Chemistry A Molecular Approach, First Canadian Edition Nivaldo J Tro Solution Manual

Chemistry A Molecular Approach, First Canadian Edition By Nivaldo J. Tro

Email: Richard@qwconsultancy.com

Chapter 1. Units of Measurement for Physical and Chemical Change Student Objectives 1.1 Physical and Chemical Changes and Physical and Chemical Properties 

Define, recognize, and understand the difference between physical and chemical changes.

1.2 Energy: A Fundamental Part of Physical and Chemical Change  

Define energy, work, kinetic energy, potential energy, and thermal energy.

State and understand the law of conservation of energy.

1.3 The Units of Measurement      

Understand the importance of reporting correct units with measurements.

Know the differences between the three most common sets of units: English system, metric system, and International System (SI).

Know the SI base units for length, mass, time, and temperature.

Know the three most common temperature scales (Fahrenheit, Celsius, and Kelvin), the freezing and boiling points of water on each scale, and the relationships between the scales.

Calculate temperature conversions between each scale.

Know and use the SI prefix multipliers for powers of ten.

1.4 The Reliability of a Measurement    

Understand that all measurements have some degree of uncertainty and that the last digit in a measurement is estimated.

Know how to determine the number of significant figures in a measurement using a set of rules.

Know how to determine the number of significant figures after calculations.

Distinguish between accuracy and precision.

1.5 Solving Chemical Problems     

Understand dimensional analysis and know how to use conversion factors.

Understand the problem‐solving strategy: sort, strategize, solve, and check.

Convert from one unit to another.

Make order‐of‐magnitude estimations without using a calculator.

Rearrange algebraic equations to solve for unknown variables.

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Chapter 1. Units of Measurement for Physical and Chemical Change Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 1. Units of Measurement for Physical and Chemical Change Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

1.1 Physical and Chemical Changes and Physical and Chemical Properties  Differences between physical and chemical changes  Examples and classifying changes

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1.2 Energy: A Fundamental Part of Physical and Chemical Change  Definitions of work and energy  Classification and types of energy o kinetic  thermal o potential  Definition and examples of the law of conservation of energy

Figure 1.1 Boiling Is a Physical Change and Rusting Is a Chemical Change

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Figure 1.2 Energy Conversions Figure 1.3 Using Chemical Energy to Do Work

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Chapter 1. Units of Measurement for Physical and Chemical Change Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

1.1 Physical and Chemical Changes and Physical and Chemical Properties  Conceptual Connection 1.1 Chemical and Physical Changes

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1.2 Energy: A Fundamental Part of Physical and Chemical Change  The examples of work being done by a person moving a box and chemical energy ultimately moving the car are consistent and simple.

Additional examples using gravitation (very familiar) are straightforward.  Several examples are cited for the law of conservation of energy; ask students to name and describe other forms of energy (solar, mechanical, chemical, electrical) and devices that convert between these forms.

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Boiling (especially) does not change a substance’s chemical identity.

Confront the confusion that can occur when a physical change accompanies a chemical one: burning liquid gasoline produces gases (physical or chemical or both?).

Work is a form of energy and thus has the same units as energy.

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Chapter 1. Units of Measurement for Physical and Chemical Change Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

1.3 The Units of Measurement  Loss of Mars Climate Orbiter because of inconsistent units  Systems of measurement and units o English system o metric system o International System (SI)  SI base units o length: metre o mass: kilogram o time: second o temperature: Kelvin  Temperature scales and conversions o Fahrenheit to Celsius and vice versa o Celsius to Kelvin and vice versa  Derived units o volume (cubic metre, cubic centimetre, litre, millilitre) o density, mass per unit volume (g mL‐1, g cm‐3)

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1.4 The Reliability of a Measurement  Significance and reporting of numerical values o estimating measurements  Counting significant figures or digits o nonzero digits o interior zeroes o leading zeroes o trailing zeroes o exact numbers  Significant figures in calculations o multiplication and division (fewest significant figures) o addition and subtraction (fewest decimal places) o rounding  Precision vs. accuracy  Scientific integrity and data reporting

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Unnumbered figure: Air Canada Flight 143 Table 1.1 SI Base Units Unnumbered figure: The CN Tower is 553.3 metres tall; a hockey player is about 2 metres tall Figure 1.4 Comparison of the Celsius and Kelvin Temperature Scales Example 1.1 Converting Between Temperature Scales Table 1.2 Metric Prefixes Example 1.2 Converting Between Different SI Prefixes Figure 1.5 The Relationship between Length and Volume Table 1.3 Some Common Derived Units Table 1.4 The Density of Some Common Substances at 20 °C Example 1.3 Calculating Density and Converting Density Units Chemistry and Medicine: Bone Density Table 1.5 Common Derived Units with Special Names

Table 1.6 Maximum August Temperatures in St. John’s, NL and Windsor, ON Figure 1.6 Estimation in Weighing Example 1.4 Reporting the Correct Number of Digits Example 1.5 Determining the Number of Significant Figures in a Number Example 1.6 Significant Figures in Calculations The Nature of Science: Integrity in Data Gathering Unnumbered figure: Accuracy and precision

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Chapter 1. Units of Measurement for Physical and Chemical Change Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

1.3 The Units of Measurement  Students are amazed and horrified that NASA could lose an expensive spacecraft because of inconsistent units.  Metric and SI units are unfamiliar to most Americans.

That a nickel has a mass of 5 g and that a yard is nearly as long as a metre gives a good frame of reference.  The practical examples of different temperatures on the Celsius scale (unnumbered figure) provide practical reference points.  Several of the large SI unit prefixes (mega, giga, tera) are already familiar from memory capacity in computers.  Conceptual Connection 1.2: Density  The Chemistry and Medicine box on bone density provides an open‐ended conceptual question about designing an experiment to measure bone density; this may be good for a brief in‐class discussion.

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1.4 The Reliability of a Measurement  Use a 400‐mL beaker and a 100‐mL graduated cylinder to measure quantities of water. Make the point about the importance of estimating measurements. Add the quantities of water together and ask the students to calculate the final volume ... to the correct precision.  Two tables present air quality data (with different precision) that might appear in a newspaper or other publication. Initiate a discussion of the certainty of digits in reported data.  Water‐quality standards have evolved substantially since the advent of instrumental methods for quantitative analysis. Ask the question: Does zero mean that a particular analyte is not present?  The number on a calculator display requires interpretation; only the user knows the certainty of the values entered.  A discussion about why integrity in data reporting is particularly important in science is appropriate. It should point out that scientists report how they did the experiments so others can try to repeat and verify the work. Use recent examples from the media.

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A common misconception is that 100 cm3 is equal to 1 m3.

Some students initially are confused that density can be used as a conversion factor even when the units are inverted.

Students presume that calculators are flawless but forget that calculators do only what the user dictates.

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Chapter 1. Units of Measurement for Physical and Chemical Change Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

1.5 Solving Chemical Problems  Converting from one unit to another o dimensional analysis o multiple approaches to any problem  General problem‐solving strategy o sort o strategize o solve o check  Calculations using units raised to a power  Order‐of‐magnitude estimations  Using equations

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Example 1.7 Problems using Equations Example 1.8 Problems using Equations

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Chapter 1. Units of Measurement for Physical and Chemical Change Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

1.5 Solving Chemical Problems  General chemistry classes at most schools have students with a wide range of math skills. A quick review of algebra may be useful.  Emphasize that watching an instructor work problems is not nearly as effective as working those same problems on one’s own. Give students time to work a problem or two in class; allow them to work in small groups.  Emphasize the good practice of writing units and keeping track of units in every calculation. Simple dimensional analysis prevents many headaches throughout the year of general chemistry.  Promote estimation as part of the problem‐solving model. Tell the students to ask themselves, “Does this answer make sense?” Reduce the reliance on blindly entering numbers into a calculator and transcribing whatever answer comes up.  Cognitive load theory says that a person can remember 7–9 items in short‐term memory. A problem loaded with unit conversions, spurious facts, and many steps does not test a person’s understanding of an underlying idea or concept. It becomes a measure of cognitive ability outside the realm of chemistry.

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Students often want to follow one particular “recipe” to solve one particular kind of problem.

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Chapter 1. Units of Measurement for Physical and Chemical Change

Procedure for Solving Unit Conversion Problems

Additional Problem

Convert 1.76 miles to metres.

Given 1.76 mi

Sort Begin by sorting the information in the problem into Given and Find.

Find m

Strategize

Conceptual Plan

Devise a conceptual plan for the problem. Begin with the given quantity and symbolize each conversion step with an arrow. Below each arrow, write the appropriate conversion factor for that step.

Focus on the units. The conceptual plan should end at the find quantity and its units. In these examples, the other information needed consists of relationships between the various units, as shown.

mi



1 km 0.6214 mi



km

m

1000 m 1 km

Relationships Used 1 km = 0.6214 mi 1 km = 1000 m

Solve

Solution

Follow the conceptual plan. Begin with the given quantity and its units. Multiply by the appropriate conversion factor(s), cancelling units, to arrive at the find quantity.

1.76 mi 

1 km 0.6214 mi



1000 m = 2832.31 m 1 km

Round the answer to the correct number of significant figures by following the rules in Section 1.4. Remember that exact conversion factors do not limit significant figures.

2832.31 m = 2830 m

Check

The units (m) are correct. The magnitude of the answer (2830) makes physical sense since a metre is a much smaller unit than a mile.

Check your answer. Are the units correct? Does the answer make physical sense?

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Chapter 1. Units of Measurement for Physical and Chemical Change

Additional Problem for Unit Conversion Involving Units Raised to a Power

Calculate the number of cubic metres of concrete necessary to support a deck if each of 14 concrete piers require 4750 cubic inches.

Sort

Given 14 piers, 4750 in3

Begin by sorting the information in the problem into Given and Find.

Find m3

Strategize

Conceptual Plan

Write a conceptual plan for the problem. Begin with the given information and devise a path to the information that you are asked to find. Notice that for cubic units, the conversion factors must be cubed.

piers  in3



14 piers

 1m   39.37 in   

m3 3

Relationships Used 1 m = 39.37 in 1 pier = 4750 in3 (Given)

Solve

Solution

Follow the conceptual plan to solve the problem.

Round the answer to three significant figures to reflect the three significant figures in the least precisely known quantity (4750). These conversion factors are all exact and therefore do not limit the number of significant figures.

4750 in3 14 piers   1 pier

1 m 

3

 39.37 in 

3

= 1.0897 m3

1.0897 m3 = 1.09 m3

The units of the answer are correct and the magnitude makes sense. The unit metres is larger than inches, so cubic metres are much larger than cubic inches.

Check

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Chapter 1. Units of Measurement for Physical and Chemical Change Additional Problem for Density as a Conversion Factor (Example 1.3)

An experimental automobile has a 100.0 litre fuel tank filled with ethanol. How many pounds does the fuel add to the mass of the car?

Sort

Given 100.0 L

Begin by sorting the information in the problem into Given and Find.

Find lb

Strategize

Conceptual Plan

Devise a conceptual plan by beginning with the given quantity, in this case the volume in litres (L).

The overall goal of this problem is to find the mass.

You can convert between volume and mass using density (g cm-3). However, you must first convert the volume to cm3. Once you have converted the volume to cm3, use the density to convert to g.

Finally, convert g to lb.

L 

mL

1000 mL 1L



cm3 

1 cm3 1 mL

0.789 g 1 cm3

g



lb

1 lb 453.59 g

Relationships Used 1000 mL = 1 L 1 mL = 1 cm3 d (ethanol) = 0.789 g cm-3 1 lb = 453.59 g

Solve

Solution

Follow the conceptual plan to solve the problem.

Round the answer to three significant figures to reflect the three significant figures in the density.

100 L 

0.789 g 1000 mL 1 cm3 1 lb    3 1 L 1 mL 453.59 g 1 cm = 173.94 lb

173.94 lb = 174 lb The units of the answer (lb) are correct. The magnitude of the answer (174) makes physical sense since a litre of water has a mass of 1 kilogram or about 2.2 pounds; 100 litres of water is about 220 lbs. Ethanol has a lower density than water (about 80% or 8/10).

Check

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Chapter 1. Units of Measurement for Physical and Chemical Change

Additional Problem for Solving Problems Involving Equations (Examples 1.7 and 1.8)

What is the mass in grams of an ice cube that is 1.1 inches per side?

Sort

Given l = 1.1 in

Begin by sorting the information in the problem into Given and Find.

Find g

Strategize

Conceptual Plan

Write a conceptual plan for the problem. Focus on the equation(s). The conceptual plan shows how the equation takes you from the given quantity (or quantities) to the find quantity. The conceptual plan may have several parts, involving other equations or required conversions. In these problems, you must use the geometrical relationships given in the problem as well as the definition of density.

l



V

V=l3

in3





cm3

 2.54 cm   1 in   

g

0.917 g 1 cm3

3

Relationships Used V = l 3 [volume of a cube with a length of l] 2.54 cm = 1 in d (ice) = 0.917 g cm-3 Solve

Solution

Follow the conceptual plan. Solve the equation(s) for the find quantity. Gather each of the quantities that must go into the equation in the correct units.

(Convert to the correct units if necessary.)

Substitute the numerical values and their units into the equation(s) and compute the answer. Round the answer to the correct number of significant figures.

V = 1.1 in  1.331 in3

3

= 1.331 in3

 2.54 cm      1 in 

3



0.917 g 1 cm3

= 20.0008 g

20.0008 g = 20 g The units (g) are correct. The magnitude of the answer (20) seems to make physical sense.

Check

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Chapter 2. Atoms and Elements Student Objectives 2.1 Imaging and Moving Individual Atoms  

Describe scanning tunnelling microscopy (STM) and how atoms are imaged on surfaces.

Define atom and element.

2.2 Early Ideas About the Building Blocks of Matter  

Describe the earliest definitions of atoms and matter (Greeks).

Know that greater emphasis on observation and the development of the scientific method led to the scientific revolution.

2.3 Modern Atomic Theory and the Laws That Led to It    

State and understand the law of conservation of mass (also from Section 1.2).

State and understand the law of definite proportions.

State and understand the law of multiple proportions.

Know the four postulates of Dalton’s atomic theory.

2.4 Atomic Structure  

Define radioactivity, nucleus, proton, and neutron.

Understand Thomson's plum‐pudding model and how Ernest Rutherford’s gold‐foil experiment refuted it by giving evidence for a nuclear structure of the atom.

2.5 Atomic Mass: The Average Mass of an Element’s Atoms  

Calculate atomic mass from isotope masses and natural abundances.

Define mass spectrometry and understand how it can be used to measure mass and relative abundance.

2.6 Molar Mass: Counting Atoms by Weighing Them    

Understand the relationship between mass and count of objects such as atoms.

Define mole and Avogadro’s number.

Calculate and interconvert between number of moles and atoms.

Calculate and interconvert between number of moles and mass.

2.7 The Periodic Table of the Elements      

Define the periodic law.

Know that elements with similar properties are placed into columns (called groups) in the periodic table.

Define and distinguish between metals, nonmetals, and metalloids.

Identify main‐group and transition elements on the periodic table.

Know the general properties of elements in some specific groups: noble gases, alkali metals, alkaline earth metals, and halogens.

Know and understand the rationale for elements that form ions with predictable charges.

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Chapter 2. Atoms and Elements Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 2. Atoms and Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

2.1 Imaging and Moving Individual Atoms  Description of scanning tunnelling microscopy (STM)  Introduction to macroscopic and microscopic perspectives  Definitions of atom and element

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Intro figure: tip of an STM moving across a surface Figure 2.1 Scanning Tunnelling Microscopy Figure 2.2 Imaging Atoms

2.2 Early Ideas About the Building Blocks of Matter  History of chemistry from antiquity (~450 BC)  Scientific revolution (1400s–1600s)

2.3 Modern Atomic Theory and the Laws That Led to It  Law of conservation of mass o Matter is neither created nor destroyed. o Atoms at the start of a reaction may recombine to form different compounds, but all atoms are accounted for at the end. o Mass of reactants = mass of products.  Law of definite proportions o Different samples of the same compound have the same proportions of constituent elements independent of sample source or size.  Law of multiple proportions  John Dalton’s atomic theory

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Unnumbered figure: models and photos of Na and Cl2 forming NaCl Example 2.1 Law of Definite Proportions Unnumbered figure: space filling models of CO and CO2 illustrating the law of multiple proportions

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Chapter 2. Atoms and Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

2.1 Imaging and Moving Individual Atoms  Other STM images can be found readily on the Internet.  It is useful to reiterate the analogies about size; the one used in the chapter compares an atom to a grain of sand and a grain of sand to a large mountain range.

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2.2 Early Ideas About the Building Blocks of Matter  The view of matter as made up of small, indestructible particles was ignored because more popular philosophers like Aristotle and Socrates had different views.  Leucippus and Democritus may have been proven correct, but they had no more evidence for their ideas than Aristotle did.  Observations and data led scientists to question models; the scientific method promotes the use of a cycle of such inquiry.

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2.3 Modern Atomic Theory and the Laws That Led to It  That matter is composed of atoms grew from experiments and observations.  Conceptual Connection 2.1 The Law of Conservation of Mass  Investigating the law of definite proportions requires preparing or decomposing a set of pure samples of a compound like water.  Investigating the law of multiple proportions requires preparing or decomposing sets of pure samples from related compounds like NO, NO2, and N2O5.

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STM is not actually showing images of atoms like one might imagine seeing with a light microscope.

Atoms are not coloured spheres; the images use colour to distinguish different atoms.

Theories are not automatically accepted and may be unpopular for long periods of time.

Philosophy and religion can be supported by arguments; science requires that theories be testable and therefore falsifiable.

Measurements to establish early atomic theories were performed at the macroscopic level. The scientists observed properties for which they could collect data (e.g., mass or volume).

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Chapter 2. Atoms and Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

2.4 Atomic Structure  Thomson’s cathode‐ray tube experiments o High voltage produced a stream of particles that travelled in straight lines. o Each particle possessed a negative charge. o Thomson measured the charge‐to‐mass ratio of the electron.  Millikan’s oil‐drop experiments o Oil droplets received charge from ionizing radiation. o Charged droplets were suspended in an electric field. o The mass and charge of each oil drop was used to calculate the mass and charge of a single electron.  Thomson’s plum‐pudding model: negatively charged electrons in a sea of positive charge  Radioactivity: alpha decay provides the alpha particles for Rutherford’s experiment.  Rutherford’s experiment o Alpha particles directed at a thin gold film deflect in all directions, including back at the alpha source. o Only a concentrated positive charge could cause the alpha particles to bounce back.  Rutherford’s nuclear theory o Most mass and all positive charge contained in a small nucleus. o Most of atom by volume is empty space. o Protons: positively charged particles. o Neutral particles with substantial mass also in nucleus.  Atomic number (number of protons): defining characteristic of an element  Isotope: same element, different mass  Properties of subatomic particles o atomic mass units (amu)  Proton, neutron: ~1 amu  Electron: ~0.006 amu o charge  Relative value: 1 for electron, +1 for proton  Absolute value: 1.6  1019 C  Ion: atom with nonzero charge o Anion: negatively charged (more electrons) o Cation: positively charged (fewer electrons)

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Figure 2.3 Cathode Ray Tube Unnumbered figure: properties of electrical charge Figure 2.4 Thomson’s Measurement of the Charge‐to‐Mass Ratio of the Electron Figure 2.5 Millikan’s Measurement of the Electron's Charge Unnumbered figure: plum‐pudding model Figure 2.6 Rutherford’s Gold Foil Experiment Unnumbered figure: Rutherford Museum, McGill University Unnumbered figure: picture of Ernest Rutherford Figure 2.7 The Nuclear Atom Unnumbered figure: baseball Table 2.1 Characteristics of Subatomic Particles Example 2.2 Atomic Numbers, Mass Numbers, and Isotope Symbols Chemistry in Your Day: Where Did Elements Come From?

Unnumbered figure: Image of gaseous pillar in a star‐forming region of the Eagle Nebula

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Chapter 2. Atoms and Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

2.4 Atomic Structure  Review the attraction, repulsion, and additivity of charges.  Discuss the physics of electric fields generated by metal plates.  A demonstration of a cathode ray tube will help students better understand Thomson’s experiments.  Demonstrate how Millikan’s calculation works and why he could determine the charge of a single electron.  It may be useful to give a brief description of radioactivity.

Rutherford’s experiment makes more sense if one knows some properties of the alpha particle and from where it comes.  Thomson identified electrons and surmised the existence of positive charge necessary to form a neutral atom. The plum‐pudding model is the simplest way to account for the observations.  The analogy of the baseball and a grain of rice to a proton and an electron is meant to illustrate the difference in mass but not size.  Electrical charge can be demonstrated with static electricity. Two balloons charged with wool or human hair will repel each other.  Names of elements come from various sources. Tom Lehrer’s “Element Song” can be found on the Internet.  Isotopic abundances are invariant in typical lab‐sized samples because of such large numbers of atoms.  Conceptual Connection 2.2 The Nuclear Atom, Isotopes, and Ions  The history of chemistry involves considerable cultural and gender diversity. Examples include both Lavoisiers (French), Dalton (English), Thomson (English), Marie Curie (Polish/French), Mendeleev (Russian), Millikan (American), Robert Boyle (Irish), Amedeo Avogadro (Italian).  The Chemistry in Your Day box gives a broad description of the origin of atoms.

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Millikan did not measure the charge of a single electron; he measured the charge of a number of electrons and deduced the charge of a single electron.

Students often don’t understand the source of alpha particles in Rutherford’s experiments.

Students sometimes confuse the mass number as being equal to the number of neutrons, not the number of neutrons plus the number of protons.

Students logically (but mistakenly) presume that the mass of an isotope is equal to the sum of the masses of the protons and neutrons in that isotope.

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Chapter 2. Atoms and Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

2.5 Atomic Mass: The Average Mass of an Element’s Atom  Average atomic mass is based on natural abundance and isotopic masses.  Mass spectrometry o Atoms converted to ions and deflected by magnetic fields to separate by mass o Output data: relative mass vs. relative abundance

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2.6 Molar Mass: Counting Atoms by Weighing Them  Mole concept and Avogadro’s number  Converting between moles and number of atoms  Converting between mass and number of moles

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Unnumbered figure: periodic table box for chlorine Example 2.3 Atomic Mass Figure 2.8 The Mass Spectrometer Figure 2.9: Mass Spectrum of Xenon

Unnumbered figure: relative sizes of Al, C, He Unnumbered figure: balance with marbles and peas Example 2.4 The Mole Concept: Converting from Mass to Moles and Number of Atoms

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Chapter 2. Atoms and Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

2.5 Atomic Mass: The Average Mass of an Element’s Atom  The masses of isotopes must be reconciled with an element having only whole number quantities of protons and neutrons; the values should be nearly integral since the mass of electrons is so small.  Mass spectrometry is an effective way to demonstrate where values of natural abundance are obtained.

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2.6 Molar Mass: Counting Atoms and Weighing Them  Review the strategy for solving numerical problems: sort, strategize, solve, check.  Estimating answers is an important skill; the number of atoms will be very large (i.e. some large power of ten) even from a small mass or small number of moles.  Conceptual Connection 2.3 The Mole

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Students are tempted to calculate average atomic mass by adding together isotopic masses and dividing by the number of isotopes.

Atomic mass on the periodic table is usually not integral even though elements have only whole numbers of protons and neutrons.

Many students are intimidated by estimating answers in calculations involving powers of ten.

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Chapter 2. Atoms and Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

2.7 The Periodic Table of the Elements  Periodic law and the periodic table o Generally arranged by ascending mass o Recurring, periodic properties; elements with similar properties arranged into columns: groups (or families)  Major divisions of the periodic table o Metals, nonmetals, metalloids o Main‐group elements, transition elements  Groups (families) o Noble gases (group 18) o Alkali metals (group 1) o Alkaline earth metals (group 2) o Halogens (group 17)  Ions with predictable charges: based on stability of noble gas electron count o Group 1: 1+ o Group 2: 2+ o Group 13: 3+ o Group 15: 3 o Group 16: 2 o Group 17: 1

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Unnumbered figure: stamp featuring Dmitri Mendeleev Unnumbered figure: Mendeleev’s periodic table from 1871 Figure 2.10 Metals, Nonmetals, and Metalloids Figure 2.11 The Periodic Table: Main‐Group and Transition Elements Unnumbered figure: the alkali metals Unnumbered figure: the halogens Figure 2.12 Elements That Form Ions with Predictable Charges

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Chapter 2. Atoms and Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

2.7 The Periodic Table of the Elements  Other displays of the periodic table can be found in journals (Schwartz, J. Chem. Educ. 2006, 83, 849; Moore, J. Chem. Educ.

2003, 80, 847; Bouma, J. Chem. Educ. 1989, 66, 741), books, and on the Internet.  Periodic tables are arranged according to the periodic law but can compare many features, e.g., phases of matter, sizes of atoms, and common ions. These are presented as a series of figures in the text.

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The periodic table is better at predicting microscopic properties, though macroscopic properties are also often illustrated.

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Chapter 2. Atoms and Elements

Additional Problem for Converting between Number of Moles and Number of Atoms (Example 2.4)

Calculate the number of moles of iron in a sample that has 3.83 x 1023 atoms of iron.

Sort

Given 3.83 x 1023 Fe atoms

You are given a number of iron atoms and asked to find the amount of iron in moles.

Strategize Convert between number of atoms and number of moles using Avogadro’s number.

Find mol Fe

Conceptual Plan atoms



mol

1 mol Fe 6.022  10 23 Fe atoms

Relationships Used 6.022 x 1023 = 1 mol (Avogadro’s number)

Solve

Solution

Follow the conceptual plan. Begin with 3.83 x 1023 Fe atoms and multiply by the ratio that equates moles and Avogadro’s number.

3.83  10 23 Fe atoms 

Check

The sample was smaller than Avogadro’s number, so the answer should be a fraction of a mole. The value of the sample has three significant figures, and the answer is provided in that form.

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1 mol Fe = 0.636 mol Fe 6.022  10 23 Fe atoms

24

Chapter 2. Atoms and Elements Additional Problem for Converting between Mass and Number of Moles (Example 2.4)

Calculate the number of grams of silver in an American Silver Eagle coin that contains 0.288 moles of silver.

Sort

Given 0.288 mol Ag

You are given the amount of silver in moles and asked to find the mass of silver.

Find g Ag

Strategize

Conceptual Plan

Convert amount (in moles) to mass using the molar mass of the element.

mol Ag



g Ag

107.87 g Ag 1 mol Ag

Relationships Used 107.87 g Ag = 1 mol Ag

Solve

Solution

Follow the conceptual plan to solve the problem.

Start with 0.288 mol, the given number, and multiply by the molar mass of silver.

0.288 mol Ag 

Check

The magnitude of the answer makes sense since we started with an amount smaller than a mole. The molar amount and answer both have three significant figures.

107.87 g Ag = 31.07 g Ag 1 mol Ag

31.07 g = 31.1 g Ag

.

25

Chapter 2. Atoms and Elements

Additional Problem for the Mole Concept— Converting Between Mass and Number of Atoms (Example 2.4)

What mass of iron (in grams) contains 1.20  1022 atoms of Fe? A paperclip contains about that number of iron atoms.

Sort

Given 1.20  1022 Fe atoms

You are given a number of iron atoms and asked to find the mass of Fe. Find g Fe

Strategize

Conceptual Plan

Convert the number of Fe atoms to moles using Avogadro’s number. Then convert moles Fe into grams of iron using the molar mass of Fe.

Fe atoms



mol Fe

1 mol Fe 6.022  10 23 Fe atoms



g Fe

55.85 g Fe 1 mol Fe

Relationships Used 6.022  1023 = 1 mol (Avogadro’s number) 55.85 g Fe = 1 mol Fe

Solve Follow the conceptual plan to solve the problem.

Begin with 1.20 x 1022 atoms of Fe, multiply by the ratio derived from Avogadro’s number, and finally multiply by the atomic mass of Fe.

Check

Solution 1.20  1022 Fe atoms 

1 mol Fe 55.85 g Fe  6.022  1023 Fe atoms 1 mol Fe

= 1.11 g Fe

The units and magnitude of the answer make sense.

The sample is smaller than a mole. The number of atoms and mass both have three significant figures.

.

26

Chapter 2. Atoms and Elements

Additional Problem for the Mole Concept (Example 2.4)

An iron sphere contains 8.55  1022 iron atoms.

What is the radius of the sphere in centimetres?

The density of iron is 7.87 g cm-3.

Sort

Given 8.55  1022 Fe atoms

You are given the number of iron atoms in a sphere and the density of iron. You are asked to find the radius of the sphere.

Find radius (r) of a sphere

Strategize

Conceptual Plan

The critical parts of this problem are density, which relates mass to volume, and the mole, which relates number of atoms to mass:

Fe atoms  mol Fe  g Fe  V (cm3)

d = 7.87 g cm-3

1 mol Fe 6.022  10 23 Fe atoms

(1) Convert from the number of atoms to the number of moles using Avogadro’s number (2) Convert from the number of moles to the number of grams using the molar mass of iron

V (cm3)



V =

4  r3 3

55.85 g Fe 1 mol Fe

1 cm3 7.87 g Fe

r (cm)

Relationships Used

(3) Convert from mass to volume using the density of iron

6.022 x 1023 = 1 mol (Avogadro’s number)

(4) Find the radius using the formula for the volume of a sphere

55.85 g Fe = 1 mol Fe d (density of Fe) = 7.87 g cm-3 V = 4/3 r3 [volume of a sphere with a radius of r]

Solve

Solution

Follow the conceptual plan to solve the problem.

Begin with 8.55 x 1022 Fe atoms and convert to moles, then to grams, and finally to a volume in cm3. Solve for the radius using the rearranged equation.

8.55  1022 atoms 

55.85 g Fe 1 mol Fe  23 6.022  10 atoms 1 mol Fe 

r = 3

1 cm3 = 1.00757 cm3 7.87 g Fe

3V 3  1.00757 cm3 = 3 = 0.622 cm 4 4

The units (cm) are correct and the magnitude of the answer makes sense compared with previous problems.

Check

.

27

Chapter 3. Molecules, Compounds, and Nomenclature Student Objectives 3.1 Hydrogen, Oxygen, and Water  

Know some chemical and physical properties of H2, O2, and H2O. Know and understand that compounds (e.g., NaCl) are different from the elements (e.g., Na and Cl2) from which they are composed.

3.2 Chemical Bonds   

Define and understand the difference between ionic and covalent bonds.

Describe and understand the formation of an ionic compound from its elements.

Describe and understand the sharing of electrons in a covalent bond.

3.3 Representing Compounds: Chemical Formulas and Molecular Models         

Define and understand empirical formula, molecular formula, and structural formula.

Write the empirical formula, molecular formula, and structural formula for simple molecules.

Recognize and understand the differences between ball‐and‐stick models and space‐filling models.

Recognize and identify characteristic colours for elements in molecular models.

Identify elements as atomic or molecular.

Differentiate between atomic or molecular elements and ionic or molecular compounds.

Know and understand that ionic compounds are composed of formula units and not discrete molecules.

Know and understand that covalent compounds tend to exist as discrete molecules.

Know and understand that a polyatomic ion is composed of atoms that are covalently bound to each other.

3.4 Formulas and Names        

Know that ionic compounds are ubiquitous in the Earth’s crust as minerals.

Know and understand the rules for writing formulas for ionic compounds.

Write formulas for ionic compounds using the charges of the ions and the principle of electrical neutrality.

Know and understand the rules for naming ionic compounds.

Write names from formulas and formulas from names of ionic compounds.

Know and understand the rules for naming molecular compounds.

Write names from formulas and formulas from names of molecular compounds.

Write names and formulas for binary acids and oxyacids.

3.5 Organic Compounds       

Define saturated hydrocarbon.

Name alkanes from their structures and write structures from names of alkanes through C10.

Know that an alkene is defined by a carbon–carbon double bond and that the rest of the molecule resembles an alkane.

Name alkenes from their structures and write structures from names of alkenes through C10.

Know that an alkyne is defined by a carbon–carbon triple bond and that the rest of the molecule resembles an alkane.

Name alkynes from their structures and write structures from names of alkynes through C10.

Write names for mono‐ and disubstituted derivatives of benzene. .

28

Chapter 3. Molecules, Compounds, and Nomenclature  

Know that functional groups are specific atoms bonded to each other that can be inserted onto a hydrocarbon.

Know and identify the common functional groups: alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine.

3.6 Formula Mass and the Mole Concept for Compounds   

Define formula mass (a.k.a. molecular weight, molecular mass) and molar mass for a compound.

Understand and calculate the molar mass of a compound.

Calculate and interconvert between mass, moles, and molecules of a compound.

3.7 Composition of Compounds    

Define and understand mass percent (mass percent composition).

Calculate mass percent from a chemical formula.

Use mass percent as a conversion factor.

Use chemical formulas as conversion factors in mole calculations.

3.8 Determining a Chemical Formula from Experimental Data     

Convert masses into moles and calculate mole ratios to determine empirical formulas.

Determine empirical formulas from experimental data.

Determine molecular formulas from empirical formulas and molecular masses.

Understand combustion analysis.

Determine an empirical formula from combustion analysis.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

29

Chapter 3. Molecules, Compounds, and Nomenclature Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

3.1 Hydrogen, Oxygen, and Water  Some chemical and physical properties of hydrogen, oxygen, and water o Boiling point o Physical state at room temperature o Flammability  Some characteristics of sodium, chlorine, and sodium chloride  Compounds are very different from the elements from which they are formed

  

3.2 Chemical Bonds  Chemical bonds o Ionic bonds  Formed between a metal and a nonmetal  Feature a cation and an anion o Covalent bonds  Formed between two nonmetals  Feature electron sharing



Figure 3.2 The Formation of an Ionic Compound Figure 3.3 The Stability of a Covalent Bond



3.3 Representing Compounds: Chemical Formulas and Molecular Models  Chemical formula o Empirical formula o Molecular formula o Structural formula  Molecular models o Colour‐coded spheres o Ball‐and‐stick o Space‐filling  Pure substances o Elements  Atomic  Molecular o Compounds  Ionic  Molecular  Ionic compounds o Formula units o Polyatomic ions

      

.

Intro figure: periodic table, reaction of H2 and O2 to form H2O Unnumbered figure: table of selected properties of hydrogen, oxygen, and water Figure 3.1 Mixtures and Compounds

Unnumbered figure: NO2 and N2O4 in test tubes, same empirical formula Unnumbered figure: colour representations of elements in models Unnumbered figure: diatomic molecule of Cl2 Table 3.1 Representations of Compounds Benzene, Acetylene, Glucose, and Ammonia Figure 3.4 A Molecular View of Elements and Compounds Figure 3.5 Molecular and Ionic Compounds Example 3.1 Classifying Substances as Atomic Elements, Molecular Elements, Molecular Compounds, or Ionic Compounds

30

Chapter 3. Molecules, Compounds, and Nomenclature Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

3.1 Hydrogen, Oxygen, and Water  Compounds are introduced by comparing the properties of the constituent atoms to the properties of the compound.



Hydrogen and oxygen are molecular elements, not compounds.

3.2 Chemical Bonds  Figure 3.1 emphasizes a difference between compounds and mixtures: compounds have fixed ratios of the elements, whereas mixtures can have any ratio.  The formation of sodium chloride is schematic only and not mechanistic.



The actual reaction of chlorine with sodium occurs at the solid sodium surface where the elemental chlorine comes in contact with the metal.

3.3 Representing Compounds: Chemical Formulas and Molecular Models  Emphasize that students think from different perspectives afforded by the different representations of both formula and structure. Which form of a compound they need to think about depends on the context.  Details of molecular shapes are discussed in Chapter 10.  The classification scheme for elements and compounds gives the molecular view of each example.



Formulas don’t show the shapes of molecules, and some formulas give no indication of atom connectivity.

Coloured spheres are an obvious convenience in graphic as well as tactile models. They are unrelated to microscopic and macroscopic properties.

Students sometimes get confused about the molecularity of some elements. They sometimes don’t realize right away that the molecularity is limited only to the elemental state.

Ions don’t exist discreetly in large amounts, i.e., one can’t pull a bottle of Na+ from a shelf.

Polyatomic ions have both covalent and ionic features.





 

.

31

Chapter 3. Molecules, Compounds, and Nomenclature Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

3.4 Formulas and Names  Distinction between common and systematic names  Writing formulas for ionic compounds o Symbol and charge for cation and anion o Subscripts: number of ions o Electrical neutrality overall  Naming ionic compounds o Metal forms one type of cation  Name of cation + base name of nonmetal with suffix ‐ide o Metal forms more than one type of cation  name of cation + charge of metal in Roman numerals + base name of nonmetal with suffix ‐ide o Compound with polyatomic ion o Hydrates: use Greek prefixes to indicate number of water molecules in formula unit  Naming binary molecular compounds o Common: water, ammonia o Systematic  Greek prefix + name of first element + prefix + base name of second element + ‐ide  Naming acids o Binary: hydro‐ + base name of nonmetal + ‐ic acid o Oxyacids  ‐ate oxyanions: base name of oxyanion + ‐ic acid  ‐ite oxyanions: base name of oxyanion + ‐ous acid

                

.

Unnumbered figure: photos of calcite and trona Example 3.2 Writing Formulas for Ionic Compounds Example 3.3 Writing Formulas for Ionic Compounds Table 3.2 Common Cations and Anions Figure 3.6 Transition Elements Table 3.3 Common Polyatomic Ions Figure 3.7 Hydrates Unnumbered figure: common hydrate prefixes Example 3.4 Naming Ionic Compounds Unnumbered figure: Greek prefixes for naming binary molecular compounds Unnumbered figure: base names for common nonmetals Example 3.5 Naming Molecular Compounds Unnumbered figure: sliced fruits (acids) Unnumbered figure: scheme for naming acids Figure 3.8 Hydrochloric Acid Dissolving Zinc Metal Unnumbered figure: binary vs. oxyacids Unnumbered figure: oxoanions ending in ‐ite and ending in ‐ate

32

Chapter 3. Molecules, Compounds, and Nomenclature Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

3.4 Formulas and Names  Students benefit from many examples and practice.

These can be done as short in‐class activities.  Conceptual Connection 3.1 Unambiguous Nomenclature  There are many additional categories or families of molecular compounds. Naming them is an ongoing learning process; for example, organic chemistry has its own system. Students should be encouraged and prepared to keep track of them throughout the course.  The figure of acid dissolving metal (Figure 3.8) is an opportunity to emphasize microscopic and macroscopic representations. It introduces the idea of an acid and demonstrates that ions can move freely in an aqueous solution.  Conceptual Connection 3.2 Nomenclature

.





Nomenclature immediately confronts students’ ability to assign charge and balance charge from the cation and the anion.

Nomenclature is not an exciting topic, but it is crucial—like vocabulary in a new language.

Being able to recognize the polyatomic ions right away is an especially important skill.

33

Chapter 3. Molecules, Compounds, and Nomenclature Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

3.5 Organic Compounds  Organic compounds o Hydrocarbons o Functional groups  Naming hydrocarbons o IUPAC system of nomenclature o Procedure for nomenclature of alkanes o Nomenclature of alkenes and alkynes  Cyclic hydrocarbons o Cyclic alkanes and alkenes  Aromatic hydrocarbons o Nomenclature of aromatic hydrocarbons  Functionalized hydrocarbons o Nomenclature of molecules containing one or more functional groups

           

Unnumbered figure: gasoline pump Unnumbered figure: structural formula and space‐filling model for butane Unnumbered figure: structural formula and space‐filling model of ethene Unnumbered figure: functional groups Table 3.4 Prefixes for Base Names of Hydrocarbon Chains up to n = 10 carbon atoms Table 3.5 Common Alkyl Groups Example 3.6 Naming Alkanes Example 3.7 Naming Alkanes Example 3.8 Naming Alkenes and Alkynes Unnumbered figure: aromatic hydrocarbons (benzene) Table 3.6 Families of Organic Compounds Example 3.9 Naming Functionalized Hydrocarbons

3.6 Formula Mass and the Mole Concept for Compounds  Define and calculate formula mass o For each different element: number of atoms  mass in amu of element; sum over all elements  Molar mass o Formula mass in amu  Interconvert between moles, mass, and number of molecules using molar mass

  

Algorithm for calculating formula mass Example 3.10 Calculating Formula Mass Example 3.11 The Mole Concept— Converting Between Mass and Number of Moles

3.7 Composition of Compounds  Mass percent of an element  Mass percent as a conversion factor  Conversion factors from chemical formulas



Unnumbered figure: ozone hole over Antarctica Example 3.12 Mass Percent Composition Example 3.13 Chemical Formulas as Conversion Factors Chemistry in Your Day: Drug Tablets

  

.

34

Chapter 3. Molecules, Compounds, and Nomenclature Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

3.5 Organic Compounds  Make and show models (or virtual models that can be rotated) of analogous or homologous structures, e.g., methane, ethane, propane.  Make and show models of functional groups.  Show examples of molecules with the same molecular formula but different functional groups.  Conceptual Connection 3.3 Organic Nomenclature



3.6 Formula Mass and the Mole Concept for Compounds



Students have difficulty with the relative size of atoms in a molecule, the size of one molecule versus another, and the size of molecules at the molecular level.

