ignored when subtracted from or added to the initial concentration.; however, by itself x is significant.

Section 15.3

A mixture of a weak acid with a salt of its conjugate base is an acidic buffer. In calculations involving an acidic buffer, the only chemical equation needed is that for the dissociation of the weak acid: HA + H2O ⇌ H3O+ + A–.

Section 15.3

A mixture of a weak base with a salt of its conjugate acid is a basic buffer. In calculations involving a basic buffer, the only chemical equation needed is that for the dissociation of the weak base: B + H2O ⇌ HB+ + OH–.

Section 15.3

Use the following approach for any neutralization reaction: (1) Determine the moles (or concentration) of each species before reaction. (2) Allow the reaction to go to completion, and determine the mole of each species “after reaction.” (3) Allow the system to equilibrate, and calculate the concentration of each species at equilibrium.

Section 15.3

This expression is the basis for the Henderson-Hasselbalch equation, which is used to calculate the pH of a buffer solution as shown in the next section.

Section 15.4

The acid and base in this expression refers to a conjugate acid/base pair, such as CH3CO2H/CH3CO2– or NH4+/NH3. CH3CO2H is the conjugate acid of CH3CO2– and CH3CO2– is the conjugate base of CH3CO2H. Similarly, NH4+ is the conjugate acid of NH3 and NH3 is the conjugate base of NH4+.

Section 15.4

This expression contains a ratio of the concentration of a conjugate base to the concentration of its conjugate acid remaining after reaction. Because the base and its conjugate acid are contained in the same volume, moles of base and conjugate acid remaining after reaction can be used instead of molarities.

Section 15.6

The change in pH seen in a titration curve is due primarily to the neutralization reaction, but it is also due to the dilution effect of mixing two solutions. Volume changes constantly during a titration.

Section 15.6

Titration curve points of particular interest are pH values (1) before any base (or acid) is added, (2) before the equivalence point (especially at the half-equivalence point), (3) at the equivalence point, and (4) beyond the equivalence point.

Section 15.7

For an acid-base titration the pH at the equivalence point will be (1) > 7 if the base is stronger than the acid, (2) = 7 if the acid and base have equal strengths, (3) < 7 if the acid is stronger than the base.

Section 15.7

The weaker the acid, the shorter is the steeply rising portion of the titration curve. If the acid is very weak, the equivalence point is difficult to detect.

Section 15.9

As noted earlier, the weaker the acid, the shorter is the steeply rising portion of the titration curve. If the acid is very weak, the equivalence point is difficult to detect.

147

Chapter 15—Applications of Aqueous Equilibria

Section 15.10

Although H3PO4 is a triprotic acid, HPO42– is such a weak acid that the third equivalence point is not observed in the H3PO4/NaOH titration curve.

The solubility product constant, Ksp, is for a saturated solution.

Section 15.11

No ionic compound is 100% insoluble. In order for an equilibrium to exist, all ionic compounds must dissolve to some extent.

Section 15.11

The relative solubilities of two ionic compounds can be determined by comparing their Ksp values only if they have the same number of ions. It is possible for a 1:1 compound, such as CaCO3, to have a larger Ksp value but a lower solubility than a 1:2 compound, such as CaF2.

Section 15.12

The addition of a common ion to a solution of a sparingly soluble ionic compound lowers the solubility of that compound.

Section 15.12

The solubility of a sparingly soluble ionic compound is increased on addition of H3O+ if the anion of the compound is the conjugate base of a weak acid.

Section 15.12

As with acid-base neutralization reactions, the best approach to doing complex ion formation problems is to allow the reaction to go to completion and then let the system equilibrate.

Section 15.12

Remember that amphoteric means able to react as an acid or as a base.

Section 15.14

Like oxides, which react with H+ to form H2O, sulfides become more soluble on addition of H+ due to the formation of the weak acid H2S.

148

Chapter 15—Applications of Aqueous Equilibria

Lecture/Laboratory Demonstration References Section 15.2

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Equilibrium: The Dissociation of Acetic Acid,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 160-161.

Section 15.2

Bassam Z. Shakhashiri, “Effect of Acetate Ion on the Acidity of Acetic Acid: The Common Ion Effect,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.155-157.

Section 15.2

Lee R. Summerlin and James L. Ealy,Jr., “The Common Ion Effect: Second Demonstration,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 93-94.

Section 15.2

Lee R. Summerlin and James L. Ealy,Jr., “The Common Ion Effect: Ammonium Hydroxide and Ammonium Acetate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 95.

Section 15.3

Bassam Z. Shakhashiri, “Buffering Action of Alka-Seltzer,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989). pp.186-187.

Section 15.3

Bassam Z. Shakhashiri, “Buffering Action and Capacity,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989). pp.173-185.

Section 15.7

Dianne N. Epp, “Teas as Natural Indicators,” J. Chem. Educ., Vol. 70, 1993, 326.

Section 15.11

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Solubility of Some Silver Compounds,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 83-85.

Section 15.11

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Red and White Precipitates in Sodium Silicate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 131.

Section 15.12

Bassam Z. Shakhashiri, “Fizzing and Foaming: Reactions of Acids with Carbonates,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp.96-99.

Section 15.12

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Colorful Complex Ions in Ammonia,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 75-76.

Section 15.12

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Green and Blue Copper Complexes,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 71-72.

Section 15.12

Bassam Z. Shakhashiri, “Acidic and Basic Properties of Oxides,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp.109-113.

149

Chapter 15—Applications of Aqueous Equilibria

Section 15.13

R. A. Pacer, “Will a precipitate form? Will it dissolve?” J. Chem. Educ., Vol. 71, 1994, 69. Principles of ionic compound precipitate formation are shown in a series of four colorful demonstrations: (1) Yellow PbI2 is formed only when Q > Ksp. (2) Upon addition of hydrochloric acid, green NiCO3(s) is dissolved due to the formation of the bluish-green complex ion [Ni(NH3)6]2+. (3) Kinetic effects are seen when MgC2O4(s) fails to form at room temperature from a solution having Q > Ksp but does form when the solution is heated with stirring to 90oC. (4) Several Ag(I) precipitates and complex ions are formed when OH–, Cl–, NH3, Br–, S2O32–, I–, and S2– are successively added to a solution containing Ag+.

Section 15.14

R. J. Stolzberg, “Shedding a New Light on Solubility Calculations,” J. Chem. Educ., Vol. 65, 1988, 65. Principles of precipitate formation, fractional precipitation, and precipitate dissolution are demonstrated from observing a series of reactions involving chloride and chromate salts of silver. Scattering of a laser beam is used to allow the observation of colloidal precipitate formation.

Literature References Section 15.3

Charles L. Bering, “A Good Idea Leads to a Better Buffer,” J. Chem. Educ., Vol. 64, 1987, 803.

Section 15.3

Edwin S. Gould, “Phosphate Buffers and Telephone Poles – A Useful Analogy with Limitations,” J. Chem. Educ., Vol. 76, 1999, 1511.

Section 15.4

Henry N. Po and N. M. Senozan, “The Henderson-Hasselbalch Equation: Its History and Limitations,” J. Chem. Educ., Vol. 78, 2001, 1499-1503.

Section 15.5

Dennis Barnum, “Predicting Acid-Base Titration Curves Without Calculations,” J. Chem. Educ., Vol 76, 1999, 938-942.

Section 15.7

Ara S. Kooser, Judith L. Jenkins, and Lawrence E. Welch, “Acid-Base Indicators: A New Look at an Old Topic,” J. Chem. Educ., Vol. 78, 2001, 1504-1506.

Section 15.7

Robert C. Mebane and Thomas R. Rybolt, “Edible Acid-Base Indicators,” J. Chem.

Educ., Vol. 62, 1985, 285.

Section 15.11

Stephen J. Hawkes, “What Should We Teach Beginners about Solubility and Solubility Product?” J. Chem. Educ., Vol. 75, 1998, 1179-1181.

Section 15.11

Roy W. Clark and Judith M. Bonicamp, “The Ksp-Solubility Conundrum,” J. Chem.

Educ., Vol. 75, 1998, 1182-1185.

Section 15.11

Andres Raviolo, “Assessing Students’ Conceptual Undertanding of Solubility Equilibrium,” J. Chem. Educ., Vol. 78, 2001, 629-631.

Section 15.11

Robert Perkins, “The Useless Tea Kettle,” J. Chem. Educ., Vol. 618, 1984, 383.

Section 15.12

Edward Koubek, Cole McWherter, and George L. Gilbert, “Acid-Base Chemistry of Aluminum Ion in Aqueous Solution,” J. Chem. Educ., Vol. 75, 1998, 60.

Section 15.15

Dale D. Clyde, “Swimming Pools, Hot Rods, and Qualitative Analysis,” J. Chem. Educ., Vol. 65, 1988, 911-913.

150

Chapter 15—Applications of Aqueous Equilibria

Media References Section 15.1

Strong Acid-Strong Base Neutralization activity from the Instructor Resource DVD

Section 15.1

Weak Acid-Strong Base Neutralization activity from the Instructor Resource DVD

Section 15.3

Buffer pH activity from the Instructor Resource DVD

Section 15.3

Buffer Solutions activity from the Instructor Resource DVD

Section 15.4

Henderson-Hasselbalch Equation activity from the Instructor Resource DVD

Section 15.4

Calculating pH Using Henderson-Hasselbalch Equation activity from the Instructor Resource DVD

Section 15.5

Acid-Base Titration movie from the Instructor Resource DVD

Section 15.7

Weak Acid-Strong Base Titration activity from the Instructor Resource DVD

Section 15.8

Weak Base-Strong Acid Titration activity from the Instructor Resource DVD

Section 15.9

Polyprotic Acid-Strong Base Titration activity from the Instructor Resource DVD

Section 15.12

Common Ion Effect movie from the Instructor Resource DVD

Section 15.12

Dissolution of Mg(OH)2 by Acid movie from the Instructor Resource DVD

Section 15.14

Precipitation Reactions movie from the Instructor Resource DVD

151

CHAPTER 16 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM Chapter Learning Goals for Students Section 16.1

Qualitatively determine whether simple chemical or physical changes are spontaneous.

Section 16.2

Qualitatively predict whether the sign of ΔS is positive or negative for a chemical or physical change.

Section 16.3

On the basis of probability, determine which of two states has the higher entropy.

Section 16.4

Compare the entropy of a substance at different temperatures.

Section 16.5

Calculate the standard entropy of reaction from the standard molar entropies of products and reactants.

Section 16.6

Determine whether a reaction is spontaneous by determining the sign of ΔStotal.

Section 16.7

Use the equation ΔG = ΔH – TΔS to calculate the free energy of reaction and to determine the temperature at which a nonspontaneous reaction becomes spontaneous.

Section 16.8

Calculate the standard free energy of reaction from enthalpies and standard entropies.

Section 16.9

Calculate the standard free energy of reaction from standard free energies of formation.

Section 16.10

Calculate the free energy of reaction for a system having nonstandard pressures and concentrations.

Section 16.11

From the standard free energy of reaction, calculate the value of the equilibrium constant and vice versa.

From the standard free energy of reaction, calculate a vapor pressure.

152

Chapter 16—Thermodynamics: Entropy, Free Energy, and Equilibrium

Lecture Outline 16.1. Spontaneous Processes1 A. Spontaneous process – one that proceeds on its own without any external influence B. A spontaneous reaction always moves a system toward equilibrium.

1. Spontaneity of forward or reverse reaction depends on: a. Temperature. b. Pressure. c. Composition of the reaction mixture.

2. Q < K – reaction proceeds in the forward direction.

3. Q > K – reaction proceeds in the reverse direction.

C. Spontaneity of a reaction is no indication of the speed of the reaction.

16.2. Enthalpy, Entropy, and Spontaneous Processes: A Brief Review2 A. Spontaneous reactions 1. Some are accompanied by conversion of potential energy to heat – reactions are exothermic.

2. Some spontaneous reactions are endothermic – system moves spontaneously to a state of higher potential energy by absorbing heat from the surroundings.

3. Enthalpy alone does not account for the direction of spontaneous change.

4. Second thermodynamic driving force for spontaneous change – nature’s tendency to move to a condition of maximum randomness or disorder B. Molecular systems tend to move spontaneously to a state of maximum randomness or disorder.

C. Entropy, S – molecular randomness or disorder 1. State function 2. ΔS = Sfinal – Sinitial 3. When ΔS is positive there is an increase in randomness or disorder of a system.

4. Examples a. Solid → liquid b. Liquid → gas c. A reaction that results in an increase in the number of gaseous molecules. d. When certain solutes dissolve 5. When ΔS is negative there is a decrease in randomness or disorder of a system.

16.3. Entropy and Probability3 A. Systems tend to a state of maximum randomness because the random state, which is more probable, can be achieved in more ways.

B. Probabilities of ordered and random states are proportional to the number of ways that the state can be achieved.

C. Boltzmann – The entropy of a particular state is related to the number of ways that the state can be achieved.

S = k ln W 1. ln W = the number of ways that the state can be achieved –23 2. k = 1.38 × 10 J/K D. Gas expands spontaneously because the state of greater volume is more probable.

1. For an ideal gas V ΔS = R ln final Vinitial P=

nRT V

1 Test Item File Questions: Multiple Choice 1-7; Short Answer 1 2 Test Item File Questions: Multiple Choice 8-15, 94-99, 101; Short Answer 2 3 Test Item File Questions: Multiple Choice 16-26; Algorithmic 1-3; Short Answer 3-7

153

Chapter 16—Thermodynamics: Entropy, Free Energy, and Equilibrium

ΔS = R ln

Pinitial Pfinal

2. The entropy of a gas increases when its pressure decreases at constant temperature.

3. The entropy of a gas decreases when the pressure increases at constant temperature.

16.4. Entropy and Temperature4 A. Entropy – also associated with molecular motion B. Random molecular motion increases as the temperature of a substance increases.

1. Increase in the average kinetic energy of the molecules a. Total energy is distributed among the individual molecules in a number of ways. b. Boltzmann – The more ways, W, that the energy can be distributed the greater the randomness of the state and the higher the entropy.

2. Plot of entropy versus temperature – at absolute zero, every substance is a solid where particles are rigidly fixed in a crystalline structure C. Third law of thermodynamics — The entropy of a perfectly ordered crystalline substance at 0 K is zero.

16.5. Standard Molar Entropies and Standard Entropies of Reaction5 A. Standard molar entropy, S° 1. The entropy of 1 mole of the pure substance at 1 atm pressure and a specified temperature, usually 25oC. 2. Absolute entropies – measured with respect to an absolute reference point.

B. Standard entropy of reaction, ΔS° 1. ΔS° = S°(products) – S°(reactants) 2. Multiply the ΔS° value for each substance by the stoichiometric coefficient of that substance in the balanced equation.

3. Entropy increases whenever a molecule breaks into two or more pieces.

16.6. Entropy and the Second Law of Thermodynamics6 A. Value of ΔG – criterion for spontaneity 1. ΔG = ΔH – TΔS 2. ΔG > 0; nonspontaneous reaction 3. ΔG < 0; spontaneous reaction 4. ΔG = 0; reaction is at equilibrium B. First law of thermodynamics – in any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant.

1. Helps keep track of energy flow between system and the surroundings 2. Does not indicate the spontaneity of the process C. Second law of thermodynamics – in any spontaneous process, the total entropy of a system and its surroundings always increases.

1. Provides a clear-cut criterion of spontaneity 2. Direction of spontaneous change is always determined by the sign of the total entropy change. ΔS total = ΔSsystem + ΔSsurroundings a. ΔStotal > 0; spontaneous reaction b. ΔStotal < 0; nonspontaneous reaction c. ΔStotal = 0; reaction at equilibrium D. All reactions proceed spontaneously in the direction that increases the entropy of the system plus surroundings.

E. To determine ΔStotal, need to know ΔSsystem and ΔSsurroundings.

4 Test Item File Questions: Multiple Choice 27-32, 100, 102; Algorithmic 4; Short Answer 4 5 Test Item File Questions: Multiple Choice 33-41; Algorithmic 5; Short Answer 9-11 6 Test Item File Questions: Multiple Choice 39-46

154

Chapter 16—Thermodynamics: Entropy, Free Energy, and Equilibrium 1. Δ Ssystem a. The entropy of reaction b. Calculate from standard molar entropies 2. At constant pressure ΔH rxn ΔSsurroundings = − T ΔH o rxn ΔS total = ΔSrxn − T 16.7. Free Energy7 A. Second law restated in terms of the thermodynamic properties of the system – chemists are more interested in the system.

B. Free energy: G = H – TS 1. TS = part of the system’s energy that is already disordered 2. H – TS = part of the system’s energy that is still ordered and therefore free to cause spontaneous change by becoming disordered 3. State function ΔG = ΔH – TΔS C. Relationship between free energy and spontaneity 1. ΔStotal = ΔS – ΔH/T; 2. – TΔStotal = ΔH – TΔS 3. – TΔStotal = ΔG 4. ΔG and ΔStotal have opposite signs.

D. In any spontaneous process at constant temperature and pressure, the free energy of the system decreases.

E. Temperature acts as a weighting factor that determines the relative importance of the enthalpy and entropy changes (textbook Table 16.2).

1. Can estimate the temperature at which ΔG changes from positive to negative with ΔG = ΔH − T ΔS o 2. To estimate the temperature at which ΔG changes from a positive to a negative value, set ΔG° = ΔH° – TΔS° = 0. ΔH ο T= ΔS ο 16.8. Standard Free-Energy Changes for Reactions8 A. Standard free-energy change, ΔG° 1. Change in free energy that occurs when reactants in their standard states are converted to products in their standard states 2. ΔG° – an extensive property; refers to the number of moles indicated in the chemical equation 3. Calculate ΔG° from standard enthalpy change, ΔH°, and standard entropy change, ΔS°. ΔG° = ΔH° – TΔS° 16.9. Standard Free Energies of Formation9 A. Standard free energy of formation, ΔG fο 1. Free-energy change for formation of 1 mole of the substance in its standard state from the most stable form of the constituent elements in their standard states 2. For an element in its most stable form at 25°C, ΔG fο = 0 B. ΔG fο measures a substance’s thermodynamic stability with respect to its constituent elements.

7 Test Item File Questions: Multiple Choice 50-65; Short Answer 16-19 8 Test Item File Questions: Multiple Choice 66-69; Short Answer 20 9 Test Item File Questions: Multiple Choice 70-75; Short Answer 21

155

Chapter 16—Thermodynamics: Entropy, Free Energy, and Equilibrium 1. ΔG fο is negative – substance is stable and does not decompose to its constituent elements 2. ΔG fο is positive – substance is thermodynamically unstable (but may not decompose to its constituent elements if the rate of decomposition is slow) C. Can use ΔG fο to calculate standard free-energy changes for reactions ΔG o = ΔGfo (products) - ΔGfo (reactants)

16.10. Free-Energy Changes and Composition of the Reaction Mixture10 A. Free-energy change for a reaction when the reactants and products are present at nonstandardstate pressures and concentrations B. ΔG = ΔG ο + RT ln Q C. Q = the reaction quotient — an expression having the same form as the equilibrium constant expression (but concentrations are not necessarily at equilibrium) 16.11. Free Energy and Chemical Equilibrium11 A. When the reaction mixture is mostly reactants 1. Q << 1 2. RT ln Q << 0 3. ΔG < 0 4. Total free energy decreases as the reaction proceeds spontaneously in the forward direction.

B. When the reaction mixture is mostly products 1. Q >> 1 2. RT ln Q > 0 3. ΔG > 0 4. Total free energy decreases as the reaction proceeds spontaneously in the reverse direction.

C. Free-energy curve – shows how the total free energy of a reaction mixture changes as the reaction progresses (textbook Figure 16.10) 1. Goes through a minimum somewhere between pure reactants and pure products 2. System is at equilibrium D. Relationship between free energy and the equilibrium constant 1. At equilibrium a. ΔG° = 0 b. Q = K 2. ΔG° = – RT lnK E. Relationship between ΔG° and K 1. ΔG° < 0, ln K > 0 – equilibrium mixture is mainly products 2. ΔG° > 0, ln K < 0 – equilibrium mixture is mainly reactants 3. ΔG° = 0, ln K = 0 – equilibrium mixture contains comparable amounts of reactions and products F. Properties of nature that determine the direction and extent of a chemical reaction 1. Value of K ο 2. K determined by ΔG rxn ο ο 3. ΔG rxn determined by ΔH fο and ΔS rxn

10 Test Item File Questions: Multiple Choice 76-83 11 Test Item File Questions: Multiple Choice 84-93, 103-113; Short Answer 22-26

156

Chapter 16—Thermodynamics: Entropy, Free Energy, and Equilibrium

CHAPTER 16 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM Teaching Tips, Points of Emphasis, and Common Misconceptions Section 16.1

Most, but not all, spontaneous chemical and physical changes are accompanied by heat transfer from system to surroundings (ΔH is negative) and a lowering of the potential energy of the system. When ΔH is positive for a spontaneous physical or chemical change, another driving force (entropy) is playing an important role.

Section 16.3

The value of R used in energy-related calculations is 8.314 J/K, not 0.0821 L⋅atm mol–1 K–1.

Section 16.3

The greater the volume (and the lower the pressure) of a gas, the greater the entropy.

Section 16.4

Entropy increases with an increase in the motion of atoms, ions, and molecules. The higher the temperature, the more rapid the motion and the greater the entropy.

Section 16.5

In general, So (gas) > So (liquid) > So (solid), and So (large molecule) > So (small molecule).

Section 16.5

Remind students to multiply the standard molar entropy of each reactant and product by the appropriate coefficient from the balanced chemical equation when calculating standard entropy of a reaction.

Section 16.6

A reaction is spontaneous only if ΔSsystem + ΔSsurroundings = ΔStotal > 0. ΔSsystem is calculated from standard molar entropies. ΔSsystem = –ΔH/T, where –ΔH is calculated from standard molar enthalpies.

Section 16.7

Two types of reactions have a spontaneity that cannot be reversed by changing the temperature (those with ΔH and ΔS both negative or both positive), and two have a spontaneity that can be reversed by changing the temperature (those with opposite signs for ΔH and ΔS).

Section 16.7

According to ΔG = ΔH –TΔS, a spontaneous reaction is favored by a decrease in enthalpy and an increase in entropy. At lower temperatures the enthalpy term usually dominates; at higher temperatures the entropy term can become dominant.

Section 16.9

For an element in its most stable form at 25oC, both the standard free-energy of formation and the standard enthalpy of formation are equal to zero.

Section 16.9

A substance with a negative standard free energy of formation with respect to decomposition to its constituent elements. However, the substance may be unstable with respect to some other reaction under the same set of conditions.

Section 16.11

When all reactants and products are under standard conditions, Q = 1 and ΔG = ΔGo.

157

Chapter 16—Thermodynamics: Entropy, Free Energy, and Equilibrium

Lecture/Laboratory Demonstration References Section 16.4

Brian B. Laird, “Entropy, Disorder, and Freezing,” J. Chem. Educ., Vol. 76, 1999, 1388-1390.

Literature References Section 16.1

Ruth Ben-Zvi, Judith Silberstein, and Rachel Mamiok, “A Model of Thermal Equilibrium: A Tool for the Introduction of Thermodynamics,” J. Chem. Educ., Vol. 70, 1993, 31-34.

Section 16.1

James N. Spencer, Richard S. Moog, and Ronald J. Gillespie, “Demystifying Introductory Chemistry. Part 4: An Approach to Reaction Thermodynamics Through Enthalpies, Entropies, and Free Energies of Atomization,” J. Chem. Educ., Vol. 73, 1996, 631-636.

Section 16.1

Raymond S. Ochs, “Thermodynamics and Spontaneity,” J. Chem. Educ., Vol. 73, 1996, 952-954.

Section 16.2

John J. Fortman, “Pictorial Analogies of Heat Flow, Thermodynamics, and Entropy,” J. Chem. Educ., Vol. 70, 1993, 102-103.

Section 16.2

Joseph H. Lechner, “Visualizing Entropy,” J. Chem. Educ., Vol. 76, 1999, 13821387.

Section 16.2

Peter Weiss, “Another Face of Entropy,” Science News, Vol. 154, 1998, 108-109.

Section 16.2

L. Glasser, “Order, Chaos, and All That!” J. Chem. Educ., Vol. 66, 1989, 997-1001.

Section 16.2

John P. Lowe, “Entropy: Conceptual Disorder,” J. Chem. Educ., Vol. 65, 1988, 403406.

Section 16.3

Douglas K. Russell, “The Boltzmann Distribution,” J. Chem. Educ., Vol. 73, 1996, 299-300.

Section 16.3

P. G. Nelson, “Derivation of the Second Law of Thermodynamics from Boltzmann’s Distribution Law,” J. Chem. Educ., Vol. 65, 1989, 390-393.

Section 16.5

Travis Thoms, “Periodic Trends for the Entropy of Elements,” J. Chem. Educ., Vol. 72, 1995, 16.

Section 16.7

Sidney Rosen, “J. Willard Gibbs (1839-1903): A Modern Genius,” J. Chem. Educ., Vol. 60, 1983, 593-594.

Section 16.7

Robert J. Deltete and David L. Thorsell, “Josiah Willard Gibbs and Wilhelm Ostwald: A Contrast in ScientificStyle,” J. Chem. Educ., Vol. 73, 1996, 289-295.

