Chemistry, Global Edition, 7E John E. McMurry Solution Manual

Type:

Solution Manual

Resource:

Chemistry

Edition:

7th Edition

Author(s):

John E. McMurry Robert C. Fay Jill Robinson

CHAPTER 1 CHEMICAL TOOLS: EXPERIMENTATION AND MEASUREMENT Chapter Learning Goals for Students Section 1.1

Identify the steps involved in the scientific method.

Differential between a qualitative and a quantitative measurement.

Section 1.2

Write numbers in scientific notation and use prefixes for multiples of SI units.

Section 1.3

Describe the difference between mass and weight.

Convert between different prefixes used in mass measurements.

Section 1.4

Convert between different prefixes used in length measurements.

Section 1.5

Convert between common units of temperature measurement.

Section 1.6

Convert between SI and metric units of volume.

Convert between different prefixes used in volume measurements.

Section 1.7

Calculate mass, volume, or density using the formula for density.

Predict whether a substance will float or sink in another substance based on density.

Section 1.8

Calculate the kinetic energy of a moving object.

Convert between common energy units.

Section 1.9

Specify the number of significant figures in a measurement.

Evaluate the level of accuracy and precision in a data set.

Report a measurement to the appropriate number of significant figures.

Section 1.10

Report the answer of a mathematical calculation to the correct number of significant figures.

Round a measurement to a specified number of significant figures.

Section 1.11

.

Change a measurement into different units using appropriate conversion factors.

1

Chapter 1—Chemical Tools: Experimentation and Measurement

Lecture Outline Introduction 1.1. The Scientific Method in a Chemical Context: Improved Pharmaceutical Insulin A. Scientific method – systematic approach to research 1. Observations – qualitative descriptions or quantitative measurements 2. Hypothesis – possible explanation of the observations 3. Experimentation – testing of the hypothesis 4. Theory – consistent explanation of known observations and logical interpretations of experimental results; has predictive power and can be modified or replaced B. Scientific method applied to Humalog® 1. Observation – aggregation of injected insulin lessens its ability to act like insulin produced by the human body 2. Hypothesis – the chemical structure of insulin led to this problem and that modifying its chemical structure could change its behavior in the body 3. Experimentation – insulin analogs produced and tested; Humalog® developed which mimics the behavior of insulin produced by the human body 4. Theory – modeling showed that the last five amino acids in the B-chain were important in aggregation but not biological activity 1.2. Experimentation and Measurement1 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 2. mega (M); factor = 106 3. kilo (k); factor = 103 4. deci (d); factor = 10–1 5. centi (c); factor = 10–2 6. milli (m); factor = 10–3 7. micro (); factor = 10–6 8. nano (n); factor = 10–9 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 1.3. Mass and Its Measurement2 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 1.4 Length and Its Measurement3 A. Meter 1. Standard SI unit of length 2. Distance traveled by light in a vacuum in 1/299,792,458th of a second 1.5. Temperature and Its Measurement4 Test Item Questions: Multiple Choice 4 – 10, Algorithmic 1 – 13, Short Answer 3 – 4 Test Item File Questions: Multiple Choice 11 – 15, Algorithmic 14 – 20, Short Answer 5 3 Test Item File Questions: Multiple Choice 16 – 19, Algorithmic 21 – 26, Short Answer 6 4 Test Item File Questions: Multiple Choice 20 – 26, Algorithmic 27 – 29, Short Answer 7 1 2

2

Chapter 1—Chemical Tools: Experimenation and Measurement A. Common unit – Celsius degree (oC) 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 æ 9 oF ö o F = ç o ´ oC÷ + 32 o F è5 C ø o

C =

5 oC  9 oF

( F - 32 F ) o

o

1.6. Derived Units: Volume and Its Measurement5 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 (m 3) 2. Commonly used measurements: a. Cubic decimeter (dm3) = metric liter (L) b. Cubic centimeter (cm3) = metric milliliter (mL) 1.7. Derived Units: Density and Its Measurement6 A. Density – intensive property relating mass to volume 1. Expressed in units of g/cm3 or g/mL 2. Temperature dependent property 3. Useful property – allows conversion of a liquid’s volume to mass 1.8. Derived Units: Energy and Its Measurement7 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 Test Item File Questions: Multiple Choice 27, Algorithmic 30 – 36, Short Answer 8 – 9 Test Item File Questions: Multiple Choice 28 – 37, Algorithmic 38 – 41, Short Answer 10 – 11 7 Test Item File Questions: Multiple Choice 38 – 40, Algorithmic 42 – 44, Short Answer 12 – 13 5 6

.

3

Chapter 1—Chemical Tools: Experimentation and Measurement 3. Nutritional calorie (Calorie) 4. 1 Cal = 1000 cal = 1 kcal = 4.184 kJ 1.9. Accuracy, Precision, and Significant Figures in Measurement8 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 1.10. Rounding Numbers9 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 1.11. Calculations: Converting from One Unit to Another 10 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 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 Unique Properties of Nanoscale Materials? 11

Test Item File Questions: Algorithmic 45 – 53, Short Answer 14 – 15 Test Item File Questions: Multiple Choice 41 – 44, Algorithmic 54 – 61, Short Answer 16 10 Test Item File Questions: Multiple Choice 45 – 49, Algorithmic 62 – 66, Short Answer 17 11 Test Item File Questions: Algorithmic 67 – 68 8 9

4

Chapter 1—Chemical Tools: Experimenation and Measurement

.

5

Chapter 1—Chemical Tools: Experimentation and Measurement

CHAPTER 1 CHEMICAL TOOLS: EXPERIMENTATION 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 1.1

A theory provides the best explanation based on available information. Additional data can result in modification or replacement of the theory.

Section 1.2

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 1.2

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 1.6

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 1.7

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 1.7

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 1.9

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

Section 1.9

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 1.10

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

Section 1.10

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 1.10

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.

6

Chapter 1—Chemical Tools: Experimenation and Measurement Section 1.11

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 1.2

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

Section 1.7

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. 126-127. Regular and diet soft drinks are identified by differences in their densities.

Section 1.7

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

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

Section 1.7

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 1.9

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

Section 1.9

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

Section 1.9

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 1.9

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

Section 1.9

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

Section 1.9

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

.

7

Chapter 1—Chemical Tools: Experimentation and Measurement

8

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

Use symbols to represent element names.

Section 2.2

Identify the location of metals, nonmetals, and semimetals on the periodic table.

Indicate the atomic number, group number, and period number for an element whose position in the periodic table is given.

Identify groups as main group, transition metal group, or inner transition metal group.

Section 2.3

Specify the location and give examples of elements in the alkali metal, alkaline earth metal, halogen, and noble gas groups.

Use the properties of an element to classify it as metal, nonmetal, or semimetal and give its location in the periodic table.

Section 2.4

Determine the mass of the products in a reaction using the law of mass conservation.

Section 2.5

Demonstrate the law of multiple proportions using mass composition of two compounds of the same elements.

Section 2.6

Describe Thomson’s cathode-ray experiment and what it contributed to the current model of atomic structure.

Describe Millikan’s oil drop experiment and what it contributed to the current model of atomic structure.

Section 2.7

Describe Rutherford’s gold foil experiment and what it contributed to the current model of atomic structure.

Describe the structure and size of the atom.

Calculate the number of atoms in a sample given the size of the atom.

Section 2.8

Determine the mass number, atomic number, and number of protons, neutrons, and electrons from an isotope symbol.

Write the isotope symbol for elements.

Section 2.9

Calculate atomic weight given the functional abundance and mass of each isotope.

Convert between grams and numbers of moles or atoms using molar mass and Avogadro’s number.

Identify an element given the mass and number of atoms or moles.

Section 2.10

.

Classify molecular representations of matter as a mixture, pure substance, element, or compound.

10

Chapter 2— Atoms, Molecules, and Ions

Convert between structural formulas, ball-and-stick models, and chemical formulas.

Section 2.11

Classify bonds as ionic or covalent.

Determine the number of electrons and protons from chemical symbol and charge.

Match the molecular representation of an ionic compound with its chemical formula.

Section 2.12

Convert between name and formula for binary ionic compounds.

Convert between formula and name for ionic compounds with polyatomic ions.

Convert between name and formula for binary molecular compounds.

Lecture Outline 2.1. Chemistry and the Elements1 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 2.2. Elements and the Periodic Table2 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)

1 2

Test Item File Questions: Multiple Choice 1 – 3, Algorithmic 1 – 5, Short Answer 1 – 2 Test Item File Questions: Multiple Choice 4 – 7, Algorithmic 9, Short Answer 3 – 5 11

Chapter 2—Atoms, Molecules, and Ions

4. Inner Transition Metal (or Rare Earth) Elements (14 groups shown separately at the bottom of the table) 2.3. Some Common Groups of Elements and Their Properties3 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 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 2.4. Observations Supporting Atomic Theory: The Conservation of Mass and the Law of Definite Proportions4 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.

2.5. The Law of Multiple Proportions and Dalton’s Atomic Theory5 A. Dalton’s Atomic Theory 1. Elements made of tiny particles called atoms 2. Each element characterized by the mass of its atoms Test Item File Questions: Multiple Choice 8 – 15, Algorithmic 10 – 33, Short Answer 6 – 8, Test Item File Questions: Multiple Choice 16 – 19, Algorithmic 34 – 38, Short Answer 9 5 Test Item File Questions: Multiple Choice 20 – 28, Algorithmic 39, Short Answer 10 3 4

.

12

Chapter 2— Atoms, Molecules, and Ions

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.

2.6. Atomic Structure: Electrons6 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. 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 2.7. Atomic Structure: Protons and Neutrons7 A. Rutherford directed a beam of alpha particles at a thin gold foil.

1. Alpha () 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  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 2.8. Atomic Numbers8 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 Test Item File Questions: Multiple Choice 29 – 30, Short Answer 11 Test Item File Questions: Multiple Choice 31 – 34, Algorithmic 40, Short Answer 12 8 Test Item File Questions: Multiple Choice 35 – 50, Algorithmic 41 – 49, Short Answer 13 – 18 6 7

13

Chapter 2—Atoms, Molecules, and Ions

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.

2.9. Atomic Weights and the Mole9 A. Unified atomic mass unit (u) previously the atomic mass unit (amu) 1. Exactly 1/12th the mass of an atom of 126 C 2. 1 u = 1.660 539  10–24 g 3. Unit also called a Dalton (Da) B. Isotopic mass 1. Mass of an atom in atomic mass units 2. Numerically close to the atom’s mass number C. Atomic massweight values – weighted averages for the naturally occurring mixtures of isotopes 1. Average atomic mass of an element (atomic weight) = (mass of each isotope  the fraction of the isotope) 2.  used for the term “the sum of” 3. Use atomic weights 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 weight.

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 weight in grams 2. Mass of 6.022  1023 atoms of an element 3. Serves as a conversion factor between numbers of atoms and mass 2.10. Mixtures and Chemical Compounds; Molecules and Covalent Bonds 10 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 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 9

Test Item File Questions: Multiple Choice 51 – 55, Algorithmic 50 – 54, Short Answer 19 – 22 Test Item File Questions: Multiple Choice 56, Algorithmic 55 – 62, Short Answer 23

10

.

14

Chapter 2— Atoms, Molecules, and Ions

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 2.15 and 2.16 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 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. 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 2.5 b. Most names end with the suffix ite or ate 11 12

Test Item File Questions: Multiple Choice 57 – 60, Algorithmic 63 – 79, Short Answer 24 – 26 Test Item File Questions: Multiple Choice 61 – 77, Algorithmic 80 – 111, Short Answer 27 – 28 15

Chapter 2—Atoms, Molecules, and Ions

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.

C. Binary molecular compounds – molecular compounds (nonmetals) containing only two elements 1. One of the elements is more cationlike – takes the name of the element 2. One of the elements is more anionlike – 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 2.6) – 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 How is the Principle of Atom Economy Used to Minimize Waste in a Chemical Synthesis?

.

16

Chapter 2— Atoms, Molecules, and Ions

CHAPTER 2 ATOMS, MOLECULES, AND IONS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 2.1

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

Section 2.2

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 2.2

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

Section 2.3

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 2.3

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 2.5

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 2.7

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 2.9

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 2.9

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

Section 2.9

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 2.10

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.

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.

17

Chapter 2—Atoms, Molecules, and Ions

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 2.2

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

Section 2.8

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

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

Section 2.1

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

Section 2.2

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

Section 2.4

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 2.5

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

Section 2.7

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

Section 2.7

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

Educ., Vol. 66, 1989, 738.

Section 2.8

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

Section 2.9

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

Section 2.9

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

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

Section 2.9

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 2.12

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.

.

18

Chapter 2— Atoms, Molecules, and Ions

19

CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Chapter Learning Goals for Students Section 3.1 and Section 3.2

Visualize bonds broken and formed in a chemical reaction and relate numbers of molecules or atoms to the balanced reaction.

Balance a chemical reaction, given the formulas of reactants and products

Section 3.3

Calculate formula weight, molecular weight, and molar mass, given a chemical formula or structure.

Interconvert between mass, moles, and molecules or atoms of a substance.

Relate the amount (moles or mass) of reactants and products in a balanced equation using stoichiometry.

Section 3.4

Calculate percent yield, given amounts of reactants and products.

Section 3.5

Visualize the relative amounts of atoms or molecules in the reactants and products of a balanced reaction.

Determine which reactant is limiting and calculate the theoretical yield of the product and the amount of excess reactant.

Calculate the percent yield when one reactant is limiting.

Section 3.6

Calculate the percent composition given a chemical formula or structure.

Determine the empirical and molecular formula given the mass percent composition and molecular weight of a compound.

Section 3.7

Determine the empirical and molecular formula, given combustion analysis data and molecular weight.

Section 3.8

Determine the molecular weight of a substance, given a mass spectrum.

Determine empirical and molecular formula using both mass and combustion analysis data.

Identify a compound using a molecular weight measured with a high accuracy mass spectrometer.

22

Chapter 3—Mass Relationships in Chemical Reactions

Lecture Outline 3.1. Representing Chemistry on Different Levels1 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 3.2. Balancing Chemical Equations2 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 3.3. 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 weight – sum of atomic weights of all atoms in a molecule b. Formula weight – sum of atomic weights 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 weight 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 1Test Item File Questions: Multiple Choice 1 – 3 2Test Item File Questions: Multiple Choice 4 – 9, Algorithmic 2 – 5, Short Answer 1 – 3 3Test Item File Questions: Multiple Choice 10 – 34, Algorithmic 6 – 36, Short Answer 4 – 12

23

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 Inquiry – Can Alternative Fuels Decrease Carbon Dioxide Emissions?

3.6. Percent Composition and Empirical Formulas6 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 4Test Item File Questions: Multiple Choice 35 – 36, Algorithmic 37 – 42, Short Answer 13 – 14 5Test Item File Questions: Multiple Choice 37 – 45, Algorithmic 43 – 51, Short Answer 15 – 17 6Test Item File Questions: Multiple Choice 46 – 49, Algorithmic 52 – 59, Short Answer 18 – 19

24

Chapter 3—Mass Relationships in Chemical Reactions

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 2. May be a multiple of the empirical formula: Molecular mass Multiple = Empirical formula mass E. Can derive percent composition from chemical formula 3.7. Determining Empirical Formulas: Elemental Analysis 7 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 CO 2 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 3.8. Determining Molecular Masses: Mass Spectrometry8 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 7Test Item File Questions: Multiple Choice 50 – 56, Algorithmic 60 – 61, Short Answer 20 8Test Item File Questions: Multiple Choice 57 – 59

25

Chapter 3—Mass Relationships in Chemical Reactions

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 weights determined to as many as seven significant figures

26

Chapter 3—Mass Relationships in Chemical Reactions

CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 3.2

Equations may be balanced using fractions. H 2(g) + ½ H2(g) → H2O(l) is a balanced equation.

Section 3.2

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

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 3.7

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.

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.

27

Chapter 3—Mass Relationships in Chemical Reactions

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 3.6

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 3.8

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 3.2

William C. Herndon, “On Balancing Chemical Equations: Past and Present (A Critical Review and Annotated Bibliography),” J. Chem. Educ., Vol. 74, 1997, 13591362.

Section 3.2

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.

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.

28

Chapter 3—Mass Relationships in Chemical Reactions

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

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

Section 3.6

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 3.6

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

29

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

Calculate the molarity of a solution given the mass of solute and total volume.

Calculate the amount of solute in a given volume of solution with a known molarity.

Describe the proper technique for preparing solutions of known molarity.

Section 4.2.

Calculate the concentration of a solution that has been diluted.

Describe the proper technique for diluting solutions.

Section 4.3

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

Calculate the concentration of ions in a strong electrolyte solution.

Section 4.4

Classify reactions as precipitation, acid-base neutralization, or oxidation-reduction (redox) reaction.

Section 4.5

Write ionic and net ionic equations and identify spectator ions given the molecular equation.

Section 4.6

Use solubility guidelines (Table 4.2) to predict the solubility of an ionic compound in water.

Predict whether a precipitation reaction will occur when aqueous salt solutions are mixed and write the ionic and net ionic equations.

Section 4.7

Convert between name and formula for an acid.

Classify acids as strong or weak based on the molecular picture of dissociation.

Write the ionic equation and net ionic equation for an acid–base neutralization reaction.

Section 4.8

Convert between solute moles, and volume using molarity in stoichiometry calculations.

Section 4.9

Determine the concentration of a solution using titration data.

Visualize the substances present in solution during a titration procedure..

Section 4.10

Assign oxidation numbers to each atom in a chemical species.

Section 4.11

Identify redox reactions, the oxidizing agents, and the reducing agents.

Section 4.12

Use the location of elements in the periodic table and activity series to predict if a redox reaction will occur.

Develop an activity series and predict if a redox reaction will occur based on experimental data provided.

32

Chapter 4—Reactions in Aqueous Solution

Section 4.13

Use a redox titration to determine the concentration of an oxidizing or reducing agent in solutions.

Section 4.14

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

.

33

Chapter 4—Reactions in Aqueous Solution

Lecture Outline 4.1. Solution Concentrations: Molarity1 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 4.2. Diluting Concentrated Solutions2 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    Vf L  L   L 

()

()

4. Note: This equation should be used only for problems that deal with dilution of a solution!

4.3. Electrolytes in Aqueous Solution3 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 4.1 4.4. Types of Chemical Reactions in Aqueous Solution 4 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 4.5. Aqueous Reactions and Net Ionic Equations5 A. Molecular equations – reactant and product formulas written as if they were molecules Test Item File Questions: Multiple Choice 1 – 4, Algorithmic 1 – 4, Short Answer 1 – 4 Test Item File Questions: Multiple Choice 5 – 8, Algorithmic 5 – 8, Short Answer 5 3 Test Item File Questions: Multiple Choice 9 – 17, Algorithmic 9 – 19, Short Answer 6 – 7 4 Test Item File Questions: Multiple Choice 18 – 21, Algorithmic 20 – 22, Short Answer 8 – 9 5 Test Item File Questions: Multiple Choice 22 – 26, Algorithmic 23 – 25, Short Answer 10 – 11 34 1 2

Chapter 4—Reactions in Aqueous Solution

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 4.6. Precipitation Reactions and Solubility Guidelines 6 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 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 4.7. Acids, Bases, and Neutralization Reactions7 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. Common acid and bases, Table 4.3 D. 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 Test Item File Questions: Multiple Choice 27 – 37, Algorithmic 26 – 38, Short Answer 12 – 13 Test Item File Questions: Multiple Choice 38 – 55, Algortihmic 39 – 47, Short Answer 14 – 18 .

35 6 7

Chapter 4—Reactions in Aqueous Solution

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 E. 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) 4.8. 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) 4.9. Measuring the Concentration of a Solution: 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 4.10. Oxidation-Reduction (Redox) Reactions10 A. Oxidation and reduction 1. Oxidation a. Loss of one or more electrons b. Increase in oxidation number 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 4.11. Identifying Redox Reactions11 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

Test Item File Questions: Multiple Choice 56 – 62, Algorithmic 48 – 52, Short Answer 19 Test Item File Questions: Algorithmic 53 10 Test Item File Questions: Multiple Choice 63 – 68, Algorithmic 54 – 60, Short Answer 20 – 21 11 Test Item File Questions: Multiple Choice 69 – 74, Algorithmic 61, Short Answer 22 – 23 36 8 9

Chapter 4—Reactions in Aqueous Solution

4.12. The Activity Series of the Elements12 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 4.13. Redox Stoichiometry Titrations13 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 B. Strategy for redox titrations – textbook Figure 4.9: 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 4.14. Some Applications of Redox Reactions14 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 Inquiry – How Do Sports Drinks Replenish the Chemicals Lost in Sweat?

Test Item File Questions: Multiple Choice 75 – 79, Algorithmic 62 – 69, Short Answer 24 – 27 Test Item File Questions: Multiple Choice 80 – 83, Algorithmic 70 – 73 14 Test Item File Questions: Multiple Choice .

37 12 13

Chapter 4—Reactions in Aqueous Solution

CHAPTER 4 REACTIONS IN AQUEOUS SOLUTION Teaching Tips, Points of Emphasis, and Common Misconceptions Section 4.1

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 4.2

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 4.3

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 4.3

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

Section 4.3

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 4.3

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 4.5

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 4.5

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 4.6

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 4.6

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

Section 4.7

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 4.7

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

Aquated proton is actually a complex mixture of hydrates, including H 9O4+.

Section 4.7

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.

38

Chapter 4—Reactions in Aqueous Solution

Section 4.7

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

Section 4.7

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.

Section 4.7

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 4.10

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 4.10

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

Section 4.10

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 4.10

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

Section 4.11

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 4.12

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 4.12

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

Lecture/Laboratory Demonstration References Section 4.3

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.

Section 4.6

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.

.

39

Chapter 4—Reactions in Aqueous Solution

Section 4.6

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 4.10

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 4.10

Lee R. Summerlin and James. L. Ealy, Jr., “Oxidation States of Manganese: Mn 7+, Mn6+, Mn4+, and Mn2+,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 133-134.

Section 4.12

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 4.12

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 4.12

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 4.12

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 4.1

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

Section 4.3

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

Section 4.3

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

Section 4.5

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

Section 4.6

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

Educ., Vol. 61, 1984, 765.

40

Chapter 4—Reactions in Aqueous Solution

Section 4.7

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

Section 4.9

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 4.10

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

.

41

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

Label the wavelength, frequency, and amplitude in an electromagnetic wave and understand their meaning.

Interconvert between wavelength and frequency of electromagnetic radiation.

Section 5.2

Calculate the energy of electromagnetic radiation in units of J/photon or kJ/mol, when given the frequency or wavelength.

Describe the photoelectric effect and explain how it supports the theory of particlelike properties of light. (Figures 5.4 and 5.5) Calculate the frequency or wavelength of radiation needed to produce the photoelectric effect given the work function of a metal.

Section 5.3

Describe the difference between a continuous and line spectrum. (Figure 5.7)

Section 5.4

Compare the wavelength and frequency of different electron transitions in the Bohr model of the atom.

Relate wavelengths calculated using the Balmer–Rydberg equation to energy levels in the Bohr model of the atom.

Section 5.5

Calculate the wavelength of a moving object using the de Broglie equation.

Explain why the wavelength of macroscopic objects is not observed.

Section 5.6

Calculate the uncertainty in the position of the moving object if the velocity is known.

Section 5.7

Identify and write valid sets of quantum numbers that describe electrons in different types of orbitals.

Section 5.8

Identify an orbital based on its shape and describe it using a set of quantum numbers.

Visualize the nodal planes in different types of orbitals and different shells.

Section 5.9

Assign a set of four quantum numbers for electrons in an atom.

Section 5.10

Explain how electron shielding gives the order of subshells from lowest to highest in energy. (Figure 5.9) Predict the order of filling of subshells based upon energy.

Section 5.11 and Section 5.12

Assign the electron configuration to atoms in the ground state.

41

Chapter 5—Periodicity and the Electronic Structure of Atoms

Section 5.13

Draw orbital filling diagrams for the ground state of an atom and determine the number of unpaired electrons.

Identify atoms from orbital filling diagrams or electron configurations.

Section 5.14

Explain the periodic trend in atomic radii.

Predict the relative size of atoms based upon their position in the periodic table.