3.7 Composition of Compounds  The box on drug tablets gives a practical example of percent by mass in everyday use.



There is sometimes confusion between the fraction (or ratio) and the percent.

 

.

Free rotation about single bonds can sometimes mislead students into thinking that slightly different representations of the same molecule are different molecules altogether. Making molecular models and rotating about the bonds helps alleviate this confusion.

Reminding students of the octet rule will help them draw better structures for organic compounds.

Without practice, some students will identify identical alkenes or alkynes simply because of the orientation of the molecule. In other words, the molecule with the double bond between the first and second carbon is the same regardless if it appears at the left or right end of a carbon chain. It is useful for students to learn to recognize symmetry in structures.

35

Chapter 3. Molecules, Compounds, and Nomenclature Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

3.8 Determining a Chemical Formula from Experimental Data  Elemental analysis  Deriving an empirical formula  Molecular formulas from empirical formulas and molar mass  Combustion analysis

   

.

Examples 3.14 and 3.15 Obtaining an Empirical Formula from Experimental Data Example 3.16 Calculating a Molecular Formula from an Empirical Formula and Molar Mass Figure 3.9 Combustion Analysis Apparatus Examples 3.17 and 3.18 Obtaining an Empirical Formula from Combustion Analysis

36

Chapter 3. Molecules, Compounds, and Nomenclature Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

3.8 Determining a Chemical Formula from Experimental Data  Elemental analysis is used to provide information about the mass of constituent elements in a compound. The goal is to convert to moles since moles can be compared directly to determine the simplest ratio.  The molecular formula is some whole number multiple of the empirical formula of the elements in a molecule. The molar mass provides that whole number.  Conceptual Connection 3.4 Chemical Formula and Mass Percent Composition

.



Students are tempted to analyze for oxygen in a combustion analysis problem in the same way they analyze for carbon and hydrogen.

37

Chapter 3. Molecules, Compounds, and Nomenclature

Additional Problem for the Mole Concept— Converting Between Mass and Number of Molecules (Example 3.11)

An ibuprofen tablet contains 200 mg of the painkiller whose formula is C13H18O2. How many ibuprofen molecules does it contain?

Sort

Given 200 mg C13H18O2

You are given the mass of ibuprofen and asked to find the number of molecules.

Find molecules of C13H18O2

Strategize

Conceptual Plan

Convert from mass to molecules by determining the number of moles. This requires converting the units of mg to g and using the molar mass of ibuprofen.

Avogadro’s number is used to change from moles to molecules.

mg C13H18O2  1g 1000 mg

g

1 mol 206.27 g



mol  molecules

6.022  1023 molecules 1 mol

Relationships Used 1 mg = 103 g C13H18O2 molar mass = 13(12.01) + 18(1.008) + 2(16.00) = 206.27 g mol-1 6.022  1023 = 1 mole

Solve

Solution

Follow the conceptual plan to solve the problem.

200 mg 

1 g 1000 mg



1 mol 6.022  1023 molecules  206.27 g 1 mol

Round the answer to the correct number of significant figures. The number 200 has only one significant figure (see Section 1.4).

= 5.838  1020 = 6  1020 molecules

Check

The units (molecules) are correct. The magnitude of the answer (6  1020 molecules) makes sense since we started with less than one molar mass of ibuprofen.

Check your answer. Are the units correct? Does the answer make physical sense?

.

38

Chapter 3. Molecules, Compounds, and Nomenclature

Additional Problem for Mass Percent Composition (Example 3.12)

Calculate the mass percent of calcium in calcium CaCO3.

Sort

Given CaCO3

You are given the molecular formula of calcium carbonate and asked to find the mass percent of Ca.

Find mass percent Ca

Strategize

Conceptual Plan

The molecular formula tells us that there is 1 mol Ca, 1 mol C, and 3 mol O for each mol calcium carbonate. Determine the mass of calcium and the mass of calcium carbonate using the equation that defines the mass percent.

mass % Ca =

1 molar mass Ca  100 molar mass CaCO3

Relationships Used mass % element X =

mass element X in 1 mol compound  100 mass of 1 mol compound

molar mass CaCO3 = 1(40.08) + 1(12.01) + 3(16.00) = 100.09 g mol-1 Solve

Solution

Follow the conceptual plan to solve the problem. Round the answer to four significant figures to reflect the four significant figures in the least precisely known quantity (40.08).

1  molar mass Ca = 40.08 molar mass CaCO3 = 100.09 g mol-1

Mass % Ca = =

1 molar mass Ca  100 molar mass CaCO3 40.08 g/mol 100.09 g/mol

 100

= 40.04%

Check

The units of the answer are correct and the magnitude makes sense. Because the mass of calcium carbonate is coincidentally about 100 g mol-1, the mass percent nearly matches the molar mass of calcium.

.

39

Chapter 3. Molecules, Compounds, and Nomenclature

Additional Problem for Mass Percent Composition as a Conversion Factor (Example 3.12)

The calcium in antacid tablets that contain calcium carbonate CaCO3 is touted as a good source of dietary calcium for bones. The mass percent of calcium in calcium carbonate is 40.04%. How much calcium does one obtain by consuming two antacid tablets that each contain 850 mg CaCO3? The recommended daily intake is 1000 mg for adults aged 18 to 50 years old.

Do the two tablets supply the minimum requirement?

Sort

Given 1700 mg CaCO3 (2 tablets  850 mg = 1700)

You are given the sample size of calcium carbonate and mass percent of calcium.

Mass % Ca = 40.04%

Find mg Ca

Strategize

Conceptual Plan

Convert between the mass of a compound and the mass of a constituent element using the mass percent relationship.

mg CaCO3 

mg Ca

40.04 mg Ca 100 mg CaCO3

Relationships Used 40.04 mg Ca = 100 mg CaCO3

Solve

Solution

Follow the conceptual plan to solve the problem.

1700 mg CaCO3 

40.04 mg Ca 100 mg CaCO3

= 680.68 mg Ca

680.68 mg = 680 mg

Check

The units of the answer (mg) are correct. The magnitude of the answer (680) makes physical sense since we expected somewhat less than half (40% is less than 50%). The two tablets do not supply the entire minimum requirement for calcium. .

40

Chapter 3. Molecules, Compounds, and Nomenclature Additional Problem for Chemical Formulas as Conversion Factors (Example 3.13)

Fuel cells convert molecules to energy without actually burning them. One technology hopes to use hydrogen taken from ethanol C2H6O, which in turn is produced from grain. What mass of hydrogen (in grams) is contained in 1.00 gallon of ethanol? (The density of ethanol is 0.789 g mL-1.)

Sort

Given 1.00 gal ethanol C2H6O dethanol = 0.789 g mL-1

You are given a volume of ethanol and its density. You are asked to find the mass of hydrogen it contains.

Find g H Strategize

Conceptual Plan

First, convert the volume of ethanol into a mass using conversion factors and the density.

gal C2H6O  L C2H6O  mL C2H6O  g C2H6O 3.785 L 1 gal

Second, convert the moles of water into moles of hydrogen and finally into grams of hydrogen.

1000 mL 1L

0.789 g 1 mL

 mol C2H6O  mol H  g H 1 mol C2H6O 46.07 g

6 mol H 1 mol C2H6O

1.008 g H 1 mol H

Relationships Used 3.785 L = 1 gal 1000 mL = 1 L molar mass C2H6O = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g mol-1 6 mol H : 1 mol C2H6O 1.008 g H = 1 mol H Solve Follow the conceptual plan.

Round the answer to the correct number of significant figures.

Solution 1 gal 

3.785 L 1000 mL 0.789 g   = 2986.4 g C2H6O 1 mL 1 gal 1 L

2986.4 g C2H6O 

1 mol 6 mol H 1.008 g H   46.06 g 1 mol C2H6O 1 mol H

= 392.13 g H = 392.1 g H

Check

The units (g) are correct. The magnitude of the answer (392 g) seems to make physical sense since hydrogen is so light.

.

41

Chapter 3. Molecules, Compounds, and Nomenclature

Additional Problem for Obtaining an Empirical Formula from Experimental Data (Examples 3.14 and 3.15)

A laboratory analysis of ascorbic acid (Vitamin C) produced the mass percent composition: C 40.98% H 4.58% O 54.50% Find the empirical formula.

1. Mass from percentages Write down or compute as given the masses of each element present in a sample of the compound. If you are given percent composition, assume a 100 g sample and compute them.

Given In a 100 g sample: 40.98 g C, 4.58 g H, 54.50 g C

Find empirical formula

1 mol C 12.01 g C

= 3.412 mol C

1 mol H 1.008 g H

= 4.544 mol H

2. Mass to moles

40.98 g C 

Convert each of the masses in step 1 to moles by using the appropriate molar mass for each element as a conversion factor.

4.58 g H 

54.50 g O 

3. Pseudoformula from coefficients

1 mol O = 3.406 mol O 16.00 g O

C3.412H4.544O3.406

Write a pseudoformula using the number of moles of each element from step 2 as subscripts.

4. Divide by lowest subscript

C 3.412 H 4.544 O 3.406 3.406

3.406

 C1H1.33O1

3.406

Divide all of the subscripts in the formula by the smallest subscript.

5. Convert to whole numbers If the subscripts are not whole numbers, multiply all subscripts by a small whole number to get whole number subscripts.

C1H1.33O1  3 = C3H4O3

The correct empirical formula for ascorbic acid is C3H4O3. .

42

Chapter 3. Molecules, Compounds, and Nomenclature Fractional Subscript Multiply by this 0.20

5

0.25

4

0.33

3

0.40

5

0.50

2

0.66

4

0.75

4

0.80

5

.

43

Chapter 3. Molecules, Compounds, and Nomenclature

Additional Problem for Calculating a Molecular Formula from an Empirical Formula and Molar Mass (Example 3.16)

Ascorbic acid (Vitamin C) has the empirical formula of C3H4O3 and a molar mass of 176.13 g mol-1. Find its molecular formula.

Sort

Given Empirical formula = C3H4O3 molar mass of ascorbic acid = 176.13 g mol-1

You are given the empirical formula and molar mass of ascorbic acid and asked to find the molecular formula.

Find molecular formula

Molecular formula = empirical formula  n

Strategize A molecular formula is always a whole-number multiple of the empirical formula. Divide the molar mass by the empirical formula mass to get the whole number.

n =

molar mass empirical formula mass

Solve

Empirical formula molar mass

Compute the empirical formula mass.

= 3(12.01) + 4(1.008) + 3(16.00) = 88.06

Divide the molar mass by the empirical formula mass to find the whole number.

n =

176.13 g/mol molar mass = = 2 empirical formula mass 88.06 g/mol

Molecular formula = C3H4O3  2 Multiply the empirical formula by the whole number to obtain the molecular formula.

= C6H8O6

Compute the molar mass using the computed formula:

Check

6(12.01) + 8(1.008) + 6(16.00) = 176.14 The computed molar mass is in agreement with the given molar mass. The answer is correct.

.

44

Chapter 3. Molecules, Compounds, and Nomenclature

Additional Problem for Obtaining an Empirical Formula from Combustion Analysis (Examples 3.17 and 3.18)

Upon combustion, a compound containing only carbon and hydrogen produced 2.59 g CO2 and 1.19 g H2O. Find the empirical formula of the compound.

1. Masses of combustion products

Given 2.59 g CO2 and 1.19 g H2O

Write the masses of each combustion product as given. If given, write the mass of the sample.

Find Empirical formula

2. Masses to Moles

2.59 g CO2 

Convert the masses of CO2 and H2O from step 1 to moles by using the appropriate molar mass for each compound as a conversion factor.

1.19 g H2O 

1 mol CO2 44.01 g CO2

= 0.0588 mol CO2

1 mol H2O 18.02 g H2O

= 0.0662 mol H2O

0.0588 mol CO2 

= 0.0588 mol C

Convert the moles in step 2 to moles of C and moles of H using the conversion factors inherent in the formulas CO2 and H2O.

1 mol C 1 mol CO2

0.0662 mol H2O 

2 mol H 1 mol H2O

= 0.132 mol H

4. Other elements

No other elements, so proceed to the next step.

3. Moles to Moles of C and H

If the compound contains an element other than C and H, find the mass of the other element by subtracting the sum of the masses of C and H from the mass of the sample. Finally, convert the mass of the other element to moles.

C0.0588H0.132

5. Write the pseudoformula Write down the pseudoformula for the compound using the number of moles of each element (in steps 3 and 4) as subscripts.

6. Divide by the smallest subscript

C 0.0588 H 0.132 0.0588

 C1H2.24

0.0588

Divide all the subscripts in the formula by the smallest subscript. (Round all subscripts that are within 0.1 of a whole number.)

7. Whole number subscripts

C1H2.24  4  C4H9

If the subscripts are not whole numbers, multiply by a small whole number to get whole-number subscripts.

This is the correct empirical formula.

.

45

Chapter 4. Chemical Reactions and Stoichiometry Student Objectives 4.1 Chemistry of Cuisine 

Define and understand the role of certain chemicals in everyday cooking and baking.

4.2 Writing and Balancing Chemical Equations  

Define and understand the difference between a chemical equation and a chemical reaction.

Define reactants and products and understand their physical states.

4.3 Solutions and Solubility     

Understand the interactions between water as a solvent and different solutes dissolved in it.

Define and understand electrolyte, strong electrolyte, weak electrolyte, and nonelectrolyte.

Define strong and weak acids.

Understand what is meant by soluble and insoluble.

Know the solubility trends for compounds made from common anions and cations.

4.4 Precipitation Reactions  

Define precipitate and precipitation reaction.

Use the solubility trends to predict how the ions from soluble compounds combine to form precipitates.

4.5 Acid–Base Reactions       

Define acid–base reaction (neutralization).

Understand the Arrhenius definitions for acids and bases.

Recognize diprotic and polyprotic acids from chemical formulas.

Know the lists of strong acids and bases.

Describe what happens in an acidbase reaction.

Understand the equivalence point of a titration, and use solution stoichiometry and equivalence point to calculate the concentration of an unknown in a titration.

Recognize gasevolution reactions and predict the identity of evolved gases.

4.6 OxidationReduction Reactions    

Define oxidation, reduction, and oxidation–reduction (redox) reactions.

Assign oxidation states to atoms in a chemical formula.

Identify the elements undergoing oxidation and reduction in a redox reaction.

Understand that combustion is a common redox reaction.

4.7 Reaction Stoichiometry: How Much Is Produced?    

Define and understand stoichiometry, the numerical relationships among chemical amounts in a balanced chemical reaction.

Understand and use a balanced chemical reaction to calculate the mole relationships between components.

Calculate the mass of a reactant needed to produce a certain mass of product.

Calculate the mass of a product formed from a certain mass of reactant.

.

46

Chapter 4. Chemical Reactions and Stoichiometry

4.8 Limiting Reactant, Theoretical Yield, and Percent Yield    

Define limiting reactant and theoretical yield.

Predict a limiting reactant using initial reactant masses and the theoretical yield.

Calculate and determine a theoretical yield and a percent yield.

Understand and describe the role of gasoline additives in the combustion of vehicle fuels.

4.9 Solution Concentration and Solution Stoichiometry     

Define solution, solvent, solute, aqueous solution, and molarity.

Calculate the molarity of a solution.

Use molarity as a conversion factor in calculating numbers of moles and volumes of solution.

Calculate the concentration after a solution has been diluted.

Calculate reaction component amounts using volume, moles, concentration, and stoichiometry.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

47

Chapter 4. Chemical Reactions and Stoichiometry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

4.1 Chemistry of Cuisine  Cooking o Examples of chemical compounds and reactions in everyday cooking and baking

 

Unnumbered figure: cake Unnumbered figure: fresh fish

4.2 Writing and Balancing Chemical Equations  Chemical reactions and chemical equations  Combustion reaction o Chemical reactions; reactants vs. products



Table 4.1 States of Reactants and Products in Chemical Equations Unnumbered figure: table of atom balance for equation of methane combustion Examples 4.1 and 4.2 Balancing Chemical Equations

4.3 Solutions and Solubility  Competing interactions o Solutesolute o Solvent–solute  Dissolving ionic compounds o Dissociation or ionization o Strong electrolyte solutions  Dissolving (most) molecular compounds o No dissociation o Nonelectrolyte solutions  Dissolving acids o Molecular compounds that dissociate o Strong electrolyte solutions o Weak electrolyte solutions  Solubility of ionic compounds o Relative terms: soluble, insoluble o Rules based on common anions and cations

 

 

       

.

Figure 4.1 Solute and Solvent Interactions Figure 4.2 Charge Distribution in a Water Molecule Figure 4.3 Solute and Solvent Interactions in a Sodium Chloride Solution Figure 4.4 Sodium Chloride Dissolving in Water Figure 4.5 Sugar and Water Interactions Unnumbered figure: photo of AgCl in water Figure 4.6 Properties of Electrolytes in Solution Unnumbered figures: illustrations of AgCl and AgNO3 in water Table 4.2 General Solubility Rules for Ionic Compounds in Water Example 4.3 Predicting Whether an Ionic Compound Is Soluble

48

Chapter 4. Chemical Reactions and Stoichiometry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

4.1 Chemistry of Cuisine  Present different examples of chemical reactions in cooking and baking  Introduce the Maillard reaction

4.2 Writing and Balancing Chemical Equations  Tips for writing chemical equations and discussion of coefficients vs. subscripts  Balancing chemical equations by inspection  Conceptual Connection 4.1 Balanced Chemical Equations



Students often get confused between coefficients indicating moles of a substance and subscripts indicating moles of atoms in a molecule

4.3 Solutions and Solubility  Solutions maximize solute–solvent interactions and diminish solute–solute interactions. That competition determines whether a solute dissolves.  Review representations of solid NaCl and aqueous NaCl, using Figure 4.4.  Try the electrolyte demonstration in class using solid NaCl, aqueous NaCl, hydrochloric acid, acetic acid, and sugar water. Distilled water likely will not cause the light bulb to illuminate, but does that mean there are no ions in the solution?  NaCl dissolves to a large extent while AgCl does not. The macroscopic AgCl mixture gives no evidence for dissolution of any of the compound; that could be measured by conductivity.  The table of solubility rules provides many examples for practice. Some ions are common to several portions of the rules, so the table is not that complex or extensive to remember.



Salts dissolve to form ions in water; molecules like sugar don’t ionize at all. Acetic acid dissolves as a mixture of molecules and ions.

Insoluble is a relative term, just as soluble is relative.

When writing chemical equations for solubility, students often are tempted to include H2O(l) in the equations.

 

.

49

Chapter 4. Chemical Reactions and Stoichiometry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

4.4 Precipitation Reactions  Define precipitate and precipitation reaction  Procedure for writing balanced precipitation reactions o Formulas for reactants o Mix‐and‐match cations and anions o Correct formulas for products  Solubility of products  Molecular equation, net ionic equation  Spectator ion

   

4.5 Acid–Base Reactions  Acid–base reactions o Arrhenius definitions o Monoprotic vs. diprotic and polyprotic acids o Common acids and bases o Reactions: production of water and a salt o Titrations  Equivalence point  Indicator

      

.

Unnumbered figure: interaction of hard water with soap Figure 4.7 Precipitation of Lead(II) Iodide Unnumbered figure: cross‐products of KI(aq) + Pb(NO3)2(aq) Examples 4.4 Writing Equations for Precipitation Reactions

Figure 4.8 The Hydronium Ion Unnumbered figures: common examples of acids and bases Table 4.3 Common Acids and Bases Figure 4.9 Acid–Base Reaction Example 4.5 Writing Equations for Acid– Base Reactions Table 4.4 Types of Compounds That Undergo Gas–Evolution Reactions Example 4.6 Writing Equations for Gas– Evolution Reactions

50

Chapter 4. Chemical Reactions and Stoichiometry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

4.4 Precipitation Reactions  A demonstration emphatically makes the point. The example in the text is especially good because of the brightly coloured precipitate.  Show an example with the three equations and the logical progression from molecular to complete ionic to net ionic.



4.5 Acid–Base Reactions  Demonstrate acidity and basicity of common foods and household goods using red cabbage juice as an indicator.  The microscopic representations of acids and bases in Figure 4.9 provide a useful conceptual understanding for later discussions about acids and bases.  Titration is an important analytical technique that can be used for a variety of quantitative analyses.

Besides colour, other properties can be measured; often, this is done by instruments for greater accuracy and automated for speed.  Gas–evolution forces reactions to proceed to completion since one of the products generally escapes.  A good demonstration is to compare the solubility of chalk in water and in hydrochloric acid. Use a flame to test the identity of the escaping gas in the latter case.



.



Students tend to forget the rules for writing electrically neutral formulas for ionic compounds when writing precipitation reactions.

The three equations are meant to convey slightly different ideas.

Students may suppose especially from net ionic equations that aqueous solutions need not be electrically neutral, e.g., Ag+(aq) combining with Cl(aq) in the absence of counterions.

A gas that evolves from a reaction may be partially soluble in water, so one may not be able to measure all of the gas escaping

51

Chapter 4. Chemical Reactions and Stoichiometry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

4.6 Oxidation–Reduction Reactions  Definitions of oxidation and reduction  Rules for assigning oxidation states  Identifying redox reactions

     

4.7 Reaction Stoichiometry: How Much Is Produced  Balanced reactions o Coefficients o Stoichiometry  Conversions o Mole‐to‐mole conversions o Mass‐to‐mass conversions

 

.

Figure 4.10 Oxidation–Reduction Reaction Figure 4.11 Oxidation–Reduction Without Oxygen Figure 4.12 Redox with Partial Electron Transfer Example 4.7 Assigning Oxidation States Example 4.8 Identifying Redox Reactions, Oxidizing Agents, and Reducing Agents Chemistry in Your Day: Bleached Blonde

Unnumbered figures: mass‐to‐mass stoichiometric conversions Example 4.9 Stoichiometry

52

Chapter 4. Chemical Reactions and Stoichiometry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

4.6 Oxidation–Reduction Reactions  Students may have some difficulty visualizing how NaCl comes to be so ordered compared with the gaseous chlorine and solid sodium.  Common oxidation states were some of the first properties related to periodicity and the order of the periodic table.  A good demonstration of redox is the reaction of zinc metal with hydrochloric acid.

Test the identity of the evolved gas with a flame. Emphasize the importance of relating the phase of each substance to its electric charge.  Bleaching one’s hair is a benign example of oxidation; others, like rust and corrosion, are more destructive.  Conceptual Connection 4.2 Oxidation and Reduction



4.7 Reaction Stoichiometry: How Much Is Produced?  Moles continue to be central to conversions and calculations. For chemical equations, stoichiometry allows those conversions to include any reaction component.  Conceptual Connection 4.3 Stoichiometry







.

Chemical reactions pair a substance undergoing oxidation with one undergoing reduction. Although we sometimes speak of them independently, they always occur together.

Students get the terminology confused: an oxidizing agent is itself reduced; a reducing agent is itself oxidized.

Masses of reactants and products are never directly equivalent. Masses can only be related through moles, and only moles can be directly interchanged through stoichiometry.

Students sometimes get confused using stoichiometry to convert from a reactant, e.g., N2, to a product that contains other elements, e.g., NH3, without having to consider the other reactant (H2).

53

Chapter 4. Chemical Reactions and Stoichiometry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

4.8 Limiting Reactant, Theoretical Yield, and Percent Yield  Difference between balanced reaction and physical quantities in a given case o Limiting reactant o Theoretical yield o Percent yield  Yield calculations o Limiting reactant from initial reactant masses  Theoretical yield from initial reactant masses

  

4.9 Solution Concentration and Solution Stoichiometry  Concentration o Molarity o Dilution  Solution stoichiometry o Volume and concentration to moles o Moles and concentration to volume

       

.

Unnumbered figures: molecular models of C3H8, O2, CO2, and H2O in stoichiometric relationships Unnumbered figure: conceptual plan for limiting reactant problems Examples 4.10 and 4.11 Limiting Reactant and Theoretical Yield

Figure 4.13 Preparing a One‐Molar NaCl Solution Example 4.12 Calculating Solution Concentration Example 4.13 Using Molarity in Calculations Figure 4.14 Preparing 3.00L of 0.500 mol L‐1 CaCl2 from 10.0 mol L‐1 Stock Solution Example 4.14 Solution Dilution Unnumbered figure: strategy for solution stoichiometry Example 4.15 Solution Stoichiometry Chemistry in Your Day: Blended Ethanol gasoline

54

Chapter 4. Chemical Reactions and Stoichiometry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

4.8 Limiting Reactant, Theoretical Yield, and Percent Yield  Show examples of problems with limiting reactants using some common pitfalls of students: forgetting to count both atoms of a molecular element or forgetting differences in coefficients within the chemical equation.  Conceptual Connection 4.4 Limiting Reactant and Theoretical Yield  Conceptual Connection 4.5 Reactant in Excess



4.9 Solution Concentration and Solution Stoichiometry  Figures 4.13 and 4.14 offer a means to integrate a laboratory technique into the lecture.  For a group activity or in‐class example, have students calculate a dilution. To encourage them to think before blindly entering numbers in a calculator ask, “How many mL of a 2.3 mol L‐1 NaCl solution does it take to make 100mL of 2.5 mol L‐1 NaCl?”  Conceptual Connection 4.6 Solutions



 



.

Some students confuse percent and fraction and interchange them. They can be converted to each other, but they are not the same.

Students sometimes confuse coefficients and subscripts within chemical formulas in stoichiometry calculations.

Some students try to rewrite balanced reactions in limiting reactant cases by including leftover reactants on the right side of the equation.

Concentration is an intensive property (similar to density), whereas both the number of moles and the volume are extensive properties.

Solutes occupy volume. One can’t begin with 1 L of solvent to get 1 L of solution.

55

Chapter 4. Chemical Reactions and Stoichiometry Additional Problem for Stoichiometry (Example 4.15)

Humans use the glucose molecule as an energy source in the body: C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(aq) How many grams of carbon dioxide are produced for every 100.0 g of glucose metabolized?

Sort

Given 100 g C6H12O6

The problem gives the mass of glucose and asks you to find the mass of carbon dioxide produced.

Find g CO2

Strategize

Conceptual Plan

The conceptual plan follows the pattern of mass A  moles of A  moles of B  mass of B. From the chemical equation you can determine the relationship between the moles of reactant and products. Use the molar masses to convert between grams and moles.

g C6H12O6  mol C6H12O6  mol CO2  g CO2 1 mol 180.2 g C6H12O6

44.01 g CO2 1 mol CO2

6 mol CO2 1 mol C6H12O6

Relationships Used molar mass CO2 = 44.01 g mol-1 1 mol C6H12O6 = 6 mol CO2 molar mass C6H12O6 = 180.2 g mol-1

Solve

Solution

Follow the conceptual plan. Begin with the grams of glucose and end with grams of carbon dioxide.

100 g C6H12O6 

1 mol C6H12O6 180.2 g C6H12O6



6 mol CO2 1 mol C6H12O6



44.01 g CO2 1 mol CO2

= 146.5 g CO2

Check

The units (g) are correct. The magnitude of the answer makes sense. The mass of glucose is much larger than that of carbon dioxide but many more CO2 molecules are produced.

.

56

Chapter 4. Chemical Reactions and Stoichiometry Nitrogen dioxide NO2 can be synthesized by the reaction:

Additional Problem for Limiting Reagent and Theoretical Yield

N2(g) + 2 O2(g)  2 NO2(g) Starting with 46.3 g N2 and 92.0 g O2, find the theoretical yield of NO2 in grams.

Given 46.3 g N2

Sort You are given the mass of each reactant in grams and asked to find the theoretical yield of product.

92.0 g O2

Find theoretical yield of NO2

Strategize

Conceptual Plan

Determine which reactant makes the least amount of product by converting from grams of reactant to moles of reactant to moles of product. Use molar masses to convert between grams and moles and use the stoichiometric relationships (from the equation) to convert between moles of reactant and moles of product. Convert the moles of limiting reactant to grams of product.

g N2  mol N2  mol NO2 1 mol N2 28.02 g N2

2 mol NO2 1 mol N2

g O2  mol O2  mol NO2 1 mol O2 32.00 g O2

2 mol NO2 2 mol O2

Smaller of mol NO2

mol NO2  g NO2 46.01 g NO2 1 mol NO2

Relationships Used molar mass of N2 = 28.02 g molar mass of O2 = 32.00 g molar mass of NO2 = 46.01 g 1 mol N2 : 2 mol NO2 2 mol O2 : 2 mol NO2

.

57

Chapter 4. Chemical Reactions and Stoichiometry Solve

Solution

Beginning with the given amount of each reactant, calculate the amount of product that can be made in moles. Convert the amount of the product made by the limiting reagent to grams—this is the theoretical yield.

46.3 g N2 

92.0 g O2 

1 mol N2 28.02 g N2 1 mol O2 32.00 g O2



2 mol NO2 = 3.30 mol NO2 1 mol N2



2 mol NO2 = 2.88 mol NO2 2 mol O2

2.88 is smaller than 3.30 mol

2.88 mol NO2 

46.01 g NO2 = 132 g NO2 1 mol NO2

The units of the answer are correct and the magnitude makes sense. The oxygen is the limiting reagent.

Check

.

58

Chapter 4. Chemical Reactions and Stoichiometry Additional Problem for Calculating Solution Concentration (Example 4.12)

If 37.6 g of LiBr is dissolved in enough water to make 2.50 L of solution, what is the molarity of the solution?

Sort

Given 37.6 g LiBr

You are given the mass of LiBr and the volume of the solution and asked to find its molarity.

2.50 L of solution

Find molarity M (in mol L-1)

Strategize

Conceptual Plan

When formulating the conceptual plan, think about the definition of molarity, the amount of solute in moles per litre of solution.

g



mol

1 mol LiBr 85.94 g LiBr

mol LiBr, L solution You are given the mass of LiBr, so first use the molar mass of LiBr to convert from g LiBr to mol LiBr.

molarity (M) =



M (in mol L-1)

amount of solute (mol) volume of solution (L)

Relationships Used

Then use the number of moles of LiBr and litres of solution to find the molarity.

molar mass of LiBr = 85.94 g mol-1

Solve

Solution

Follow the conceptual plan to solve the problem.

Round the answer to three significant figures to reflect the three significant figures in the density.

37.6 g LiBr 

1 mol LiBr 85.94 g LiBr

molarity (M) =

= 0.4375 mol LiBr

-1 0.4375 mol LiBr = 0.175 M (mol L ) 2.50 L solution

The units of the answer (mol L-1) are correct. The answer is reasonable—mol L-1 should be less than 1. The number of moles is < 1, while the volume is > 1 L.

Check

.

59

Chapter 4. Chemical Reactions and Stoichiometry Additional Problem for Using Molarity in Calculations (Example 4.13)

How many litres of 1.50 M KOH solution contains 0.455 mol of KOH?

Sort

Given 1.50 M KOH

You are given the concentration of KOH solution and asked to find the volume of solution that contains a given amount of KOH in moles.

0.455 mol KOH

Find volume of KOH solution in L

Strategize

Conceptual Plan

The conceptual plan begins with mol KOH and converts to L of solution using the molarity as a conversion factor.

mol KOH



L solution

1 L solution 1.50 mol KOH

Relationships Used 1.50 M KOH =

1.50 mol KOH 1 L solution

Solve

Solution

Follow the conceptual plan. Begin with mol KOH and convert to L solution.

0.455 mol KOH 

Check

The units (L) are correct. The magnitude of the answer (0.303) seems reasonable. Less than 1 mol KOH was used for molarity greater than 1. The volume should be less than 1 L.

.

1 L solution = 0.303 L 1.50 mol KOH

60

Chapter 4. Chemical Reactions and Stoichiometry Additional Problem for Solution Dilution (Example 4.14)

To what volume should you dilute 0.300 L of a 6.00 M KOH solution to obtain a 2.50 M KOH solution?

Sort

Given V1 = 0.300 L

You are given the initial volume, initial concentration, and final concentration of a solution, and you need to find the final volume.

M1 = 6.00 M M2 = 2.50 M

Find V2

Strategize

Conceptual Plan

A volume times a concentration equals the number of moles. The initial and final number of moles is the same during a dilution, so M1V1 = M2V2. You are given V1, M1, and M2. You are asked to find V2.

V1, M1, M2



V2

M1V1 = M2V2

Relationships Used M1V1 = M2V2

Solve

Solution

Begin with the equation. Substitute V1, M1, and M2.

Solve for V2.

M1V1 = M2V2 V2 =

M1V1 V2

6.00 M  0.300 L 2.50 M = 0.720 L

V2 =

Begin with a volume of 0.300 L and make a final volume of 0.720 L.

The units (L) are correct. The magnitude of the answer (20.0) seems reasonable since the concentration is diluted from 6 mol L-1 to 2.5 mol L-1, a factor of between 2 and 3. The volume must increase by a factor of between 2 and 3.

Check

.

61

Chapter 4. Chemical Reactions and Stoichiometry Additional Problem for Solution Stoichiometry (Example 4.15)

What volume (in L) of 2.50 M NaCl solution is required to completely react with 0.150 L of a 0.150 M AgNO3 solution according to the reaction: NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)

Sort You are given the volume and concentration of a AgNO3 solution. You are asked to find the volume of NaCl solution required to react with it.

Given 0.150 L of 0.150 M AgNO3 0.250 M NaCl

Find volume NaCl solution (in L)

Strategize

Conceptual Plan

The conceptual plan has the form:

L AgNO3 soln  mol AgNO3  mol NaCl  L NaCl soln

volume A  mol A  mol B  volume B. The molar concentrations of both solutions can be used as conversion factors between volume and number of moles.

The stoichiometric coefficients from the equation are used to adjust for different numbers of moles of reactant and product.

0.150 mol AgNO 3 1 L solution

1 mol NaCl 1 mol AgNO3

1 L solution 0.250 mol NaCl

Relationships Used M [AgNO 3 ] =

0.150 mol AgNO 3 1 L solution

1 mol AgNO3 = 1 mol NaCl M [NaCl] =

Solve

Solution

Begin with L AgNO3 solution and follow the conceptual plan to arrive at L NaCl solution.

0.150 L 

0.250 mol NaCl 1 L solution

0.150 mol AgNO3 1 L



1 mol NaCl 1L  1 mol AgNO3 0.250 mol NaCl

= 0.0900 L

Check

The units (g) are correct. The magnitude of the answer (0.0900) seems to make physical sense. Since the NaCl is more concentrated than the AgNO3, a smaller volume of NaCl solution than of AgNO3 solution is required.

.

62

Chapter 4. Chemical Reactions and Stoichiometry Additional Problem for Writing Equations for Precipitation Reactions (Examples 4.4)

Write an equation for the precipitation reaction that occurs (if any) when solutions of calcium chloride and potassium carbonate are mixed.

1. Write formulas from two compounds Write the formulas of the two compounds being mixed as reactants in a chemical reaction.

2. Write formulas of two products Write the formulas of the products that could form from the reactants. Combine the cation from each reactant with the anion from the other.

Write correct formulas for these ionic compounds.

Possible Products CaCO3

3. Check solubility rules Use the solubility rules to determine whether either of the products is insoluble.

KCl

CaCO3 is insoluble. Compounds containing carbonate CO32 are generally insoluble and Ca2+ is not an exception.

KCl is soluble. Both K+ and Cl are listed as generally soluble.

4. Both products are soluble If all of products are soluble, there will be no precipitate. Write NO REACTION after the arrow.

Since one product is not soluble, we proceed to the next step.

5. At least one product insoluble If any of the possible products are insoluble, write its formula as the product of the reaction using “(s)” to indicate solid. Write any soluble products with “(aq)” to indicate aqueous.

.

63

Chapter 4. Chemical Reactions and Stoichiometry Additional Problem for Acid–Base Titration

The titration of a 20.0 mL sample of an HCl solution of unknown concentration requires 8.32 mL of a 0.150 mol L-1 NaOH solution to reach the equivalence point. What is the concentration of unknown HCl in mol L-1?

Sort

Given 8.32 mL of 0.150 M NaOH

You are given the volume and concentration of NaOH solution required to titrate a given volume of HCl solution. You are asked to find the concentration of the HCl solution.

20.0 mL of HCl

Find concentration of HCl solution

Strategize

Conceptual Plan

Start by writing a balanced equation for the acid– base neutralization reaction.

mL NaOH  L NaOH  mol NaOH  mol HCl 1L 1000 mL

The conceptual plan has the form: volume A  moles A  moles B. The concentration of the NaOH solution is the conversion factor between moles and volume of NaOH. The balanced equation provides the relationship between the number of moles of base and acid.

0.150 mol NaOH 1L

1 mol HCl 1 mol NaOH

mol HCl, volume HCl  molarity M =

In the second part, use the number of moles of HCl and the given volume to calculate the molarity of the HCl solution.

mol L

Relationships Used 1 L = 1000 mL M (NaOH) =

0.150 mol NaOH L

1 mol HCl : 1 mol NaOH Molarity (M) =

Solve

moles of solute (mol) liters of solution (L)

Solution

Begin with the volume of NaOH, convert it to L, and solve for the number of moles using the concentration of NAOH. Don’t forget to consider the ratio of moles of HCl to NaOH.

8.32 mL 

1 L 0.150 mol NaOH 1 mol HCl   1 mol NaOH 1000 mL 1 L

= 0.001248 mol HCl .

64

Chapter 4. Chemical Reactions and Stoichiometry Convert from mol HCl to molarity.

Molarity =

0.001248 mol 1000 mL  1L 20 mL

= 0.0624 mol L-1

Round the answer to the correct number of significant figures.

The units (mol L-1) are correct. The magnitude of the answer (0.0624) seems reasonable; about 2.5 times as much HCl solution as base was used. The HCl concentration should be about 2.5 times less concentrated.

Check

.

65

Chapter 5. Gases Student Objectives 5.1 Breathing: Putting Pressure to Work 

Define pressure and understand how differences in pressure lead to the act of breathing.

5.2 Pressure: The Result of Molecular Collisions    

Understand pressure from a molecular point of view.

Understand examples of pressure: wind and pressure imbalance on the eardrum.

Understand why pressure can be measured in mm of Hg, and convert between the different pressure units.

Understand the origin of the two numbers for blood pressure.

5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law      

Know and be able to rationalize Boyle’s law, the inverse relationship between volume and pressure.

Use the inverse mathematical relationship between pressure and volume to solve initial and final states problems at constant temperature and amount.

Know and be able to rationalize Charles’s law, the direct relationship between volume and temperature.

Use the direct mathematical relationship between volume and temperature to solve initial and final states problems at constant pressure and amount.

Know and be able to rationalize Avogadro’s law, the direct relationship between volume and amount (moles).

Use the direct mathematical relationship between volume and amount to solve initial and final states problems at constant pressure and temperature.

5.4 The Ideal Gas Law  

Know and understand how the ideal gas law combines the three simple gas laws into one equation.

Calculate using the ideal gas law and the gas constant, R, with the appropriate value and units.

5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas   

Define standard temperature and pressure and molar volume of an ideal gas.

Know and understand the relationship between molar volume, molar mass, and density.

Calculate using density, molar mass, and molar volume.

5.6 Mixtures of Gases and Partial Pressures     

Define and understand partial pressure of a gaseous component in a mixture.

Define and determine mole fraction of a component in a mixture.

Know and understand that Dalton’s law of partial pressures relates to mole fraction of a gas to the partial pressure of the gas.

Understand how the total pressure affects the partial pressures of gases in blood, especially during deep‐sea diving.

Know and understand the technique of collecting gases over water.

.

66

Chapter 5. Gases 5.7 Gases in Chemical Reactions: Stoichiometry Revisited  

Understand how stoichiometry applies to gases via the number of moles in the ideal gas law.

Understand how stoichiometry relates to molar volume.

5.8 Kinetic Molecular Theory: A Model for Gases      

Define kinetic molecular theory for gases.

Understand each of the three postulates/assumptions of the kinetic molecular theory.

Understand how the kinetic molecular theory explains Boyle’s, Charles’s, Avogadro’s, and Dalton’s laws.

Follow the derivation of the ideal gas law from the kinetic molecular theory.

Understand that all gases at the same temperature have the same kinetic energy, and understand the relationship between speed and molar mass.

Understand the graphical representation of the distribution of molecular speeds.

5.9 Mean Free Path, Diffusion, and Effusion of Gases  

Define and understand mean free path.

Define diffusion and effusion and understand how they are related to the kinetic molecular theory.

5.10 Real Gases: The Effects of Size and Intermolecular Forces   

Understand that the ideal gas law is an approximation that works well under certain circumstances and not so well at low temperature and/or high pressure.

Understand that nonideal behaviour arises from the finite volume of gas particles and the intermolecular forces between particles.