Section 16.11

Zheng Xianmin, “The Free Energy Prediction and the Principle of Le Châtelier,” J. Chem. Educ., Vol. 66, 1989, 401-402

158

Chapter 16—Thermodynamics: Entropy, Free Energy, and Equilibrium

Media References Section 16.1

Spontaneous Process activity from the Instructor Resource DVD

Section 16.1

Nitrogen Triiodide movie from the Instructor Resource DVD

Section 16.2

Changes of State movie from the Instructor Resource DVD

Section 16.3

Entropy and Probability activity from the Instructor Resource DVD

Section 16.4

Entropy and Temperature activity from the Instructor Resource DVD

Section 16.5

Thermite movie from the Instructor Resource DVD

Section 16.8

Airbags movie from the Instructor Resource DVD

Section 16.11

Free Energy and Chemical Equilibrium activity from the Instructor Resource DVD

159

CHAPTER 17 ELECTROCHEMISTRY Chapter Learning Goals for Students Section 17.1

Sketch a galvanic cell, identifying the anode and cathode half-reactions, the sign of each electrode, and the direction of electron and ion flow.

Write balanced chemical equations for reactions occurring in a galvanic cell.

Section 17.2

Write and interpret shorthand notations for galvanic cells.

Section 17.3

Interconvert cell potential and free-energy change for a reaction.

Section 17.4

Use a table of standard reduction potentials to calculate standard cell potentials.

Section 17.5

Use a table of standard reduction potentials to rank substances in order of increasing oxidizing strength or reducing strength and to determine whether a reaction is spontaneous.

Section 17.6

Use the Nernst equation to calculate cell potentials for reactions occurring under nonstandard-state conditions.

Section 17.7

From a measured cell potential for a reaction involving hydrogen ion and a reference cell potential, calculate the pH of the solution.

Section 17.8

Calculate equilibrium constants from standard cell potentials and vice versa.

Section 17.9

Write balanced chemical equations for reactions occurring in common batteries.

Section 17.10

Compare fuel cells and batteries.

Calculate standard potentials for fuel cells.

Section 17.11

Write balanced equations for the reactions that occur when iron rusts in the presence of Mg and in the absence of Mg.

Section 17.12

Describe half-cell and overall reactions occurring in electrolytic processes.

Section 17.13

Sketch electrochemical cells used in commercial applications of electrolysis.

Section 17.14

Perform electrolytic cell calculations interconverting current and time, charge, moles of electrons, and moles (or grams) of product.

160

Chapter 17—Electrochemistry

Lecture Outline 17.1. Galvanic Cells1 A. Galvanic cell – a spontaneous chemical reaction generates an electric current B. Electrolytic cell – an electric current drives a nonspontaneous reaction C. Redox reaction 1. Oxidation – a loss of electrons (an increase in oxidation number) 2. Reduction – a gain of electrons (a decrease in oxidation number) 3. Represent oxidation and reduction aspects of the reaction with half-reactions 4. Oxidizing agent – species that causes oxidation to occur and is itself reduced 5. Reducing agent – species that causes reduction to occur and is itself oxidized 6. If a spontaneous reaction is carried out in a beaker: a. Oxidizing agent and reducing agent are in direct contact. b. Electrons are directly transferred. c. Enthalpy of reaction is lost to the surroundings for an exothermic reaction.

7. If a spontaneous reaction is carried out in a galvanic cell – chemical energy released by the reaction is converted to electrical energy.

D. For the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 1. Use a Daniell cell, a type of galvanic cell, to carry out the reaction (textbook Figure 17.2). a. Consists of two half-cells i. A beaker with a strip of Zn in a solution of ZnSO4 ii. A beaker with a strip of Cu in a solution of CuSO4 b. Electrodes – strips of zinc and copper c. Salt bridge – a U-shaped tube that contains a gel permeated with a solution of an inert electrolyte 2. The electrons can be transferred only through the wire. a. Oxidation and reduction half-reactions occur at separate electrodes. b. Electric current flows through the wire.

3. Anode – the electrode at which oxidation takes place a. The negative (–) electrode b. Produces electrons 4. Cathode – the electrode at which reduction takes place a. The positive (+) electrode b. Consumes electrons 5. Anode and cathode half-reactions must add to give the overall cell reaction.

6. Salt bridge maintains electrical neutrality by a flow of ions. a. Anions flow through the salt bridge from the cathode to the anode compartment. b. Cations migrate through the salt bridge from the anode to the cathode compartment.

7. Electrons move through the external circuit from the anode to the cathode.

17.2. Shorthand Notation for Galvanic Cells2 A. Single vertical line (|) represents a phase boundary.

B. Double vertical line (||) represents a salt bridge.

C. Shorthand for the anode half-cell is always written on the left of the salt-bridge symbol, followed on the right of the symbol by the shorthand for the cathode half-cell.

1. Reactants in each half-cell are written first, followed by products.

2. Electrons move through the external circuit from left to right.

3. For Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu(s): Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) D. Cell involving a gas 1 Test Item File Questions: Multiple Choice 1-13, 111-116, 118-120. 122-125; Algorithmic 1-2; Short

Answer 1 2 Test Item File Questions: Multiple Choice 14-25, 117, 121, 126; Algorithmic 3-7; Short Answer 2-6 161

Chapter 17—Electrochemistry 1. Additional vertical line due to presence of additional phase 2. List the gas immediately adjacent to the appropriate electrode.

E. Detailed notation includes ion concentrations and gas pressures.

17.3. Cell Potentials and Free-Energy Changes for Cell Reactions3 A. Electromotive force (emf) – the driving force (electrical potential) that pushes the negatively charged electrons away from the anode and pulls them toward the cathode.

1. Also called cell potential (E) or cell voltage 2. Potential of a galvanic cell is a positive quantity.

B. Coulomb (C) – amount of charge transferred when a current of 1 ampere (A) flows for 1 s.

1J=1C×1V C. Cell potential – measured with a voltmeter 1. Gives a positive reading when the + and – terminals of the voltmeter are connected to cathode (+) and anode (–), respectively 2. Can use voltmeter-cell connections to determine which electrode is the anode and which is the cathode D. Two driving forces of a chemical reaction: cell potential, E and free-energy change, ΔG 1. Related by ΔG = – nFE a. n = number of moles of electrons transferred in the reaction b. F (faraday) – the electrical charge on 1 mol of electrons c. 1 F = 96,500 C/mol e– 2. ΔG and E have opposite signs.

3. Spontaneous reaction has a positive cell potential but negative ΔG. E. Standard cell potential, E°– the cell potential when both reactants and products are in their standard states 1. Solutes at 1 M concentration 2. Gases at a partial pressure of 1 atm 3. Solids and liquids in pure form 4. T = 25°C F. ΔG° = –nFE° 17.4. Standard Reduction Potentials4 A. Standard cell potential is the sum of anode and cathode half-cell potentials. o o o Ecell = Eanode + Ecathode 1. Potential of a single electrode not measurable 2. Potential difference measured by placing a voltmeter between two electrodes 3. Set of standard half-cell potentials developed a. Arbitrary standard half-cell chosen as reference point – assigned an arbitrary potential b. Potential of all other half-cells expressed relative to the reference half-cell B. Standard hydrogen electrode (S.H.E.) – reference half-cell 1. Corresponding half-reaction – assigned an arbitrary potential of exactly 0 V o E = 0.00 V 2 H+(aq, 1 M) + 2 e– → H2(g, 1 atm) 2. Shorthand notation for S. H. E. H+(1 M) | H2(1 atm) | Pt(s) C. Standard potentials for half-cells determined by constructing a galvanic cell with the half-cell of interest paired with the standard hydrogen electrode 1. Standard oxidation potential – the corresponding half-cell potential for an oxidation halfreaction 2. Standard reduction potential – the corresponding half-cell potential for a reduction halfreaction 3.Whenever half-reaction is reversed, the sign of Eo must be reversed.

3 Test Item File Questions: Multiple Choice 26-34; Short Answer 7-9 4 Test Item File Questions: Multiple Choice 35-40, 127; Algorithmic 8; Short Answer 10-12

162

Chapter 17—Electrochemistry 4. Standard oxidation potential and standard reduction potential always have the same magnitude but opposite signs.

5. Table of standard reduction potentials – textbook Appendix D D. Conventions used in constructing a table of half-cell potentials 1. Half-reactions written as reductions a. Oxidizing agents and electrons shown on reactant side b. Reducing agents shown on product side 2. Half-cell potentials are standard reduction potentials, also known as standard electrode potentials.

3. Half-reactions are listed in order of decreasing standard reduction potential. a. Strongest oxidizing agents are located in the upper left of the table. b. Strongest reducing agents are in the lower right of the table.

4. Half-reaction ordering corresponds to ordering of oxidation reactions in the activity series.

5. The more active metals at the top of the activity series have the more positive oxidation potential (more negative reduction potential).

17.5. Using Standard Reduction Potentials5 A. Table of standard reduction potentials – summarizes much chemical information 1. Two or more oxidizing or reducing agents arranged in order of increasing strength 2. Use to predict the spontaneity or nonspontaneity of thousands of redox reactions a. Combine half-reactions of interest and use o o o Ecell = Eanode + Ecathode b. May need to multiply half-reactions by some factor to ensure that electrons cancel c. Values of E° for the half-reactions not multiplied by that factor B. E° values independent of amount of reaction 1. ΔG° = – nFE° 2. ΔG° is an extensive property (depends on amount of substance). a. When amount of substance that reacts changes, ΔG° changes by the same amount as does n, the number of electrons transferred. b. But E° = – ΔG°/nF remains constant.

C. Can predict reaction spontaneity by knowing the location of oxidizing and reducing agents in the table.

1. An oxidizing agent can oxidize any reducing agent that lies below it in the table.

2. E° for the overall reaction must be positive.

17.6. Cell Potentials and Composition of the Reaction Mixture: The Nernst Equation6 A. Cell potentials depend on temperature and on the composition of the reaction mixture.

1. ΔG = ΔG° + RT ln Q 2. ΔG = – nFE; ΔG° = – nFE° 3. – nFE = – nFE° + RT ln Q B. Nernst equation in volts at 25°C used to calculate cell potentials under nonstandard-state conditions 0.0592 E = Eo − log Q n 17.7. Electrochemical Determination of pH7 A. Important application of Nernst equation 1. Electrochemical determination of pH 2. Measured by a pH meter B. Consider a cell with a hydrogen electrode anode and a second reference electrode cathode.

5 Test Item File Questions: Multiple Choice 41-54; Algorithmic 9-12; Short Answer 13-14 6 Test Item File Questions: Multiple Choice 55-66, 128-135; Algorithmic 13-14; Short Answer 15 7 Test Item File Questions: Multiple Choice 67-72; Short Answer 16

163

Chapter 17—Electrochemistry 1. Pt(s) | H2(1 atm) | H+(? M) || reference cathode 2. Ecell = 0.0592 pH + Eref 3. pH is a linear function of the cell potential.

E − Eref pH = cell 0.0592 C. pH of a solution measured by measuring Ecell D. Actual pH measurements use a glass electrode with a calomel electrode as the reference.

17.8. Standard Cell Potentials and Equilibrium Constants8 A. Standard free-energy change for a reaction is related to both the standard cell potential and the equilibrium constant.

1. ΔG° = – nFE° 2. ΔG° = – RT ln K B. Combining the two equations: RT 2.303RT Eo = ln K = log K nF nF 0.0592 Eo = log K n C. Most common use – calculating equilibrium constants from standard cell potentials D. Equilibrium constants for redox reactions tend to be either very large or very small in comparison with equilibrium constants for acid-base reactions.

1. Positive value of E° corresponds to K > 1.

2. Negative value of E° corresponds to K < 1.

E. Three different ways to determine the value of an equilibrium constant K 1. K from concentration data: [C]c [D]d K= [A]a [B]b 2. K from thermochemical data: −ΔG o ln K = RT 3. K from electrochemical data: nFE o ln K = RT 17.9. Batteries9 A. Most important practical application of galvanic cells B. Battery features depend on application C. General features 1. Compact and lightweight 2. Physically rugged and inexpensive 3. Provide a stable source of power for relatively long periods of time D. Lead storage battery 1. Used as a reliable power source for starting automobiles for over 75 years 2. 12 V battery – six 2 V cells connected in series 3. Anode – series of grids packed with spongy lead 4. Cathode – series of grids packed with lead dioxide, dipped into an aqueous solution of H2SO4 (38% w/w) 5. Electrode half-reactions and the overall cell reaction

8 Test Item File Questions: Multiple Choice 73-77; Short Answer 17 9 Test Item File Questions: Multiple Choice 78-86; Short Answer 18

164

Chapter 17—Electrochemistry Anode: Pb(s) + HSO4–(aq) → PbSO4(s) + H+(aq) + 2 e–

E° = 0.296 V

Cathode: PbO2(s) + 3 H+(aq) + HSO4–(aq) + 2 e– → 2 PbSO4(s) + 2 H2O(l) E° = 1.628 V

____________________________________________________________________________________________

Overall: Pb (s) + PbO2(s) + 2 H+(aq) + 2 HSO4–(aq) → 2 PbSO4(s) + 2 H2O(l)

E° = 1.924 V

6. PbSO4 adheres to the surface of the electrodes. a. Recharge by using an external source of direct current to drive the cell reaction in the reverse, nonspontaneous direction E. Dry-cell batteries (Leclanché cells) – common household batteries 1. Anode – Zn metal can 2. Cathode – inert graphite rod surrounded by a paste of solid MnO2 and carbon black 3. Electrolyte – a moist paste of NH4Cl and ZnCl2 in starch a. Surrounds the MnO2 containing paste b. Acidic – causes corrosion of the Zn anode (Zn → Zn2+) 4. Alkaline dry cell – modified version of Leclanché a. Replace NH4Cl (acidic) with NaOH or KOH b. Electrode reactions – oxidation of zinc and reduction of manganese dioxide i. Produces ZnO due to basic conditions ii. Zinc corrodes more slowly iii. Battery has a longer life c. Produces higher power and more stable current and voltage – more efficient ion transport in the alkaline electrolyte 5. Mercury battery – used in watches, heart pacemakers, and other devices a. Small size b. Anode – Zn (same as dry cell) c. Cathode — steel in contact with HgO in an alkaline medium of KOH and Zn(OH)2 F. Nickel-cadmium batteries – used in calculators and portable power tools 1. Rechargeable 2. Anode – cadmium metal 3. Cathode – NiO(OH) supported on nickel metal 4. Solid products of electrode reaction adhere to the surface of the electrodes – allows battery to be recharged G. Nickel-metal hydride batteries – used in consumer electronic products and in hybrid gaselectric automobliles 1. Less expensive than nickel-cadmium batteries 2. Light weight 3. Environmentally friendly 4. Anode – metal alloy, LaNi5 5. Cathode – NiO(OH) supported on nickel metal H. Lithium-ion batteries – used in portable electronic products o 1. Li has a higher E value for oxidation and lower atomic mass than any other metal.

2. Anode – lithium metal or lithium atoms in a graphite electrode + 3. Cathode – metal oxide or metal sulfide that can incorporate Li 4. Electrolyte – lithium salt in an organic solvent 5. Used in cell phones, laptop computers, and digital cameras, power tools, headlamps, hybrid electric cars 17.10. Fuel Cells10 A. Fuel cell – a galvanic cell in which one of the reactants is a traditional fuel B. Reactants not self-contained within the cell – supplied from an external reservoir C. Hydrogen-oxygen fuel cell 10 Test Item File Questions: None

165

Chapter 17—Electrochemistry 1. Best known 2. Used in space vehicles as a source of electric power 3. Porous carbon electrodes containing metallic catalysts 4. Hot, aqueous KOH electrolyte 5. Overall reaction 2 H2(g) + O2(g) → 2 H2O(l) D. Proton-exchange membrane (PEM) fuel cell 1. Designed to power electric vehicles 2. Similar to hydrogen-oxygen fuel cell, except KOH electrolyte replaced by a plastic membrane that conducts H+ but not e– 3. More environmentally friendly than the hydrogen–oxygen fuel cell E. Direct methanol fuel cell (DMFC) 1. Similar to PEM fuel cell 2. Uses aqueous CH3OH instead of gaseous H2 3. Lighter and higher energy density than conventional batteries 4. Fuel is more readily available, safer, and easier to store than H2 17.11. Corrosion11 A. Corrosion – oxidative deterioration of a metal B. Well-known example of corrosion – conversion of iron to rust 1. Requires both oxygen and water 2. Involves pitting of the metal surface 3. Rust is deposited at a location physically separated from the pits C. Proposed mechanism for formation of rust – an electrochemical process in which iron is oxidized in one region of the surface and oxygen is reduced in another region 1. Anode region: Fe(s) → Fe2+(aq) + 2 e– E° = 0.45 V 2. Cathode region: O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l) E° = 1.23 V 3. Electrons flow from anode to cathode through the metal.

4. Ions migrate through the water droplets. a. Fe2+ reacts with O2 and is oxidized to Fe3+ b. Fe3+ reacts with H2O to form Fe2O3 ⋅ xH2O(s) (rust) 5. Dissolved salt in water greatly increases electrolyte conductivity.

6. Explains why cars rust more rapidly when road salt is used to melt snow and ice n+ 7. O2 able to oxidize all metals except a few – O2/H2O half-reaction lies above the M /M half-reaction D. Prevention of corrosion – shield the metal surface from oxygen and moisture 1. Durable surface coating – metals such as chromium, tin, or zinc 2. Galvanizing – coating by dipping into a bath of molten zinc 3. Cathodic protection – protecting a metal from corrosion by connecting it to a second metal that is more easily oxidized 17.12. Electrolysis and Electrolytic Cells12 A. Electrolytic cell – electric current used to drive a nonspontaneous reaction B. Processes occurring in galvanic and electrolytic cells are the reverse of each other.

C. Electrolysis – the process of using an electric current to bring about chemical change D. Electrolytic cell 1. Two electrodes in an electrolyte connected to a battery or some other source of direct electric current 2. Battery – an electron pump, pushing electrons into one electrode and pulling them out of the other electrode 3. Anode – electrode where oxidation takes place a. Positive sign 11 Test Item File Questions: Multiple Choice 87-91 12 Test Item File Questions: Multiple Choice 92-98, 136-138

166

Chapter 17—Electrochemistry b. Battery pulls electrons out of it 4. Cathode – electrode where reduction takes place a. Negative sign b. Battery pushes electrons into it E. Electrolysis of molten NaCl + 1. Cathode – attracts Na Na+ + e– → Na(l) 2. Anode – attracts Cl– 2 Cl– → Cl2(g) + e– F. Electrolysis of aqueous NaCl 1. Electrode reactions in an aqueous solution may differ from those for a molten salt.

2. Cathode reaction can involve the reduction of Na+ or the reduction of water. o a. Reduction of water preferred – E less negative for H2O than Na+ b. 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq) 3. Anode reaction can involve the oxidation of Cl– or the oxidation of water. a. Actual reaction – oxidation of Cl– due to overvoltage b. 2 Cl–(aq) → Cl2(g) + 2 e– 4. Overvoltage – amount of voltage needed above the calculated standard reduction (or oxidation) potential for electrolysis to occur a. Needed when the half-reaction has a substantial barrier for electron transfer (slow rate) i. Surmounts barrier ii. Reaction proceeds at satisfactory rate. b. Small overvoltage needed for solution for deposition of metals c. Large overvoltage needed for formation of O2 or H2 d. Can’t predict – need experimental evidence if cell potentials are similar 5. Overall cell reaction 2 Cl–(aq) + 2 H2O(l) → Cl2(g) + 2 H2(g) + 2 OH–(aq) 6. Na+ is a spectator ion and reacts with the OH– to form NaOH G. Electrolysis of water 1. Electrolysis of any aqueous solution requires the presence of an electrolyte to carry the current in solution.

2. If the ions of the electrolyte are less easily oxidized and reduced than water is, then water will react at both electrodes.

3. Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– 4. Cathode: 4 H2O(l) + 4 e– → 2 H2(g) + 4 OH–(aq) 5. Overall cell reaction: 2 H2O(l) → O2(g) + 2 H2(g) 17.13. Commercial Applications of Electrolysis13 A. Manufacture of sodium 1. Produced commercially in a Downs cell by electrolysis of a molten mixture of NaCl and CaCl2 2. Liquid Na produced at the cylindrical steel cathode is less dense than the molten salt and thus floats to the top part of the cell, where it is drawn off into a suitable container.

B. Manufacture of chlorine and sodium hydroxide 1. Electrolysis of aqueous NaCl 2. Basis of chlor-alkali industry 3. Anode and cathode electrolysis reactions of aqueous NaCl carried out in a membrane cell 4. Membrane keeps Cl2 and OH– apart but allows a current of Na+ to flow.

C. Manufacture of aluminum 1. Hall–Heroult process

13 Test Item File Questions: Multiple Choice 99-102

167

Chapter 17—Electrochemistry o

2. Electrolysis of a molten mixture of Al2O3 and cryolite (Na3AlF6) at 1000 C in a cell with graphite electrodes 3. Success of process – use of cryolite as a solvent 4. Electrode reactions involve the formation of complex ions. a. Ions are reduced at the cathode to produce Al(l). b. Ions are oxidized at the anode to produce O2(g). i. O2(g) reacts with the graphite electrode to produce CO2(g) ii. Requires frequent replacement of the anodes 5. Largest single consumer of electricity in the U.S. 6. One mole of electrons produces only 9 g of Al. D. Electrorefining – purification of a metal by means of electrolysis E. Electroplating – coating of one metal on the surface of another using electrolysis 1. Cathode – object to be plated (carefully cleaned) 2. Electrolytic cell contains a solution of ions of the metal to be deposited.

17.14. Quantitative Aspects of Electrolysis14 A. Amount of substance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell.

B. Follows directly from the stoichiometry of the reaction and the atomic mass of the product.

C. Moles of electrons passed through a cell are determined from the electric current and the time that the current flows.

1 mol e Moles of e - = charge (C) × 96,500 C D. Sequence of conversion used to calculate the mass or volume of product produced by passing a known current for a fixed period of time.

Current × time → Charge → Moles of e– → Moles of product → Grams of liters of product 1. Current in amperes (A) 2. Time in seconds (s) 3. Charge in coulombs (C) E. Think of electrons as reactants in a balanced equation, and proceed as with any other stoichiometry problem.

14 Test Item File Questions: Multiple Choice 103-110; Algorithmic 15-18; Short Answer 19-20

168

Chapter 17—Electrochemistry

CHAPTER 17 ELECTROCHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 17.1

Anions flow through the salt bridge into the anode compartment while cations flow into the cathode compartment to maintain electroneutrality.

Section 17.1

The anode of a galvanic cell is negative (like an anion), and the cathode is positive (like a cation). Electrons move from the negative anode through an external circuit to the positive cathode.

Section 17.3

The driving force of a cell reaction, measured in volts and called cell voltage, cell potential, or electromotive force, is a function of the natural tendency of one substance to lose electrons and of a second substance to gain electrons. The greater these tendencies, the higher the cell voltage.

Section 17.3

When a chemical equation is balanced by the half-reaction method, the number of electrons that occur on both sides of the balanced equation (before canceling) is equal to the value of n.

Section 17.3

For a spontaneous reaction, ΔG is negative but E is positive.

Section 17.3

Standard conditions for Eo, the standard cell potential, are identical with standard conditions for ΔGo, ΔHo, and ΔSo.

Section 17.4

Every reduction half-cell reaction has a corresponding standard reduction potential, Eored. If the reaction is reversed, it becomes an oxidation half-cell with a corresponding standard oxidation half-cell potential, Eoox. For any given half-cell reaction, Eoox = –Eored.

Section 17.4

The more positive the half-cell potential, the greater the tendency for the reaction to occur as written. The more negative the half-cell potential, the greater the tendency for the reverse reaction to occur.

Section 17.5

Half-cell potentials are never multiplied by a coefficient when used in the equation Eocell = Eoox + Eored.

Section 17.5

When predicting whether a reaction is spontaneous using Table 17.1, some students find it useful to circle the two potential reactants and connect them with a line. If the line has a negative slope, the reaction is spontaneous; if the line has a positive slope, the reaction is nonspontaneous. If both potential reactants are on the same side of the arrows, no reaction can occur because an oxidation requires a simultaneous reduction.

Section 17.6

The general form of the Nernst equation is used to correct the standard cell potential for concentrations other than 1 M, partial pressures other than 1 atm, or a temperature other than 25oC. This form of the Nernst equation applies only to reactions occurring at 25oC.

Section 17.7

The Nernst equation relates the concentration of any chemical species, including H+, to cell potential. This relationship is utilized in a pH meter which measures a cell potential and displays its value in terms of pH. 169

Chapter 17—Electrochemistry

Section 17.8

The more positive the standard cell potential, the larger the value of the equilibrium constant, and the farther the equilibrium lies to the right.

Section 17.9

A multicell battery is comprised of galvanic cells wired in series, so the battery voltage equals the sum of the voltages of the individual cells.

Section 17.10

Although many metals react more readily than iron to form oxides, their oxides provide a protective coating against further corrosion. The corrosion product of iron (“rust”) flakes off and exposes a fresh surface to further corrosion.

Section 17.11

In an electrolytic cell a constant input of energy from an external direct current (dc) source, such as a battery, is used to drive a nonspontaneous reaction. The voltage of the battery must be more positive than the absolute value of the voltage for the electrolysis reaction.

Section 17.11

Overvoltage is additional voltage that must be applied in excess of the calculated voltage to initiate an electrolytic reaction. It is most important for those systems involving gases that impede the flow of electrons between electrode surfaces and reactants in solution.

Section 17.12

This procedure and the Hall–Heroult process discussed below are examples of commercial uses of a colligative property (freezing point depression) to enable electrolytic reactions to be carried out more economically.

Section 17.12

Originally, chlorine was a useless byproduct of this process. The search for uses of chlorine resulted in many commercial chlorinated compounds, including plastic film used to wrap food.

Section 17.13

Stress the use of this flowchart along with the dimensional analysis method of problem solving.