42

Chapter 5—Periodicity and the Electronic Structure of Atoms

Lecture Outline 5.1. The Nature of Radiant Energy 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 () – the number of peaks (maxima) that pass by a fixed point per unit time (s–1 or Hz) 2. Wavelength () – 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) 4. Intensity of radiant energy is proportional to amplitude 2 D. Speed of light (c) – rate of travel of all electromagnetic radiation in a vacuum 1. c = 3.00  108 m/s 2. Wavelength  Frequency = Speed (m)   (s–1) = c (m/s) 3. Frequency and wavelength inversely related: c c  = or ν = ν  a. long ; low ν b. short ; high ν 5.2. Particlelike Properties of Radiant Energy: The Photoelectric Effect and Planck’s Postulate2 A. The photoelectric effect can be seen when light of a minimum frequency strikes some metal surfaces resulting in electron emission.

B. Einstein postulated that light behaves like a stream of particles called photons.

1. Based on Planck’s Postulate, the energy of a photon is quantized—occurs in a discrete quantity. hc a. E = h ; E = ; h = 6.626 x 10–34 Js (Planck’s constant)



b. High-energy radiation – higher , shorter  c. Low-energy radiation – lower , longer  C. The work function, , is the minimum amount of energy required to cause electron emission.

1. Low intensity, high frequency light results in few electrons emitted.

2. High intensity, high frequency light results in many electrons emitted.

D. Einstein concluded that the nature of light is more complex than previously believed.

1. Light can behave as waves.

2. Light can behave as small particles—photons with a quantized or discrete amount of energy.

5.3. The Interaction of Radiant Energy with Atoms: Line Spectra3 A. Individual energetically-excited atoms emit light 1. Provides clue to atomic makeup 2. Consists of only a few  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 æ1 1ö 1 = Rç 2 - 2 ÷ l è2 n ø Test Item File Questions: Multiple Choice 1 – 6, Algorithmic 1, Short Answer 1 – 6 Test Item File Questions: Multiple Choice 7 – 10, Algorithmic 2, Short Answer 7 – 9 3 Test Item File Questions: Multiple Choice 11 – 14, Algorithmic 3, Short Answer 10 – 11 1 2

Chapter 5—Periodicity and the Electronic Structure of Atoms 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 æ1 1ö 1 = R ç 2 - 2 ÷ where n > m l è2 n ø 5.4. The Bohr Model of the Atom: Quantized Energy 4 A. The hydrogen atom has a small, highly charged nucleus with electrons in specific orbitals, like planets.

1. Each orbital has a specific energy.

2. Electron energies are quantized – have only specific energy values associated with the orbital.

B. Addition of energy to an atom causes an electron to transition from a lower to a higher energy orbital.

1. Absorption of energy results in an excited atom.

2. Excited atom is unstable. a. Electron returns to a lower energy level by emission of an amount of energy equal to the energy difference between the higher and lower energy levels. b. Since energy levels are quantized, the energy of the emitted light is of only certain frequencies 3. Calculate the energy differences between levels by measuring the frequencies or wavelengths emitted.

C. The Bohr model failed to predict line spectra of other atoms.

D. The model does not provide an accurate depiction of the electron’s location.

5.5. Wavelike Properties of Matter: deBroglie’s Hypothesis5 A. Einstein – relationship between mass and  E hc / l h m= 2 = 2 = lc c c B. de Broglie – matter can behave in some respects like light 1. Both light and matter wavelike as well as particlelike 2. Relationship between  of an electron or of any other particle or object of mass m moving at velocity v h h m= ; l= lv mv C. Dual wave/particle description of light and matter 1. Wave properties of macroscopic objects have not been observed.

2. On the atomic scale, matter behaves differently. Diffraction (wave characteristic) of electrons has been observed.

5.6. The Quantum Mechanical Model of the Atom: Heisenberg Uncertainty Principle6 A. 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 B. Heisenberg Uncertainty Principle – both position (x) and momentum (mv) of an electron cannot be known beyond a certain level of precision h (Dx)(Dmv) ³ 4p 1. Cannot know both the position and the momentum of an electron with a high degree of certainty Test Item File Questions: Multiple Choice 15 – 22, Algorithmic 4 Test Item File Questions: Multiple Choice 23 – 26, Algorithmic 5 – 6, Short Answer 12 6 Test Item File Questions: Multiple Choice 27 – 30, 44 4 5

Chapter 5—Periodicity and the Electronic Structure of Atoms

2. If momentum is known with a high degree of certainty, a. mv is small b.  x (position of the electron) is large 3. If the exact position of the electron is known, a. x is small b. mv is large 5.7. The Quantum Mechanical Model of the Atom: Orbitals and Quantum Numbers7 A. Quantum mechanical model of atomic structure 1. Mathematical form is a wave equation 2. Wave function () or orbital – solution to wave equation a. Has a specific energy b. Contains information about an electron’s position in 3-D space c. 2 – 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,…)

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 5.10 1. Hydrogen – energy levels depend only on n 2. Multielectron atoms – energy levels depend on both n and l 5.8. The Shapes of Orbitals8 A. s orbitals – textbook Figure 5.11 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 5.12) 7 8

Test Item File Questions: Multiple Choice 31 – 41, Algorithmic 7 – 12, Short Answer 13 – 18 Test Item File Questions: Multiple Choice 42 – 44, Algorithmic 13 – 15, Short Answer 19 – 21

Chapter 5—Periodicity and the Electronic Structure of Atoms

B. p orbitals – textbook Figure 5.13 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 5.14 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. 9. Electron Spin and the Pauli Exclusion Principle 9 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 Test Item File Questions: Multiple Choice 45, Algorithmic 16 – 19 Test Item File Questions: Multiple Choice 46 – 49, Algorithmic 20 11 Test Item File Questions: Multiple Choice 50 – 51, Algorithmic 21 46 9

10

Chapter 5—Periodicity and the Electronic Structure of Atoms

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. Anomalous Electron Configurations12 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 Table 13 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 Inquiry – How Does Knowledge of Atomic Emission Spectra Help Us Build More Efficient Light Bulbs?

Test Item File Questions: Multiple Choice 52, Algorithmic 22 – 23, Short Answer 22 Test Item File Questions: Multiple Choice 53 – 57, Algorithmic 24 – 35, Short Answer 23 – 24 14 Test Item File Questions: Multiple Choice 58, Algorithmic 36 – 41, Short Answer 25 12 13

Chapter 5—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 5.2

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 5.2

A photon is a quantum of light energy.

Section 5.4

The Bohr model does not fit with the current understanding of atomic structure.

Section 5.4

For a single atom, transition of an electron from a higher to a lower energy level results in the emission of one photon of light with an energy that corresponds to the difference in energy of the two levels.

Section 5.7

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 5.7

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 5.7

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 5.7

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

Section 5.8

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.

48

Chapter 5—Periodicity and the Electronic Structure of Atoms

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.

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 5.3

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 5.2

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

Analogies for quantized states.

Section 5.6

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

Section 5.6

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

Section 5.6

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

Section 5.6

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

Chapter 5—Periodicity and the Electronic Structure of Atoms

Section 5.7

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

Section 5.7

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.

Section 5.7

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 5.7

John J. Fortman, “Pictorial Analogies VII: Quantum Numbers and Orbitals,” J. Chem. Educ., Vol. 70, 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.

50

CHAPTER 6 IONIC COMPOUNDS: PERIODIC TRENDS AND BONDING THEORY Chapter Learning Goals for Students Section 6.1

Predict the ground-state electron configuration for main group and transition metal ions.

Determine the number of unpaired electrons in a transition metal ion.

Section 6.2

Predict the relative size of anions, cations, and atoms.

Predict the relative size of isoelectronic ions.

Section 6.3

Order elements from lowest to highest ionization energy.

Explain the periodic trend in ionization energy.

Section 6.4

Compare successive ionization energies (higher second, third, fourth, etc.) for different elements.

Identify elements based on values of successive ionization energies.

Section 6.5

Compare the value of the electron affinity for different elements.

Explain the periodic trend in electron affinity.

Section 6.6

Use the octet rule to predict charges on main group ions, electron configurations of main group ions and formulas for ionic compounds.

Section 6.7

Visualize ionic compounds on the molecular level.

Draw a Born-Haber cycle, identifying the energies involved in the calculation of lattice energy for an ionic compound formed from its elements.

Use the Born-Haber cycle to solve for the energy change associated with one of the steps.

Section 6.8

Predict the relative magnitude of lattice energy given the formula or molecular representation of an ionic compound.

50

Chapter 6 — Ionic Compounds: Periodic Trends and Bonding Theory

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 C. Large jumps in successive ionization energies of the elements – textbook Table 6.2 Test Item File Questions: Multiple Choice 1 – 6, Algorithmic 4 – 9, Short Answer 1 – 12 Test Item File Questions: Multiple Choice 7 – 12, Algorithmic 10, Short Answer 13 3 Test Item File Questions: Multiple Choice 13 – 21, Algorithmic 11 – 15, Short Answer 14 4 Test Item File Questions: Multiple Choice 22 – 28, Algorithmic 16 – 21, Short Answer 15 – 16 51 1 2

Chapter 6—Ionic Compounds: Periodic Trends and Bonding Theory 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 ii. Can gain electrons because of Zeff

5 6

Test Item File Questions: Multiple Choice 29 – 35, Algorithmic 22 – 23, Short Answer 17 – 18 Test Item File Questions: Multiple Choice 36, Algorithmic 24 – 31, Short Answer 19 52

Chapter 6 — Ionic Compounds: Periodic Trends and Bonding Theory 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 Solids 7 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 Inquiry – How Has an Understanding of Ionic Compounds Led to the Production of Safer Solvents?

7 8

Test Item File Questions: Multiple Choice 37 – 45, Algorithmic 32 Test Item File Questions: Multiple Choice 46 – 47, Algorithmic 33 – 39, Short Answer 20 53

Chapter 6—Ionic Compounds: Periodic Trends and Bonding Theory

CHAPTER 6 IONIC COMPOUNDS: PERIODIC TRENDS AND BONDING THEORY 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 every 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 6.7

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 6.7

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 6.7

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 6.7

The Born-Haber cycle is an example of Hess’s law, which will be introduced in Chapter 9. It is used to calculate properties that are impossible to measure directly.

54

Chapter 6 — Ionic Compounds: Periodic Trends and Bonding Theory Section 6.8

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

55

CHAPTER 7 COVALENT BONDING AND ELECTRON-DOT STRUCTURES Chapter Learning Goals for Students Section 7.1

Describe the difference between an ionic and a covalent bond.

Describe the changes in energy that occur as two nuclei approach to form a covalent bond (Figure 7.2).

Name a covalent compound given the chemical formula.

Section 7.2

Predict trends in bond length and bond dissociation energy based on bond order and atomic size.

Section 7.3

Rank elements by increasing value of electronegativity.

Classify bonds as nonpolar covalent, polar covalent, or ionic.

Visualize regions of high and low electron density in a polar covalent bond.

Predict trends in bond dissociation energy based on both atomic size and polarity of the bond.

Section 7.4

Explain the different physical properties of ionic and covalent compounds.

Section 7.5

Draw electron-dot structures by using valence electrons to give all atoms (except H) an octet.

Section 7.6

Use the five-step procedure for drawing electron-dot structures for all molecules including those with expanded octets and those containing multiple bonds.

Section 7.7

Draw electron-dot structures for radicals.

Section 7.8

Draw electron-dot structures for molecules with more than one central atom.

Section 7.9

Draw resonance structures and use curved arrows to depict how one structure can be converted to another.

Section 7.10

Calculate the formal charge on each atom in an electron-dot structure.

Use the formal charge to evaluate the contribution of different resonance structures to the resonance hybrid.

59

Chapter 7—Covalent Bonding and Electron-Dot Structures

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. Bonds between the same pairs of atoms in different compounds have similar bond lengths. c. Can predict from atomic radii 7.2. Strengths of Covalent Bonds 2 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, H = – 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.

D. Shorter bonds are typically stronger bonds.

1. Bond energies decrease on going down the column for different members of the same family bonded to a given atom – as atomic size increases, bond length increases and bond strength decreases 2. Multiple bonds are shorter and stronger than single bonds. a. Typically D (single bond) < D (double bond) < D (triple bond) 7.3. Polar Covalent Bonds: Electronegativity3 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: – ( represents partial charge formed.)

3. Less electronegative atom: + 4. The more polar the bond, the greater the bond dissociation energy.

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 (EN  0.4) – form nonpolar bonds 2. Atoms whose electronegativities differ by more than two (EN > 2 – form ionic bonds 3. Atoms whose electronegativities differ by less than two (EN < 2) – form polar covalent bonds 7.4. A Comparison of Ionic and Covalent Compounds 4 A. Ionic compounds 1. High-melting solids Test Item File Questions: Multiple Choice 1 – 3 Test Item File Questions: Multiple Choice 4 – 10, Algorithmic 1, Short Answer 1 3 Test Item File Questions: Multiple Choice 11 – 23, Algorithmic 3 – 9, Short Answer 2 – 6, 4 Test Item File Questions: Multiple Choice 24 – 26, Algorithmic 10 – 11, Short Answer 7 – 8 60 1 2

Chapter 7—Covalent Bonding and Electron-Dot Structures 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.

7.5. Electron-Dot Structures: The Octet Rule5 A. Electron-dot structure – atom’s valence-electron distribution in a molecule B. Sharing electrons in a bond allows atoms to achieve a complete valence shell or noble gas electron configuration.

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. Procedure for Drawing Electron-Dot Structures6 A. Use lines to represent shared electrons in a covalent bond.

1. One line represents the two shared electrons of a single bond.

2. Two lines represent the four shared electrons of a double bond.

3. Three lines represent the six shared electrons of a triple bond.

B. Guidelines.

1. Find the number of valence electrons in the molecule or ion. a. Find the sum of the valence electrons for all atoms. b. Add an additional electron for each negative charge. c. Subtract an electron for each positive charge.

2. Connect the atoms with single lines using the following guidelines. a.The central atom is usually the one with the lowest electronegativity (except hydrogen which is never a central atom).

Test Item File Questions: Multiple Choice 27 – 29, Short Answer 9 Test Item File Questions: Multiple Choice 30 – 37, Algorithmic 12 – 20, Short Answer 10 – 11 .

61 5 6

Chapter 7—Covalent Bonding and Electron-Dot Structures b. Hydrogen and the halogens usually only form a single bond. c. Elements in the second row form the number of bonds given in Table 7.4 d. Elements in the third row and lower can have expanded octets. e. Elements in Group 3A often have less than eight electrons (are electron deficient).

3. Subtract the number of electrons used to make bonds from the total above. Assign remaining electrons to terminal atoms, giving each an octet (except for H).

4. Place any remaining electrons on the central atom.

5. If the central atom does not have an octet, form multiple bonds to terminal atoms.

7.7. Drawing Electron-Dot Structures for Radicals A. Radicals are compounds that have an odd number of electrons; one electron is in a half-filled orbital.

1. Very reactive.

2. Found in a large number of important processes, both commercial and physiological.

7.8. Electron-Dot Structures of Compounds Containing Only Hydrogen and Second-Row Elements7 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. For larger molecules containing several second-row elements, multiple electron-dot structures are often possible. Additional information about atom connectivity will be needed in order to draw these structures 7.9. Electron-Dot Structures and Resonance8 A. Sometimes more than one electron-dot structure needed for molecular description: resonance 1. These electron-dot structures do not accurately represent electronic structure 2. Limitation of Lewis theoryB. 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 a. The number of valence electrons remains constant b. Atom connectivities do not change c. Bonding and non-bonding electrons can shift 3. Double bonds are delocalized 7.10. Formal Charges9 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 3. Like formal charges should not be on adjacent atoms Inquiry – How Do We Make Organophosphate Insecticides Less Toxic to Humans?

Test Item File Questions: Multiple Choice 38 – 39, Algorithmic 21 – 22, Test Item File Questions: Multiple Choice 40 – 42, Algorithmic 23 – 25, Short Answer 12 9 Test Item File Questions: Multiple Choice 43 – 51, Algorithmic 26 – 27, Short Answer 13 – 14 62 7 8

Chapter 7—Covalent Bonding and Electron-Dot Structures

CHAPTER 7 COVALENT BONDING AND ELECTRON-DOT STRUCTURES 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.

There is a general correlation between bond length and bond strength. The shorter the bond, the stronger the bond.

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 electron-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 7.10

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.

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.

.

63

Chapter 7—Covalent Bonding and Electron-Dot Structures 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 7.3

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

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

Section 7.3

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 7.3

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.

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 7.9

Todd P. Silverstein, “The ‘Big Dog-Puppy Dog’ Analogy for Resonance,” J. Chem.

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

Section 7.9

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

Section 7.9

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

64

Chapter 7—Covalent Bonding and Electron-Dot Structures

Section 7.9

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

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

Section 7.10

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

Section 7.10

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

.

65

CHAPTER 8 COVALENT COMPOUNDS: BONDING THEORIES AND MOLECULAR STRUCTURE Chapter Learning Goals for Students Section 8.1

Use the VSEPR model to predict geometry from the total number of charge clouds and lone pairs of electrons around an atom.

Use the VSEPR model to predict bond angles and overall shape of a molecule or ion with one central atom.

Use the VSEPR model to predict bond angles and overall shape of a molecule with more than one central atom.

Section 8.2 Section 8.3 and Section 8.4

Describe the difference between a sigma, , and a pi, , bond.

Determine the type of hybrid orbitals based upon the number of charge clouds around an atom.

Write an electron-dot structure for a molecule and determine hybridization and bond angles for nonterminal atoms.

Identify which orbitals overlap to form  and  bonds in molecules.

Section 8.5

Predict whether a given molecule has a dipole moment and draw its direction.

Interpret electrostatic potential maps of molecules.

Calculate percent ionic character in a bond.

Section 8.6

Identify the types of intermolecular forces experienced by a molecule.

Relate the strength of intermolecular forces to physical properties such as melting point and boiling point.

Sketch the hydrogen bonding that occurs between two molecules.

Section 8.7

Interpret the molecular orbital diagram for first row diatomic molecule or ion.

Section 8.8

Interpret the molecular orbital diagram for a second row diatomic molecule or ion.

Calculate the bond order and predict magnetic properties.

Section 8.9

Draw orbital overlap diagrams for molecules and describe the use of both valence bond theory and molecular orbital theory.

59

Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure

Lecture Outline 8.1. Molecular Shapes: The VSEPR Model1 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 8.1 8.2. Valence Bond Theory2 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 8.3. Hybridization and sp3 Hybrid Orbitals3 A. Hybrid orbitals – combination of wave functions for atomic orbitals that forms a new set of equivalent wave functions B. Combine one s with three p – form four equivalent new orbitals, each named sp3 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.

8.4. Other Kinds of Hybrid Orbitals4 A. Geometries related to hybrid orbitals used 1. Textbook Table 8.2 2. Each electron-cloud geometry has a corresponding set of hybrid orbitals B. Sigma () bond 1. Bond formed by head-on overlap 2. Has shared electrons centered about the axis between the two nuclei C. Pi () 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 8.5. Polar Covalent Bonds and Dipole Moments5 A. Bond dipole 1. A bond with partial positive (+) and partial negative (–) ends 2. Due to an electronegativity difference of the atoms in the bond 3. Represented by +→; indicates direction of electron displacement a. Point of arrow represents – end of dipole b. Crossed end represents + 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 Test Item File Questions: Multiple Choice 1 – 10, Algorithmic 1 – 14, Short Answer 1 – 4 Test Item File Questions: Multiple Choice 11 3 Test Item File Questions: Multiple Choice 12 – 14, Algorithmic 15, Short Answer 5 4 Test Item File Questions: Multiple Choice 15 – 23, Algorithmic 16 – 17, Short Answer 6 – 7 5 Test Item File Questions: Multiple Choice 24 – 30, Algorithmic 18 – 26, Short Answer 8 – 14 60 1 2

Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure 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 () 1. Magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges 2.  = Q  r D. Predict molecular polarity, see Figure 8.8 8.6. Intermolecular Forces6 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. + end of molecule near the anion b. – 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) 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

6

Test Item File Questions: Multiple Choice 31 – 33, Algorithmic 27 – 37, Short Answer 15 – 17 61

Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure 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 8.7. Molecular Orbital Theory: The Hydrogen Molecule7 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  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 * 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 = ½(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 8.8. Molecular Orbital Theory: Other Diatomic Molecules8 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 2s and *2s molecular orbitals 2. Two 2p orbitals on the internuclear axis a. Interact head-on b. Form 2p and *2p molecular orbitals 3. Remaining 2p orbitals perpendicular to the internuclear axis a. Interact in sideways manner b. Form 2p and *2p molecular orbitals 4. Energy levels of molecular orbitals – textbook Figure 8.21 D. Molecular orbital diagrams require some experience to generate.

7 8

Test Item File Questions: Multiple Choice 34 – 35 Test Item File Questions: Multiple Choice 36 – 47, Short Answer 18 – 20 62

Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure 8.9. Combining Valence Bond Theory and Molecular Orbital Theory 9 A. Molecules with resonance structures 1. -bond framework a. Localized b. Describe with valence bond theory 2. -bond framework a. Delocalized b. Describe with molecular orbital theory Inquiry – Why Do Different Drugs Have Different Physiological Responses?

9

Test Item File Questions: Multiple Choice 48 – 50, Short Answer 21 63

Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure

CHAPTER 8 COVALENT COMPOUNDS: BONDING THEORIES AND MOLECULAR STRUCTURE Teaching Tips, Points of Emphasis, and Common Misconceptions Section 8.1

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 8.1

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

Section 8.1

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 8.1

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).

Section 8.1

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

Section 8.1

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

Section 8.2

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 8.4

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 8.5

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 8.6

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 8.7

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 8.5

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.

64

Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure Section 8.6

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

Section 8.6

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

Section 8.6

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

Section 8.6

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 8.6

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 8.6

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.

Literature References Section 8.1

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

Section 8.1

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.

Section 8.1

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 8.1

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

Section 8.3

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

Section 8.3

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

Section 8.5

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

Section 8.5

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

Section 8.5

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

65

Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure Section 8.5

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

Section 8.6

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 8.6

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

Section 8.6

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

Section 8.6

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

Section 8.6

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

Section 8.6

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

Section 8.7

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

Section 8.8

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.

66

CHAPTER 9 THERMOCHEMISTRY: CHEMICAL ENERGY Chapter Learning Goals for Students Section 9.1

Calculate the kinetic energy of an object in motion.

Convert between common units for energy.

Differentiate between the concepts of heat and temperature.

Section 9.2

Identify state functions.

Identify the sign of heat and work.

Section 9.3

Calculate PV work.

Section 9.4

Calculate the internal energy change (E) for a reaction.

Section 9.5 and Section 9.6

Given a thermochemical equation and the amount of reactant or product, calculate the amount of heat transferred.

Classify endo- and exothermic reactions.

Section 9.7

Calculate heat capacities, temperature changes, or heat transfer, using the equation for heat capacity (C), specific heat (c) or molar heat capacity (Cm).

Calculate enthalpy changes in a calorimetry experiment.

Section 9.8

Use Hess’s law to find H for an overall reaction, given reaction steps and their H values.

Section 9.9

Identify standard states of elements.

Write standard enthalpy of formation reactions (Hf) for compounds from their elements.

Use values of standard heats of formation (Hf) for elements and compounds to calculate H for a reaction.

Section 9.10

Use bond dissociation energies to estimate H for a reaction.

Section 9.11

Calculate Hc for various fuels using thermochemical principles such as Hess’s law, calorimetry and/or bond dissociation enthalpies.

Section 9.12 and Section 9.13

Predict the sign of the entropy change (S) given the chemical equation or a molecular diagram.

Using the relationship between the Gibbs free energy (G) and spontaneity, predict the sign of G, H, S. 70

Chapter 9—Thermochemistry: Chemical Energy

Use H and S to determine the temperature at which a reversible system is at equilibrium.

71

Chapter 9—Thermochemistry: Chemical Energy

Lecture Outline 9.1. 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.