Recognize and identify the components of the van der Waals equation.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

67

Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

5.1 Breathing: Putting Pressure to Work  Pressure o Practical definition o Breathing as a result of pressure difference

 

Intro figure: pressure in a champagne bottle Figure 5.1 Gas Pressure

5.2 Pressure: The Result of Molecular Collisions  Pressure o Molecular view o Units: millimetres of mercury, torr, atmosphere, pascal, pounds per square inch o Measurements  Mercury barometer  Manometer o Blood pressure



Unnumbered figure: regions of high and low atmospheric pressure Figure 5.2 Pressure and Particle Density Figure 5.3 Pressure Imbalance Figure 5.4 The Mercury Barometer Table 5.1 Common Units of Pressure Example 5.1 Converting Between Pressure Units Figure 5.5 The Manometer Chemistry and Medicine: Blood Pressure

      

5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law  Simple gas laws o Boyle’s law: volume and pressure  V  1/P  P1V1 = P2V2 o Charles’s law: volume and temperature  VT  V1/T1 = V2/T2 o Avogadro’s law: volume and amount  Vn  V1/n1 = V2/n2

           

5.4 The Ideal Gas Law  Derived from the simple gas laws: PV = nRT o Units: Kelvin for T, moles for n o Gas constant R: function of P, V units

    .

Figure 5.6 The J‐Tube Figure 5.7 Volume versus Pressure Figure 5.8 Molecular Interpretation of Boyle’s Law Figure 5.9 Increase in Pressure with Depth Example 5.2 Boyle’s Law Figure 5.10 Volume versus Temperature Figure 5.11 Molecular Interpretation of Charles’s Law Unnumbered figure: hot‐air balloon Unnumbered figure: balloon in liquid N2 Example 5.3 Charles’s Law Figure 5.12 Volume versus Number of Moles Example 5.4 Avogadro’s Law

Unnumbered figure: relationship between ideal gas law and simple gas laws Unnumbered figure: label on aerosol can Example 5.5 Ideal Gas Law I Example 5.6 Ideal Gas Law II 68

Chapter 5. Gases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

5.1 Breathing: Putting Pressure to Work  All students should have experience with a bottle of carbonated liquid, though it is more likely with soda pop than champagne.

5.2 Pressure: The Result of Molecular Collisions  Recognizing pressure as related to force will likely help most students. The difference can be explained by talking about snowshoes and ice skates.  Large numbers of people have experienced their ears “popping” in an airplane or during another ascent in altitude.  Tornados, hurricanes, and cyclones are extreme examples of differential pressures in weather.  Blood pressure is an interesting alternative since the previous examples deal with gases.



Barometers and manometers combine liquids and gases. The end of the tube can be open, under a vacuum, or filled with air. The outcomes are different and can be confusing.

5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law  The easiest example of Boyle’s law is a cylinder with a sliding piston. A plastic syringe works well as long as one has a convenient connection to a gauge.  Limited numbers of students will have personal experience with diving, though most students likely will have experience swimming underwater.  Charles’s law can be demonstrated using balloons and dry ice/acetone or liquid nitrogen.  Avogadro’s law can be demonstrated by producing a gas over water to show displacement. An interesting challenge question comes from burning a candle within a sealed container connected to a pressure gauge. Oxygen is consumed, but carbon dioxide is produced.



Breathing at increased depths of water becomes more difficult but is possible even to substantial depths.

Hot air and helium balloons rise because of decreased density compared with the surrounding air.

Some students memorize the different equations for initial and final states problems, but all of them can be derived easily from the ideal gas law.

In initial and final states problems with T, the ratio of the values must be in Kelvin and not in degrees Celsius.

 



5.4 The Ideal Gas Law  A series of R values is possible with different units, especially for pressure.  Common problems include using mixtures of unit conversions: oC to K, mL to L, mmHg to atm. .

69

Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas  Standard temperature and pressure o Molar volume  Density: molar volume/molar mass



o 

d

 

Unnumbered figure: molar volumes of different gases at STP Example 5.7 Density Example 5.8 Molar Mass of a Gas

PM , M  molar mass RT

Finding the molar mass of a gas

5.6 Mixtures of Gases and Partial Pressures  Mixtures of gases o Partial pressure o Mole fraction o Dalton’s law  Deep‐sea diving o Hypoxia o Oxygen toxicity o Nitrogen narcosis  Collecting gases over water o Vapour pressure of water

        

5.7 Gases in Chemical Reactions: Stoichiometry Revisited  Calculations: moles from the ideal gas law o P, V, T of gas A  moles A  moles B  P, V, T of gas B  Calculations: molar volume o V of gas A  moles A  moles B  mass of gas B

 

.

Table 5.3 Composition of Dry Air Unnumbered figure: pie chart of composition of dry air Example 5.9 Total Pressure and Partial Pressures Figure 5.13 Oxygen Partial Pressure Limits Unnumbered figure: gas pressure in lungs Example 5.10 Partial Pressures and Mole Fractions Table 5.4 Vapour Pressure of Water versus Temperature Figure 5.14 Collecting a Gas over Water Example 5.11 Collecting Gases over Water

Example 5.12 Gases in Chemical Reactions Example 5.13 Using Molar Volume in Gas Stoichiometric Calculations

70

Chapter 5. Gases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas  Conceptual Connection 5.1 Molar Volume  Density can be calculated from molar volume and molar mass or from P, T, and M.



5.6 Mixtures of Gases and Partial Pressures  The partial pressure of oxygen in air is important in physiology. Oxygen available in the lungs for transfer to the blood can be changed by a change in the total air pressure or by a change in the percent oxygen in air.  Partial pressure of oxygen plays a role in diving and in climbing at high altitude. Too little or too much can be dangerous.  The vapour pressure of water must be considered when collecting gases over water, a common laboratory technique. The solubility of gases in water is generally small except for CO2.  The vapour pressure of water is a function of temperature and is covered in more detail in Chapter 11.



5.7 Gases in Chemical Reactions: Stoichiometry Revisited  The ideal gas law is used to obtain moles rather than using mass or volume and concentration of an aqueous solution.  At the same pressure and temperature, the mole ratio from a balanced reaction is the same as the volume ratio for gaseous components.  At the same volume and temperature, the mole ratio from a balanced reaction is the same as the pressure ratio for gaseous components.  Conceptual Connection 5.2 Pressure and Number of Moles







.

Ideal gases have a common molar volume but not a common mass.

Standard temperature for gases is not the same standard as in thermochemistry and thermodynamics.

That gases behave ideally, independent of identity, gives rise to partial pressures.

The sum of the mole fractions is 1.

Stoichiometry problems involving gases are inherently the same as other stoichiometry problems. Gases simply present another means of obtaining moles of substances.

71

Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

5.8 Kinetic Molecular Theory: A Model for Gases  Kinetic molecular theory o Postulates/assumptions  Particles negligibly small  Average kinetic energy proportional to T  Elastic collisions  Relevance to gas laws o Pressure as a force per unit area o Boyle’s law o Charles’s law o Avogadro’s law o Dalton’s law o Ideal gas law  Temperature and molecular speeds

 

o o 



Figure 5.15 A Model for Gas Behaviour Figure 5.16 Elastic versus Inelastic Collisions Figure 5.17 Pressure on the Wall of a Container Figure 5.18 Velocity Distribution for Several Gases at 25 oC Figure 5.19 Velocity Distribution for Nitrogen at Several Temperatures Example 5.14 Root Mean Square Velocity

  

Figure 5.20 Mean Free Path Figure 5.21 Effusion Example 5.15 Graham’s Law of Effusion

  

1 2 3 mv  RT 2 2 3RT urms  M KE 

Distribution of molecular speeds o Dependence on T o Dependence on M

5.9 Mean Free Path, Diffusion, and Effusion of Gases  Mean free path  Diffusion  Effusion o

Graham’s law:

rateA  rateB

MB MA

.

72

Chapter 5. Gases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

5.8 Kinetic Molecular Theory: A Model for Gases  The physics in the kinetic molecular theory is remarkably simple given all of the behaviours it explains. The ideal gas law can be derived from the definition of pressure as force per unit area.  Two of the three assumptions in the kinetic molecular theory give rise to nonideal behaviour.  Conceptual Connection 5.3 Kinetic Molecular Theory



5.9 Mean Free Path, Diffusion, and Effusion of Gases  Diffusion is easy for students to understand, as they have experience with odours (especially bad ones) spreading throughout a room.  An analogy for diffusion is thermal equilibration, as they both are illustrations of the second law of thermodynamics.







.

One of the main uses of going through a derivation is pointing out that knowing where an equation comes from and the assumptions required to derive it gives clues concerning its range of applicability.

The analogy of billiard balls to atoms in an elastic collision is partially flawed since the balls lose some energy to friction with the table.

Students are likely to have difficulty with distributions of velocities within a collection of molecules.

Most students are surprised by the high speeds of atoms and molecules at room temperature.

73

Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

5.10 Real Gases: The Effects of Size and Intermolecular Forces  Real gases o High pressures: finite particle volume o Low temperatures: intermolecular forces  Van der Waals equation

 

o

  

2  n   P  a    V  nb  nRT  V   

 

.

Figure 5.22 Molar Volumes of Real Gases Figure 5.23 Particle Volume and Ideal Behaviour Figure 5.24 The Effect of Particle Volume Table 5.5 Van der Waals Constants for Common Gases Figure 5.25 The Effect of Intermolecular Forces Figure 5.26 Real versus Ideal Behaviour Chemistry in Your Day: Pressure in Outer Space

74

Chapter 5. Gases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

5.10 Real Gases: The Effects of Size and Intermolecular Forces  Two of the three assumptions from the kinetic molecular theory break down and give rise to nonideal behaviour.  Each of the effects can be demonstrated independently.  Conceptual Connection 5.4 Real Gases



.

Why do we use the ideal gas law if we know it is “wrong”? For some gases under some conditions, the corrections are unnecessary.

75

Chapter 5. Gases

Additional Problem for Converting between Pressure Units (Example 5.1)

A normal systolic blood pressure value is 110 mmHg. What is the pressure in pascal?

Sort

Given 110 mmHg

The problem gives a pressure in mmHg and asks you to find the equivalent in pascal.

Find Pa

Strategize

Conceptual Plan

Since Table 5.1 does not have a direct conversion factor between mmHg and Pa, but does provide relationships between both of these units and atmospheres, you can convert to atm as an intermediate step.

mmHg



1 atm 760 mmHg

atm



Pa

101,325 Pa 1atm

Relationships Used 1 atm = 760 mmHg 101 325 Pa = 1 atm (both from Table 5.1)

Solve

Solution

Follow the conceptual plan to solve the problem. Begin with 110 mmHg and use the conversion factors to arrive at the pressure in Pa.

110 mmHg ×

1 atm 101,325 Pa × = 14,700 Pa 760 mmHg 1 atm

The units (Pa) are correct. The magnitude of the answer (14 700) makes physical sense since the blood pressure is a fraction of one atm.

Check

.

76

Chapter 5. Gases

Additional Problem for Ideal Gas Law (Example 5.5)

Calculate the volume in L occupied by 1.43 mol of oxygen gas at a pressure of 3.25 atm and a temperature of 298 K.

Sort

Given n = 1.43 mol

The problem gives you the number of moles of oxygen gas, the pressure, and the temperature. You are asked to find the volume in L.

P = 3.25 atm T = 298 K

Find V in L

Strategize

Conceptual Plan

You are given three of the four variables (n, P, T) in the ideal gas law and asked to find the fourth (V). The conceptual plan shows how the ideal gas law provides the relationship between the given quantities and the quantity to be found.

n, P, T PV



V

 nRT

Relationships Used PV = nRT (ideal gas law)

Solve

Solution

To solve the problem, first solve the ideal gas law for V.

PV  nRT nRT V  P

Then substitute the given quantities to compute V.

1.43 mol  V

=

0.08206 L atm  298 K mol K 3.25 atm

= 10.8 L

The units of the answer are correct and the magnitude makes sense.

Check

.

77

Chapter 5. Gases

Additional Problem for Density (Example 5.7)

Calculate the density of oxygen gas at 37.4 oC and a pressure of 720 mmHg.

Sort

Given T = 37.4 oC

The problem gives you the temperature and pressure of a gas and asks you to find its density. The problem also states that the gas is oxygen.

P = 720 mmHg

Find d

Strategize

Conceptual Plan

Equation 5.6 provides the relationship between the density of a gas and its temperature, pressure, and molar mass. The temperature and pressure are given, and you can compute the molar mass from the formula of the gas, which we know is O2.

P, T,   d

d

P RT

Relationships Used d

P RT

molar mass O2 = 32.00

Solve

Solution

To solve the problem, you must gather each of the required quantities in the correct units. Convert the temperature to kelvins and the pressure to atmospheres.

T(K) = 37.4 oC + 273 = 310.4 K P  720 mmHg 

d

Now substitute the quantities into the equation to compute density.

1 atm = 0.947 atm 760 mmHg

P RT

 32.00 g  0.947 atm    1 mol  = 0.08206 L atm  310.4 K mol K

= 1.19 g/L

The units of the answer (g L-1) are correct. The magnitude of the answer (1.19) makes physical sense compared to the density of gases calculated earlier.

Check

.

78

Chapter 5. Gases Additional Problem for Molar Mass of a Gas (Example 5.8)

A sample of gas has a mass of 0.205 g. Its volume is 0.112 L at a temperature of 25 oC and a pressure of 740 mmHg. Find its molar mass.

Sort

Given

The problem gives you the mass of a gas sample, along with its volume, temperature, and pressure. You are asked to find the molar mass.

m = 0.205 g

V = 0.112 L T(oC) = 25 oC P = 740 mmHg

Find molar mass (g mol-1)

Strategize

Conceptual Plan

The conceptual plan has two parts: 1) Use the ideal gas law to find the number of moles of gas.

P, V, T



PV

 nRT

n, m 2) Use the definition of molar mass to find the molar mass.



molar mass =

n

molar mass mass (m ) moles (n )

Relationships Used PV = nRT molar mass =

mass (m ) moles (n )

Solve

Solution

Convert the temperature to K and pressure to atm. Then find the number of moles, solving the ideal gas law for n.

T(K) = 25 oC + 273 = 298 K P

 740 mmHg 

1 atm = 0.974 atm 760 mmHg

Use the number of moles (n) and the given mass (m) to find the molar mass.

.

79

Chapter 5. Gases PV  nRT n



n

=

PV RT 0.974 atm  0.112 L = 4.46  103 mol 0.08206 L atm  298 K mol K

mass (m ) moles (n ) 0.205 g = = 46.0 g/mol 4.46  10 3 mol

molar mass =

The units (g mol-1) are correct. The magnitude of the answer (46.0) seems to make sense for a gas. For example, it could be NO2.

Check

.

80

Chapter 5. Gases Additional Problem for Total Pressure and Partial Pressures (Example 5.9)

A 1.50 L mixture of helium, neon, and argon has a total pressure of 754 mmHg at 310 K. If the partial pressure of helium is 431 mmHg and partial pressure of neon is 211 mmHg, what mass of argon is present in the mixture?

Sort

Given PHe = 431 mmHg

The problem gives you the partial pressures of two of the three components in a gas mixture, along with the total pressure, the volume, and the temperature, and asks you to find the mass of the third component.

PNe = 211 mmHg Ptotal = 754 mmHg V

= 1.50 L

T

= 310 K

Find mAr

Strategize

Conceptual Plan

You can find the mass of argon from the number of moles of argon, which you can calculate from the partial pressure of argon and the ideal gas law.

Ptotal, PHe, PNe



PAr

Ptotal  PHe  PNe  PAr

PAr, V, T  Begin by finding the partial pressure of argon from Dalton’s law of partial pressures. nAr

nAr

PV

 nRT



mAr

1 mol Ar 39.95 g Ar

Then use the partial pressure of argon together with the volume of the sample and the temperature to find the number of moles of argon.

Relationships Used Ptotal = PHe + PNe + PAr (Dalton’s law)

Finally, use the molar mass of argon to compute the mass of argon from the number of moles of argon.

PV = nRT (ideal gas law) molar mass Ar = 39.95 g mol-1

Solve

Solution

Follow the conceptual plan. To find the partial pressure of argon, solve the equation for PAr, and substitute the values of the other partial pressures.

Ptotal = PHe + PNe + PAr PAr = Ptotal - PHe - PNe = 754 mmHg  431 mmHg  211 mmHg = 112 mmHg

.

81

Chapter 5. Gases

Convert the partial pressure from mmHg to atm and use it in the ideal gas law to compute the amount of argon in moles.

112 mmHg 

n

Use the molar mass of argon to convert from amount of argon in moles to mass of argon.

1 atm = 0.147 atm 760 mmHg

PV 0.147 atm 1.50 L  RT 0.08206 L atm  310 K mol K

8.67  10 3 mol Ar 

= 8.67  10 3 mol

39.94 g Ar = 0.346 g Ar mol Ar

The units of the answer are correct and the magnitude makes sense.

Check

.

82

Chapter 5. Gases Additional Problem for Partial Pressures and Mole Fractions (Example 5.10)

A 10 L scuba diving tank is filled with a mixture of helium and oxygen containing 20.4 g He and 3.52 g O2 at 298 K. Calculate the mole fraction and partial pressure of each component in the mixture and calculate the total pressure.

Sort

Given mHe = 20.4 g

The problem gives the masses of two gases in a mixture and the volume and temperature of the mixture. You are asked to find the mole fraction and partial pressure of each component, as well as the total pressure.

mO2 = 3.52 g

V = 10 L T = 298 K

Find He,  O , PHe, PO , Ptotal 2

2

Strategize

Conceptual Plan

The conceptual plan has several parts:

mHe

1) To calculate the mole fraction of each component, first find the number of moles of each component. To accomplish this, convert masses to moles.

2) Compute the mole fraction of each component using the mole fraction definition. Calculate the total pressure from the sum of the moles of both components. Alternatively, you could calculate the partial pressure of each component and sum them.

3) Use the mole fractions of each component and the total pressure to calculate the partial pressure of each component.



nHe

1 mol He 4.00 g He



=

He

Ptotal





mO2

nO2

1 mol O 2 32.00 g O 2

n He ; n +n He O 2



n O

O

= 2

2 n +n He O

2

(nHe + nO2 )  RT V

PHe = He Ptotal ; PO2 =  O2 Ptotal

Relationships Used

a = na / ntotal (mole fraction definition) PtotalV = ntotalRT (ideal gas law) Pa = a Ptotal

Solve

Solution

Follow the conceptual plan to solve the problem.

Begin by converting each of the masses to amounts in moles.

20.4 g He 

.

1 mol He = 5.10 mol He 4.00 g He

83

Chapter 5. Gases

3.52 g O2 

He =

Compute each of the mole fractions. O = 2

Ptotal 

Compute the total pressure.

1 mol O2 32.00 g O2

= 0.110 mol O2

nHe 5.10 = = 0.979 nHe + nO2 5.10 + 0.110 nO2 nHe + nO2

=

0.110 = 0.0211 0.110 + 5.10

(nHe + nO2 )  RT V 0.08206 L atm  298 K mol K 10.0 L

(5.10 + 0.110) mol  =

Compute the partial pressure of each component.

Check

= 12.7 atm

PHe = He Ptotal = 0.979  12.7 atm = 12.5 atm PO2 =  O2 Ptotal = 0.021 12.7 atm = 0.267 atm

The units of the answer are correct and the magnitudes are reasonable.

.

84

Chapter 5. Gases Additional Problem for Gases in Chemical Reactions (Example 5.12)

Methanol (CH3OH) can be synthesized by the reaction: CO(g) + 2 H2(g)  CH3OH(g) What volume (in litres) of methanol gas, measured at a temperature of 473 K and a pressure of 820 mmHg, is produced from 100.0 g of carbon monoxide CO?

Sort You are given the mass of carbon monoxide, a reactant of a chemical reaction. You are asked to find the volume of the product produced (methanol) at a specified temperature and pressure.

Given 100.0 g CO T = 473 K P = 820 mmHg

Find Vmethanol

Strategize

Conceptual Plan

The volume of methanol produced can be calculated from the number of moles of methanol, which can be obtained from the number of moles of carbon monoxide and the stoichiometry of the reaction.

g CO



mol CO

1 mol CO 28.01 g CO

1 mol CH3OH 1 mol CO

n (mol CH3OH), P, T PV

 mol CH3OH



Vmethanol

 nRT

Find the number of moles of carbon monoxide from its mass by using the molar mass.

Relationships Used Use the stoichiometric relationship from the balanced equation to find the number of moles of methanol formed from that quantity of carbon monoxide.

PV = nRT (ideal gas law) 1 mol CO : 1 mol CH3OH (from balanced equation) molar mass CO = 28.01 g

Substitute the moles of methanol, together with the pressure and temperature into the ideal gas law to find the volume of hydrogen.

Solve

Solution

Follow the conceptual plan to solve the problem.

Begin by using the mass of carbon monoxide to get the number of moles of CO.

100.0 g CO 

Convert the number of moles of carbon monoxide to moles of methanol.

1 mol CO = 3.570 mol CO 28.01 g CO

3.570 mol CO = 3.570 mol CH3OH

.

85

Chapter 5. Gases

Convert the pressure to atmospheres and use the ideal gas law to find the volume of methanol.

P = 820 mmHg 

VCH3OH =

nCH3OHRT P

3.570 mol  =

Check

1 atm = 1.08 atm 760 mmHg

0.08206 L atm  473 K mol K 1.08 atm

= 128 L

The units of the answer are correct. The magnitude makes sense. You start with about 3.5 mol of reactant. If the reaction was conducted at STP, you would expect about 80 L. Since the temperature is much higher than 298 K and the pressure is somewhat higher, the result should be greater.

.

86

Chapter 5. Gases

Additional Problem for Molar Volume and Stoichiometry (Example 5.13)

How many grams of water form when 2.41 L of oxygen gas at STP completely react with an appropriate amount of H2?

2 H2(g) + O2(g)  2 H2O(g)

Sort

Given 2.41 L O2

You are given the volume of oxygen gas (a reactant) at STP and asked to determine the mass of water that forms upon complete reaction.

Find g H2O

Strategize

Conceptual Plan

Since the reaction occurs under standard conditions, you can convert directly from the volume (in L) of hydrogen gas to the amount in moles. Then use the stoichiometric relationship from the balanced equation to find the number of moles of water formed. Finally, use the molar mass of water to obtain the mass of water formed.

L O2  mol O2  1 mol O2 22.4 L O2

mol H2O  g H2O

2 mol H2O 1 mol O2

18.02 g H2O 1 mol H2O

Relationships Used 1 mol = 22.4 L (at STP) 1 mol O2 : 2 mol H2O (from balanced equation) molar mass H2O = 18.02 g mol-1

Solve Follow the conceptual plan to solve the problem.

Check

Solution

2.41 L O2 

1 mol O2 22.4 L O2



2 mol H2O 1 mol O2



18.02 g H2O = 3.88 g 1 mol H2O

The units of the answer are correct. The magnitude makes sense since you start with about one-tenth of a molar volume (22.4  10 = 2.2) and make twice as many moles of water as oxygen. The mass of water is about one-fifth of a molar mass (18  5 = 3.6).

.

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Chapter 5. Gases

Additional Problem for Root Mean Square Velocity (Example 5.14)

Calculate the root mean square velocity of oxygen molecules at 100 oC.

Sort

Given O2 T = 100 oC

You are given the kind of molecule and the temperature and are asked to find the root mean square velocity.

Find urms Strategize

Conceptual Plan

The conceptual plan for this problem simply shows how the molar mass of oxygen and the temperature (in kelvins) can be used with the equation that defines the root mean square velocity.

, T



urms 3 RT 

urms =

Relationships Used 1 J = 1 kg m-2 s-2 3 RT 

urms =

Solve

Solution

The quantities must be in the correct units—temperature in K and molar mass in kg mol-1.

T = 100 oC + 273 = 373 K



=

32.00 g O2 1 mol O2



1 kg = 3.200  102 kg/mol O2 1000 g

Substitute the quantities into the equation. urms =

3 RT  8.314 J  373 K mol K 3.200  10 2 kg/ mol O2 3

=

kg m2 =

2.91 105

J = kg

2.91 105

s2 kg

= 538 m/s

The units of the answer (m s-1) are correct, and the magnitude seems reasonable compared with the velocity previously calculated for 25 oC.

Check

.

88

Chapter 5. Gases

Additional Problem for Graham’s Law of Effusion (Example 5.15)

An unknown gas effuses at a rate that is 0.578 times that of nitrogen gas at the same temperature. Calculate the molar mass of the unknown gas in g mol-1.

Sort

Given

You are given the ratio of effusion rates for the unknown gas and nitrogen and asked to find the molar mass of the unknown gas.

Rateunk = 0.578 RateN2

Find unk

Strategize

Conceptual Plan

The conceptual plan uses Graham’s law of effusion. You are given the ratio of rates and you know the molar mass of nitrogen. You can use Graham’s law to find the molar mass of the unknown gas.

rate ratio, nitrogen



Rateunk = RateN2

 N2

unk

 unk

Relationships Used Rateunk = RateN2

Solve

 N2 (Graham’s law)  unk

Solution

Solve the equation for unk and substitute the correct values to compute it.

Rateunk = RateN2  unk =

=

 N2  unk  N2 2

 Rateunk     RateN  2   28.01 g/mol

 0.578 

2

= 83.8 g/mol

The units of the answer (g mol-1) are correct, and the magnitude makes sense. The unknown gas appears to be krypton.

Check

.

89

Chapter 6. Thermochemistry Student Objectives 6.1 Chemical Hand Warmers  

Define thermochemistry.

Understand the idea of heat exchange as a flow of energy.

6.2 The Nature of Energy: Key Definitions     

Define energy, work, heat, kinetic energy, thermal energy, potential energy, and chemical energy.

Understand the difference between kinetic and potential energy and know why thermal energy and chemical energy are examples of each, respectively.

Know the law of conservation of energy.

Understand the difference between the system and the surroundings.

Know and convert between the common units of energy: joule, calorie, kilocalorie, kilowatt‐hour.

6.3 The First Law of Thermodynamics: There Is No Free Lunch       

Define and understand thermodynamics and the first law of thermodynamics.

Understand why a perpetual motion machine violates the first law of thermodynamics.

Define and understand internal energy and state function.

Understand the flow of energy, as work or heat, from the standpoint of the system and the surroundings.

Understand that energy lost by the surroundings is equal to the energy gained by the system and vice versa.

Understand the mathematical definition of the first law of thermodynamics in terms of the change in internal energy, heat, and work.

Understand the sign conventions for heat, work, and the change in internal energy.

6.4 Quantifying Heat and Work      

Define and understand thermal equilibrium.

Define and understand heat capacity, specific heat capacity, and molar heat capacity.

Understand the equation that relates heat flow to the amount of substance, the specific heat capacity, and the temperature change.

Calculate using the equation relating heat flow and temperature change in terms of both the system and the surroundings.

Define and understand pressure–volume work.

Understand and use the equation for work in terms of the pressure and volume change.

6.5 Measuring ΔrU for Chemical Reactions: Constant‐Volume Calorimetry  

Understand how constant‐volume calorimetry is used to measure the exchange of energy between the system and surroundings in terms of the heat flow.

Calculate the heat released from reactions in a bomb calorimeter.

.

90

Chapter 6. Thermochemistry 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure      

Define enthalpy in terms of internal energy, pressure, and volume.

Understand that the change in enthalpy for a reaction or process is equal to the heat flow under constant pressure.

Define and understand endothermic and exothermic reactions.

Understand the molecular view of endothermic and exothermic reactions.

Define enthalpy of reaction.

Understand and calculate enthalpy changes with respect to the stoichiometry of chemical equations.

6.7 Constant‐Pressure Calorimetry: Meassuring rH   

Describe and understand the coffee‐cup calorimeter and how it can measure the enthalpy change for a chemical reaction or physical process.

Understand and calculate enthalpy changes for reactions in a coffee‐cup calorimeter.

Understand the difference between constant‐volume and constant‐pressure calorimetry.

6.8 Relationships Involving rH  

Know, understand, and calculate using reaction enthalpies: multiplying a chemical equation by a factor, reversing a chemical equation, and summing a series of chemical equations.

Understand the use of Hess’s law to calculate the enthalpy change for a reaction from a series of steps.

6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation    

Define and understand standard state and standard enthalpy of formation.

Write thermochemical equations for the formation of compounds.

Understand the equation for calculating the enthalpy of reaction from enthalpies of formation as an illustration of Hess’s law.

Calculate the enthalpy of reaction using enthalpies of formation of products and reactants.

6.10 Energy Use and the Environment    

Know about energy consumption: sources and uses.

Know about environmental problems associated with fossil fuel combustion.

Understand and perform calculations involving the combustion of fossil fuels and the formation of carbon dioxide.

Know about potential alternative fuels, especially renewable energy sources.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

91

Chapter 6. Thermochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

6.1 Chemical Hand Warmers  Thermochemistry  Heat flow



6.2 The Nature of Energy: Key Definitions  Energy o Work o Heat o Kinetic energy  Thermal energy o Potential energy  Chemical energy  Law of conservation of energy o System o Surroundings  Energy units o Joule (J) o Calorie (cal) o Calorie or kilocal (Cal or kcal) o Kilowatt‐hour (kWh)



Figure 6.1 Transfer of energy, curling stone used as example Figure 6.2 Energy Transformation: Potential and Kinetic Energy I Figure 6.3 Energy Transformation: Potential and Kinetic Energy II Figure 6.4 Energy Transfer Unnumbered figure: 100‐W light bulb Table 6.1 Energy Conversion Factors Table 6.2 Energy Uses in Various Units

     

6.3 The First Law of Thermodynamics: There Is No Free Lunch  First Law of Thermodynamics o Definition o Perpetual motion machine  State function  Internal energy o System o Surroundings o Relationship to heat and work  U = q + w

     

.

Intro figure: chemical hand warmers

Chemistry in Your Day: Perpetual Motion Machines Figure 6.5 Altitude as a State Function Unnumbered figures: energy diagram of CO2 formation and decomposition Table 6.3 Sign Conventions for q, w, and U Figure 6.6 Energy, Work, and Heat; curling used as example Example 6.1 Internal Energy, Heat, and Work

92

Chapter 6. Thermochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

6.1 Chemical Hand Warmers  Chemical hand warmers are easy to get and provide a simple but useful demonstration.

6.2 The Nature of Energy: Key Definitions  Many students will have experience with a billiard table. Some examples can be reproduced on a countertop in a classroom.  Drop a tennis ball from waist‐high and ask the students to observe and explain all of its behaviour, including coming to rest on the floor, in terms of kinetic and potential energy.



Energy is always conserved, but one might not be able to identify easily the form(s) into which it is transformed.

6.3 The First Law of Thermodynamics: There Is No Free Lunch  Paths to the top of a hill or mountain are common examples of path functions, whereas the change in elevation is a state function.  Conceptual Connection 6.1 System and Surroundings  Conceptual Connection 6.2 Heat and Work



Billiard balls don’t keep moving forever since friction creates the heat loss that accounts for some of the energy.

.

93

Chapter 6. Thermochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

6.4 Quantifying Heat and Work  Thermal equilibrium  Heat flow and temperature change o Heat capacity  Specific heat capacity (Cs, in J(goC)‐1)  Molar heat capacity (J(moloC)‐1) o q=mCsT  Pressure–volume work o w = PV

         

6.5 Measuring ΔrU for Chemical Reactions: Constant‐ Volume Calorimetry  Internal energy change o w = 0 at constant V o q = U at constant V  Calorimetry o Bomb calorimeter

Table 6.4 Specific Heat Capacities of Some Common Substances Unnumbered figure: photo of Victoria, BC Example 6.2 Temperature Changes and Heat Capacity Unnumbered figure: illustration of metal block in water Example 6.3 Thermal Energy Transfer Unnumbered figure: illustration of internal combustion engine Figure 6.7 Piston Moving Within a Cylinder Against an External Pressure Example 6.4 Pressure–Volume Work Figure 6.8 The Bomb Calorimeter Example 6.5 Measuring rU in a Bomb Calorimeter

6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure  Enthalpy o Definition: H = U + PV o H = q at constant pressure  Reactions or processes o Endothermic: H > 0 o Exothermic: H < 0  Thermochemical equations o H and stoichiometry



6.7 Constant‐Pressure Calorimetry: Measuring rH  Calorimetry o Coffee‐cup calorimeter  q = H at constant pressure

 

.

 

Unnumbered figures: illustrations of endothermic and exothermic processes Example 6.6 Exothermic and Endothermic Processes Example 6.7 Stoichiometry Involving rH

Figure 6.9 The Coffee‐Cup Calorimeter Example 6.8 Measuring rH in a Coffee‐ Cup Calorimeter

94

Chapter 6. Thermochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

6.4 Quantifying Heat and Work  A good conceptual check concerning heat capacity and temperature is to ask a student whether a chalk tray and eraser at the front of the room are the same temperature.  Specific heat capacity is a measure of how much energy is required to change the temperature of a substance.  Conceptual Connection 6.3 The Heat Capacity of Water  Conceptual Connection 6.4 Thermal Energy Transfer



  

Temperature and heat are not the same entity. Temperature is a measure of average kinetic energy, whereas heat is a flow of energy.

Heat is a flow of energy. Substances do not contain heat; they lose or absorb heat.

“Hot” and “cold” are only relative terms.

Students sometimes forget that  in this context always means final  initial.

6.5 Measuring ΔrU for Chemical Reactions: Constant‐ Volume Calorimetry  Heat capacities and heats of reaction are measured quantities; the values in tables come from experiments using calorimeters.  Microcalorimeters are used for small quantities, especially for very valuable samples or biological samples. Ask the students to estimate the limits of this technique.



The bomb calorimeter must be calibrated. Even though the heat capacity for water is well known, the best way to get the values for all the other parts is to measure it. This can be done by electrical heating.

6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure  The direction of heat flow and conventions for numerical signs of exothermic and endothermic processes should be emphasized.  Conceptual Connection 6.5 The Difference between rH and rU  Conceptual Connection 6.6 Exothermic and Endothermic Reactions



Practice with thermochemical equations will convince students that H can be used as a conversion factor just like the stoichiometric coeffecients from the balanced reaction.

6.7 Constant‐Pressure Calorimetry: Measuring rH  The coffee‐cup calorimeter can be used for reactions and physical processes, e.g., dissolving a solid and phase changes.  Conceptual Connection 6.7 Constant‐Pressure versus Constant‐Volume Calorimetry



Students may initially miss the fact that an exothermic reaction done in a coffee‐ cup calorimeter will result in an increased temperature for the aqueous solution even though the reaction itself is decreasing in enthalpy.

.

95

Chapter 6. Thermochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

6.8 Relationships Involving rH  Chemical equations and rH o Multiplying equations by a factor o Reversing equations o Summing a series of equations  Hess’s law o Illustration of state function

 

6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation  Definitions o Standard state o Standard enthalpy change, Ho o Standard enthalpy of formation, fHo  Calculating the standard enthalpy change for a reaction  r H o   np  f H o  products   nr  f H o  reactants

Figure 6.10 Hess’s Law Example 6.9 Hess’s Law

     

6.10 Energy Use and the Environment  Energy consumption o Energy use o Energy sources  Environmental problems related to energy use o Carbon dioxide production from fossil fuel combustion

    

.

Table 6.5 Standard Enthalpies (or Heats) of Formation (∆fH°) at 298 K Example 6.10 Standard Enthalpies of Formation Figure 6.11 Calculating the Enthalpy Change for the Combustion of Methane Example 6.11 ∆rH° and Standard Enthalpies of Formation Unnumbered figure: photo of thermite reaction Example 6.12 ∆rH° and Standard Enthalpies of Formation

Unnumbered figure: pie chart of energy consumption Unnumbered figure: pie chart of world oil reserves Figure 6.12 Energy Consumption by Source Figure 6.13 Consumption of Renewable Energy Example 6.13 Portable Energy

96

Chapter 6. Thermochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

6.8 Relationships Involving rH  Quantitative manipulations of thermochemical equations require changes in rH: multiplying by a factor, reversing sign, and summing enthalpies.  Hess’s law is a powerful and flexible way to generate new thermodynamic data from chemical equations.



In order to apply Hess’s law properly, all the other species except the desired ones must cancel.

6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation  Absolute enthalpy cannot be measured, so the enthalpies of formation of the elements (in their standard states) are set to 0.  Each enthalpy of formation value must be multiplied by the stoichiometric coefficient from the balanced reaction.



Students sometimes fail to see that the equation for the standard enthalpy change is just an application of Hess’s law.

6.10 Energy Use and the Environment  Review the uses and sources of energy in the world, and examine trends in consumption.  Performing simple calculations concerning CO2 emission makes the emission problem far more obvious.  Discuss alternative fuels and sources of energy, both the positives and negatives.



Renewable fuels like ethanol still produce carbon dioxide.

.

97

Chapter 6. Thermochemistry Additional Problem for Pressure–Volume Work (Example 6.4)

Inflating one’s lungs requires the inflator to do pressure–volume work on the surroundings. If the lungs are inflated from a volume of 2.5 L to 5.5 L against an external pressure of 1.10 bar, how much work is done?

Sort

Given V1 = 2.5 L

You are given the initial and final volumes of the lungs and the pressure against which they expand.

The lungs and their contents are the system.

V2 = 5.5 L P = 1.10 bar

Find w

Strategize

Conceptual Plan

The equation w = PV specifies the amount of work done during a volume change against an external pressure.

P, V 

A conversion factor for converting to joules is necessary.

Relationships Used

w

w =  P V

V = V2 – V1 1bar L = 100J

Solve

Solution

Enter the values into the equation and calculate the value for work.

V = 5.5 – 2.5 L = 3.0 L

w   PV  1.10 bar  3.0 L 

100 J 1bar L

  330 J

The units of the answer (J) are correct. The sign is negative as it should be for expansion; work is being done on the surroundings by expanding the lungs.

Check

.

98

Chapter 6. Thermochemistry Additional Problem for Measuring rU in a Bomb Calorimeter (Example 6.5)

When 1.10 g of ethanol (C2H6O) undergoes combustion in a bomb calorimeter, the temperature rises from 22.32 oC to 29.48 oC. Find rU for the combustion of ethanol in kJ mol-1. The heat capacity of the bomb calorimeter, determined in a separate experiment is 4.70 kJ(oC)-1.

Given 1.10 g ethanol Ti = 22.32oC Tf = 29.48oC Ccal = 4.70 kJ(oC)-1

Find rU

Strategize

Conceptual Plan

The conceptual plan has three parts:

Ccal, T

1) Use the temperature change and heat capacity of the calorimeter to find qcal.





qcal

qrxn

qrxn  qcal

qcal  Ccal  T

2) Use qcal to find qrxn, a matter of changing signs.

Relationships Used

3) Divide qrxn by the number of moles of ethanol to get rU per mole of ethanol.

Erxn =

qrxn mol C3H6O

qcal = Ccal  T = qrxn T = Tf – Ti molar mass (C2H6O) = 46.07 g mol-1

Solve

Solution

Enter the values into the equations for T, qcal and rU.

T = 29.48oC – 22.32oC = 7.16oC qcal  Ccal  T   qrxn 

.

4.70 kJ o

C

 7.16 o C = 33.64 kJ

99

Chapter 6. Thermochemistry rE

= =

qrxn mol C3H6O 33.64 kJ mol C2H6O 1.10 g C2H6O  46.07 g C2H6O

= 1409 kJmol1 = 1410 kJmol1

The units of the answer (kJ mol-1) are correct. The sign is negative and is correct for a reaction that gives off energy.

Check

.

100

Chapter 6. Thermochemistry Additional Problem for Stoichiometry involving H (Example 6.7)

A cigarette lighter contains 2.90 g of butane, C4H10.

Calculate the heat (in kJ) associated with the complete combustion of the butane in the lighter.

C4H10(g) + 13/2 O2(g)  4 CO2(g) + 5 H2O(g) rH= 2658 kJ

Sort You are given the mass of propane and asked to find the heat evolved in its combustion.

Given 2.90 g C4H10

Find q

Strategize

Conceptual Plan

Starting with mass of butane, convert to moles using the molar mass as a conversion factor.

Next, use the stoichiometric relationship between C4H10 and kJ to find the heat evolved.

g C4H10

mol C4H10



1 mol C4H10 58.12 g C4H10



kJ

2658 kJ 1 mol C4H10

Relationships Used molar mass C4H10 = 58.12 g mol-1 1 mol C4H10: 2658 kJ (from equation)

Solve

Solution

Follow the conceptual plan to solve the problem.

Begin with the mass and multiply with the appropriate conversion factors to arrive at kJ.

2.90 g C 4H10 

Check

1 mol C 4H10 58.12 g C 4H10



2658 kJ 1 mol C 4H10

=  133 kJ

The units (kJ) are correct. The sign is negative, as it should be for heat evolved by the reaction.

.