Lecture/Laboratory Demonstration References Section 17.1

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Visible OxidationReduction in Electrochemical Cells,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 205206.

Section 17.4

J. J. Fortman and R. Battino, “Determining the Metal Activity Series Using a Potato Porcupine,” J. Chem. Educ., Vol. 70, 1993, 939. Pieces of metal are inserted into a potato. A voltmeter is attached to a given metal and a copper reference electrode.

With the potato serving as a salt bridge and source of electrolytes, the observed cell potentials are used to rank the metals in an activity series.

Section 17.11

Bassam Z. Shakhashiri, “Electrolytic Cells in Series: A Red, White, and Blue Electrolysis,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp.170-173.

Section 17.12

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Electroplating Copper,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 199-200.

170

Chapter 17—Electrochemistry

Literature References Section 17.1

Ron DeLorenzo, “Electrochemical Errors,” J. Chem. Educ., Vol. 62, 1985, 424-425.

Section 17.1

Penelope Ann Huddle, Margaret Dawn White, and Fiona Rogers, “Using a Teaching Model to Correct Known Misconceptions in Electrochemistry,” J. Chem. Educ., Vol. 77, 2000, 104-110.

Section 17.5

Richard S. Treptow, “Dental Filling Discomforts Illustrate Electrochemical Potentials of Metals,” J. Chem. Educ., Vol. 55, 1978, 189.

Section 17.9

Michael J. Smith and Colin A. Vincent, “Structure and Content of Some Primary Batteries,” J. Chem. Educ., Vol. 78, 2001, 519-521.

Section 17.9

Gerald Ondrey, Charlene Crabb, and Takeshi Kamiya, “Batteries: Full Speed Ahead,” Chem. Eng., Vol. 106 (2), 1999, 47-51.

Section 17.10

A. I. Onuchukwu, “Evaluation of Corrosion Susceptibility of a Metal: Student Corrosion Experiment II,” J. Chem. Educ., Vol. 65, 1988, 934.

Media References Section 17.1

Oxidation-Reduction Reactions Part II movie from the Instructor Resource DVD

Section 17.1

Galvanic Cells activity from the Instructor Resource DVD

Section 17.1

Copper–Zinc Cell movie from the Instructor Resource DVD

Section 17.2

Shorthand Notation for Galvanic Cells activity from the Instructor Resource DVD

Section 17.3

Redox Chemistry of Iron and Copper movie from the Instructor Resource DVD

Section 17.4

Voltaic Cells II: The Zinc–Hydrogen Cell movie from the Instructor Resource DVD

Section 17.5

Standard Reduction Potentials movie from the Instructor Resource DVD

Section 17.8

Equilibrium Constant activity from the Instructor Resource DVD

Section 17.12

Electrolysis of Water movie from the Instructor Resource DVD

Section 17.13

Electroplating movie from the Instructor Resource DVD

Section 17.14

Quantitative Aspects of Electrolysis activity from the Instructor Resource DVD

171

CHAPTER 18 HYDROGEN, OXYGEN, AND WATER Chapter Learning Goals for Students Section 18.1

Describe the properties of elemental hydrogen, including its appearance, molecular structure, and occurrence in nature.

Apply the gas laws to problems involving hydrogen, including calculation of its density and calculation of the number of grams of solid reactants needed to produce a certain volume of hydrogen.

Section 18.2

Describe the isotopes of hydrogen, and compare and contrast the properties of the isotopes and compounds containing the isotopes.

Explain and give examples of the isotope effect.

Section 18.3

Write balanced chemical equations for the laboratory synthesis, industrial synthesis, and industrial uses of hydrogen.

Assign oxidation numbers and identify the oxidizing agent and reducing agent for redox reactions of hydrogen. Balance equations for these reactions, using either the oxidation-number method or the half-reaction method.

Section 18.4

Write chemical formulas of hydrogen compounds of elements, and predict which compounds are covalent and which are ionic.

Section 18.5

Classify binary hydrides as ionic, covalent, or interstitial, determine their formulas, and predict their properties.

Write equations for reactions of ionic hydrides with water.

Perform stoichiometry calculations for reactions of ionic hydrides.

Section 18.6

Describe the properties of elemental oxygen, including its appearance, molecular structure, and occurrence in nature.

Section 18.7

Write balanced chemical equations for the laboratory synthesis, industrial synthesis, and industrial uses of hydrogen.

Apply the gas laws to problems involving oxygen, including the calculation of its density and the calculation of the number of grams of solid reactants needed to produce a certain volume of oxygen.

Assign oxidation numbers and identify the oxidizing agent and reducing agent for redox reactions of oxygen. Balance equations for these reactions, using either the oxidation-number method or the half-reaction method.

Section 18.8

Write chemical formulas of oxygen compounds of elements, and predict which compounds are covalent and which are ionic.

172

Chapter 18 — Hydrogen, Oxygen, and Water

Section 18.9

Classify oxides as basic, acidic, or amphoteric, determine their formulas, and predict their properties.

Section 18.10

Identify compounds as oxides, peroxides, and superoxides.

Draw molecular orbital energy-level diagrams, determine bond orders and magnetic behavior of peroxide ion and superoxide ion.

Section 18.11

Draw an electron-dot structure for hydrogen peroxide.

Write balanced chemical equations for reactions of hydrogen peroxide, and tell whether hydrogen peroxide is acting as an oxidizing agent or a reducing agent in the reactions.

Section 18.12

Use structure and bonding concepts to explain the physical and chemical properties of elemental oxygen and ozone.

Section 18.13

Explain hard water and how hard water can be softened by ion exchange.

Section 18.14

Identify and describe the general properties of a hydrate.

Determine the empirical formula of a hydrate.

173

Chapter 18—Hydrogen, Oxygen, and Water

Lecture Outline 18.1. Hydrogen1 A. Henry Cavendish – English chemist who isolated hydrogen in pure form Acid + Metal → H2(g) + Metal ion B. Lavoisier – French chemist who named hydrogen 1. Hydrogen – water former 2. Combines with oxygen to form water C. Properties of hydrogen 1. Colorless, odorless, and tasteless gas 2. Nonpolar, diatomic molecule 3. Weak intermolecular forces 4. Low boiling and melting points 5. Bond dissociation energy = 436 kJ/mol D. Natural occurrence 1. 75% of the mass of the universe 2. Atmospheric abundance – 0.53 ppm by volume 3. Ninth most abundant element in terms of mass in the earth’s crust and oceans 18.2. Isotopes of Hydrogen2 A. Three isotopes of hydrogen 1. Protium a. Ordinary hydrogen ( 11H ) b. 99.985% of atoms in naturally occurring hydrogen 2. Deuterium – heavy hydrogen ( 21H or D) 3. Tritium – ( 31H or T) B. Similar properties – textbook Table 18.1 1. Same electron configuration for all three isotopes – determines chemical behavior 2. Isotope effects – arise from differences in isotope mass a. Greater for hydrogen than for any other element – percentage differences between the masses of the isotopes are largest for hydrogen b. D2 – higher boiling and melting point and greater heat of dissociation than H2 c. D2O – higher boiling and melting point than H2O i. Smaller equilibrium constant for dissociation ii. Used as a coolant and a moderator in nuclear reactors C. D2 and D2O manufactured by electrolysis of an inert electrolyte in ordinary water 1. Electrolysis – process of using an electric current to bring about chemical change 2. Kinetic isotope effect – effect of isotopic mass on chemical reaction rate a. Heavier isotopes – slower reaction faster b. 2 H2O(l) → 2 H2(g) + O2(g) 2 D2O(l) → 2 D2(g) + O2(g) slower 18.3. Preparation and Uses of Hydrogen3 A. Purest hydrogen prepared by electrolysis of water 1. Requires a large amount of energy 2. Not economical for large-scale production ΔH° = + 572 kJ 3. 2 H2O(l) → 2 H2(g) + O2(g) B. Laboratory synthesis – reaction of dilute acid with an electropositive metal n+ M(s) + 2n H+(aq) → n H2(g) + M (aq) C. Large-scale industrial preparation methods – reducing agent extracts oxygen from steam 1 Test Item File Questions: Multiple Choice 1-3 2 Test Item File Questions: Multiple Choice 4-9; Short Answer 1-5 3 Test Item File Questions: Multiple Choice 10-16; Algorithmic 1-5; Short Answer 6

174

Chapter 18 — Hydrogen, Oxygen, and Water 1. Steam-hydrocarbon re-forming process 2. First step – produce synthesis gas a. H2O(g) + CH4(g) → CO(g) + 3 H2(g) b. High temperature, moderately high pressure, and a nickel catalyst 3. Second step – water-gas shift reaction a. Shifts the composition of synthesis gas b. Removes CO and produces more H2 c. CO(g) + H2O(g) → CO2(g) + H2(g) d. Metal oxide catalyst, T = 400°C 4. Third step – removal of CO2(g) – a. CO2(g) + 2 OH (aq) → CO32–(aq) + H2O(l) b. CO32– remains in the aqueous phase D. Uses of hydrogen 1. 95% of H2 produced – used in synthesis and consumed in industrial plants 2. Largest single consumer a. Haber process b. N2(g) + 3 H2(g) → 2 NH3(g) 3. Synthesis of methanol CO(g) + H2(g) → CH3OH(l) 18.4. Reactivity of Hydrogen4 A. Hydrogen has properties similar to both the alkali metals and the halogens.

1. Lose electron → H+ a. Ei = 1312 kJ/mol b. Doesn’t completely transfer its valence electron in ordinary chemical reactions c. Shares electron with a nonmetallic element to produce a covalent compound d. Gas phase – complete ionization of a hydrogen atom to produce a bare proton e. Liquids and solids – bare proton too reactive to exist by itself 2. Gain electron → H– a. Eea = – 73 kJ/mol b. Will accept an electron from an active metal to give an ionic hydride B. H2 – relatively unreactive 1. Strong H–H bond 2. Reaction of H2 and O2 – highly exothermic a. ΔH = – 572 kJ b. Mixtures of H2 in O2 – highly flammable and can explode 18.5. Binary Hydrides5 A. Binary hydride – compound that contains hydrogen and just one other element B. Ionic hydride – formed by alkali metals and heavier alkaline-earth metals 1. Saltlike, high-melting, white crystalline compounds 2. Prepared by direct reaction of the elements.

3. Alkali-metal hydrides – face-centered cubic crystal structure a. Ionic in the liquid state b. Electrical conductors when molten 4. H– – good proton acceptor a. Ionic hydrides react with water to give H2 gas and OH– ions b. Redox reaction 5. Good reducing agents C. Covalent hydrides – hydrogen attached to another element by a covalent bond 1. Most common – hydrides of nonmetallic elements 2. Discrete, small molecules 4 Test Item File Questions: Multiple Choice 17-18; Algorithmic 6 5 Test Item File Questions: Multiple Choice 19-32, 92-109; Short Answer 7-14

175

Chapter 18—Hydrogen, Oxygen, and Water a. Relatively weak intermolecular forces b. Bases or volatile liquids at ordinary temperatures D. Metallic hydrides – lanthanide, actinide and certain d–block transition metals.

1. Interstitial hydrides – crystal lattice of metal atoms with smaller hydrogen atoms occupying holes (interstices) between larger metal atoms 2. Nonstoichiometric compounds – atomic composition can’t be expressed as a ratio of small whole numbers 3. Properties composition dependent – function of the partial pressure of H2 gas in the surroundings 4. Potential hydrogen-storage devices – can contain large amounts of hydrogen 18.6. Oxygen6 A. Joseph Priestley (English) and Karl Wilhelm Scheele (Swedish) – first isolated and characterized oxygen 1. Odorless, tasteless gas 2. Supports combustion better than air 3. Priestley – called gas “dephlogisticated air” 4. Lavoisier – named element oxygen B. Properties of oxygen 1. Pale blue liquid and solid 2. Paramagnetic in all three phases 3. O2 – double bonded a. Bond length – 121 pm b. Bond dissociation energy = 498 kJ C. Additional information 1. Most abundant element on the planet’s surface 2. Crucial to human life 3. Oxidizing agent in metabolic “burning” of food 4. Occurrence in nature a. 23% of atmosphere – primarily as O2 b. 46% of lithosphere c. 85% of hydrosphere – in the form of H2O D. Photosynthesis process – green plants use solar energy to produce O2 and glucose from carbon dioxide and water 1. Replaces O2 removed by combustion and respiration processes 2. 6 CO2 + 6 H2O → 6 O2 + C6H12O6 E. Metabolism of carbohydrates – reverse of photosynthesis 18.7. Preparation and Uses of Oxygen7 A. Laboratory preparation 1. Electrolysis of water 2. Decomposition of aqueous hydrogen peroxide in presence of catalyst 3. Thermal decomposition of an oxoacid salt B. Industrial synthesis 1. Fractional distillation of liquefied air 2. Third ranking industrial chemical produced in the U.S. C. Industrial uses 1. Steelmaking – removes impurities from iron by forming oxides 2. Sewage treatment – destroys malodorous compounds by oxidation 3. Paper bleaching – oxidizes compounds that cause unwanted colors 4. Oxyacetylene torches – provide high temperatures needed for cutting and welding metals 5. Inexpensive and readily available oxidizing agent 6 Test Item File Questions: Multiple Choice 33-36 7 Test Item File Questions: Multiple Choice 37-43; Algorithmic 7-12; Short Answer 15-16

176

Chapter 18 — Hydrogen, Oxygen, and Water 18.8. Reactivity of Oxygen8 A. High electronegativity B. Needs only two electrons to achieve an octet C. Reacts with active metals to produce an ionic oxide – gains two electrons and achieves a noble gas configuration D. Forms covalent oxides with nonmetals 1. Achieves a noble gas configuration by sharing two additional electrons 2. Forms two single bonds or one double bond. a. Double bonds are formed between oxygen and other small atoms b. Good overlap between oxygen pπ orbital and pπ orbitals of small atoms E. Reacts directly with nearly all elements in the periodic table 1. Doesn’t react with noble gases, platinum, or gold 2. Reactions slow at room temperature 3. Extremely reactive at higher temperatures 18.9. Oxides9 A. Classified on the basis of the oxygen’s oxidation state B. Oxidation number = – 2, oxides C. Categorize in terms of acid-base character 1. Basic oxides a. Ionic b. Formed by metals on left side of periodic table 2. Acidic oxides (acid anhydrides) a. Covalent b. Formed by nonmetals on right side of periodic table 3. Amphoteric oxides a. Exhibit both acidic and basic properties b. Formed by elements with intermediate electronegativity c. Strongly polar covalent character d. Bonds have intermediate ionic-covalent character 4. Acid-base properties and ionic-covalent character dependencies a. Position of element in the periodic table i. Acidic and covalent character increases left to right across the periodic table ii. Basic and ionic character increases top to bottom down the periodic table b. Oxidation state of element – acidic and covalent character increases with increasing oxidation state 5. As the bonding in oxides changes from ionic to covalent, a change in structure and physical properties occurs.

D. Uses determined by properties 1. High thermal stability, mechanical strength, and electrical resistance – use in hightemperature electrical insulators 2. Amphoteric oxides – SiO2 used in optical fibers 3. Acidic oxides – used as precursors to industrial acids 18.10. Peroxides and Superoxides10 A. Oxidation number = –1, peroxides (O22–) B. Oxidation number = –1/2, superoxides (O2–) C. Formed by heating the heavier group 1A and 2A metals in an excess of air.

D. Ionic solids E. MO Theory 1. Explains trends in bond lengths and magnetic properties of O2, O22–, and O2– 8 Test Item File Questions: Multiple Choice 44, 110; Short Answer 17 9 Test Item File Questions: Multiple Choice 45-51, 111-128; Short Answer 18-20 10 Test Item File Questions: Multiple Choice 52-62, 129-132; Short Answer 21-22

177

Chapter 18—Hydrogen, Oxygen, and Water 2. Due to number of electrons in π2p* orbital (textbook Table 18.2) F. Peroxide ion 1. Diamagnetic 2. O–O single bond 3. Basic properties – reacts with strong acids to form hydrogen peroxide, H2O2 G. Superoxide ion 1. One unpaired electron 2. Paramagnetic 3. Bond order = 1.5 4. Evolves oxygen when dissolved in water a. Disproportionation reaction – a substance is both oxidized and reduced in a reaction b. 2 KO2(s) + H2O(l) → O2(g) + 2 K+(aq) + HO2–(aq) + OH–(aq) ↑ ↑ ↑ –1/2 0 –1 18.11. Hydrogen Peroxide11 A. Uses 1. Due to oxidizing properties 2. Antiseptic, bleach, starting material for synthesis of other peroxide containing compounds B. Properties 1. Colorless, syrupy liquid o o 2. mp = – 0.4 C; bp = 150 C 3. Explodes when heated 4. Strong hydrogen bonding present in liquid H2O2 (indicated by high bp) 5. Weak acid in aqueous solutions C. Both an oxidizing and a reducing agent 1. Oxidizing agent: O22– → O2– 2– 2. Reducing agent: O2 → O2 3. Can oxidize and reduce itself (unstable with respect to disproportionation to water and oxygen) a. 2 H2O2(l) → 2 H2O(l) + O2(g) ΔH° = –196 kJ b. Slow at room temperature c. Heat, light, and many catalysts can cause rapid, exothermic, and potentially explosive decomposition.

18.12. Ozone12 A. Two allotropes of oxygen 1. Ordinary dioxygen, O2 2. Ozone, O3 B. Ozone 1. Toxic, pale blue gas 2. Sharp, penetrating odor 3. Produced by passing an electric discharge through O2 C. Bonding in ozone 1. Two resonance structures that are bent 2. Central atom surrounded by two σ bonds and one lone pair of electrons 3. π bond is delocalized over all three oxygen atoms 4. Net bond order = 1.5 D. Powerful oxidizing agent – used to kill bacteria in drinking water 18.13. Water13 A. Most familiar and abundant compound on earth 11 Test Item File Questions: Multiple Choice 63-67; Short Answer 23 12 Test Item File Questions: Multiple Choice 68-71; Algorithmic 13; Short Answer 24 13 Test Item File Questions: Multiple Choice 72-80; Algorithmic 14

178

Chapter 18 — Hydrogen, Oxygen, and Water B. Natural occurrence 24 1. 97.3% – oceans: volume of oceans = 1.35 × 10 mL (twice Avogadro’s number) 2. 2.0% – polar ice caps and glaciers 3. 0.6% – underground fresh water 4. 0.01% – freshwater lakes and rivers C. Seawater 1. Major constituent – sodium chloride 2. Unlimited source of chemicals 3. Three substances obtained from seawater commercially: NaCl, Mg, and Br2 D. Freshwater lakes purification process 1. Preliminary filtration 2. Sedimentation 3. Sand filtration 4. Aeration 5. Sterilization E. Hard water – contains appreciable concentrations of doubly charged cations 1. Soap scum and boiler scale formed by reaction of cations + 2. Softened by ion exchange – process by which Na replaces doubly charged ions 18.14. Hydrates14 A. Solid compounds that contain water molecules B. Complex or unknown structures C. Use a dot in the formula to specify the composition.

D. Formation is common with salts that contain +2 and +3 cations.

E. Metal cation-water bonding interactions – increase with increasing charge on cation F. Hygroscopic compounds – absorb water from the air 1. Anhydrous compounds that have a tendency to form hydrates 2. Useful as drying agents

14 Test Item File Questions: Multiple Choice 81-91; Algorithmic 15-17; Short Answer 25-26

179

Chapter 18—Hydrogen, Oxygen, and Water

CHAPTER 18 HYDROGEN, OXYGEN, AND WATER Teaching Tips, Points of Emphasis, and Common Misconceptions Section 18.3

Review bond dissociation energies and Hess’s law to show that the formation of hydrogen from the decomposition of water requires energy, mainly due to the relative strength of the O–H bonds broken compared to the strength of H–H and O=O bonds formed.

Section 18.4

The properties of molecular hydrogen compounds are best predicted if H is assigned a position in the periodic table based on its electronegativity (between B and C).

Section 18.4

Chemical species with high charge-to-radius ratios, such as H+, are unstable and tend to form new species with lower charge and/or charge dispersed over a greater volume.

Section 18.5

Like all ionic compounds, ionic hydrides are crystalline solids with high melting points. They are formed by direct reaction with only the most active metals.

Section 18.5

Like all molecular compounds which have weak intermolecular forces of attraction, covalent hydrides are gases, liquids, or low-melting solids.

Section 18.6

Remember that double bonds are shorter and stronger than single bonds but longer and weaker than triple bonds formed from the same two atoms.

Section 18.8

Oxygen forms single and double (occasionally triple) bonds with other period 2 nonmetals in order to attain a complete octet. Less commonly does it form multiple bonds with period 3 and higher-period elements.

Section 18.9

The most basic oxides are ionic solids formed by the reaction of oxygen with elements at the left and bottom of the periodic table; the most acidic oxides are covalent and formed by the reaction of oxygen with elements at the right and top of the periodic table.

Section 18.9

The higher the oxidation state of an element, the more acidic its oxide.

Section 18.9

Like ionic hydrides, ionic oxides are crystalline solids with high melting points.

Section 18.9

Like covalent hydrides, most covalent oxides are gases, liquids, or low-melting solids.

Section 18.10

Note that the presence of an O–O bond indicates that the compound is a peroxide (or, less commonly, a superoxide).

Section 18.11

Because the oxygen atom in a peroxide is in an oxidation state (–1) intermediate between two more stable oxidation states (0 and –2), hydrogen peroxide can be either an oxidizing agent or a reducing agent.

Section 18.14

Notice that many hydrates form when the metal cation is very small relative to the anion. Compared with the bare cation, the hydrated cation is larger, packs better with the anion, and forms a more stable crystalline solid.

180

Chapter 18 — Hydrogen, Oxygen, and Water

Lecture/Laboratory Demonstration References Section 18.3

Bassam Z. Shakhashiri, “Preparation and Properties of Hydrogen,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp.128-130. Hydrogen is produced by the reaction of zinc with hydrochloric acid and collected by displacing water from an inverted Erlenmeyer flask. The flame of a burning candle is extinguished when inserted into the flask, but an almost imperceptible flame is seen at the mouth of the flask, demonstrating the flammability of hydrogen in the presence of oxygen from the air.

Section 18.3

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Making Hydrogen Gas from an Acid and a Base,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 33-34. Hydrogen gas is produced in flasks and collected in balloons.

Section 18.7

Bassam Z. Shakhashiri, “Preparation and Properties of Oxygen,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp.137-141. Oxygen is produced by the reaction of hydrogen peroxide potassium permanganate and collected by displacing water from an inverted Erlenmeyer flask. The oxygen is shown to support vigorous combustion when a glowing wooden ember and hot steel wool burst into flames when placed in an atmosphere of oxygen gas.

Section 18.8

Bassam Z. Shakhashiri, “Combining Volume of Oxygen with Sulfur,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp.190-192. SO2, not SO3, produced in the combustion reaction of sulfur and oxygen.

Literature References Section 18.3

Alton Banks, “What’s the Use? Hydrogen,” J. Chem. Educ., Vol 66, 1989, 801.

Section 18.6

Foil A. Miller, “Joseph Priestley, Preeminent Amateur Chemist,” J. Chem. Educ., Vol 66, 1989, 801.

Section 18.10

Michael Laing, “The Three Forms of Molecular Oxygen,” J. Chem. Educ., Vol 66, 1989, 453-455.

Media References Section 18.2

Isotopes of Hydrogen activity from the Instructor Resource DVD

Section 18.3

Electrolysis of Water movie from the Instructor Resource DVD

Section 18.3

Oxidation–Reduction Reactions – Part II movie from the Instructor Resource DVD

Section 18.3

Oxidation Numbers activity from the Instructor Resource DVD

Section 18.4

Formation of Water movie from the Instructor Resource DVD 181

Chapter 18—Hydrogen, Oxygen, and Water

Section 18.8

Reactions with Oxygen movie from the Instructor Resource DVD

Section 18.8

Reactivity of Oxygen activity from the Instructor Resource DVD

Section 18.7

Sodium and Potassium in Water movie from the Instructor Resource DVD

Section 18.9

Periodic Trends: Acid–Base Behavior of Oxides movie from the Instructor Resource DVD

Section 18.9

Carbon Dioxide Acts as an Acid in Water movie from the Instructor Resource DVD

Section 18.9

Melting Point of Oxides activity from the Instructor Resource DVD

Section 18.9

Oxidation State activity from the Instructor Resource DVD

182

CHAPTER 19 THE MAIN-GROUP ELEMENTS Chapter Learning Goals for Students Section 19.1

Determine which of two main-group elements has: a) the more metallic character, b) the higher ionization energy, c) the larger atomic radius, d) the higher electronegativity e) the more acidic oxide, f) the more ionic hydride, and g) the more ionic oxide.

Section 19.2

Contrast the chemical and physical properties of the second-row main-group elements with the properties of the heavier members in the same groups.

Section 19.3

Compare the properties of the group 3A elements. Include valence electron configurations, common oxidation states, and trends in atomic radii, ionic radii, first ionization energies, and electronegativities.

Section 19.4

Draw the structure of diborane and describe its bonding.

Section 19.5

Give the formulas of products formed when aluminum reacts with halogens, nitrogen, oxygen, acids, or bases. Balance the equations.

Section 19.6

Compare the properties of the group 4A elements. Include valence electron configurations, common oxidation states, and trends in atomic radii, first ionization energies, and electronegativities. Describe how each is found in nature, and give one method of preparation and one commercial use.

Demonstrate how germanium, tin, and lead illustrate the gradual transition from semimetallic to metallic character.

Section 19.7

Describe the carbon allotropes.

Briefly describe the chemistry of carbon oxides, carbonates, cyanides, and carbides, including commercial uses of these compounds.