9.2. 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. E = 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. E = 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 9.3. Expansion Work3 A. Work Test Item File Questions: Multiple Choice 1, Algorithmic 1 – 3, Short Answer 1 – 7 Test Item File Questions: Multiple Choice 2, Algorithmic 4 – 8, Short Answer 8 – 12 3 Test Item File Questions: Multiple Choice 3 – 6, Algorithmic 9 72 1 2

Chapter 9—Thermochemistry: Chemical Energy

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  V) 3. Expansion of the system. a. System does work on surroundings b. E is negative c. w = – PV i. Work is negative ii. V is positive 4. Contraction of the system a. Surroundings do work on the system b. E is positive c. w = – PV i. Work is positive ii. V is negative 9.4. Energy and Enthalpy4 A. Total energy change of a system 1. E = q + w (q = heat) 2. E = q + (– PV) B. Amount of heat transferred: q = E + PV C. Reactions carried out with constant volume 1. V = 0 2. No PV work done 3. qv = E D. Reactions carried out at constant pressure 1. V  0 2. Energy change due to both heat transfer and PV work 3. qp = E + PV E. Heat of reaction, H 1. Enthalpy change of a system 2. H = E + PV 3. State function – value depends only on the current state of the system 4. H = Hproducts – Hreactants 5. Amount of heat released in a specific reaction depends on actual amounts of reactants.

9.5. Thermodynamic Equations and the Thermodynamic Standard State5 A. Value of enthalpy change, H, 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.

B. Temperature and pressure also must be reported. 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, H C. Standard enthalpy of reaction – enthalpy change measured under standard conditions 1. H values for a given equation a. Equation balanced for number of moles of reactants and products b. All substances are in standard states 4 5

Test Item File Questions: Multiple Choice 7 – 15, Algorithmic 10 – 12, Short Answer 13 – 17 Test Item File Questions: Multiple Choice 16 – 21, Algorithmic 13 – 15 73

Chapter 9—Thermochemistry: Chemical Energy

c. Physical state of each substance specified d. Refers to reaction going in direction written e. Reversing direction of reaction changes sign of H 9.6. Enthalpies of Physical and Chemical Changes6 A. 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. H is positive 3. Exothermic reactions a. Hproducts < Hreactants b. Heat flows to surroundings from system c. H is negative B. 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 9.7. 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= DT 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)  (T) 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)  (T) 9.8. 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 B. Combine individual reactions so that their sums will be the desired reaction – textbook example 9.7

Test Item File Questions: Multiple Choice 22 – 33, Algorithmic 16 – 23, Test Item File Questions: Multiple Choice 34 – 41, Algorithmic 24 – 31, Short Answer 18 – 20 8 Test Item File Questions: Multiple Choice 42 – 44, Algorithmic 32 – 34, Short Answer 21 74 6 7

Chapter 9—Thermochemistry: Chemical Energy

9.9. Standard Heats of Formation9 A. Standard heat of formation 1. Enthalpy change Hof for the hypothetical formation of 1 mole of a substance in its standard state from the most stable forms of its constituent elements in their standard states 2. The most stable forms of all elements in their standard state have Hof = 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 DH o = S DH products - S DH reactants

C. Textbook Appendix B 9.10. Bond Dissociation Enthalpies10 A. Bond dissociation enthalpies 1. Enthalpy changes, H, for corresponding bond-breaking reactions 2. H = D = bond dissociation energy 3. Always positive 4. Always need energy to break a bond B. H = D(bonds broken) – D(bonds formed) C. Textbook Table 7.2 9.11. Fossil Fuels, Fuel Efficiency, and Heats of Combustion11 A. Heat of combustion (Hoc) – amount of energy released on burning a substance B. Fuel efficiency 1. Calculate Hoc 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 9.12. 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. S = Sfinal – Sinitial 4. Sfinal > Sinitial a. S is positive b. System has become more random 5. Sfinal < Sinitial a. S is negative b. System has become less random C. Spontaneous process 1. Favored by decrease in H (negative H) 2. Favored by increase in S (positive S) 9.13. An Introduction to Free Energy13 A. Gibbs free-energy change (G); G = H – TS. Test Item File Questions: Multiple Choice 45 – 49, Algorithmic 35 – 37 Test Item File Questions: Multiple Choice 50 – 53, Algorithmic 38 – 39 11 Test Item File Questions: Multiple Choice 54 – 56 12 Test Item File Questions: Multiple Choice 57 – 64, Algorithmic 40 – 43, Short Answer 22 13 Test Item File Questions: Multiple Choice 65 – 75, Algorithmic 44 – 47, Short Answer 23 – 25 75 9

10

Chapter 9—Thermochemistry: Chemical Energy B. Sign of G used as a criterion for determining spontaneity of a process 1. G negative – spontaneous 2. G positive ≠ nonspontaneous C. Temperature dependence (TS) term for G. 1. Spontaneity of some processes depends on temperature 2. Low temperatures – H dominates and controls spontaneity 3. High temperatures – TS dominates and controls spontaneity D. G = 0 1. Process at equilibrium 2. Balanced between spontaneous and nonspontaneous 3. Temperature at which a nonspontaneous reaction becomes spontaneous DH T= DS Inquiry – How is the Energy Content of New Fuels Determined?

76

Chapter 9—Thermochemistry: Chemical Energy

CHAPTER 9 THERMOCHEMISTRY: CHEMICAL ENERGY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 9.2

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 9.2

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 9.3

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 9.4

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 E is negative.

Section 9.4

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 9.4

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 9.4

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

Section 9.5

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 9.5

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 9.6

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

Section 9.6

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

Section 9.7

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 9.7

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 77

Chapter 9—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 9.7

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

Section 9.8

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

Section 9.8

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

Section 9.9

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 9.10

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

Section 9.10

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

Section 9.10

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 9.12

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

Section 9.13

H has units of kJ, S has much smaller units of J/K. Thus, reaction spontaneity as measured by G = H –TS is usually dominated by H 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 9.6

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 9.7

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 9.7

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.

Section 9.7

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.

78

Chapter 9—Thermochemistry: Chemical Energy

Section 9.7

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 9.7

Lee R. Summerlin 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 9.11

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 9.7

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

Section 9.7

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

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

Section 9.7

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

Section 9.7

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

Section 9.11

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

Section 9.11

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

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

Section 9.11

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

Section 9.11

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

Section 9.12

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

79

CHAPTER 10 GASES: THEIR PROPERTIES AND BEHAVIOR Chapter Learning Goals for Students Section 10.1

Perform calculations converting between different units of pressure.

Describe how a barometer and a manometer measure pressure.

Section 10.2

Use the individual gas laws to calculate pressure, volume, molar amount, or temperature for a gas sample when conditions change.

Section 10.3

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

Section 10.4

Calculate volumes of gases in chemical reactions.

Calculate the density or molar mass of a gas using the formula for gas density.

Section 10.5

Calculate the partial pressure, mole fraction, or amount of each gas in a mixture.

Section 10.6

Use the assumptions in the kinetic-molecular theory of gases to predict behavior.

Calculate the average molecular speed of a gas particle at a given temperature.

Section 10.7

Visualize the process of effusion and diffusion.

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

Section 10.8

Understand the conditions when gases deviate the most from ideal behavior.

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

Section 10.9

Convert between different units used to express the concentrations of pollutants.

Use the gas laws, Dalton’s law and stoichiometry to calculate amounts of pollutant gases in the atmosphere.

Identify the components and causes of photochemical smog.

Section 10.10

Explain the principle of the greenhouse gas effect.

Section 10.11

Describe the trends in greenhouse gas concentrations over time and measured and predicted effects of climate change.

81

Chapter 10—Gases: Their Properties and Behavior

Lecture Outline 1

10.1 Gases and Gas Pressure 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) c. Table 10.2 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 2 10.2. The Gas Laws 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 1Test Item File Questions: Multiple Choice 1 – 11, Algorithmic 1 – 6, Short Answer 1 – 3 2Test Item File Questions: Multiple Choice 12 – 23, Algorithmic 7 – 12, Short Answer 4 – 6

82

Chapter 10—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 – Standard Molar Volume 3 10.3 The Ideal Gas Law 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 4 10.4. Stoichiometric Relationships with Gases 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 5 10.5. Mixtures of Gases: Partial Pressure and Dalton’s Law 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 6 10.6. The Kinetic–Molecular Theory of Gases 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 7 10.7. Gas Diffusion and Effusion: Graham’s Law A. Consequence of constant motion and high velocities of gas particles – gases mix rapidly 1. Diffusion – mixing of different gases a. Random molecular motion 3Test Item File Questions: Multiple Choice 24 – 34, Algorithmic 13 – 25, Short Answer 7 – 9 4Test Item File Questions: Multiple Choice 35 – 47, Algorithmic 26 – 34, Short Answer 10 – 12 5Test Item File Questions: Multiple Choice 48 – 55, Algorithmic 35 – 41, Short Answer 13 – 14 6Test Item File Questions: Multiple Choice 56 – 64, Algorithmic 42 – 51, Short Answer 15 – 18 7Test Item File Questions: Multiple Choice 65 – 68, Algorithmic 52 – 56, Short Answer 19 – 20

83

Chapter 10—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 µ 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 8 10.8. The Behavior of Real Gases 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: æ an2 ö ç P + 2 ÷ V - nb = nRT V ø è 1. a = correction for intermolecular attractions 2. b = correction for molecular volume 9 10.9. The Earth’s Atmosphere and Air Pollution A. Troposphere – region nearest earth’s surface 1. Easily disturbed by human activity 2. EPA sets air quality standards a. Units of measure are ppm and ppb b. Very low levels of concentration B. Primary pollutants 1. Carbon monoxide a. Colorless, odorless, poisonous gas emitted from combustion processes b. Limits the human body’s ability to absorb oxygen 2. Lead a. Toxic metal that accumulates in the bones b. Sources include leaded gasoline and processing of metal ores 3. Particulate pollution a. Extremely small particles and liquid droplets b. Pass into the lungs with adverse health effects c. Contribute to reduced visibility (haze) 4. Sulfur dioxide a. Reactive gas that contributes to particulate pollution and acid rain b. Byproduct of the combustion of fossil fuels.

C. Secondary pollutants

(

)

8Test Item File Questions: Multiple Choice 69 – 70, Short Answer 21 9Test Item File Questions: Multiple Choice 71 – 75, Algorithmic 57, Short Answer 22 - 25

84

Chapter 10—Gases: Their Properties and Behavior

1. Nitrogen dioxide, ozone and photochemical smog a. Photochemical smog is the hazy brown layer over many cities b. Nitrogen dioxide (NO2) and ozone (O3) are harmful compounds whose production is related to combustion process and sunlight.

10 10.10. The Greenhouse Effect A. Absorption of infrared (IR) radiation by the atmosphere resulting in an increase in planetary temperature 1. Naturally occurring phenomenon that affects climate by regulating temperature 2. Enables life as we know it on Earth B. Principle of the greenhouse effect 1. Sun emits ultraviolet (UV) and visible radiation 2. Ozone absorbs much of the UV radiation 3. Absorption of visible radiation results in heating of Earth’s surface 4. IR radiation re-emitted a. Some escapes into space b. Some absorbed by greenhouse gases in the atmosphere 5. Greenhouse gases return some of the IR radiation to Earth, warming it 11 10.11. Climate Change A. Increasing concentrations of greenhouse gases change the thermal balance of the Earth 1. Other factors also part of this complex process 2. Not all areas of the Earth affected the same B. Combustion of fossil fuels has increased the concentrations of greenhouse gases in the atmosphere significantly since the onset of the industrial revolution C. Atmospheric levels of CO2 and CH4 are currently at historically high levels D. A historical correlation is seen between high CO 2 levels and high Antarctic temperatures E. Average global surface temperatures are increasing 1. Arctic sea ice coverage is decreasing 2. Causes significant impact to various ecosystems, possibly leading to species extinction 3. Sea levels are rising F. Climate scientists agree that warming trends are due to human activity Inquiry – Which Gases Are Greenhouse Gases?

10Test Item File Questions: Multiple Choice 76 – 78, Algorithmic 58 - 59 11Test Item File Questions: Algorithmic 60

85

Chapter 10—Gases: Their Properties and Behavior

CHAPTER 10 GASES: THEIR PROPERTIES AND BEHAVIOR Teaching Tips, Points of Emphasis, and Common Misconceptions Section 10.1

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

Section 10.2

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 10.2

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 10.1

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 10.2

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 10.2

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 10.2

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 10.2

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

Section 10.2

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

Section 10.2

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 10.2

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 10.4

Bassam Z. Shakhashiri, “Determination of the Molecular Mass of a Volatile Liquid,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 51-54. A boiling-water bath is 86

Chapter 10—Gases: Their Properties and Behavior

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 10.4

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 10.7

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 10.7

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 10.7

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

Section 10.7

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 10.1

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

Section 10.2

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

72, 1995, 257.

Section 10.2

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

Section 10.2

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

Section 10.4

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

Section 10.4

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

Section 10.6

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.

87

Chapter 10—Gases: Their Properties and Behavior Section 10.6

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

88

CHAPTER 11 LIQUIDS, SOLIDS, AND PHASE CHANGES Chapter Learning Goals for Students Section 11.1

Predict which substance has higher viscosity or surface tension based on molecular structure.

Section 11.2

Use the equation for Gibbs free energy to calculate the temperature or S for a phase change.

Calculate the amount of heat associated with phase changes and draw heating curves.

Section 11.3

Use the Clausius-Clapeyron equation to calculate vapor pressure at varying temperatures or (Hvap).

Section 11.4

Classify types of solids based on their chemical composition and properties.

Section 11.5

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

Section 11.6

Identify a unit cell from a pattern.

Identify the four kinds of spherical packing arrangements in crystalline solids and the three kinds of cubic unit cells.

Calculate the density of a substance, atomic radii of its atoms or molecular mass given its unit cell dimensions.

Section 11.7 Section 11.8 and Section 11.9

Use the unit cell to determine the formula and geometry of ionic compounds.

Use a phase diagram to identify the phase of a substance at a given temperature and pressure.

Sketch a phase diagram given the appropriate data.

88

Chapter 11—Liquids, Solids, and Phase Changes

Lecture Outline 1

11.1. Properties of Liquids 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 2 11.2. Phase Changes between Solids, Liquids, and Gases 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, G 1. G = H – TS 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: H and S both positive 5. Gas → liquid, gas → solid, liquid → solid: H and S both negative 6. Knowing H and S for a phase transition can calculate temperature at which two phases are in equilibrium. a. At equilibrium, G = 0 b. T = H/S C. Heating curve – graphical display of results of adding heat to a sample (See textbook Figure 11.3.)

1. Heating solid H2O a. Added heat increases the temperature of the solid b. Molar heat capacity relates the amount of added heat to the temperature change 2. Melting solid H2O (fusion) a. Melting point - 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 (Hfusion) – amount of energy required to overcome enough intermolecular forces to convert solid into liquid c. Temperature remains constant 3. Heating liquid H2O 4. Vaporizing liquid H2O a. Boiling point - 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 (Hvap) – amount of energy necessary to convert liquid into gas 5. Heating H2O vapor 6. Hvap >> Hfusion 1Test Item File Questions: Multiple Choice 1 – 3, Algorithmic 1 – 3, Short Answer 1 – 2 2Test Item File Questions: Multiple Choice 4 – 12, Algorithmic 4 – 10, Short Answer 3 – 4,

89

a. For vaporization, must overcome all intermolecular forces in compound b. For fusion, must overcome fewer intermolecular forces 3 11.3. Evaporation, Vapor Pressure, and Boiling Point 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  - H  vap  + C, where C = a constant characteristic of each substance ln Pvap =   R T  

( )

2. The vapor pressure at some temperature can be calculated if the vapor pressure at another temperature and Hvap are known.

 P   H vap   1 1  ln  1  =   −   P2   R   T2 T1  3. Knowing vapor pressure at several temperatures, can calculate Hvap of a liquid from H 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 4 11.4. Kinds of Solids 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 11.4 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 3Test Item File Questions: Multiple Choice 13 – 22, Algorithmic 11 – 14, Short Answer 5 4Test Item File Questions: Multiple Choice 23 – 24, Algorithmic 15 – 24, Short Answer 6 – 7

90

Chapter 11—Liquids, Solids, and Phase Changes

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 5 11.5. Probing the Structure of Solids: X–Ray Crystallography 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  of the radiation 3. Interference between two ’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 nl 2. Bragg equation: nl = 2d ´ sin q; d = 2 sin q a.  = known b. Sin  = angle at which incoming rays are reflected (can be measured) c. n = an integer (usually 1) 6 11.6. The Packing of Spheres in Crystalline Solids: Unit Cells 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 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 5Test Item File Questions: Multiple Choice 25 – 26, Short Answer 8 – 9 6Test Item File Questions: Multiple Choice 27 – 40, Algorithmic 25 – 31, Short Answer 10 – 15

91

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 7 11.7. Structures of Some Ionic Solids 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 8 11.8. Structure of Some Covalent Network Solids 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 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 9 11.9. Phase Diagrams A. Change any one state of matter spontaneously into either of the other two, depending on temperature and pressure 7Test Item File Questions: Multiple Choice 41 – 46, Short Answer 16 8Test Item File Questions: Multiple Choice 47, Algorithmic 32 – 35 9Test Item File Questions: Multiple Choice 48 – 51, Algorithmic 36

92

Chapter 11—Liquids, Solids, and Phase Changes

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 Inquiry – How Is Caffeine Removed from Coffee?

93

CHAPTER 11 LIQUIDS, SOLIDS, AND PHASE CHANGES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 11.2

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 11.2

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 11.5

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 11.6

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 11.6

Primitive cubic is also known as simple cubic.

Section 11.6

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 11.6

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 11.6

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 11.7

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 11—Liquids, Solids, and Phase Changes

Lecture/Laboratory Demonstration References Section 11.3

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 11.3

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 11.9

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 11.1

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

Section 11.4

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

Section 11.6

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 11.6

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

Section 11.6

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 11.7

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

Section 11.8

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

Section 11.9

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

Section 11.9

Stephen J. Hawkes, “There Is No Perceptible Inflection at the Triple Point,” J. Chem.

Educ., Vol. 76, 1999, 226.

95

CHAPTER 12 SOLUTIONS AND THEIR PROPERTIES Chapter Learning Goals for Students Section 12.1 and Section 12.2

Describe the types of intermolecular forces involved when a solute dissolves in a solvent.

Predict solubility based on the chemical structure of the solute and solvent.

Describe the effect of ion size and lattice energy on the solubility of ionic compounds.

Section 12.3 Calculate the concentration of a solution in units of mass percent.\ Calculate the concentration of a solution in units of parts per million (ppm) or parts per billion (ppb).

Calculate the concentration of a solution in units of molality.

Convert from one unit of concentration to another.

Section 12.4

Analyze a solubility graph to determine the solubility of a solute at a given temperature.

Use Henry’s Law to calculate the solubility of a gas.

Section 12.5 and Section 12.6

Qualitatively predict changes to the vapor pressure of a solution when a solute is added.

Calculate the vapor pressure of a solution containing a nonvolatile solute.

Calculate the vapor pressure of a solution containing a volatile solute.

Section 12.7

Calculate the amount of boiling point elevation and freezing-point depression for a solution.

Section 12.8

Calculate osmotic pressure for a solution.

Use osmotic pressure or other colligative property to calculate the molecular weight of a solute.

Section 12.9

Interpret the liquid-liquid phase diagram of a solution and apply the diagram to fractional distillation.

99

Chapter 12 – Solutions and Their Properties

Lecture Outline 1

12.1. Solutions 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 12.1 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 2 12.2. Energy Changes and the Solution Process 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. G is negative – spontaneous process and the substance dissolves 2. G is positive – nonspontaneous process and the substance does not dissolve 3. Hsoln – enthalpy of solution a. Difficult to predict b. Can be either negative (exothermic) or positive (endothermic) 4. Ssoln – entropy of solution a. Usually positive b. Increase in molecular randomness during dissolution G. Variations in heats of solution due to interplay of three kinds of interactions 1. Solvent-solvent interactions – positive H a. need to overcome intermolecular forces between solvent molecules b. separate molecules to make room for solute molecules 2. Solute-solute interactions – positive H a. Need to overcome intermolecular forces between solute molecules b. Lattice energy for ionic solids 3. Solvent-solute interactions – negative H 1Test Item File Questions: Multiple Choice 1 – 4, Algorithmic 1 – 2, Short Answer 1 2Test Item File Questions: Multiple Choice 5 – 19, Algorithmic 3 – 10, Short Answer 2 - 3 .

100

Chapter 12 — 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 Hsoln is endothermic or exothermic textbook Figure 12.4 3 12.3. Concentration Units for Solutions 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: a.

b.

2. For dilute aqueous solutions at room temperature, 1 kg solution = 1 L solution 1 mg solute a. 1 ppm = 1 L soln 1 mg 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 Moles of solute Molality (m) = Mass of solvent (kg) 3Test Item File Questions: Multiple Choice 20 – 50, Algorithmic 11 – 35, Short Answer 4 - 10

101

Chapter 12 – Solutions and Their Properties

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 4 12.4. Some Factors That Affect Solubility 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 (unstable) 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 5 12.5. Physical Behavior of Solutions: Colligative Properties 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 6 12.6. Vapor-Pressure Lowering of Solutions: Raoult’s Law 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 D. Solute-solvent intermolecular forces > solvent intermolecular forces; vapor pressure less than predicted by Raoult’s law

4Test Item File Questions: Multiple Choice 51 – 55, Algorithmic 36 – 40, Short Answer 11 – 13, 5Test Item File Questions: Multiple Choice 56, Short Answer 14 – 16, 6Test Item File Questions: Multiple Choice 57 – 62, Algorithmic 41 – 44, Short Answer 17 - 21 .

102

Chapter 12 — Solutions and Their Properties

E. The van’t Hoff factor, i, corrects for the incomplete dissociation of many ionic solutes in solution.

1. Ideal i = 1 for molecular solutes or number of ions the chemical formula of an ionic compound.

2. For actual ionic solutes: i=

moles of ions in solution moles of solute dissolved

F. 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 ) G. Free-energy change accompanies vapor pressure lowering – more negative G, easier the vaporization process H. Liquid → gas: 1. H is positive a. Need energy to overcome intermolecular forces b. Similar value for both solvent and solution 2. S is positive a. Molecular disorder increases b. S for liquid → vapor is smaller for solution than for pure solvent 3. Larger G for solution (subtracting a smaller TS from H) a. Vaporization more difficult b. Lower vapor pressure 7 12.7. Boiling-Point Elevation and Freezing-Point Depression of Solutions 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. Tb = Kb  m  i 2. Tf = Kf  m  i 3. Use molality as the concentration unit because numbers of solute and solvent particles are independent of temperature.

B. Common uses of freezing point depression and boiling point elevation.

1. Sprinkling of salt to melt snow 2. Antifreeze in automobiles 3. De-icing of airplane wings C. Due to an entropy difference between pure solvent and solvent in a solution 1. At equilibrium DH vap Tb = DS vap 2. Svap smaller for a solution, Tb larger 3. At equilibrium DH fus Tf = DSfus 4. Sfusion larger for a solution, Tf smaller.

7Test Item File Questions: Multiple Choice 63 – 73, Algorithmic 45 – 56, Short Answer 22 – 23

103

Chapter 12 – Solutions and Their Properties

8

12.8. Osmosis and Osmotic Pressure 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 C. Osmotic pressure of a solution (  ) – pressure needed to prevent osmotic flow of solvent through a semipermeable membrane 1.  = MRTi 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 D. Molar mass determinations using osmosis 1. Can use any of the four colligative properties 2. Most accurate is osmotic pressure – magnitude of osmosis effect is so great E. Desalination of seawater by reverse osmosis – increase osmotic pressure to force molecules from solution side to solvent side 9 12.9. Fractional Distillation of Liquid Mixtures 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 12.20 B. Represent fractional distillation with a liquid/vapor phase diagram 1. Plot T vs. composition – textbook Figure 12.19 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 Inquiry – How Does Hemodialysis Cleanse the Blood of Patients with Kidney Failure?

8Test Item File Questions: Multiple Choice 74 – 81, Algorithmic 57, Short Answer 24 - 26 9Test Item File Questions: Short Answer 27 .

104

Chapter 12 — Solutions and Their Properties

CHAPTER 12 SOLUTIONS AND THEIR PROPERTIES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 12.2

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

Section 12.2

Ssoln is usually positive; Hsoln can be either negative or positive.