101

Chapter 6. Thermochemistry Additional Problem for Measuring rH in a Coffee-Cup Calorimeter (Example 6.8)

Calcium metal reacts with hydrochloric acid according to the balanced equation: Ca(s) + 2 HCl(aq)  CaCl2(aq) + H2(g) A 0.150 g sample of Ca metal is combined with enough HCl to make 100.0 mL of solution in a coffee-cup calorimeter.

The calcium reacts and the temperature of the solution rises from 23.5 oC to 25.4 oC. Find rH for the reaction. Use 1.00 g mL-1 as the density of the solution and Cs,soln = 4.18 J(g oC)-1 as the specific heat capacity of the solution.

Sort You are given the mass of calcium, the volume of the solution, the initial and final temperatures, the density of the solution, and the heat capacity of the solution. You are asked to find the enthalpy change for the reaction.

Given 0.158 g Ca 100.0 mL solution Ti = 23.5 oC Tf = 25.4 oC d = 1.00 g mL-1 Cs, soln = 4.18 J(goC)-1

Find rH

Strategize

Conceptual Plan

The conceptual plan has three parts:

Cs, soln, msoln, T

1) Use the temperature changes, specific heat capacity, and mass of the solution to solve for qsoln.

2) Use qsoln to find qrxn, which is just a change of sign. At constant pressure qrxn is equivalent to rH. 3) Divide qrxn by the number of moles of calcium to get rH per mole of calcium.



q  m  Cs  T

ΔHrxn 

qsoln



qrxn

qrxn  qsoln

qrxn mol Ca

Relationships Used q = m  Cs  T qrxn = qsoln

Solve Enter the values and determine the mass of the solution and the temperature change. Solve for qsoln using the equation.

Solution msoln = 100 mL soln 

1.00 g = 100 g 1 mL soln

T = Tf – Ti = 25.4 oC – 23.5 oC = 1.9 oC .

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Chapter 6. Thermochemistry

Convert qsoln to qrxn with the sign change.

Solve for the enthalpy change by entering the values into the equation.

qsoln = msoln  Cs, soln  T = 100 g 

4.18 J  1.9 o C = 779.9 J g oC

qrxn = qsoln = 779.9 J

r H 

qrxn mol Ca 779.9 J 1 mol Ca 0.150 g Ca  40.08 g Ca

=

=  2.084 105 J =  2.1105 J

Check

The units (J) are correct. The sign is negative as expected for an exothermic reaction.

.

103

Chapter 6. Thermochemistry

Additional Problem for rHo and Standard Enthalpies of Formation (Example 6.11)

Use the standard enthalpies of formation to determine rHo for the reaction

Sort

Given

You are given the balanced equation and asked to find the enthalpy of reaction.

Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)

Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)

Find rHo

Strategize

Conceptual Plan

To calculate rHo from standard enthalpies of formation, subtract the heats of formations of the reactants multiplied by their stoichiometric coefficients from the heats of formation of the products multiplied by their stoichiometric coefficients.

 r H o =  np  f H o (products)   nr  f H o (reactants)

Solve

Solution

Begin by looking up (in Appendix IIB) the standard enthalpy of formation for each reactant and product. Remember that the standard enthalpy of formation of pure elements in their standard state is zero. Compute rHo by substituting into the equation.

Reactant or product

fHo (kJ mol-1, Appendix IIB)

Zn(s)

0

HCl(aq)

167.2

ZnCl2(aq)

153.39

H2(g)

0

 r H o =  np  f H o (products)   nr  f H o (reactants) o = [1( f H oZnCl (aq) ) + 1( f HHo (g) )]  [1( f HZn(s) ) + 2( f H oHCl(aq) )] 2

2

= [1( 153.39 kJ) + 1(0 kJ)]  [1(0 kJ) + 2( 167.2 kJ)] = -153.39 kJ  (  234.4 kJ) = +81.0 kJ

Check

The units (kJ) are correct. The answer is positive, which means that the reaction is endothermic.

.

104

Chapter 6. Thermochemistry Additional Problem for Fossil Fuels and Climate Change (Example 6.13)

In Brazil, vehicles are fuelled almost entirely by ethanol C2H6O, produced from sugar cane. Calculate the heat (in kJ) produced by ethanol for each 1.00 kg of CO2 produced.

Sort

Given 1.00 kg CO2

You are given the mass of CO2 emitted and asked to find the energy output for ethanol.

Strategize

Find kJ C2H6O(l) + 7/2 O2(g)  2 CO2(g) + 3 H2O(g) rHo = 1409 kJ

Write the thermochemical equation for the combustion of ethanol.

The conceptual plan has two parts: 1) Use the molar mass of CO2 to convert from mass of CO2 to moles of CO2.

2) Use the stoichiometric relationship between moles of CO2 produced and kJ of energy released to calculate the energy output.

Conceptual Plan kg CO2  g CO2  mol CO2  1000 g 1 kg

1 mol CO2 44.01 g

kJ

1409 kJ 1 mol C2H6O

Relationships Used For C2H6O: 2 mol CO2: 1409 kJ 1 kg = 1000 g molar mass CO2 = 44.01 g mol-1

Solve Convert kg CO2 to mol CO2 and then to kJ.

Solution 1 kg CO2 

1000 g CO2 1 kg CO2

22.72 mol CO2 

Check



1 mol CO2 44.01 g CO2

1409 kJ 2 mol CO2

= 22.72 mol CO2

=  1.60  104 kJ

The units (kJ) are correct. The value is negative, as expected for exothermic combustion reactions. For comparison, other fuels have the following values: For C:

8.94 x 103 kJ

For CH4: 1.82 x 104 kJ For C8H18: 1.44 x 104 kJ Ethanol compares favourably.

.

105

Chapter 7. The Quantum‐Mechanical Model of the Atom Student Objectives 7.1 Quantum Mechanics: The Theory That Explains the Behaviour of the Absolutely Small  

Know that the behaviour of macroscopic objects like baseballs is strikingly different from the behaviour of microscopic objects like electrons.

Know that the quantum‐mechanical model provides the basis for the organization of the periodic table and our understanding of chemical bonding.

7.2 The Nature of Light       

Define and understand electromagnetic radiation.

Define and understand amplitude, wavelength, and frequency.

Use the speed of light to convert between wavelength and frequency.

Know the electromagnetic spectrum and its different forms of radiation.

Know and understand interference and diffraction and how they demonstrate the wave nature of light.

Know and explain the photoelectric effect and how it demonstrates the particle nature of light.

Use equations to interconvert energy, wavelength, and frequency of electromagnetic radiation.

7.3 Atomic Spectroscopy and the Bohr Model  

Define and understand atomic spectroscopy and emission spectrum.

Understand how the Bohr model explains the emission spectrum of hydrogen.

7.4 The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy      

Know that electrons and photons behave in similar ways: both can act as particles and as waves.

Know that photons and electrons, even when viewed as streams of particles, still display diffraction and interference patterns in a double‐slit experiment.

Use de Broglie’s relation to interconvert wavelength, mass, and velocity.

Know the complementarity of position and velocity through Heisenberg’s uncertainty principle.

Know the similarities and differences in classical and quantum‐mechanical concepts of trajectory.

Differentiate between deterministic and indeterminacy.

7.5 Quantum Mechanics and the Atom       

Define orbital and wave function.

Know that the Schrödinger equation is the ultimate source of energies and orbitals for electrons in atoms.

Know the properties and allowed values of the principal quantum number, n.

Know the properties, allowed values, and letter designations of the angular momentum quantum number, l.

Know the properties and allowed values of the magnetic quantum number, ml.

Know and understand how atomic spectroscopy defines the energy levels of electrons in the hydrogen atom.

Calculate the energies and wavelengths of emitted and absorbed photons for hydrogen.

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Chapter 7. The Quantum‐Mechanical Model of the Atom 7.6 The Shapes of Atomic Orbitals      

Define probability density and radial distribution function.

Define and understand node.

Identify the number of nodes in a radial distribution function for an s orbital.

Know the shapes of s, p, d, and f orbitals and the relationships to quantum numbers.

Know that the shape of an atom is dictated by the combined shapes of the collection of orbitals for that atom.

Define and understand phase.

7.7 Electron Configurations: How Electrons Occupy Orbitals Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 7. The Quantum‐Mechanical Model of the Atom Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

7.1 Quantum Mechanics: The Theory That Explains the Behaviour of the Absolutely Small  Difference between macroscopic world and microscopic world



Intro figure: photos and illustrations of the galaxies to electron‐density maps

7.2 The Nature of Light  Electromagnetic radiation o Waves o Amplitude o Wavelength o Frequency o Speed  c = /  Electromagnetic spectrum o Radio waves o Microwaves o Infrared o Visible o Ultraviolet o X‐ray o Gamma  Wave behaviour o Interference  Constructive  Destructive o Diffraction  Particle behaviour o Photoelectric effect o Photon or quantum of light  E = h = hc/

  

Figure 7.1 Electromagnetic Radiation Unnumbered figure: photo of lightning Unnumbered figure: wavelength and amplitude Figure 7.2 Wavelength and Amplitude Figure 7.3 Components of White Light Figure 7.4 The Colour of an Object Figure 7.5 The Electromagnetic Spectrum Example 7.1 Wavelength and Frequency Unnumbered figures: photos of medical X‐ray, sunburned torso Unnumbered figure: thermal image Chemistry and Medicine: Radiation Treatment for Cancer Unnumbered figures: constructive and destructive interference Unnumbered figure: photo of water wave interference Figure 7.6 Diffraction Figure 7.7 Interference from Two Slits Figure 7.8 The Photoelectric Effect Figure 7.9 Blackbody Radiation Curves Example 7.2 Photon Energy Example 7.3 Wavelength, Energy, and Frequency Example 7.4 The Photoelectric Effect

                

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Chapter 7. The Quantum‐Mechanical Model of the Atom Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

7.1 Quantum Mechanics: The Theory That Explains the Behaviour of the Absolutely Small

7.2 The Nature of Light  Water waves provide a practical example of some of the properties of waves, especially wavelength and amplitude.  Standing waves is a useful demonstration.  Sound waves are compared to the light from an exploding firework.  The use of colour (wavelength) and brightness (amplitude) in Figure 7.2 can be compared to water and air (sound) analogies.  The radiation treatment for cancer is a positive use of radiation. Its use is possible because delivery can be controlled and focused.  Diffraction patterns are very interesting; a demonstration with white and laser light can use masks and patterns that can be printed onto slides or overheads [Lisensky, George C.; Kelly, Thomas F.; Neu, Donald R.; Ellis, Arthur B. J. Chem. Educ. 1991, 68, 91.]  Albert Einstein won the Nobel Prize in Physics for the photoelectric effect and not for relativity.  Point out to students that we calculate the energy of a particle (photon) using a wave property (wavelength or frequency).  Conceptual Connection 7.1 The Photoelectric Effect







.

Wavelengths of electromagnetic radiation, like the signals for a radio or cell phones, have very little energy. Gamma rays and X‐rays are much more likely to damage cells and tissue.

Radiation may have a negative connotation, but radiation keeps the Earth warm enough to inhabit, and ionizing radiation from gamma rays and X‐rays can be useful.

The analogy with water waves can go too far, especially with respect to the significance of amplitude.

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Chapter 7. The Quantum‐Mechanical Model of the Atom Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

7.3 Atomic Spectroscopy and the Bohr Model  Atomic spectroscoy o Emission spectrum o Bohr model

          

7.4 The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy  Interference o Electron diffraction o Particle beam o de Broglie wavelength:  = h/mv  Complementary properties and uncertainty o Heisenberg’s uncertainty principle: x × m  v ≥ h/4  Determinacy, indeterminacy, and probability o Classical concept of trajectory o Quantum mechanical probability

     

.

Unnumbered figure: photo of neon sign Figure 7.10 Mercury, Helium, and Hydrogen Figure 7.11 Emission Spectra Figure 7.12 The Bohr Model and Emission Spectra.

Figure 7.13 Excitation and Radiation Example 7.5 Wavelength of Light for a Transition in the Hydrogen Atom Chemistry in Your Day: Atomic Spectroscopy, a Bar Code for Atoms Figure 7.14 Emission Spectra of Oxygen and Neon Figure 7.15 Flame Tests for Sodium, Potassium, Lithium, and Barium Figure 7.16 Emission and Absorption Spectrum of Mercury Figure 7.17 Hydrogen Energy Transition and Radiation

Figure 7.18 Electron Diffraction Unnumbered figure: illustration of double‐ slit experiment with electrons Unnumbered figure: photo of Werner Heisenberg Figure 7.19 Trajectory of a Macroscopic Object Figure 7.20 Trajectory versus Probability in Quantum Mechanics Figure 7.21 The Quantum‐Mechanical Five‐ Hole

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Chapter 7. The Quantum‐Mechanical Model of the Atom Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

7.3 Atomic Spectroscopy and the Bohr Model  Demonstrations of emission spectra are relatively simple. Lamps often show single colours (sodium vapour lamps). Mercury lamps (low pressure) or even mercury arc lamps used for lighting (medium and high pressure) have many lines and colours. Red and green laser pointers are effective.  Emission spectra are used in astronomy to detect elemental makeup of stars, planets, and other glowing bodies.  The colours in flame tests can be demonstrated easily.  Conceptual Connection 7.2 Emission Spectra



7.4 The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy  An electron exhibits wave properties, but each electron has the same mass and the same charge regardless of wavelength.  Heisenberg’s uncertainty principle in particular challenges the centuries‐old scientific tenet that two experiments arranged the same way should give identical results.  Schrödinger’s cat is a popular illustration of the absurdity of quantum mechanics at the macroscopic level.  Conceptual Connection 7.3 The de Broglie Wavelength of Macroscopic Objects (illustrates the insignificance of the wavelengths of macroscopic objects)





 

.

The Bohr model predicts emission spectra for H and for other single‐electron atoms, but it is not a correct model.

The most common misconception about electrons comes from the Bohr model: that electrons move in orbits as planets do around the sun.

Electron interference patterns occur even when the electrons go through the double slits singly and cannot interact with each other.

Students have a hard time visualizing what the wavelength of a particle means.

Students are misled by the probabilistic nature of quantum mechanics in much the same way that Einstein was. They presume that probabilities must be invoked because of an incompleteness in the theory.

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Chapter 7. The Quantum‐Mechanical Model of the Atom Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

7.5 Quantum Mechanics and the Atom  Orbital, a probability map o Schrödinger equation o Hydrogen energy levels







 

 1  En  2.18  10 18 J 2  n 

o Principal quantum number, n o Angular momentum quantum number, l o Magnetic quantum number, ml Atomic spectroscopy o Excitation and radiation o Hydrogen atom 

 

Unnumbered figure: diagram of principal energy levels Unnumbered figure: letter designations of l quantum number Unnumbered figure: table of n, l, and ml quantum numbers for n = 1–3 Example 7.6 Quantum Numbers I Example 7.7 Quantum Numbers II

 1 1 E  2.18  1018 J 2  2  n   f ni 

7.6 The Shapes of Atomic Orbitals  Electron configuration o Ground state orbitals and electron count  Electron spin o Spin quantum number o Pauli exclusion principle o Consequences of Pauli principle  Energy splitting for multielectron atoms o By principal energy level  Shielding o By sublevels  Penetration  Writing electron configurations o Minimize energy o Two electrons per orbital  Drawing orbital diagrams  Hund’s rule: electrons first occupy degenerate orbitals singly with paired spins

             

.

Figure 7.22 The 1s Orbital: Two Representations Figure 7.23 The 1s Orbital Surface Figure 7.24 The Radial Distribution Function for the 1s Orbital Unnumbered figure: illustration of nodes in vibrating strings Figure 7.25 Probability Densities and Radial Distribution Functions for the 2s and 3s Orbitals Figure 7.26 The 2p Orbitals and the Radial Distribution Functions for the 2p and 3p Orbitals Unnumbered figure: total radial probability Figure 7.27 The 3d Orbitals Figure 7.28 The 4f Orbitals Unnumbered figure: illustration of waves and phase Unnumbered figures: illustrations of phases of s and p orbitals Figure 7.29 Why Atoms Are Spherical Table 7.1 Solutions to the Shrödinger Equation for a One‐Electron Atom Unnumbered figure: graph of exponential function

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Chapter 7. The Quantum‐Mechanical Model of the Atom Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

7.5 Quantum Mechanics and the Atom  The Schrödinger equation is a mathematical model beyond the scope of the course.

Quantum numbers are results of the application of the Schrödinger equation. The names of the quantum numbers may be confusing, especially the angular momentum and magnetic quantum numbers, as most students will have no connection to the meanings of those terms.



The exact values of n, l, and ml are obtained only through tedious mathematics. The logic of the mathematics is the only guiding light in quantum theory since its basic principles are so counterintuitive in the macroscopic world.

7.6 The Shapes of Atomic Orbitals  The dot representation of probability in Figure 7.22 is much more realistic than the solid sphere in Figure 7.23.  A standing wave version of a vibrating string can be demonstrated with a Slinky toy. Two students can be used to demonstrate zero, one, or two nodes without much difficulty.  Conceptual Connection 7.4 The Shapes of Atoms  Conceptual Connection 7.5 Radial Probabilities



Orbitals define the probability of finding an electron. The shape and graphical representation may suggest a physical container.

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Chapter 7. The Quantum‐Mechanical Model of the Atom Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

7.7 Electron Configurations: How Electrons Occupy Orbitals  Orbital diagram  Pauli exclusion principle  Degenerate orbitals  Shielding and penetration  Electronic configuration for multielectron atoms  Electronic configuration for ions

              

.

Figure 7.30 The Stern–Gerlach Experiment Unnumbered figure: quantum numbers of helium electrons Figure 7.31 Shielding Figure 7.32 Radial Distribution Functions for the 1s, 2s, and 2p orbitals Figure 7.33 Penetration Figurem7.34 Radial Distribution Functions for 3s, 3p, and 3d orbitals Figure 7.35 Change in the energy ordering of orbitals from a one‐electron atom to a multielectron atom Unnumbered figures: orbital diagrams for Li and C Unnumbered figures: orbital diagrams for elements 3–10 Example 7.8 Electron Configuration Example 7.9 Writing Orbital Diagrams Table 7.2 First‐Row Transition Metal Orbital Occupancy Examples 7.10 and 7.11 Writing Electronic Configurations for Transition Metals Unnumbered figure: orbital diagram for Ag and Zn, zinc ion Example 7.12 Electron Configurations and Magnetic Properties for Ions

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Chapter 7. The Quantum‐Mechanical Model of the Atom Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

7.7 Electron Configurations: How Electrons Occupy Orbitals  Students need help understanding the reasons behind the order of orbitals, especially for those like 4s and 5s that appear to be out of order. Many will prefer to memorize a list instead.  For periods 5 and higher, students often forget about the f orbitals in writing configurations.  Conceptual Connection 7.6 Penetration and Shielding  Conceptual Connection 7.7 Electron Configurations and Quantum Numbers



.

The order of adding electrons (s, p, d, f) to predict the electron configuration of neutral atoms is relatively simple with a few exceptions.

Each of the periods does not have the same total number of electrons.

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Chapter 7. The Quantum‐Mechanical Model of the Atom Additional Problem for Photon Energy (Example 7.2)

A one-second pulse of a red laser pointer with a wavelength of 635 nm contains 5.0 mJ of energy. How many photons does it contain?

Sort

Given Epulse = 5.0 mJ

 = 635 nm

You are given the wavelength and total energy of a light pulse and asked to find the number of photons it contains.

Find number of photons

Strategize

Conceptual Plan

In the first part of the conceptual plan, calculate the energy of an individual photon from its wavelength.



In the second part, divide the total energy of the pulse by the energy of each photon to get the number of photons in a pulse.



Ephoton hc

E 



Epulse Ephoton

= number of photons

Relationships Used E = hc/  (Equation 7.3)

Solve

Solution

Convert wavelength to metres and substitute the values into the energy equation.

Convert the energy of the pulse to joules J. Then divide by the energy of a single photon.

 = 635 nm 

Ephoton =

hc



10 9 m = 6.35  10 7 m 1 nm

(6.626  10 34 J  s )(3.00  108 =

m ) s

6.35  10 7 m

= 3.13  10 19 J

5.0 mJ 

10 3 J = 5.0  10 3 J 1 mJ

number of photons =

.

Epulse Ephoton

=

5.0  103 J = 1.6  1016 3.13  1019 J

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Chapter 7. The Quantum‐Mechanical Model of the Atom Check

The magnitude of the answer makes physical sense since the pulse was much larger than the individual photon energy.

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117

Chapter 7. The Quantum‐Mechanical Model of the Atom Additional Problem for Wavelength of Light for a Transition in the Hydrogen Atom (Example 7.5)

Determine the wavelength of light emitted when an electron in a hydrogen atom makes a transition from an orbital in n = 5 to n = 4.

Sort

Given n = 5  n = 4

You are given the energy levels of an atomic transition and asked to find the wavelength of emitted light.

Find 

Strategize

Conceptual Plan

Calculate the energy of the electron in the n = 5 and n = 4 orbitals using Equation 7.7 and subtract to find the difference.

n = 5, n = 4

The negative value of the difference indicates that energy is being emitted. Convert the energy value to a wavelength using Equation 7.3.

Eatom  Ephoton





Eatom = Ephoton

E 

hc



Eatom

E = E4 – E5



Relationships Used En = 2.18 x 1018 J (1 / n2) (Equation 7.7) E = hc /  (Equation 7.3)

Solve Follow the conceptual plan to solve the problem.

Round the answer to three significant figures to reflect the three significant figures in the least precisely known quantity (4750). These conversion factors are all exact and therefore do not limit the number of significant figures.

Solution

Eatom = E4  E5   1  1  =  2.18  1018 J  2    2.18  1018 J  2   4   5   1  1 =  2.18  1018 J  2  2  5  4 20 =  4.90  10 J

Ephoton = Eatom = +4.90 x 1020 J

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Chapter 7. The Quantum‐Mechanical Model of the Atom

hc

E





hc E (6.626  1034 J  s )(3.00  108 m/ s ) = 4.90  1020 J



or



= 4.06  106 m

Check

The units of the answer are correct (m). A comparison with the values in Figure 7.17 indicates that the answer should be in the infrared region. Figure 7.5 confirms that the answer is within that region.

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Chapter 8. Periodic Properties of the Elements Student Objectives 8.1 Nerve Signal Transmission  

Know that the ions of sodium and potassium, two group 1 elements, play major roles in nerve‐signal transmission.

Know that the sodium‐potassium pumps in cell membranes can distinguish between Na+ and K+ because of each ion’s radius, one of the most prevalent periodic trends.

8.2 The Development of the Periodic Table   

Know that the rapid discovery of new elements in the nineteenth century resulted in several scientists’ efforts to categorize all known elements.

Know that Mendeleev organized the modern form of the periodic table to group together elements with similar characteristics.

Know and understand that the periodic law summarizes the behaviour of the elements—arranging them by atomic number results in strong correlation with elemental properties.

8.3 Electron Configurations, Valence Electrons, and the Periodic Table   

Define valence and core electrons.

Know the s, p, d, and f blocks of the periodic table.

Use the periodic table to predict electron configurations.

8.4 The Explanatory Power of the Quantum‐Mechanical Model 

Understand that many of the chemical properties of elements are due to the number of valence electrons and that elements in the same group have the same number of valence electrons.

8.5 Periodic Trends in the Size of Atoms and Effective Nuclear Charge    

Know the definitions and differences among van der Waals, covalent, and atomic radii.

Know and predict trends in atomic radius down a group (larger radius) and to the right across a period (smaller radius).

Use the ideas of screening and effective nuclear charge to explain the trends for atomic radius.

Know that the radii of transition elements remain approximately constant across each period.

8.6 Ionic Radii   

Know how to write electron configurations for ions. For anions, extra electrons are simply filled in.

For cations, electrons are removed from the highest sublevel of the highest principal energy level.

Identify and distinguish between paramagnetic and diamagnetic atoms/ions.

Know the relationship between the radius of a neutral atom and its ions: cations are smaller while anions are larger than the corresponding neutral atom.

8.7 Ionization Energy   

Know and predict trends in first ionization energy down a group (smaller ionization energy) and to the right across a period (larger ionization energy).

Use the ideas of screening and effective nuclear charge to explain the trends for ionization energy.

Understand trends in second and successive ionization energies with respect to the noble‐gas core.

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Chapter 8. Periodic Properties of the Elements 8.8 Electron Affinities and Metallic Character  

Define and understand the basic trend for electron affinity: it generally gets less exothermic down a group and more exothermic to the right across a period.

Know the periodic trends in metallic character: it increases down a group and decreases to the right across a period.

8.9 Some Examples of Periodic Chemical Behaviour: The Alkali Metals, the Halogens, and the Noble Gases 

Know the names, periodic trends, and representative chemical reactions of the elements of a few groups: group 1 (alkali metals), group 2 (alkaline earth metals), group 17 (halogens), and group 18 (noble gases).

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 8. Periodic Properties of the Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

8.1 Nerve Signal Transmission  Elements in a group o Na+ and K+ roles in nerve transmission  Periodic property: radius



8.2 The Development of the Periodic Table  History of element discovery and development of the periodic table o Dobereiner, Mendeleev, Meyer, Moseley  Periodic law o Atomic mass, atomic number



 

  

.

Intro figure: cell membrane and flux of Na+, K+ ions Unnumbered figure: group 1 Unnumbered figure: radii of Na+ and K+

Unnumbered figure: portrait of Dmitri Mendeleev Unnumbered figure: portrait of Henry Moseley (discovered the atomic number) Figure 8.1 Eka‐Aluminum and Eka‐Silicon Figure 8.2 Graph Showing a Portion of Moseley’s X‐Ray Data

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Chapter 8. Periodic Properties of the Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

8.1 Nerve Signal Transmission  The body requires large quantities of sodium and potassium for various purposes, especially the sodium‐ potassium pump that facilitates nerve‐signal transmission. Emphasize the sizes of these two ions and the membrane proteins that regulate their passage into and out of cells.



Essential minerals like sodium and potassium are present as ions and not as the neutral form of the element, particularly since these minerals exist in an aqueous environment in the body.

8.2 The Development of the Periodic Table  Ask the students to arrange the elements for the first two periods by atomic mass instead of atomic number.

Point out where the problems arise and why Mendeleev was so successful with his approach.  Find some alternative periodic tables to show in class: Monroe, C.J.; Turner, W.D.; J. Chem. Educ. 1926, 3, 1058.

Emerson, Edgar I.; J. Chem. Educ. 1944, 21, 111.

Sanderson, R.T.; J. Chem. Educ. 1964, 41, 187.



The periodic law predicts and summarizes why certain elements have similar properties. The quantum theory explains the basis of the law.

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Chapter 8. Periodic Properties of the Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures and Solved Problems

8.3 Electron Configurations, Valence Electrons, and the Periodic Table  Electron configurations o Core electrons o Valence electrons  Outermost principal energy level  Same valence electron count for elements in a group  Periodic table o Orbital blocks for s, p, d, f o Electron configuration by position in periodic table o Exceptions for transition elements

       

8.4 The Explanatory Power of the Quantum‐ Mechanical Model  Periodicity o Elements in the same group have similar properties o Elements in the same group have same valence electron count  Periodic table and ions with predictable charge o Explanation of the concept from Section 2.7

   

8.5 Periodic Trends in the Size of Atoms and Effective Nuclear Charge  Radius definitions o Van der Waals o Covalent o Nonbonding  Radius trend o Decreases to the right across a period o Increases down a group  Effective nuclear charge  Difference in trends for main‐group versus transition elements

         

.

Figure 8.3 Outer Electron Configurations of the First 18 Elements in the Periodic Table Example 8.1 Valence Electrons and Core Electrons Figure 8.4 The s, p, d, and f Blocks of the Periodic Table Unnumbered figure: electron configuration of Cl Unnumbered figure: valance electrons of the noble gases Unnumbered figure: electron configurations of first and second row transition metals Unnumbered figure: valance electrons of the alkali metals Example 8.2 Writing Electron Configurations from the Periodic Table

Figure 8.5 Elements That Form Ions with Predictable Charges Unnumbered figure: electron configurations of elements in group 18 Unnumbered figure: electron configurations of elements in groups 1 and 2 unnumbered figure: electron configurations of elements in group 17

Unnumbered figure: van der Waals radius of Kr Unnumbered figure: covalent radius of Br Figure 8.6 Atomic Radius versus Atomic Number Figure 8.7 Trends in Atomic Radius Figure 8.8 Screening and Effective Nuclear Charge Unnumbered figure: shielding contribution, f Example 8.3 Calculating Zeff Using Slater’s Rules Unnumbered figure: Zeff versus radius Figure 8.9 Trends in Atomic Radius Example 8.4 Atomic Size

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Chapter 8. Periodic Properties of the Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

8.3 Electron Configurations, Valence Electrons, and the Periodic Table  The distinction between core and valence electrons is an important concept that explains periodicity and represents the basis for bonding and structure.  The organization of the periodic table by orbital blocks (Figure 8.4) is graphical and simplifies writing electron configurations based on position in the periodic table.



The number of valence electrons is easy to predict for main‐group elements but more difficult for transition elements.

8.4 The Explanatory Power of the Quantum‐Mechanical Model  Valence electrons determine chemical properties and the formation of common ions. Groups 1 and 2 form cations as do the transition elements, but the former are restricted to a single characteristic charge especially because of the ease of attaining a noble‐gas configuration.



Some exceptions to the chemical inertness of the noble gases will be introduced in later chapters.

8.5 Periodic Trends in the Size of Atoms and Effective Nuclear Charge  The values used to determine atomic radii are experimental values obtained from several kinds of sources, e.g., van der Waals, covalent, and nonbonding; atomic radii are average values. In solving problems and homework, students need to use the values from a single source, like the book, to get the answers in the solution manual.  The pattern of the plot of radius vs. atomic number (Figure 8.9) is easy to remember. The height of the peaks (the alkali metals) increases; then the values decrease across each period.  Effective nuclear charge explains the periodic trends, especially the left‐to‐right trend.  Conceptual Connection 8.1 Atomic Size



Atoms are not solid spheres as they often appear to be in graphical representations. Their outer surfaces are the combination of many orbitals. These issues, combined with the use of experimental values from bonding and nonbonding data, give rise to average atomic radii.

.

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Chapter 8. Periodic Properties of the Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures and Solved Problems

8.6 Ionic Radii  Electron configurations of ions o Anions  Add electrons indicated by charge  Fill according to increasing energy o Cations  Remove electrons indicated by charge  Remove electrons first from orbitals with highest n  Often not the reverse of filling order  Ionic radii o Cations: smaller than neutral atom o Anions: larger than neutral atom

  

Figure 8.10 Sizes of Atoms and Their Cations Figure 8.11 Sizes of Atoms and Their Anions Example 8.5 Ion Size

8.7 Ionization Energy  Definition: energy required to remove an electron from a gas‐phase atom  Increases to the right across a period  Decreases down a group  Successive ionization energies are larger  Large jump in IE when core electrons are removed



Figure 8.12 First Ionization Energy versus Atomic Number for the Elements Through Xenon Figure 8.13 Trends in First Ionization Energy Example 8.6 Ionization Energy Unnumbered figure: first, second, and third ionization energies of Na and Mg Figure 8.14 Trends in First Ionization Energy Unnumbered figure: ionization energies of Na and Mg Table 8.1 Successive Values of Ionization Energies for the Elements Sodium Through Argon

     

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Chapter 8. Periodic Properties of the Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

8.6 and 8.7 Ions: Electron Configurations, Magnetic Properties, Ionic Radii, and Ionization Energy  Anions follow the same general procedure for writing electron configurations. Cations will often confuse students. For example, the 4s orbital is filled before the 3d orbitals in a neutral atom, but one removes a 4s electron before a 3d electron to make the cation.  The size of ions compared to their neutral atom is easy to predict based on the charge. In cations, the greater positive charge in the nucleus increases the average attraction for each electron. In anions, there is a lower average attraction between nucleus and electrons compared to the neutral atom.  The trend for first ionization energy is exactly opposite that of radius but for the same reasons. The pattern of the plot of first IE vs. atomic number is somewhat of a mirror image of the plot of radius vs. atomic number; in the former, the noble gases are the peaks, whereas the alkali metals are the peaks in the latter.  Successive ionizations make a steep increase when core electrons are removed.  Use isoelectronic pairs or triads as examples to emphasize thinking about periodic trends rather than simply memorizing them.  Conceptual Connection 8.2 Ions, Isotopes and Atomic Size  Conceptual Connection 8.3 Ionization Energies and Chemical Bonding



.

A common approach is to add an electron to the previous atom when writing an electron configuration.

An anion follows the same procedure as for neutral atoms, but cations require electron removal from the highest n‐value orbitals.

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Chapter 8. Periodic Properties of the Elements Lecture Outline Terms, Concepts, Relationships, Skills

Figures and Solved Problems

8.8 Electron Affinities and Metallic Character  Electron affinity o Most elements release energy when gaining an electron o Trends are not as simple as the other periodic trends  Metallic character o Decreases to the right across a period o Increases down a group



8.9 Some Examples of Periodic Chemical Behaviour: The Alkali Metals, the Halogens, and the Noble Gases  Alkali metals o Properties o Reactions  Halogens o Properties o Reactions  Noble gases o Properties o Reactions

 



      

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Figure 8.15 Electron Affinities of Selected Main‐Group Elements Figure 8.16 Trends in Metallic Character

Table 8.2 Properties of the Alkali Metals Figure 8.17 Reaction of Sodium and Chlorine to Form Sodium Chloride Figure 8.18 Reactions of the Alkali Metals with Water Table 8.3 Properties of the Halogens Unnumbered figure: photos of elemental chlorine, bromine, and iodine Example 8.7 Alkali Metals and Halogen Reactions Chemistry and Medicine: Potassium Iodide in Radiation Emergencies Table 8.4 Properties of the Noble Gases Unnumbered figure: photo of liquid helium boiling

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Chapter 8. Periodic Properties of the Elements Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

8.8 Electron Affinities and Metallic Character  Electron affinities generally do not follow such simple and regular trends as radius and first ionization energy. Group 1 shows a good trend, but most periods do not.  A graphic showing the trends in metallic character in the periodic table makes this trend easy to remember.

Metallic character is a macroscopic property unlike the microscopic properties outlined in the rest of the chapter.  Conceptual Connection 8.3 Periodic Trends



The characterization of elements as metal, metalloid, or nonmetal refers to macroscopic properties, not microscopic ones.

8.9 Some Examples of Periodic Chemical Behaviour: The Alkali Metals, the Halogens, and the Noble Gases  Since the periodic table often characterizes the behaviour of groups of elements, and since their common valence‐electron counts dictate chemical properties, it makes sense to introduce some descriptive chemistry at this stage. Demonstrations, real or virtual, will capture attention since they are exothermic and generally involve flames.

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129

Chapter 8. Periodic Properties of the Elements

Write the electron configurations of these elements:

Additional Problem for Electron Configurations (Example 8.1)

a) S b) Ar c) Ca d) Zn

Solution S 1s22s22p63s23p4

Sulfur has 16 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, and 3s, and distribute four electrons into the three 3p orbitals.

[Ne] 3s23p4 Solution Ar 1s22s22p63s23p6

Argon has 18 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, 3s, and each of the three 3p orbitals.

[Ar] Solution Ca 1s22s22p63s23p64s2

Calcium has 20 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, 3s, each of the three 3p, and 4s orbitals.

[Ar] 4s2

Solution Zn 1s22s22p63s23p64s23d10

Zinc has 30 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, 3s, each of the three 3p, 4s, and each of the five 3d orbitals.

[Ar] 4s23d10

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Chapter 8. Periodic Properties of the Elements Write the electron configurations of these elements and identify the core and valence electrons.

Additional Problem for Valence Electrons and Core Electrons (Example 8.1)

a) Sr b) Ar c) Ca d) Zn Solution Sr 1s22s22p63s23p64s23d104p65s2

Strontium has 38 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, 3s, each of the three 3p, 4s, each of the five 3d, each of the three 4p, and 5s orbitals. For strontium, the n = 1, 2, 3, and 4 principal quantum levels are complete or full.

The n = 5 level is outermost.

core = 36 electrons, [Kr] valence = 2 electrons

Solution Br 1s22s22p63s23p64s23d104p5

Bromine has 35 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, 3s, each of the three 3p, 4s, and each of the five 3d, and distribute five electrons into the three 4p orbitals. For bromine, the n = 1, 2, and 3 principal quantum levels are complete or full. The n = 4 level is outermost.

core = 28 electrons, [Ar] 3d10 valence = 7 electrons

Solution Ca 1s22s22p63s23p64s2

Calcium has 20 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, 3s, each of the three 3p, and 4s orbitals. For calcium, the n = 1, 2, and 3 principal quantum levels are complete or full.

The n = 4 level is outermost.

core = 18 electrons, [Ar] valence = 2 electrons

Solution Zn 1s22s22p63s23p64s23d10

Zinc has 30 electrons. Distribute two electrons into the 1s, 2s, each of the three 2p, 3s, each of the three 3p, 4s, and each of the five 3d orbitals. For zinc, the n = 1, 2, and 3 principal quantum levels are complete or full. The n = 4 level is outermost.

core = 28 electrons [Ar] 3d10 valence = 2 electrons

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Chapter 8. Periodic Properties of the Elements On the basis of periodic trends, choose the SMALLER atom from each pair:

Additional Problem for Atomic Size (Example 8.4)

a) Sr or Ca b) Ar or Kr c) Ca or K d) O or F

Solution

1

2

Strontium is below calcium in group 2. As you move down a group, orbitals are occupied in higher principal quantum number levels, resulting in a larger radius.

H Be Mg Ca

Calcium is smaller.

Sr Ba

Solution

17

18

Argon is above krypton in group 18. As you move down a group, orbitals are occupied in higher principal quantum number levels, resulting in a larger radius.

F

He Ne Ar Kr

Argon is smaller.

Xn Rn

Solution

1

Calcium (group 2) is to the right of potassium (group 1). As you move right across a period, the effective nuclear charge experienced by outermost electrons increases, resulting in a smaller radius.

H

2

Be

K

Ca

Calcium is smaller.

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132

Chapter 8. Periodic Properties of the Elements 6

Solution Oxygen (group 6) is to the left of fluorine (group 7).

As you move right across a period, the effective nuclear charge experienced by outermost electrons increases, resulting in a smaller radius.

7

8 He

O

F

Ne

Fluorine is smaller.

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133

Chapter 8. Periodic Properties of the Elements Write a balanced equation for each of these cases:

Additional Problem for Alkali Metal and Halogen Reactions (Example 8.7)

a) The reaction between sodium metal and bromine gas. b) The reaction between potassium metal and water. c) The reaction between hydrogen gas and chlorine gas. d) The reaction between calcium and water.

Solution 2 Na(s) + Br2(g)  2 NaBr(s)

Alkali metals react with halogens to form metal halides. Write the reagents and products using their correct forms. Balance the equation.

Solution 2 K(s) + 2 H2O(l)  2 K+(aq) + 2 OH(aq) + H2(g)

Alkali metals react with water to form the alkali metal cation, hydroxide, and hydrogen gas. Write the reagents and products using their correct forms.

Balance the equation.

Solution H2(g) + Cl2(g)  2 HCl(g)

Halogens react with each other or with hydrogen to form the interhalogen compound or hydrogen halide, respectively. Write the reagents and products using their correct forms. Balance the equation.

Solution Ca(s) + 2 H2O(l)  Ca2+(aq) + OH(aq) + H2(g)

Alkali earth metals react much like the alkali metals.

They react with water to form the alkali earth metal cation, hydroxide, and hydrogen gas. Write the reagents and products using their correct forms.

Balance the equation.

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Chapter 9. Chemical Bonding I: Lewis Theory Student Objectives 9.1 Bonding Models and AIDS Drugs  

Know that X‐ray crystallography was used to characterize the structure of HIV‐protease, a biomolecule critical to the reproduction of HIV. Know that Lewis structures are simple predictors of how atoms combine to form ionic compounds and molecules.

9.2 Types of Chemical Bonds  

Know and understand that chemical bonds form because they lower the potential energy between the charged particles in the constituent atoms.

Define and understand ionic bond, covalent bond, and metallic bonding.

9.3 Representing Valence Electrons with Dots    

Know that valence electrons can be represented with dots around an element symbol.

Identify and draw atoms with their valence electrons represented as dots.

Know that Lewis theory involves the sharing or transfer of electrons.

Define and know the octet rule.

9.4 Ionic Bonding and Lattice Energies      

Draw Lewis structures of ionic compounds containing main‐group elements.

Understand that the formation of an ionic compound from neutral atoms is exothermic—the amount of energy released is largely caused by lattice energy.

Know that the Born–Haber cycle is a way of accounting for the energetics of each of the steps in the formation of an ionic compound from its constituent elements.

Use a Born–Haber cycle to calculate the lattice energy of an ionic compound.

Know that lattice energy decreases for larger ions and increases with increasing charge.

Understand why ionic solids are poor electrical conductors while ionic liquids and aqueous solutions of ionic compounds are good electrical conductors.