Describe the conditions under which carbon dioxide becomes a supercritical fluid, and give some applications of supercritical carbon dioxide.

Section 19.8

Given the formula of a silicate anion, determine the charge, the number of shared oxygens, and the structure of the silicate anion.

Section 19.9

Compare the properties of the group 5A elements. Include valence electron configurations, common oxidation states, and trends in atomic radii, first ionization energies, and electronegativities. Describe how each is found in nature, and give one commercial use.

Section 19.10

Give an example of a nitrogen-containing compound for each common oxidation state exhibited by nitrogen. For each compound, sketch its electron-dot structure, assign formal charges, indicate which resonance structure(s) is (are) paramagnetic, and describe its geometry.

Briefly describe the chemistry of ammonia, hydrazine, and the nitrogen oxides.

183

Chapter 19—The Main-Group Elements

Section 19.11

Give an example of a phosphorus-containing compound for each common oxidation state exhibited by phosphorus. For each compound, sketch its electron-dot structure and describe its geometry.

Show how phosphoric acids can be interconverted by removing or adding water molecules.

Section 19.12

Compare the properties of the group 6A elements. Include valence electron configurations, common oxidation states, and trends in atomic radii, ionic radii, first ionization energies, electron affinities, electronegativities, and redox potentials for X + 2 H+ + 2 e– → H2X.

Section 19.13

Give an example of a sulfur-containing compound for each common oxidation state exhibited by sulfur. For each compound, write a balanced chemical equation for its preparation, sketch its Lewis electron-dot structure, and describe its geometry.

Write formulas for oxoanions, determine which of two sulfur-containing oxoanions is the strongest acid and which is the strongest base.

Section 19.14

Give an example of a halogen oxoacid, HXOn, for n = 1, 2, 3, and 4. Name each acid, sketch its Lewis electron-dot structure, and describe its geometry.

184

Chapter 19—The Main-Group Elements

Lecture Outline 19.1. A Review of General Properties and Periodic Trends1 A. Zeff increases across the periodic table.

1. Each additional valence electron does not completely shield the additional nuclear charge.

2. Atom’s electrons are more strongly attracted to the nucleus.

3. Ionization energy increases, atomic radius decreases.

4. Electronegativity increases.

5. Metallic character decreases, and nonmetallic character increases across the table.

B. Atomic radius increases from the top to the bottom of a group in the periodic table.

1. Additional shells of electrons are occupied.

2. Valence electrons are farther from the nucleus. a. Ionization energy and electronegativity generally decrease. b. Metallic character increases. c. Nonmetallic character decreases.

C. Metals form ionic compounds with nonmetals.

1. Metallic hydrides 2. Ionic solids with high melting points D. Nonmetals tend to form covalent molecular compounds with each other.

1. Nonmetallic hydrides 2. Covalent molecules – either gases or volatile liquids at room temperature 19.2. Distinctive Properties of the Second-Row Elements2 A. Properties of second-row elements differ markedly from those of heavier elements in the same periodic group.

B. Second-row atoms have especially small sizes and especially high electronegativities.

1. Accentuate nonmetallic behavior 2. BeO – amphoteric (other group 2A element oxides are basic) 3. Boron – forms mainly covalent molecular compounds 4. Hydrogen bonding interactions – restricted to compounds of highly electronegative second-row elements N, O, and F C. Second-row elements have only four valence orbitals – form a maximum of four covalent bonds D. Small size of second-row atoms – allows formation of multiple bonds involving π overlap of the 2p orbitals 1. 3p orbitals are more diffuse. a. Longer bond distances b. Poor π overlap 2. π bonds involving p orbitals – rare for elements of the third and higher rows 19.3. The Group 3A Elements3 A. Gallium 1. Largest liquid range of any metal 2. Used in making GaAs, a semiconductor B. Indium – also used in making semiconductor devices C. Thallium 1. Extremely toxic 2. No commercial uses D. Valence electron configuration: ns2np1 1. Most stable oxidation state for Ga and In: +3 1 Test Item File Questions: Multiple Choice 1-15, 128-138, 146-148; Algorithmic 1-5; Short Answer

1-2 2 Test Item File Questions: Multiple Choice 16-23, 139-142; Algorithmic 6-8; Short Answer 3 3 Test Item File Questions: Multiple Choice 24-34, 149; Short Answer 4

185

Chapter 19—The Main-Group Elements 2. Most stable oxidation state for Tl: +1 E. Properties – consistent with increasing metallic character down the group 1. Metal (except B) 2. Boron – much higher electronegativity and much smaller atomic radius a. Shares valence electrons b. Has nonmetallic character 4 19.4. Boron A. Occurs in concentrated deposits of borate minerals B. Obtained from BBr3 2 BBr3(g) + 3 H2(g) → 2 B(s) + 6 HBr(g) C. Crystalline boron – strong, hard, high-melting substance 1. Chemically inert at room temperature 2. Desirable component in high-strength composite materials D. Boron halides – highly reactive, volatile covalent compounds 1. BX3 molecules 2. Behave as Lewis acids – uses vacant 2p orbitals in accepting a share in a pair of electrons from a Lewis base E. Boron hydrides (boranes) – volatile, molecular compounds with formulas BnHm 1. Simplest – diborane (B2H6) 2. Two BH2 groups connected by two bridging H atoms a. Geometry around B atoms – tetrahedral b. Bridging B–H bonds significantly longer than terminal B–H bonds 3. Diborane – 12 valence electrons; electron deficient a. B atoms use sp3 hybrid orbitals to bond to four neighboring H atoms. b. Bond formed from overlap of a boron sp3 hybrid orbital and hydrogen 1s orbital c. Three–center, two–electron bond — joins each bridging H atom to both B atoms d. Electron density between adjacent atoms less than in an ordinary 2c-2e bond e. Two electrons in B–H–B bridge spread out over three atoms 19.5. Aluminum5 A. Most abundant metal in earth’s crust B. Name comes from alum, KAl(SO4)2 ⋅ 12 H2O, a medicinal salt C. Occurs in many common minerals and gemstones 1. Sapphire and ruby – impure forms of Al2O3 2. Impurity creates color a. Cr – impurity in ruby b. Fe and Ti – impurity in sapphire D. Commercially obtained from bauxite, Al2O3. xH2O E. Reducing agent 1. Loses all three valence electrons 2. Common ion: Al3+ F. Reactions of aluminum 1. Halogens → AlX3 a. Vigorous at room temperature b. Releases large amount of heat 2. Oxygen → Al2O3 – vigorous at room temperature only on surface 3. Acids and bases → Al3+ + H2 19.6. The Group 4A Elements6 A. Especially important, both in industry and in living organisms B. Increase in metallic character down a group in the periodic table 4 Test Item File Questions: Multiple Choice 35-40, 153; Short Answer 5-6 5 Test Item File Questions: Multiple Choice 41-42, 150-151 6 Test Item File Questions: Multiple Choice 43-55; Short Answer 7

186

Chapter 19—The Main-Group Elements C. Valence electron configuration: ns2np2 D. Most common oxidation state: + 4; covalent compounds E. +2 oxidation state 1. Occurs for Sn and Pb 2. Most stable for Pb 3. Ionic compounds F. No simple M4+(aq) ions for any of the group 4A elements 19.7. Carbon7 A In uncombined form, carbon found as diamond and graphite B. Diamond 1. Has a covalent network structure in which each C atom uses sp3 hybrid orbitals to form a tetrahedral array of σ bonds a. Interlocking three-dimensional network of strong bonds b. Hardest known substance c. Highest known melting point 2. Electrical insulator – valence electrons localized in σ bonds C. Graphite 1. Two-dimensional sheetlike structure 2. Uses sp2 hybrid orbitals a. Forms trigonal planar σ bonds with three neighboring C atoms b. Remaining p orbital (⊥ plane of sheet) used to form a π bond c. π bonds delocalized and free to move in plane of sheet d. Electrical conductivity parallel to the sheets 1020 greater than diamond 3. Useful as electrode material 4. Carbon sheets held together by London dispersion forces a. Can easily slide over each other b. Slippery feel c. Used as a lubricant d. Difficult for electrons to hop from one sheet to another D. Fullerene – third crystalline allotrope of carbon; found in soot formed by vaporizing graphite 1. Spherical C60 molecules with shape of a soccer ball 2. Prepared by electrically heating a graphite rod in a helium atmosphere 3. Molecular substance – soluble in nonpolar, organic solvents 4. Derivatives of fullerene a. Compounds in which other atoms are attached to C60 cage; example – C60F36 b. Compounds in which metal atoms are trapped in the C60 cage; example – La@C60 5. Other carbon clusters a. Egg shaped – C70 b. Nanotubes – tube-shaped molecules E. Forty amorphous forms that resemble graphite 1. Coke a. Heat coal in absence of air b. Used as reducing agent in the manufacture of steel 2. Charcoal a. Wood heated in absence of air b. Porous spongelike structure c. Strong adsorbent properties d. Used in filters for removing foul-smelling molecules from water 3. Carbon black a. Heat hydrocarbons in a limited supply of oxygen b. Used in manufacturing of tires and printing ink 7 Test Item File Questions: Multiple Choice 56-70, 155, 157, 159; Short Answer 8-11

187

Chapter 19—The Main-Group Elements F. Oxides of carbon 1. CO a. Colorless, odorless, toxic gas b. Forms when C or hydrocarbon fuels burned in a limited supply of oxygen c. Used for industrial synthesis of methanol d. Toxicity i. Due to its ability to bond strongly to the Fe2+ atom of hemoglobin ii. Impairs the ability of hemoglobin to carry O2 to the tissues 2. CO2 a. Colorless, odorless, nonpoisonous gas b. Formed when fuels burn in an excess of O2 c. By-product of yeast-catalyzed fermentation of sugar in the manufacture of alcoholic beverages d. Produced when metal carbonates react with acids e. Used in beverages i. CO2 solutions – mildly acidic ii. CO2(g) + H2O(l) → H+(aq) + HCO3–(aq) iii. Gives bite to carbonated beverages f. Used in fire extinguishers i. Nonflammable and 1.5 times more dense than air ii. Settles over fire and cuts off source of O2 G. Carbonates 1. Carbonic acid – forms two series of salts: carbonates and hydrogen carbonates 2. Na2CO3 – soda ash a. Used in making glass b. Na2CO3⋅10 H2O – washing soda i. Used in laundering textiles ii. CO32– removes cations from hard water iii. Produces OH–, which removes grease from fabrics 3. NaHCO3 – baking soda a. Reacts with acidic substances in food b. Produces CO2 c. Causes dough to rise H. Hydrogen cyanide and cyanides 1. HCN a. Highly toxic, volatile substance b. CN–(aq) + H+(aq) → HCN(aq) 2. CN– a. Pseudohalide ion i. Behaves like halides ii. Forms insoluble AgCN b. Acts as a Lewis base – bonds through the lone pair of electrons on carbon 3. Toxicity of HCN and cyanides a. Due to strong bonding of CN– to Fe3+ in cytochrome oxidase (important enzyme in food metabolism) b. Enzyme is unable to function c. Halts cellular energy production d. Rapid death 4. CN– used to extract Au and Ag from their ores a. Produces M(CN)2– ion b. M(CN)2– reduced by Zn to produce M I. Carbides 1. Binary compounds of carbon 2. Carbon atom has a negative oxidation state 188

Chapter 19—The Main-Group Elements a. Carbides of active metals b. Interstitial carbides of transition metals c. Covalent network carbides 19.8. Silicon8 A. Hard, gray, semiconducting solid that melts at 1410°C 1. Diamondlike structure 2. Relatively poor overlap of π orbitals – no graphitelike allotrope 3. Naturally found combined with oxygen (SiO2 and various silicate minerals) B. Ultrapure silicon needed for making solid-state semiconductor devices. Purification methods: 1. Conversion to SiCl4 a. Si(s) + 2 Cl2(g) → SiCl4(l) b. Separate by fractional distillation 2. Conversion of SiCl4 back to elemental silicon by reduction with hydrogen SiCl4(g) + 2 H2(g) → Si(s) + 4 HCl(g) 3. Further purification – zone refining (textbook Figure 19.4) C. Silicates – 90% of earth’s crust 1. Silicon oxoanions 2. Basic structural building block a. SiO4 tetrahedron b. Occurs as simple orthosilicate ion, SiO44– 3. Common minerals – anions in which two or more O atoms bridge between Si atoms to give rings, chains, layers, and extended three-dimensional structures.

D. Sharing of two oxygen atoms per SiO4 tetrahedron → cyclic anions or infinitely extended chain anions 1. Repeating unit Si2O64– (Textbok Figure 19) 2. Si6O1812– – cyclic anion a. Present in beryl, Be3Al2Si6O18 b. Emerald gemstone – 2% of Al3+ in beryl is replaced with Cr E. Additional sharing of O atoms gives 1. (Si4O116–)n – double-stranded chain anions a. Found in asbestos minerals b. Tremolite – Ca2Mg5(Si4O11)2(OH)2 c. Fibrous material d. Bonds between chains relatively weak and easily broken 2. (Si4O104–) n – infinitely extended two-dimensional layer anions a. Found in clay minerals, micas, and talc b. Talc – Mg3(OH) 2(Si4O10) c. Mica i. Sheetlike material ii. Bonds between two-dimensional layers relatively weak and easily broken F. Silica – an infinitely extended three-dimensional structure in which the layer anions (Si4O104–)n are stacked on top of one another, and SiO4 and AlO4 tetrahedra share all four of their corners with neighboring tetrahedra 1. Quartz – crystalline form of SiO2 2. Aluminosilicates (feldspars) – most abundant of all minerals a. Partially substitute the Si4+ in SiO2 with Al4+ b. Orthoclase – KAlSi3O8 c. Zeolites – SiO4 and AlO4 tetrahedra joined together in an open structure that has a three-dimensional network of cavities linked by channels i. Act as molecular sieves for separating small molecules from larger ones ii. Used as catalysts in the manufacture of gasoline 8 Test Item File Questions: Multiple Choice 71-77, 163-169; Short Answer 12-15

189

Chapter 19—The Main-Group Elements 19.9. The Group 5A Elements9 A. Down the periodic group – increasing atomic size, decreasing ionization energy, and decreasing electronegativity B. Increasing metallic character of heavier elements – evident in the acid–base properties of their oxides 1. N and P oxides – acidic 2. As and Sb oxides – amphoteric 3. Bi2O3 oxides – basic C. Electron configuration: ns2np3 1. Maximum oxidation state: + 5 2. Minimum oxidation state: – 3 3. N and P – exhibit all oxidation states between – 3 and + 5 4. As and Sb – most important oxidation states: + 3 and + 5 D. Metallic character – increases down the group 1. Sb3+ and Bi3+ – found in salts 2. No simple cations in compounds of N or P E. As, Sb, and Bi found in sulfide ores and used in making various metal alloys 19.10. Nitrogen10 A. Colorless, odorless, tasteless gas B. Makes up 78% of earth’s atmosphere C. Separated out of liquid air by fractional distillation D. Uses of N2 1. Gas – protective inert atmosphere in manufacturing processes 2. Liquid – refrigerant 3. Most important use – Haber process (manufacture of NH3) E. N2 unreactive due to N–N triple bond 1. Reactions involving N2 have a high activation energy and/or an unfavorable equilibrium constant.

2. Haber process requires high temperatures, high pressures, and a catalyst.

N2(g) + 3 H2(g) → 2 NH3(g) F. Classify nitrogen compounds by oxidation state (textbook Table 19.6) G. Ammonia – gateway to nitrogen chemistry 1. Starting material for industrial synthesis of other important nitrogen compounds 2. Colorless, pungent-smelling gas 3. Polar, trigonal pyramidal NH3 molecules 4. Very soluble in water (due to hydrogen bonding) 5. Easily condensed 6. Excellent solvent for ionic compounds 7. Reacts with acids to yield ammonium salts – resemble alkali metal salts in their solubility 8. Brønsted-Lowry base – produces weakly alkaline aqueous solutions H. Hydrazine (H2NNH2) – derivative of NH3 1. Replace one H atom with an NH2 group 2. Prepared by reaction of ammonia with OCl– 3. Poisonous, colorless liquid 4. Explosive in presence of air or other oxidizing agents 5. Used as a rocket fuel 6. Weak base and versatile reducing agent in aqueous solutions I. Oxides of nitrogen 1. N2O – colorless, sweet-smelling gas a. Laughing gas – small doses are mildly intoxicating 9 Test Item File Questions: Multiple Choice 78-82; Short Answer 16 10 Test Item File Questions: Multiple Choice 83-96, 143, 152; Short Answer 17-18

190

Chapter 19—The Main-Group Elements b. Used as a dental anesthetic c. Propellant for dispensing whipped cream 2. NO – colorless gas a. Produced in laboratory by reacting Cu with dilute HNO3 b. Large quantities prepared by catalytic oxidation of NH3 c. Important in biological processes 3. NO2 – highly toxic, reddish brown gas that forms rapidly when NO is exposed to air a. Paramagnetic b. Dimerizes to form N2O4 c. N2O4 predominates at lower temperatures and NO2 predominates at higher temperatures.

4. HNO2 – produced when NO2 reacts with water a. Disproportionation reaction – NO2 undergoes both oxidation and reduction + – b. 2 NO2(g) + H2O(l) → HNO2(aq) + H (aq) + NO3 (aq) J. Nitric acid – one of the most important inorganic acids 1. Used in making ammonium nitrate for fertilizers 2. Used in manufacturing explosives, plastics, and dyes 3. Produced by the Ostwald process a. Air oxidation of NH3 to nitric oxide b. Oxidation of nitric oxide to nitrogen dioxide c. Disproportionation of NO2 in water 4. Yellow color of concentrated HNO2 – due to presence of NO2 produced by a slight amount of decomposition 5. Strong acid – 100% dissociated in water 6. Strong oxidizing agent a. Stronger oxidizing agent than H+(aq) b. Can oxidize inactive metals c. Product depends upon the nature of reducing agent and reaction conditions 7. Aqua regia – mixture of concentrated HCl and concentrated HNO3 (3:1) a. More potent oxidizing agent than HNO3 b. NO3– – serves as the oxidizing agent 3+ c. Cl– – converts Au to AuCl4– 19.11. Phosphorus11 A. Found in phosphate rock, Ca3(PO4)2, and in fluorapatite, Ca5(PO4)3F – 1. Apatites – phosphate minerals with the formula 3 Ca3(PO4)2 ⋅ CaX2 (X– = F or OH–) 2. Sixth most abundant element in human body a. Bones – Ca3(PO4)2 b. Tooth enamel – Ca5(PO4)3OH c. Found in DNA and RNA B. Exists in two common allotropic forms 1. White phosphorus – toxic, waxy, white solid a. Discrete tetrahedral P4 molecules b. Low melting point c. Soluble in nonpolar solvents d. Highly reactive – due to unusual bonding e. P–O bonds are “bent,” relatively weak, and highly reactive 2. Red phosphorus a. Nontoxic b. Polymeric structure C. Phosphorus compounds 1. Most important oxidation states: + 3 and + 5 11 Test Item File Questions: Multiple Choice 97-104, 144; Short Answer 19-22

191

Chapter 19—The Main-Group Elements 2. More likely to be found in a positive oxidation state because of its electronegativity D. Phosphine (PH3) – colorless, extremely poisonous gas 1. Most important hydride of phosphorus 2. Neutral aqueous solutions 3. Poor proton acceptor 4. Easily oxidized: burns in air to form H3PO4 E. Phosphorous halides (PX3 or PX5) 1. Product depends on the relative amounts of reactants 2. Gases, volatile liquids, or low-melting solids F. Oxides and oxoacids of phosphorus 1. Burn P in presence of oxygen → P4O6 or P4O10 a. Molecular compounds with a tetrahedral array of P atoms b. Acidic oxides react with water → H3PO3 or H3PO4 c. P4O10 – used as a drying agent 2. H3PO3 a. Weak diprotic acid b. Only two of the three H atoms are bonded to oxygen 3. H3PO4 – low-melting, colorless, crystalline solid a. Phosphoric acid used in the laboratory – syrupy aqueous solution (82% H3PO4 by mass) b. Manufacturing method – depends on use i. Food additive – begin with molten phosphorus ii. Fertilizers begin with Ca3(PO4)2 c. Sometimes called orthophosphoric acid 4. Other phosphoric acids a. Diphosphoric acid (pyrophosphoric acid) H4P2O7 – combine 2 molecules of H3PO4 and eliminate H2O b. Triphosphoric acid, H5P3O10 c. Polymetaphosphoric acid (HPO3)n – infinitely long chain of phosphate groups 5. Na5P3O10 a. Component of some synthetic detergents b. P3O105– acts as a water softener 19.12. The Group 6A Elements12 A. Periodic trends 1. Oxygen and sulfur – typical nonmetals 2. Gray selenium a. Most stable allotrope of selenium b. Lustrous semiconducting solid 3. Tellurium – semiconductor; classified as a semimetal 4. Polonium a. Radioactive element that occurs in trace amounts in uranium ores b. Silvery white metal B. Electron configuration: ns2np4 1. Common oxidation state: – 2 2. Stability of state decreases with increasing metallic character a. Oxygen – powerful oxidizing agent b. H2Se and H2Te – reducing agents 3. S, Se and Te – less electronegative than oxygen a. Found in positive oxidation states b. Common oxidation state: + 4 and + 6 C. Commercial uses of Se, Te, and Po – limited 1. Se – making red-colored glass and in photocopiers 12 Test Item File Questions: Multiple Choice 105-110, 154; Short Answer 23

192

Chapter 19—The Main-Group Elements 2. Te – used in alloys 3. Po – heat source in space equipment and a source of alpha particles in research 19.13. Sulfur13 A. Many allotropic forms 1. Rhombic sulfur – yellow crystalline solid a. Most stable form of sulfur b. Contains crown-shaped S8 rings c. Above 95°C converts to monoclinic sulfur (cyclic S8 molecules packed differently in the crystal) o d. Melts at 113 C when heated at an ordinary rate 2. Above melting point, sulfur is a fluid, straw-colored liquid.

3. Between 160°C and 195°C a. Dark, reddish brown, viscous liquid b. S8 ring opens and forms S8 chains that form long polymers with more than 200,000 S atoms in the chain.

4. Above 195°C a. More fluid liquid b. Chains fragment 5. Boils at 445°C a. Pour boiling liquid into water – forms plastic sulfur b.Chains freeze in disordered array B. Hydrogen sulfide – colorless gas with strong, foul odor of rotten eggs 1. Extremely toxic 2. Prepared by treating FeS with dilute H2SO4 3. Generated in solution by hydrolysis of thioacetamide for use in qualitative analysis 4. Weak diprotic acid and mild reducing agent C. Oxides and oxoacids of sulfur 1. SO2 – colorless, toxic gas formed when sulfur burns in air a. Slowly oxidized in the atmosphere to SO3 b. Used for sterilizing wine and dried fruit c. SO3 dissolves in rainwater to give sulfuric acid d. Aqueous solutions contain dissolved SO2 and little H2SO3 D. Sulfuric acid – world’s most important industrial chemical 1. Manufactured by contact process a. S burns in air to give SO2. b. SO2 is oxidized to SO3. c. SO3 reacts with water to give H2SO4.

2. Strong acid in dissociation of first proton 3. Forms two series of salts: sulfates and hydrogen sulfates 4. Oxidizing properties depend on concentration and temperature a. Dilute solutions at room temperatures – behave like HCl b. Hot, concentrated H2SO4 can oxidize metals not oxidized by H+.

5. Used in manufacturing soluble phosphate and ammonium sulfate fertilizers 19.14. The Halogens: Oxoacids and Oxoacid Salts14 A. Electron configuration: ns2np5 1. Most electronegative group of elements in the periodic table 2. Gain an electron when forming ionic compounds 3. Share an electron with nonmetals to form molecular compounds B. Oxoacids and oxoacid salts 1. Most important compounds of halogens in positive oxidation states 13 Test Item File Questions: Multiple Choice 111-120, 145, 156, 158, 160-162; Short Answer 24 14 Test Item File Questions: Multiple Choice 121-127; Short Answer 25

193

Chapter 19—The Main-Group Elements 2. General formula – HXOn 3. Oxidation state = + 1, + 3, + 5, or + 7 (depends on value of n) C. Acid strength increases with increasing oxidation state of the halogen 1. Acidic hydrogen bonded to oxygen 2. Strong oxidizing agents D. Hypohalous acids formed from halogen disproportionation reactions 1. X2(g, l, or s) + H2O(l) → HOX(aq) + H+(aq) + X–(aq) a. Equilibrium lies to the left b. Shifts right in basic solution 2. NaOCl a. Strong oxidizing agent b. 5% solution – chlorine bleach E. React Cl2 with hot NaOH → NaClO3 1. Chlorate salts 2. Used as weed-killers and strong oxidizing agents F. Anhydrous perchloric acid – colorless, shock-sensitive liquid that decomposes explosively on heating 1. Powerful and dangerous oxidizing agent 2. Perchlorate salts – strong oxidants G. Iodine forms more than one perhalic acid 1. Paraperiodic acid, H5IO6 a. White crystals obtained from evaporating HIO4 solutions b. Weak, polyprotic acid 2. Metaperiodic acid, HIO4(s) a. Produced when H5IO6 loses water b. Strong monoprotic acid

194

Chapter 19—The Main-Group Elements

CHAPTER 19 THE MAIN-GROUP ELEMENTS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 19.2

Due to small size, high electronegativity, and the ability to form π bonds, the first member of each s- and p-block group differs substantially in its chemistry from the other members of the group.

Section 19.2

Compounds of period 2 elements have a maximum of four covalent bonds and an octet of electrons; elements beyond period 2 may have more than four bonds and more than eight electrons.