Section 12.2

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 12.3

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 12.3

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 12.3

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 12.4

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 12.6

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 12.1

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 12.2

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 12.2

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

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

Section 12.2

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

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

105

Chapter 12 – Solutions and Their Properties

Section 12.4

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 12.4

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 12.4

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 12.8

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

Section 12.8

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 12.8

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 12.8

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 12.2

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

Section 12.2

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

Educ., Vol. 71, 1994, 156.

Section 12.2

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 12.3

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

Section 12.3

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

Section 12.3

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 12.4

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

106

.

Chapter 12 — Solutions and Their Properties

Section 12.4

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

Section 12.4

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

Section 12.8

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

Section 12.8

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

Section 12.9

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

107

CHAPTER 13 CHEMICAL KINETICS Chapter Learning Goals for Students Section 13.1

Determine the average reaction rate over a specified period of time and estimate instantaneous reaction rate from a graph of concentration versus time.

Relate the rate of consumption of any reactant to the rate of formation of any product using reaction stoichiometry.

Section 13.2

Find the reaction order with respect to each reactant, the overall reaction order, and the units of the rate constant from a given rate law.

Predict the change in reaction rate when the concentration of a reactant changes by a specified amount.

Section 13.3

Determine the rate law and rate constant using initial rate and concentration data.

Section 13.4

Use the integrated rate law to determine the half-life and the concentrations remaining at various times for a zeroth order reaction.

Section 13.5

Use the integrated rate law to determine the half-life and the concentrations remaining at various times for a first order reaction.

Section 13.6

Use the integrated rate law to determine the half-life and the concentrations remaining at various times for a second order reaction.

Determine the reaction order and rate constant graphically.

Section 13.7

Interpret potential energy diagrams and suggest geometries for successful collisions and transition states using collision theory.

Section 13.8

Calculate rate constants, temperatures, and activation energy using the Arrhenius equation.

Section 13.9

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

Section 13.10 and Section 13.11

For mechanisms with an initial slow step (a) predict the rate law for the overall reaction or (b) propose a mechanism that agrees with the experimental rate law.

For mechanisms with an initial fast equilibrium step, predict the rate law.

Section 13.12

Use a molecular diagram for a catalyzed reaction to determine the rate law and propose a mechanism consistent with the rate law.

Identify catalysts and intermediates in the mechanism of a catalyzed reaction.

Interpret a potential energy diagram representing the mechanism in a catalyzed reaction.

108

Chapter 13—Chemical Kinetics

Section 13.13

Classify catalysts as homogeneous or heterogeneous.

109

Chapter 13—Chemical Kinetics

Lecture Outline 1

13.1. Reaction Rates A. Rate of a reaction – how fast the concentration of a reactant or a product changes per unit time D concentration Rate = D 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.

3. General rate of a reaction is defined as the rate of consumption of a reactant or formation of a product divided by its coefficient in the balanced chemical equation.

D. Plot concentration (y-axis) versus time (x-axis).

1. [concentration] and time 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) 2 13.2. Rate Laws and Reaction Order 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 m n a. Rate law: Rate = k  A   B 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 3 13.3. Method of Initial Rates: Experimental Determination of a Rate Law 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.

1Test Item File Questions: Multiple Choice 1 – 10, Algorithmic 1 – 8, Short Answer 1 – 3 2Test Item File Questions: Multiple Choice 11 – 17, Algorithmic 9 – 20, Short Answer 4 - 10 3Test Item File Questions: Multiple Choice 18 – 27, Short Answer 11

110

Chapter 13—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 2 2 = 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.

4 13.4. Integrated Rate Law: Zeroth–Order Reactions 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. Integrated rate law for a zeroth-order reaction, A → products [A] = - kt + [A]o C. Rate is independent of the concentration of the reactant.

D. Plot of [A] versus time gives a straight line if zeroth–order.

1. Slope = – k 2. Intercept = [A]o E. Relatively uncommon; occurs under special circumstances.

5 13.5. Integrated Rate Law: for a First–Order Reactions A. 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) B. Half-life = t1/2 1. Time required for reactant concentration to drop to one-half its initial value 0.693 t1/2 = k 2. For a first-order reaction half-life is constant.

3. Depends only on the rate constant

4Test Item File Questions: Multiple Choice 28 – 29, Algorithmic 21 – 23, Short Answer 12 5Test Item File Questions: Multiple Choice 30 – 41, Algorithmic 24 – 31, Short Answer 13 – 14

111

Chapter 13—Chemical Kinetics

6

13.6. Integrated Rate Law: Second-Order Reactions 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.

1 t1/2 = k[A]o 7

13.7. Reaction Rates and Temperature: The Arrhenius Equation 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 f =e 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.

1. Reactants won’t react unless the orientation of the reaction partners is correct for formation of the transition state.

- Ea / RT

6Test Item File Questions: Multiple Choice 42 – 49, Algorithmic 32 – 35, Short Answer 15 7Test Item File Questions: Multiple Choice 50 – 58, Algorithmic 36 – 37

112

Chapter 13—Chemical Kinetics

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 8

13.8. Using the Arrhenius Equation A. Can determine the activation energy if values of the rate constant are known at different temperatures.

B. Rearrange the equation to get æ -E ö æ 1 ö ln k = çç a ÷÷ ç ÷ + ln A è R øèT ø 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. æ k ö æ -E ö æ 1 1 ö ln çç 2 ÷÷ = çç a ÷÷ çç - ÷÷ è k ø è R ø è T2 T1 ø 9

13.9. Reaction Mechanisms 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. Thermolecular 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 10 13.10. Rate Laws for Elementary Reactions 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.

8Test Item File Questions: Multiple Choice 59 – 65, Algorithmic 38, Short Answer 16 – 18 9Test Item File Questions: Multiple Choice 66 – 70, Algorithmic 39 – 40, Short Answer 19 - 20 10Test Item File Questions: Multiple Choice 71, Algorithmic 41 – 43

113

Chapter 13—Chemical Kinetics

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 13.5 11 13.11. Rate Laws for Overall Reactions 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 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 13.12. Catalysis 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.

13 13.13. Homogeneous and Heterogeneous Catalysts 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 11Test Item File Questions: Multiple Choice 72 – 73, Algorithmic 44, Short Answer 21 12Test Item File Questions: Multiple Choice 74 – 76, Algorithmic 45 – 47, Short Answer 22 13Test Item File Questions: Multiple Choice, Algorithmic 48 – 49, Short Answer 23 – 24

114

Chapter 13—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 14 13.14. Enzyme Catalysis A. Enzymes are large protein molecules that catalyze biological reactions.

1. Larger and more complex than inorganic catalysts.

2. Often catalyze a single reaction involving a single substrate.

3. Activity is measured in turnover number – number of substrate molecules acted on by one molecule of enzyme per second.

B. Lock and key model for enzyme activity.

1. Substrate migrates to the active site in the enzyme.

2. Enzyme-substrate complex forms involving hydrogen bonds and/or other weak intermolecular attractions.

3. Positioning of the substrate in the active site facilitates the chemical reaction.

4. At low substrate concentrations, the reaction is first order in substrate; at higher concentrations the reaction is zero order in substrate.

Inquiry – What Causes the Ozone Hole?

14Test Item File Questions: Algorithmic 50

115

Chapter 13—Chemical Kinetics

CHAPTER 13 CHEMICAL KINETICS Teaching Tips, Points of Emphasis, and Common Misconceptions None

Lecture/Laboratory Demonstration References Section 13.1

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 13.1

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 13.1

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 13.1

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 13.2

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 13.3

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 13.7

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.

Section 13.7

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 13.11

Jose E. Cortes-Figueroa and Deborah A. Moore, “Using CBL Technology and a 116

Chapter 13—Chemical Kinetics

Graphing Calculator to Teach the Kinetics of Consecutive First-Order Reactions,” J. Chem. Educ., Vol. 76, 1999, 635-638.

Section 13.14

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 13.14

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. 101-102. Oxygen gas is catalytically produced in the presence of detergent resulting in the formation of a large quantity of foam.

Section 13.14

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 13.2

Lionello Pobliani and Mario N. Berberan-Santos, “Inflation Rates, Car Devaluation, and Chemical Kinetics,” J. Chem. Educ., Vol. 73, 1996, 950-952.

Section 13.2

Charles J. Marzzacco, “An Analogy to Help Students Understand Reaction Orders,” J. Chem. Educ., Vol. 75, 1998, 482.

Section 13.4

Francisco J. Arnaiz, “Mice in the Box for Zero-Order Kinetics,” J. Chem. Educ., Vol. 76, 1999, 1458.

Section 13.5

Edward Todd Urbansky, “Don’t Be Tricked by Your Integrated Rate Plot,” J. Chem.

Educ., Vol. 78, 2001, 921-923.

Section 13.7

Ilya A. Leenson, “The Arrhenius Law and Storage of Food,” J. Chem. Educ., Vol. 76, 1999, 504-505.

Section 13.7

Keith L. Laidler, “Just What Is a Transition State?” J. Chem. Educ., Vol. 65, 1988, 540-542.

Section 13.7

Kent W. Piepgrass, “Audience-Appropriate Analogies: Collision Theory,” J. Chem.

Educ., Vol. 75, 1998, 72.

Section 13.7

Lee A. Krug, “The Collision Theory and an American Tradition,” J. Chem. Educ., Vol. 64, 1987, 1000.

Section 13.9

John J. Fortman, “Pictorial Analogies XIII: Kinetics and Mechanisms,” J. Chem.

Educ., Vol. 71, 1994, 848-849.

Section 13.9

Arthur M. Last, “Doing the Dishes: An Analogy for Use in Teaching Reaction Kinetics,” J. Chem. Educ., Vol. 62, 1985, 1015-1016.

117

Chapter 13—Chemical Kinetics

Section 13.11

Richard A. Potts, “Auto Analogies,” J. Chem. Educ., Vol. 62, 1985, 579.

Section 13.11

David W. Ball, “Another Auto Analogy: Rate-Determining Steps,” J. Chem. Educ., Vol. 64, 1987, 486-487.

Section 13.12

Doris Kolb, “Catalysis,” J. Chem. Educ., Vol. 56, 1979, 743-747.

Section 13.12

Albert Haim, “Catalysis: New Reaction Pathways, Not Just a Lowering of the Activation Energy,” J. Chem. Educ., Vol. 66, 1989, 935-937.

Section 13.14

Cynthia M. Friend, “Catalysis on Surfaces,” Scientific American, April 1993, 74-79.

Section 13.14

Sir John Meurig Thomas, “Solid Acid Catalysts,” Scientific American, April 1992, 112-118.

Section 13.14

Mitch Jacoby, “Getting Auto Exhausts to Pristine,” Chem. Eng. News, January 25, 1999, 36-44.

Section 13.14

Robert J. Farrauto, Ronald M. Heck, and Barry K. Speronello, “Environmental Catalysts,” Chem. Eng. News., September 7, 1992, 34-44.

Section 13.14

Reed A. Howald, “The Fizz Keeper, a Case Study in Chemical Education, Equilibrium, and Kinetics,” J. Chem. Educ., Vol. 76, 1999, 208-209.

118

CHAPTER 14 CHEMICAL EQUILIBRIUM Chapter Learning Goals for Students Section 14.1

Describe characteristics of a reaction in chemical equilibrium.

Section 14.2

Write an equilibrium constant expression Kc for gas and solution phase reactions.

Determine a new equilibrium constant expression when reactions are manipulated or added.

Calculate the value of the equilibrium constant Kc given concentrations of reactant and products.

Identify a reaction in equilibrium from a molecular representation and evaluate the equilibrium constant.

Section 14.3

Write the equilibrium expression Kp in terms of partial pressures of reactants and products. Relate the equilibrium constants Kp and Kc.

Section 14.4

Write equilibrium constant expressions for reactions involving heterogeneous equilibria.

Section 14.5

Determine the extent of a reaction given the equilibrium constant.

Predict the direction a reaction will shift to reach equilibrium given initial concentrations.

Calculate concentrations or partial pressures or products and reactants in equilibrium.

Calculate concentrations of reactants and products in equilibrium when initial concentrations are unequal.

Section 14.6 and Section 14.7

Predict the direction a reaction in equilibrium will shift as a result of changes in concentration.

Section 14.8

Predict the direction a reaction in equilibrium will shift as a result of changes volume or pressure.

Section 14.9

Predict the direction a reaction in equilibrium will shift as a result of changes in temperature.

Section 14.10

Relate the equilibrium constant to the rate of the forward and reverse reactions.

Describe the effect of a catalyst on the equilibrium position of a mixture.

121

Chapter 14—Chemical Equilibrium

Lecture Outline 1

14.1. The Equilibrium State 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 2 14.2. The Equilibrium Constant Kc 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 F. For a chemical equation multiplied by a common factor, n.

1. Equilibrium expression = original expression raised to the n power 2. Kc = (Kc)n G. Whenever two or more chemical equations are added to give the equation for an overall reaction 1. Equilibrium expression (overall reaction) = product of the individual equilibrium expressions.

2. Kc (overall) = Kc1  Kc2 3 14.3. The Equilibrium Constant Kp 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)n 2. n = the sum of the coefficients of gaseous products minus the sum of the coefficients of gaseous reactants

1Test Item File Questions: Multiple Choice 1, Short Answer 1 – 2 2Test Item File Questions: Multiple Choice 2 – 12, Algorithmic 1 – 5, Short Answer 3 – 4 3Test Item File Questions: Multiple Choice 13 – 27, Algorithmic 6 – 7, Short Answer 5 – 10

122

Chapter 14—Chemical Equilibrium

4

14.4. Heterogeneous Equilibria 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 3. The concentrations are constant and included in value of equilibrium constant.

5 14.5. Using the Equilibrium Constant A. Judging extent of reaction (how much reactant is converted to product) 1. Large value of Kc ( > 103) a. Products predominate over reactants. b. Reaction proceeds almost to completion. c. Position of reaction is to the right.

2. Small value of Kc ( < 10–3) a. Reactants predominate over products. b. Reaction hardly proceeds. c. Position of the reaction is to the left.

3. Intermediate value of Kc (10–3 < Kc < 103) 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 from left to right (products favored). b. Qc > Kc; reaction goes from right to left (reactants favored). c. Qc = Kc; reaction is at equilibrium.

C. Calculating equilibrium concentrations 1. If the equilibrium constant is known and all but one equilibrium concentration is known, calculate the unknown concentration directly from the equilibrium constant expression.

2. If the equilibrium constant is known and initial concentrations are known, calculate equilibrium concentrations using the steps outlined in Figure 14.6.

6 14.6. Factors That Alter the Composition of an Equilibrium Mixture: LeChâtelier’s Principle 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.

7 14.7. Altering an Equilibrium Mixture: Changes in Concentration 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) 4Test Item File Questions: Multiple Choice 28 – 35, Algorithmic 8 – 11, Short Answer 11 5Test Item File Questions: Multiple Choice 36 – 61, Algorithmic 12 – 26, Short Answer 12 – 17 6Test Item File Questions: Multiple Choice 62, Algorithmic 27 – 29, 7Test Item File Questions: Multiple Choice 63 – 64, Algorithmic 30 – 31, 37, Short Answer 18 – 19

123

Chapter 14—Chemical Equilibrium

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 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.

8 14.8. Altering an Equilibrium Mixture: Changes in Pressure and Volume 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 9 14.9. Altering an Equilibrium Mixture: Changes in Temperature A. Disturb an equilibrium by a change in concentration, pressure, or volume.

1. Qc  Kc 8Test Item File Questions: Algorithmic 32 – 35, Short Answer 20 – 22 9Test Item File Questions: Multiple Choice 65 – 67, Algorithmic 36, 38 – 40, Short Answer 23 – 24

124

Chapter 14—Chemical Equilibrium

2. Value of Kc remains constant with constant T. B. Change in temperature always changes the equilibrium constant.

1. Depends on sign of H for the reaction.

2. Exothermic reaction (H = –) – equilibrium constant decreases as the temperature increases.

3. Endothermic reaction (H= +) – equilibrium constant increases as the temperature increases.

C. Le Châtelier’s 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.

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.

10 14.10. The Link Between Chemical Equilibrium and Chemical Kinetics 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 C. The effect of a catalyst on equilibrium.

1. A catalyst provides a new, lower energy pathway, increasing the rate of the chemical reaction.

2. Forward and reverse reactions pass through the same transition state.

3. Rates for forward and reverse reactions increase by the same factor.

4. The catalyst 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 Inquiry – How Does Equilibrium Affect Oxygen Transport in the Bloodstream?

10Test Item File Questions: Multiple Choice 68 – 75, Algorithmic 41, Short Answer 25

125

Chapter 14—Chemical Equilibrium

CHAPTER 14 CHEMICAL EQUILIBRIUM Teaching Tips, Points of Emphasis, and Common Misconceptions Section 14.2

The subscript c in Kc means that the amounts of reactants and products are expressed as molar concentrations.

Section 14.2

Both the equilibrium equation and the value of the equilibrium constant depend on how the chemical equation is written and balanced.

Section 14.5

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 14.5

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 14.5

Students should apply these criteria results not by rote memorization but by analysis of what must occur when Qc  Kc.

Section 14.5

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 14.5

Mathematical tricks are not apparent to many students. Point out the value of setting up the problem to obtain a perfect square.

Section 14.5

Mathematics sometimes yields results that have no physical meaning. Students should learn to analyze answers to problems to determine whether they make sense.

Section 14.7

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 14.7

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 14.10

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 14.1

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 intensity of the color due to an increase in the intensity in the amount of NO 2. The 126

Chapter 14—Chemical Equilibrium

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 NO 2. 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 14.7

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), pp. 77-78.

Section 14.8

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 14.9

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), pp. 60-61.

Concentrated hydrochloric acid is added to pink [Co(H 2O)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 14.1

Vladimir E. Fainzilberg and Stewart Karp, “Chemical Equilibrium in the General Chemistry Curriculum,” J. Chem. Educ., Vol. 71, 1994, 769-770.

Section 14.1

Carl W. David, “An Elementary Discussion of Chemical Equilibrium,” J. Chem.

Educ., Vol. 65, 1988, 407-409.

Section 14.1 Section 14.1

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 14.1

Audrey H. Wilson, “Equilibrium: A Teaching/Learning Activity,” J. Chem. Educ., Vol 75, 1998, 1176-1177.

Section 14.1

Penelope A. Huddle, Margie W. White, and Fiona Rogers, “Simulations for Teaching Chemical Equilibrium,” J. Chem. Educ., Vol 77, 2000, 920-926.

Section 14.5

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

Section 14.5

William R. Smith and Ronald W. Munson, “Chemical Equilibrium and Polynomial Equations: Beware of Roots,” J. Chem. Educ., Vol. 66, 1989, 489-490.

Section 14.5

E. Weltin, “Calculating Equilibrium Concentration by Iteration: Recycle Your Approximations,” J. Chem. Educ., Vol. 72, 1995, 36-38.

Section 14.6

Martin R. Feldman and Monica L. Tarver, “Fritz Haber,” J. Chem. Educ., Vol. 60, 1983, 463-464.

127

Chapter 14—Chemical Equilibrium

Section 14.8

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 14.8

Reed A. Howard, “The Fizz Keeper, a Case Study in Chemical Education,” J. Chem.

Educ., Vol. 76, 1999, 208-209.

128

CHAPTER 15 AQUEOUS EQUILIBRIA: ACIDS AND BASES Learning Goals for Students Section 15.1

Identify Arrhenius acids and bases and Brønsted–Lowry acids and bases and conjugate acid–base pairs.

Section 15.2

Predict the direction of a reaction based upon relative strengths of the conjugate acid–base pairs.

Section 15.3

Predict the relative strengths of binary acids (HA) and oxoacids (HnYOm) based on their chemical structure.

Section 15.4

Calculate the concentration of [H3O+] or [OH-] using the value of Kw.

Section 15.5 and Section 15.6

Calculate the pH of a solution given [H3O+] or [OH-].

Calculate [H3O+] or [OH-] given the pH of a solution.

Section 15.7

Calculate the pH of a strong acid or base solution.

Section 15.8

Calculate the value for Ka given the initial concentration of a weak acid and its equilibrium pH.

Section 15.9 and Section 15.10

Section 15.11

Given the Ka value for a weak acid and its initial concentration, calculate the pH, percent dissociation, and the concentration of all species present in solution.

Calculate the pH and the concentration of all species present in a solution of a diprotic acid.

Visualize the species present when a diprotic acid dissociates.

Section 15.12

Calculate the pH and equilibrium concentrations in a solution of a weak base.

Section 15.13

Relate Ka, Kb, pKa, and pKb for a conjugate acid–base pair.

Section 15.14

Predict whether a salt is acidic, basic, or neutral and calculate the pH of the salt solution.

Section 15.15

Identify the Lewis acid and Lewis base and use curved arrow notation to indicate donation of a lone pair of elections in Lewis acid–base reaction.

130

Chapter 15—Aqueous Equilibria: Acids and Bases

Lecture Outline 1

15.1. Acid–Base Concepts: The Brønsted–Lowry Theory 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.

G. Hydrated protons and hydronium ions 1. Hydronium ion, H3O+– ion in which H+ is bonded to the oxygen atom of a solvent water molecule.

2. Simplest hydrate of the proton – [H(H2O)]+ 3. Produces higher hydrates through hydrogen bonding with other water molecules 4. General formula – [H(H2O)n]+ 5. Symbols H+(aq) and H3O+(aq) used to represent a proton hydrated by an unspecified number of water molecules 2 15.2. Acid Strength and Base Strength 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 15.1 3 15.3. Factors That Affect Acid Strength A. The extent of dissociation of an acid, HA, is often determined by the strength and polarity of the H–A bond.

1Test Item File Questions: Multiple Choice 1 – 8, Algorithmic 1 – 12, Short Answer 1 2Test Item File Questions: Multiple Choice 9 – 20, Short Answer 2 3Test Item File Questions: Multiple Choice 21 – 25, Algorithmic 13 – 20, Short Answer 3 – 4, 23

131

Chapter 15—Aqueous Equilibria: Acids and Bases

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.

4 15.4. Dissociation of Water 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.

5. In an aqueous solution, [H3O+][OH–] = 1.0  10–14.

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 ] 3. [H3O ] = [OH–], neutral solution and [H3O+] = 1.0  10–7 M 4. [H3O+] > [OH–], acid solution and [H3O+] > 1.0  10–7 M 5. [H3O+] < [OH–], basic solution and [H3O+] < 1.0  10–7M 5 15.5. The pH Scale A. pH scale – logarithmic scale used to express the hydronium ion concentration 1. pH = –log[H3O+] 2 [H3O+] = antilog(–pH) = 10–pH 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. +

4Test Item File Questions: Multiple Choice 26 – 33, Algorithmic 21 – 25, Short Answer 5 – 8 5Test Item File Questions: Multiple Choice 34 – 44, Algorithmic 26 – 32, Short Answer 9 – 10

132

Chapter 15—Aqueous Equilibria: Acids and Bases

1. Acidic solutions, pH < 7 2. Basic solutions, pH > 7 3. Neutral solutions, pH = 7 6 15.6. Measuring pH 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 7 15.7. The pH in Solutions of Strong Acids and Strong Bases 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 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 8 15.8. Equilibria in Solutions of Weak Acids 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

D. The larger the value of Ka, the stronger the acid.

E. pKa = -log Ka F. Ka values can be determined from pH measurements.

9 15.9. Calculating Equilibrium Concentrations in Solutions of Weak Acids 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:

6Test Item File Questions: Multiple Choice 45 – 47, Short Answer 11 – 12 7Test Item File Questions: Multiple Choice 48 – 54, Algorithmic 33 – 41, Short Answer 13 8Test Item File Questions: Multiple Choice 55 – 62, Short Answer 14 – 17 9Test Item File Questions: Multiple Choice 63 – 69, Algorithmic 42 – 46

133

Chapter 15—Aqueous Equilibria: Acids and Bases

10

15.10. Percent Dissociation in Solutions of Weak Acids [HA] dissociated ´ 100 A. Percent dissociation = [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.

11 15.11. Polyprotic Acids 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: H 2A, HA–, and H2O. 1. Strongest acid – H2A 2. Principal reaction – dissociation of H2A 3. All of H3O+ comes from first dissociation step.