9.5 Covalent Bonding: An Introduction to Lewis Structures of Molecules    

Know that most nonmetal atoms prefer to be surrounded by eight valence electrons, but hydrogen requires only two.

Understand that in Lewis theory, a pair of electrons, one from each of two atoms, forms a bond or bonding pair that helps each atom achieve an octet. The two atoms can also share two pairs of electrons (a double bond) or three pairs of electrons (triple bond).

Identify and draw covalent compounds with single, double, and triple bonds between constituent atoms.

Draw Lewis structures for molecular compounds and polyatomic ions.

9.6 Bond Energies, Bond Lengths, and Bond Vibrations   

Define bond energy.

Estimate reaction enthalpies using average bond energies for all bonds broken and formed in a chemical reaction.

Understand the inverse relationship between bond length and bond strength.

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Chapter 9. Chemical Bonding I: Lewis Theory 9.7 Electronegativity and Bond Polarity    

Know and understand that a pair of electrons does not have to be shared equally between two atoms.

Unequal sharing results in a polar covalent bond.

Define electronegativity and know its periodic trends.

Understand that bonds can range from a nonpolar covalent bond to a polar covalent bond to an ionic bond depending on the difference in electronegativity between the two atoms.

Define dipole moment and percent ionic character.

9.8 Resonance and Formal Charge  

Define resonance structures and understand how Lewis structures represent the individual and the hybrid structures.

Define formal charge and understand how to calculate it for the atoms in a Lewis structure.

9.9 Exceptions to the Octet Rule: Drawing Lewis Structures for Odd‐Electron Species and Incomplete Octets    

Draw Lewis structures for odd‐electron species.

Draw Lewis structures for molecules containing atoms with incomplete octets.

Draw Lewis structures for molecules containing atoms with expanded octets.

Understand why the second‐period elements cannot have expanded octets.

9.10 Lewis Structures for Hypercoordinate Compounds 9.11 Bonding in Metals: The Electron Sea Model 

Understand how the electron sea model accounts for the general macroscopic properties of metals.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 9. Chemical Bonding I: Lewis Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

9.1 Bonding Models and AIDS Drugs  X‐ray crystallography o Structure of HIV‐protease  Bonding theory o Lewis theory o Lewis structure

 

Intro figure: HIV‐protease and Indinavir Unnumbered figure: photo of G. N. Lewis

9.2 Types of Chemical Bonds  Types of bonds o Ionic o Covalent o Metallic  Electrons in bonds o Transferred (ionic) o Shared (covalent) o Pooled (metallic)



Figure 9.1 Ionic, Covalent, and Metallic Bonding Unnumbered table: kinds of bonds Figure 9.2 Possible Configurations of One Negatively Charged Particle and Two Positively Charged Ones

9.3 Representing Valence Electrons with Dots  Lewis structure o Elemental symbol surrounded by valence electrons o Octet rule  Chemical bond o Sharing electrons o Octet rule: complete octet by sharing



 

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Unnumbered figures: Lewis dot structures of period 2 elements

138

Chapter 9. Chemical Bonding I: Lewis Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

9.1 Bonding Models and AIDS Drugs  Proteins are complex molecules, but building models of them starts with a single bond. Student understanding of X‐ray crystallography can start with a gross simplification that begins with an analogy to X‐rays in radiology. In protein structures, X‐rays diffract as a result of the nuclei of atoms.  Lewis structures were proposed in a J. Am. Chem. Soc. paper in 1916 and have served as a simple but effective bonding model since.

9.2 Types of Chemical Bonds  A brief review of electrostatic potential energy should point out that the energy is proportional to the magnitude of the charges and inversely proportional to the distance between them.  Figure 9.2 shows the possibilities of one negative charge and two positive charges. Ask the students to imagine possibilities for two negative charges and two positive charges.

9.3 Representing Valence Electrons with Dots  The placement of the electrons in the Lewis dot structures for the period 2 elements gives a starting point for students in terms of the number of bonds often formed by these elements. The students will realize with more study that these are not always the case.

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Chapter 9. Chemical Bonding I: Lewis Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

9.4 Ionic Bonding and Lattice Energies  Lewis structures of ionic compounds o Complete octet by addition or subtraction of electrons  Energetics o Lattice energy o Born–Haber cycle o Trends  Ion size  Ion charge  Electrical conductivity



Unnumbered figures: Lewis dot structures of KCl and Na2S Figure 9.3 Lattice Energy Figure 9.4 Born–Haber Cycle for Sodium Chloride Unnumbered figure: bond lengths of the group 1 metal chlorides Unnumbered figure: lattice energy of group 1 metal chlorides Unnumbered figures: bond lengths of NaF and CaO Unnumbered figure: lattice energy of NaF and CaO Example 9.1 Predicting Relative Lattice Energies Unnumbered figure: melting of NaCl Unnumbered figures: electrical conductivity of NaCl(s) and NaCl(aq) Chemistry and Medicine: Ionic Compounds in Medicine

         

9.5 Covalent Bonding: An Introduction to Lewis Structures of Molecules  Lewis structures of covalent compounds o Bonding pairs o Lone pairs o Single bonds o Double, triple bonds  Skeletal structure  Total number of valence electrons  Distribution of electrons to satisfy octet rule  Covalent bonds (intramolecular) versus intermolecular forces

      

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Unnumbered figures: Lewis structures for H 2O Unnumbered figures: Lewis structures for Cl2 and H2 Unnumbered figures: Lewis structures for O2 and N2 Unnumbered figures: Lewis structures for H3O+ and H2O2 Examples 9.2 and 9.3 Drawing Lewis Structures Figure 9.5 Intermolecular and Intramolecular Forces Example 9.4 Drawing Lewis Structures for Polyatomic Ions

140

Chapter 9. Chemical Bonding I: Lewis Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

9.4 Ionic Bonding and Lattice Energies  The electron configuration of K+ but not K demonstrates a complete octet.  Lattice energy cannot be measured directly in the laboratory but is obtained through a thermodynamic cycle.  Ion size and ion charge discussions about the trend in lattice energy are an application of Coulomb’s law.  The “reality” section brings focus on the electrical conductivity of ionic compounds.

Electricity can be conducted only when the charged particles can move.  Conceptual Connection 9.1 Melting Points of Ionic Solids



Despite especially the group 1 and group 2 elements’ willingness to ionize, that ionization is still endothermic. Lattice energy is the main reason that ionic solids form from their elements.

9.5 Covalent Bonding: An Introduction to Lewis Structures of Molecules  Introduce the idea of the common numbers of bonds, e.g., 2 for O, 3 for N, 4 for C, etc. It may be useful to use colours or other symbols to differentiate electrons as you assign them in the first few Lewis structure examples.  The discussion of intramolecular (microscopic) versus intermolecular (macroscopic) forces is an important one.

The latter are always much weaker than bond energies (Section 9.10).  Conceptual Connection 9.2 Energy and the Octet Rule



Lewis structures and octets are simple schemes to understand how atoms combine to form molecules. They are neither the forces nor the justification for why they do.

The overall lower energy of the molecule drives its formation.

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Chapter 9. Chemical Bonding I: Lewis Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

9.6 Bond Energies, Bond Lengths, and Bond Vibrations  Bond energies o Breaking bonds: endothermic o Forming bonds: exothermic o Higher for multiple bonds o Average bond energy o Estimation of rH  Bond lengths o Centre‐to‐centre distance between nuclei shorter for multiple bonds  Bond vibrations o Stretching vibration o IR spectroscopy

          

9.7 Electronegativity and Bond Polarity  Covalent bonds o Nonpolar covalent o Polar covalent  Electronegativity o Pauling values o Bond polarity, EN  Covalent, EN = 0–0.4  Polar covalent, EN = 0.4–2.0  Ionic, EN = 2.0+  Dipole moment o  = qr o Percent ionic character

        

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Unnumbered table: carbon–carbon bond energies Table 9.1 Average Bond Energies Figure 9.6 Estimating rH from Bond Energies Example 9.5 Calculating rH from Bond Energies Table 9.2 Average Bond Lengths Unnumbered figures: molecular models, bond lengths of halogens Unnumbered table: carbon–carbon bond lengths Unnumbered table: nitrogen–halogen bond lengths Figure 9.7 How Bond Strength Affects the Energy of IR Absorptions of H—X Stretching Vibrations Figure 9.8 Some Vibrational Motions of Polyatomic Molecules Figure 9.9 IR Absorption Spectra of Ethanol and Ethanal

Figure 9.10 Orientation of Gaseous Hydrogen Fluoride in an Electric Field Figure 9.11 Electron Density Plot for the HF Molecule Figure 9.12 Electronegativities of the Elements Unnumbered figures: molecular models of Cl2, NaCl, and HCl Table 9.3 The Effect of Electronegativity Difference on Bond Type Figure 9.13 Electronegativity Difference (EN) and Bond Type Table 9.4 Dipole Moments of Several Molecules in the Gas Phase Figure 9.14 Percent Ionic Character versus Electronegativity Difference for Some Compounds Example 9.6 Classifying Compounds as Pure Covalent, Polar Covalent, or Ionic

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Chapter 9. Chemical Bonding I: Lewis Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

9.6 Bond Energies, Bond Lengths, and Bond Vibrations  Average bond energies are useful for providing approximate values for rH.  Bond energy for a bond being broken is positive, while that for a bond being formed is negative.  Introduction to bond stretching vibrations and IR absorption spectra  Conceptual Connection 9.3 Bond Energies and rH



When estimating rH, students are sometimes tempted to use “products minus reactants,” as they did using enthalpies of formation.

9.7 Electronegativity and Bond Polarity  Polar bonds in diatomic molecules can be detected by placing the molecules in an electric field.  Electronegativity has several predictive uses, but differences in electronegativity and the resulting bonds must not be too rigid, particularly values between two types, e.g., 1.9 and 2.0.  A dipole moment is a physical measurement, so comparison with the predictions from EN is instructive.  Conceptual Connection 9.4 Percent Ionic Character



Bonds can be described as a continuum ranging from equal sharing of a pair of electrons between two atoms through degrees of polar covalent bonds to an ionic bond in which the electrons are completely transferred.

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Chapter 9. Chemical Bonding I: Lewis Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

9.8 Resonance and Formal Charge  Resonance o Equivalent Lewis structures o Hybrid structure  Formal charge o Calculation o Minimization o Negative formal charges on electronegative atoms

    

Unnumbered figure: Lewis structure of O3 Figure 9.15 Hybridization Example 9.7 Writing Resonance Structures Unnumbered figures: formal charges of OCN Example 9.8 Assigning Formal Charges and Predicting Preferred Lewis Structures

9.9 Exceptions to the Octet Rule: Drawing Lewis Structures for Odd‐Electron Species and Incomplete Octets  Odd‐electron species o Radicals or free radicals  Incomplete octet o Especially Be, B



Chemistry and the Environment: Free Radicals and the Atmospheric Vacuum Cleaner Unnumbered figures: Lewis structures of NO and NO2 Unnumbered figures: Lewis structures of BF3, BH3, NH3, and BF3NH3

9.10 Lewis Structures for Hypercoordinate Compounds  Expanded octet o 10, 12, or even 14 electrons around the central atom is impossible for second‐period elements



9.11 Bonding in Metals: The Electron Sea Model  Microscopic properties o Low ionization energies o Electrons not attached to particular atoms  Macroscopic properties o Thermal and electrical conductivity o Malleability o Ductility

 

 

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Unnumbered figures: Lewis structures of AsF5, SF6, and H2SO4 Examples 9.9 and 9.10 Writing Lewis Structures for Hypercoordinate Compounds

Figure 9.16 The Electron Sea Model for Sodium Chemistry in the Environment: The Lewis Structure of Ozone Unnumbered figure: photo of ductility of Cu

144

Chapter 9. Chemical Bonding I: Lewis Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

9.8 Resonance and Formal Charge  Resonance forms indicate that the actual structure can be represented by several equivalent Lewis structures.  The analogy of resonance forms to dogs is one that overcomes the urge to think of Lewis structures as rapidly changing forms.  Formal charges enable students to discriminate among different Lewis structures for a given molecule to determine which is better.



Resonance does not arise from two or more interconverting forms. Two or more forms indicate a special situation in which the hybrid structure has special stability or properties, e.g., the different properties of benzene from other alkenes.

Formal charges on constituent atoms in a Lewis structure can be confused with oxidation numbers.



9.9 Exceptions to the Octet Rule: Drawing Lewis Structures for Odd‐Electron Species and Incomplete Octets  Many radicals exist, especially if one considers organic examples. The depletion of ozone by chlorine atoms is initiated by breaking a C—Cl bond in CF2Cl2 or a similar compound.  Incomplete octets are common for boron compounds, but the formal charge for boron when it makes three bonds is 0.

9.10 Lewis Structures for Hypercoordinate Compounds  Expanded octets are common for structures with P and S as well as for oxyanions of the halogens.



Expanded octets apply only to central atoms and never to terminal atoms.

9.11 Bonding in Metals: The Electron Sea Model  Metals require a unique bonding model to account for many of their macroscopic properties, e.g., electrical conductivity.

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Chapter 9. Chemical Bonding I: Lewis Theory

Determine whether the bond between each pair of atoms will be pure covalent, polar covalent, or ionic:

Additional Problem for Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic (Example 9.6)

a) S and O b) Al and F c) C and Br d) Mg and Cl

Solution EN = 1.0

Find the electronegativity values in Figure 9.12: S = 2.5 O = 3.5

polar covalent

Solution EN = 2.5

Find the electronegativity values in Figure 9.12: Al = 1.5 F = 4.0

ionic

Solution EN = 0.3

Find the electronegativity values in Figure 9.12: C = 2.5 Br = 2.8

pure covalent

Solution EN = 1.8

Find the electronegativity values in Figure 9.12: Mg = 1.2 Cl = 3.0

polar covalent, near ionic

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Chapter 9. Chemical Bonding I: Lewis Theory

Additional Problem for Drawing Lewis Structures (Examples 9.2, 9.3)

Draw a Lewis structure for CS2.

Draw a Lewis structure for CH3Cl.

Correct Skeletal Structure

C is the same electronegativity as S but we put it in the centre since there are two S.

C is less electronegative than Cl; H is always terminal.

Write the correct skeletal structure for the molecule.

Number of Electrons

C=4

C=4

Calculate the total number of electrons for the Lewis structure by summing the valence electrons of each atom in the molecule.

S=6

H=1

S=6

H=1

total = 16

H=1 Cl = 7 total = 14

Distribute Electrons

First, bonding electrons.

First, bonding electrons.

Next, lone pairs.

Next, lone pairs.

Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms as possible. Begin with the bonding electrons and proceed to lone pairs on the central atom.

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147

Chapter 9. Chemical Bonding I: Lewis Theory Form Double or Triple Bonds

Move lone pairs from S to C.

If any atom lacks an octet, form double or triple bonds as necessary to give them octets.

.

All H have duets. All other atoms have octets. The structure above is complete.

148

Chapter 9. Chemical Bonding I: Lewis Theory

Additional Problem for Drawing Lewis Structures for Polyatomic Ions (Example 9.4)

Draw a Lewis structure for NO2.

Correct Skeletal Structure

N is lower electronegativity than O and is at the centre.

Write the correct skeletal structure for the molecule.

Number of Electrons

N=5

Calculate the total number of electrons for the Lewis structure by summing the valence electrons of each atom in the molecule.

O=6 O=6 charge = 1 total = 18

First, bonding electrons.

Distribute Electrons Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms as possible. Begin with the bonding electrons and proceed to lone pairs on the central atom.

Next, lone pairs.

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149

Chapter 9. Chemical Bonding I: Lewis Theory Move lone pairs from O to N.

Form Double or Triple Bonds If any atom lacks an octet, form double or triple bonds as necessary to give them octets.

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150

Chapter 9. Chemical Bonding I: Lewis Theory

Additional Problem for Calculating rH from Bond Energies (Example 9.5)

Hydrogen is considered an optimal fuel since it burns without forming any carbon dioxide.

2 H2(g) + O2(g)  2 H2O(g) Use bond energies to calculate rH.

Equation Begin by writing the reaction using the Lewis structures of the molecules involved.

H–H + H–H + O=O  H–O–H + H–O–H

Bonds Broken

Bond broken, H

Determine which bonds are broken in the reaction and sum the bond energies of these.

2 H–H = 2(436 kJ) 1 O=O = 1(498 kJ) = 1370 kJ

Bonds Formed

Bond formed, H

Determine which bonds are formed in the reaction and sum the negatives of the bond energies of these.

4 H–O = 4(464 kJ) = 1856 kJ

Result Find rH by summing the results of the previous two steps.

rH = 1370 – 1856 = 486 kJ

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Student Objectives 10.1 Artificial Sweeteners: Fooled by Molecular Shape   

Know and understand that energy content and taste are due to microscopic properties related to structure but are independent of each other.

Know that taste depends a great deal on the three‐dimensional structures of food molecules.

Know that a simple model to determine and predict molecular shapes is VSEPR theory: valence shell electron pair repulsion theory.

10.2 VSEPR Theory: The Five Basic Shapes    

Know and understand that VSEPR theory is based on electron groups that repel each other.

Know that VSEPR predicts five basic shapes according to the number of electron groups surrounding a central atom: linear (2), trigonal planar (3), tetrahedral (4), trigonal bipyramidal (5), and octahedral (6).

Know the bond angles for each basic shape.

Recognize molecules in their correct shapes based on their number of electron groups.

10.3 VSEPR Theory: The Effect of Lone Pairs   

Understand the difference between electron geometry and molecular geometry.

Know and understand the effect of lone pair electrons on molecular geometry with respect to shape and bond angle.

Know the different molecular geometries that arise from tetrahedral, trigonal bipyramidal, and octahedral electron geometries.

10.4 VSEPR Theory: Predicting Molecular Geometries  

Know the procedures for predicting and drawing molecular geometries.

Predict and draw the electron and molecular geometries for molecules, including molecules with more than one central atom.

10.5 Molecular Shape and Polarity    

Identify polar bonds in molecules based on EN. Understand how polar bonds translate into net dipole moments for molecules.

Know and understand how vector addition is used to predict net dipole moments.

Understand how microscopic polarity results in macroscopic properties of molecules, e.g., the immiscibility of water and oil.

10.6 Valence Bond Theory: Orbital Overlap as a Chemical Bond  

Understand an interaction energy diagram for the formation of bonds with respect to internuclear distance.

Know and understand how the overlap of atomic orbitals leads to bonds and how this is explained by valence bond theory.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory 10.7 Valence Bond Theory: Hybridization of Atomic Orbitals    

Define and understand hybridization and the role of atomic orbitals.

Know and understand the common types of hybridization: sp3, sp2, and sp.

Know the hybridizations for expanded octets: sp3d and sp3d2.

Know how to predict hybridization and draw valence bond models of molecules.

10.8 Molecular Orbital Theory: Electron Delocalization     

Know the basis for molecular orbital theory.

Know and understand how linear combinations of atomic orbitals (LCAO) form molecular orbitals.

Define bonding orbital and antibonding orbital and understand the differences between the two.

Predict and draw molecular orbital diagrams.

Understand that molecular orbital theory provides the best explanation of the paramagnetism of O2 and provides the best model for electron delocalization in molecules.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

10.1 Artificial Sweeteners: Fooled by Molecular Shape  Microscopic properties o Energy from combustion o Molecular shape  Sense of taste: active site  Valence shell electron pair repulsion theory (VSEPR)



Intro figure: shapes of sugar and aspartame

10.2 VSEPR Theory: The Five Basic Shapes  Kinds of electron groups  Basic shapes (geometry) o Linear  Two electron groups  Examples: BeCl2, CO2 o Trigonal planar  Three electron groups  Examples: BF3, H2CO o Tetrahedral  Four electron groups  Example: CH4 o Trigonal bipyramidal  Five electron groups  Example: PCl5 o Octahedral  Six electron groups  Example: SF6



Figure 10.1 Repulsion Between Electron Groups Unnumbered figures: Lewis structures and geometries of BeCl2 and CO2 Unnumbered figures: Lewis structures and geometries of BF3 and H2CO Figure 10.2 Representing Electron Geometry with Balloons Unnumbered figure: tetrahedral shape Unnumbered figure: Lewis structure and geometry of CH4 Unnumbered figure: trigonal bipyramidal shape Unnumbered figure: Lewis structure and geometry of PCl5 Unnumbered figure: octahedral shape Unnumbered figure: Lewis structure and geometry of SF6 Example 10.1 VSEPR Theory and the Basic Shapes

         

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

10.1 Artificial Sweeteners: Fooled by Molecular Shape  Molecules have a variety of microscopic and macroscopic properties. Two microscopic properties include energy (heat) released during combustion and shape as interpreted by taste receptors. Ask the students for other potential microscopic properties, e.g., molar mass, solubility in water. What about macroscopic properties, e.g., melting point, density, conductivity?

10.2 VSEPR Theory: The Five Basic Shapes  A fundamental idea is that electron groups repel each other.  Models are extremely useful for this topic.

One can demonstrate the electron groups in class using balloons.  Conceptual Connection 10.1 Electron Groups and Molecular Geometry



.

Double bonds can shrink bond angles but otherwise have no effect on geometry.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

10.3 VSEPR Theory: The Effect of Lone Pairs  Difference between electron and molecular geometries  Four electron groups with lone pairs o Trigonal pyramidal  One lone pair  Smaller bond angle than tetrahedral  Example: NH3 o Bent  Two lone pairs  Smaller bond angle than tetrahedral or trigonal bipyramidal  Example: H2O  Five electron groups with lone pairs o Seesaw  One lone pair; goes in trigonal plane  Example: SF4 o T‐shaped  Two lone pairs, both in trigonal plane  Example: BrF3 o Linear  Three lone pairs, all in trigonal plane  Example: XeF2  Six electron groups with lone pairs o Square pyramidal  Example: BrF5 o Square planar  Lone pairs 180o apart  Example: XeF4

.

           

Unnumbered figure: Lewis structure and geometry of NH3 Unnumbered figure: ideal and actual bond angles for NH3 Figure 10.3 Nonbonding versus Bonding Electron Pairs Unnumbered figure: Lewis structure and geometry of H2O Unnumbered figure: ideal and actual bond angles for H2O Figure 10.4 The Effect of Lone Pairs on Molecular Geometry Unnumbered figure: Lewis structure and geometry of SF5 Unnumbered figure: Lewis structure and geometry of BrF3 Unnumbered figure: Lewis structure and geometry of XeF2 Unnumbered figure: Lewis structure and geometry of BrF5 Unnumbered figure: Lewis structure and geometry of XeF4 Table 10.1 Electron and Molecular Geometries

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

10.3 VSEPR Theory: The Effect of Lone Pairs  Models are extremely useful for showing both kinds of geometry.  Students may be tempted to memorize Table 10.1 (using it for the first few examples is reasonable), but show them how the information can be obtained by logic.  Conceptual Connection 10.2 Molecular Geometry and Electron Groups

 



.

Lone pair electrons affect molecular shape.

VSEPR theory refers to all electron groups, not just to atoms.

Drawing good Lewis structures is essential to predicting geometry. Laziness with Lewis structures leads to omitted lone pairs, leading to incorrect shapes.

Students are tempted to place lone pairs at the axes of the trigonal bipyramid rather than in the trigonal plane.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

10.4 VSEPR Theory: Predicting Molecular Geometries  Procedure o Draw Lewis structure. o Determine total number of electron groups on central atom. o Determine number of bonding groups and number of lone pairs on central atom. o Use Table 10.1 to identify electron and molecular geometry.  Drawing three‐dimensional shapes on paper  Shapes of molecules with more than one central atom

    

10.5 Molecular Shape and Polarity  Polar bonds  Net dipole moment  Adding dipoles: vector addition o One dimension o Two dimensions o Three dimensions o Common cases (Table 10.2)  Macroscopic consequences of molecular polarity

         

10.6 Valence Bond Theory: Orbital Overlap as a Chemical Bond  Valence bond theory o Interaction energy diagram o Overlap of atomic orbitals o Shape determined by geometry of overlapping orbitals

 

.

Examples 10.2 and 10.3 Predicting Molecular Geometries Unnumbered figure: geometries of carbon atoms in glycine Unnumbered figure: Lewis structure and geometry of glycine Unnumbered figure: representing molecular geometries on paper Example 10.4 Predicting the Shape of Larger Molecules

Unnumbered figure: model and electron density plot of HCl Unnumbered figure: Lewis structure, model, and electron density plot of CO2 Unnumbered figure: Lewis structure, model, and electron density plot of H2O Unnumbered figure: vector addition Table 10.2 Common Cases of Adding Dipole Moments to Determine whether a Molecule Is Polar Example 10.5 Determining Whether a Molecule Is Polar Figure 10.5 Interaction of Polar Molecules Unnumbered figure: photo of oil and water Unnumbered figure: photo of marbles illustrating interaction of polar molecules with nonpolar molecules Chemistry in Your Day: How Soap Works Figure 10.6 Energy Diagram for H2 Unnumbered figure: orbital diagram of H and S and orbital overlap illustration of H2S

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

10.4 VSEPR Theory: Predicting Molecular Geometries  Show several examples using the procedure; emphasize the importance of beginning with a good Lewis structure.  Introduce the conventions for demonstrating directionality on paper.  Shapes of larger molecules are approached by solving the shape for each atom that can be at the centre of a set of electron groups.



All pairs of electrons are critical to the procedure.

10.5 Molecular Shape and Polarity  Molecular modelling and rendering software (e.g. CHIME, RASMOL, JMOL) can show surface charge maps and structures that can be rotated.  Net dipole moments require determining the sum of dipole moments along polar bonds.

Review the vector addition rules and examples.  Polarity or nonpolarity affects macroscopic properties like boiling point (see Chapter 11).



Molecules with polar bonds can have a zero net dipole moment. Individual dipoles often add to zero in the basic geometries.

10.6 Valence Bond Theory: Orbital Overlap as a Chemical Bond  Electrostatic attraction and repulsion account for the curve when energy is plotted versus internuclear distance in an interaction energy diagram.  This section also reviews orbital diagrams from Chapter 8.  Conceptual Connection 10.3 What Is a Chemical Bond, Part I?



Valence bond theory uses atomic orbital overlap to create bonds in molecules. For some, the difficulty may simply involve being able to visualize the overlap, i.e., sort out the portions of the figures.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

10.7 Valence Bond Theory: Hybridization of Atomic Orbitals  Hybridization and hybrid orbitals o sp3  Four equivalent orbitals  Tetrahedral arrangement 2 o sp  Three equivalent orbitals with a p orbital remaining  Trigonal planar arrangement o Sigma bonds and pi bonds  Rotation restricted for pi bonds o sp hybridization  Two equivalent orbitals + 2p orbitals remaining o sp3d hybridization  Five equivalent hybrid orbitals 3 2 o sp d hybridization  Six equivalent hybrid orbitals  Determining hybridization and drawing valence bond models o Lewis structure o VSEPR geometry o Hybridization o Molecular sketch  Hybrid orbitals  Sigma and pi bonds

                     

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Unnumbered figure: orbital diagram of H and C and geometry of CH4 Unnumbered figure: hybridization of carbon’s atomic orbitals Figure 10.7 sp3 Hybridization Unnumbered figure: valence bond model of CH4 Unnumbered figure: orbital overlap model of NH3 Unnumbered figure: orbital diagrams showing sp2 hybridization Unnumbered figure: orbital diagrams of H, C, and O and hybridization of carbon’s atomic orbitals Figure 10.8 sp2 Hybridization Unnumbered figure: valence bond model of H2CO Figure 10.9 Sigma and Pi Bonding Unnumbered figure: valence bond model of H2CO Unnumbered figure: valence bond models of 1,2‐ dichloroethane and 1,2‐dichloroethene Unnumbered figure: Lewis structures and models of cis‐1,2‐dichloroethene and trans‐1,2‐ dichloroethene Chemistry in Your Day: The Chemistry of Vision Figure 10.10 sp Hybridization Unnumbered figure: orbital diagrams showing sp hybridization Unnumbered figure: orbital diagrams of H and C and sp hybridization of C in acetylene Unnumbered figure: Lewis structure and valence bond model of C2H2 Unnumbered figure: Lewis structure and valence bond model of SF6 Table 10.3 Hybridization Scheme from Electron Geometry Examples 10.6 and 10.7 Hybridization and Bonding Scheme Example 10.8 Hybridization and Bonding Scheme

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

10.7 Valence Bond Theory: Hybridization of Atomic Orbitals  s and p orbitals are not equivalent, but the combination of s and p creates equivalent hybrid orbitals.  Sigma bonds allow free rotation around the bond axis, but pi bonds restrict rotation. The box about the chemistry of vision is an excellent example.  Table 10.3 summarizes the hybridization scheme from the electron geometry of hybrid orbitals.  Conceptual Connection 10.4 Single and Double Bonds





.

Hybrid orbitals are a means to create a set of equivalent orbitals for bonding, but they are not meant to exist until the atoms form the bond.

The shapes of the hybrid orbitals may not be obvious from the shapes of the pure s and p orbitals. For example, the small portion of a hybrid orbital may be confusing since it does not appear to get involved with the bond.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

10.8 Molecular Orbital Theory: Electron Delocalization  Linear combinations of atomic s orbitals o Constructive: bonding o Destructive: antibonding o Molecular orbital diagrams  H2, He2, He2+  Bond order  Linear combinations of atomic p orbitals o Shapes of bonding and antibonding orbitals o Period 2 homonuclear diatomics o 2s–2p mixing  Paramagnetism and diamagnetism  Liquid oxygen o Period 2 heteronuclear diatomic molecules o Polyatomic molecules  Electron delocalization in ozone, benzene

                        

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Unnumbered figure: bonding MO for H2 Unnumbered figure: antibonding MO for H2 Figure 10.11 Formation of Bonding and Antibonding Orbitals Unnumbered figure: MO diagram for H2 Unnumbered figure: MO diagram for He2 Unnumbered figure: MO diagram for He2+ Example 10.9 Bond Order Unnumbered figure: MO diagram for Li2 Unnumbered figure: MO diagram for Be2 Unnumbered figure: sigma bonding and antibonding interactions of 2p orbitals Unnumbered figures: pi bonding and antibonding interactions of 2p orbitals Unnumbered figure: MO diagram for B2, C2, N2 Unnumbered figure: MO diagram for O2, F2, Ne2 Figure 10.12 The Effect of 2s–2p Mixing on the Molecular Orbital Diagrams for Second‐ Period Diatomic Molecules Figure 10.13 Molecular Orbital Energy Diagrams for Second‐Row p‐Block Homonuclear Diatomic Molecules Unnumbered figure: photo and illustration of O2 paramagnetism Unnumbered figure: valence bond model of O2 Example 10.10 Molecular Orbital Theory Unnumbered figure: MO diagram for NO Figure 10.14 Shape of 2s Bonding Orbital in NO Unnumbered figure: MO diagram for HF Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic Molecules and Ions Unnumbered figure: Lewis structure and valence bond model of O3 Unnumbered figure: molecular orbital model of O3 Unnumbered figure: resonance structures and molecular orbital model of benzene

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

10.8 Molecular Orbital Theory: Electron Delocalization  Molecular orbital theory gives rise to a bonding pair orbital (lower in energy) and a nonbonding orbital (higher in energy). The stability of a molecule or ion composed of two atoms depends on the number of valence electrons that must be used and the orbital energy levels into which those electrons are populated. A stable species must have an overall lower energy relative to that of the constituent atoms.  Both sigma and pi have bonding and antibonding orbitals with corresponding appropriate energy levels. Nonbonding molecular orbitals don't help or hinder since they have the same energy as the corresponding atomic orbital.  The 2s–2p mixing argument explains the paramagnetism of molecular oxygen.  Heteronuclear diatomic molecules demonstrate the effect of different electronegativities and atomic orbital energies on MO formation.  MO theory works for more complex molecules and explains the electron delocalization in ozone, benzene, nitrate, and other molecules.  Conceptual Connection 10.5 What Is a Chemical Bond, Part II?

 

.

MO theory is presented from a pictorial point of view, but the details are very mathematical.

Oxygen is paramagnetic, illustrating that while Lewis theory is useful, it does have limitations.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

Additional Problems for Predicting Molecular Geometries (Examples 10.2 and 10.3)

Predict the geometry and bond angles of PCl5.

Predict the geometry and bond angles of SOCl2.

Lewis Structure

PCl5 has 40 electrons.

SOCl2 has 26 electrons.

Central atom P has 5 electron groups (single bonds).

Central atom S has 4 electron groups (2 single bonds, 1 double bond, 1 lone pair).

Central atom P has…

Central atom S has…

5 bonding groups

3 bonding groups

0 lone pairs

1 lone pair

The electron geometry is trigonal bipyramidal and molecular geometry is trigonal bipyramidal.

The electron geometry is tetrahedral and molecular geometry is trigonal pyramidal.

Bond angles: 120o equatorial

Bond angles: <109.5o

Draw a Lewis structure for the molecule.

Electron Groups Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, triple bonds, and single electrons each count as one group.

Bonding Groups and Lone Pairs Determine the number of bonding groups and number of lone pairs around the central atom.

Electron and Molecular Geometries Determine the geometries using the list in Table 10.1.

90o axial

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

Additional Problems for Hybridization and Bonding Scheme (Examples 10.6 and 10.7)

Write a hybridization and bonding scheme for COCl2.

Write a hybridization and bonding scheme for SO42-.

Lewis Structure

COCl2 has 24 electrons.

SO42- has 32 electrons.

The central atom C has three electron groups and therefore a trigonal pyramidal electron geometry.

The central atom S has four electron groups and therefore a tetrahedral electron geometry.

The trigonal pyramidal electron geometry corresponds to sp2 hybridization.

The tetrahedral electron geometry corresponds to sp3 hybridization.

Draw a Lewis structure for the molecule.

Electron Geometry from VSEPR Use VSEPR theory to predict the electron geometry about the central atom.

Central Atom Hybridization Select the correct hybridization for the interior atom(s) based on electron geometry.

Sketch Sketch the molecule, beginning with the central atom and its orbitals.

Label Label all bonds using the s or p notation and identify the type of overlapping orbitals.

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Use molecular orbital theory to predict the bond order in HeH+, the cation of a helium and hydrogen atom.

Additional Problem for Bond Order (Example 10.9)

Count Electrons Begin by sorting the information in the problem into Given and Find.

H: 1 electron He: 2 electrons + charge = 1 electrons total

= 2 electrons

Energy Diagram HeH+ has two electrons. Assign the electrons to the molecular orbitals, filling lower energy orbitals first and proceeding to higher orbitals.

Calculate Bond Order

bonding  antibonding 2 2  0 = =1 2

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and dividing by two.

HeH+ bond order =

Result Since the bond order is positive, HeH+ should be stable.

Is it stable?

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

Draw an MO energy diagram and determine the bond order for O22. Is it diamagnetic or paramagnetic?

Additional Problem for Molecular Orbital Theory (Example 10.10)

Energy Level Diagram Write an energy level diagram for the molecular orbitals in O22. Use the energy ordering for O2.

Count Valence Electrons Count the valence electrons for the O22 ion.

O atom = 6 electrons O atom = 6 electrons 2 charge = 2 electrons total

.

= 14 electrons

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Chapter 10. Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Fill Molecular Orbitals The O22 ion has 14 valence electrons. Assign the electrons to molecular orbitals with the lowest energy, following Hund’s rule.

Calculate Bond Order

bonding  antibonding 2 8  6 = =1 2

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and dividing by two.

O22 bond order =

Results The bond order is 1. The O22 ion has no unpaired electrons and is neither diamagnetic nor paramagnetic.

The ion results from the removal of two protons H+ from hydrogen peroxide H–O–O–H.

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Chapter 11. Liquids, Solids, and Intermolecular Forces Student Objectives 11.1 Climbing Geckos and Intermolecular Forces  

Know that intermolecular forces are attractive forces between individual molecules.

Know that the gecko’s ability to climb smooth surfaces is due to intermolecular forces.

11.2 Solids, Liquids, and Gases: A Molecular Comparison   

Know the properties that differentiate the phases of matter: density, molar volume, molecular shape, and strength of intermolecular forces.

Define crystalline and amorphous and recognize the difference in solids.

Know that both temperature and pressure can affect phase changes.

11.3 Intermolecular Forces: The Forces That Hold Condensed States Together          

Know and understand that intermolecular forces originate from the interactions between charges, partial charges, and temporary charges on molecules, atoms, and ions.

Know how Coulomb’s law describes the mathematical relationship between energy of attraction, magnitude of charge, and distance.

Know and understand that dispersion (London) forces result from fluctuations of electron distribution within molecules and atoms.

Identify and predict how the shape and sizes of molecules or atoms affects the magnitude of dispersion forces the particles exhibit as well as macroscopic physical properties like boiling point.

Know and understand that polar molecules have permanent dipoles that attract each other through dipole–dipole interactions.

Know and understand the phenomenon of hydrogen bonding.

Predict the ability of molecules to exhibit hydrogen bonding.

Recognize hydrogen bonding as the force that holds double‐stranded DNA together.

Rank a series of molecular compounds with respect to boiling point.

Know and understand that the interaction of ions and dipoles leads to the dissolution and solvation of ions by water and other polar liquids.

11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action    

Know and understand that surface tension is due to intermolecular forces.

Describe examples of surface tension.

Know and understand that viscosity is due to intermolecular forces, mass, shape, and length.

Know and understand that capillary action is the result of both cohesive and adhesive forces.

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Chapter 11. Liquids, Solids, and Intermolecular Forces 11.5 Vaporization and Vapour Pressure        

Understand the process of vaporization and how it changes with temperature, surface area, and the degree of intermolecular forces.

Understand that molecules or atoms have a distribution of thermal energies that change as a function of temperature.

Know that the heat of vaporization, vapH, is a quantitative measure for the process of vaporization.

Calculate and interconvert mass, moles, and energy using the heat of vaporization.

Know and understand how vapour pressure and dynamic equilibrium dictate vaporization and condensation.

Know that the vapour pressure of a liquid depends on temperature and that the boiling point of a liquid depends on the external pressure.

Use the Clausius–Clapeyron equation to relate temperature and vapour pressure.

Define critical temperature and critical pressure.

11.6 Sublimation and Fusion   

Define and understand sublimation and deposition.

Define and understand fusion in the context of phase changes.

Use the heat of fusion, fusH in calculations involving energy, masses, and moles.

11.7 Heating Curve for Water  

Understand the different segments in the heating curve for H2O that ranges from below the melting point to above the boiling point.

Calculate the energy changes associated with heating a substance (like H2O) through a series of temperature changes and phase changes.

11.8 Phase Diagrams   

Know that a phase diagram relates the states of matter for a substance to temperature and pressure.

Identify the main regions and significant points in a phase diagram.

Understand the effect of changes in temperature and changes in pressure on the phase of a substance as shown by its phase diagram.

11.9 Water: An Extraordinary Substance  

Know that water has unique properties compared with similar molecules based on size, constituent atoms, and molar mass.

Know that the unique properties of water are attributable to hydrogen bonding.

11.10 Crystalline Solids: Determining Their Structure by X‐Ray Crystallography    

Recall that waves can interfere constructively and destructively.

Know that X‐rays diffract when interacting with the atoms in crystalline solids, forming diffraction patterns.

Know that diffraction patterns can be analyzed and used to identify the three‐dimensional structure of the atoms or molecules in a crystalline solid.

Use Bragg’s law to calculate the relationship between the distance between crystalline layers, the wavelength of electromagnetic radiation, and the angle of reflection.

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Chapter 11. Liquids, Solids, and Intermolecular Forces 11.11 Crystalline Solids: Unit Cells and Basic Structures     

Define and identify unit cells.

Know and identify the cubic crystalline lattice types: simple, body‐centred, and face‐centred.

Identify the kind of unit cell, the coordination number, and the edge length for the three cubic crystalline lattice types.

Use the kind of unit cell and the radius of an atom to calculate the density of a metal.

Identify the hexagonal and cubic closest‐packing structures, and know their unit cells and component layers.

11.12 Crystalline Solids: The Fundamental Types   

Know the organization of crystalline solids—molecular, ionic, and atomic—including basic properties and examples.

Know and identify constituent atoms, lattice types, and unit cells for some common ionic solids: CsCl, NaCl, ZnS, and CaF2.

Know and identify atomic solid types—nonbonding, metallic, and network covalent—and some of their properties and examples.

11.13 Crystalline Solids: Band Theory 

Know that the organization of conduction and valence bands of molecular orbitals forms the basis for conductors, semiconductors, and insulators.

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Chapter 11. Liquids, Solids, and Intermolecular Forces Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 11. Liquids, Solids, and Intermolecular Forces Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

11.1 Climbing Geckos and Intermolecular Forces  Intermolecular forces defined  Microhairs on a gecko’s feet and intermolecular forces



11.2 Solids, Liquids, and Gases: A Molecular Comparison  Phases of matter o Gas o Liquid o Solid  Phase properties o Density, volume, mass o Shape  Phase changes o Temperature o Pressure

  



   

.