Section 19.2

Molecules of period 2 p-block elements contain single, double, or triple bonds; molecules of higher period elements generally contain only single bonds.

Section 19.3

Heavier p-block elements tend to exhibit a stable oxidation state equal to the group number minus 2.

Section 19.4

2c–2e terminal B–H bonds are stronger and shorter than bridging 3c–2e bonds in B2H6.

Section 19.6

Students tend to equate ionic character with high charge. On the contrary, the higher the oxidation state, the more covalent the bonds formed by an element. Bonds in SnCl4 are more covalent than those in SnCl2.

Section 19.7

Diamond is an electrical insulator, but graphite is an electrical conductor.

Section 19.7

CN– and CO have the same number of electrons and the same Lewis electron-dot structure (:C≡N:– and :C≡O:).

Section 19.8

The number of terminal (unshared) oxygen atoms equals the charge on the silicate.

Section 19.10

Due to the high strength of the nitrogen–nitrogen bond, N2, the most common molecule in the earth’s atmosphere is practically unreactive under ordinary conditions. An exception is its use by plants in nitrogen fixation, a process yet to be duplicated in the laboratory.

Section 19.10

Any highly exothermic reaction that generates gaseous products has the potential to occur with explosive force.

Section 19.10

When an element occurs in two extremely different oxidation states in a compound, as does nitrogen in NH4NO3, the compound may decompose violently.

Section 19.10

The equilibrium for an exothermic reaction shifts to the left as the temperature is raised.

Section 19.10

A change in reaction conditions may result in a change in the reaction products.

Section 19.11

Note that both allotropes of period 3 phosphorus have only single bonds, whereas period 2 nitrogen has a triple bond in N2.

Section 19.11

In oxoacids, only hydrogen atoms attached to oxygen are acidic.

195

Chapter 19—The Main-Group Elements

Section 19.13

Note that all allotropes of period 3 sulfur have only single bonds, whereas period 2 oxygen has a double bond in O2.

Section 19.13

Expecting to see a yellow solid, students frequently do not recognize finely divided sulfur, which appears white.

Section 19.13

A small segment of the population experiences a severe allergic reaction to foods treated with sulfites, the presence of which is noted on wine-bottle labels but not always on labels of dried fruits and nuts.

Section 19.14

The greater the number of oxygen atoms, the higher the oxidation state of the halogen and the stronger the oxoacid.

Lecture/Laboratory Demonstration References Section 19.1

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Halogens Compete for Electrons,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washingon DC, 1988), pp. 60-61

Section 19.1

Bassam Z. Shakhashiri, “Acidic and Basic Properties of Oxides,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp.109-113.

Section 19.10

Bassam Z. Shakhashiri, “Preparation and Properties of Ammonia,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp. 202-204. Ammonia gas is produced from the reaction of ammonium chloride with calcium hydroxide and is collected in an Erlenmeyer flask by the displacement of air. The ammonia gas turns moistened litmus paper blue, demonstrating the basic properties of ammonia dissolved in water.

When the mouth of an ammonia-filled Erlenmeyer flask is placed below the surface of a pool of water, water rapidly enters the flask to replace ammonia that dissolves in the water. White smoke forms when ammonia vapor is passed over the top of a beaker of concentrated hydrochloric acid forming solid ammonium chloride.

Section 19.13

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Plastic Sulfur,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 53.

Section 19.14

C. E. Ophardt, “Redox Demonstrations and Descriptive Chemistry: Part 2.

Halogens,” J. Chem. Educ., Vol. 64, 1987, 807. Using an abbreviated table of reduction potentials as a predictive tool, reactions of bromine and iodine in various oxidation states are demonstrated. When excess I– is reacted with BrO3– in the presence of starch, a blue color forms immediately, the final products being I2 and Br–. When I– is reacted with excess BrO3– in the presence of starch, within minutes a blue color turns purple then yellow with the final products being Br2 and IO3–. A second series of reactions demonstrates the instability of certain oxidation states of iodine due to reaction with oxygen and disproportionation under acidic and basic conditions.

Section 19.14

Robert C. Hansen, “An Overhead Demonstration of Some Descriptive Chemistry of the Halogens and Le Châtelier’s Principle,” J. Chem. Educ., Vol. 65, 1988, 264-265.

196

Chapter 19—The Main-Group Elements

Literature References Section 19.1

George Gorin, “Mendeleev and Moseley: The Principal Discoverers of the Periodic Law,” J. Chem. Educ., Vol. 73, 1996, 490-493.

Section 19.1

Eric R. Scerri, “The Evolution of the Periodic System,” Scientific American., September 1998, 8-83.

Section 19.1

N. K. Goh and L. S. Chia, “Using the Learning Cycle to Introduce Periodicity,” J. Chem. Educ., Vol. 66, 1989, 747.

Section 19.1

Robert H. Goldsmith, “Metalloids,” J. Chem. Educ., Vol. 59, 1982, 526-527.

Section 19.1

Kimberley A. Waldron, Eric M. Fehringer, Amy E. Streeb, Jennifer E. Trotsky, and Joshua J. Pearson, “Screen Percentages Based on Slater Effective Nuclear Charge as a Versatile Tool for Teaching Periodic Trends,” J. Chem. Educ., Vol. 78, 2001, 635639.

Section 19.1

Joan Mason, “Periodic Contractions Among the Elements; Or, on Being the Right Size,” J. Chem. Educ., Vol. 65, 1988, 17-20.

Section 19.1

Gabriel Pinto, “Using Balls from Different Sports to Model the Variation of Atomic Sizes,” J. Chem. Educ., Vol. 75, 1998, 725-726.

Section 19.1

Ronald L. Rich, “Periodicity in the Acid-Base Behavior of Oxides and Hydroxides,” J. Chem. Educ., Vol. 62, 1985, 44.

Section 19.13

Miriam C. Nagel, “Herman Frasch, Sulfur King,” J. Chem. Educ., Vol. 58, 1981, 6061.

Media References Section 19.1

Periodic Trends: Acid–Base Behavior of Oxides movie from the Instructor Resource DVD

Section 19.2

Carbon and Silicon Pi Overlap activity from the Instructor Resource DVD

Section 19.4

Boron Halides activity from the Instructor Resource DVD

Section 19.7

Carbon Dioxide Behaves as an Acid in Water movie from the Instructor Resource DVD

Section 19.8

Silicon Zone Refining activity from the Instructor Resource DVD

Section 19.8

Silicates activity from the Instructor Resource DVD

Section 19.10

Nitrogen Dioxide and Dinitrogen Tetraoxide movie from the Instructor Resource DVD

Section 19.11

Reactions with Oxygen movie from the Instructor Resource DVD 197

Chapter 19—The Main-Group Elements

Section 19.11

Reactions with Oxygen movie from the Instructor Resource DVD

Section 19.11

Diphosphoric Acid activity from the Instructor Resource DVD

Section 19.14

Physical Properties of the Halogens movie from the Instructor Resource DVD

198

CHAPTER 20 TRANSITION ELEMENTS AND COORDINATION CHEMISTRY Chapter Learning Goals for Students Section 20.1

Write valence electron configurations for transition metal atoms and ions. Given electron configurations for transition metals and their ions, locate the transition metal on the periodic table.

Section 20.2

Compare the properties of the first-transition-series elements. Include appearance, valence electron configurations for atoms and ions, common oxidation states, and trends in melting points, atomic radii, densities, ionization energies, and standard oxidation potentials.

Section 20.3

Assign oxidation states to transition metals.

Based on variations in Zeff, predict which of two transition metal cations or oxoanions is the strongest oxidizing or reducing agent.

Section 20.4

Balance equations for redox reactions of chromium, iron, and copper.

Determine the products of reactions occurring when dilute H2SO4 or aqueous NaOH is added to compounds of chromium, iron, and copper.

Section 20.5

Write formulas of coordination complexes. Identify the ligands and their donor atoms. Determine the coordination number and the oxidation state of the metal and the charge on any complex ion.

Section 20.6

Given the electron dot structure of a molecule or ion, determine whether it can serve as a chelate ligand.

Draw structures for transition metal compounds containing chelate ligands, identifying the Lewis acid, Lewis base, ligand donor atoms, and chelate rings, and determine the coordination number, coordination geometry, and oxidation state of the metal.

Section 20.7

Name coordination compounds from their formulas and vice versa.

Section 20.8

Identify linkage isomers and ionization isomers.

Determine the number and structures of diastereoisomers possible for a given coordination complex.

Section 20.9

Determine which coordination complexes are chiral, and draw the structure of the enantiomers.

Section 20.10

Given the visible absorption spectrum of a transition metal complex, determine the most likely color of the complex, and calculate the energy difference between ground and excited states that corresponds to an absorption peak wavelength.

Section 20.11

Give a valence bond theory description of a coordination complex, and show the number of unpaired electrons and the hybrid orbitals used by the metal ion.

199

Chapter 20—Transition Elements and Coordination Chemistry

Section 20.12

Give a crystal field theory description of octahedral, tetrahedral, and square planar coordination complexes, and predict the number of unpaired electrons.

200

Chapter 20—Transition Elements and Coordination Chemistry

Lecture Outline 20.1. Electron Configurations1 A. 4s subshell filled first; 3d subshell filled according to Hund’s rule 1. Add one electron to each of the five 3d orbitals before adding a second electron.

2. Two exceptions: Cr and Cu. a. Cr: 4s13d5 b. Cu: 4s13d10 B. Electron configurations depend on both orbital energies and electron-electron repulsions.

1. If two valence subshells have similar energies, can’t always predict the configurations.

2. Exceptions from the expected orbital-filling pattern result in either half-filled or completely filled subshells.

3. For Cr and Cu: 3d and 4s have similar energies, and as a consequence, one electron shifts from 4s to 3d. a. Shift decreases the electron-electron repulsions. b. Gives either a filled or half-filled subshell C. Easy to predict electron configurations of transition metal cation 1. All valence electrons occupy d orbitals.

2. Neutral atom loses one or more electrons; Zeff increases. a. Remaining electrons more strongly attracted to the nucleus b. Orbital energies decrease c. 3d orbitals experience a steeper drop in energy with increasing Zeff than does the 4s d. 3d orbitals in cations lower in energy than the 4s orbitals D. Groups 1B – 8B 1. Electron configuration similar to analogous elements in groups 1A – 8A a. Groups 1B, 2B: electron configuration similar to groups 1A, 2A b. Groups 3B – 8B: electron configuration similar to groups 3A – 8A 20.2. Properties of Transition Elements2 A. Properties can be understood in terms of electron configurations.

B. Metallic properties 1. Malleable, ductile, lustrous, and good conductors of heat and electricity 2. Sharing of d, as well as s, electrons gives rise to stronger metallic bonding.

3. Transition metals are harder, have higher melting and boiling points, and are more dense than the group 1A and 2A metals.

4. Melting points increase as the number of unpaired d electrons available for metallic bonding increases and then decrease as the d electrons pair up and become less available for bonding.

C. Atomic radii and densities 1. Decrease in radii with increasing atomic number a. Added d electrons only partially shield the added nuclear charge. b. Increase in radii toward the end of each series c. Due to more effective shielding and increasing electron-electron repulsion as d orbitals become doubly occupied 2. Similar radii – accounts for ability to blend together in forming alloys 3. Second and third series a. Group 4B on have nearly identical radii b. Smaller than expected size of the third-series atoms due to lanthanide contraction 4. Lanthanide contraction – the general decrease in atomic radii of the f-block lanthanide elements between the second and third transition series a. Increase in effective nuclear charge as 4f subshell is filled 1 Test Item File Questions: Multiple Choice 1-16, 97-102; Algorithmic 1-5; Short Answer 1-8 2 Test Item File Questions: Multiple Choice17-25, 105-128; Short Answer 9-10

201

Chapter 20—Transition Elements and Coordination Chemistry b. Size decrease due to a larger Zeff almost exactly compensates for the expected size increase due to an added quantum shell of electrons. c. Atoms of third series have radii very similar to those of the second series.

5. Densities – inversely related to atomic radii a. Second and third series have nearly the same atomic volume. b. Third series has unusually high densities.

D. Ionization energies and standard oxidation potentials.

1. Ionization energies a. Increase left to right across a transition series b. Due to increase in Zeff and a decrease in atomic radius 2. E°s for oxidation potentials of first series transition metals are positive (except Cu). a. Solid metal is oxidized to aqueous cation more readily than H2 gas is oxidized to H+. b. M(s) + 2 H+(aq) → M2+(aq) + H2(g) c. Better reducing agents than H2 d. Can be oxidized by “nonoxidizing” acids (lack an oxidizing anion) e. Cu(s) requires a stronger oxidizing agent.

3. General trend for E° correlates with the general trend in ionization energies.

4. Ease of oxidation of the metal decreases as the ionization energies increase across the first series.

20.3. Oxidation States of Transition Elements3 A. Exhibit a variety of oxidation states – have oxidation states less than their group number B. First series forms + 2 cation (except Sc) – due to loss of the two 4s electrons C. First series – can also lose 3d electrons 1. 3d and 4s energies are similar.

2. Energy required to remove the third electron is provided by the larger ΔG° of hydration of the more highly charged +3 cation.

D. Highest oxidation state for the groups 3B–7B metals is the group number.

1. Corresponds to loss of all the valence s and d electrons.

2. Later transition metals – loss of all valence electrons prohibited by the increasing Zeff value E. Transition metal ions in a high oxidation state – good oxidizing agents F. Early transition metal ions in low oxidation state – good reducing agents G. Divalent ions of the later metals – poor reducing agents because of the larger value of Zeff H. Stability of the higher oxidation states increases down a periodic group.

20.4. Chemistry of Selected Transition Metals4 A. Chromium 1. Obtained from chromite (FeO.Cr2O3) a. Reduce chromite with carbon → ferrochrome (Fe(s) + Cr(s)) b. Used in making stainless steel 2. Pure form obtained from reduction of Cr2O3 with Al 3. Used to electroplate metallic objects with an attractive, protective coating 4. Hard, lustrous, takes a high polish 5. Resistant to corrosion – microscopic film of Cr2O3 protects surface 6. Systemize aqueous chemistry by oxidation states and the species that exist under acidic and basic conditions. (textbook Table 20.3) 7. Cr2+ a. In aqueous solution – Cr(H2O)62+ is a beautiful blue solution. b. Rapidly oxidized by O2 to Cr3+ 8. Cr3+ a. Cr(H2O)63+ – violet

3 Test Item File Questions: Multiple Choice 26-31, 103-104; Algorithmic 6; Short Answer 11-12 4 Test Item File Questions: Multiple Choice 32-46

202

Chapter 20—Transition Elements and Coordination Chemistry b. Cr3+ solutions are green – anions replace some of the bound water molecules to give green complex ions 9. Cr(OH)n a. Acid strength increases with increasing polarity of O–H bond. b. Polarity increases with increasing oxidation state of Cr atom. i. Cr(OH)2 – basic ii. Cr (OH)3 – amphoteric iii. CrO2(OH)2 – chromic acid (H2CrO4) 10. +6 oxidation state a. Acidic solution – Cr2O72– i. Powerful oxidizing agent ii. Used as an oxidant in analytical chemistry b. Basic solution – CrO42–, a weaker oxidizing agent B. Iron – very important in human civilization and in living systems 1. Relatively soft and easily corroded 2. Combines with carbon and other metals to make alloys 3. Most important ores a. Hematite (Fe2O3) and magnetite (Fe3O4) a. Reduce with coke in a blast furnace → Fe(s) 4. Most important oxidation states: +2 and +3 5. Oxidize with an acid that lacks a nonoxidizing anion in absence of air: a. Fe(s) → Fe2+(aq) b. Oxidation of Fe2+ has a negative standard potential. c. In presence of air: Fe2+ oxidized to Fe3+ 6. Oxidize with an acid that has an oxidizing anion: Fe(s) → Fe3+(aq) 7. Fe3+(aq) + base → Fe(OH)3(s) a. Very insoluble b. Forms if pH > 2 c. Reaction accounts for red-brown rust stains in sinks C. Copper – found in the elemental state 1. Most important ores a. Sulfides b. Chalcopyrite, CuFeS2 2. High electrical conductivity and negative oxidation potential – used to make electrical wiring and corrosion-resistant water pipes 3. Less reactive than other first-series transition metals 4. Prolonged exposure to moist air and CO2: a. Cu → Cu2(OH)2CO3 b. Continues to react with acid rain forming Cu2 (OH) 2SO4, the green patina seen on bronze monuments 5. Two common oxidation states: +1 and +2 ο ο less negative than E Cu/Cu 6. E Cu + + / Cu 2 + a. Any oxidizing agent strong enough to oxidize copper to the Cu(I) ion is also able to oxidize the Cu(I) ion to the Cu(II) ion. b. Cu+(aq) can undergo a disproportionation reaction. i. 2 Cu+(aq) → Cu(s) + Cu2+(aq) E° = + 0.37 V ii. Large equilibrium constant, K = 1.8 × 106 c. Cu+ – not an important species in aqueous solution 7. Cu+ exists in solid compounds. a. Disproportionation equilibrium is reversed in presence of Cl–. b. CuCl precipitation shifts the preceding reaction left.

2+ 8. Cu – more common a. Most common compound – blue–colored CuSO4⋅5 H2O 203

Chapter 20—Transition Elements and Coordination Chemistry b. Four H2O’s bound to the Cu2+ ; fifth H2O hydrogen bonded to SO42– c. Loses color on heating – suggests that blue color is due to the bonding of Cu2+ to the water molecules 20.5. Coordination Compounds5 A. Coordination compound – a compound in which a central metal ion is attached to a group of surrounding molecules or ions by coordinate covalent bonds B. Ligands – molecules or ions that surround the central metal ion in a complex C. Donor atoms – atoms that are attached directly to the metal ion D. Complex formation is a Lewis acid–base interaction in which ligands act as Lewis bases and the central metal ion behaves as a Lewis acid.

E. Some coordination compounds are salts which contain a complex cation or anion along with enough ions of opposite charge to give a compound that is electrically neutral overall.

1. Enclose the complex ion in brackets in the formula.

2. Indicates that the complex ion is a discrete structural unit F. Metal complex – refers both to neutral molecules and to complex ions G. Coordination number – the number of ligand donor atoms that surround a central metal ion in a complex 1. Most common coordination numbers: 4 and 6 2. Depends on the metal ion’s size, charge, and electron configuration and the size and shape of the ligands H. Characteristic shape of metal complex is determined by the metal ion’s coordination number.

1. Two-coordinate complexes – linear 2. Four-coordinate complexes – tetrahedral and square planar 3. Six-coordinate complexes – octahedral I. Charge on a metal complex is equal to the charge on the metal ion plus the sum of the charges on the ligands.

20.6. Ligands6 A. All ligands are Lewis bases.

1. Have at least one unshared pair of electrons.

2. Forms a coordinate covalent bond to a metal ion 3. Classified as monodentate or polydentate a. Depends on number of ligand donor atoms that bond to the metal b. Use the electron pair of a single donor atom – monodentate c. Use electron pairs on more than one donor atom – polydentate B. Chelating agents – polydentate ligands 1. Multipoint attachment to a metal ion resembles the grasping of an object by the claws of a crab.

2. Forms a chelate ring 3. Metal chelate – a complex that contains one or more chelate rings 20.7. Naming Coordination Compounds7 A. If the compound is a salt, name the cation first and then the anion.

B. In naming a complex ion or a neutral complex, name the ligands first, in alphabetical order, and then the metal.

1. Anionic ligands end in –o. (textbook Table 20.5) 2. Complex name is one word.

C. If more than one ligand of a particular type – use Greek prefixes to indicate number D. If the name of a ligand itself contains a Greek prefix, put the ligand name in parentheses and use an alternate prefix.

5 Test Item File Questions: Multiple Choice 47-53; Short Answer 13-16 6 Test Item File Questions: Multiple Choice 54-59, 129-133; Algorithmic 7-8; Short Answer 17 7 Test Item File Questions: Multiple Choice 60-67; Short Answer 18

204

Chapter 20—Transition Elements and Coordination Chemistry E. Use a Roman numeral in parentheses immediately following the name of the metal, to indicate the metal’s oxidation state.

F. In naming the metal, use the ending –ate if the metal is in an anionic complex.

20.8. Isomers8 A. Isomers 1. Compounds with the same formula but a different arrangement of constituent atoms 2. Different compounds with different physical and chemical properties B. Constitutional isomers – isomers that have different connections among their constituent atoms 1. Linkage isomers – arise when a ligand can bond to a metal through either of two different donor atoms 2. Ionization isomers – isomers that differ in the anion that is bonded to the metal ion C. Stereoisomers 1. Isomers that have the same connections among atoms but have a different arrangement of the atoms in space 2. Diastereoisomers (geometric isomers) – have different relative orientations of metal-ligand bonds a. cis isomer – identical ligands occupy adjacent corners of the square in a square planar complex b. trans isomer – identical ligands are across from one another in a square planar complex c. Have different properties i. cis–Pt(NH3)2Cl2 – polar molecule; more soluble in water ii. trans–Pt(NH3) 2Cl2 – nonpolar; two Pt–Cl and Pt–NH3 dipoles point in opposite directions and cancel out d. Square planar complexes of the type MA2B2 and MA2BC (M= metal ion; A, B, C = ligands) can exist as cis-trans isomers. e. No cis-trans isomers for four coordinate tetrahedral complexes – all four corners of a tetrahedron adjacent to one another f. Octahedral complexes of the type MA4B2 can also exist as diastereoisomers – two B ligands can be either on adjacent or on opposite corners of the octahedron g. Easy to distinguish – various bonds in cis and trans isomers point in different directions 20.9. Enantiomers and Molecular Handedness9 A. Enantiomers – molecules or ions that are nonidentical mirror images of one another 1. Differ from one another because of their handedness a. Chiral – objects that have a handedness to them b. Achiral – objects that lack handedness B. An object is not chiral if it has a symmetry plane cutting through its middle so that one half of the object is a mirror image of the other half.

C. Certain molecules and ions are chiral.

1. Tris(ethylenediamine)cobalt(III) ion, [Co(en)3]3+ (textbook Figure 20.21) – two nonidentical mirror image forms a. “Right-handed” enantiomer – the three ethylenediamine ligands spiral to the right (clockwise) b. “Left-handed” enantiomer – the ethylenediamine ligands spiral to the left (counterclockwise) 2. [Co(NH3) 6] 3+ – achiral; has several symmetry planes D. Enantiomers have identical properties except for their reactions with other chiral substances and their effect on plane-polarized light.

8 Test Item File Questions: Multiple Choice 68-71, 134-136, 142; Short Answer 19-20 9 Test Item File Questions: Multiple Choice 72-75, 137-141

205

Chapter 20—Transition Elements and Coordination Chemistry 1. Plane-polarized light – light in which the electric vibrations of the light wave are restricted to a single plane 2. Pass ordinary light through a polarizing filter then through a solution of one of the enantiomers a. Plane of polarization rotated either to the right or to the left b. Pass through the solution of the other enantiomer – plane of polarization rotated through an equal angle, but in the opposite direction 3. Optical isomers – another term used for enantiomers because of their effect on planepolarized light 4. Label enantiomers (+) or (–) depending on the direction of rotation of the plane of polarization 5. Racemic mixture a. 50:50 mixture of the (+) and (–) isomers that produces no net optical rotation b. Rotations produced by the individual enantiomers exactly cancel.

20.10. Color of Transition Metal Complexes10 A. Color of transition metal complexes depends on the identity of the metal and the ligands.

B. Metal complex can absorb light by undergoing an electronic transition from its lowest energy state (E1) to a higher energy state (E2).

1. Wavelength of absorbed light depends on the energy separation ΔE = E2 – E1 between the two states.

2. Absorbance – the measure of the amount of light absorbed by a substance a. Absorption spectrum – plot of absorbance versus wavelength b. Color that we see is complementary to the color absorbed.

20.11. Bonding in Complexes: Valence Bond Theory11 A. Valence bond theory 1. Bonding results when a filled ligand orbital containing a pair of electrons overlaps a vacant hybrid orbital on the metal ion.

2. Produces a coordinate covalent bond B. If you know the geometry of a complex, you know which hybrid orbitals the metal ion uses. (textbook Table 20.7) C. Magnetic behavior of transition metal complexes 1. Paramagnetic – substance containing unpaired electrons and attracted by magnetic fields 2. Diamagnetic substance – contains only paired electrons and weakly repelled by magnetic fields 3. Number of unpaired electrons in a transition metal complex can be determined by quantitative measurement of the force exerted on the complex by a magnetic field.

D. High-spin complex – one in which the d electrons are arranged according to Hund’s rule to give the maximum number of unpaired electrons E. Low-spin complex – one in which the d electrons are paired up to give a maximum number of doubly occupied d orbitals and a minimum number of unpaired electrons 20.12. Crystal Field Theory12 A. Crystal field theory – a model that views bonding in complexes as arising from electrostatic interactions and considers the effect of ligand charges on energies of metal ion d orbitals.

1. Accounts for color and magnetic properties of transition metal complexes 2. No covalent bonds, no shared electrons, and no hybrid orbitals 3. Electrostatic interactions within an array of ions 4. Neutral, dipolar ligand – the electrostatic interactions are of the ion–dipole type B. Octahedral complexes 1. Metal positively charged 10 Test Item File Questions: Multiple Choice 76, 143-146; Short Answer 21 11 Test Item File Questions: Multiple Choice 77-84; Algorithmic 9; Short Answer 22 12 Test Item File Questions: Multiple Choice 85-96, 147-150; Algorithmic 10; Short Answer 23-25

206

Chapter 20—Transition Elements and Coordination Chemistry 2. Ligands negatively charged a. Ligands repel each other. b. Minimize repulsion by getting as far apart from one another as possible c. Metal-ligand attractions > ligand-ligand repulsions d. Complex is more stable than the separated ions.