12 15.12. Equilibria in Solutions of Weak Bases 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

10Test Item File Questions: Multiple Choice 70 – 72, Short Answer 18 – 20 11Test Item File Questions: Multiple Choice 73 – 78, Algorithmic 47 – 52, 12Test Item File Questions: Multiple Choice 79 – 84, Algorithmic 53 – 56, Short Answer 21

134

Chapter 15—Aqueous Equilibria: Acids and Bases

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.

13 15.13. Relation Between Ka and Kb 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 15.14. Acid–Base Properties of Salts 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 2. Neither the cation nor the anion reacts with water to produce H 3O+ 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 H 3O+ 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

13Test Item File Questions: Multiple Choice 85 – 89, Algorithmic 57 14Test Item File Questions: Multiple Choice 90 – 98, Algorithmic 58 – 69, Short Answer 22

135

Chapter 15—Aqueous Equilibria: Acids and Bases

15

15.15. Lewis 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 Inquiry – What Is Acid Rain and What Are Its Effects?

15Test Item File Questions: Multiple Choice 99 – 103, Algorithmic 70 – 73, Short Answer 24 – 25

136

Chapter 15—Aqueous Equilibria: Acids and Bases

CHAPTER 15 AQUEOUS EQUILIBRIA: ACIDS AND BASES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 15.5

Typically pH measurements fall within the range 0–14; however, the scale extends beyond those values in both directions and includes negative pHs.

Section 15.11

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 15.13

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 15.3

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 15.3

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.

Section 15.6

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 15.14

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 15.14

Edward Koubek, “Acid-Base Chemistry of the Aluminum Ion in Aqueous Solution,” J. Chem. Educ., Vol. 75, 1998, 60.

Literature References Section 15.1

Doris Kolb, “Acids and Bases,” J. Chem. Educ., Vol. 55, 1978, 459-464.

Section 15.1

George B. Kauffman, “The Brønsted-Lowry Acid-Base Concept,” J. Chem. Educ., Vol. 65, 1988, 28-31.

Section 15.1

Jamie L. Adcock, “Teaching Brønsted-Lowry Acid-Base Theory in a Direct Comprehensive Way,” J. Chem. Educ., Vol. 78, 2001, 1495-1496.

Section 15.3

Manus Monroe and Karl Abrams, “The Relative Strength of Oxyacids and Its Application,” J. Chem. Educ., Vol. 62, 1985, 41-43.

137

Chapter 15—Aqueous Equilibria: Acids and Bases

Section 15.5

Doris Kolb, “The pH Concept,” J. Chem. Educ., Vol. 56, 1979, 49-53.

Section 15.5

Stephen J. Hawkes, “Teaching the Truth about pH,” J. Chem. Educ., Vol. 71, 1994, 747-749.

Section 15.5

Addison Ault, “Do pH in Your Head,” J. Chem. Educ., Vol. 76, 1999, 936-938.

Section 15.8

Todd P. Silverstein, “Weak vs Strong Acids and Bases,” J. Chem. Educ., Vol. 77, 2000, 849-850.

Section 15.8

John J. Fortman, “Pictoral Analogies XI: Concentrations and Acidity of Solutions,” J. Chem. Educ., Vol. 71, 1994, 430-432.

Section 15.15

Glenn T. Seaborg, “The Research Style of Gilbert N. Lewis: Acids and Bases,” J. Chem. Educ., Vol. 61, 1984, 93-100.

138

CHAPTER 16 APPLICATIONS OF AQUEOUS EQUILIBRIA Chapter Learning Goals for Students Section 16.1

Write a balanced equation for a neutralization reaction, calculate the equilibrium constant, and determine whether the pH after neutralization is greater than, equal to, or less than 7.

Section 16.2

Calculate the effect of a common-ion on concentrations, pH, and percent dissociation in a solution of a weak acid.

Visualize the common ion effect at the molecular level and predict the effect on pH and percent dissociation of a weak acid.

Section 16.3

Calculate the pH of a buffer solution and the change in pH on addition of a strong acid or a strong base.

Section 16.4

Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution and to prepare a buffer solution that has a given pH.

Section 16.5 and Section 16.6

Calculate the pH at various points in a strong acid–strong base titration.

Section 16.7

Calculate the pH at various points in a weak acid–strong base titration

Section 16.8

Calculate the pH at various points in a weak base–strong acid titration.

Section 16.9

Calculate the pH at various points in a diprotic acid–strong base titration.

Visualize the molecular species present during a titration and interpret titration curves.

Section 16.10

Write the equilibrium-constant expression for dissolution of an ionic compound, and calculate the value of its Ksp.

Section 16.11

Calculate ion concentrations and the solubility of an ionic compound from its Ksp.

Section 16.12

Describe how the presence of molecular and ionic species will affect the solubility of an ionic compound.

Calculate solubility in a solution that contains a common ion.

Use the formation constant Kf to calculate ion concentrations in a solution that contains a complex ion.

Use the formation constant Kf to calculate the solubility of an ionic compound when the cation forms a complex ion.

Section 16.13

Calculate the ion product IP for an ionic compound, and determine whether a precipitate will form when various solutions are mixed.

140

Chapter 16—Applications of Aqueous Equilibria

Section 16.14 and Section 16.15

Design a scheme for separating ions is a mixture based on selective precipitation.

141

Chapter 16—Applications of Aqueous Equilibria

Lecture Outline 1

16.1. Neutralization Reactions A. Strong acid–strong base 1. Net ionic equation for the neutralization reaction: H 3O+(aq) + OH–(aq) ⇌ 2 H2O(l) 2. When equal numbers of moles of acid and base are mixed together, [H3O+] = [OH–] = 1  10–7 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).

B(aq) + H2O(l) ⇌ BH+(aq) + OH–(aq) Kb H3O+(aq) + OH–(aq) ⇌ 2 H2O(l) 1/Kw 3. Neutralization of any weak base with a strong acid goes 100% to completion because H 3O+ 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 B(aq) + H2O(l) ⇌ BH+(aq) + OH–(aq) Kb – + H3O (aq) + OH (aq) ⇌ 2 H2O(l) 1/Kw 3. Less tendency to proceed to completion than neutralizations involving strong acids or strong bases 2 16.2. The Common-Ion Effect 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 of the pH of a solution prepared from a weak acid and a salt of its conjugate base 1. Identify the acid–base properties of the various species in solution.

1Test Item File Questions: Multiple Choice 1 – 4, Algorithmic 1 – 10, Short Answer 1 – 4 2Test Item File Questions: Multiple Choice 5 – 8, Algorithmic 11 – 17, Short Answer 5 – 8

142

Chapter 16—Applications of Aqueous Equilibria

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 – 3 16.3. Buffer Solutions 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+ ] = Ka [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 [H 3O+] 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.

4 16.4. The Henderson-Hasselbalch Equation A. Explains how pH affects the percent dissociation of a weak acid. [base] pH = pKa + log [acid]

[base] = pH - pKa [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. 3. Adjust the [base]/[acid] ratio. log

3Test Item File Questions: Multiple Choice 9 – 21, Algorithmic 18 – 22 4Test Item File Questions: Multiple Choice 22 – 31, Short Answer 9 – 15

143

Chapter 16—Applications of Aqueous Equilibria

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.

16.5. pH Titration Curves 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 5 16.6. Strong Acid–Strong Base Titrations 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 6 16.7. Weak Acid–Strong Base Titrations 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] 2.

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 [A - ] [HA] C. Halfway to the equivalence point pH = pK a + log

5Test Item File Questions: Multiple Choice 32 – 33, Algorithmic 23 – 24, Short Answer 16 – 18 6Test Item File Questions: Multiple Choice 34 – 40, Algorithmic 25 – 32, Short Answer 19

144

Chapter 16—Applications of Aqueous Equilibria

1. Exactly half of the weak acid has been neutralized.

2. [A-] = [HA] 3. pH = pKa of the weak acid D. Addition of strong base at the equivalence point 1. Basic salt solution.

2 mmol A– after neutralization = mL of strong base  [strong 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.

E. 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 [H 3O + ] =

[ Kw ]

[OH - ] F. Key features of the weak acid–strong base titration curve compared to the strong acid–strong base itration curve: 1. The initial rise in pH is greater for the weak acid–strong base curve but is more level in the region midway to the equivalence point.

2. The increase in the pH near the equivalence point is smaller for the weak acid–strong base curve.

3. The pH at the equivalence point is greater than 7.

G. Can use pH range at the equivalence point to select a suitable indicator.

7 16.8. Weak Base–Strong Acid Titrations 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+] 8

16.9. Polyprotic Acid–Strong Base Titrations 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 7Test Item File Questions: Multiple Choice 41 – 43, Algorithmic 33 – 37, Short Answer 20 8Test Item File Questions: Multiple Choice 44 – 48, Short Answer 21

145

Chapter 16—Applications of Aqueous Equilibria

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.

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

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 A–(aq) + H2O(l) ⇌ HA(aq) + OH–(aq) Kb = Kw/Ka2 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–] 9 16.10. Solubility Equilibria for Ionic Compounds 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 1. MaXb(s) ⇌ a Mb+(aq) + b Xa–(aq) 2. Ksp = [Mb+]a[Xa–]b 10 16.11. Measuring Ksp and Calculating Solubility from Ksp 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 11 16.12. Factors That Affect Solubility 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 9Test Item File Questions: Algorithmic 38 – 39, 10Test Item File Questions: Multiple Choice 49 – 50, Algorithmic 40 – 47, Short Answer 22 – 24 11Test Item File Questions: Multiple Choice 51 – 60, Algorithmic 48 – 51, Short Answer 25

146

Chapter 16—Applications of Aqueous Equilibria

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 D. Amphoterism 1. Amphoteric oxides – soluble both in strongly acidic and in strongly basic solutions 12 16.13. Precipitation of Ionic Compounds 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 13 16.14. Separation of Ions by Selective Precipitation 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] Kspa = [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.

16.15. 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 16.18 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 Inquiry – What is Causing a Decrease in the pH of the Oceans?

12Test Item File Questions: Multiple Choice 61 – 67, Algorithmic 52 – 54, Short Answer 26 13Test Item File Questions: Algorithmic 55 – 56

147

Chapter 16—Applications of Aqueous Equilibria

CHAPTER 16 APPLICATIONS OF AQUEOUS EQUILIBRIA Teaching Tips, Points of Emphasis, and Common Misconceptions Section 16.2

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 16.3

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 16.3

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 16.3

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 16.3

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 16.4

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 16.4

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 16.6

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 16.6

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 16.7

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 16.7

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 16.9

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.

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.

148

Chapter 16—Applications of Aqueous Equilibria

Section 16.10

The solubility product constant, Ksp, is for a saturated solution.

Section 16.11

No ionic compound is 100% insoluble. In order for an equilibrium to exist, all ionic compounds must dissolve to some extent.

Section 16.11

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 16.12

The addition of a common ion to a solution of a sparingly soluble ionic compound lowers the solubility of that compound.

Section 16.12

The solubility of a sparingly soluble ionic compound is increased on addition of H 3O+ if the anion of the compound is the conjugate base of a weak acid.

Section 16.12

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 16.12

Remember that amphoteric means able to react as an acid or as a base.

Section 16.14

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 H 2S.

149

Chapter 16—Applications of Aqueous Equilibria

Lecture/Laboratory Demonstration References Section 16.2

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 16.2

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 16.2

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 16.2

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 16.3

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 16.3

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 16.7

Dianne N. Epp, “Teas as Natural Indicators,” J. Chem. Educ., Vol. 70, 1993, 326.

Section 16.11

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 16.11

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 16.12

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 16.12

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 16.12

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 16.12

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.

Section 16.13

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 150

Chapter 16—Applications of Aqueous Equilibria

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 16.14

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 16.3

Charles L. Bering, “A Good Idea Leads to a Better Buffer,” J. Chem. Educ., Vol. 64, 1987, 803.

Section 16.3

Edwin S. Gould, “Phosphate Buffers and Telephone Poles – A Useful Analogy with Limitations,” J. Chem. Educ., Vol. 76, 1999, 1511.

Section 16.4

Henry N. Po and N. M. Senozan, “The Henderson-Hasselbalch Equation: Its History and Limitations,” J. Chem. Educ., Vol. 78, 2001, 1499-1503.

Section 16.5

Dennis Barnum, “Predicting Acid–Base Titration Curves Without Calculations,” J. Chem. Educ., Vol 76, 1999, 938-942.

Section 16.7

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 16.7

Robert C. Mebane and Thomas R. Rybolt, “Edible Acid–Base Indicators,” J. Chem.

Educ., Vol. 62, 1985, 285.

Section 16.11

Stephen J. Hawkes, “What Should We Teach Beginners about Solubility and Solubility Product?” J. Chem. Educ., Vol. 75, 1998, 1179-1181.

Section 16.11

Roy W. Clark and Judith M. Bonicamp, “The Ksp-Solubility Conundrum,” J. Chem.

Educ., Vol. 75, 1998, 1182-1185.

Section 16.11

Andres Raviolo, “Assessing Students’ Conceptual Understanding of Solubility Equilibrium,” J. Chem. Educ., Vol. 78, 2001, 629-631.

Section 16.11

Robert Perkins, “The Useless Tea Kettle,” J. Chem. Educ., Vol. 618, 1984, 383.

Section 16.12

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 16.15

Dale D. Clyde, “Swimming Pools, Hot Rods, and Qualitative Analysis,” J. Chem. Educ., Vol. 65, 1988, 911-913.

151

Chapter 16—Applications of Aqueous Equilibria

152

CHAPTER 17 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM Chapter Learning Goals for Students Section 17.1

Define a spontaneous process and classify various physical processes and chemical reactions as spontaneous or nonspontaneous

Section 17.2

Predict the sign of ΔS for various physical processes and chemical reactions.

Section 17.3

Determine the number of arrangements in a system and use Boltzmann’s equation to calculate the entropy.

Predict which state of substance has higher entropy.

Calculate ΔS for the expansion or compression of an ideal gas at constant temperature.

Section 17.4 and Section 17.5

Predict the relative entropies of two substances.

Use standard molar entropies (S°) to calculate the standard entropy of reaction (ΔS°).

Section 17.6

Describe how the change in entropy and enthalpy of a system affects spontaneity.

Calculate values of ΔSsys, ΔSsurr, and ΔStotal for a reaction, and use these values to determine whether the reaction is spontaneous. Estimate the temperature at which a reaction changes between spontaneous and nonspontaneous.

Section 17.7

Describe how the sign and magnitude of ΔS, ΔH, and temperature affect the value of ΔG and the spontaneity of the reaction.

Calculate values of ΔS, ΔH, and ΔG for a reaction, and use these values to determine whether the reaction is spontaneous. Estimate the temperature at which a reaction changes between spontaneous and nonspontaneous.

Section 17.8

Define the standard free-energy change (ΔG°) for a reaction, and calculate ΔG° from ΔH° and ΔS°.

Section 17.9

Use values of ΔG°f to determine if a compound is thermodynamically stable.

Calculate ΔG° for a reaction from values of ΔG°f and determine if the reaction is spontaneous under standard conditions and how temperature will affect spontaneity.

Section 17.10

Calculate ΔG for a reaction under nonstandard-state conditions.

Section 17.11

Use the relationship between ΔG° and the equilibrium constant for a reaction, and be able to calculate one from the other.

153

Chapter 17—Thermodynamics: Entropy, Free Energy, and Equilibrium

Lecture Outline 1

17.1. Spontaneous Processes 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.

2 17.2. Enthalpy, Entropy, and Spontaneous Processes: A Brief Review 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. S = Sfinal – Sinitial 3. When S 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 S is negative there is a decrease in randomness or disorder of a system.

3 17.3. Entropy and Probability 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 2. k = 1.38  10–23 J/K D. Gas expands spontaneously because the state of greater volume is more probable.

1. For an ideal gas V DS = R ln final Vinitial

P=

nRT V

1Test Item File Questions: Multiple Choice 1 – 6, Algorithmic 1 – 2, Short Answer 1 2Test Item File Questions: Multiple Choice 7 – 11, Algorithmic 3 – 5, Short Answer 2 3Test Item File Questions: Multiple Choice 12 – 21, Algorithmic 6 – 9, Short Answer 3 – 7

154

Chapter 17—Thermodynamics: Entropy, Free Energy, and Equilibrium

DS = 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.

4 17.4. Entropy and Temperature 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.

5 17.5. Standard Molar Entropies and Standard Entropies of Reaction A. Standard molar entropy, S 1. The entropy of 1 mole of the pure substance at 1 atm pressure and a specified temperature, o usually 25 C. 2. Absolute entropies – measured with respect to an absolute reference point.

B. Standard entropy of reaction, S 1. S = S(products) – S(reactants) 2. Multiply the S 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.

6 17.6. Entropy and the Second Law of Thermodynamics A. Value of G – criterion for spontaneity 1. G = H – TS 2. G > 0; nonspontaneous reaction 3. G < 0; spontaneous reaction 4. G = 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.

DStotal = DSsystem + DSsurroundings a. Stotal > 0; spontaneous reaction b. Stotal < 0; nonspontaneous reaction c. Stotal = 0; reaction at equilibrium D. All reactions proceed spontaneously in the direction that increases the entropy of the system plus surroundings.

E. To determine Stotal, need to know Ssystem and Ssurroundings.

4Test Item File Questions: Multiple Choice 22 – 23, Algorithmic 10 – 14, Short Answer 8 5Test Item File Questions: Multiple Choice 24 – 27, Algorithmic 15 – 19, 23, Short Answer 9 – 11 6Test Item File Questions: Multiple Choice 28 – 32, Algorithmic 20 – 22, Short Answer 12 – 15

155

Chapter 17—Thermodynamics: Entropy, Free Energy, and Equilibrium 1.  Ssystem a. The entropy of reaction b. Calculate from standard molar entropies 2. At constant pressure DH rxn DSsurroundings = T DH rxn o DStotal = DSrxn T 7 17.7. Free Energy and the Spontaneity of Chemical Reactions 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 G = H – TS C. Relationship between free energy and spontaneity 1. Stotal = S – H/T; 2. – TStotal = H – TS 3. – TStotal = G 4. G and Stotal 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 17.2).

1. When H and S have the same sign, the spontaneity is temperature dependent, as the sign of G depends on the temperature.

2. Can estimate the temperature at which G changes from positive to negative with

DG = DH - TDS

3. To estimate the temperature at which G changes from a positive to a negative value, set G = H – TS = 0.

T=

H S

8

17.8. Standard Free-Energy Changes for Reactions A. Standard free-energy change, G 1. Change in free energy that occurs when reactants in their standard states are converted to products in their standard states 2. G – an extensive property; refers to the number of moles indicated in the chemical equation 3. Calculate G from standard enthalpy change, H, and standard entropy change, S. G = H – TS 9 17.9. Standard Free Energies of Formation A. Standard free energy of formation, DG f 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

7Test Item File Questions: Multiple Choice 33 – 45, Algorithmic 24 – 26, Short Answer 16 – 19 8Test Item File Questions: Multiple Choice 46 – 49, Algorithmic 27, Short Answer 20 9Test Item File Questions: Multiple Choice 50 – 56, Algorithmic 28, Short Answer 21

156

Chapter 17—Thermodynamics: Entropy, Free Energy, and Equilibrium

2. For an element in its most stable form at 25C, DG f = 0 B. DG f measures a substance’s thermodynamic stability with respect to its constituent elements.

1. DG f is negative – substance is stable and does not decompose to its constituent elements 2. DG f is positive – substance is thermodynamically unstable (but may not decompose to its constituent elements if the rate of decomposition is slow) C. Can use DG f to calculate standard free-energy changes for reactions

DG o = DGfo (products) - DGfo (reactants) 10

17.10. Free-Energy Changes for Reactions under Nonstandard-State Conditions A. Free-energy change for a reaction when the reactants and products are present at nonstandardstate pressures and concentrations B. C. Q = the reaction quotient — an expression having the same form as the equilibrium constant expression (but concentrations are not necessarily at equilibrium) 11 17.11. Free Energy and Chemical Equilibrium A. When the reaction mixture is mostly reactants 1. Q << 1 2. RT ln Q << 0 3. G < 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. G > 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 17.11) 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. G = 0 b. Q = K 2. G = – RT lnK E. Relationship between G and K 1. G < 0, ln K > 0 – equilibrium mixture is mainly products 2. G > 0, ln K < 0 – equilibrium mixture is mainly reactants 3. G = 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 2. K determined by DG rxn 3. DG rxn determined by DH f and DS rxn Inquiry – Does Entropy Prevent the Evolution of Biological Complexity?

10Test Item File Questions: Multiple Choice 57 – 63, Algorithmic 29 11Test Item File Questions: Multiple Choice 64 – 73, Algorithmic 30 – 31, Short Answer 22 – 26

157

Chapter 17—Thermodynamics: Entropy, Free Energy, and Equilibrium

CHAPTER 17 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM Teaching Tips, Points of Emphasis, and Common Misconceptions Section 17.1

Most, but not all, spontaneous chemical and physical changes are accompanied by heat transfer from system to surroundings (H is negative) and a lowering of the potential energy of the system. When H is positive for a spontaneous physical or chemical change, another driving force (entropy) is playing an important role.

Section 17.3

The value of R used in energy-related calculations is 8.314 J/K, not 0.0821 Latm mol–1 K–1.

Section 17.3

The greater the volume (and the lower the pressure) of a gas, the greater the entropy.

Section 17.4

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 17.5

In general, So (gas) > So (liquid) > So (solid), and So (large molecule) > So (small molecule).

Section 17.5

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 17.6

A reaction is spontaneous only if Ssystem + Ssurroundings = Stotal > 0. Ssystem is calculated from standard molar entropies. Ssystem = –H/T, where –H is calculated from standard molar enthalpies.

Section 17.7

Two types of reactions have a spontaneity that cannot be reversed by changing the temperature (those with H and S both negative or both positive), and two have a spontaneity that can be reversed by changing the temperature (those with opposite signs for H and S).

Section 17.7

According to G = H –TS, 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 17.9

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 17.9

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 17.11

When all reactants and products are under standard conditions, Q = 1 and G = Go.

158

Chapter 17—Thermodynamics: Entropy, Free Energy, and Equilibrium

Lecture/Laboratory Demonstration References Section 17.4

Brian B. Laird, “Entropy, Disorder, and Freezing,” J. Chem. Educ., Vol. 76, 1999, 1388-1390.

Literature References Section 17.1

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 17.1

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 17.1

Raymond S. Ochs, “Thermodynamics and Spontaneity,” J. Chem. Educ., Vol. 73, 1996, 952-954.

Section 17.2

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

Section 17.2

Joseph H. Lechner, “Visualizing Entropy,” J. Chem. Educ., Vol. 76, 1999, 13821387.

Section 17.2

Peter Weiss, “Another Face of Entropy,” Science News, Vol. 154, 1998, 108-109.

Section 17.2

L. Glasser, “Order, Chaos, and All That!” J. Chem. Educ., Vol. 66, 1989, 997-1001.

Section 17.2

John P. Lowe, “Entropy: Conceptual Disorder,” J. Chem. Educ., Vol. 65, 1988, 403406.

Section 17.3

Douglas K. Russell, “The Boltzmann Distribution,” J. Chem. Educ., Vol. 73, 1996, 299-300.

Section 17.3

P. G. Nelson, “Derivation of the Second Law of Thermodynamics from Boltzmann’s Distribution Law,” J. Chem. Educ., Vol. 65, 1989, 390-393.

Section 17.5

Travis Thoms, “Periodic Trends for the Entropy of Elements,” J. Chem. Educ., Vol. 72, 1995, 16.

Section 17.7

Sidney Rosen, “J. Willard Gibbs (1839-1903): A Modern Genius,” J. Chem. Educ., Vol. 60, 1983, 593-594.

Section 17.7

Robert J. Deltete and David L. Thorsell, “Josiah Willard Gibbs and Wilhelm Ostwald: A Contrast in Scientific Style,” J. Chem. Educ., Vol. 73, 1996, 289-295.

Section 17.11

Zheng Xianmin, “The Free Energy Prediction and the Principle of Le Châtelier,” J. Chem. Educ., Vol. 66, 1989, 401-402

159

Chapter 17—Thermodynamics: Entropy, Free Energy, and Equilibrium

160

CHAPTER 18 ELECTROCHEMISTRY Chapter Learning Goals for Students Section 18.1

Assign oxidation numbers and classify half-reactions as oxidation or reductionreactions.