Intro figure: geckos and intermolecular forces Unnumbered figure: electron micrograph of microhairs on a gecko’s feet

Table 11.1 The Three States of Water Table 11.2 Properties of the States of Matter Figure 11.1 Liquids Assume the Shapes of Their Containers Figure 11.2 Gases Are Compressible Figure 11.3 Crystalline and Amorphous Solids Unnumbered figure: molecular representation of phase changes Unnumbered figure: molecular illustration of propane

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Chapter 11. Liquids, Solids, and Intermolecular Forces Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

11.1 Climbing Geckos and Intermolecular Forces  The gecko’s feet provide yet another illustration of practical chemistry.

11.2 Solids, Liquids, and Gases: A Molecular Comparison  The phases of matter provide a meaningful connection between the microscopic level and the macroscopic level.  Conceptual Connection 11.1 State Changes



.

Students easily grasp that steam is still H2O, but they sometimes don’t recognize for larger molecular compounds that boiling concerns only intermolecular forces rather than intramolecular.

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Chapter 11. Liquids, Solids, and Intermolecular Forces Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

11.3 Intermolecular Forces: The Forces That Hold Condensed States Together  Basics of intermolecular forces o Coulomb’s law  

 

E

   

1

qq  1 2 4 o r

  

Types of intermolecular forces o Dispersion (London) forces  Exhibited by all molecular compounds  Mass  Shape, length o Dipole–dipole forces  Exhibited by polar compounds  Permanent dipoles  Dipole moments o Hydrogen bonding  H—X, where X = N, O, F  Extreme case of dipole–dipole o Ion–dipole  Solution and solvation Boiling point as a measure of intermolecular forces Hydrogen bonding in DNA o Base pairing  A‐T  G‐C

                    

.

Unnumbered figure: floating candles Unnumbered figure: polarization of He atom Figure 11.4 Ion‐Induced Dipole Interactions Unnumbered figure: helium electrons frame by frame Figure 11.5 Dispersion Interactions Table 11.3 Boiling Points of the Noble Gases Unnumbered figure: molecular models of n‐pentane and neopentane Figure 11.6 Dispersion Force and Molecular Shape Figure 11.7 Boiling Points of the n‐Alkanes Figure 11.8 Dipole–Dipole Interaction Unnumbered table: formula, molar mass, structure, melting point, and boiling point of formaldehyde and ethene Figure 11.9 Dipole Moment and Boiling Point Figure 11.10 Polar and Nonpolar Compounds Example 11.1 Dipole–Dipole Forces Figure 11.11 Dispersion Forces and Melting Points for the Polar Trihalomethanes Figure 11.12 Hydrogen Bonding in HF Unnumbered table: comparison of ethanol and dimethyl ether Figure 11.13 Hydrogen Bonding in Ethanol Figure 11.14 Hydrogen Bonds versus Dipole– Dipole Interactions Figure 11.15 Boiling Points of Group 14, 15 and 16 Hydrides Example 11.2 Hydrogen Bonding Figure 11.16 Dipole‐Induced Dipole Forces Figure 11.17 Ion–Dipole Forces Table 11.4 Types of Intermolecular Forces Chemistry and Medicine: Hydrogen Bonding in DNA Figure 11.18 Nucleotides Figure 11.19 Complementary Base Pairing via Hydrogen Bonds Figure 11.20 Copying DNA

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Chapter 11. Liquids, Solids, and Intermolecular Forces Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

11.3 Intermolecular Forces: The Forces That Hold Condensed States Together  Intermolecular forces are based on the distribution of charge, ranging from fluctuations in electron distribution to polar bonds and permanent dipoles to ions.  Several tables and figures warrant exploration: o Table 11.3 Boiling Points of the Noble Gases o Figure 11.7 Boiling Points of the n‐Alkanes o Figure 11.9 Dipole Moment and Boiling Point o Unnumbered table: comparison of ethanol and diethyl ether  Hydrogen bonding accounts for many of the properties of water and aqueous solutions as well as important behaviours in organic and biochemical systems.









.

All intermolecular forces are due to the distribution of charge. Even in the case of dispersion forces, Coulomb’s law is applicable.

Hydrogen bonding is an example of a relatively strong intermolecular force, but it is still a weak interaction when compared with covalent bond energies.

Specific conditions must be met for hydrogen bonding. Not every compound containing hydrogen nor every compound containing hydrogen and oxygen/nitrogen/fluorine exhibits this intermolecular force.

The two molecules involved in a hydrogen bonding interaction can be identical or different.

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Chapter 11. Liquids, Solids, and Intermolecular Forces Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action  Surface tension o Model o Examples  Viscosity o Motor oil  Capillary action o Meniscus of water o Meniscus of mercury

         

11.5 Vaporization and Vapour Pressure  Vaporization and condensation o Distribution of thermal energy o Energetics, vapH o Calculations using vapH as a conversion factor  Vapour pressure and dynamic equilibrium o Closed systems o Piston: volume and temperature dependence  Temperature dependence of vapour pressure o Boiling point definition o Clausius–Clapeyron equation

   

 H vap  1     ln  R T 

  

 ln Pvap  

    

Critical point o Critical temperature o Critical pressure o Supercritical fluid

 

  .

Unnumbered figure: photo of trout fly and trout Figure 11.21 The Origin of Surface Tension Figure 11.22 Surface Tension in Action Figure 11.23 Spherical Water Droplets Table 11.5 Viscosity of Several Hydrocarbons at 20 oC Table 11.6 Viscosity of Liquid Water at Several Temperatures Chemistry in Your Day: Viscosity and Motor Oil Unnumbered figure: photo of drawing blood using capillary action Figure 11.24 Capillary Action Figure 11.25 Meniscuses of Water and Mercury

Figure 11.26 Vaporization of Water Figure 11.27 Distribution of Thermal Energy Unnumbered figure: sweating person Table 11.7 Heats of Vaporization of Several Liquids at Their Boiling Points and at 25 oC Example 11.3 Using the Heat of Vaporization in Calculations Figure 11.28 Vaporization in a Sealed Flask Figure 11.29 Dynamic Equilibrium Figure 11.30 Dynamic Equilibrium in Pentane Figure 11.31 Vapour Pressure of Several Liquids at Different Temperatures Figure 11.32 Boiling Figure 11.33 Temperature during Boiling Table 11.8 Boiling Points of Water at Several Locations of Varied Altitudes Figure 11.34 A Clausius–Clapeyron Plot for Diethyl Ether (CH3CH2OCH2CH3) Example 11.4 Using the Clausius–Clapeyron Equation to Determine Heat of Vaporization from Experimental Measurements of Vapour Pressure Example 11.5 Using the Two‐Point Form of the Clausius–Clapeyron Equation to Predict the Vapour Pressure at a Given Temperature Figure 11.35 Critical‐Point Transition 178

Chapter 11. Liquids, Solids, and Intermolecular Forces Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action  Students can provide other examples of surface tension (undissolved powder on a water surface, water bugs that walk on the surface of a pond); viscosity (maple syrup, corn syrup); capillary action (the wick of a candle, a sponge or paper towel to soak up a spill).



Surface tension, viscosity, and capillary action are macroscopic properties attributable to intermolecular forces.

11.5 Vaporization and Vapour Pressure  Vaporization occurs at the surface interface between a liquid and the atmosphere in part because it is easier for molecules to break free.  Like molecules in a gas, molecules in a liquid are characterized by a distribution of energies that change with temperature.

Evaporation has a cooling effect because the highest energy molecules escape, leaving behind a collection of molecules with a lower average kinetic energy.  Vaporization and dynamic equilibrium can be demonstrated with sealed containers.

Students experience this with water bottles.

Ask for other examples.  The supercritical fluid is a unique state of matter, combining aspects of liquid and gas phases. Supercritical carbon dioxide has a low critical temperature and can provide an exciting demonstration.  Supercritical carbon dioxide is now used in many commercial applications: as a solvent for reactions, extraction of food components (e.g., caffeine), and as an alternative to halogenated organic solvents in dry cleaning.  Conceptual Connection 11.2 Vapour Pressure



The difference between vaporization and boiling is that vaporization is a phase transition that occurs only at the surface, whereas boiling is a phase transition that can occur at any point within the liquid.

Vaporization is affected by temperature and pressure; the mathematical dependence on temperature is given by the Clausius– Clapeyron equation.

The rate of vaporization increases with surface area, but vapour pressure is a constant for a given liquid at a given temperature, regardless of the container size or liquid volume (greater than 0).

The two‐point form, especially of the Clausius–Clapeyron, presents some challenges for the student with weak algebra skills.

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Chapter 11. Liquids, Solids, and Intermolecular Forces Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

11.6 Sublimation and Fusion  Sublimation and deposition o Definitions o Examples  Fusion o Melting point o Energetics, fusH o vapH > fusH

 

11.7 Heating Curve for Water  Calculations o T < mp o T = mp o bp > T > mp o T = bp o T > bp



Figure 11.39 Heating Curve for Water

11.8 Phase Diagrams  Major features o s, l, g regions o Phase changes o Triple point o Normal mp, bp o Critical point  Navigation  Comparison of water, iodine, carbon dioxide

 

Figure 11.40 Phase Diagram for Water Figure 11.41 Navigation on the Phase Diagram for Water Figure 11.42 Phase Diagrams for Other Substances

11.9 Water: An Extraordinary Substance  Boiling point comparison of main‐group hydrides  Hydrogen bonding  Environmental solvent  Freezing water and biological systems

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   



   

.

Figure 11.36 The Sublimation of Ice Unnumbered figure: photo of ice crystals on frozen peas Unnumbered figure: photo of dry ice Figure 11.37 Temperature during Melting Table 11.9 Heats of Fusion of Several Substances Figure 11.38 Heat of Fusion and Heat of Vaporization

Unnumbered figure: photo of Phoenix Mars Lander Figure 11.43 Boiling Points of Main‐Group Hydrides Figure 11.44 Hydrogen Bonding in Water Unnumbered figure: photo of frozen and fresh lettuce Chemistry in the Environment: Water Pollution

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Chapter 11. Liquids, Solids, and Intermolecular Forces Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

11.6 Sublimation and Fusion  Provide some examples of sublimation and deposition: dry ice, the freezer burn example of frozen peas, formation of frost in winter, freeze‐dried foods, etc.  Sublimation and deposition are phase transitions between gas and solid without involvement of the liquid phase.



11.7 Heating Curve for Water  The heating curve for water plots temperature as a function of added heat. The largest requirement comes from the heat of vaporization. Figure 11.39 shows not only the plot but also microscopic views of each segment.  Conceptual Connection 11.3 Cooling of Water with Ice



11.8 Phase Diagrams  Compare the phase diagrams of water, iodine, and carbon dioxide. Iodine’s diagram is similar to that of many substances, but water's and carbon dioxide’s are relatively unique. For water, the slope of the fusion curve is opposite that of most substances, and for carbon dioxide the liquid exists only at relatively high pressures.  Conceptual Connection 11.4 Phase Diagrams







Students often are surprised to learn that freezing is an exothermic process, likely because their experience with freezing has mostly to do with water. Ask them what happens when molten iron turns to solid.

Students sometimes interpret fusion as going from liquid to solid rather than the reverse.

The heat of fusion and heat of vaporization are much larger than the molar heat capacity of any of the phases.

Students often are surprised to learn that melting/freezing, sublimation/deposition, and boiling/condensation are constant temperature processes.

These are relatively simple phase diagrams for pure substances. A more detailed phase diagram for water shows the different phases of ice. Mixtures also have phase diagrams.

11.9 Water: An Extraordinary Substance  The unique properties of water are attributable mostly to hydrogen bonding.  Water is the solvent for the environment.

Changes and negative impacts accompany the positive aspects of its role.  Ask students to suggest advantages and disadvantages of the lower density of solid water versus liquid water.

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Chapter 11. Liquids, Solids, and Intermolecular Forces Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

11.10 Crystalline Solids: Determining Their Structure

by X‐Ray Crystallography  Review of constructive and destructive interference  X‐ray diffraction o Contact with atomic nuclei o Interaction with multiple crystal layers: constructive interference  Bragg’s law o n = 2d sin 



11.11 Crystalline Solids: Unit Cells and Basic Structures  Solids and unit cells o Cubic structures  Equal edge lengths  90o angles  Types of cubic lattice structures o Simple o Body‐centred o Face‐centred  Lattice properties o Atoms per unit cell o Unit cell composition o Edge lengths o Density and unit cell structure  Closest‐packed structures o Hexagonal o Cubic o Unit cells and layer composition



    

           

.

Unnumbered figure: photo of crystalline mineral Unnumbered figure: photo of snowflake Unnumbered figure: illustration of wave interference Figure 11.45 Diffraction from a Crystal Figure 11.46 X‐Ray Diffraction Analysis Unnumbered figure: diffraction pattern of DNA

Unnumbered figure: illustration of simple cubic unit cell Unnumbered table: properties of the three cubic unit cells Figure 11.47 The Cubic Crystalline Lattices Figure 11.48 Simple Cubic Crystal Structure Unnumbered figure: illustration of body‐ centred cubic unit cell Figure 11.49 Body‐Cntered Cubic Crystal Structure Figure 11.50 Face‐Centred Cubic Crystal Structure Unnumbered figure: illustration of face‐ centred cubic unit cell Example 11.6 Relating Density to Crystal Structure Unnumbered figures: illustrations of closest‐ packed structures Figure 11.51 Hexagonal Closest‐Packing Crystal Structure Unnumbered figures: cubic closest‐packing structures Figure 11.52 Cubic Closest‐Packing Crystal Structure

182

Chapter 11. Liquids, Solids, and Intermolecular Forces Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

11.10 Crystalline Solids: Determining Their Structure

by X‐Ray Crystallography  Layers of atoms or molecules in a crystalline solid diffract X‐rays in manners that cause constructive as well as destructive interference. The diffraction patterns can be interpreted to yield information about atom distances, radii, and bond lengths. Ultimately, the structures of molecules, including large, complex molecules such as proteins, can be determined.

11.11 Crystalline Solids: Unit Cells and Basic Structures  Use models that can be moved and rotated to optimize students’ understanding of the cubic crystalline types.  Collections of crystal structure models are viewable in CHIME and RASMOL.  Cubic lattice structures have a common edge length, enabling calculations involving density, edge length, and mass.  Closest‐packing models are well illustrated. In particular, the layers and how they stack are shown in detail for the hexagonal and cubic closest‐packed structures. Models are likely to be beneficial; for a large class, they can be placed below a document camera.

.

 

Visualizing unit cells and the atoms per unit cell requires several perspectives.

Closest‐packed geometries are determined by how one layer of atoms is oriented with respect to the layer above and below.

183

Chapter 11. Liquids, Solids, and Intermolecular Forces Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

11.12 Crystalline Solids: The Fundamental Types  Solid types and properties o Molecular o Ionic  CsCl  NaCl  ZnS  CaF2 o Atomic  Nonbonding  Metallic  Network covalent

         

11.13 Crystalline Solids: Band Theory  Band theory  Energy levels of lithium molecules  Band gap o Conductors: no band gap o Semiconductors: small band gap o Insulators: large band gap

 

.

Figure 11.53 Types of Crystalline Solids Figure 11.54 Cesium Chloride Unit Cell Figure 11.55 Sodium Chloride Unit Cell Figure 11.56 Zinc Sulfide (Zinc Blende) Unit Cell Unnumbered figure: illustration of a tetrahedral hole Figure 11.57 Calcium Fluoride Unit Cell Figure 11.58 The Electron Sea Model Figure 11.59 Closest‐Packed Crystal Structures in Metals Figure 11.60 Network Covalent Atomic Solids Figure 11.61 The Structure of Quartz

Figure 11.62 Energy Levels of Molecular Orbitals in Lithium Molecules Figure 11.63 Band Gap

184

Chapter 11. Liquids, Solids, and Intermolecular Forces Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

11.12 Crystalline Solids: The Fundamental Types  The taxonomy of crystalline solid types includes many that are new to most students.  Crystalline solids can be composed of molecules, ions, or atoms. Atomic solids can be nonbonding, metallic, or network covalent.

Ionic solids and silicates are most commonly encountered previously by students.

11.13 Crystalline Solids: Band Theory  The organization of MO energy levels is shown for lithium molecules.  The behaviour of atoms in a solid can be organized by the relationship between occupied and unoccupied orbitals. The band gap classifies the material as a conductor, semiconductor, or insulator.

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Chapter 11. Liquids, Solids, and Intermolecular Forces

A polar molecule will exhibit dipole–dipole forces.

Which molecules will show dipole–dipole forces?

Additional Problem for Dipole–Dipole Forces (Example 11.1)

a) CHF3 b) OCS c) H2S

Solution 1) Electronegativity values from Figure 9.12: H = 2.1 C = 2.5 F = 4.0 2) There are three polar C–F bonds. The net dipole moment points down along the H–C bond.

Solution 1) Electronegativity values from Figure 9.12: O = 3.5 C = 2.5 S = 2.5 2) The C=S bond is not polar; the C=O bond is polar. The net dipole moment points along the C=O bond.

Solution 1) Electronegativity values from Figure 9.12: S = 2.5 H = 2.1 2) Each H–S bond is only weakly polar. A very small net dipole exists and points toward S bisecting the H–S–H angle.

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Chapter 11. Liquids, Solids, and Intermolecular Forces

Additional Problem for Using the Heat of Vaporization in Calculations (Example 11.3)

Calculate the energy in kJ necessary to vaporize 100 g of water at its boiling point.

Sort

Given 100 g water

You are given a certain mass of water and asked what energy is necessary to vaporize it.

Find kJ

Strategize

Conceptual Plan

The heat of vaporization gives the relationship between heat absorbed and moles of water vaporized. Begin with the grams of water and convert to moles using the molar mass. Then use vapH as a conversion factor to obtain kJ of energy.

g



mol

1 mol H2O 18.02 g H2O



kJ

40.7 kJ 1 mol H2O

Relationships Used Hvap

= 40.7 kJ mol-1 (at 100 oC)

1 mol H2O = 18.02 g H2O

Solve

Solution

Follow the conceptual plan to solve the problem.

100 g H2O 

1 mol H2O 18.02 g H2O



40.7 kJ = 226 kJ 1 mol H2O

The units of the answer are correct and the magnitude makes sense. The 100 g sample is slightly more than 5 mol.

Check

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Chapter 11. Liquids, Solids, and Intermolecular Forces

Additional Problem for Using the TwoPoint Form of the Clausius–Clapeyron Equation to Predict the Vapour Pressure at a Given Temperature (Example 11.5)

Octane has a normal boiling point of 125.5 oC and a heat of vaporization (vapH) of 38.8 kJ mol-1. What is the vapour pressure of octane at 20 oC?

Sort

Given

The problem gives you the normal boiling point of octane (the temperature at which the vapour pressure is 760 mmHg) and the heat of vaporization. You are asked to find the vapour pressure at a different temperature.

T1 = 125.5 oC

P1 = 760 mmHg vapH = 38.8 kJ mol-1 T2 = 20 oC

Find P2

Strategize

Conceptual Plan

Use the Clausius–Clapeyron equation to find P2.

ln

Hvap  1 P2 1 =    P1 R  T1 T2 

Relationships Used Clausius–Clapeyron equation (two-point form) T (K) = oC + 273.15

Solve

Solution

Convert T1 and T2 from oC to K.

T1 (K)

= 125.5 + 273.15 = 398.65 K

T2 (K)

= 20 + 273.15 = 293.15 K

Substitute into the equation and solve for P2.

ln

P2 P1

=

P2 ln = 760 torr

H vap  1 1    R  T1 T2  J  1 1 mol     =  4.21 J 293.15 K   398.65 K 8.314 mol K

38.8  103

= 760 torr (e 4.21 )

P2

= 760 torr (0.0148) = 11.3 torr

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188

Chapter 11. Liquids, Solids, and Intermolecular Forces Check

The units of the answer (Torr) are correct. The magnitude of the answer (11.3) makes physical sense since octane is not a very volatile molecule.

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189

Chapter 11. Liquids, Solids, and Intermolecular Forces

Additional Problem for Relating Density to Crystal Structure (Example 11.6)

A form of iron crystallizes with a body-centred unit cell. The radius of iron is 124 pm. Calculate the density of this form of iron in g cm-3.

Sort

Given r = 124 pm, body-centred cubic

You are given the radius of an iron atom and its crystal structure. You are asked to find the density of solid iron.

Find d

Strategize

Conceptual Plan

The conceptual plan is based on the definition of density.

l = 4r / 3 V = l3 [volume of a cube with a length of l] m = mass of unit cell

For a body-centred cubic unit cell, the atoms touch along the centre diagonal. The cell length can be derived in terms of the atom radius.

= number of atoms in unit cell x mass of each atom

d =m/V

Solve

Solution

First, find the length of the unit cell using the equations relating the length to the radius of an atom in the body-centred cell.

l = 4r / 3 = 4  1.24 x 108 cm / 1.73 = 2.87  10-8 cm

Second, convert the length to a volume by cubing the value.

V = l3 = 23.6  1024 cm3

Find the mass of the iron atoms in the cell. The body-centred cell contains two iron atoms. Use the molar mass and Avogadro’s number to find the mass of each atom.

m = 2 atoms Fe 

55.85 g Fe 1 mol Fe  1 mol Fe 6.022  10 23 atoms Fe

= 1.85  10 22 g

The density is the mass divided by the volume.

Compute the value.

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190

Chapter 11. Liquids, Solids, and Intermolecular Forces m V 1.85  1022 g = 23.6  10 24 cm3 = 7.84 g/cm3

d =

The units (g) are correct. The magnitude of the answer (7.84) seems to make physical sense.

Check

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191

Chapter 12. Solutions Student Objectives 12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater  

Define solution, solute, and solvent.

Understand that water in tissue will combine with saltwater to dilute it, causing dehydration.

12.2 Types of Solutions and Solubility     

Know and understand that common solutions can have a gas, liquid, or solid as either the solvent or solute.

Define entropy and know that it causes the mixing that results in solution formation.

Know that the common types of intermolecular forces determine whether solutions will form when components are mixed.

Identify some common laboratory solvents.

Identify organic substances as water soluble or fat soluble.

12.3 Energetics of Solution Formation  

Know and understand the components of solution formation and the energy changes associated with them.

Define and understand heat of hydration and heat of solution.

12.4 Solution Equilibrium and Factors Affecting Solubility    

Know and understand the differences among the states that describe solution equilibrium and dissolution: unsaturated, saturated, and supersaturated.

Know the temperature dependence of the solubility of solids.

Know that the solubility of gases is a function of both temperature and pressure.

Use Henry’s law to calculate molar concentrations of gases in solution.

12.5 Expressing Solution Concentration   

Define the different expressions of solution concentration: molarity, molality, parts by mass, parts by volume, mole fraction, and mole percent.

Know how to prepare a solution of known concentration.

Convert between the different units of concentration.

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192

Chapter 12. Solutions 12.6 Colligative Properties: Vapour Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure            

Define colligative property.

Understand dynamic equilibrium with respect to vapour pressure in solutions and the effect of a solute on the rate of vaporization.

Use Raoult’s law to calculate the vapour pressure of a solution.

Understand vapour pressure for solutions containing two volatile components, and understand deviations from Raoult’s law for nonideal solutions.

Calculate the vapour pressure of a solution containing two volatile components.

Understand the basis for freezing point depression.

Calculate the freezing point depression of a solution from its molality and vice versa.

Understand the basis for boiling point elevation.

Calculate the boiling point elevation of a solution from its molality and vice versa.

Understand osmosis and osmotic pressure.

Calculate the osmotic pressure of a solution.

Use colligative properties to calculate the molar mass of an unidentified solute.

12.7 Colligative Properties of Strong Electrolyte Solutions   

Understand the difference between colligative properties of nonelectrolytes and electrolytes.

Calculate the van’t Hoff factor from deviations in freezing point depression, boiling point elevation, and osmotic pressure.

Define hyperosmotic, hyposmotic, and isosmotic in relation to biological cells.

12.8 Colloids     

Define colloidal dispersion or colloid.

Know the differences between the different kinds of colloid: aerosol, solid aerosol, foam, emulsion, and solid emulsion.

Understand Brownian motion in solutions.

Know the structure of a soap molecule and a micelle.

Know and understand the Tyndall effect.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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193

Chapter 12. Solutions Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater  Life of Pi by Yann Martel  Solution o Solvent o Solute  Spontaneous mixing o Two liquids o Solutions of different concentration o Saltwater in biological tissues



12.2 Types of Solutions and Solubility  Common types of solutions o Gas–gas o Liquid  Gas–liquid  Liquid–liquid  Solid–liquid o Solid–solid  Solubility o Entropy o Intermolecular forces o Solvents

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        

12.3 Energetics of Solution Formation  Energetics o soluteH o solventH o solnH = soluteH + solventH + mixH  Aqueous solutions o Heat of solution o Heat of hydration  Hhydration = Hsolvent + mixH  solnH = soluteH + hydH

     

.

Intro figure: rowboat on an ocean, osmotic pressure, Life of Pi (book) Figure 12.1 A Typical Solution Unnumbered figure: illustration of water flow across a cell membrane Figure 12.2 The Tendency to Mix

Unnumbered figure: photo and illustration of dissolved CO2 in soda water Table 12.1 Common Types of Solutions Figure 12.3 Spontaneous Mixing of Two Ideal Gases Figure 12.4 Intermolecular Forces Involved in Solutions Figure 12.5 Forces in a Solution Table 12.2 Relative Interactions and Solution Formation Table 12.3 Common Laboratory Solvents Example 12.1 Solubility Unnumbered table: solubility in water of first five straight‐chain alcohols Chemistry in Your Day: Bisphenol A Unnumbered figure: separation of solute particles Unnumbered figure: separation of solvent particles Unnumbered figure: mixing of solute and solvent particles Figure 12.6 Energetics of the Solution Process Figure 12.7 Heat of Hydration and Heat of Solution Figure 12.8 Ion–Dipole Interactions

194

Chapter 12. Solutions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater  Mentioning Life of Pi helps explain why drinking seawater is detrimental despite the fact that the mixture is mostly water.  Spontaneous mixing is closely related to effusion and diffusion in gases.



The flow of water out of tissue or cells may not appear to be the same principle as spontaneous mixing.

12.2 Types of Solutions and Solubility  Solutions involve the intermolecular forces between solute particles, between solvent particles, and between the two kinds. This is a good opportunity to have the class predict the magnitude of these interactions.  Predicting the solubility of vitamins (Example 12.1) will require a consideration of which bonds and (functional) groups are polar.  Conceptual Connection 12.1 Solubility



Solutions can involve solute and solvent from each of the three phases. Many students may only have thought about liquids as solvents.

Solubility is a continuum. Many “insoluble” compounds are soluble to a small extent.

12.3 Energetics of Solution Formation  Have the students predict which energies are endothermic and exothermic.  Solubility can be predicted from knowledge of the heat of hydration and the lattice energy.





.

Heats of solution can be exothermic or endothermic, though students may presume only the former from practical experience.

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Chapter 12. Solutions Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

12.4 Solution Equilibrium and Factors Affecting Solubility  Equilibrium and solubility o Dissolution o Recrystallization o Solution states  Unsaturated  Saturated  Supersaturated  Factors affecting solubility o Temperature o Pressure  Henry’s law

        

12.5 Expressing Solution Concentration  Solution concentration units o Molarity (M) o Molality (m) o Mole fraction () o Mole percent (mol %) o Parts by mass  Percent (%)  Parts per million (ppm)  Parts per billion (ppb) o Parts by volume  Percent (%)  Parts per million (ppm)  Parts per billion (ppb)  Making a solution of known concentration  Interconverting units

      

.

Figure 12.9 Dissolution of NaCl Figure 12.10 Precipitation from a Supersaturated Solution Figure 12.11 Solubility and Temperature Unnumbered figure: photo of rock candy Unnumbered figure: photos of cold and warm soda pop Figure 12.12 Pop Fizz Unnumbered figures: illustration of gas solubility as a function of pressure Table 12.4 Henry’s Law Constants for Several Gases in Water at 25 oC Example 12.2 Henry’s Law

Chemistry in the Environment: Lake Nyos Table 12.5 Solution Concentration Terms Figure 12.13 Preparing a Solution of Known Concentration Example 12.3 Using Parts by Mass in Calculations Chemistry in the Environment: Pharmaceuticals and Personal Care Products Example 12.4 Calculating Concentrations Example 12.5 Converting Between Concentration Units

196

Chapter 12. Solutions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

12.4 Solution Equilibrium and Factors Affecting Solubility  The representation of solubility as a dynamic equilibrium helps explain saturation and the process of recrystallization. This requires a balance between dissolving and precipitating.  Solubility can be manifested in two distinct properties: the amount of solute that dissolves and the rate at which the solute dissolves.  The phenomenon of supersaturation is related to other ‘super’ phenomena: heating, cooling, etc.  The solubility of gases is anomalous relative to that of most substances but logically so when considering soluteH.  Conceptual Connection 12.2 Solubility and Temperature  Conceptual Connection 12.3 Henry’s Law



Saturation especially is a dynamic process in which solute can reversibly enter and leave the solution.

12.5 Expressing Solution Concentration  The most common solution concentration terms are presented. It should be pointed out that molarity is most widely used by chemists, but the others are common in other contexts and disciplines. Medicine uses a variety of units including ones not included in the text.  Parts per million and parts per billion don’t make much sense for large concentrations.

Give the students practical examples of these, such as in the expression of groundwater contaminants.  Environmental pollutants are often evaluated at low or very low concentrations. Every water treatment plant must report certain solutes and pollutants; the data for locality makes for an engaging discussion, as does detection limits and the meaning of ‘zero’ concentration.



Students sometimes question the need for molarity and molality instead of one or the other. The issue is conservation of mass and thus concentration for molality but the opposite for molarity.

.

197

Chapter 12. Solutions Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

12.6 Colligative Properties: Vapour Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure  Solute effect on vapour pressure o Dynamic equilibrium o Solute effect on solvent–solvent interactions  Raoult’s law o o Psolution   solvent Psolvent 

  

     

Vapour pressure of solution with two volatile components o Ideal solutions o Nonideal solutions Freezing point depression o Tf = m × Kf Boiling point elevation o Tb = m × Kb Osmosis o  = MRT

         

12.7 Colligative Properties of Strong Electrolyte Solutions  Van’t Hoff factor, i  Medical solutions o Isosmotic o Hyperosmotic o Hyposmotic

     

.

Unnumbered figure: photo of road salting Figure 12.14 Sodium Chloride Dissolved in Water Unnumbered figure: illustration of dynamic equilibrium between liquid and vapour Unnumbered figures: illustration of the effect of solute on vapour pressure of a liquid Unnumbered figures: illustrations of osmosis Example 12.6 Calculating the Vapour Pressure of a Solution Containing a Nonelectrolyte and a Nonvolatile Solute Figure 12.15 Behaviour of Nonideal Solutions Example 12.7 Calculating the Vapour Pressure of a Two‐Component Solution Unnumbered figure: phase diagrams of a pure liquid and a solution Unnumbered figure: photo of antifreeze/coolant Table 12.8 Freezing Point Depression and Boiling Point Elevation Constants for Several Liquid Solvents Example 12.8 Calculating the Freezing Point Depression Example 12.9 Calculating the Boiling Point Elevation Chemistry in Your Day: Antifreeze in Frogs Figure 12.16 An Osmosis Cell Example 12.10 Determining Osmotic Pressure

Table 12.8 Van’t Hoff Factors at 0.05 m Concentration in Aqueous Solution Figure 12.17 Ion Pairing Example 12.11 Van’t Hoff Factor and Freezing Point Depression Example 12.12 Calculating the Vapour Pressure of a Solution Containing an Ionic Solute Figure 12.18 Red Blood Cells and Osmosis Unnumbered figure: photos of a nurse and 0.9% sodium chloride solution

198

Chapter 12. Solutions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

12.6 Colligative Properties: Vapour Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure  Models show the effect of a solute on solvent–solvent interactions.  Solutes reduce the vapour pressure of a solution relative to the pure solvent. Vapour pressure lowering is the basis for the other colligative properties.  Freezing point depression and boiling point elevation are easy to demonstrate and explain with examples from the students’ experience. The equations provide a quantitative measure of the effects.  Plants and animals use several mechanisms to protect themselves from freezing. In addition to producing higher solute concentrations that take advantage of freezing point depression, some also have proteins that inhibit ice crystal formation, a different kinetic mechanism.  Osmosis is an important concept in biology and medicine. The last section shows the form of cells when being subjected to different concentrations.  Conceptual Connection 12.4 Raoult’s Law



12.7 Colligative Properties of Strong Electrolyte Solutions  Freezing point depression, boiling point elevation, and osmotic pressure depend on the number of particles produced by the solute upon dissociation.  Conceptual Connection 12.5 Colligative Properties



 



.

The vapour pressure of a solution is lower than the vapour pressure of the pure solvent.

The actual pressure is given by Raoult’s law.

Solutions can be ideal or nonideal depending on the strength of solute–solvent interactions.

The equations for freezing point depression and boiling point elevation give the change in temperature but not the final temperature.

For strong electrolytes, the equations for boiling point elevation, freezing point depression, and osmotic pressure are modified only by including the multiplicative van’t Hoff factor, i.

Strong electrolytes are considered to dissociate completely only in very dilute solutions, hence the variation in values of i.

199

Chapter 12. Solutions Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

12.8 Colloids  Types of colloidal dispersions o Aerosol o Solid aerosol o Foam o Emulsion o Solid emulsion  Properties o Brownian motion o Specific structure  Soap structure  Micelle structure o Tyndall effect

       

.

Figure 12.19 A Colloid Table 12.9 Types of Colloidal Dispersions Figure 12.20 Brownian Motion Figure 12.21 Structure of a Soap Figure 12.22 Micelle Structure Figure 12.23 The Tyndall Effect Unnumbered figure: photo of the Tyndall effect Figure 12.24 Micelle Repulsions

200

Chapter 12. Solutions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

12.8 Colloids  Colloids can be complex, but there are a number of common examples. It is not possible to show the molecular‐level representations since colloidal materials involve macromolecular‐sized particles.  The soap example is one that everyone has experienced.  The Tyndall effect can be demonstrated in several contexts.



.

A colloid is a substance that is not a solution but has a solute‐like substance evenly distributed or dispersed through the solvent.

201

Chapter 12. Solutions

Additional Problem for Henry’s Law (Example 12.2)

What pressure of nitrogen is required to keep the nitrogen concentration in a bottle of water at 0.12 M at 25 °C? Is this a reasonable value?

Sort

Given Snitrogen = 0.012 M

You are given the solubility of nitrogen and asked to find the pressure required to achieve this solubility.

Find Pnitrogen

Strategize

Conceptual Plan

Use Henry’s law to find the required pressure from the solubility. You will need the Henry’s law constant for nitrogen.

Snitrogen



Pnitrogen

Snitrogen = kH Pnitrogen

Relationships Used Sgas

= kH Pgas (Henry’s law)

kH (nitrogen) = 6.1  104 M atm-1 (from Table 12.4) Solve

Solution

Solve the Henry’s law equation for Pnitrogen and substitute the other quantities to compute it.

SN2  kH, N2  PN2 PN2 

SN2 kH, N2

0.12 M 6.1 104 M/atm  197 atm 

The units (atm) are correct. The magnitude of the answer (197) seems correct, because the concentration requested is the same as the one in a previous example for carbon dioxide, but CO2 is far more soluble than nitrogen. The outcome is not a reasonable pressure of nitrogen to use because it is dangerously high.

Check

.

202

Chapter 12. Solutions Additional Problem for Calculating Concentrations (Example 12.4)

A solution is prepared by dissolving 114 g of glucose (C6H12O6) in 0.500 kg of water. The final volume of the solution is 590 mL. Calculate each value for this solution: a) molarity b) molality c) percent by mass d) mole fraction e) mole percent.

Molarity To calculate molarity, first find the amount of glucose in moles from the mass and molar mass.

mol C6H12O6 = 114 g C6H12O6 

1 mol C6H12O6 = 0.633 mol 180 g C6H12O6

amount solute (in mol) volume solution (in L) 0.633 mol C6H12O6 = 0.590 L H2O

Molarity (M) =

Then divide the amount in moles by the volume of the solution in litres.

= 1.07 M

Molality To calculate molality, use the amount of glucose in moles and divide by the mass of water in kilograms.

amount solute (in mol) amount solvent (in kg) 0.633 mol C6H12O6 = 0.500 kg

Molality (m) =

= 1.27 m

Percent by Mass To calculate the percent by mass, divide the mass of the solute glucose by the sum of the masses of the solute and the solvent (water) and multiply by 100%.

mass solute  100% mass solution 114 g =  100% 114 g + 500 g = 18.6 %

Percent by mass =

Mole Fraction To calculate the mole fraction, first determine the amount of water in moles from the mass of water and its molar mass.

mol H2O = 500 g H2O 

.

1 mol H2O = 27.75 mol 18.02 g H2O

203

Chapter 12. Solutions Then divide the amount of glucose in moles by the total number of moles.

 solute =

nsolute nsolute + nsolvent

0.633 mol 0.633 mol + 27.75 mol = 0.0223 =

Mole Percent To calculate the mole percent, multiply the mole fraction by 100%.

mol % = nsolute x 100% = 0.0223 x 100% = 2.23%

.

204

Chapter 12. Solutions Additional Problem for Calculating the Boiling Point Elevation (Example 12.9)

A solution is prepared from 445 g of ethylene glycol (C2H6O2) and 500 g of water. This solution represents one that is 50% by volume ethylene glycol. At what temperature will the water in this solution boil?

Sort

Given 445 g of ethylene glycol

You are given the mass of the solute ethylene glycol and the mass of the water solvent. You are asked to find the new boiling point.

500 g of water

Find boiling point of water in solution

Strategize

Conceptual Plan

You need the molality of the solution to calculate the boiling point elevation. First, calculate the number of moles of ethylene glycol using its mass and molar mass.

g C2H6O2

Next, calculate the molality of the solution using the moles of solute and mass of solvent water.

molality

Calculate the boiling point elevation and add it to the normal boiling point of water.



mol C2H6O2

1 mol C2H6O 2 62.07 g C2H6O2



molality

mol C2H6O2 kg H2O

Tb

Tb = m  Kb





boiling point

bp = 100 °C + Tb

Relationships Used 1 mol C2H6O2 = 62.07 g C2H6O2 Tb

= m  Kb

Kb (water) bp

= 0.512 °C m-1

= 100 °C + Tb

Solve

Solution

Follow the conceptual plan.

mol C2H6O2 = 445 g C2H6O2 

Calculate the number of moles of ethylene glycol and then the molality.

Molality (m) =

1 mol C2H6O2 = 7.17 mol 62.07 g C2H6O2

amount solute (in mol) amount solvent (in kg) 7.17 mol C2H6O2 = 0.500 kg = 14.3 m

.

205

Chapter 12. Solutions

Determine the boiling point elevation and add it to the normal boiling point of water.

Tb = m  K b = 17.3 m 

0.512 oC m

= 8.86 oC

bp = 100 °C + 8.86 °C = 109 °C

Check

The units are correct. The magnitude of the answer (109) seems to make physical sense.

.

206

Chapter 12. Solutions

Additional Problem for Calculating the Vapour Pressure of a Solution Containing an Ionic Solute (Example 12.12)

A solution contains 0.481 mol of Na2SO4 and 10.0 mol water. Calculate the vapour pressure of the solution at 25 °C. The vapour pressure of pure water at 25 °C is 23.8 torr.

Sort

Given 0.481 mol Na2SO4

You are given the number of moles of each component of a solution and asked to find the vapour pressure of the solution. You are also given the vapour pressure of pure water at the desired pressure.

10.0 mol water Powater = 23.8 torr at 25 °C

Find Psolution Strategize

Conceptual Plan

Use Raoult’s law to solve this problem. Calculate solvent from the given amounts of solute and solvent.

water & Powater



Psolution

Psolution = water  Powater Relationships Used

The key to the problem is to understand the dissociation of sodium sulfate.

Psolution = water  Powater

Solve

Solution

Write an equation for the dissociation of sodium sulfate.

Na2SO4(s)  2 Na+(aq) + SO42(aq)

Since one mole of sodium sulfate dissociates into three moles of ions, the total number of moles of solute must be multiplied by three when computing the mole fraction.

H O 2

=

nH2O nH2O + (3  nNa2SO4 )

10.0 mol 10.0 mol + (3  0.481 mol) = 0.874 =

Use the mole fraction of water and vapour pressure of pure water to compute the vapour pressure of the solution.

Psolution = water  Powater = 0.874  23.8 torr = 20.8 torr The units of the answer (torr) are correct. The magnitude of the answer (20.8) makes physical sense since water dominates the mole fraction. The vapour pressure of the solution is very similar to pure water.

Check

.

207

Chapter 13. Chemical Kinetics Student Objectives 13.1 Catching Lizards  

Know that physiological processes (like a lizard’s metabolism) depend on temperature.