C. Effect of ligand charges on the energies of d orbitals 1. Negatively charged d electrons are repelled by the negatively charged ligands. a. Orbital energies are higher in the complex. b. Not all obitals raised in energy by the same amount 2. d z 2 , d x 2 − y 2 are raised higher in energy than the d xy , d xz , d yz (point between ligands).

3. Crystal field splitting (Δ) – the energy splitting between the two sets of d orbitals (textbook Figure 20.29) D. Crystal field splitting energy, Δ – corresponds to λ’s in the visible region of the spectrum 1. Color of complexes due to electronic transitions between the lower and higher energy sets of d orbitals 2. Size depends on the nature of the ligands a. Spectrochemical series – changes in the crystal field splitting as the ligand varies b. Weak-field ligands i. Produce a relatively small value of Δ ii. High-spin complexes c. Strong-field ligands i. Produce a relatively large value of Δ ii. Low-spin complexes E. Spin-state depends on the relative values of Δ and P, the spin-pairing energy.

1. Spin-pairing energy – energy needed to overcome repulsion that exists when an electron is placed into an orbital that already contains another electron.

2. Δ > P; low-spin arrangement has lower energy.

3. Δ < P; high-spin arrangement has higher energy.

F. Choice of spin-states arises only for d4-d7 complexes.

G. Tetrahedral and square planar complexes – have different energy splitting for the d orbitals (textbook Figure 20.30) H. Energy splitting in tetrahedral complexes is just the opposite of that in octahedral complexes.

1. Δ(tetrahedral complexes) ≈ ½ Δ(octahedral complexes) 2. None of the orbitals points directly at the ligands.

3. Four ligands instead of six 4. Δ < P 5. All complexes are high-spin.

I. Square planar complexes – two trans ligands along the z-axis are missing 1. Large energy gap between the x2-y2 orbital and the four lower-energy orbitals.

2. Most common for metal ions with d8 electronic configurations 3. Favors low-spin complexes in which all four lower-energy orbitals are filled and the higher energy x2-y2 orbital is vacant

207

Chapter 20—Transition Elements and Coordination Chemistry

CHAPTER 20 TRANSITION ELEMENTS AND COORDINATION CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 20.1

Anomalous electron configurations frequently occur if the atom can attain half-filled or completely filled subshells.

Section 20.1

There are no anomalous electron configurations for transition metals in a positive oxidation state. The ns orbital is vacant, and the remaining valence electrons occupy (n –1)d orbitals, where n is the period.

Section 20.2

Within any transition series, the melting points increase for the first three or four elements, then show a general decrease.

Section 20.2

Due to the lanthanide contraction, second- and third-series transition elements have similar atomic radii, so the heavier third-series elements are much denser than the lighter second-series elements.

Section 20.2

Of the first-series metals, only copper cannot be oxidized by hydrochloric acid.

Section 20.3

Transition metals tend to have many more oxidation states than main-group metals.

Section 20.3

All first series-transition metals except scandium form +2 ions.

Section 20.3

Electrons generally become more difficult to remove from left to right across a transition series. The first elements in a transition series tend to occur in relatively high oxidation states, and later elements tend to occur in low oxidation states.

Section 20.4

Simple transition-metal ions and complex ions are soluble in water, but uncharged transition-metal compounds are often insoluble in water.

Section 20.4

The higher the oxidation state of the central atom, the more acidic the hydroxo species.

Section 20.5

Students should review Lewis acid-base theory in preparation for the sections in this chapter on coordination chemistry.

Section 20.5

Ions present in a coordination compound to balance the charge of a complex ion are sometimes called counterions.

Section 20.5

Most transition metal complexes with a coordination number 4 are tetrahedral.

Metal ions with eight d electrons usually form complexes that are square planar.

Section 20.6

Chelate ligands usually form five- or six-membered rings with a metal ion.

Section 20.7

Older literature named ligand water aquo instead of aqua. The ending -o implies an anionic ligand.

Section 20.7

Ammonia as a ligand has the name ammine, with two ms. If one or more of the hydrogen atoms are replaced by groups of atoms bonded through carbon, the resulting ligand is an organic amine, with one m.

208

Chapter 20—Transition Elements and Coordination Chemistry

Section 20.8

Isomers have the same chemical formula but different physical and chemical properties.

Section 20.8

When counting diastereoisomers, students frequently arrive at too many because they draw the same isomer in different orientations.

Section 20.10

The color of a complex corresponds to wavelengths of light that are not absorbed by the complex. The observed color is usually the complement of the color absorbed.

If all wavelengths of light are absorbed, a complex appears black. If no wavelengths of light are absorbed, a complex appears white (colorless).

Section 20.11

The number of hybrid orbitals used by a metal ion in a coordination complex equals the number of atomic orbitals from which the hybrid orbitals were formed and equals the number of ligand donor atoms.

Section 20.11

Although both have a coordination number of four, the metal ions in square planar and tetrahedral complexes require different hybrid orbitals because of the differences in the orientation of the ligands.

Section 20.12

Crystal field theory is an electrostatic model. Valence bond theory is an orbital overlap model.

Section 20.12

The stronger the interaction between a ligand and a metal ion, the greater the crystal field splitting, Δ, and the higher the energy (and the lower the wavelength) of light absorbed by the complex.

Section 20.12

The strong-field ligands CN– and CO are toxic to humans because of their strong interactions with iron in cytochrome oxidase and hemoglobin, respectively.

Section 20.12

If Δ > P, the complex is predicted to be low-spin. If Δ < P, the complex is predicted to be high-spin.

Section 20.12

Almost all tetrahedral complexes are high-spin, and square planar complexes are low-spin. Octahedral complexes can be either low-spin or high-spin.

Lecture/Laboratory Demonstration References Section 20.3

Doris Kolb, “Oxidation States of Manganese,” J. Chem. Educ., Vol. 65, 1988, 1004.

To demonstrate the numerous oxidation states attainable by manganese, several aquated ions, oxoanions, and oxides are formed in beakers positioned on an overhead projector: Mn2+(pale pink), Mn3+(rose-colored), MnO2(brown), MnO43– (blue), MnO42–(green), and MnO4–(purple).

Section 20.4

J. M. Anderson, “Oxidation-reduction in Blood Analysis: Demonstrating the Reaction in a Breathalyzer,” J. Chem. Educ., Vol. 67, 1990, 263. The oxidation of a primary alcohol by yellow dichromate ion is shown to first form an aldehyde, then a carboxylic acid, and green chromium(III) ion. The use of this reaction, principles of spectrometry, and gas laws in a commercial device for measuring blood-alcohol content is discussed.

209

Chapter 20—Transition Elements and Coordination Chemistry

Section 20.4

Bassam Z. Shakhashiri, “Precipitates and Complexes of Copper(II),” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 318-323.

Section 20.4

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Green and Blue Copper Complexes,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 71-72.

Section 20.4

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Copper Sulfate: Blue to White,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 69-70.

Section 20.5

Lee R. Summerlin and James L. Ealy, Jr., “Cobalt Complexes: Changing Coordination Numbers,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 41-42.

Literature References Section 20.2

Peter P. Edwards and M. J. Sienko, “On the Occurrence of Metallic Character in the Periodic Table of the Elements,” J. Chem. Educ., Vol. 60, 1983, 691-696.

Section 20.2

Frank Rioux, “An Ionic Model for Metallic Bonding,” J. Chem. Educ., Vol. 62, 1985, 383-384.

Section 20.5

R. Bruce Martin, “A Stability Ruler for Metal Ion Complexes,” J. Chem. Educ., Vol. 64, 1987, 402.

Section 20.5

D. Venkataraman, Yahua Du, Scott R. Wilson, A. Hirsch, Peng Zhang, and Jeffrey S. Moore, “A Coordination Geometry Table of the d-Block Elements and Their Ions,” J. Chem. Educ., Vol. 74, 1997, 915-918.

Section 20.6

Daniel T. Haworth, “Some Linguistic Detail on Chelation,” J. Chem. Educ., Vol. 75, 1998, 47.

Section 20.6

Mark Conway, Smallwood Holoman, Ladell Jones, Ray Leenhouts, and Gerald Williamson, “Selecting and Using Chelating Agents,” Chem. Eng., Vol. 106(3), 1999, 86-90.

Section 20.6

J. Roger Hart, “EDTA-Type Chelating Agents in Everyday Consumer Products,” J. Chem. Educ., Vol. 61, 1984, 1060-1061.

Section 20.6

N. M. Senozan and R. L. Hunt, “Hemoglobin: Its Occurrence, Structure, and Adaptation,” J. Chem. Educ., Vol. 59, 1982, 173-178.

Section 20.6

Colin J. Rix, “The Biochemistry of Some Iron Porphyrin Complexes,” J. Chem.

Educ., Vol. 59, 1982, 389-392.

Section 20.6

Ei-Lchiro Ochiai, “Toxicity of Heavy Metals and Biological Defense: Principles and Applications in Bioinorganic Chemistry, Part VII,” J. Chem. Educ., Vol. 72, 1995, 479-484.

Section 20.8

John J. Fortman, “Pictorial Analogies VIII: Types of Formulas and Structural Isomers,” J. Chem. Educ., Vol. 70, 1993, 755.

210

Chapter 20—Transition Elements and Coordination Chemistry

Section 20.8

Hannia Lujan-Upton, “Introducing Stereochemistry to Non-science Majors,” J. Chem. Educ., Vol. 78, 2001, 475-477.

Section 20.9

Steven C. Stinson, “Chiral Drugs,” Chem. Eng. News, September 19, 1994, 38-57.

Section 20.9

Karen F. Schmidt, “Mirror Image Molecules: New Techniques Promise More Potent Drugs and Pesticides,” Science News, May 29, 1993, 348-350.

Section 20.10

Laurence Poncini and Franz L. Wimmer, “Color Classification of Coordination Compounds,” J. Chem. Educ., Vol. 64, 1987, 1001-1002.

Media References Section 20.1

Electron Configurations activity from the Instructor Resource DVD

Section 20.3

Oxidation States of Transition Elements activity from the Instructor Resource DVD

Section 20.5

Coordination Complex Geometry activity from the Instructor Resource DVD

Section 20.7

Naming Coordination Compounds activity from the Instructor Resource DVD

Section 20.8

Isomerism movie from the Instructor Resource DVD

Section 20.8

Isomers Classification activity from the Instructor Resource DVD

Section 20.9

Chirality movie from the Instructor Resource DVD

Section 20.9

Optical Activity movie from the Instructor Resource DVD

Section 20.10

Color Wheel activity from the Instructor Resource DVD

Section 20.10

Color Observation activity from the Instructor Resource DVD

Section 20.12

Crystal Field Theory activity from the Instructor Resource DVD

Section 20.12

Spin Pairing activity from the Instructor Resource DVD

211

CHAPTER 21 METALS AND SOLID-STATE MATERIALS Chapter Learning Goals for Students Section 21.1

Predict whether a given metal is likely to be found in nature as an oxide, a sulfide, a carbonate, a silicate, a chloride, or a free metal.

Section 21.2

Describe the three steps in the metallurgical processes for isolating and purifying a metal from its ore.

Describe the Mond process.

Write balanced chemical equations for the production of metals.

Section 21.3

Describe the processes by which iron ore is converted to steel.

Classify oxides as acidic or basic.

Section 21.4

Explain the properties of a metal according to the electron-sea and the band theory models.

Use band theory to explain transition metal melting-point trends.

Section 21.5

Classify doped semiconductors as n-type or p-type.

Distinguish a metal from a semiconductor by comparing their electrical conductivity temperature dependencies.

Section 21.6

Describe semiconductor applications, including diodes, light-emitting diodes, diode lasers, photovoltaic (solar) cells, and transistors.

Section 21.7

Define superconducting transition temperature (Tc) and the Meissner effect.

Determine the number and types of atoms in a superconductor unit cell.

Section 21.8

Define the terms ceramics, sintering, sol, and gel.

Describe the sol–gel method for the production of ceramic powders.

Section 21.9

Classify composite materials as ceramic–ceramic, ceramic–metal, or ceramic– polymer.

Section 21.10

Compare properties of nanoparticles and their corresponding bulk materials.

Describe quantum dots and nanotubes.

212

Chapter 21—Metals and Solid-State Materials

Lecture Outline 21.1. Sources of the Metallic Elements1 A. Minerals 1. Crystalline, inorganic constituents of rocks that make up the earth’s crust 2. Source of most metals B. Most abundant minerals – silicates and aluminosilicates 1. Difficult to concentrate and reduce 2. Not important sources of metals C. More important minerals – sulfides and oxides 1. Hematite (Fe2O3), rutile (TiO2), and cinnabar (HgS) 2. Ores – mineral deposits from which metals can be produced economically D. Chemical composition of most common ores correlates with metal periodic table location 1. Early transition metals a. Occur as oxides b. Less electronegative metals form compounds by losing electrons to highly electronegative nonmetals.

2. Late transition metals a. Occur as sulfides b. The more electronegative metals tend to form compounds with more covalent character by bonding to less electronegative nonmetals 3. Electronegative p-block metals – sulfide ores 4. s-Block metals a. Occur as carbonates, silicates, and chlorides (for Na and K) b. s–Block oxides are strongly basic and too reactive to exist with acidic oxides.

21.2. Metallurgy2 A. Ore 1. Complex mixture of metal-containing mineral and economically worthless material called gangue 2. Gangue – consists of sand, clay, and other impurities B. Metallurgy 1. The science and technology of extracting metals from their ores 2. Three-step process a. Concentration of the ore – chemical treatment prior to reduction b. Reduction of the mineral to the free metal c. Refining or purification of the metal 3. Used for making alloys – metallic materials composed of two or more elements C. Concentration and chemical treatment of ores 1. Concentrate by separating the mineral from the gangue – ores have different properties that are exploited in various separation methods 2. Metal sulfide ores a. Concentrated by flotation b. Process exploits the differences in the ability of water and oil to wet the surfaces of the mineral and the gangue 3. Bayer process a. Uses chemical treatment to concentrate the mineral b. Separate Al2O3 in bauxite from Fe2O3 impurities by treatment with hot aqueous NaOH 4. Roasting a. A process that involves heating the mineral in air b. Used to convert minerals to compounds that are more easily reduced 1 Test Item File Questions: Multiple Choice 1-17, 85-88; Short Answer 1-3 2 Test Item File Questions: Multiple Choice 18-32, 89-91; Short Answer 4-7

213

Chapter 21—Metals and Solid-State Materials D. Reduction – concentrated ore is reduced to free metal, either by chemical reduction or by electrolysis.

1. Method used – depends on the activity of the metal as measured by E° for reduction (textbook Table 21.2) a. Most active metals i. Most negative standard reduction potentials ii. Most difficult to reduce b. Least active metals i. Most positive standard reduction potentials ii. Easiest to reduce 2. Au and Pt – inactive; found in nature in uncombined form 3. Cu, Ag, and Hg a. Occur as sulfide ores b. Easily reduced by roasting 4. Cr, Zn, and W a. More active metals b. Reduce metal oxides with a chemical reducing agent (C, H2, or a more electropositive metal) 5. Most active metals a. Produced by electrolytic reduction b. No chemical reducing agent strong enough to reduce compounds E. Refining – purification methods used include distillation, chemical purification, and electrorefining 1. Zn – volatile enough to be refined by distillation 2. Ni a. Purified by the Mond process b. Chemical method involving formation and subsequent decomposition of Ni(CO)4 3. Cu a. Purified by electrorefining b. Electrolytic process in which Cu is oxidized to Cu2+ at an impure Cu anode, and Cu2+ from aqueous CuSO4 is reduced to Cu at a pure Cu cathode 21.3. Iron and Steel3 A. Metallurgy of iron 1. Special technological importance 2. Major constituent of steel B. Production of iron 1. Carbon monoxide reduction of Fe ore in a blast furnace (textbook Figure 21.5) 2. Overall reaction: Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2 (g) C. Cast iron (pig iron) obtained from a blast furnace – brittle material containing 4% elemental carbon and smaller amounts of other impurities.

D. Basic oxygen process – used for the purification and conversion of iron to steel 1. Expose molten iron from blast furnace to a jet of pure oxygen gas in a furnace lined with basic oxides. a. Oxidizes impurities b. Acidic oxides produced react with CaO and form a slag that is poured off.

2. Produces steel with about 1% carbon and very small amounts of P and S E. Composition of liquid steel – monitored by chemical analysis 21.4. Bonding in Metals4 A. Metals – malleable, ductile, lustrous, and good conductors of heat and electricity 1. Malleable – can be hammered into sheets 3 Test Item File Questions: Multiple Choice 33-39; Short Answer 8 4 Test Item File Questions: Multiple Choice 40-46, 92-115; Algorithmic 1-2; Short Answer 9-10

214

Chapter 21—Metals and Solid-State Materials 2. Ductile – can be drawn into wires 3. Lustrous – shiny appearance B. Can understand properties by examining bonding in metals C. Valence electrons 1. Can’t be localized in a bond between any particular pair of atoms 2. Delocalized and belong to the crystal as a whole D. Electron-sea model – crystal is a three-dimensional array of metal cations immersed in a sea of delocalized electrons free to move throughout the crystal 1. Delocalized, mobile valence electrons act as electrostatic glue holding metal cations together.

2. Simple qualitative explanation for electrical and thermal conductivity of metals a. Mobile electrons free to move from a negative electrode to a positive electrode when subjected to an electrical potential b. Mobile electrons conduct heat by carrying kinetic energy from one part of the crystal to another. c. Electrons extend in all directions, therefore, when a metallic crystal is deformed, no localized bonds are broken. d. Explains malleability and ductility E. Molecular orbital theory for metals 1. Number of molecular orbitals formed equals number of atomic orbitals combined 2. Difference in energy between successive MOs decreases as number of atoms increases. a. MOs merge into an almost continuous band of energy levels. b. MO theory for metals is called band theory. c. Bottom half of band consists of bonding MOs. d. Top half of band consists of antibonding MOs. 3. Each electron in a metal has a discrete kinetic energy and a discrete velocity. a. Depends on the particular MO b. Increases from bottom to the top of a band 4. One-dimensional metal wire or three-dimensional crystal a. Energy levels within a band occur in degenerate pairs (textbook Figure 21.8). i. One set – electrons moving to the right ii. Other set – electrons moving to the left b. No electrical potential i. No net electric current in either direction ii. Two sets equally populated c. Electrical potential i. Electrons moving right (toward the + battery terminal) are accelerated while electrons moving left (toward the – battery terminal) are decelerated. ii. Electrons moving very slowly left, change direction iii. Number of electrons moving right > number of electrons moving left iv. Net electric current 5. Electrical potential can shift electrons from one set of energy levels to the other only if the band is partially filled. a. Electrical insulators — materials that have only completely filled bands b. Metals — materials that have partially filled bands c. Can have s and p subshells sufficiently close in energy so that the s and p bands overlap, resulting in a partially filled composite band 6. Transition metals – d band overlaps the s band a. Composite band has six MOs per metal atom. b. Maximum bonding for metals i. Six valence electrons per metal atom ii. Near group 6B c. Causes melting points of transition metals to be at a maximum at group 6B 215

Chapter 21—Metals and Solid-State Materials 21.5. Semiconductors5 A. Semiconductor – a material with an electrical conductivity intermediate between that of a metal and that of an insulator B. Valence band – the bonding MOs C. Conduction band – the higher-energy, antibonding MOs D. Band gap – the separation of the valence band and the conduction band in terms of energy E. Insulator 1. No vacant MOs in the valence band; large band gap 2. Can’t excite electrons within the valence band 3. Can’t excite electrons to vacant MOs in the conduction band due to the large band gap F. Metal conductors 1. No energy gap between the highest occupied and lowest unoccupied MOs 2. Electrical conductivity decreases with increasing temperature – increased vibrational motion of cations interferes with flow of electrons G. Semiconductor – smaller band gap than insulators 1. A few electrons can jump the gap and occupy the higher-energy conduction band 2. Partially filled conduction band and valence band – unoccupied MOs in valence band result from electrons that jumped the gap 3. Electrical potential can accelerate the electrons in the partially filled bands – semiconductor conducts a small current 4. Electrical conductivity increases with increasing temperature – electrons with enough energy to jump the band gap increase H. Best criterion for distinguishing a metal from a semiconductor – temperature-dependence of electrical conductivity I. Doping – a process in which conductivity of a semiconductor is increased by adding certain impurities in small amounts 1. Add an impurity with more electrons than the semiconductor a. n-type semiconductor b. Charge carriers are negative electrons.

2. Example of n-type semiconductor: Si doped with P a. Each P atom occupies a Si position and introduces an extra electron. b. Extra electron occupies the conduction band. c. More electrons in conduction band cause the conductivity to be higher.

3. Add an impurity with fewer electrons than the semiconductor a. p-type semiconductor b. Creates positive holes in the valence band 4. Essential components in modern solid–state electronic devices.

21.6. Semiconductor Applications6 A. Diodes 1. Convert alternating current to direct current 2. Permit electron flow in only one direction 3. Constructed of a p-type semiconductor in contact with an n-type semiconductor – called a p–n-junction B. Light-emitting diodes – LEDs 1. LED comparison to incandescent lightbulbs a. Smaller b. Brighter c. Longer lived d. Faster switching times 2. Electrons and holes combine at p–n-junction a. Energy released as light 5 Test Item File Questions: Multiple Choice 47-58, 116-119; Algorithmic 3-4; Short Answer 1-13 6 Test Item File Questions: Multiple Choice 59-60; Short Answer 14-19

216

Chapter 21—Metals and Solid-State Materials b. Wavelength of light roughly equal to band gap energy 3. Light sources a. Traffic signals b. Vehicle brake lights c. Digital clocks d. Hemoglobin oxygen monitors C. Diode lasers 1. Mechanism of light production similar to LEDs a. Laser light more directional b. Laser light more intense 2. Uses a. Laser pointers b. Bar code scanners E. Photovoltaic cells 1. Solar cells 2. p–n-junctions 3. Covert sunlight into energy F. Transistors 1. Control and amplify electrical signals in integrated circuits 2. n–p–n or p–n–p junctions 3. Uses a. Cell phones b. Computers 21.7. Superconductors7 A. Superconductor 1. A material that loses all electrical resistance below a characteristic temperature, the superconducting transition temperature (Tc) 2. Below Tc, once an electric current is started, it will flow indefinitely without loss of energy.

B. 1986 – BaxLa2–xCuO4 found to have a Tc = 35 K 1. Soon after, other copper-containing oxides with even higher Tc discovered 2. Ceramics a. Nonmetallic inorganic solids b. Most are electrical insulators C. No generally accepted theory of superconductivity in ceramic superconductors – infinitely extended layers of Cu and O and the fractional oxidation number of Cu appear to play a role in the current flow D. Superconductors can levitate a magnet – cool a superconductor below Tc and lower a magnet toward it, the magnet and superconductor repel each other.

E. Present applications of conventional superconductors require the use of liquid helium (expensive and requires cryogenic equipment).

1. Can use liquid N2 with new higher-temperature superconductors 2. Abundant refrigerant that is cheaper than milk F. Problems with high-temperature superconductors 1. Brittle powders with high melting points 2. Not easily made into wires and coils needed for electrical equipment 3. Currents carried at 77 K (boiling point of liquid N2) are too low for practical applications.

G. High-temperature superconductors based on fullerene – the allotrope of carbon that contains C60 molecules 1. K3C60 – metallic conductor at room temperature; superconductor at 18 K 2. Fullerides may prove to be better materials than the Cu–O ceramics for making superconducting wires.

3. Fullerides are three-dimensional superconductors.

7 Test Item File Questions: Multiple Choice 61-70, 120; Short Answer 20

217

Chapter 21—Metals and Solid-State Materials 21.8. Ceramics8 A. Ceramics – inorganic, nonmetallic, nonmolecular solids, including both crystalline and amorphous materials, such as glasses 1. Traditional silicate ceramics – made by heating aluminosilicate clays to high temperatures 2. Advanced ceramics – materials that have high-tech engineering, electronic, and biomedical applications a. Oxide ceramics – alumina (Al2O3) b. Nonoxide ceramics – silicon carbide and silicon nitride B. Properties of ceramics – superior to those of metals 1. Higher melting points 2. Stiffer, harder, and more resistant to wear and corrosion 3. Maintain their strength at high temperatures 4. Less dense than steel – attractive lightweight, high-temperature materials for replacing metal components in aircraft, space vehicles, and automobiles C. Strong chemical bonding 1. Covalent network solid a. Highly directional covalent bonds prevent planes of atoms from sliding over one another when the solid is subjected to the stress of a load or an impact. b. Solid can’t deform to relieve the stress. c. Maintains its shape up to a point, but then the bonds give way suddenly 2. Oxide ceramics – bonding largely ionic a. Behave similarly to covalent network solids 3. Metals – deform under stress because planes of metal cations slide easily in the electron sea 4. Metals dent; ceramics shatter D. Ceramic processing – the series of steps that leads from raw material to the finished ceramic object 1. Determines product strength and resistance to fracture 2. Sintering –particles of powder welded together without completely melting a. Occurs below the melting point b. Crystal grains grow larger and the material density increases as void spaces between particles disappear. c. Need high-purity, fine powders that are tightly compacted prior to sintering – impurities and remaining voids can lead to microscopic cracks E. Sol–gel method – method used for preparing high-purity, fine powders that can be tightly compacted 1. Synthesis of a metal oxide powder from a metal alkoxide (a compound derived from a metal and an alcohol) 2. Sol – colloidal dispersion consisting of extremely fine particles having a diameter of only 0.001 to 0.1 μm 3. Gel – a rigid, gelatin-like material 21.9. Composites9 A. Ceramic composite – a hybrid material in which a ceramic powder is mixed, prior to sintering, with fibers of a second ceramic material 1. Combines advantageous properties of both components 2. Whiskers – tiny fiber-shaped particles that are very strong because they are single crystals B. Fibers and whiskers increase the strength and fracture toughness of composite materials.