Use the half-reaction method to balancea redox reaction.

Section 18.2

Draw and interpret the sketch of agalvanic cell. Label the anode and cathode,indicate the direction of electron and ionflow, and write balanced equations for theelectrode and overall cell reactions.

Section 18.3

Use shorthand notation to represent agalvanic cell.

Section 18.4

Interconvert cell potential and standard free-energy change for electrochemical reactions.

Section 18.5

Use tabulated values of standard reductionpotentials for half-reactions(Appendix D) to order substances by increasingstrength as an oxidizing or reducingagent.

Use a table of standard reduction potentials to calculate standard cell potentials anddetermine spontaneity.

Section 18.6 and Section 18.7

Calculate the cell potential undernonstandard-state conditions using theNernst equation.

Use the cell potential to calculate solution concentration.

Section 18.8 and Section 18.9

Use standard reduction potentials tocalculate the equilibrium constant.

Section 18.10

Describe batteries, and calculate values of E°, G°, and K for the cell reaction.

Section 18.11

Describe how corrosion occurs and methods used to prevent it.

Section 18.12

Draw and interpret the sketch of anelectrolytic cell. Label the anode and cathode,indicate the direction of electron andion flow, and write balanced equations forthe electrode and overall cell reactions.

Predict the products formed and the electrode and overall cell reactions when an aqueous solution of an ionic compound is electrolyzed.

Section 18.13 and Section 18.14

Relate the current, time, and amount of product produced in an electrolytic cell.

170

Chapter 18—Electrochemistry

Lecture Outline 1

18.1. Balancing Redox Reactions by the Half-Reaction Method 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 18.1 2 18.2.Galvanic Cells 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 18.2). a. Consists of two half-cells i. A beaker with a strip of Zn in a solution of ZnSO 4 ii. A beaker with a strip of Cu in a solution of CuSO 4 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.

3 18.3. Shorthand Notation for Galvanic Cells A. Single vertical line (|) represents a phase boundary.

B. Double vertical line (||) represents a salt bridge.

1Test Item File Questions: Multiple Choice 1 – 13, Algorithmic 1 – 14, Short Answer 1 – 5 2Test Item File Questions: Multiple Choice 14 – 23, Algorithmic 15 – 19, Short Answer 6 3Test Item File Questions: Multiple Choice 24 – 32, Algorithmic 20 – 28, Short Answer 7 – 11

171

Chapter 18—Electrochemistry

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. Electrodes written on extreme left (anode) or extreme right (cathode).

3. Electrons move through the external circuit from left to right (anode to cathode).

4. For Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu(s): Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) D. Cell involving a gas 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.

4 18.4. Cell Potentials and Free-Energy Changes for Cell Reactions 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 with units of volts (V).

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, G 1. Related by G = – 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. G and E have opposite signs.

3. Spontaneous reaction has a positive cell potential but negative G. 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. G = –nFE 5 18.5. Standard Reduction Potentials 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 2 H+(aq, 1 M) + 2 e–→H2(g, 1 atm) Eo = 0.00 V 2. Shorthand notation for S. H. E. H+(1 M) | H2(1 atm) | Pt(s) 4 Test Item File Questions: Multiple Choice 33 – 39, Algorithmic 29 – 30, Short Answer 12 – 14 5 Test Item File Questions: Multiple Choice 40 – 42, Algorithmic 31 – 34, Short Answer 15 – 17

172

Chapter 18—Electrochemistry

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.

4. Standard oxidation potential and standard reduction potential always have the same magnitude but opposite signs.

5. Table of standard reduction potentials – textbook Table 18.1 and 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).

6 18.6. Using Standard Reduction Potentials 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. Valuesof E for the half-reactions not multiplied by that factor B. E values independent of amount of reaction 1. G = – nFE 2. G is an extensive property (depends on amount of substance). a. When amount of substance that reacts changes, G changes by the same amount as does n, the number of electrons transferred. b. But E = – G/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.

7 18.7. Cell Potentials under Nonstandard-State Conditions: The Nernst Equation A. Cell potentials depend on temperature and on the composition of the reaction mixture.

1. G = G + RT ln Q 2. G = – nFE; G = – 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 6Test Item File Questions: Multiple Choice 43 – 54, Algorithmic 35 – 42, Short Answer 18 – 19 7 Test Item File Questions: Multiple Choice 55 – 66, Algorithmic 43 – 45, Short Answer 20

173

Chapter 18—Electrochemistry

8

18.8. Electrochemical Determination of pH 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.

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.

9 18.9. Standard Cell Potentials and Equilibrium Constants A. Standard free-energy change for a reaction is related to both the standard cell potential and the equilibrium constant.

1. G = – nFE 2. G = – 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: −G o ln K = RT 3. K from electrochemical data: nFE o ln K = RT 10 18.10.Batteries 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 almost 100 years 2. 12 V battery – six 2 V cells connected in series 8 Test Item File Questions: Multiple Choice 67 – 71, Algorithmic 46, Short Answer 21 9Test Item File Questions: Multiple Choice 72 – 75, Algorithmic 47 – 48, Short Answer 22 10Test Item File Questions: Multiple Choice 76 – 82, Algorithmic 49 – 50, Short Answer 23

174

Chapter 18—Electrochemistry

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 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 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 automobiles 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 and lithium-ion batteries – used in portable electronic products 1. Li has a higher Eo 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 11 18.11.Corrosion A. Corrosion – oxidative deterioration of a metal 11Test Item File Questions: Multiple Choice 83 – 86, Algorithmic 51

175

Chapter 18—Electrochemistry

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 O2and 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 7. O2 able to oxidize all metals except a few – O2/H2O half-reaction lies above the Mn+/M half-reaction 8. Some metals (Al, Mg, Ti, Cr, Zn) form a metal oxide coating that adheres to the surface of the metal, protecting it from further contact and reaction with oxygen.

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 12 18.12. Electrolysis and Electrolytic Cells 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 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. a. Reduction of water preferred – Eo 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–

12Test Item File Questions: Multiple Choice 87 – 91, Algorithmic 52 – 53

176

Chapter 18—Electrochemistry

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 O 2 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) 13 18.13. Commercial Applications of Electrolysis 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 2. Electrolysis of a molten mixture of Al2O3 and cryolite (Na3AlF6) at 1000o 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 O 2(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.

14 18.14.Quantitative Aspects of Electrolysis 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.

13Test Item File Questions: Multiple Choice 92 – 95 14Test Item File Questions: Multiple Choice 96 – 101, Algorithmic 54 – 57, Short Answer 24 – 25

177

Chapter 18—Electrochemistry

C. Moles of electrons passed through a cell are determined from the electric current and the time that the current flows.

Moles of e − = charge (C) 

1 mol e − 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 or 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.

15 Inquiry – How Do Hydrogen Fuel Cells Work?

15Test Item File Questions: Multiple Choice 102, Algorithmic 58

178

Chapter 18—Electrochemistry

CHAPTER 18 ELECTROCHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 18.1

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 18.1

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 18.1

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.

Section 18.2

Anions flow through the salt bridge into the anode compartment while cations flow into the cathode compartment to maintain electroneutrality.

Section 18.2

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 18.4

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 18.4

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 18.4

For a spontaneous reaction, G is negative but E is positive.

Section 18.4

Standard conditions for Eo, the standard cell potential, are identical with standard conditions for Go, Ho, and So.

Section 18.5

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 18.5

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 18.6

Half-cell potentials are never multiplied by a coefficient when used in the equation Eocell = Eoox + Eored.

Section 18.6

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 179

Chapter 18—Electrochemistry

the arrows, no reaction can occur because an oxidation requires a simultaneous reduction.

Section 18.7

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 18.8

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.

Section 18.9

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 18.10

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 18.10

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 18.11

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 18.11

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 18.12

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 18.12

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 18.13

Stress the use of this flowchart along with the dimensional analysis method of problem solving.

Lecture/Laboratory Demonstration References Section 18.2

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. 205-206.

Section 18.5

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.

180

Chapter 18—Electrochemistry

Section 18.11

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 18.12

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.

Literature References Section 18.1

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

Section 18.2

Ron DeLorenzo, “Electrochemical Errors,” J. Chem. Educ., Vol. 62, 1985, 424-425.

Section 18.2

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 18.6

Richard S. Treptow, “Dental Filling Discomforts Illustrate Electrochemical Potentials of Metals,” J. Chem. Educ., Vol. 55, 1978, 189.

Section 18.10

Michael J. Smith and Colin A. Vincent, “Structure and Content of Some Primary Batteries,” J. Chem. Educ., Vol. 78, 2001, 519-521.

Section 18.10

Gerald Ondrey, Charlene Crabb, and Takeshi Kamiya, “Batteries: Full Speed Ahead,” Chem. Eng., Vol. 106 (2), 1999, 47-51.

Section 18.11

A. I. Onuchukwu, “Evaluation of Corrosion Susceptibility of a Metal: Student Corrosion Experiment II,” J. Chem. Educ., Vol. 65, 1988, 934.

181

Chapter 19—Nuclear Chemistry

CHAPTER 19 NUCLEAR CHEMISTRY Chapter Learning Goals for Students Section 19.1 and Section 19.2

Give the different kinds of radioactive decay and the results of each.

Write balanced nuclear reactions.

Section 19.3

Predict the type of radioactive decay for a given isotope.

Write a balanced nuclear reaction for a radioactive decay series.

Section 19.4

Relate half-life and decay constant.

Calculate the amount of radioactive isotope remaining after a given amount of time.

Relate decay rates to decay constant, half-life, or amount remaining.

Section 19.5

Calculate mass defects and binding energies for nuclides.

Calculate the mass and energy change in a nuclear reaction.

Calculate the mass and energy change of a chemical reaction.

Section 19.6

Identify key aspects of fission and fusion reactions and their role in nuclear power and weapons.

Calculate the energy change associated with a fission or fusion reaction.

Section 19.7

Write balanced equations for nuclear transmutations.

Section 19.8

Calculate the radiation intensity in a radioactive sample and convert between units of radiation intensity.

Section 19.9

Calculate the age of an object using radioisotopic dating.

222

Chapter 19—Nuclear Chemistry

Lecture Outline 19.1Nuclear Reactions and Their Characteristics1 A. Nuclear reaction – a reaction that changes the nucleus 238 U → 42 He + 23490Th 92 B. Differences between nuclear reactions and chemical reactions 1. Nuclear reaction – a change in an atom’s nucleus; oftentimes, the identity of the atom changes.

2. Chemical reaction – a change in the way that atoms combine; the atom’s identity doesn’t change.

3. Different isotopes of an element a. Have essentially the same behavior in chemical reactions b. Have a different behavior in nuclear reactions 4. The energy changes accompanying nuclear reactions are far greater than those accompanying chemical reactions.

19.2. Radioactivity2 A. 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 1. Radioisotope – an unstable isotope 2. Three common types of radiation with different properties: alpha (), beta (), gamma () B. Alpha () 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 He2+ =  2 4. Emission of an 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) C. Beta () radiation 1. Attracted by a positively charged electrode 2. Repelled by a negatively charged electrode 3. In nuclear reactions electrons are beta particles 0 −1

e=

4. Superscript 0 – the mass of an electron is essentially zero when compared with that of a proton or neutron 5. Subscript (–1) – the relative 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 + -10e 6 D. Gamma () 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 1 2

Test Item File Questions: Multiple Choice 1- 3, Algorithmic 1 – 5, Short Answer 1 – 2 Test Item File Questions: Multiple Choice 4 – 20, Algorithmic 6 – 11, Short Answer 3 – 6

223

5. Almost always accompanies  and  emission – mechanism for release of energy 6. emission often not shown in nuclear equations – doesn’t change either the mass number or the atomic number of the product nucleus E. 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 + 0 + 1e =  1

e=

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 F. 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 (nuclear particles: neutrons and protons) on both sides are equal.

3. Sums of charges on nuclei and any elementary particles on both sides are equal 19.3. NuclearStability3 A. The ratio of neutrons to protons in the nucleus affects nuclear stability, see Figure 19.2.

1. 3600 known isotopes found in the band of nuclear stability 2. 264 isotopes are nonradioactive 3. Others undergo spontaneous radioactive decay B. Generalizations regarding nuclear stability 1. Every element has at least one radioactive isotope 2. Hydrogen is the only element whose most abundant isotope contains more protons than neutrons.

3. As atomic number increases, the ratio of neutrons to protons gradually increases 4. All isotopes heavier than Bi-209 are radioactive C. Within the band of nuclear stability, radioactive nuclei undergo decay reactions related to the isotopic neutron/proton ratio 1. Nuclei on the top side of the band are neutron-rich and undergo  emission which decreases the ratio 2. Nuclei on the bottom side of the band are neutron-poor and undergo positron emission, electron-capture or  emission which increase the ratio D. Many radioactive nuclei decay via a series of nuclear reactions, called a radioactive decay chain, until a stable state is achieved.

19.4. Radioactive Decay Rates4 A. Decay of radioactive nuclei – a first-order process 1. Integrated rate law – analogous to that given in Section 13.5.

ln

Nt = −kt N0

a. N0 is the number of radioactive nuclei initially present. b. Ntis the number of radioactive nuclei present at time t. c. k is the decay constant.

2. Decay rate – characterized by half-life 0.693 t1/2 = k

3 4

Test Item File Questions: Multiple Choice 21 – 30, Algorithmic 12 – 20, Short Answer 7 Test Item File Questions: Multiple Choice 31 – 39, Algorithmic 21 – 23, Short Answer 8

224

Chapter 19—Nuclear Chemistry 3. Calculation of the ratio of remaining and initial amounts of radioactive sample Nt/N0 at any time t

N   t  ln  t  = 0.693    N0   t1/2  19.5. Energy Changes During Nuclear Reactions5 A. Can calculate the energy change on formation of the nucleus from isolated protons and neutrons 1. E = 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: eV = 1.60 x 10−19J 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.

19.6. Nuclear Fission and Fusion6 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 800 different fission products a. Frequently occurring pathway 1 0

142 91 1 n + 235 92 U → 56 Ba + 36 Kr + 30 n

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.

5 6

Test Item File Questions: Multiple Choice 40 – 50, Algorithmic 24 – 26, Short Answer 9 – 17 Test Item File Questions: Multiple Choice 51 – 62, Algorithmic 27 – 28, Short Answer 18 – 20

225

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 19.7. Nuclear Transmutation7 A. Nuclear transmutation 1. The change of one element into another, often 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.

19.8. Detecting and Measuring Radioactivity8 A. Detect radiation by measuring its ionizing properties B. Devices such as Geiger counters are used for measuring radiation C. 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: Bq = 1 disintegration/s 2. Curie (Ci): 1 Ci = 3.7 1010 disintegrations/s a. Larger unit than the becquerel and more commonly used b. Often used in medicine and biochemistry 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 rem (roentgen equivalent for man) = 0.01 Sv D. The effects of ionizing radiation on the human body depend on several factors.

1. Kind of radiation and its energy 2. Length of exposure 3. Whether the source is outside or inside the body a. From an external source, x-rays and  radiation are more harmful than  and  particles b. From an internal source,  and  particles are more dangerous 4. The average radiation dose received annually by most people is only 120 mrem a. About 70% is from natural sources b. About 30% is from medical procedures c. Exposure to emissions from nuclear power plants and nuclear weapons fallout is barely detectable 19.9. Applications of Nuclear Chemistry9 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 14 7

N + 10n → 146 C + 11H

2. Carbon-14 reacts to form 14CO2, which mixes with 12CO2and is taken up by plants during photosynthesis.

Test Item File Questions: Multiple Choice 63 – 67, Algorithmic 29 – 31, Short Answer 21 – 22 Test Item File Questions: Multiple Choice 68 – 85, Algorithmic 32 – 36, Short Answer 23 – 24 9 Test Item File Questions: Multiple Choice 86 – 92, Algorithmic 37, Short Answer 25 – 29 7 8

226

Chapter 19—Nuclear Chemistry 14

C + O2 → 14CO2

3. Ratio of14C to 12C in a living organism is the same as that in the atmosphere.

4. 14C/12C decreases when the organism dies because14C undergoes radioactive decay 14 6

C → 147 N + 10e−

5. For14C, t½ = 5730 y 6. Can determine how long ago the organism died by measuring amount of 14C remaining 7. As the amount of 14C diminishes, accuracy decreases with an outer limit of about 60,000 y B. Use other radionuclides for dating rocks.

1. 238U a. For 238U, t½ = 4.46 109 y b. 238U decays through a series to give206Pb c. Measurement of 238U/206Pb ratio can determine the age of the uranium-containing rock 2. 40K a. For 40K, t½ = 1.28 109 y b. Potassium-40 decays through electron capture and positron emission 40 19

40 K + −01e → 18 Ar

40 19

40 K → 18 Ar + −01e

c. Compare amount of40Ar to 40K remaining in sample C. Medical uses of radioactivity – grouped into three classes 1. In vivo procedures 2. Therapeutic procedures 3. 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. 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 F. 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.

Inquiry – Are there any naturally occurring nuclear reactors?

CHAPTER 19 NUCLEAR CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 19.1

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

Section 19.2

An alpha () particle ( 2 He ) is a helium nucleus. The left subscript indicates the charge on the particle, and the +2 charge (right superscript) is not used in writing nuclear equations.

4

2+

227

Section 19.2

0 A beta () particle ( −1 e or –) is an electron.

Section 19.2

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 19.2

Electron capture: proton + core electron → neutron.

Section 19.2

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

Section 19.3

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

Section 19.5

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, E = mc2.

Section 19.5

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 19.8

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 19.8

High-energy photons ( and  rays) are the most penetrating forms of radiation.

Particulate forms of radiation ( and  particles) are less penetrating.

Lecture/Laboratory Demonstration References None

Literature References Section 19.1

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

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

Section 19.1

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

Section 19.1

S. G. Hutchinson and F. I. Hutchinson, “Radioactivityin Everyday Life,” J. Chem.

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

Section 19.1

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

Section 19.2

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

Section 19.2

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

228

Chapter 19—Nuclear Chemistry Section 19.4

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 19.4

M. W. Rowe, “Radioactive Dating: A Method for Geochronology,” J. Chem. Educ., Vol. 62, 1985, 580-584.

Section 19.4

M. W. Rowe, “Archeological Dating,” J. Chem. Educ., Vol. 63, 1986, 16-20.

Section 19.4

Ron DeLorenzo, “California Earthquakes: Predicting the Next Big One Using Radiocarbon Dating,” J. Chem. Educ., Vol. 57, 1980, 601.

Section 19.6

Ruth Lewin Sime, “Lise Meitner and the Discovery of Fission,” J. Chem. Educ., Vol. 66, 1989, 373-378.

Section 19.6

Ruth Lewin Sime, “Lise Meitner and the Discovery of Fission,” Scientific American, January 1998, 80-85.

Section 19.6

Michael Freemantle, “Ten Years After Chernobyl: Consequences Are Still Emerging,” Chem. Eng. News, April 29, 1996, 18-28.

Section 19.6

Gregory R. Choppin, “Aspects of Nuclear Waste Disposal of Use in Teaching Basic Chemistry,” J. Chem. Educ., Vol. 71, 1994, 826-829.

Section 19.6

Torkil H. Jensen, “Fusion – A Potential Power Source,” J. Chem. Educ., Vol. 71, 1994, 820-823.

Section 19.6

Charles H. Atwood, “How Much Radon Is Too Much?” J. Chem. Educ., Vol. 69, 1992, 351-353.

Section 19.9

Kent J. Crippen and Robert D. Curtright, “Modeling Nuclear Decay: A Point of Integration Between Chemistry and Mathematics,” J. Chem. Educ., Vol. 75, 1998, 1434-1436.

Section 19.9

M. W. Rowe, “Radioactive Dating: A Method for Geochronology,” J. Chem. Educ., Vol. 62, 1985, 580-584.

Section 19.9

M. W. Rowe, “Archeological Dating,” J. Chem. Educ., Vol. 63, 1986, 16-20.

Section 19.9

Ron DeLorenzo, “California Earthquakes: Predicting the Next Big One Using Radiocarbon Dating,” J. Chem. Educ., Vol. 57, 1980, 601.

Section 19.9

Rosalyn S. Yalow, “Development and Proliferation of Radioimmunoassay Technology,” J. Chem. Educ., Vol. 76, 1999, 767-768.

Section 19.9

Rosalyn S. Yalow, “Radioactivity in the Service of Man,” J. Chem. Educ., Vol. 59, 1982, 735-738.

Section 19.9

Mairin B. Brennan, “Positron Emission Tomography Merges Chemistry with Biological Imaging,” Chem. Eng. News, February 19, 1996, 26-33.

Section 19.9

Marcus E. Raichle, “Visualizing the Mind,” Scientific American,April 1994, 64.

Section 19.9

Timothy J. McCarthy, Sally W. Schwarz, and Michael J. Welch, “Nuclear Medicine

229

and Positron Emission Tomography: An Overview,” J. Chem. Educ., Vol. 71, 1994, 830-836.

230

CHAPTER 20 TRANSITION ELEMENTS AND COORDINATION CHEMISTRY Chapter Learning Goals for Students Section 20.1

Write electron configurations for transition metal atoms and ions and predict the number of unpaired electrons.

Section 20.2

Describe the trends in physicaland chemical properties oftransition metal elements acrossa period. Explain the reasons fortrends.

Section 20.3

Relate the oxidizing and reducing strength of a transition metal ion to its oxidation state and location in the periodic table.

Section 20.4

Predict the products and write balanced equations for the common laboratory reactions of first-series transition metals, their ions, and their compounds

Section 20.5 and Section 20.6

Identify Lewis acids, Lewis bases, ligands, and donor atoms in a coordination complex.

Determine the formula of a coordination complex, identify the oxidation state and coordination number of the metal atom, and draw the structure of the complex.

Classify a ligand as mono, bi, tri, tetra, or hexadentate based on its chemical structure.

Section 20.7

Determine the systematic name of a coordination complex from its formula or structure.

Determine the formula of a coordination complex from its systematic name or structure.

Section 20.8

Classify isomers according the scheme given in Figure 20.16.

Draw diastereomers for octahedral and square planar complexes.

Section 20.9

Classify a coordination complex as chiral or achiral and draw enantiomers for chiral compounds.

Section 20.10

Relate the color of a metal complex, the wavelength of light it absorbs, and the energy difference between its ground and excited state.

Section 20.11

Describe the bonding in a metal complex in terms of valence bond theory, includingorbital diagrams, the hybridorbitals used by the metal atom,and the number of unpairedelectrons.

Section 20.12

Draw crystal field energylevel diagrams for octahedral, tetrahedral, and square complexes. Classify ligands as strong or weak-field and predict (oraccount for) color and magnetic properties of complexes in terms of crystal field theory.

210

Chapter 20—Transition Elements and Coordination Chemistry

Lecture Outline 1

20.1. Electron Configurations 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 2 20.2.Properties of Transition Elements 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

1Test Item File Questions: Multiple Choice 1 – 12, Algorithmic 1 – 11, Short Answer 1 – 8 2Test Item File Questions: Multiple Choice 13 – 15, Algorithmic 12 – 17, Short Answer 9 – 10

211

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 H 2 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.

3 20.3. Oxidation States of Transition Elements 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 G 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 Z eff 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 Z eff 4 20.4. Chemistry of Selected Transition Metals 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 Cr 2O3 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 b. Cr3+ solutions are green – anions replace some of the bound water molecules to give green complex ions

3Test Item File Questions: Multiple Choice 16 – 21, Algorithmic 18 – 20, Short Answer 11 - 12 4Test Item File Questions: Multiple Choice 22 – 36, Algorithmic 21

212

Chapter 20—Transition Elements and Coordination Chemistry

9. Cr(OH)n a. Acid strength increases with increasing polarity ofO–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. Cr (VI) compounds must be handled with care because they are toxic and are known to cause cancer.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 steel 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 an oxidizing 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 Cu 2 (OH) 2SO4, the green patina seen on bronze monuments 5. Two common oxidation states: +1 and +2 6. ECu+ / Cu2+ less negative than ECu/Cu+ 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.

8. Cu2+– more common a. Aqueous blue-colored solution ofCu(H2O)2+ b. Numerous other solid state compounds and complex ions 213

Chapter 20—Transition Elements and Coordination Chemistry

5

20.5. Coordination Compounds 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 (electron pair donors) and the central metal ion behaves as a Lewis acid (electron pair acceptor).