Know that the rate of chemical reactions generally depends on the concentration of the reactants and the temperature of the reaction.

13.2 The Rate of a Chemical Reaction     

Know that a rate is the change of a particular property with respect to time.

Know and understand that the rate of a chemical reaction is a change in concentration measured during a change in time.

Know the difference between average and instantaneous rates, and understand how each can be measured or estimated from a plot of concentration versus time (or from a table of data used to produce such a plot).

Calculate an average rate of reaction and predict a change in concentration from a plot of concentration versus time (or from a table of data used to produce such a plot).

Know that the rate of a reaction is measured experimentally and that instruments like a spectrometer or gas chromatograph are used to measure concentrations.

13.3 The Rate Law: The Effect of Concentration on Reaction Rate      

Know the general definition of a rate law and the meaning and significance of the rate order and rate constant.

Know that order of the reaction predicts the dependence of concentration on time.

Understand that the order of a reaction must be determined experimentally and not from a balanced reaction.

Know that the plot of rate (often in units of mol L‐1 s‐1) versus concentration produces a straight horizontal line for zero‐order reactions, a straight line with a positive slope for first‐order reactions, and a curved line with a positive slope for higher‐order reactions.

Use the method of initial rates to determine the order of a reaction from a table of concentration and initial rates.

Determine and identify the order of a reaction with multiple reactants by observing the effect of a concentration change on the rate of reaction for each independent reactant.

.

208

Chapter 13. Chemical Kinetics 13.4 The Integrated Rate Law: The Dependence of Concentration on Time            

Understand the difference between rate law and integrated rate law.

Know and understand the first‐order integrated rate law and that a plot of the natural logarithm of the reactant concentration versus time is linear.

Determine the rate constant from the slope of a plot of natural logarithm of reactant concentration versus time for a first‐order reaction.

Use the first‐order integrated rate law to find the reactant concentration at a given time or the time elapsed for a given concentration change.

Know and understand the second‐order integrated rate law and that a plot of the inverse reactant concentration versus time is linear.

Determine the rate constant from the slope of a plot inverse reactant concentration versus time for a second‐order reaction.

Use the second‐order integrated rate law to find the reactant concentration at a given time or the time elapsed for a given concentration change.

Know and understand the zero‐order integrated rate law and that a plot of the reactant concentration versus time is linear.

Determine the rate constant from the slope of a plot of reactant concentration versus time for a zero‐ order reaction.

Use the zero‐order integrated rate law to find the reactant concentration at a given time or the time elapsed for a given concentration change.

Define half‐life and understand how it can be identified from a plot of reactant concentration versus time.

Calculate the half‐life or predict the concentration of a reactant using the mathematical expressions for first‐ and second‐order reactions.

13.5 The Effect of Temperature on Reaction Rate        

Know that the temperature dependence of reaction rate is modelled by the Arrhenius equation, an equation that describes the rate constant k in terms of temperature, a frequency factor, and an activation energy.

Define activation energy and frequency factor.

Know that a transition state or activated complex represent a transient arrangement of atoms that occurs as reactants form products.

Draw and interpret a potential energy diagram, a plot of energy versus reaction progress, including the energy levels of the reactants, transition state, and products.

Know that atoms and molecules contain a distribution of energy that depends on temperature and that those exceeding the activation energy can undergo reaction.

Calculate the activation energy from a plot of the natural logarithm of the rate constant versus inverse temperature, ln k versus T‐1.

Know and understand the collision model: atoms or molecules need to come in contact or collide in order to react.

Know that the collision model accounts for the frequency factor in the Arrhenius equation and that effective collisions require the proper orientation between two reacting molecules.

.

209

Chapter 13. Chemical Kinetics 13.6 Reaction Mechanisms        

Know that most reactions do not occur in a single step on a molecular level and that the series of steps is called the reaction mechanism.

Know that the sum of the elementary steps must be equal to the overall chemical reaction.

Know that reaction intermediates are species formed by one elementary step and consumed by another.

Recognize unimolecular and bimolecular reactions.

Know that the rate‐determining step of a mechanism determines the rate law for the overall reaction.

Draw and interpret a potential energy diagram, a plot of energy versus reaction progress, for a reaction that involves multiple steps.

Understand the rate law resulting from a mechanism in which an intermediate is formed in a fast initial step.

Show that a given mechanism is consistent with an observed rate law.

13.7 Catalysis      

Know that a catalyst increases the rate of a reaction without being consumed by the reaction.

Know that catalysts lower the activation energy of a reaction, often by changing the mechanism or the nature of the transition state.

Know that an automobile’s catalytic converter modifies exhaust molecules by catalyzing reactions to produce less harmful emissions.

Know the definitions of homogeneous and heterogeneous catalysis and understand how they differ.

Know that catalysts are involved in important atmospheric reactions including the depletion of the ozone layer and the interconversion of other pollutants.

Know the general model for the behaviour of enzymes and understand their role as biological catalysts.

.

210

Chapter 13. Chemical Kinetics Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

211

Chapter 13. Chemical Kinetics Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

13.1 Catching Lizards  Dependence of activity/metabolism on temperature o Lizards and ectotherms o Chemical reactions



Intro figure: illustration of a lizard, glass of ice water, and the Arrhenius equation

13.2 The Rate of a Chemical Reaction  Rate examples o Speed (change in distance per time) o Weight loss (change in weight per time) o Chemical reactions (change in amount or concentration per time)  Expressions of rate o Mathematical relationships o Plots of concentration versus time o Average rate o Instantaneous rate  Measuring rates o Spectrometer o Gas chromatograph

 

Figure 13.1 The Rate of a Chemical Reaction Figure 13.2 Concentrations of Reactant and Product As a Function of Time for the Reaction of Gaseous H2 and I2, Forming HI Unnumbered table: data from Figure 13.2 Example 13.1 Expressing Reaction Rates Figure 13.3 The Spectrometer Figure 13.4 The Gas Chromatograph

   

13.3 The Rate Law: The Effect of Concentration on Reaction Rate  Rate laws o Rate = k[A]n o Rate constant, k o Order, n  Determining order o Plot of [A] versus time for n = 0, 1, 2 o Plot of rate versus [A] for n = 0, 1, 2 o Initial rates  Zero order: no change in rate  First order: linear dependence  Second order: square dependence

rate 2  [ A] 2    General:  rate 1  [A]1 

    

Figure 13.5 Reactant Concentration as a Function of Time for Different Reaction Orders Figure 13.6 Reaction Rate as a Function of Reactant Concentration for Different Reaction Orders Figure 13.7 Sublimation Unnumbered tables: initial rate data for examples Example 13.2 Determining the Order and Rate Constant of a Reaction

n

 Multiple reactants

.

212

Chapter 13. Chemical Kinetics Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

13.1 Catching Lizards  The discussion of lizard activity and metabolism related to temperature is an interesting example of chemical kinetics, though some students may not see the connection immediately. Ask the students to suggest other examples, like the souring of milk left at room temperature.

13.2 The Rate of a Chemical Reaction  There are many common examples of rates besides speed and weight loss. Solicit additional ones.  Figure 13.1 shows different representations of a fast and slow A + B  C reaction using symbols to represent the reactants and products. This can serve as a segue to a more mathematical representation, a plot of concentration versus time for reactants and products (Figure 13.2).  Spectrometry and gas chromatography are just two ways to measure quantities of molecules.



13.3 The Rate Law: The Effect of Concentration on Reaction Rate  The rate law for the general reaction A  products is rate = k[A]n.  First‐order reactions are much more intuitive for most students than the other reaction orders.  Reaction orders are most commonly positive integers, but they can be negative and fractional as well. Fractional orders make little sense to most students until the relevant mechanisms are explored.  The method of initial rates can be used for any reaction regardless of order because the rate is relatively constant in the first 5–10% of a reaction.  Conceptual Connection 13.1 Rate and Concentration







.

The language of kinetics can be confusing: rate, rate law, instantaneous rate, average rate, and rate constant all have different meanings.

Students often presume that energetically favourable reactions (e.g., formation of water from H2 and O2) are always fast.

The concept of reaction order is initially lost on some students. Reactions of the same order behave kinetically the same, as shown especially in Section 13.4.

The order must be determined experimentally. Students are used to placing much emphasis on stoichiometry from previous studies and naturally want to extend that emphasis to reaction order.

213

Chapter 13. Chemical Kinetics Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

13.4 The Integrated Rate Law: The Dependence of Concentration on Time  Integrated rate laws o First order



[A]t  kt [A]0



 plot of ln [A] versus t  slope = k  intercept = ln [A]0 Second order



1 1  kt  [A]t [A]0



 ln

o





 plot of 1/[A] versus t  slope = k  intercept = a/[A]0 o Zero order  [A]t = kt + [A]0 Half‐life o First order: t1/2 = 0.693/k o Second order: t1/2 = 1/k[A]0 o Zero order: t1/2 = [A]0/2k



13.5 The Effect of Temperature on Reaction Rate  Arrhenius equation





   



 Ea

o o

o

o

k  A  e RT

Potential energy diagram  Activation energy, Ea  Transition state (activated complex) Plot of ln k versus 1/T  Slope = Ea/R  Intercept = ln A Two‐point equation

 

k 2 Ea  1 1      k1 R  T1 T2 

 

 ln o

o

Exponential factor  Thermal energy distribution  Temperature dependence Frequency factor  Collision model  Collision orientation

.

 

Table 13.1 Atmospheric Lifetimes of Several CFCs Figure 13.8 First‐Order Integrated Rate Law Example 13.3 The First‐Order Integrated Rate Law: Using Graphical Analysis of Reaction Data Example 13.4 The First‐Order Integrated Rate Law: Determining the Concentration of a Reactant at a Given Time Figure 13.9 Second‐Order Integrated Rate Law Example 13.5 The Second‐Order Integrated Rate Law: Using Graphical Analysis of Reaction Data Figure 13.10 Zero‐Order Integrated Rate Law Figure 13.11 Half‐Life: Concentration versus Time for a First‐Order Reaction Example 13.6 Half‐Life, Lifetime, and Decay Time Table 13.2 Rate Law Summary Table Figure 13.12 The Activation Energy Barrier Unnumbered figures: molecular models of CH3NC, CH3CN Figure 13.13 The Activated Complex Figure 13.14 Thermal Energy Distribution Example 13.7 Using an Arrhenius Plot to Determine Kinetic Parameters Example 13.8 Using the Two‐Point Form of the Arrhenius Equation Figure 13.15 The Collision Model Unnumbered figures: illustrations of molecular collisions

214

Chapter 13. Chemical Kinetics Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

13.4 The Integrated Rate Law: The Dependence of Concentration on Time  Integrated rate laws are provided for the first‐order, second‐order, and zero‐order cases. In addition, plots of data are shown that yield straight lines from which the rate constant can be determined.  Use of the integrated rate laws gives better understanding of the importance of reaction order.  It is useful to demonstrate plotting reaction data using Excel or a similar spreadsheet program.  The linearity (or lack thereof) of the data discussed in this section can also be determined by comparing the slopes between the first two points of data and the last two points.  The integrated rate laws can be derived quickly using elementary calculus. Students with good math backgrounds may appreciate seeing the connection.  The half‐lives are derived for the three cases.

Encourage students to understand these simple derivations rather than memorizing the equations.  Conceptual Connection 13.2 Kinetics Summary



13.5 The Effect of Temperature on Reaction Rate  The activation barrier is shown in a plot of energy versus reaction progress (sometimes called reaction coordinate).  The thermal energy distribution is very similar to the distribution of kinetic energies from the kinetic molecular theory of gases.  The collision model gives a good conceptual basis for how simple reactions occur.  The orientation factor can be modelled very well for simple molecules but not so easily for complex molecules.  Conceptual Connection 13.3 Collision Theory

.

 



Students sometimes confuse the terms rate law and integrated rate law.

This section in particular is very challenging to students with weak algebra skills.

Since the first‐order half‐life equation is used so commonly (especially for nuclear processes), students tend to use the equation for all half‐life cases.

Some of the sharper students may question why the pre‐exponential factor A is independent of temperature. Strictly speaking, the collision frequency is a function of temperature, but that dependence is smaller than the temperature dependence of the exponential part of the Arrhenius equation.

215

Chapter 13. Chemical Kinetics Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

13.6 Reaction Mechanisms  Elementary steps o Rate law for elementary step  Reaction intermediate  Molecularity o Unimolecular o Bimolecular o Termolecular  Rate‐determining step  Predicting rate laws from mechanisms

 

13.7 Catalysis  Catalyst o Increases rate  Smaller Ea o Not consumed by the reaction o Example: catalytic converter  Process and phase o Homogeneous o Heterogeneous  Hydrogenation catalyst  Enzymes o Active site o Substrate o Enzyme–substrate binding o Kinetic model o Examples  Sucrase  Chymotrypsin



  

               .

Table 13.3 Rate Laws for Elementary Steps Unnumbered figure: illustration of traffic analogy to rate‐limiting step Figure 13.16 Energy Diagram for a Two‐Step Mechanism Unnumbered figure: steady‐state approximation Example 13.9 Steady‐State Approximation

Figure 13.17 Catalyzed and Uncatalyzed Decomposition of Ozone Unnumbered figure: illustration of a catalytic converter Table 13.4 Change in Vehicular Emissions in Canada Unnumbered figure: photo of polar stratospheric cloud Figure 13.18 Homogeneous and Heterogeneous Catalysis Figure 13.19 Ozone Depletion in the Antarctic Spring Figure 13.20 Catalytic Hydrogenation of Ethene Figure 13.21 Enzyme–Substrate Binding Unnumbered figure: molecular models of sucrose, glucose, and fructose Figure 13.22 An Enzyme‐Catalyzed Reaction Figure 13.23 Michaelis–Menten Behaviour of the Sucrase‐Catalyzed Conversion of Sucrose to Glucose and Fructose Chemistry and Medicine: Enzyme Catalysis and the Role of Chymotrypsin in Digestion Figure 13.24 The Structure of a Protein Figure 13.25 Protein Digestion Figure 13.26 Chymotrypsin, a Digestive Enzyme Figure 13.27 The Action of Chymotrypsin 216

Chapter 13. Chemical Kinetics Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

13.6 Reaction Mechanisms  This section provides the second illustration of the importance of reaction order. Knowing the order of a reaction can lead to the development of a mechanism for the reaction, leading to a greater understanding of the reaction on a molecular level.  The observed rate law provides evidence about the rate‐determining elementary step.  For most reactions, creating the actual elementary steps requires much work far beyond the scope of the class.





 

13.7 Catalysis  Catalysis is a crucial process industrially (e.g., cracking and reforming gasoline molecules), environmentally (e.g., catalytic converters and pollution reduction), and biochemically (enzymes).  Provide additional examples of catalysts.

Examples with comparative rate constants dramatically illustrate the difference between catalyzed and uncatalyzed reactions.  There are several simple demonstrations of catalysis: decomposition of hydrogen peroxide with MnO2; reaction of H2 and O2 catalyzed by Pd (Owens, G. S.; Richmond, T. G. Chemical Educator 1(4): S 1430–4171 (96), 1996, 04045–9).  Catalysts decrease the activation barrier by altering the transition state or providing a different mechanism/pathway for the reaction.





.

The rate law for an overall reaction cannot be written from the balanced reaction, but the rate law for an elementary step is written from the balanced reaction. This in part defines an elementary step.

A reaction intermediate is not the same as a transition state. An intermediate can be observed (e.g., carbocations) and exists as a local energy minimum on a potential energy diagram.

In a multistep mechanism, the rate‐ determining step has the largest Ea and the smallest k.

Mechanisms can be disproven but cannot be proven to be true.

Catalysts are not consumed by the reaction at hand, but they can be consumed by unwanted side reactions and rendered ineffective after some amount of time.

Kinetics (and catalysis) is path dependent, whereas the thermodynamics and extent of a reaction are path independent. Catalysis does not affect the amount of energy released or absorbed by a reaction, nor does it affect the product yield.

217

Chapter 13. Chemical Kinetics

Additional Problem for The First-Order Integrated Rate Law: Determining the Concentration of a Reactant at a Given Time (Example 13.4)

Hydrogen peroxide decomposes according to the reaction: H2O2(aq)  H2O(l) + 0.5 O2(g)

Sort

Given

The reaction follows first-order decay with a rate constant of 3.68 h1. If the initial concentration is 30.0 ppm, what is peroxide concentration after 2.00 hours?

You have been given the rate of a first-order reaction and the initial concentration of the reactant.

You are asked to find the concentration at 2.00 hours.

k = 3.68 h1

[H2O2]0 = 30.0 ppm

Find [H2O2] at t = 2.00 h

Strategize

Conceptual Plan

Use the first-order integrated rate law to get from the given information to that which you are asked to find.

k, t, [A]0  [A]t

Relationships Used ln [A]t = -kt + ln [A]0

Solve

Solution

Substitute the rate constant, the initial concentration, and the time into the integrated rate law.

ln [H2O2 ]t =  kt + ln [H2O2 ]0 3.68  2.00 h ) + ln (30.0) h ln [H2O2 ]t =  7.36 + 3.40 ln [H2O2 ]t = (

[H2O2 ]t

Solve the integrated rate law for the concentration of [H2O2]t.

= exp(  3.96) = 0.0191 ppm = 19.1 ppb

The concentration has decreased a great deal. Only a very small concentration of peroxide remains.

Check

.

218

Chapter 13. Chemical Kinetics

The decomposition of ozone is given by the equation:

Additional Problem for Using the Two-Point Form of the Arrhenius Equation (Example 13.8)

O3(g)  O2(g) + O(g) The rate constant at 700 K was measured as 4.85  104 mol L1 s1 and at 800 K was 3.58  105 mol L1 s1. Find the activation energy in kJ mol-1.

Given T1 = 700 K

Sort

k1 = 4.85  104 mol L1 s1

You are given the rate constant of a reaction at two different temperatures. You are asked to find the activation energy.

T2 = 800 K k2 = 3.58  105 mol L1 s1 Find Ea

Strategize

Conceptual Plan

Use the two-point form of the Arrhenius equation, which relates activation energy to the given information and R (a constant).

T1, k1, T2, k2 

Ea

Relationships Used ln

E 1 k2 1 = a   k1 R  T1 T2 

Solve

Solution

Substitute the two rate constants and two temperatures into the equation.

ln

Ea  1 1    R  T1 T2  E  1 3.58  105 M1s1 1   ln = a 1 1 4 4.85  10 M s 800 K  R  700 K

Solve the equation for Ea, the activation energy, and convert to kJ mol-1.

k2 k1

=

7.38 =

Ea  1.78  104    K R  

  J K  8.314 Ea = 7.38  4  mol  K  1.78  10  J = 3.45  105 mol = 345 kJ/mol

The units of the answer are correct, and the magnitude makes sense.

Check

.

219

Chapter 14. Chemical Equilibrium Student Objectives 14.1 Fetal Hemoglobin and Equilibrium  

Know that hemoglobin is the bloodstream’s carrier for oxygen and that the binding of oxygen to hemoglobin is an equilibrium reaction.

Know that the fetal–maternal oxygen equilibrium is maintained via the placenta where oxygen is exchanged.

14.2 The Concept of Dynamic Equilibrium  

Know and understand that in a dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

Understand the concept of dynamic equilibrium through the population analogy.

14.3 The Expression for the Equilibrium Constant     

Know and understand the relationship between equilibrium expressions written in terms of pressure or concentration.

Write and interpret the equilibrium expressions given by Kp and Kc for a chemical reaction.

Show the relationship between Kp and Kc for a gas‐phase reaction.

Know and understand that the concentrations of solids and liquids in a reaction do not change over the course of a reaction and do not appear in the equilibrium expression.

Write and interpret an equilibrium expression for a reaction involving a solid or liquid.

14.4 The Equilibrium Constant (K)    

Define and understand the law of mass action.

Write and interpret equilibrium expressions using concentrations.

Understand the significance of numerical values of the equilibrium constant, especially very large and very small values.

Know and understand the mathematical relationships between chemical equations and equilibrium constants.

14.5 Calculating the Equilibrium Constant from Measured Quantities  

Know and understand that the equilibrium constant will be identical for a given reaction at a given temperature; the equilibrium can be established at an infinite combination of concentrations.

Write and compute the table of values for an equilibrium reaction at the initial, change in, and equilibrium concentrations. Know that this is often called the ICE table.

14.6 The Reaction Quotient: Predicting the Direction of Change   

Know that the reaction quotient Q is defined in the same way as the equilibrium constant K except that Q can be defined for a state other than equilibrium.

Know and understand how Q can be compared with K and used to determine in which direction a reaction will proceed in order to establish equilibrium.

Predict the direction of a reaction by comparing the values of Q and K.

.

220

Chapter 14. Chemical Equilibrium 14.7 Finding Equilibrium Concentrations   

Calculate equilibrium concentrations from the equilibrium constant K and all but one of the equilibrium concentrations or pressures of the reactants and products.

Calculate equilibrium concentrations from the equilibrium constant K and initial concentrations or pressures of the reactants and products.

Calculate equilibrium concentrations from initial concentrations or pressures in cases in which the equilibrium constant is small.

14.8 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances     

Define Le Châtelier’s Principle.

Know and understand the effect of changing concentration on a system at equilibrium.

Know and understand the effect of changing volume or pressure of a system at equilibrium that involves gases.

Know and understand the effect of a temperature change on a system at equilibrium.

Predict the effect of temperature change on a reaction for which the heat flow is known.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

221

Chapter 14. Chemical Equilibrium Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

14.1 Fetal Hemoglobin and Equilibrium  Hemoglobin–oxygen equilibrium  Oxygen transfer from maternal to fetal hemoglobin

  

Intro figure: illustration of umbilical cord, red blood cells, and hemoglobin Unnumbered figure: Hemoglobin is the oxygen‐carrying protein in red blood cells Figure 14.1 Oxygen Exchange Between the Maternal and Fetal Circulation

14.2 The Concept of Dynamic Equilibrium  Equilibrium o Reversible reactions o Dynamic o Population analogy



Figure 14.2 Dynamic Equilibrium

14.3 The Expression for the Equilibrium Constant  Equilibrium constant types o Concentration: Kc o Pressure: Kp o Kp = Kc(RT)n  Units of K  Liquids and solids o No change in concentration during reaction o Not included in K expressions



Example 14.1 Determining Kp from Pressure Measurement and Relating Kp and Kc Figure 14.3 Heterogeneous Equilibrium Example 14.2 Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid

14.4 The Equilibrium Constant (K)  Representations o Chemical equation o Equilibrium constant, K o Equilibrium expression: law of mass action  Values of K o Small K: reactants favoured o Large K: products favoured  Rules for manipulating K values o Reverse equation: invert K o Multiply equation: raise K to power o add equations together: multiply K values



 

  

.

Figure 14.4 The Meaning of a Large Equilibrium Constant Figure 14.5 The Meaning of a Small Equilibrium Constant Chemistry and Medicine: Life and Equilibrium Example 14.3 Manipulating the Equilibrium Constant to Reflect Changes in the Chemical Equation

222

Chapter 14. Chemical Equilibrium Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

14.1 Fetal Hemoglobin and Equilibrium  More oxygen in the blood means a higher hemoglobin saturation level.  The fetal–maternal oxygen‐sharing model is a practical example of an equilibrium that occurs via the placenta. Ask the students to suggest other related equilibrium systems.



The placenta exchanges oxygen but not blood between the mother and fetus.

14.2 The Concept of Dynamic Equilibrium  The forward and reverse reactions in an equilibrium are different chemical reactions that can be characterized by a rate expression for each. At equilibrium, these reaction rates are equal.



The rates of the forward and reverse reactions are equal at equilibrium, but the rate constants are almost never equal.

14.3 The Expression for the Equilibrium Constant  Equilibrium expressions can be written in terms of pressure or concentration.  The formal definition of K from thermodynamics references concentrations and pressures to 1 M and 1 bar, respectively, cancelling all the units and rendering K dimensionless.  Liquids and solids do not appear in equilibrium expressions because their concentrations do not change.  Conceptual Connection 14.1 Heterogeneous Equilibria, Kp, and Kc



The numerical value of K expressed as Kc is different from that expressed as Kp. Most chemistry disciplines use no units for K; some biochemists do use them.

There is some misunderstanding between a solid or liquid concentration not changing versus having no value. The concentration of water has a value (55.5 M at 25 oC), but that concentration does not change when water participates in a chemical reaction.

14.4 The Equilibrium Constant (K)  The equilibrium expression is written by inspection of the chemical equation. Students can be asked to discuss qualitatively what is expected for each case: Large or small K? Large or small concentration?  Practise using the rules for manipulating K values.  Consistent communication requires concentrations in K expressions to be in M and pressures to be in atm (or bar, formally).  Conceptual Connection 14.2 Equilibrium Constants



 



.

Many students are initially confused by the rules for manipulating K values because of their experience with manipulating H values. Working through several examples will ease this confusion.

A recurring confusion for this and the next few chapters is k (rate constant) versus K (equilibrium constant).

223

Chapter 14. Chemical Equilibrium Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

14.5 Calculating the Equilibrium Constant from Measured Quantities  Law of mass action o Same value of K for given reaction at a given T o Particular equilibrium position dependent on initial condition  ICE table o Initial o Change o Equilibrium



Table 14.1 Initial and Equilibrium Concentrations for the Reaction H2(g) + I2(g)

14.6 The Reaction Quotient: Predicting the Direction of Change  Q expression o Same form as K expression o Concentrations (or pressures) often not at equilibrium  Predicting direction of reaction o Q > K: reverse o Q < K: forward o Q = K: equilibrium



Examples 14.4 and 14. 5 Finding Equilibrium Constants from Experimental Partial Pressure Measurements and Concentration Measurements



Figure 14.6 Q, K, and the Direction of a Reaction Example 14.6 Predicting the Direction of a Reaction by Comparing Q and K



.

2 HI(g) at 640 K

224

Chapter 14. Chemical Equilibrium Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

14.5 Calculating the Equilibrium Constant from Measured Quantities  Equilibrium concentrations are calculated using the ICE table. Warn students about missing coefficients from balanced reactions, especially in the change or equilibrium concentrations.



A reaction can have an infinite number of equilibrium positions (combinations of concentrations/pressures), but a given chemical reaction has only one value of K at a given temperature.

14.6 The Reaction Quotient: Predicting the Direction of Change  The reaction quotient Q is the same as K but under nonequilibrium conditions.  Conceptual Connection 14.3 Q and K



Students sometimes think the Q expression is another type of calculation altogether, but it is just the value of the concentration expression under nonequilibrium conditions.

.

225

Chapter 14. Chemical Equilibrium Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

14.7 Finding Equilibrium Concentrations  Problem scenarios o Finding a missing equilibrium concentration given K and other concentrations o Finding equilibrium concentrations given K and initial concentrations o Finding equilibrium partial pressures given K and initial partial pressures o Finding equilibrium concentrations given initial concentrations and a small K



14.8 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances  Le Châtelier’s Principle o Definition o Population analogy  Equilibrium disturbances o Concentration change o Volume change in gaseous equilibrium o Temperature change  Exothermic reactions  Endothermic reactions



   

     

.

Example 14.7 Finding Equilibrium Amounts When You Know the Equilibrium Constant and All but One Equilibrium Amounts of Reactants and Products Unnumbered table: ICE table Examples 14.8 and 14.9 Finding Partial Pressures at Equilibrium from Initial Pressures and the Equilibrium Constant Example 14.10 Finding Equilibrium Concentrations from Equilibrium Constant and Initial Concentrations Examples 14.11 and 14.12 Finding Equilibrium Amounts in Cases with a Small Equilibrium Constant

Figure 14.7 Le Châtelier’s Principle: The Effect of Changing the Amount of Reactant or Product Figure 14.8 Le Châtelier’s Principle: Changing the Amount of Reactant or Product Example 14.13 The Effect of Changing the Amount of a Reactant or Product on Equilibrium Figure 14.9 Le Châtelier’s Principle: The Effect of a Pressure Change Example 14.14 The Effect of a Volume Change on Equilibrium Figure 14.10 Le Châtelier’s Principle: The Effect of a Temperature Change Example 14.15 The Effect of a Temperature Change on Equilibrium

226

Chapter 14. Chemical Equilibrium Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

14.7 Finding Equilibrium Concentrations  Finding a missing equilibrium concentration simply involves the use of the other values and the equilibrium expression.  Finding equilibrium concentrations from initial conditions often involves the quadratic equation. A review of algebra and the solutions of the quadratic equation are appropriate. Emphasize that while both roots are mathematically relevant, one of the roots will always be physically meaningless.  A small value of K allows the use of an approximation that obviates the need for the quadratic equation. The trick is knowing when the approximation is valid.  Conceptual Connection 14.4 The x Is Small Approximation



These problems particularly represent the downfall of ‘spectator’ students. Proficiency in solving these types of problems comes with lots of practice and not through watching someone else work the problems.

14.8 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances  A qualitative use of the Q expression to evaluate the effect of a concentration change eliminates the need to memorize particular effects.



There are only two ways to change the value of K: write a different reaction or change the temperature.

Volume changes for gaseous equilibria are perhaps the toughest for students to understand. Problem 88 could be used as an excellent sample problem that illustrates all the relevant principles.



.

227

Chapter 14. Chemical Equilibrium

Consider the reaction:

Additional Problem for Finding Equilibrium Constants from Experimental Concentration Measurements (Examples 14.4 and 14.5)

CO(g) + 2 H2(g)

CH3OH(g)

A reaction mixture initially contains [CO] = 0.600 M and [H2] = 1.20 M. At equilibrium, the CO concentration is found to be 0.100 M. Find the equilibrium constant for the reaction.

(1)

CO(g) + 2 H2(g)

Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations and equilibrium concentrations of the reagents and products.

Initial

CH3OH(g)

[CO]

[H2]

0.600

1.20

[CH3OH] 0.000

Change Equil (2)

0.100

CO(g) + 2 H2(g)

For the reactant or product whose concentration is known both initially and at equilibrium, calculate the change in concentration of all other reactants and products.

(3)

[CO]

[H2]

[CH3OH]

Initial

0.600

1.20

0.000

Change

0.500

Equil

0.100

CO(g) + 2 H2(g)

Use the change calculated in step 2 and the stoichiometric relationships from the balanced chemical equation to determine the changes in concentration of all other reactants and products.

CH3OH(g)

CH3OH(g)

[CO]

[H2]

[CH3OH]

Initial

0.600

1.20

0.000

Change

0.500

1.00

+0.500

Equil

0.100

(4)

CO(g) + 2 H2(g)

Sum each column for each reactant and product to determine the equilibrium concentrations.

CH3OH(g)

[CO]

[H2]

[CH3OH]

Initial

0.600

1.20

0.000

Change

0.500

1.00

+0.500

Equil

0.100

0.20

0.500

(5)

.

228

Chapter 14. Chemical Equilibrium Use the balanced equation to write an expression for the equilibrium constant and substitute the equilibrium concentrations to compute K.

Kc =

[CH3OH] [CO] [H 2 ]2

0.500 (0.100)(0.20)2 = 125 =

.

229

Chapter 14. Chemical Equilibrium

Consider the reaction:

Additional Problem for Finding Equilibrium Amounts When You Know the Equilibrium Constant and All but One of the Equilibrium Amounts of the Reactants and Products (Example 14.7)

2 COF2(g)

CO2(g) + CF4(g) Kc = 2.00

In an equilibrium mixture, the concentration of COF2 is 0.35 M and the concentration of CO2 is 0.144 M. What is the equilibrium concentration of CF4?

Given [COF2] = 0.35 M

Sort You are given the equilibrium constant of a chemical reaction, together with the equilibrium concentrations of the reactant and one product. You are asked to find the equilibrium concentration of the other product.

[CO2] = 0.144 M Kc = 2.00

Find [CF4]

Strategize

Conceptual Plan

You can calculate the concentration of the product using the given quantities and the expression for Kc.

[COF2], [CO2], Kc  Kc =

[CF4]

[CO2 ] [CF4 ] [COF2 ]

Relationships Used [CO2 ] [CF4 ] [COF2 ]

Kc =

Solve

Solution

Solve the equilibrium expression for [CF4] and then substitute in the appropriate values to compute it.

Kc

=

[CO2 ] [CF4 ] [COF2 ]

[CF4 ] =

K c [COF2 ] [CO2 ]

(2.00)(0.35) 0.144 = 4.9 M =

The units of the answer are correct, and the magnitude makes sense just by estimating the answer.

Check

.

230

Chapter 14. Chemical Equilibrium

Additional Problem for Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant (Examples 14.9 and 14.10)

Consider the reaction: N2O4(g)

2 NO2(g) Kc = 0.36

A reaction mixture initially contains [N2O4] = 0.100 M. Find the equilibrium concentrations of N2O4 and NO2.

(1) Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations of the reactants and products.

N2O4(g)

2 NO2(g) Kc = 0.36 [N2O4]

Initial

0.100

[NO2] 0.000

Change Equil (2) Use the initial concentrations to calculate the reaction quotient Q for the initial concentrations.

Compare Q to K and predict the direction in which the reaction will proceed.

Qc =

Q < K, so the reaction should proceed right.

(3) Represent the change in the concentration of one of the reactants or products with the variable x.

Define the changes in the concentrations of the other reactants or products in terms of x.

[NO2 ]2 (0)2 = = 0 [N2O4 ] 0.100

N2O4(g)

2 NO2(g) Kc = 0.36 [N2O4]

[NO2]

Initial

0.100

0.000

Change

x

+2x

Equil (4) Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

(5) Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Using the given value of the equilibrium constant, solve the expression for the variable x.

N2O4(g)

2 NO2(g) Kc = 0.36 [N2O4]

[NO2]

Initial

0.100

0.000

Change

x

+2x

Equil

0.100  x

+2x

Kc =

[NO2 ]2 [N2O4 ]

=

(2x )2 0.100  x

4x 2 0.100  x 0.36(0.100  x ) = 4x 2 or 0.036  0.36x = 4x 2 0.36 =

4x 2 + 0.36x  0.036 = 0

.

231

Chapter 14. Chemical Equilibrium

0.36  b 2  4ac  2a 0.36  0.84  8 x   0.15 or x   0.060 b 

x=

(0.36)2  4(4)( 0.036) 2(4)

(6)

The negative x gives unrealistic concentrations.

Substitute x into the expression for the equilibrium concentrations of the reactants and products (from step 4) and compute the concentrations.

[N2O4] = 0.100 – x = 0.100 – 0.060 = 0.040 M [NO2] = +2x = 2(0.060) = 0.12 M

(7)

Kc =

Check your answer by substituting the computed equilibrium values into the equilibrium expression.

The computed value of K should match the given value of K.

[NO2 ]2 [N2O4 ]

=

(0.120)2 0.040

= 0.36

.

232

Chapter 14. Chemical Equilibrium Consider the reaction:

Additional Problem for Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant (Examples 14.11 and 14.12)

2 H2(g) + S2(g) Kc = 1.67 × 107

2 H2S(g)

A reaction mixture initially contains [H2S] = 0.010 M. Find the equilibrium concentrations of H2 and S2.

(1)

2 H2(g) + S2(g) Kc = 1.67×107

2 H2S(g)

Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations of the reactants and products.

Initial

[H2S]

[H2]

[S2]

0.010

0.000

0.000

Change Equil (2)

Qc =

Use the initial concentrations to calculate the reaction quotient Q for the initial concentrations. Compare Q to K and predict the direction in which the reaction will proceed.

[H2 ]2 [S 2 ] = [H2S]2

(0)2 0

 0.010 

2

= 0

Q < K, so the reaction should proceed right.

(3)

2 H2(g) + S2(g) Kc = 1.67×107

2 H2S(g)

Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products in terms of x.

[H2S]

[H2]

[S2]

Initial

0.010

0.000

0.000

Change

2x

+2x

+x

Equil (4)

2 H2(g) + S2(g) Kc = 1.67×107

2 H2S(g)

Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

(5)

[H2S]

[H2]

[S2]

Initial

0.010

0.000

0.000

Change

2x

+2x

+x

Equil

0.0102x

+2x

+x

Kc =

Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Using the given value of the equilibrium constant, solve the expression for the variable x. Make the approximation for small equilibrium constants.

[H2 ]2 [S2 ] (2 x )2 x = 2 [H2S] 0.010  2x (2 x )2 x 0.010  2x

1.67  107 

(0.010  2x  0.010)

 0.010  1.67  107   4 x 3 2

2



x  1.61 10

4



x 3   0.010  1.67  107 / 4  4.18  1012

.

233

Chapter 14. Chemical Equilibrium (6) [H2S] = 0.0100 – 2(1.61  104)

Substitute x into the expression for the equilibrium concentrations of the reactants and products (from step 4) and compute the concentrations.

= 0.0100 M [H2] = +2x = 2(1.61  104) = 3.22  104 M [S2] = +x = 1.61  104 = 1.61  104 M

(7)

Kc =

Check your answer by substituting the computed equilibrium values into the equilibrium expression. The computed value of K should match the given value of K.

[H2 ]2 [S2 ] (3.22  10  )2 (1.61 10 4 ) = 2 2 [H2S]  0.010 

= 1.67  10 7

.

234

Chapter 15. Acids and Bases Student Objectives 15.1 Heartburn  

Know that acid reflux or heartburn is due to irritation of the esophagus by stomach acid, a strong mineral acid.

Know that the potency of an acid (or base) is measured by the concentration of H+ (or H3O+) ions, an equilibrium phenomenon.

15.2 The Nature of Acids and Bases   

Know the names and uses of some common acids.

Know that some organic acids are responsible for the sour taste of some foods, including fruits.

Know the names and uses of some common bases.

15.3 Definitions of Acids and Bases    

Define Arrhenius acid and Arrhenius base.

Know and understand that a proton in water is solvated and is often represented as H3O+.

Define Brønsted–Lowry acid and Brønsted–Lowry base.

Understand and identify conjugate acid–base pairs.

15.4 Acid Strength and the Acid Dissociation Constant (Ka)    

Know and understand that strong acids dissociate completely to protons and the conjugate base.

Know and understand that weak acids dissociate to a small extent, leaving the molecular form of the acid the predominant species.

Understand that the degree of dissociation of an acid is related to the attraction between the proton and the conjugate base.

Know and understand that the acid ionization constant, Ka, is the equilibrium constant for the dissociation of an acid to a proton and the conjugate base.

15.5 Base Solutions     

Know and understand that strong bases dissociate completely in solution and that many strong bases are the soluble metal hydroxides.

Know and understand that weak bases dissociate water to a small extent to form OH.

Identify amines as weak bases.

Know and understand that the base ionization constant, Kb, is the equilibrium constant for the reaction of a weak base with water to form OH and the conjugate acid.

Calculate and interconvert among [OH], [H3O+], and pH in solutions of strong and weak bases.

.

236

Chapter 15. Acids and Bases 15.6 Autoionization of Water and pH       

Know and understand that water itself can act as an acid, producing H+ and OH in a process called autoionization.

Know that the ion product constant for water (also called the dissociation constant for water) is 1  1014 at 25 oC and can be expressed as the product of [H+] and [OH].

Know that the acidity or basicity of a solution is defined by the balance between proton and hydroxide concentrations.

Define pH in terms of [H3O+].

Calculate and interconvert pH and proton concentration.

Define pOH in terms of [OH].

Define pKa in terms of Ka.

15.7 Finding the [H3O+], [OH], and pH of Acid or Base Solutions    

Calculate the pH of a strong acid solution from [H3O+].

Calculate the pH of a weak acid solution using the Ka and the initial acid concentration.

Calculate the percent ionization of a weak acid using the Ka and the initial acid concentration.

Calculate the pH of a mixture of weak acids by comparing the sources of proton.

15.8 The Acid–Base Properties of Ions and Salts     

Know the active ingredients of common antacids.

Know and understand how the scales of acid and base strength complement each other for a series of conjugate acid–base pairs.

Know the mathematical relationship between Ka, Kb, and Kw. Calculate the pH of a solution containing an anion acting as a base.

Determine whether an anion, cation, or salt is basic, acidic, or pH‐neutral.

15.9 Polyprotic Acids    

Know the structures of some polyprotic acids and understand that each proton is governed by its own dissociation constant.

Calculate the pH of a polyprotic acid solution using acid dissociation data and an ICE table.

Calculate the pH of a dilute sulfuric acid solution.

Calculate the concentration of the anions from weak diprotic acids in solution.

15.10 Lewis Acids and Bases   

Define Lewis acid and Lewis base.

Know that molecules with an incomplete octet of electrons can act as Lewis acids and that molecules with lone electron pairs can act as Lewis bases.

Know that small cations can act as Lewis acids.

15.11 Strengths of Acids and Bases and Molecular Structure  

Know and understand that both bond polarity and bond strength influence the acidity of binary acids.

Know that the strength of oxyacids depends on the electronegativity of the central atom (Y) and the number of oxygens attached to that atom.

.

237

Chapter 15. Acids and Bases 15.12 Acid Rain   

Know that SO2 and NO2, byproducts of fuel combustion reactions, combine with water and oxygen to form sulfuric acid and nitric acid, the components of acid rain.