1. Most of the chemical bonds are aligned along the fiber axis, giving the fibers great strength.

2. Can deflect cracks – prevents cracks from moving cleanly in one direction 3. Can bridge cracks – holds two sides of a crack together 8 Test Item File Questions: Multiple Choice 71-81; Short Answer 21-22 9 Test Item File Questions: Multiple Choice 82-83; Short Answer 23-24

218

Chapter 21—Metals and Solid-State Materials C. Composites in which the two phases are different types of materials 1. Ceramic-metal composites (cermets) 2. Ceramic-polymer composites 3. High strength-to-weight ratios D. Ceramic fibers used in composites usually made by high-temperature methods Inquiry – Why Is It Said That the Next Big Thing Will Be Really Small?10 A. The study and production of materials and structures with at least one 1-100 nm dimension B. Classification 1. One 1-100 nm dimension – nanolayer 2. Two 1-100 nm dimensions a. Nanotube b. Nanowire 3. Three 1-100 nm dimensions – nanoparticles C. Comparison with corresponding bulk materials 1. Lower melting points 2. Different colors a. Bulk gold – yellow b. Gold nanoparticles – red 3. Greater reactivity a. Greater surface area b. More particles on surface c. Surface particles held less strongly than interior particles D. Example – quantum dot 1. Emits light when irradiated with uv 2. Wavelength of light depends on particle size a. Large particles – red b. Small particles – violet 3. Smaller the particle, larger the band gap E. Uses 1. Chemically modified quantum dots that target and bond to a specific biomolecule – biomolecule labels 2. Aluminum oxide nanoparticles a. Sunscreens b. Cosmetics c. Stain-resistant fabrics d. Dirt-repellent paints e. Coatings for self-cleaning windows 3. Carbon nanotubes a. Reinforcing rods b. Toughening elements i. Structural composite materials ii. Concrete 4. Gold nanoparticles – highly selective oxidation catalysts

10 Test Item File Questions: Multiple Choice 84; Short Answer 25

219

Chapter 21—Metals and Solid-State Materials

CHAPTER 21 METALS AND SOLID-STATE MATERIALS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 21.2

In general, the ease of reducing an ore to the free metal increases from left to right of the periodic table. The s-block and aluminum metal ores are reduced with difficulty.

Heavy transition metal ores at the right of the d-block are easiest to reduce.

Section 21.2

A driving force in the reduction of an ore by roasting or chemical treatment with carbon or hydrogen is the formation of gaseous byproducts, which can be continuously removed from the system.

Section 21.2

Metal carbonyls are nonpolar, easily purified, volatile liquids or solids. They are convenient sources of metals in the zero oxidation state and readily converted to the free metal.

Section 21.3

According to the Lux-Flood acid-base theory, an acid is an oxide ion acceptor, and a base is an oxide ion donor. This theory is useful in the steel and ceramic industries, where oxides are commonly employed.

Section 21.4

In the electron-sea model, electrons are delocalized and free to move throughout the entire metal.

Section 21.4

The number of molecular orbitals equals the number of atomic orbitals from which the molecular orbitals were formed.

Section 21.4

Metals have electrons in a partially filled band or in a filled band that overlaps with a second, empty band to give a partially filled composite band. Insulators have electrons in completely filled bands that do not overlap with empty bands.

Section 21.4

Population of the lower (bonding) portions of a band by electrons strengthens metallic bonding, whereas population of the upper (antibonding) portion of a band weakens metallic bonding.

Section 21.5

Silicon and germanium are the two most common semiconductors.

Section 21.5

A semiconductor has a band gap between its valence and conduction bands that is small enough to allow a small equilibrium thermal electron population in the conduction band, leaving holes in the valence band. These electrons and holes can then move when a potential is applied.

Section 21.5

The electrical conductivity of a metal decreases with an increase in temperature, whereas the electrical conductivity of a semiconductor increases with an increase in temperature.

Section 21.5

Doping a semiconductor enhances its electrical conductivity by increasing the population of electrons and/or positive holes in the partially filled bands.

Section 21.8

A sol is a colloid of finely divided metal hydroxide particles. Removing water from a sol converts some hydroxide groups to oxide bridges, resulting in a gel. Further removal of water yields a metal oxide ceramic powder.

220

Chapter 21—Metals and Solid-State Materials

Section 21.9

Composites are hybrid materials of two ceramics, a ceramic and a metal, or a ceramic and a polymer.

Lecture/Laboratory Demonstration References None

Literature References Section 21.1

James K. Beattie, “A New Perspective on Rutile,” J. Chem. Educ., Vol. 75, 1998, 641.

Section 21.2

Daniel Bartet and Eugenia Aguilia, “Teaching the Concepts of Metallurgy Through the Use of Postage Stamps,” J. Chem. Educ., Vol. 64, 1987, 526-528.

Section 21.4

Peter P. Edwards and M. J. Sienko, “On the Occurrence of Metallic Character in the Periodic Table of the Elements,” J. Chem. Educ., Vol. 60, 1983, 691-696.

Section 21.4

Joseph F. Lomax, “Conducting Midshipmen – A Classroom Activity Modeling Extended Bonding in Solids,” J. Chem. Educ., Vol. 69, 1992, 794-795.

Section 21.4

Frank Rioux, “An Ionic Model for Metallic Bonding,” J. Chem. Educ., Vol. 62, 1985, 383-384.

Media References Section 21.1

Common Metal Ores activity from the Instructor Resource DVD

Section 21.2

Formation of Silver Crystals movie from the Instructor Resource DVD

Section 21.3

Blast Furnace activity from the Instructor Resource DVD

Section 21.3

Thermite movie from the Instructor Resource DVD

Section 21.5

Conductivity Band Structure activity from the Instructor Resource DVD

Section 21.7

Resistance Curve activity from the Instructor Resource DVD

221

CHAPTER 22 NUCLEAR CHEMISTRY Chapter Learning Goals for Students Section 22.1

Calculate mass defects and binding energies for nuclides. Use values of binding energy per nucleon to compare the relative stabilities of two nuclides.

Calculate the mass defect in a chemical reaction, and compare it to the mass defect in a nuclear reaction.

Section 22.2

Classify nuclear reactions as fission or fusion.

Calculate the energy released by a nuclear fission or fusion reaction.

Section 22.3

Write balanced equations for nuclear transmutations.

Section 22.4

Perform calculations interconverting units used for measuring radiation.

Compare properties of types of ionizing radiation and their biological effects.

Relate biological effects of short-term radiation to doses.

Section 22.5

Show how radiocarbon dating is used to determine the age of an object.

List medical uses of radioactivity.

222

Chapter 22—Nuclear Chemistry

Lecture Outline Introduction – A Summary of Chapter 22 Nuclear Chemistry Topics A. Nuclear Reactions and Their Characteristics B. Nuclear Reactions and Radioactivity C. Radioactive Decay Rates 22.1. Energy Changes During Nuclear Reactions1 A. Can calculate the energy change on formation of the nucleus from isolated protons and neutrons.

1. ΔE = mc2 2. Relates energy change of a nuclear process to corresponding mass change B. Mass defect 1. Loss in mass that occurs when protons and neutrons combine to form a nucleus 2. Lost mass converted into energy known as the binding energy a. Binding energy – energy that holds the nucleons together b. Calculate from the Einstein equation c. Usually expressed on a per-nucleon basis using the electron volts as the energy unit: 1 eV = 1.60 x 10–19 J C. Plot binding energy per nucleon – higher binding energy corresponds to higher stability D. Mass and energy are interconvertible.

1. Laws of conservation of mass and conservation of energy must be combined.

2. Neither mass nor energy is conserved separately.

3. Mass and energy are interconverted in a nuclear reaction, but the combination of the two is conserved.

22.2. Nuclear Fission and Fusion2 A. Lighter and heavier elements are less stable than mid-mass elements near iron-56.

1. Heavy nuclei – gain stability and release energy if they fragment to yield mid-mass elements 2. Light nuclei – can gain stability and release energy if they fuse together B. Fission – the fragmentation of heavy nuclei 1. Nuclei break into fragments when struck by neutrons 2. Doesn’t occur in exactly the same way each time 3. Uranium-235 – more than 100 different fission pathways a. Frequently occurring pathway 1 91 n + 235 U → 142 Ba + 36 Kr + 3 01 n 0 92 56 b. Three neutrons released induce nine more fissions → 27 neutrons c. Chain reaction – a reaction that continues to occur even if the supply of neutrons from outside is cut off d. Small sample size – many neutrons escape before initiating additional fission events e. Critical mass – a sufficient amount of radioactive nuclide that allows the chain reaction to become self-sustaining 4. Using mass defect and the Einstein equation, can calculate the amount of energy released during nuclear fission – use the masses of the atoms corresponding to the relevant nuclei C. Nuclear reactors.

1. Principle of a nuclear reactor – a subcritical amount of uranium fuel is placed in a containment vessel surrounded by circulating coolant, and control rods are added 2. Control rods are made of boron and cadmium. a. Absorb and regulate the flow of neutrons b. Raised and lowered to maintain fission at a controlled rate 3. Energy from the controlled fission heats the circulating coolant.

1 Test Item File Questions: Multiple Choice 1-12; Short Answer 1-9 2 Test Item File Questions: Multiple Choice 13-26; Short Answer 10-12

223

Chapter 22—Nuclear Chemistry 4. Heated circulating coolant produces steam to drive a turbine and produce electricity.

D. Nuclear fusion – the joining together of light nuclei 1. Releases enormous amounts of energy 2. Fusion of hydrogen nuclei – a potential power source a. Hydrogen isotopes are cheap and plentiful. b. Fusion products are nonradioactive and nonpolluting.

3. Technical problems to achieving a practical and controllable fusion method – High initiation temperature: T = 40 × 106 K 22.3. Nuclear Transmutation3 A. Nuclear transmutation 1. The change of one element into another, brought about by bombardment of an atom with a high-energy particle 2. Creates an unstable nucleus, which causes a nuclear change to occur B. All transuranium elements have been produced by bombardment reactions.

22.4. Detecting and Measuring Radioactivity4 A. Detect radiation by measuring its ionizing properties B. Ionizing radiation – high-energy radiation of all kinds 1. Interaction of radiation with a molecule knocks an electron from the molecule. adiation ⎯→ ion + e Molecule ⎯r⎯ 2. Includes α particles, β particles, γ particles, as well as X rays and cosmic rays 3. X rays – high-energy photons 4. Cosmic rays a. Energetic particles from interstellar space b. Primarily protons C. Devices used for measuring radiation 1. Photographic film badge 2. Geiger counter 3. Scintillation counter D. Expression of radiation intensity depends on what is being measured.

1. Becquerel – SI unit for measuring the number of radioactive disintegrations occurring each second in a sample: 1 Bq = 1 disintegration/s 10 2. Curie (Ci): 1 Ci = 3.7 × 10 disintegrations/s 3. Gray – SI unit for measuring the amount of energy absorbed per kilogram of tissue exposed to a radiation source a. 1 Gy = 1 J/kg b. 1 rad = 0.01 Gy 4. Sievert (Sv) – SI unit that measures the amount of tissue damage caused by radiation 1 rem (roentgen equivalent for man) = 0.01 Sv E. Biological Effects of Radiation 1. Ionizing radiation can be harmful biologically.

2. The effects of ionizing radiation on the human body depend on several factors. a. Kind of radiation and its energy b. Length of exposure c. Whether the source is outside or inside the body 22.5. Applications of Nuclear Chemistry5 A. Radiocarbon dating 1. Depends on the slow and constant production of radioactive carbon-14 in the upper atmosphere by neutron bombardment of nitrogen atoms 3 Test Item File Questions: Multiple Choice 27-33; Short Answer 13-14 4 Test Item File Questions: Multiple Choice 34-56; Short Answer 15-16 5 Test Item File Questions: Multiple Choice 57-63; Algorithmic 1; Short Answer 17-21

224

Chapter 22—Nuclear Chemistry

14 7

N + 01 n → 146 C + 11 H

2. Carbon-14 reacts to form CO2, which mixes with 12CO2 and is taken up by plants during photosynthesis.

14 C + O 2 → 14 CO 2 3. Ratio of 14C to 12C in a living organism is the same as that in the atmosphere.

4. 14C /12C decreases when the organism dies because 14C undergoes radioactive decay.

14 C → 147 N + −10 e 6 5. For 14C, t1/2 = 5730 y 6. Can determine how long ago the organism died by measuring amount of 14C remaining B. Use other radionuclides for dating rocks.

1. 238U a. For 238U, t1/2 = 4.46 × 109 y b. 238U decays through a series to give 206Pb c. Measurement of 238U /206Pb ratio can determine the age of uranium-containing rock.

40 2. K a. For 40K, t1/2 = 1.28 × 109 y b. Potassium-40 decays through electron capture and positron emission 40 Ar K + −01e - → 40 19 18 40 19

K → 4108 Ar + 01e -

c. Compare amount of 40Ar to 40K remaining in sample C. Medical uses of radioactivity – grouped into four classes 1. In vivo procedures 2. In vitro procedures 3. Radiation therapy 4. Imaging procedures D. In vivo procedures 1. Studies that take place inside the body 2. Assess the functioning of a particular organ or body system 3. A radiopharmaceutical agent is administered, and its path in the body is determined by analysis of blood or urine samples.

E. In vitro procedures 1. Studies that take place outside the body.

2. Radioimmunoassay – techniques which measure small concentrations of substances in body fluids 3. Take advantage of antibodies to bind to antigens F. Therapeutic procedures 1. Radiation used as a weapon to kill diseased tissue 2. Involve either external or internal sources of radiation 3. Internal radiation therapy a much more selective technique than external therapy G. Imaging procedures 1. Give diagnostic information about the health of body organs by analyzing the distribution pattern of radionuclides introduced into the body 2. A radiopharmaceutical agent known to concentrate in a specific tissue or organ is injected into the body and external radiation detectors monitor its distribution pattern.

H. Magnetic resonance imaging (MRI) 1. Uses radio waves to stimulate certain nuclei in the presence of an extremely powerful magnetic field.

2. Stimulated nuclei give off a signal that can be measured, interpreted, and correlated with their environment in the body.

225

Chapter 22—Nuclear Chemistry

CHAPTER 22 NUCLEAR CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 22.1

When a nucleus if formed from protons and neutrons, some mass (mass defect) is converted to energy (binding energy), as related by the Einstein equation, ΔE = Δmc2.

Section 22.1

The variation in slope of a plot of the binding energy versus mass number shows that a typical fusion process releases far more energy than the fission processes employed in nuclear reactors.

Section 22.4

Ionizing radiation is any particle or photon with sufficient energy to remove an electron on collision with an atom or molecule, thus creating an ion.

Section 22.4

High-energy photons (γ and χ rays) are the most penetrating forms of radiation.

Particulate forms of radiation (α and β particles) are less penetrating.

Lecture/Laboratory Demonstration References None

Literature References Section 22.2

Ruth Lewin Sime, “Lise Meitner and the Discovery of Fission,” J. Chem. Educ., Vol. 66, 1989, 373-378.

Section 22.2

Ruth Lewin Sime, “Lise Meitner and the Discovery of Fission,” Scientific American, January 1998, 80-85.

Section 22.2

Michael Freemantle, “Ten Years After Chernobyl: Consequences Are Still Emerging,” Chem. Eng. News, April 29, 1996, 18-28.

Section 22.2

Gregory R. Choppin, “Aspects of Nuclear Waste Disposal of Use in Teaching Basic Chemistry,” J. Chem. Educ., Vol. 71, 1994, 826-829.

Section 22.2

Torkil H. Jensen, “Fusion – A Potential Power Source,” J. Chem. Educ., Vol. 71, 1994, 820-823.

Section 22.4

Charles H. Atwood, “How Much Radon Is Too Much?” J. Chem. Educ., Vol. 69, 1992, 351-353.

Section 22.5

Kent J. Crippin and Robert D. Cutright, “Modeling Nuclear Decay: A Point of Integration Between Chemistry and Mathematics,” J. Chem. Educ., Vol. 75, 1998, 434-436.

Section 22.5

M. W. Rowe, “Radioactive Dating: A Method for Geochronology,” J. Chem. Educ., Vol. 62, 1985, 580-584.

Section 22.5

M. W. Rowe, “Archeological Dating,” J. Chem. Educ., Vol. 63, 1986, 16-20.

226

Chapter 22—Nuclear Chemistry

Section 22.5

Ron DeLorenzo, “California Earthquakes: Predicting the Next Big One Using Radiocarbon Dating,” J. Chem. Educ., Vol. 57, 1980, 601.

Section 22.5

Rosalyn S. Yakow, “Development and Proliferation of Radioimmunoassay Technology,” J. Chem. Educ., Vol. 76, 1999, 767-768.

Section 22.5

Rosalyn S. Yakow, “Radioactivity in the Service of Many,” J. Chem. Educ., Vol. 59, 1982, 735-738.

Section 22.5

Mairin B. Brennan, “Positron Emission Tomography Merges Chemistry with Biological Imaging,” Chem. Eng. News, February 19, 1996, 26-33.

Section 22.5

Marcus E. Raichle, “Visualizing the Mind,” Scientific American., April 1994, 64.

Section 22.5

Timothy J. McCarthy, Sally W. Schwarz, and Michael J. Welch, “Nuclear Medicine and Positron Emission Tomography: An Overview,” J. Chem. Educ., Vol. 71, 1994, 830-836.

Inquiry

V. E. Viola, “Teaching Nuclear Science: A Cosmological Approach,” J. Chem.

Educ., Vol. 71, 1994, 840-844.

Media References Section 22.1

Nuclear Stability activity from the Instructor Resource DVD

Section 22.3

Half-Life activity from the Instructor Resource DVD

Section 22.4

Separation of Rays activity from the Instructor Resource DVD

Section 22.5

Carbon Isotopes activity from the Instructor Resource DVD

Section 22.5

First Order Process movie from the Instructor Resource DVD

227

CHAPTER 23 ORGANIC and BIOLOGICAL CHEMISTRY Chapter Learning Goals for Students Section 23.1

Draw electron dot structures for isomers of simple alkanes.

Write condensed structures of organic molecules.

Give the geometry around carbon atoms bonded to two, three, and four other atoms.

Determine which structures represent different molecules and which are merely different conformations of the same molecule.

Section 23.2

Identify functional groups in molecules. Draw structures of molecules containing functional groups listed in textbook Table 23.1.

Section 23.3

Given the structure of an alkane, determine its IUPAC name and vice versa.

Section 23.4

Given the structure of an alkene or alkyne, determine its IUPAC name and vice versa.

Predict the products and write balanced chemical equations for alkene addition reactions.

Section 23.5

Given the structure of a cycloalkane, determine its IUPAC name and vice versa.

Section 23.6

Given the structure of an aromatic compound, determine its IUPAC name and vice versa.

Predict the products and write balanced chemical equations for aromatic substitution reactions.

Section 23.7

Classify molecules as alcohols, ethers, or amines. State the general properties of these classes of molecules.

Section 23.8

Classify carbonyl-containing molecules as aldehydes, ketones, carboxylic acids, esters, or amides, and state the general properties of these classes of molecules.

Give systematic names of carboxylic acids, esters, and amides.

Predict the products and write balanced chemical equations for carbonyl substitution reaction and reactions of carboxylic acids, esters, and amides with water.

Section 23.9

Explain the metabolic reactions between a molecule containing the –OH functional group and ATP to form a phosphate.

Section 23.10

Identify functional groups in α-amino acids. Draw structures that show the geometry about each atom.

Classify amino acid side chains as acidic, basic, or neutral, hydrophilic or hydrophobic.

229

Chapter 23—Organic and Biological Chemistry

Section 23.11

Draw the structure and give the three-letter shorthand notation for simple proteins.

Classify a carbohydrate as simple or complex.

Classify a monosaccharide as a ketose or aldose.

Differentiate between α- and β- forms of D-glucose.

Describe common disaccharides and polysaccharides, including lactose, sucrose, and glycogen.

Section 23.12

Draw the structures of simple fats and oils.

Section 24.13

Draw the structures of simple nucleic acids.

Show the complementary base pairs in strands of DNA. Show the RNA base sequence complementary to a given DNA base sequence.

230

Chapter 23—Organic Chemistry

Lecture Outline 23.1. Organic Molecules and Their Structures: Alkanes1 A. Hydrocarbons – molecules that contain only carbon and hydrogen B. Alkanes 1. Hydrocarbons that have only single bonds 2. Saturated hydrocarbons C. Straight-chain alkanes – compounds with all their carbons connected in a row D. Branched-chain alkanes – those with a branching connection of carbons E. Isomers 1. Molecules that have the same molecular formula but different structures 2. Different isomers are different chemical compounds.

3. Have different structures and different physical properties F. Drawing organic structures 1. Condensed structures – carbon–hydrogen and carbon–carbon single bonds aren’t shown.

2. Bonds are understood.

3. The horizontal bonds between carbons aren’t shown.

G. The shapes of organic molecules 1. A molecule can be arbitrarily shown in a great many ways.

2. Rotation around carbon–carbon single bonds is possible.

3. Conformations – a number of possible three-dimensional structures resulting from the ability of parts of a molecule to spin around a carbon–carbon single bond 4. Geometry about each individual carbon atom is predicted using the VSEPR model a. Carbon having only single bonds – tetrahedral with 109.5o bond angles b. Carbon having one double bond and two single bonds – trigonal planar with 120o bond angles c. Carbon with one triple bond and one single bond – linear with 180o bond angles 23.2. Families of Organic Molecules: Functional Groups2 A. Can classify organic compounds into families according to their structural features – chemical behavior of the members of a family is predictable B. Functional group – a part of a larger molecule consisting of an atom or group of atoms that has characteristic chemical behavior – a given functional group undergoes the same kinds of reactions in every molecule of which it is a part C. The chemistry of an organic molecule, regardless of its size and complexity, is determined by the functional groups it contains.

D. Most common functional groups – textbook Table 23.1 23.3. Naming Organic Compounds3 A. Three parts to a chemical name 1. Parent name – specifies the overall size of the molecule by telling how many carbon atoms are present in the longest continuous chain 2. Suffix – identifies the family to which the molecule belongs 3. Prefix – specifies the location of various substituent groups attached to the parent chain B. Straight-chain alkanes – named by counting the number of carbon atoms in the chain and adding the family suffix –ane 1. First four compounds a. Methane – CH4 b. Ethane – CH3CH3 c. Propane – CH3CH2CH3 d. Butane – CH3CH2CH2CH3 1 Test Item File Questions: Multiple Choice 1-15, 100-101; Algorithmic 1-2; Short Answer 1-13 2 Test Item File Questions: Multiple Choice 16-24, 106-107, 118; Short Answer 14 3 Test Item File Questions: Multiple Choice 25-35, 102-105, 111-117; Short Answer 15-18

231

Chapter 23—Organic and Biological Chemistry 2. All other alkanes are named from Greek numbers according to the number of carbons present – textbook Table 23.2.

C. Four steps to naming branched-chain alkanes 1. Name the parent chain. a. The longest continuous chain of carbons present b. Use the name of that chain as the parent name 2. Number the carbon atoms in the parent chain. a. Begin at the end near the first branch point b. Branches should have the lowest number combination 3. Identify and number the branching substituent. a. Alkyl group – the part of an alkane that remains when a hydrogen is removed b. Name by replacing the –ane ending of the parent alkane with an –yl ending 4. Write the name as a single word. a. Separate different prefixes with hyphens. b. Separate numbers with commas. c. Cite substituent names in alphabetical order. d. Don’t use the prefixes for alphabetizing.

D. Alkyl groups 1. Hydrogens in methane and ethane are equivalent – can form only one alkyl group 2. Hydrogens in propane are not equivalent. a. Six H’s on each end and two H’s in the middle b. Forms two different propyl groups c. Remove an end H – n-propyl d. Remove a middle H – isopropyl 3. Four different kinds of butyl groups a. Butyl and sec-butyl i Are derived from straight-chain alkanes ii n-butyl: remove H from end C iii sec-butyl: remove H from a C atom bonded to two other C atoms b. Isobutyl and tert-butyl i Derived from branched-chain isobutane ii Isobutyl: remove H from end C on the branched-chain isobutane iii tert-butyl: remove H from a C bonded to three other C atoms 4. Alkyl groups are not compounds.

23.4. Unsaturated Organic Compounds: Alkenes and Alkynes4 A. Alkenes – hydrocarbons that contain a carbon–carbon double bond B. Alkynes – hydrocarbons that contain a carbon–carbon triple bond C. Unsaturated hydrocarbons – hydrocarbons that have fewer hydrogens per carbon than the related saturated alkanes – alkenes and alkynes are unsaturated D. Name alkenes by counting the longest chain of carbons that contains the double bond, and adding the family suffix –ene E. Alkene isomers that exist because of double-bond position; for butene and higher specify the position of the double bond with a numerical prefix a. Start numbering from the chain end nearer the double bond b. Cite only the first of the double-bonded carbons c. Identify and number the position of any substituent present d. If the double bond is equidistant from both ends of the chain, numbering starts at the end nearer the substituent F. Alkene isomers that exist because of double-bond geometry 1. cis isomer – functional groups are on the same side of the double bond 2. trans isomer – function groups are on different sides of the double bond G. Alkynes are similar to alkenes.