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.

6 20.6. Ligands 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 4. Naturally occurring chelating ligands are essential constituents of many important biomolecules.

7 20.7.Naming Coordination Compounds 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.

E. Use a Roman numeral in parentheses immediately following the name of the metal, to indicate the metal’s oxidation state.

5Test Item File Questions: Multiple Choice 37 – 41, Algorithmic 22 – 23, Short Answer 13 – 16 6Test Item File Questions: Multiple Choice 42 – 43, Algorithmic 24 – 29, Short Answer 17 7Test Item File Questions: Multiple Choice 44 – 50, Algorithmic 30, Short Answer 18

214

Chapter 20—Transition Elements and Coordination Chemistry

F. In naming the metal, use the ending –ate if the metal is in an anionic complex.

8 20.8. Isomers 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 9 20.9. Enantiomers and Molecular Handedness 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.

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 8Test Item File Questions: Multiple Choice 51 – 52, Algorithmic 31 – 32, Short Answer 19 – 20 9Test Item File Questions: Multiple Choice 53, Algorithmic 33 – 35

215

Chapter 20—Transition Elements and Coordination Chemistry

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.

10 20.10. Color of Transition Metal Complexes 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 E = 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.

11 20.11. Bonding in Complexes: Valence Bond Theory 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 12 20.12. Crystal Field Theory 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 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.

10Test Item File Questions: Algorithmic 36, Short Answer 21 11Test Item File Questions: Multiple Choice 54 – 60, Algorithmic 37 – 38, Short Answer 22 12Test Item File Questions: Multiple Choice 61 – 70, Algorithmic 39 – 41, Short Answer 23 – 25

216

Chapter 20—Transition Elements and Coordination Chemistry

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 orbitals 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 () – the energy splitting between the two sets of d orbitals (textbook Figure 20.31) D. Crystal field splitting energy,  – corresponds to ’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  ii. High-spin complexes c. Strong-field ligands i. Produce a relatively large value of  ii. Low-spin complexes E. Spin-state depends on the relative values of  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. >P; low-spin arrangement has lower energy.

3. <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 20.32) H. Energy splitting in tetrahedral complexes is just the opposite of that in octahedral complexes.

1. (tetrahedral complexes) ½ (octahedral complexes) 2. None of the orbitals points directly at the ligands.

3. Four ligands instead of six 4. <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 Inquiry – How Does Cisplatin Kill Cancer Cells?

217

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.

218

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, , 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 >P, the complex is predicted to be low-spin. If <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.

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.

219

Chapter 20—Transition Elements and Coordination Chemistry

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-LchiroOchiai, “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.

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.

220

Chapter 20—Transition Elements and Coordination Chemistry

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.

221

CHAPTER 21 METALS AND SOLID-STATE MATERIALS Chapter Learning Goals for Students Section 21.1

Relate the formulas of metal-containing minerals and the location of the metals in the periodic table.

Section 21.2

Predict the relative magnitude of standard reduction potentials from an element’s position in the periodic table and determine the best method for extracting a metal from an ore.

Describe the concentration, reduction, and refining aspects of metallurgical processes, and write balanced equations for the reactions that occur when a metal is produced from its ores.

Section 21.3

Describe and balance reactions that occur in a blast furnace and the basic oxygen process in the production of steel.

Apply principles of thermodynamics to various reactions in the production of iron.

Section 21.4

Describe the electron-sea model of metals, and explain how it accounts for their properties.

Describe the molecular orbital theory for metals (band theory), draw MO energylevel diagrams for metals, and use band theory to account for their properties.

Describe the advantages of the two bonding theories in predicting properties of metals.

Section 21.5

Describe the bonding in semiconductors and account for their properties in terms of band theory.

Classify doped semiconductors as n-type or p-type.

Describe some semiconductor applications.

Predict relative band-gapenergies based on periodic trends inatomic size and electronegativity.

Relate the wavelength of lightemitted or absorbed by a semiconductorto the band gap in its band theoryenergy-level diagram.

Section 21.6 and Section 21.7

Write and balance a reaction forsynthesizing a superconductor byhydrolysis of metal ethoxides.

Describe the appearance of aplot of electrical resistance versustemperature for a superconductor.

Identify coordination numbersof bonds in superconductors.

224

Chapter 21—Metals and Solid-State Materials

Section 21.8

Describe the bonding in ceramicsand account for their properties interms of the bonding model.

Describe the synthesis of ceramics,and write balanced equations forthe reactions that occur in the sol–gelsynthesis method.

Section 21.9

Classify composite materials as ceramic–ceramic, ceramic–metal, or ceramic– polymer, and account for some of theirproperties.

225

Chapter 21—Metals and Solid-State Materials

Lecture Outline 1

21.1. Sources of the Metallic Elements 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 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.

2 21.2. Metallurgy 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

1Test Item File Questions: Multiple Choice 1 – 4, Algorithmic 1 – 13, Short Answer 1 – 3, 2Test Item File Questions: Multiple Choice 5 – 16, Algorithmic 14 – 18, Short Answer 4 – 7,

226

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, H 2, 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 Cu 2+ at an impure Cu anode, and Cu2+ from aqueous CuSO4 is reduced to Cu at a pure Cu cathode 3 21.3. Iron and Steel 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 4 21.4. Bonding in Metals A. Metals – malleable, ductile, lustrous, and good conductors of heat and electricity 1. Malleable – can be hammered into sheets 2. Ductile – can be drawn into wires 3Test Item File Questions: Multiple Choice 17 – 22, Algorithmic 19 – 20, Short Answer 8 4Test Item File Questions: Multiple Choice 23 – 29, Algorithmic 21 – 22, Short Answer 9 – 10

227

Chapter 21—Metals and Solid-State Materials

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

228

Chapter 21—Metals and Solid-State Materials

5

21.5. Semiconductors 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.

6 21.6. Semiconductor Applications 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 5Test Item File Questions: Multiple Choice 30 – 41, Algorithmic 23 – 25, Short Answer 11 – 13 6Test Item File Questions: Multiple Choice 42 – 43, Short Answer 14 – 19,

229

Chapter 21—Metals and Solid-State Materials

b. Wavelength of light roughly equal to band gap energy 3. Light sources a. Sports stadium video screens b. Vehicle brake lights 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 c. Fiber optic data transmission D. Photovoltaic cells 1. Solar cells 2. p–n-junctions 3. Covert sunlight into energy E. 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 7 21.7. Superconductors 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 Tcdiscovered 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 N2with 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. Commercial applications are emerging.

G. High-temperature superconductors based on fullerene – the allotrope of carbon that contains C60 molecules 1. K3C60 – metallic conductor at room temperature; superconductor at 19.5K, related fullerides have even higher Tc values 2. Fullerides are three-dimensional superconductors.

7Test Item File Questions: Multiple Choice 44 – 51, Algorithmic 26 – 27, Short Answer 20

230

Chapter 21—Metals and Solid-State Materials

8

21.8. Ceramics 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 m 3. Gel – a rigid, gelatin-like material 9 21.9. Composites 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 C. Composites in which the two phases are different types of materials 1. Ceramic–metal composites (cermets) 8Test Item File Questions: Multiple Choice 52 – 59, Algorithmic 28 – 30, Short Answer 21 - 22 9Test Item File Questions: Multiple Choice 60 – 61, Short Answer 23 – 24

231

Chapter 21—Metals and Solid-State Materials

2. Ceramic–polymer composites 3. High strength-to-weight ratios D. Ceramic fibers used in composites usually made by high-temperature methods 10 Inquiry – What Are Quantum Dots and What Controls Their Color?

10Test Item File Questions: Multiple Choice 62, Short Answer 25 – 26

232

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. 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.

233

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.

234

CHAPTER 22 THE MAIN-GROUP ELEMENTS Chapter Learning Goals for Students Section 22.1

Use the periodic table topredict the properties of themain-group elements and theircompounds.

Section 22.2

Account for differences in the structure and properties of second-row elements and the structure and properties of third- and higher-row elements.

Section 22.3

Describe the preparation of elemental hydrogen.

Write and balance a chemical equation for the reaction of ionic hydrides with water.

Classify binary hydrides and describe their bonding and properties.

Section 22.4 and Section 22.5

Use periodic trends to predict properties of Group 1A and 2A elements.

Write and balance reactions of alkali and alkaline earth metals with halogens, oxygen, and water.

Write electrode reactions and perform calculations in the electrolysis of molten salts to produce alkali and alkalineearth metals.

Section 22.6

Explain why the properties of boron differ from other 3A elements.

Describe the structure and bonding in diborane.

Section 22.7

Draw electron-dot structures and predict hybrid orbitals and geometry for molecules and ions in main-group compounds containing elements from Group 4A. Describe the structure and properties of the allotropes of carbon.

Section 22.8

Use the shorthand notation for a silicate anion to represent its structure and interpret the notation to find the formula and charge of a silicate anion.

Section 22.9

Write electron-dot structures and describe the structure, bonding, and properties of molecules and ions in main group compounds containing nitrogen and phosphorus.

Section 22.10

Relate compound names, formulas, and oxidation states in compounds.

Section 22.11

Classify the oxide of an element as ionic or covalent, and as acidic, basic, or amphoteric.

Section 22.12

Write reactions of oxides and sulfur compounds.

Write electron-dot structures and describe the structure, bonding, and properties of molecules in main-group compounds containing oxygen and sulfur.

194

Chapter 22—The Main-Group Elements

Section 22.13

Write balanced equations for the formation of halogens and reactions of their compounds.

Relate compound names, formulas, and oxidation states in compounds.

Predict the relative acid strengths of oxoacids.

Section 22.14

Assign oxidation numbers to elements in noble gas compounds.

195

Chapter 22—The Main-Group Elements

Lecture Outline 1

22.1. A Review of General Properties and Periodic Trends 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.

2 22.2. Distinctive Properties of the Second-Row Elements 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 secondrow 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  overlap of the 2p orbitals 1. 3p orbitals are more diffuse. a. Longer bond distances b. Poor  overlap 2.  bonds involving p orbitals – rare for elements of the third and higher rows 3 22.3. Group. 1A: Hydrogen A. 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 B. 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 4. Third most abundant element on an atom basis C. Laboratory synthesis – reaction of dilute acid with an active metal M(s) + 2n H+(aq) →nH2(g) + Mn+(aq) D. Hydrogen compounds 1. Hydrogen has properties similar to both the alkali metals and the halogens. a. Lose electron → H+ i. Ei = 1312 kJ/mol 1Test Item File Questions: Multiple Choice 1 – 9, Algorithmic 1 – 9, Short Answer 1 – 2 2Test Item File Questions: Multiple Choice 10 – 12, Algorithmic 10 – 16, Short Answer 3 3Test Item File Questions: Multiple Choice 13 – 16, Algorithmic 17 – 26

196

Chapter 22—The Main-Group Elements

ii. Doesn’t completely transfer its valence electron in ordinary chemical reactions iii. Shares electron with a nonmetallic element to produce a covalent compound iv. Gas phase – complete ionization of a hydrogen atom to produce a bare proton v. Liquids and solids – bare proton too reactive to exist by itself b. Gain electron → H– i. Eea = – 73 kJ/mol ii. Will accept an electron from an active metal to give an ionic hydride E. Binary hydride – compound that contains hydrogen and just one other element F. Ionic hydride – formed by alkali metals and heavier alkalineearth 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 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 G. Covalent hydrides – hydrogen attached to another element by a covalent bond 1. Most common – hydrides of nonmetallic elements 2. Discrete, small molecules a. Relatively weak intermolecular forces b. Gases or volatile liquids at ordinary temperatures H. 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 22.4. Group 1A: Alkali Metals 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 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 4Test Item File Questions: Multiple Choice 17 – 21, Algorithmic 27 – 28, Short Answer 4 – 5

197

Chapter 22—The Main-Group Elements

2. 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) 3. 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 5 22.5. Group 2A: Alkaline-Earth Metals 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 alkalineearth metals 1. Same kinds of redox reactions as alkali metals 2. Less reactive than alkali metals – Ei (alkalineearth) >Ei (alkali) 3. Halogens → ionic halide salts, MX2 4. Oxygen → oxides, MO; Sr, Ba → peroxides, MO2 5. Water → metal hydroxides, M(OH)2 6 22.6. Group 3A A. Group 3A is the first of the p-block elements with valence electron configurations of ns2np1 B. Except for boron these elements are metals.

1. Aluminum has many commercial applictions.

2. Gallium has the largest liquid range of any metal and is used in making GaAs, a semiconductor.

3. Indium is also used in making semiconductor devices.

4. Thallium is extremely toxic and has no commercial uses.

5. Metallic character increases upon moving down the periodic table.

6. Boron a. Smaller and more electronegative than the other members of the group and forms covalent bonds rather than ionic bonds. b. Semimetal.

C. Boron 1. Occurs in concentrated deposits of borate minerals 2. Obtained from BBr3 2 BBr3(g) + 3 H2(g) → 2 B(s) + 6 HBr(g) 3. Crystalline boron – strong, hard, high-melting substance a. Chemically inert at room temperature b. Desirable component in high-strength composite materials D. Boron compounds 1. Boron halides – highly reactive, volatile covalent compounds a. BX3molecules 5Test Item File Questions: Multiple Choice 22 – 26, Algorithmic 29 – 30, Short Answer 6 6Test Item File Questions: Multiple Choice 27 – 40, Algorithmic 31 – 35, Short Answer 7 – 9

198

Chapter 22—The Main-Group Elements

b. Behave as Lewis acids – uses vacant 2p orbitals in accepting a share in a pair of electrons from a Lewis base 2. Boron hydrides (boranes) – volatile, molecular compounds with formulas BnHm a. Simplest – diborane (B2H6) b. Two BH2 groups connected by two bridging H atoms i. Geometry around B atoms – tetrahedral ii. Bridging B–H bonds significantly longer than terminal B–H bonds c. Diborane – 12 valence electrons; electron deficient i. B atoms use sp3 hybrid orbitals to bond to four neighboring H atoms. ii. Bond formed from overlap of a boron sp3 hybrid orbital and hydrogen 1s orbital iii. Three-center, two-electron bond — joins each bridging H atom to both B atoms iv. Electron density between adjacent atoms less than in an ordinary two-center, twobond bond v. Two electrons in B–H–B bridge spread out over three atoms 7 22.7. Group 4A: Carbon A. Group members include C, Si, Ge, Sn, and Pb 1. Especially important, both in industry and in living organisms 2. Increase in metallic character down a group in the periodic table 3. Valence electron configuration: ns2np2 4. Most common oxidation state: + 4; covalent compounds 5. +2 oxidation state a. Occurs for Sn and Pb b. Most stable for Pb c. Ionic compounds 6. No simple M4+(aq) ions for any of the group 4A elements B. Carbon 1. In uncombined form, carbon found as diamond and graphite 2. Diamond a. Has a covalent network structure in which each C atom uses sp3 hybrid orbitals to form a tetrahedral array of  bonds i. Interlocking three-dimensional network of strong bonds ii. Hardest known substance iii. Highest known melting point b. Electrical insulator – valence electrons localized in  bonds 3. Graphite a. Two-dimensional sheetlike structure. b. Uses sp2 hybrid orbitals i. Forms trigonal planar  bonds with three neighboring C atoms ii. Remaining p orbital (⊥ plane of sheet) used to form a  bond iii. bonds delocalized and free to move in plane of sheet iv. Electrical conductivity parallel to the sheets 10 20 greater than diamond c. Useful as electrode material d. Carbon sheets held together by London dispersion forces i. Can easily slide over each other ii. Slippery feel iii. Used as a lubricant iv. Difficult for electrons to hop from one sheet to another 4. Graphene: two-dimensional array of hexagonally arranged carbon atoms that is just one atom thick a. Strong and flexible b. Superb electrical conductor

7Test Item File Questions: Multiple Choice 41 – 63, Algorithmic 36 – 40, Short Answer 10 – 14

199

Chapter 22—The Main-Group Elements

5. Fullerene – third crystalline allotrope of carbon; found in soot formed by vaporizing graphite a. Spherical C60 molecules with shape of a soccer ball b. Prepared by electrically heating a graphite rod in a helium atmosphere c. Molecular substance – soluble in nonpolar, organic solvents d. Other carbon clusters i. Egg shaped – C70 6. Oxides of carbon a. CO i. Colorless, odorless, toxic gas ii. Forms when C or hydrocarbon fuels burned in a limited supply of oxygen iii. Toxicity a. Due to its ability to bond strongly to the Fe2+ atom of hemoglobin b. Impairs the ability of hemoglobin to carry O 2 to the tissues b. CO2 i. Colorless, odorless, nonpoisonous gas ii. Formed when fuels burn in an excess of O 2 7. Carbonates a. Carbonic acid – forms two series of salts: carbonates and hydrogen carbonates b. Na2CO3 – soda ash i. Used in making glass ii. Na2CO310 H2O – washing soda a. Used in laundering textiles b. CO32– removes cations from hard water c. Produces OH–, which removes grease from fabrics c. NaHCO3 – baking soda i. Reacts with acidic substances in food ii. Produces CO2 iii. Causes dough to rise 8. Hydrogen cyanide and cyanides a. HCN i. Highly toxic, volatile substance ii. CN–(aq) + H+(aq) →HCN(aq) b. CN– i. Aqueous solutions are very weakly acidic. ii. Acts as a Lewis base – bonds to transition metals through the lone pair of electrons on carbon c. CN– used to extract Au and Ag from their ores. i. Produces M(CN)2– ion ii. M(CN)2– reduced by Zn to produce M iii. No longer used in U.S. due to toxicity of CN– 9. Carbides a. Binary compounds of carbon b. Carbon atom has a negative oxidation state. i. Carbides of active metals ii. Interstitial carbides of transition metals iii. Covalent network carbides 8 22.8. Group 4A: Silicon A. Hard, gray, semiconducting solid that melts at 1410C 1. Diamondlike structure 2. Relatively poor overlap of  orbitals – no graphitelike allotrope 3. Naturally found combined with oxygen (SiO2 and various silicate minerals) 8Test Item File Questions: Multiple Choice 64 – 68, Algorithmic 41 – 42, Short Answer 15 – 18

200

Chapter 22—The Main-Group Elements

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 22.8) 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– (textbook Figure 22.9) 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 AlO4tetrahedra 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 AlO4tetrahedra joined together in an open structure that has a threedimensional 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 9 22.9.Group 5A: Nitrogen A. Group members are nitrogen, phosphorus, arsenic, antimony, and bismuth 1. Down the periodic group – increasing atomic size, decreasing ionization energy, and decreasing electronegativity 2. Increasing metallic character of heavier elements – evident in the acid–base properties of their oxides a. N and P oxides – acidic b. As and Sb oxides – amphoteric c. Bi2O3 oxides – basic 9Test Item File Questions: Multiple Choice 69 – 85, Algorithmic 43 – 44,Short Answer 19 – 21

201

Chapter 22—The Main-Group Elements

3. Electron configuration: ns2np3 a. Maximum oxidation state: + 5 b. Minimum oxidation state: – 3 c. N and P – exhibit all oxidation states between – 3 and + 5 d. As and Sb – most important oxidation states: + 3 and + 5 4. Metallic character – increases down the group a. Sb3+ and Bi3+ – found in salts b. No simple cations in compounds of N or P 5. As, Sb, and Bi found in sulfide ores and used in making various metal alloys B. Nitrogen 1. Colorless, odorless, tasteless gas 2. Makes up 78% of earth’s atmosphere 3. Separated out of liquid air by fractional distillation 4. Uses of N2 a. Gas – protective inert atmosphere in manufacturing processes b. Liquid – refrigerant c. Most important use – Haber process (manufacture of NH3) 5. N2 unreactive due to N–N triple bond a. Reactions involving N2 have a high activation energy and/or an unfavorable equilibrium constant. b. Formation of nitric oxide from N2 and O2 has anunfavorable large equilibrium constant c. Formation of nitric oxide more favorable at high temperatures such as a car engine— major source of air pollution C. 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. Reacts with acids to yield ammonium salts – resemble alkali metal salts in their solubility D. 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 E. Oxides of nitrogen 1. N2O – colorless, sweet-smelling gas a. Laughing gas – small doses are mildly intoxicating b. Used as a dental anesthetic c. Propellant for dispensing whipped cream 2. NO – colorless gas a. Produced in small quantities by reacting Cu with dilute HNO 3(aq) b. Important in biological processes 3. NO2 – highly toxic, reddish brown gas that forms rapidly when NO is exposed to air a. Dimerizes to form N2O4 b. N2O4 predominates at lower temperatures and NO2 predominates at higher temperatures.

F. 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 202

Chapter 22—The Main-Group Elements

c. Disproportionation of NO2in water 4. Strong acid – 100% dissociated in water 5. 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 6. 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 c. Cl– – converts Au3+ to AuCl4– 10 22.10.Group 5A: Phosphorus A. Most abundant Group 5A element 1. Found in phosphate rock, Ca3(PO4)2, and in fluorapatite, Ca5(PO4)3F 2. Apatites – phosphate minerals with the formula 3 Ca3(PO4)2 CaX2 (X– = F– or OH–) 3. Sixth most abundant element in human body a. Bones – Ca3(PO4)2 b. Tooth enamel – Ca5(PO4)3OH c. Found in DNA and RNA 4. Phosphoric acid is one of the ten most important commercial chemicals B. Exists in two common allotropic forms 1. White phosphorus – toxic, waxy, white solid existing as 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. More stable than white phosphorus C. Phosphorus compounds 1. Most important oxidation states: + 3 and + 5 2. More likely to be found in a positive oxidation state because of its electronegativity D. Phosphine (PH3) – colorless, extremely poisonous gas 1. Simplest hydride of phosphorus 2. Neutral aqueous solutions 3. Poor proton acceptor 4. Easily oxidized: burns in air to form H 3PO4 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% H3PO4by mass) b. Manufacturing method – depends on use 10Test Item File Questions: Multiple Choice 86 – 92, Algorithmic 45 – 46, Short Answer 22 – 25

203

Chapter 22—The Main-Group Elements

i. Food additive – begin with molten phosphorus ii. Fertilizers begin with Ca3(PO4)2 c. Sometimes called orthophosphoric acid 11 22.11.Group 6A: Oxygen A. Group 6A members are oxygen, sulfur, selenium, tellurium, and polonium 1. Periodic trends follow the usual patterns a. Oxygen and sulfur – typical nonmetals b. Gray selenium i. Most stable allotrope of selenium ii. Lustrous semiconducting solid c. Tellurium – semiconductor; classified as a semimetal d. Polonium i. Radioactive element that occurs in trace amounts in uranium ores ii. Silvery white metal 2. Electron configuration: ns2np4 a. Common oxidation state: – 2 b. Stability of state decreases with increasing metallic character i. Oxygen – powerful oxidizing agent ii. H2Se and H2Te – reducing agents c. S, Se, and Te – less electronegative than oxygen i. Found in positive oxidation states ii. Common oxidation state: + 4 and + 6 3. Commercial uses of Se, Te, and Po – limited a. Se – making red-colored glass and in photocopiers b. Te – used in alloys c. Po – heat source in space equipment and a source of alpha particles in research B. Oxygen 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 C. 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 D.Preparation and uses of oxygen 1. Laboratory preparation - thermal decomposition of an oxoacid salt 2. Industrial synthesis a. Fractional distillation of liquefied air b. Third ranking industrial chemical produced in the U.S. 3. Industrial uses a. Steelmaking – removes impurities from iron by forming oxides b. Sewage treatment – destroys malodorous compounds by oxidation c. Paper bleaching – oxidizes compounds that cause unwanted colors d. Oxyacetylene torches – provide high temperatures needed for cutting and welding metals e. Inexpensive and readily available oxidizing agent 11Test Item File Questions: Multiple Choice 93 – 107, Algorithmic 47 – 59, Short Answer 26 – 28

204

Chapter 22—The Main-Group Elements

E. Oxygen compounds 1. High electronegativity 2. Needs only two electrons to achieve an octet 3. Reacts with active metals to produce an ionic oxide – gains two electrons and achieves a noble gas configuration 4. Forms covalent oxides with nonmetals a. Achieves a noble gas configuration by sharing two additional electrons b. Forms two single bonds or one double bond. i. Double bonds are formed between oxygen and other small atoms ii. Good overlap between oxygen p orbital and p orbitals of small atoms F. Acid–base properties of oxides 1. Oxidation number = – 2, oxides 2. Categorize in terms of acid–base character a. Basic oxides i. Ionic ii. Formed by metals on left side of periodic table b. Acidic oxides (acid anhydrides) i. Covalent ii. Formed by nonmetals on right side of periodic table c. Amphoteric oxides i. Exhibit both acidic and basic properties ii. Formed by elements with intermediate electronegativity iii. Bonds have intermediate ionic–covalent character d. Acid–base properties and ionic–covalent character dependencies i. Position of element in the periodic table a. Acidic and covalent character increases left to right across the periodic table b. Basic and ionic character increases top to bottom down the periodic table ii. Oxidation state of element – acidic and covalent character increases with increasing oxidation state 12 22.12.Group 6A: Sulfur A. One of the ten most abundant elements on earth.

1. Occurs in elemental form in underground deposits.

2. Present in numerous minerals such as pyrite, galena, cinnabar, and gypsum.

3. Found in natural gas and crude oil as H2S. 4. Incorporated into various proteins in plants and animals.

5. Used in the manufacture of sulfuric acid, a commercially important chemical.

B. 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) d. Melts at 113o 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. S8ring 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 12Test Item File Questions: Multiple Choice 108 – 116, Algorithmic 60, Short Answer 29

205

Chapter 22—The Main-Group Elements

a. Pour boiling liquid into water – forms plastic sulfur b.Chains freeze in disordered array C. 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 D. Oxides 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 E. 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 13 22.13.Group 7A: The Halogens A. General properties 1. Exist as diatomic molecules 2. Too reactive to exist in nature in their elemental form 3. Valence-shell electron configuration – ns2np5 a. Gain electrons b. Large positive ionization energies c. Large negative Eea 3. Occur in nature as salts and minerals B. Produced by oxidation of their anions C. Halogen compounds 1. Most reactive elements in the periodic table 2. Less reactive going down the periodic table a. Flourine often reacts violently b. Chlorine and bromine less reactive c. Iodine often reacts sluggishly 3. Metals → metal halides, MXn D.Hydrogen halides 1. HX – behave as acids when dissolved in water 2. HCl used in a large number of commercial applications 3. HF used to etch glass; one of only a few substances that react with glass E. Oxoacids and oxoacid salts 1. Most important compounds of halogens in positive oxidation states 2. General formula – HXOn 3. Oxidation state = + 1, + 3, + 5, or + 7 (depends on value of n) 4. Acid strength increases with increasing oxidation state of the halogen a. Acidic hydrogen bonded to oxygen b. Strong oxidizing agents 14 22.14.Group 8A: Noble Gases 13Test Item File Questions: Multiple Choice 117 – 132, Short Answer 30 – 32

206

Chapter 22—The Main-Group Elements

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. No known compounds of helium or neon.

2. Only one compound of argon known.

3. Kr and Xe react only with fluorine a. Xenon fluorides – powerful oxidizing agents 4. Lack of reactivity due to valence-shell configuration a. Large Ei b. Small Eea Inquiry – What Are the Barriers to a Hydrogen Economy?

14Test Item File Questions: Multiple Choice 133 – 134,

207

Chapter 22—The Main-Group Elements

CHAPTER 22 THE MAIN-GROUP ELEMENTS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 22.2

Due to small size, high electronegativity, and the ability to form  bonds, the first member of each s- and p-block group differs substantially in its chemistry from the other members of the group.

Section 22.2

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 22.2

Molecules of period 2 p-block elements contain single, double, or triple bonds; molecules of higher period elements generally contain only single bonds.

Section 22.3

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 22.3

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 22.3

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 22.3

Like all molecular compounds which have weak intermolecular forces of attraction, covalent hydrides are gases, liquids, or low-melting solids.

Section 22.6

2c–2e terminal B–H bonds are stronger and shorter than bridging 3c–2e bonds in B2H6.

Section 22.7

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 22.7

Diamond is an electrical insulator, but graphite is an electrical conductor.

Section 22.7

CN– and CO have the same number of electrons and the same Lewis electron-dot structure (:CN:– and :CO:).

Section 22.8

The number of terminal (unshared) oxygen atoms equals the charge on the silicate.

Section 22.9

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 22.9

Any highly exothermic reaction that generates gaseous products has the potential to occur with explosive force.

Section 22.9

When an element occurs in two extremely different oxidation states in a compound, as does nitrogen in NH4NO3, the compound may decompose violently.

208

Chapter 22—The Main-Group Elements

Section 22.9

The equilibrium for an exothermic reaction shifts to the left as the temperature is raised.

Section 22.9

A change in reaction conditions may result in a change in the reaction products.

Section 22.9

Note that both allotropes of period 3 phosphorus have only single bonds, whereas period 2 nitrogen has a triple bond in N2.

Section 22.9

In oxoacids, only hydrogen atoms attached to oxygen are acidic.

Section 22.11

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 22.11

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 22.11

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 22.11

Note that all allotropes of period 3 sulfur have only single bonds, whereas period 2 oxygen has a double bond in O2.

Section 22.12

Expecting to see a yellow solid, students frequently do not recognize finely divided sulfur, which appears white.

Section 22.12

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 22.13

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 22.1

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 22.1

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 22.9

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 209

Chapter 22—The Main-Group Elements

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 22.11

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 22.11

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.

Section 22.12

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 22.13

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 I 2 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 Br 2 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 22.13

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.

Literature References Section 22.1

George Gorin, “Mendeleev and Moseley: The Principal Discoverers of the Periodic Law,” J. Chem. Educ., Vol. 73, 1996, 490-493.

Section 22.1

Eric R. Scerri, “The Evolution of the Periodic System,” Scientific American, September 1998, 8-83.

Section 22.1

N. K. Goh and L. S. Chia, “Using the Learning Cycle to Introduce Periodicity,” J. Chem. Educ., Vol. 66, 1989, 747.

Section 22.1

Robert H. Goldsmith, “Metalloids,” J. Chem. Educ., Vol. 59, 1982, 526-527.

Section 22.1

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.

210

Chapter 22—The Main-Group Elements

Section 22.1

Joan Mason, “Periodic Contractions Among the Elements; Or, on Being the Right Size,” J. Chem. Educ., Vol. 65, 1988, 17-20.

Section 22.1

Gabriel Pinto, “Using Balls from Different Sports to Model the Variation of Atomic Sizes,” J. Chem. Educ., Vol. 75, 1998, 725-726.

Section 22.1

Ronald L. Rich, “Periodicity in the Acid-Base Behavior of Oxides and Hydroxides,” J. Chem. Educ., Vol. 62, 1985, 44.

Section 22.11

Foil A. Miller, “Joseph Priestly, Preeminent Amateur Chemist,” J. Chem. Educ., Vol 66, 1989, 801.

Section 22.11

Michael Laing, “The Three Forms of Molecular Oxygen,” J. Chem. Educ., Vol 66, 1989, 453-455.

Section 22.12

Miriam C. Nagel, “Herman Frasch, Sulfur King,” J. Chem. Educ., Vol. 58, 1981, 6061.

211

CHAPTER 23 ORGANIC AND BIOLOGICAL CHEMISTRY Chapter Learning Goals for Students Section 23.1

Represent the chemical structure of alkanes as condensed structures or line drawings.

Identify and draw isomers of alkanes.

Section 23.2

Identify and name functional groups in organic molecules.

Represent molecules with functional groups using line drawings..

Section 23.3

Convert between the IUPAC name and condensed formula or line drawing of alkanes and unsaturated compounds.

Section 23.4

Classify a monosaccharide as an aldose or ketose and draw a typical open-chain structure.

Classify the type of isomerism in monosaccharides.

Section 23.5

Draw orbital overlap pictures to describe bonding in organic compounds.

Identify and draw cis–trans isomers.

Section 23.6

Recognize and draw the structure of triacylglycerols.

Convert between chemical structure and name of lipids.

Relate lipid structure tophysical properties..

Section 23.7

Used curved arrows to depict electron movement to draw resonance structures and specify formal charges.

Section 23.8

Draw orbital overlap pictures for conjugated systems and identify localized and delocalized electrons.

Identify conjugated systems and the hybridization of atoms included in the system..

Section 23.9

Identify amino acids, show how they join together to form peptides, and name small peptides.

Section 23.10

Predict if an organic compound is aromatic.

Section 23.11

Identify the bases found in the nucleic acids, draw structures showing how different components of the nucleic acid are joined, and write the DNA sequence that is complementary to a specified strand.

Draw resonance structures to show that DNA bases are aromatic.

235

Chapter 23—Organic and Biological Chemistry

Lecture Outline 1

23.1. Organic Molecules and Their Structures: Alkanes A. Hydrocarbons – molecules that contain only carbon and hydrogen B. Alkanes 1. Hydrocarbons that have only single bonds 2. Methane, CH4, is the simplest alkane 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. Condensed structures – carbon–hydrogen and carbon–carbon single bonds aren’t shown.

1. Bonds are understood.

2. The horizontal bonds between carbons aren’t shown.

G. A molecule can be arbitrarily shown in a great many ways.

1. Rotation around carbon–carbon single bonds is possible.

2. 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 H. Line drawings are another shorthand representation of molecular structure.

1. Only the bonds in the carbon backbone are shown 2. The end of each line segment represents a carbon atom 3. Mentally supply the proper number of hydrogen atoms needed to fill out carbon’s valency.

I. Acylic hydrocarbons – open-chain hydrocarbons J. Cycloalkanes – alkanes that contain rings of carbon atoms K. Condensed structures can be awkward so line structures used 1. Structures are represented by polygons.

2. Rings larger than four carbons have a puckered shape.

2 23.2. Families of Organic Molecules: Functional Groups 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 3 23.3. Naming Organic Compounds 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. The 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 1Test Item File Questions: Multiple Choice 1 – 17, Algorithmic 1 – 2, Short Answer 1 – 12, 18 2Test Item File Questions: Multiple Choice 18 – 35, Algorithmic 3 – 21, Short Answer 13 – 17 3Test Item File Questions: Multiple Choice 36 – 57, Algorithmic 22 – 32, Short Answer 19 – 24

236

Chapter 23—Organic and Biological Chemistry d. Butane – CH3CH2CH2CH3 2. All other alkanes are named from Greek numbers according to the number of carbons present.

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 main 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 –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. 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.

E. Unsaturated Organic Compounds: Alkenes and Alkynes 1. Alkenes – hydrocarbons that contain a carbon–carbon double bond 2. Alkynes – hydrocarbons that contain a carbon–carbon triple bond 3. Unsaturated hydrocarbons a. Alkanes are said to be saturated having the maximum possible number of hydrogens bonded to carbon b. Alkenes and alkynes are said to be unsaturated having fewer hydrogens per carbon than the related alkanes 4. Name alkenes by counting the longest chain of carbons that contains the double bond, and adding the family suffix –ene 5. 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 6. Alkene isomers that exist because of double-bond geometry

237

Chapter 23—Organic and Biological Chemistry

4

23.4. Carbohydrates: A Biological Example of Isomers A. Carbohydrates are a large class of polyhydroxylated aldehydes and ketones—commonly called sugars 1. Simple sugars or monosaccharides cannot be broken down into smaller molecules by hydrolysis 2. Complex carbohydrates or polysaccharides consist of many simple sugars linked together and can be broken down by hydrolysis B. Monosaccharides 1. Classified as aldoses, which contain an aldehyde carbonyl group, or ketoses, which contain a ketone carbonyl group 2. The -ose suffix indicates a sugar 3. The number of carbon atoms in the sugar is indicated by the appropriate numerical prefix 4. Often depicted as having open-chain structures 5. Exist primarily as cyclic molecules in which an –OH group near one end of the chain adds to the carbonyl group at or near the other end of the chain to form a ring a. In the  cyclic form, the –OH group at C1 is on the bottom of the ring b. In the  cyclic form, the –OH group at C1 is on the top of the ring 6. Three types of isomers possible a. Constitutional isomers which have the same chemical formula but different atom connections such as linear and branched. b. Enantiomers which are nonidentical mirror images c. Anomers which have a –OH group on different sides of the ring C. Polysaccharides 1. Sucrose or table sugar is a common disaccharide composed of one molecule of glucose joined to one molecule of fructose 2. Cellulose is a fibrous polysaccharide composed of several thousand -glucose molecules joined in a chain 3. Starch is an edible polysaccharide that contains -glucose units 5 23.5. Valence Bond Theory and Orbital Overlap Pictures A. Simple representations of organic molecules include electron dot and line structures.

1. Indicate atom connections and the location of multiple bonds and electron lone pair 2. Limited in that they are do not adequately represent differences in structure or reactivity.

B. Valence bond theory describes a covalent bond as arising from overlapping orbitals on different atoms.

1. Hybridized and unhybridized orbitals can overlap 2. Overlap type leads to sigma, , and pi, , bonds C. Many organic molecules have a carbon atom with sp2 hybridization which leads to  bonds.

1.Energetically unfavorable rotation around the double bond.

2. Geometric or cis/trans isomers are possible a. cis arrangement has groups on the same side of the double bond. b. trans arrangement has groups on opposite sides of the double bond.

6 23.6. Lipids: A Biological Example of cis-transIsomerism A. Lipids are naturally occurring organic molecules that dissolve in nonpolar solvents.

B. There are a great many different kinds of lipids, all contain large hydrocarbon portions 1. Animal fats are usually solid, for example, butter and lard 2. Vegetable oils are usually liquid, for example, peanut and corn oils 3. Both have similar structures C. Fats and oils are triacylglycerols or triglycerides

4Test Item File Questions: Multiple Choice 58 – 68, Algorithmic 33 – 34, Short Answer 25 5Test Item File Questions: Multiple Choice 69 – 72, Algorithmic 35 – 37, 6Test Item File Questions: Multiple Choice 73 – 79, Short Answer 26 – 27

238

Chapter 23—Organic and Biological Chemistry 1. Esters of glycerol (1,2,3-propanetriol) with three long-chain carboxylic acids called fatty acids 2. Fatty acids are usually unbranched and have an even number of carbon atoms ranging from 12-22 3. The three fatty acids of a given triacylglycerol need not be the same 4. A fat or oil is a mixture of many different triacylglycerols D. About 40 different fatty acids occur naturally 1. Saturated fatty acids include palmitic acid (C16) and stearic acid (C18) 2. Monounsaturated fatty acids have one double bond, for example, oleic acid 3. Polyunsaturated fatty acids have more than one double bond, for example, linoleic acid and linolenic acid E. Vegetable oils generally have a higher proportion of unsaturated fatty acids than do animal fats.

1. Fats have saturated fatty acids that can line up with one another to maximize their London dispersion forces leading to a solid.

2. Oils with unsaturated fatty acids in the cis confirmation have a geometry which disrupts the intermolecular forces leading to liquids.

F. Oils can be processed by hydrogenation which converts unsaturated oils to saturated fats.

1. The amount of unsaturation affects the melting point.

2. Hydrogenation improves shelf-life as double bonds are susceptible to reaction with oxygen.

3. Isomerization of the double bonds can occur leading to trans fats.

G. Unsaturated are considered to be healthier.

H. Some lipids are considered to be essential and must be taken in through diet as they cannot be biosynthesized.

I. Numbering lipids 1. Number indicates the length of the carbon chain in the fatty acid, starting from the carboxylic acid functional group.

2. Omega designation numbers the first carbon of the double bond from the last carbon atom in the fatty acid.

23.7. Formal Charge and Resonance in Organic Compounds A. Formal charge and resonance play important roles in understanding structure and reactivity.

1. Most organic reactions occur between regions of high and low electron density.

2. Resonance helps predict stability and sites of high and low electron density.

B. Benzene has two resonance structures which contribute to the resonance hybrid.

1. Six electrons in the double bonds are delocalized over the six-ring carbon atoms. a. Reduces electron–electron repulsions. b. Stabilizes the molecule.

2. Resonance hybrid is an average of the resonance structures.

C. Common patterns that indicate resonance 1. Lone pair on an atom one  bond away from a  bond.

2. Two  bonds separated by a  bond.

3. A positive formal charge on an atom one  bond from a  bond.

7 23.8. Conjugated Systems A. Can combine a line drawing with orbital overlap picture 1. Provides a good representation of the bonding in organic molecules.

2. Lines represent  bonds.

3. p orbitals superimposed on the line structure to show multiple bonds B. Geranial has a  system with more than two adjacent p orbitals that can overlap 1. Four sp2 hybridized carbon atoms in a row 2. The four atoms are conjugated. a. Alternating single and double bonds. b. Two or more  bonds separated by  bonds. c. Double bonds are linked and do not act as separate, independent double bonds.

7Test Item File Questions: Algorithmic 38

239

Chapter 23—Organic and Biological Chemistry

C. Conjugation and resonance 1. Line drawing localizes electron in a bond.

2. In a conjugated system, the electrons are best described as being delocalized or associated with more than one or two atoms at the same time.

3. Conjugation provides a structural basis for resonance theory.

4. Resonance theory says that a  bond doesn’t fluctuate between two locations but is in both at the same time.

5. Simplified orbital picture shows electrons are localized in  bonds but delocalized in  bonds in a conjugated system. a. Overlap of the p orbitals involving the  bond and lone pair spread the four electrons over three atoms. b. Delocalization leads to resonance stabilization.

D. Conjugation and hybridization 1. According to valence bond theory atoms will adopt the hybridization that makes the molecule most stable by lowering the overall energy. a. Conjugation is stabilizing. b. An atom that is not conjugated will adopt a stable hybridization according to the VSEPR model.

8 23.9. Proteins: A Biological Example of Conjugation A. Proteins are a group of biologically important molecules 1. Make up 50% of human body’s dry weight 2. Catalyze biological reactions.

B. Proteins have many different biological functions 1. Keratin is a structural protein found in skin, hair, and fingernails 2. Hormones serve as chemical messengers such as insulin that controls carbohydrate metabolism 3. Enzymes are biological catalysts such as DNA polymerase (FYI Chapter 12) C. Proteins are long chains of amino acids 1. Amino acids contain a basic amino functional group (−NH2) and an acidic functional group (−CO2H) 2. Amino acids are linked by a peptide bond involving the –NH2 group of one and the –CO2H group of another to form an amide a. A dipeptide results when two amino acids are linked by one amide bond b. A tripeptide results when three amino acids are linked by two amide bonds c. Short chains of up to 50 linked amino acids are called peptides d. Polypeptides and proteins involve longer chains 3. Structure of peptide bond best understood in terms of conjugation and hybridization. a. Nitrogen lone pair is one  bond away from a C—O  bond. b. Conjugated system. c. C—N peptide bond has significant double bond character even though it is represented in a line structure as a single bond.

4. Twenty different amino acids are commonly found in proteins a. Each referred to by a three letter code, Figure 23.12 b. In all twenty, the amine nitrogen is connected to the carbon atom next to or  to the carboxylic acid group c. The twenty amino acids differ in the nature of the group, called the side chain, attached to the  carbon d. Nine amino acids are essential amino acids which must be obtained from the diet as the body cannot synthesize them 5. Amino acids classified as neutral, basic, or acid depending upon the structure of the side chain a. Two acidic amino acids with an additional carboxylic acid group in the side chain 8Test Item File Questions: Multiple Choice 80 – 97, Algorithmic 39 – 60, Short Answer 28

240

Chapter 23—Organic and Biological Chemistry b. Three basic amino acids with an additional amine group in the side chain c. Fifteen neutral amino acids i. Hydrophobic (water fearing) amino acids have nonpolar side chains ii. Hydrophilic (water loving) amino acids have polar side chains 6. All noncyclic proteins have an N-terminal amino acid with a free –NH2 on one end and a Cterminal amino acid with a free –CO2H on the other end a. By convention, a protein is written with the free –NH2 on the left and the free –CO2H on the right. b. The name of the protein is indicated using the three-letter abbreviations of the component amino acids 9 23.10. Aromatic Compounds and Molecular Orbital Theory A. Aromatic compounds 1. Class of compounds containing a six-membered ring with alternating single and double bonds 2. Simplest aromatic – benzene 3. Many biological and pharmaceutical agents contain an aromatic ring.

B. Benzene’s stability is a consequence of its electronic structure.

1. Conjugation and resonance explain structure of benzene but not its stability.

2. Molecular orbital theory is more useful.

C. Molecular orbital theory says that electrons are in molecular orbitals that cover the entire molecule.

D. In benzene, six molecular orbitals are formed from the mathematical combination of the six p orbitals.

1. Three bonding molecular orbitals which are lower in energy than the atomic p orbitals.

2. Three antibonding molecular orbitals which are higher in energy than the atomic p orbitals.

3. Six electrons are paired in the three bonding molecular orbitals, stabilizing the molecule.

E. The number of electrons is important.

1. Four electrons in the molecular orbitals result in two unpaired electrons—a destabilizing arrangement.

2. Eight electrons in the molecular orbitals result in two electrons in an antibonding orbital—a destabilizing arrangement.

3. Hückel’s Rule says that the number of electrons in an aromatic system is 4n + 2 where n is zero or a positive integer.

F. Aromatic compounds have certain characteristics.

1. Must be cyclic.

2. Must be planar so that p orbitals can overlap and mix mathematically.

3. Must have p orbitals on adjacent atoms around the ring.

4. Must have (4n + 2)  electrons in the ring.

10 23.11. Nucleic Acids: A Biological Example of Aromaticity A. Deoxyribosenucleic acid (DNA) and ribonucleic acid (RNA) are chemical carriers of genetic information 1. Nucleic acids are made of nucleotides linked together in a chain 2. Each nucleotide is composed of a nucleoside plus phosphoric acid, H 3PO4 3. Each nucleoside is composed of an aldopentose sugar plus an amine base a. Ribose is the sugar component of RNA b. 2-deoxyribose is the sugar of DNA c. Four different cyclic amine bases occur in DNA: adenine, guanine, cytosine, and thymine d. Four different cyclic amine bases occur in RNA: adenine, guanine, cytosine, and uracil 4. In both DNA and RNA the cyclic amine base is bonded to the C1’ of the sugar and the phosphoric acid is bonded to the C5’ of the sugar

9Test Item File Questions: Multiple Choice 98 – 103, Short Answer 29 10Test Item File Questions: Multiple Choice 104 – 115, Algorithmic 61, Short Answer 30 – 32

241

Chapter 23—Organic and Biological Chemistry

5. Nucleotides join together by forming a bond between the phosphate group of one nucleotide and the hydroxyl group on the sugar component of another B. The structure of a nucleic acid depends on the sequence of individual nucleotides 1. Abbreviations are used for each nucleotide 2. A for adenosine, G for guanine, C for cytosine, T for thymidine, and U for uracil C. DNA consists of two polynucleotide strands coiled around each other in a double helix.

1. The sugar–phosphate backbone is on the outside of the helix 2. Amine bases are on the inside 3. A base on one strand is directly opposite a base on the other strand 4. The two strands run in opposite directions 5. Strands are held together by hydrogen bonds between opposite base pairs a. Adenine and thymine form two strong hydrogen bonds to each other but not to guanine or cytosine b. Guanine and cytosine form two strong hydrogen bonds to each other but not to adenine or thymine D. DNA strands are complementary, not identical 1. Whenever guanine occurs in one strand, cytosine occurs in the other 2. Whenever adenine occurs in one strand, thymine occurs in the other E. The bases in the DNA must be planar so that they can stack in the strands.

1. Base stacking is key to DNA double-helix stability.

2. Tight stacking of bases leads to a structure with strong intermolecular attractions between adjacent bases.

F. Conjugation and resonance show that all of the DNA bases are aromatic.

Inquiry – Which Is Better: Natural or Synthetic?

242

Chapter 23—Organic and Biological Chemistry

CHAPTER 23 ORGANIC AND BIOLOGICAL CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions 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 23.1

Although cycloalkanes are usually drawn as planar polygons, most are not planar, but puckered.

Section 23.2

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 23.3

In an alkene, the double-bonded carbon atoms and atoms attached to them all lie in the same plane.

243

Chapter 23—Organic and Biological Chemistry

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 23.6

Ellin Doyle, “Trans Fatty Acids,” J. Chem. Educ., Vol. 74, 1997, 1030-1032.

244