4 Test Item File Questions: Multiple Choice 36-59, 119-120; Algorithmic 3; Short Answer 19-20

232

Chapter 23—Organic Chemistry 1. Names using the family suffix –yne 2. Simplest alkyne – acetylene 3. Isomers are possible depending on the position of the triple bond in the chain.

H. Reactions of Alkenes and Alkynes 1. Addition reactions: X–Y adds to the multiple bond of the unsaturated reactant to yield a saturated product.

2. Hydrogenation a. Addition of hydrogen b. Alkene + H2 → Alkane 3. Halogenation a. Addition of Cl2 and Br2 b. Alkene + X2 → 1,2–Dihaloalkane 4. Hydrolysis a. Alkenes react with water in the presence of H2SO4 to produce an alcohol. b. Adding an H+ and OH– + c. H goes to the carbon containing the most H’s.

23.5. Cyclic Organic Compounds5 A. Acylic alkanes – open-chain alkanes B. Cycloalkanes – alkanes that contain rings of carbon atoms C. Structures are represented by polygons – textbook Figure 23.3 1. Carbon atom is understood to be at every junction of lines.

2. Mentally supply the proper number of hydrogen atoms needed to fill out carbon’s valency.

D. Name substituted cycloalkanes, using the cycloalkane as the parent name and identifying the positions on the ring where substituents are attached.

1. Start numbering at the group that has alphabetical priority.

2. Proceed in the direction that gives the second substituent the lowest possible number.

23.6. Aromatic Compounds6 A. Aromatic compounds 1. Class of compounds containing a six-membered ring with three double bonds 2. Simplest aromatic – benzene B. Benzene and other aromatic compounds are much less reactive than alkenes and don’t normally undergo addition reaction.

C. Benzene’s stability is a consequence of its electronic structure.

1. Six π electrons are spread around the entire ring.

2. Two resonance structures

3. Often depicted as:

D. Name substituted aromatic compounds by using the suffix –benzene.

E. Disubstituted aromatic compounds — use one of the prefixes ortho–, meta–, or para–.

1. ortho (o) – two substituents in a 1,2 relationship on the ring 2. meta (m) – two substituents in a 1,3 relationship on the ring 5 Test Item File Questions: Multiple Choice 60-64; Short Answer 21 6 Test Item File Questions: Multiple Choice 65-70, 121; Short Answer 22

233

Chapter 23—Organic and Biological Chemistry 3. para (p) – two substituents in a 1,4 relationship on the ring F. Phenyl – the benzene ring itself is a substituent G. Substitution reactions – a group substitutes for one of the hydrogen atoms on the aromatic ring without changing the ring itself 1. All six H’s are equivalent.

2. Nitration a. Substitution of a nitro group (–NO2) in the presence of H2SO4 b. Key step in synthesis of explosives c. Important pharmaceutical agents d. Nitrobenzene used as starting material for dyes 3. Halogenation a. Substitution of either Br or Cl in the presence of FeBr3 or FeCl3 b. Step used in the synthesis of numerous pharmaceutical agents 4. Sulfonation a. Substitution of a sulfonic acid group (–SO3H) b. Result of benzene reacting with H2SO4 and SO3 c. Key step in synthesis of aspirin and the sulfa-drug family of antibiotics 23.7. Alcohols, Ethers, and Amines7 A. Alcohols can be considered in two ways.

1. Derivatives of water in which one of the H’s is replaced by an organic substituent 2. Derivatives of alkanes in which one of the H’s is replaced by an –OH group 3. Structural resemblance to water – simple alcohols are water-soluble 4. Name by a. Specifying the point of attachment of the –OH group to the hydrocarbon chain b. Use the suffix –ol to replace the terminal –e in the alkane name c. Number chain at the end nearer to the –OH group 5. Simple alcohols – most important and commonly encountered organic chemicals 6. Methanol – wood alcohol a. Produced by catalytic reduction of CO with hydrogen gas b. Important industrial starting material for preparing formaldehyde, acetic acid, and other chemicals 7. Ethanol a. One of the oldest known pure organic chemicals b. Alcohol present in wine, beer, and distilled liquors 8. 2-Propanol (rubbing alcohol) – used primarily as a solvent 9. Other important alcohols a. Ethylene glycol – principal constituent of automobile antifreeze b. Glycerol – moisturizing agent in many foods and cosmetics c. Phenol – used in preparing nylon, epoxy adhesives, and heat-setting resins B. Ethers – compounds that have two organic groups bonded to the same oxygen atom 1. Inert chemically – used as reaction solvents 2. Diethyl ether – used for many years as a surgical anesthetic agent C. Amines – organic derivatives of NH3 1. One or more NH3 hydrogens are replaced by an organic substituent.

2. Use suffix –amine in naming these compounds 3. Amines are bases. a. Can use the lone pair of electrons on nitrogen to accept H+ from an acid b. Produce ammonium salts 4. Ammonium salts are more soluble in water than are neutral amines – practical consequences in drug delivery 23.8. Carbonyl Compounds8

7 Test Item File Questions: Multiple Choice 71-79, 124, 126, 129; Short Answer 23-24

234

Chapter 23—Organic Chemistry A. Carbonyl group – group with a carbon-oxygen double bond 1. C=O 2. Polar bond.

B. Aldehydes and ketones — CO group is bonded to atoms (H and C) that are not strongly electronegative.

1. Aldehyde – hydrogen atom bonded to CO group 2. Ketone – two carbon atoms bonded to CO group 3. The two groups of compounds have similar properties.

4. Industrial preparation of simple aldehydes and ketones involves oxidation of the related alcohol.

5. Aldehyde and ketone functional groups are present in many biologically important compounds.

C. Carboxylic Acids, Esters, and Amides 1. CO group is bonded to an atom (O or N) that is strongly electronegative.

2. All three families undergo carbonyl group substitution reactions.

3. A group substitutes for the –OH, –OC, or –N group of the starting material.

D. Carboxylic acids contain O OH C 1. Occur widely throughout the plant and animal kingdoms 2. Name by replacing the e ending of the corresponding alkane with –oic acid 3. Dissociate slightly in aqueous solution to give H3O+ and a carboxylate anion 4. Undergo acid-catalyzed reaction with an alcohol to yield an ester E. Esters contain O

C O C 1. Many uses in medicine, in industry, and in living systems 2. Important pharmaceutical agents – aspirin and benzocaine 3. Undergo hydrolysis a. Splits the ester molecule into a carboxylic acid and an alcohol b. Can be either acid- or base-catalyzed 4. Saponification – base-catalyzed ester hydrolysis a. Soap – mixture of sodium salts of long-chain carboxylic acids b. Produced by aqueous NaOH hydrolysis of the naturally occurring esters in animal fat 5. Named by identifying the alcohol-related part and the acid-related part using the –ate ending.

F. Amides contain O

C N 1. Amide bond between nitrogen and a carbonyl-group carbon is the fundamental link used by organisms for forming proteins.

2. Neutral – do not act as proton acceptors and do not form ammonium salts 3. Can be prepared by the reaction of a carboxylic acid with ammonia or an amine 4. Name by first citing the N–alkyl group on the amine part and then identifying the carboxylic acid part using the –amide ending 23.9. An Overview of Biological Chemistry9 A. Metabolism – the sum of the many organic reactions that go on in cells 1. Extract and release energy from food 8 Test Item File Questions: Multiple Choice 80-99, 123, 125, 127-128, 130-134; Algorithmic 4; Short

Answer 25 9 Test Item File Questions: Multiple Choice 135-138; Short Answer 26 235

Chapter 23—Organic and Biological Chemistry 2. End products – carbon dioxide, water and energy 3. Occur in either linear or cyclic long sequences a. Linear sequence – product of one reaction serves as the starting material for the next b. Cyclic sequence – a series of reactions regenerates the first reactant B. Catabolism – reaction sequences that break molecules apart 1. Generally releases energy that is used to power living organisms 2. First stage: digestion – bulk food is broken down into small molecules 3. Second stage: small molecules are broken down into two-carbon acetyl groups attached to coenzyme A a. Produces acetyl coenzyme A b. An intermediate in the breakdown of all main classes of food molecules 4. Third stage: citric acid cycle a. Acetyl groups are oxidized – produces carbon dioxide and water b. Releases a great deal of energy – used in the fourth stage 5. Fourth stage: the respiratory chain. a. Produces adenosine triphosphate (ATP) b. Plays a pivotal role in the production of biological energy C. Anabolism 1. Reaction sequences that put building blocks back together to assemble larger molecules 2. Absorb energy D. The entire process of energy production revolves around the ATP–ADP interconversion.

1. Catabolic reactions “pay off” in ATP by synthesizing it from adenosine diphosphate (ADP) plus hydrogen phosphate ion.

2. Anabolic reactions “spend” ATP by transferring a phosphate group to other molecules, thereby regenerating ADP. E. ATP releases a large amount of energy when its P–O–P bonds are broken and a phosphate group is transferred.

1. Usefulness is due to ATP’s ability to drive otherwise unfavorable reactions.

2. Overall free-energy change for the two reactions together is favorable.

23.10. Amino Acids, Peptides, and Proteins10 A. Protein – biological polymers made up of many amino acids linked together to form long chains 1. A group of biological molecules that are of primary importance to all living organisms 2. Different biological functions a. Some serve a structural purpose. b. Some act as hormones to regulate specific body processes. c. Some are enzymes – biological catalysts that carry out body chemistry B. Amino acids – molecules that contain two functional groups 1. A basic amino group (–NH2) 2. An acidic group (–COOH) C. Peptide bond – an amide bond that forms when two amino acids link together D. Dipeptide – the molecule that results when two amino acids are linked together by formation of a peptide bond between the –NH2 group of one and the –COOH group of the second 1. Tripeptide – linkage of three amino acids by two peptide bonds 2. Polypeptides – chains of up to 100 amino acids 3. Protein – chains with more than 100 amino acids E. 20 different amino acids commonly found in proteins 1. Refer to each by a three-letter shorthand code (textbook Figure 23.8).

2. α-Amino acids – the amino group is connected to the carbon atom alpha to (next to) the carboxylic acid group 3. Differ in the nature of the group (side chain, R) attached to the α-carbon 10 Test Item File Questions: Multiple Choice 139-169, 198-236, 239-240; Algorithmic 5-9; Short

Answer 27-28 236

Chapter 23—Organic Chemistry 4. Classified as neutral, basic, or acidic a. Depends on side chain b. Neutral – divide into hydrophobic (nonpolar side chains) and hydrophilic (polar) side chains c. Acidic – an additional carboxylic acid on side chain d. Basic – additional amine function on side chain F. Residues – individual amino acids that are linked together by peptide bonds to form proteins G. Backbone – the repeating chain of amide linkages to which the side chains are attached H. The number of possible isomeric peptides increases rapidly as the number of amino acid residues increases.

I. All noncyclic proteins have an N-terminal amino acid and a C-terminal amino acid.

1. N-terminal amino acid – has a free –NH2 group on one end 2. C-terminal amino acid – has a free –COOH group on the other end 3. Protein is written with the N-terminal residue on the left and the C-terminal residue on the right.

4. Use three-letter abbreviations to indicate the name (textbook Figure 23.8).

24.11. Carbohydrates11 A. Carbohydrates – a large class of polyhydroxylated aldehydes and ketones 1. Monosaccharides – carbohydrates that can’t be broken down into smaller molecules by hydrolysis with aqueous acid (simple carbohydrates) 2. Polysaccharides – compounds that are made of many simple sugars linked together and that cleave into many molecules of simple sugars upon hydrolysis (complex carbohydrates) B. Monosaccharides 1. Aldose – contains an aldehyde carbonyl group 2. Ketose – contains a ketone carbonyl group 3. The “–ose” suffix – used to indicate a sugar 4. Use one of the Greek prefixes to indicate the number of carbon atoms in the sugar C. Monosaccharides are shown as having open-chain structures.

D. Exist as cyclic molecules 1. –OH group near the bottom of the chain adds to the –C=O group near the top of the chain.

2. Cyclic α form – the C1 –OH group is on the bottom side of the ring 3. Cyclic β form – the C1 –OH group is on the top side of the ring E. Some Common Disaccharides and Polysaccharides 1. Sucrose a. Most common pure organic chemical in the world b. Hydrolysis yields one molecule of glucose and one molecule of fructose. c. Invert sugar i. 50:50 mixture of sugar produced from the hydrolysis of sucrose ii. Commonly used as a food additive 2. Cellulose a. Consists of several thousand β-glucose molecules joined together by 1,4 links to form an immense polysaccharide b. Used as a structural material in plants 3. Starch a. Consists of several thousand α-glucose molecules joined by 1,4 links b. More structurally complex than cellulose c. Two types i. Amylose – several hundred to1000 α-glucose units joined together in a long chain by 1,4 links ii. Amylopectin – larger than amylose and has branches every 25 units along its chain 11 Test Item File Questions: Multiple Choice 170-180, 237-238; Short Answer 29

237

Chapter 23—Organic and Biological Chemistry 24.12. Lipids12 A. Lipids 1. Naturally occurring organic molecules that dissolve in nonpolar organic solvents 2. Defined by solubility rather than by chemical structure 3. Contain large hydrocarbon portions – accounts for their solubility behavior B. Fats and oils 1. Most plentiful lipids in nature 2. Triacylglycerols (triglycerides) – triesters of glycerol with three long-chain carboxylic acids (fatty acids) a. Fatty acids – unbranched and have an even number of carbon atoms b. Three fatty acids can be different.

3. Monounsaturated – fatty acid with only one double bond 4. Polyunsaturated fatty acids – have more than one carbon–carbon double bond 24.13. Nucleic Acids13 A. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the chemical carriers of an organism’s genetic information.

B. Nucleic acids – polymers made up of nucleotide units linked together to form a long chain 1. Nucleotide – consists of nucleoside plus phosphoric acid 2. Nucleoside – consists of an aldopentose sugar plus an amine base a. RNA – sugar = ribose b. DNA – sugar = 2–deoxyribose C. Four different cyclic amine bases in DNA: adenine, guanine, cytosine, and thymine D. Four different cyclic amine bases in RNA: adenine, guanine, cytosine and uracil E. Cyclic amine base is bonded to C1′ of the sugar, and the phosphoric acid is bonded to the C5′ sugar position.

1. Numbers with a prime superscript refer to positions on the sugar component of a nucleotide.

2. Numbers without a prime superscript refer to positions on the cyclic amine base.

F. Nucleotides link through a phosphate ester bond between the phosphate group at the 5′ end of one nucleotide and the hydroxyl group on the sugar component at the 3′ end of another nucleotide.

G. Structure of a nucleic acid depends on its sequence of individual nucleotides – an alternating sugar-phosphate backbone with different amine base side chains attached.

H. Describing the sequence of nucleotides by starting at the 5′ phosphate and identifying the bases in order – use first letter of the base name as an abbreviation for each nucleotide.

I. Base Pairing in DNA: The Watson–Crick Model 1. DNA consists of two polynucleotide strands coiled around each other in a double helix. a. Sugar-phosphate backbone is on the outside of the helix. b. Heterocyclic bases are on the inside of the helix. c. Base on one strand points directly at a base on the other strand. d. Two strands run in opposite directions – held together by hydrogen bonds between pairs of bases. i. A and T ii. G and C 2. Two strands are complementary – whenever a G base occurs in one strand, a C base occurs opposite it in the other strand.

J. Nucleic Acids and Heredity 1. Chromosomes – threadlike strands that are coated with proteins and wound into complex assemblies

12 Test Item File Questions: Multiple Choice 181-184; Short Answer 30 13 Test Item File Questions: Multiple Choice 185-197, 241-244; Short Answer 31-32

238

Chapter 23—Organic Chemistry 2. Gene – segment of a DNA chain that contains the instructions necessary to make a specific protein 3. DNA functions as a storage medium for an organism’s genetic information.

4. RNA reads, decodes, and uses the information received from DNA to make proteins.

5. Three processes in the transfer and use of genetic information a. Replication – the means by which identical copies of DNA are made i. Forms additional molecules ii. Preserves genetic information for passing on to offspring b. Transcription – means by which information in the DNA is transferred to and decoded

by RNA c. Translation – means by which RNA uses the information to build proteins K. Replication – an enzyme-catalyzed process that begins with a partial unwinding of the double helix 1. New nucleotides line up on each strand in a complementary manner 2. Two new strands begin to grow L. Transcription 1. Similar to replication except that new ribonucleotides line up 2. Uracil lines up opposite adenine.

M. Translation – protein biosynthesis, directed by messenger RNA (mRNA) 1. Occurs on ribosomes – knobby protuberances within a cell 2. Ribonucleotide sequence in mRNA specifies the order of different amino acid residues – three ribonucleotides that are specific for a given amino acid 3. Read by transfer RNA (tRNA) a. Contains a complementary base sequence that allows it to recognize a three-letter word on mRNA b. Acts as a carrier to bring a specific amino acid into place for transfer to the peptide chain

239

Chapter 23—Organic and Biological Chemistry

CHAPTER 23 ORGANIC and BIOLOGICAL CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 23.1

The molecular geometry about a carbon atom in an organic molecule depends on the number of atoms bonded to the carbon: tetrahedral for four, trigonal planar for three, and linear for two atoms bonded to carbon.

Section 23.1

For σ bond formation carbon uses sp3 hybrid orbitals in tetrahedral molecules, sp2 hybrid orbitals in trigonal planar molecules, and sp hybrid orbitals in linear molecules. These molecules generally also have zero, one, and two π bonds, respectively.

Section 23.1

An alkane is a hydrocarbon in which each carbon atom is bonded to four other carbon or hydrogen atoms. Because each carbon atom is bonded to its maximum number of atoms and cannot bond to additional atoms, the alkane is said to be saturated.

Section 23.1

C–H and most C–C bonds are not shown in condensed structures.

Section 23.1

An electron-dot structure does not give the shape of a molecule. The shape is defined by applying the VSEPR model to each carbon, nitrogen, and oxygen atom present in the molecule.

Section 23.4

To emphasize the role of a functional group in the chemistry of a molecule, it is sometimes useful to represent nonreacting portions of the molecule with the letter R. For example, both CH3OH and CH3CH2OH are represented by the general formula ROH, because the chemistry of both depends primarily on the –OH group and little or not at all on the nature of R.

Section 23.4

In an alkene, the double-bonded carbon atoms and atoms attached to them all lie in the same plane.

Although cycloalkanes are usually drawn as planar polygons, most are not planar but puckered.

Section 23.5 Section 23.6

Aromatic compounds undergo substitution reactions, whereas alkenes undergo addition reactions.

Section 23.7

Nonbonded electrons are usually not shown in condensed structures. Atoms in organic molecules (except hydrogen) generally obey the octet rule, so any atom with fewer than four bonds must have a number of nonbonded electron pairs equal to four minus the number of bonds. For example, the oxygen in an alcohol has 4 – 2 = 2 nonbonded electron pairs.

Section 23.7

Glycerol is a good moisturizing agent because it attracts water molecules, which hydrogen bond with glycerol’s three –OH groups.

Section 23.8

This hydrolysis reaction can be considered the reverse of the carbonyl substitution reaction: carboxylic acid + alcohol → ester + water.

Oxidation of an organic molecule usually involves an increase in the number of carbon–oxygen bonds. When the acetyl group is oxidized to carbon dioxide, the number of carbon–oxygen bonds increases from two to four.

Section 23.9

240

Chapter 23—Organic Chemistry

Section 23.9

Although we say that energy is stored in bonds and released when the bonds are broken, in actuality bond breaking requires energy. Energy is released when new bonds are formed that are stronger than the bonds broken.

Section 23.9

To reinforce the importance of functional groups, point out the functional groups present in molecules discussed in this chapter, for example, the aldehyde and alcohol groups in glucose.

Section 23.9

This is a reaction between an alcohol (R–OH) and an acid (HPO42–) to produce an ester (a phosphate ester). Note that glucose 6-phosphate has a phosphate group attached to C6, where C1 is the aldehyde carbon.

Section 23.10

Although written as –COOH, there is no bond between the two oxygen atoms. Since carbon must have four bonds, –COOH is actually a shorthand notation for O || –C–OH

Section 23.10

The formation of the peptide bond from reaction of the acid group on one amino acid with the amine group on a second amino acid is a carbonyl-group substitution. The peptide bond is part of the amide group formed by this reaction.

Section 23.10

All 20 α-amino acids found in proteins have an amine group and a side chain, R, on the carbon adjacent to the carboxylic acid group. The side chain influences the chemistry of the amino group.

Section 23.10

The side chains of neutral amino acids are hydrophilic if R is polar (for example, if R = –CH2OH), and hydrophobic if R is nonpolar (for example, if R = –H).

Section 23.10

Proteins are always drawn as H2N–(amino acid sequence)–CO2H and named according to the left-to-right sequence of the amino acids.

Section 23.11

A carbohydrate contains many hydroxyl groups and either an aldehyde or ketone functional group. A simple carbohydrate (monosaccharide or sugar) is an aldose if it contains an aldehyde group, or a ketose if it contains a ketone group.

Section 23.11

Common sugars have five or six carbon atoms.

Section 23.11

A sugar usually exists as a cyclic structure in equilibrium with a small amount of its open-end structure.

Section 23.11

The ending –ase indicates that the α-glycoside is an enzyme.

Section 23.12

Alcohol + acid → ester + water. In fats and oils, glycerol is the alcohol, and the acids are fatty acids that contain long hydrocarbon chains, 12 to 22 carbon atoms long. Lipids with unsaturated fatty acid groups tend to be oils; those with saturated fatty acid groups tend to be solids.

Section 23.13

The double-helix structure is a result of hydrogen bonding between pairs of bases, one located on one polynucleotide strand and a complementary base located on the second polynucleotide strand. The sequence of bases on one strand requires a specific, complementary sequence on the second strand.

241

Chapter 23—Organic and Biological Chemistry

Lecture/Laboratory Demonstration References Section 23.8

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “The Disappearing Coffee Cup,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 98. A polystyrene coffee cup is dissolved in acetone.

Section 23.8

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Making Canned Heat,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 111-112. Stearic acid is saponified to produce a solid fuel–canned heat.

Section 23.11

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “A Variation of the Starch-Iodine Clock Reaction,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 147148.

Literature References Section 23.1

Raymond B. Seymour, “Alkanes: Abundant, Pervasive, Important, and Essential,” J. Chem. Educ., Vol. 66, 1989, 59-63.

Section 23.1

A. Haudrechy, “A Simple Method of Drawing Stereoisomers from Complicated Symmetrical Structures,” J. Chem. Educ., Vol. 77, 2000, 864-866.

Section 23.8

Ellin Doyle, “Trans Fatty Acids,” J. Chem. Educ., Vol. 74, 1997, 1030-1032.

Section 23.8

Michele Clarke, Ann Brown, Dianne N. Epp, Mary Gallup, Jeffrey R. Wilson, and Judith A. Wuerthele, “Ester, What's in My Food?” J. Chem. Educ., Vol. 634, 1986, 1050-1051, Gil Down, “Why Teach Biochemistry?” J. Chem. Educ., Vol. 64, 1987, 339.

Section 23.9 Section 23.9

Sheldon J. Kopperi, “Icie Macy Hoobler: Pioneer Woman Biochemist,” J. Chem.

Educ., Vol. 65, 1988, 97-98.

Section 23.10

C. M. Drain and Barry B. Corden, “Reversible Oxygenation of Oxygen Transport Proteins,” J. Chem. Educ., Vol. 64, 1987, 441-443.

Section 23.11

Charles L. Bering, “The Biochemistry of Brewing,” J. Chem. Educ., Vol 65, 1988, 519-521.

Section 23.11

James M. Garrett, “Brand the Name with the Linkage of the Same,” J. Chem. Educ., Vol. 61, 1984, 665.

Section 23.13

Maureen M. Julian, “Rosalind Franklin: From Coal to DNA to Plant Viruses,” J. Chem. Educ., Vol. 60, 1983, 660-662.

Section 23.13

A. B. Wolbarst, “DNAmonic,” J. Chem. Educ., Vol. 56, 1979, 773.

Section 23.13

Michael F. Bruist, Wayne L. Smith, and Galen Mell, “A Simple Demonstration of How Internuclear Forces Make DNA Helical,” J. Chem. Educ., Vol. 75, 1998, 5355.

242

Chapter 23—Organic Chemistry

Media References Section 23.1

Carbon Hybrid Orbitals activity from the Instructor Resource DVD

Section 23.1

Butane Condensed Structure activity from the Instructor Resource DVD

Section 23.2

Halogenation of Methane activity from the Instructor Resource DVD

Section 23.4

Addition Reaction activity from the Instructor Resource DVD

Section 23.4

Surface Reaction – Hydrogenation movie from the Instructor Resource DVD

Section 23.4

Testing for Unsaturated Hydrocarbons with Bromine movie from the Instructor Resource DVD

Section 23.4

Cis-Trans Isomerism activity from the Instructor Resource DVD

Section 23.6

Synthesis of Nylon 610 movie from the Instructor Resource DVD

Section 23.7

Energy and Enzymes movie from the Instructor Resource DVD

Section 23.10

Chirality movie from the Instructor Resource DVD

Section 23.10

Charality of d-Glyceraldehyde activity from the Instructor Resource DVD

Section 23.10

Aspartame activity from the Instructor Resource DVD

Section 23.10

Peptide Bond Formation activity from the Instructor Resource DVD

Section 23.10

Proteins and Amino Acids movie from the Instructor Resource DVD

Section 23.13

Nucleic Acids activity from the Instructor Resource DVD

Section 23.13

DNA Replication movie from the Instructor Resource DVD

Section 23.13

Protein Synthesis movie from the Instructor Resource DVD

243