Know that normal rainfall is not neutral (i.e., pH = 7) since it combines with CO2 to form carbonic acid.

Know that acid rain adversely affects statues, buildings, trees, and forests.

.

238

Chapter 15. Acids and Bases Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

239

Chapter 15. Acids and Bases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

15.1 Heartburn  Observation: acid reflux (heartburn)  Cause: stomach acid o Strong mineral acid, HCl o Concentrated



Intro figure: illustration of milk of magnesia

15.2 The Nature of Acids and Bases  Common acids o Examples  Strong and weak  Inorganic and organic o Uses/occurrences  Manufacturing, laboratory  Sulfuric, nitric  Sour taste in foods  Acetic, citric, malic  Common bases o Examples o Uses/occurrences

 

Table 15.1 Some Common Acids Unnumbered figures: molecular models of HCl, H2SO4, HNO3, HC2H3O2 Unnumbered figure: photo of vinegar bottle Unnumbered figures: photos of fruits and molecular models of citric acid, malic acid Unnumbered figure: photo of household products Table 15.2 Common Bases

15.3 Definitions of Acids and Bases  Arrhenius o Acid: produces H+ o Base: produces OH  Brønsted–Lowry o Acid: H+ donor o Base: H+ acceptor  Conjugate acid–base pairs

   

Figure 15.1 Arrhenius Acid Figure 15.2 Arrhenius Base Figure 15.3 Conjugate Acid–Base Pairs Example 15.1 Identifying Brønsted– Lowry Acids and Bases and Their Conjugates

15.4 Acid Strength and the Acid Dissociation Constant (Ka)  Acid strength o Strong o Weak  Measurement o Acid ionization constant, Ka o Equilibrium expression

   

Figure 15.4 Ionization of a Strong Acid Table 15.3 Strong Acids Figure 15.5 Ionization of a Weak Acid Figure 15.6 Ionic Attraction and Acid Strength Table 15.4 Some Weak Acids Table 15.5 Acid Ionization Constants (Ka) for Some Monoprotic Weak Acids at 25 oC

   

H3O   A    Ka  HA  

 

Examples (Table 15.5, Ka values ranging from 102 to 1010) .

240

Chapter 15. Acids and Bases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

15.1 Heartburn  Acid reflux or heartburn is caused by stomach acid leaking into the esophagus. Antacids treat the symptoms but not the underlying cause.



Many students underestimate how strong an acid is found in the stomach and to what levels or concentrations it occurs.

15.3 Definitions of Acids and Bases  Arrhenius was the first to give a definition of acids and bases. Subsequent definitions (Brønsted–Lowry and Lewis) were more general. Show examples that illustrate how all Arrhenius acids and bases fit the Brønsted–Lowry definition, but the opposite is not true.  The conjugate base is important in determining the acidity of an acid. The stronger the attraction of the conjugate base for the proton, the weaker the acid.  Conceptual Connection 15.1 Conjugate Acid–Base Pairs



Students sometimes overcomplicate the relationship between an acid and its conjugate base and vice versa. They are different only by an H+.

15.4 Acid Strength and the Acid Dissociation Constant (Ka)  The ability of an acid to dissociate is a function of the interaction between H+ and the conjugate base.

Strong acids produce stable conjugate bases, e.g., HCl produces Cl.  Strong acids also have Ka values, but they are so large that the acids are considered to dissociate completely.  The sum of [HA] and [A] at equilibrium represents all the initial acid.  Conceptual Connection 15.2 Conjugate Bases



The strength of an acid is indicated by its Ka, the degree to which the molecular form of the acid dissociates.

Students sometimes confuse the two pairs of terms strong/weak and concentrated/dilute.

15.2 The Nature of Acids and Bases  For some of the common acids, students may be more familiar with the uses or occurrences than the identities of the acids, e.g., phosphoric acid in colas.  The common acids include many examples in which the proton is attached to an oxygen, which in turn is attached to another atom: H–O–Y.

.



241

Chapter 15. Acids and Bases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

15.5 Base Solutions  Strong bases o Complete dissociation o Examples: soluble hydroxides  Weak bases o Base ionization constant, Kb o Equilibrium expression

    

Figure 15.7 Ionization of a Strong Base Table 15.6 Strong Bases Figure 15.8 Ionization of a Weak Base Table 15.7 Some Common Weak Bases Figure 15.9 Lone Pairs in Weak Bases

 

Example 15.2 Using Kw in Calculations Table 15.8 The pH of Some Common Substances Figure 15.10 The pH Scale Example 15.3 Calculating pH from [H3O+] or [OH] Example 15.4 Calculating [H3O+] from pH Figure 15.11 pH and pOH Chemistry and Medicine: Ulcers

BH  OH    Kb  B



o Amines Calculations o Finding [OH] and pH of strong base solutions o Finding [OH] and pH of weak base solutions

15.6 Autoionization of Water and pH  Acids and bases o Water as an acid and base o Kw = [H+][ OH]  Aqueous solutions o Neutral o Acidic o Basic  Scales o pH = log [H3O+] o pOH = log [OH‐] o pKa = log Ka

    

15.7 Finding the [H3O+], [OH], and pH of Acid or Base Solutions  pH of acid solutions o Strong: complete dissociation o Weak: partial dissociation governed by initial concentration and Ka  Calculations o [H3O+] of a weak acid o x is small approximation o Percent ionization of a weak acid  Acid contributions from a mixture o Finding [OH] and pH of strong base solutions o Finding [OH] and pH of weak base solutions

.

       

Example 15.5 Finding the [H3O+] of a Weak Acid Solution Example 15.6 Finding the pH of a Weak Acid Solution Example 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Example 15.8 Finding the Equilibrium Constant from pH Example 15.9 Finding the Percent Ionization of a Weak Acid Example 15.10 Mixtures of Weak Acids Example 15.11 Finding the [OH‐] and pH of a Strong Base Solution Example 15.12 Finding the [OH‐] and pH of a Weak Base Solution 242

Chapter 15. Acids and Bases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

15.5 Base Solutions  Remind students that strong bases form OH directly since those ionic compounds dissociate completely. Weak bases remove a proton from water to produce OH.  The pH basic solutions can be obtained from [H+] = Kw/[OH] or from pH = pKw  pOH.

15.6 Autoionization of Water and pH  Conceptually, the autoionization of water matches the process of any acid dissociation.  The contribution of water to the overall [H+] of a solution can often be overlooked.  Higher acidity occurs with lower pH, higher pOH, and lower pKa.  The pH scale is logarithmic: each pH unit is a 10‐fold difference in [H+].



15.7 Finding the [H3O+], [OH], and pH of Acid or Base Solutions  Remind students that proton concentrations from strong acids at equilibrium are identical to the initial acid concentration. For weak acids, the Ka and an ICE table are used to determine the equilibrium [H+].  Conceptual Connection 15.3 The x is small Approximation  Conceptual Connection 15.4 Percent Ionization  Conceptual Connection 15.5 Strong and Weak Acids





.

The equation Kw = [H+][OH] is applicable to every aqueous solution, not just solutions of acids and bases and not just at 298 K. Students often are tempted to memorize that Kw is always 1.0 × 1014, but this is true only at 298 K. Similarly, neutral pH is 7.00 only at 298 K; 7.00 is not neutral even for body temperature The equilibrium problems in this section are exactly the same as those encountered in Section 14.8.

The equilibrium problems in this section are exactly the same as those encountered in Section 14.8.

243

Chapter 15. Acids and Bases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

15.8 The Acid–Base Properties of Ions and Salts  Antacid ingredients o Aluminum and magnesium o Hydroxides o Carbonate  Anions as weak bases o Strength of conjugate acid–base pairs o Acid vs. base scales o Ka × Kb = Kw  Cations as weak acids o Counterions of strong bases: neutral o Conjugate acids of weak bases: acidic o Small, highly charged metals: acidic  pH of salt solutions o Cations o Anions  Conjugate base of strong acid: neutral  Conjugate base of weak acid: basic



15.9 Polyprotic Acids  Table of acids, Table 15.10 o Structures o Multiple Ka values  Calculations o Finding pH of polyprotic acid solution o Finding pH of dilute sulfuric acid solution o Finding concentration of the anions for a weak acid solution



       

   

15.10 Lewis Acids and Bases  Definitions o Acid: electron pair acceptor o Base: electron pair donor  Examples o Molecules  Acids: contain Al or B  Bases: contain lone pair of electrons  Cations: Al3+, H+



.

Unnumbered table: conjugate acids and bases Chemistry and Medicine: What’s in My Antacid?

Figure 15.12 Strength of Conjugate Acid– Base Pairs Example 15.13 Determining Whether an Anion Is Basic or pH‐Neutral Example 15.14 Determining the pH of a Solution Containing an Anion Acting as a Base Unnumbered figure: molecular models of Al(H2O)63+ and Al(H2O)5(OH)2+ Example 15.15 Determining Whether a Cation is Acidic or pH‐Neutral Table 15.9 pH of Salt Solutions Example 15.16 Determining the Overall Acidity or Basicity of Salt Solutions

Table 15.10 Common Polyprotic Acids and Ionization Constants Example 15.17 Finding the pH of a Polyprotic Acid Solution Example 15.18 Dilute H2SO4 Solutions Example 15.19 Finding the Concentration of the Anions for a Weak Diprotic Acid Solution Figure 15.13 Dissociation of a Polyprotic Acid

Unnumbered figures: Lewis structures of Lewis acids and bases

244

Chapter 15. Acids and Bases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

15.8 The Acid–Base Properties of Ions and Salts  Most common antacids have aluminum and magnesium hydroxides listed as the active ingredients. It should be pointed out that these materials do not dissociate into strong bases like their Group 1 counterparts.

Aluminum and magnesium form oxides with complex structures, not discreet cations and anions.  Anions can serve as bases with a range of base strength. That strength is related to the acidity of the corresponding conjugate acid.  Aqueous ionic compounds may produce neutral, acidic, or basic solutions depending on their constituent ions.



15.9 Polyprotic Acids  Polyprotic acids cover the range from strong to very weak acids, and each of their protons has a different dissociation constant.  Ka values for polyprotic acids are successively smaller, usually by three or more orders of magnitude.  The pH of a polyprotic acid is usually governed by only the largest Ka.





Cations can serve as weak acids because of the relative acidity of their hydration spheres.

The conjugate bases of weak acids form basic solutions because of the relatively high attraction of those bases for protons, even from water.

The dissociation of the first proton of sulfuric acid is complete, but the second dissociation is only partial.

15.10 Lewis Acids and Bases  The Lewis definition is the most general acid– base definition.  General chemistry tends to focus more on the Brønsted–Lowry definition, but the Lewis definition is crucial in organic chemistry.

.

245

Chapter 15. Acids and Bases Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

15.11 Strengths of Acids and Bases and Molecular Structure  Binary acids o Bond polarity o Bond strength o Examples: Group 16, 17 hydrides  Oxyacids o Electronegativity o Number of oxygen atoms on central atom

    

15.12 Acid Rain  Atmospheric acid o SO2  sulfurous acid, sulfuric acid o NO2  nitric acid  Normal rainfall o CO2  carbonic acid  Effects o Marble statues o Trees/forests o Lakes  Legislation

   

.

Figure 15.14 Acidity Trends in the Periodic Table Unnumbered table: acidity of HOI, HOBr, and HOCl Unnumbered table: acidity of HClO, HClO2, HClO3, and HClO4 Unnumbered figure: acidity of methanoic, acetic, and chloroacetic acid Unnumbered figure: basicity of ammonia, methylamine, and hydroxylamine

Figure 15.15 Sources of Energy in North America by Country and the Demand for Col in Million Metric Tonnes Figure 15.16 The Effects of Acid Rain Figure 15.17 The Potential of Canadian Soils to Reduce the Effects of Acid Rain Figure 15.18 Effect of Legislation to Reduce Pollutant Emissions

246

Chapter 15. Acids and Bases Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

15.11 Acid Strength and Molecular Structure  The relative acidity of binary acids and oxyacids can be predicted from bond or atom properties within the molecules.  Conceptual Connection 15.6 Acid Strength and Molecular Structure  Conceptual Connection 15.7 Amine Base Strength and Molecular Structure

Binary acids have acidity that depends on the electronegativity difference and the strength of the bond. Students sometimes consider only the former and ignore the latter.

15.12 Acid Rain  Atmospheric SO2 arises mostly from coal power‐plant emissions, while NO2 comes from internal combustion engines, especially automobiles.



.

Emission levels, particularly of SO2, have been reduced by scrubbing the emission from power plants.

247

Chapter 15. Acids and Bases

Calculate the pH of each solution and indicate whether the solution is acidic or basic at 25 °C.

Additional Problem for Calculating pH from [H3O+] or [OH] (Example 15.3)

a) [H3O+] = 0.012 mol L-1 b) [H3O+] = 3.7  107 mol L-1 c) [OH] = 0.0046 mol L-1 d) [OH] = 8.1  108 mol L-1

1) pH = log [H3O+]

To calculate pH, simply substitute the given [H3O+] into the pH equation. Since pH < 7, the solution is acidic.

= log(0.012) = 1.92

2) pH = log [H3O+]

To calculate pH, simply substitute the given [H3O+] into the pH equation. Since pH < 7, the solution is acidic.

= log(3.7  107) = 6.43

3) First use Kw to find [H3O+] from [OH]. Since pH > 7, the solution is basic.

[H3O+] [OH] = Kw = 1.0  1014 [H3O+] = 1.0  1014 / [OH-] = 1.0  1014 / 0.0046 = 2.17  1012 mol L-1

pH = log [H3O+] = log(2.17  1012) = 11.7 4) .

248

Chapter 15. Acids and Bases Use Kw to find [H3O+] from [OH]. Since pH < 7, the solution is acidic (but nearly neutral).

[H3O+] [OH] = Kw = 1.0  1014 [H3O+] = 1.0  1014 / [OH] = 1.0  10-14 / 8.1  108 = 1.23  107 mol L-1

pH = log [H3O+] = log(1.23  107) = 6.91

.

249

Chapter 15. Acids and Bases Additional Problem for Finding [H3O+] of a Weak Acid Solution (Example 15.5)

Find the [H3O+] and pH of a 1.00 mol L-1 formic acid solution.

1)

HCO2H(aq) + H2O(l)

Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration.

Initial

H3O+(aq) + HCO2(aq)

[HCO2H ]

[H3O+]

[HCO2]

1.00

0.00

0.00

Change Equil

2)

H3O+(aq) + HCO2(aq)

HCO2H(aq) + H2O(l)

Represent the change in the concentration of H3O+ with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x, always keeping in mind the stoichiometry of the reaction.

[HCO2H ]

[H3O+]

[HCO2]

Initial

1.00

0.00

0.00

Change

x

+x

+x

Equil

3)

H3O+(aq) + HCO2(aq)

HCO2H(aq) + H2O(l)

Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

4)

[HCO2H ]

[H3O+]

[HCO2-]

Initial

1.00

0.00

0.00

Change

x

+x

+x

Equil

1.00x

+x

+x

Ka =

Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant Ka. Substitute the value of Ka and solve for x. Confirm that x is small.

[H3O+ ] [HCO2  ] [HCO2H]

1.8  10 4 

x2 1.00  x



1.8  104

x2

( x is small )

x  1.34  102

.

250

Chapter 15. Acids and Bases 5) [H3O+] = 1.34  102 M

Determine the H3O+ concentration from the computed value of x and compute the pH.

pH = log (1.34  102) = 1.87 6)

Ka =

Check your answer by substituting the computed equilibrium values into the acid ionization expression.

[H3O+ ] [HCO2  ] (1.34  102 )2 = [HCO2H] 1.00

= 1.80  10 4

.

251

Chapter 15. Acids and Bases Additional Problem for Finding the Percent Ionization of a Weak Acid (Example 15.9)

Find the percent ionization of a 1.00 mol L-1 formic acid solution.

1)

HCO2H(aq) + H2O(l)

Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration.

Initial

H3O+(aq) + HCO2(aq)

[HCO2H ]

[H3O+]

[HCO2]

1.00

0.00

0.00

Change Equil

2)

H3O+(aq) + HCO2(aq)

HCO2H(aq) + H2O(l)

Represent the change in the concentration of H3O+ with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x, always keeping in mind the stoichiometry of the reaction.

[HCO2H ]

[H3O+]

[HCO2]

Initial

1.00

0.00

0.00

Change

x

+x

+x

Equil

3)

H3O+(aq) + HCO2(aq)

HCO2H(aq) + H2O(l)

Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

4)

[HCO2H ]

[H3O+]

[HCO2]

Initial

1.00

0.00

0.00

Change

x

+x

+x

Equil

1.00x

+x

+x

Ka =

Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant Ka. Substitute the value of Ka and solve for x. Confirm that x is small.

[H3O+ ] [HCO2  ] [HCO2H]

1.8  10 4 

x2 1.00  x

x2 

1.8  104

( x is small )

x  1.34  102 = [H3O+ ]

.

252

Chapter 15. Acids and Bases 5)

% ionization = Use the definition of percent ionization. (Since the percent ionization is less than 5 percent, the x is small approximation is valid.)

[H3O+ ]equil [HA]initial

 100%

1.34  102  100% 1.00 = 1.34%

=

.

253

Chapter 15. Acids and Bases Additional Problem for Finding the [OH-] and pH of a Weak Base Solution (Example 15.12)

Find the [OH] and pH of a 0.500 mol L-1 NH3 solution.

1)

NH3(aq) + H2O(l)

Write the balanced equation for the reaction of the base, and use it as a guide to prepare an ICE table showing the given concentration of the weak base as its initial concentration.

Initial

NH4+(aq) + OH(aq)

[NH3]

[NH4+]

[OH]

0.500

0.00

0.00

Change Equil

2)

NH3(aq) + H2O(l)

Represent the change in the concentration of OH with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x, always keeping in mind the stoichiometry of the reaction.

NH4+(aq) + OH(aq)

[NH3]

[NH4+]

[OH]

Initial

0.500

0.00

0.00

Change

x

+x

+x

Equil

3)

NH3(aq) + H2O(l)

Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

4)

[NH3]

[NH4+]

[OH]

Initial

0.500

0.00

0.00

Change

x

+x

+x

Equil

0.500x

+x

+x

Kb =

Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the base ionization constant Kb. Substitute the value of Kb and solve for x. Confirm that x is small.

NH4+(aq) + OH(aq)

[NH4+ ] [HO- ] [NH3 ]

1.76  105  x2 

x2 0.500  x

( x is small )

0.500(1.76  105 )

x  2.96  103

.

254

Chapter 15. Acids and Bases [OH] = 2.96  103

5)

Use Kw to find [H3O+] from [OH], and use the definition of pH to find the pH of the solution.

[H3O+] [OH] = Kw = 1.0  1014 [H3O+] = 1.0  1014 / [HO-] = 1.0  1014 / 2.96  103 = 3.38  1012

pH = log [H3O+] = log(3.38  1012) = 11.5

.

255

Chapter 15. Acids and Bases Additional Problem for Determining the pH of a Solution Containing an Anion Acting as a Base (Example 15.14)

Find the pH of a 0.500 mol L-1 NaCN solution. The salt dissociates completely into Na+ and CN, and the Na+ has no acid or base properties.

1)

CN(aq) + H2O(l)

Write the balanced equation for the reaction of the base, and use it as a guide to prepare an ICE table showing the given concentration of the weak base as its initial concentration.

Initial

HCN(aq) + OH(aq)

[CN]

[HCN]

[OH]

0.500

0.00

0.00

Change Equil CN(aq) + H2O(l)

2) Represent the change in the concentration of OH with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x, always keeping in mind the stoichiometry of the reaction.

HCN(aq) + OH(aq)

[CN]

[HCN]

[OH]

Initial

0.500

0.00

0.00

Change

x

+x

+x

Equil CN(aq) + H2O(l)

3) Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

HCN(aq) + OH(aq)

[CN]

[HCN]

[HO]

Initial

0.500

0.00

0.00

Change

x

+x

+x

Equil

0.500x

+x

+x

K a  Kb = K w

4) Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the base ionization constant Kb. Substitute the value of Kb and solve for x. Confirm that x is small.

Kb =

Kw 1.0  1014 = = 2.04  105 4.9  1010 Ka

Kb =

[HCN] [OH ] [CN ]

2.04  105  x2 

x2 0.500  x

( x is small )

0.500(2.04  105 )

x  3.19  103

.

256

Chapter 15. Acids and Bases [OH] = 3.19  103

5) Use Kw to find [H3O+] from [OH], and use the definition of pH to find the pH of the solution.

[H3O+] [OH] = Kw = 1.0  1014 [H3O+] = 1.0  1014 / [OH] = 1.0  1014 / 3.19  103 = 3.13  1012

pH = log [H3O+] = log(3.13  1012) = 11.5

.

257

Chapter 16. Aqueous Ionic Equilibrium Student Objectives 16.1 The Danger of Antifreeze  

Know that the pH of blood is regulated by carbonic acid (H2CO3) and its anion, bicarbonate (HCO3), both of which are derived from carbon dioxide and water.

Know that ethylene glycol (antifreeze) is metabolized to glycolic acid, which lowers the blood pH and lowers the ability of hemoglobin to transport oxygen.

16.2 Buffers: Solutions That Resist pH Change       

Define buffer and know that a buffer typically consists of a weak acid and its conjugate base.

Know that the common ion effect is an example of Le Châtelier’s principle.

Calculate the pH of a buffer solution starting with initial concentrations of a weak acid and its conjugate base.

Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution from the pKa of the weak acid and the initial concentrations of the weak acid and conjugate base.

Know and understand that a buffer has limits; the buffering action is related to a stoichiometric consumption of either the weak acid or the conjugate base.

Calculate the pH change in a buffer after the addition of a small amount of strong acid or base.

Know that buffers can also be made from a weak base and its conjugate acid, and calculate the pH of such a buffer using the Henderson–Hasselbalch equation and the pKa of the conjugate acid.

16.3 Buffer Effectiveness: Buffer Range and Buffer Capacity    

Know and understand that the buffer capacity is optimal when the ratio of acid and conjugate base differs by less than or equal to a factor of 10.

Know and understand that a buffer is more resistant to pH changes when the concentrations of acid and base are relatively large.

Know and understand that an effective buffer covers a pH range one unit above and one unit below the pKa of the weak acid.

Use the pKa and the buffer range to choose appropriate buffer systems for specific applications.

16.4 Titrations and pH Curves         

Calculate the pH of a solution at various points along a titration of a strong acid with a strong base.

Calculate the pH of a solution at various points along a titration of a weak acid with a strong base.

Recognize the titration curve for the titration of a weak base with a strong acid.

Recognize the titration curve for the titration of a polyprotic acid with a strong base.

Define equivalence point for an acid–base titration.

Know that the equivalence point of an acid–base titration can be determined using a pH meter.

Define endpoint for an acid–base indicator.

Know that the equivalence point of an acid–base titration can be approximated by observing the endpoint of an acid–base indicator.

Know that indicators can only be used for the specific pH ranges in which the indicator changes colour.

.

258

Chapter 16. Aqueous Ionic Equilibrium 16.5 Solubility Equilibria and the Solubility Product Constant     

Know that the solubility product constant, Ksp, defines the equilibrium constant for the dissolution of an ionic compound into its constituent ions.

Calculate the molar solubility of an ionic compound in pure water using the Ksp expression and an ICE table.

Know that the solubility of an ionic compound is lower in a solution containing a common ion than in pure water.

Calculate the molar solubility of an ionic compound in the presence of a common ion using the Ksp expression and an ICE table.

Know that the solubility of an ionic compound containing a basic anion increases with decreasing pH.

16.6 Precipitation  

Know that the precipitation of an ionic compound upon mixing aqueous solutions can be predicted using the Q expression and comparing Q to Ksp. Define selective precipitation, and know and understand how cations can be selectively precipitated based on their differential solubility with various anions.

16.7 Qualitative Chemical Analysis   

Know and understand that differential precipitation is the basis of the qualitative analysis of cations in solution.

Know that the qualitative analysis scheme is based on, sequentially, insoluble chlorides, acid‐ insoluble sulfides, base‐insoluble sulfides and hydroxides, insoluble phosphates, and alkali metals and ammonium.

Know that alkali metals give a characteristic colour upon heating in an open flame.

16.8 Complex–Ion Equilibria    

Define complex ion and ligand.

Know that metal cations form complex ions with a series of anionic and neutral ligands.

Calculate complex–ion equilibria using formation constants, Kf, and cation and ligand concentrations.

Know that amphoteric metal hydroxides, like Al(OH)3, are more soluble in acidic and in basic solutions than in pure water.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

259

Chapter 16. Aqueous Ionic Equilibrium Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

16.1 The Danger of Antifreeze  Blood pH control o Carbonic acid and bicarbonate  Antifreeze (ethylene glycol) o Sweet taste o Glycolic acid product  Dangerous pH decrease

 

16.2 Buffers: Solutions That Resist pH Change  Combination of weak acid and conjugate base o Neutralizes strong acid and strong base o Calculations  pH from Ka, [HA], and [A]  Henderson–Hasselbalch equation

pH  pK a  log



Intro figure: illustration of red blood cells and pH scale Unnumbered figure: molecular model of ethylene glycol

[base] [acid]

   

 

 Buffer action on strong acid, strong base  Stoichiometry calculation  Equilibrium calculation Combination of weak base and conjugate acid o Same behaviour as reverse o Example: NH3/NH4+

  

16.3 Buffer Effectiveness: Buffer Range and Buffer Capacity  Buffer effectiveness o Relative concentrations of [HA] and [A] o Absolute concentrations of [HA] and [A]  Buffer range o  1 pH unit o pKa  1  Buffer capacity o Absolute concentrations .

   

Figure 16.1 A Buffer Solution Figure 16.2 The Common Ion Effect Example 16.1 Calculating the pH of a Buffer Solution Example 16.2 Calculating the pH of a Buffer Solution as an Equilibrium Problem and with the Henderson–Hasselbalch Equation Figure 16.3 Buffering Action Unnumbered table: ICE table for a buffer problem Example 16.3 Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base Figure 16.4 Buffer Containing a Base Example 16.4 Using the Henderson– Hasselbalch Equation to Calculate the pH of a Buffer Solution Composed of a Weak Base and Its Conjugate Acid

Unnumbered figures: ICE tables for buffer problem calculations Unnumbered figures: illustrations of concentrated and dilute buffers Example 16.5 Preparing a Buffer Chemistry and Medicine: Buffer Effectiveness in Human Blood

260

Chapter 16. Aqueous Ionic Equilibrium Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

16.1 The Danger of Antifreeze  The example of antifreeze poisoning includes several important concepts: blood is maintained in a narrow pH range using a carbonic acid/bicarbonate buffer system; ethylene glycol is oxidized to an acid in the body and acts as a weak acid, exceeding buffer capacity; hemoglobin is a poorer oxygen carrier at low pH.  The body uses carbon dioxide in the form of carbonic acid and bicarbonate to maintain blood plasma pH within a narrow range.  The acidity of carbon dioxide can be measured by adding dry ice to a solution of universal indicator.

16.2 Buffers: Solutions That Resist pH Change  Buffer solutions contain relatively large amounts of a weak acid and its conjugate base.  Buffering is easily demonstrated by adding a few drops of 1 mol L‐1 HCl and 1 mol L‐1 NaOH to two solutions containing universal indicator, one buffered and one not.  Emphasize to students that buffer components are different only by a proton; they tend to forget this when writing reactions for buffer solutions. Reactions of buffers always convert one component into the other.  Conceptual Connection 16.1 pH of Buffer Solutions  Conceptual Connection 16.2 Adding Acid or Base to a Buffer



16.3 Buffer Effectiveness: Buffer Range and Buffer Capacity  The Chemistry and Medicine box revisits the buffer capacity issue mentioned in the chapter opener.  Conceptual Connection 16.3 Buffer Capacity



.



When the x is small approximation is valid, the Henderson–Hasselbalch equation gives exactly the same results as obtained through the equilibrium expression.

Buffer problems, especially when strong acid or base is added to a buffer, are notoriously difficult for students. Working through several examples is key.

Buffers are not infinitely resistant to pH changes.

261

Chapter 16. Aqueous Ionic Equilibrium Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

16.4 Titrations and pH Curves  pH changes during titrations o Strong acid with strong base  Initial pH  pH before equivalence point  pH at equivalence point  pH past equivalence point  Plot of pH vs. volume of added base o Weak acid with strong base  Initial pH  pH before equivalence point  pH at equivalence point  pH past equivalence point  Plot of pH vs. volume of added base o Curve of weak base/strong acid titration o Curve of polyprotic acid/strong base titration  Indicators o Endpoint o Examples  Phenolphthalein  Methyl red  Table of others

 

16.5 Solubility Equilibria and the Solubility Product Constant  Solubility product constant, Ksp o Reaction: solid dissolves to ions o Equilibrium expression  Calculating molar solubility o Pure water o Relative solubility o Common ion effect o Effect of pH  Hard water

           

      

.

Figure 16.5 Acid–Base Titration Figure 16.6 Titration Curve: Strong Acid + Strong Base Unnumbered figures: graphs of pH vs. volume of base added in an strong acid– strong base titration Figure 16.7 Titration Curve: Strong Base + Strong Acid Example 16.6 Strong Acid–Strong Base Titration pH Curve Unnumbered figures: graphs of pH vs. volume of base added in an weak acid– strong base titration Example 16.7 Weak Acid–Strong Base Titration pH Curve Figure 16.8 Titration Curve: Weak Base with Strong Acid Figure 16.9 Titration Curve: Diprotic Acid with Strong Base Figure 16.10 Monitoring the pH During a Titration Figure 16.11 Monitoring the Colour Change During a Titration Figure 16.12 Phenolphthalein Figure 16.13 Indicator Colour Change Table 16.1 Ranges of Colour Changes for Several Acid–Base Indicators

Table 16.2 Selected Solubility Product Constants (Ksp) Example 16.8 Calculating Molar Solubility from Ksp Chemistry in Your Day: Hard Water Example 16.9 Calculating Ksp from Molar Solubility Example 16.10 Calculating Molar Solubility in the Presence of a Common Ion Unnumbered figure: photo of stalactites and stalagmites Example 16.11 The Effect of pH on Solubility

262

Chapter 16. Aqueous Ionic Equilibrium Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

16.4 Titrations and pH Curves  Calculating the various pH values that are obtained during a titration is tedious but illustrative of important concepts. After demonstrating one or two of these, have the students perform additional ones in small groups.  Students typically find titration calculations to be just as challenging as, if not more so than, buffer calculations. Emphasize the importance of writing the relevant balanced reaction for each calculation.  Students again write odd reactions for titrations. Emphasize that protons are changing places, nothing more.  Indicators enable one to conduct a titration without instrumentation like a pH meter. A demonstration is useful to show how quickly the colours change near the equivalence point.  Conceptual Connection 16.4 Acid–Base Titrations

 



16.5 Solubility Equilibria and the Solubility Product Constant  Solving these problems becomes routine since it requires only obtaining the square root of a Ksp value for ionic compounds with a 1:1 cation to anion ratio. For cases involving a 1:2 ratio, one must solve Ksp = 4s3. A common mistake is to forget to square both the s and the 2.  The common ion effect can be demonstrated by adding concentrated hydrochloric acid or concentrated sodium hydroxide to a saturated solution of NaCl.  The increased solubility in acid of an ionic compound containing a base can be easily demonstrated by adding concentrated hydrochloric acid to a mixture containing pure water and a piece of chalk.



.

Students initially presume that equivalence point is synonymous with neutral pH. Calculating the pH at the equivalence point of a weak acid–strong base titration is the biggest challenge. Viewing this point as a series of two steps, complete reaction with the strong base followed by partial dissociation, makes the calculations easier.

Students are sometimes surprised to see huge pH changes with very little added titrant near the equivalence point. Remind them that pH is logarithmic while addition of titrant is not.

The reaction for Ksp is the dissolution of the ionic solid to the dissolved ions, not the other way around.

263

Chapter 16. Aqueous Ionic Equilibrium Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

16.6 Precipitation  Reaction quotient, Q o Q > Ksp: precipitation o Q < Ksp: no precipitation o Q = Ksp: saturated solution  Relative ion concentrations o Unsaturated o Saturated o Supersaturated  Selective precipitation

    

16.7 Qualitative Chemical Analysis  Selective precipitation of cations o Insoluble chlorides o Insoluble sulfides under acidic conditions o Insoluble sulfides under basic conditions o Insoluble hydroxides o Insoluble phosphates o Alkali metals and NH4+ (flame test)

 

16.8 Complex–Ion Equilibria  Complex ion o Metal cation o Ligands  Anions  Molecules  Formation constant, Kf o Equilibrium reaction o Effect on solubility o Calculations  Amphoteric metal hydroxides o Acidic: aluminum complex with water molecules (soluble) o pH‐neutral: aluminum complex with water molecules and hydroxide ions (insoluble) o Basic: aluminum complex primarily with hydroxide ions (soluble)



 

  

.

Unnumbered figure: photo of silver(I) chromate precipitation Figure 16.14 Precipitation from a Supersaturated Solution Example 16.12 Predicting Precipitation Reactions by Comparing Q and Ksp Example 16.13 Finding the Minimum Required Reagent Concentration for Selective Precipitation Example 16.14 Finding the Concentrations of Ions Left in Solution after Selective Precipitation

Figure 16.15 Qualitative Analysis Figure 16.16 A General Qualitative Analysis Scheme Unnumbered figure: photo of sulfides Figure 16.17 Flame Tests

Table 16.3 Formation Constants of Selected Complex Ions in Water at 25 oC Example 16.15 Complex–Ion Equilibria Figure 16.18 Complex–Ion Formation Figure 16.19 Solubility of an Amphoteric Hydroxide

264

Chapter 16. Aqueous Ionic Equilibrium Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

16.6 Precipitation  Predicting precipitate formation involves the same problem the students learned how to solve in Chapter 14: comparing Q with K.  Remind students that one does not simply add an anion to an aqueous solution. The anion is accompanied by a countercation.  Selective precipitation makes use of the differences in Ksp of ionic compounds.



For ionic compounds with a 1:2 cation to anion ratio, students may forget to account for the stoichiometry in the Q expression.



In problems involving solubility and complex ion formation, one often has to combine two reactions, one for Ksp and one for Kf. Students may forget to multiply those equilibrium constants rather than adding them together.

16.7 Qualitative Chemical Analysis  Qualitative analysis is a simple yet powerful method to identify cations present in a solution. Most of the scheme uses selective precipitation.  Flame tests are used for the alkali metals since they do not generally form precipitates.  Qualitative schemes develop logic and problem‐solving skills.

16.8 Complex Ion Equilibria  Complex ions are formed between metal cations and a variety of anionic and neutral ligands. Their formation constants vary but tend to be generally large.  Complex formation can be viewed as another example of Le Châtelier’s principle.  Aluminum (and other cations such as Zn) can form amphoteric hydroxides, ones that have increased solubility in both acidic and basic solutions.  Conceptual Connection 16.5 Solubility and Complex–Ion Equilibria

.

265

Chapter 16. Aqueous Ionic Equilibrium Additional Problem for Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base (Example 16.3)

A 1.0 L buffer solution contains 0.15 mol acetic acid and 0.10 mol sodium acetate. The Ka for acetic acid is 1.8 × 105. Calculate the pH of the solution before and after a 0.010 mol sample of NaOH is added to the buffer. (Ignore any change in volume upon adding the NaOH.)

Initial

Ka = 1.8 × 105

The solution involves 1.0 L, so mol quantities convert directly to molar quantities.

pKa = log(1.8 × 105) = 4.74 [base] 0.10 = 4.74 + log [acid] 0.15 = 4.74  0.176 = 4.564

Find the pH of the buffer using the pKa and the Henderson–Hasselbalch equation.

pH = pK a + log

Stoichiometry

OH(aq) + HA(aq)

The addition of the NaOH base converts a stoichiometric amount of acid to the conjugate base. Write the equation for the neutralization and set up a table to track the changes.

Before

H2O(l) + A(aq)

[OH]

[HA]

[A]

0 mol

0.15 mol

0.10 mol

0.14 mol

0.11 mol

Addition 0.010 mol After

Equilibrium

0 mol

HA(aq) + H2O(l)

H3O+(aq) + A(aq)

Write the equation for the balanced equation and the Henderson–Hasselbalch expression. [base] 0.11 = 4.74 + log [acid] 0.14 = 4.74  0.105 = 4.635

pH = pK a + log

At pH = 4.564, [H3O+] = 2.73 × 105 M At pH = 4.635, [H3O+] = 2.32 × 105 M

Confirm that the x is small approximation is valid: hydronium ion concentration divided by the afteraddition HA concentration.

2.32  105  100% = 0.016% 0.14

.

266

Chapter 16. Aqueous Ionic Equilibrium The buffer pH changed about 0.071 of a pH unit after the hydroxide addition. If the hydroxide was added to pure water, the pH of pure water would change from 7 to 12:

Check

pOH = log[OH] = log(0.010) = 2.00 pOH + pH = 14.00 pH = 14 – 2.00 = 12.00 The pH change after the hydroxide addition was to a larger pH or lower [H3O+] concentration, just as expected.

.

267

Chapter 16. Aqueous Ionic Equilibrium Calculate the molar solubility of iron carbonate FeCO3 in pure water.

Additional Problem for Calculating Molar Solubility from Ksp (Example 16.8)

1) Begin by writing the reaction by which the ionic compound dissolves in water.

Fe2+(aq) + CO32(aq)

FeCO3(s)

Ksp = [Fe2+] [CO32] Write the corresponding expression for Ksp.

2)

Fe2+(aq) + CO32(aq)

FeCO3(s)

Use the stoichiometry of the reaction to write an ICE table, showing the equilibrium concentrations of the ions.

[Fe2+]

[CO32]

Initial

0

0

Change

+s

+s

Equil

s

s

3) Ksp = [Fe2+] [CO32]

Substitute the equilibrium expression with the concentration values from the ICE table.

3.07 × 1011 = (s)(s) s2 = 3.07 × 1011

Substitute the numerical value of Ksp and solve for s.

s = (3.07 × 1011) = 5.54 × 105

[Fe2+] = 5.54 × 105 M [CO32] = 5.54 × 105 M

.

268

Chapter 16. Aqueous Ionic Equilibrium Calculate the molar solubility of calcium sulfate CaSO4 in a solution containing 0.20 mol L-1 Na2SO4.

Additional Problem for Calculating Molar Solubility in the Presence of a Common Ion (Example 16.10) 1) Begin by writing the reaction by which the ionic compound dissolves in water.

Ca2+(aq) + SO42(aq)

CaSO4(s)

Ksp = [Ca2+] [SO42] Write the corresponding expression for Ksp. 2) Use the stoichiometry of the reaction to write an ICE table, showing the equilibrium concentrations of the ions.

Ca2+(aq) + SO42(aq)

CaSO4(s)

Add a column for the common ion.

[Ca2+]

[SO42]

Initial

0

0

Change

+s

+s

Equil

s

0.20 + s

Ksp = [Ca2+] [SO42]

3)

7.10 × 105 = s(0.20 + s) s is small

Substitute the equilibrium expression with the concentration values from the ICE table.

0.20s = 7.10 × 105 s = 7.10 × 105 / 0.20

Substitute the numerical value of Ksp and solve for s.

= 3.55 × 104 [Ca2+] = 3.55 × 104 mol L-1 [SO42] = 0.200355 mol L-1 = 0.20 mol L-1 In the absence of the added sulfate,

Check

Ksp = [Ca2+] [SO42] 7.10 × 105 = (s)(s) s2 = 7.10 × 105 s = (7.10 × 105) = 8.43 × 103 Solubility of [Ca2+] in pure water = 8.43 × 103 mol L-1 (84.3 × 104 mol L-1), a factor of about 25 greater than the solubility with the added sulfate.

.

269

Chapter 16. Aqueous Ionic Equilibrium Additional Problem for Complex–Ion Equilibria (Example 16.15)

A 250 mL portion of a solution that contains 1.5 mol L-1 Cu(NO3)2 is mixed with a solution that is 0.100 mol L-1 NaCN. After equilibrium is reached, what concentration of Cu2+(aq) remains?

1) Begin by writing the balanced equation for the complex ion equilibrium.

Cu2+(aq) + 4CN(aq)

Cu(CN)42(aq)

1.5  10 3 mol 1 L = 7.50  10 4 M 0.250 L + 0.250 L

0.250 L  [Cu2+ ]initial =

Calculate the initial concentrations of the cation and the ligand.

[CN ]initial =

Write the corresponding expression for Kf.

2) Use the stoichiometry of the reaction to write an ICE table showing the initial concentrations of the ions.

0.100 mol 1 L = 0.050 M 0.250 L + 0.250 L

0.250 L  

Kf =

[Cu(CN)4 2 ] = 1.0  1025 [Cu2+ ] [CN ]

Cu2+(aq) + 4CN(aq)

Initial

Cu(CN)42(aq)

[Cu2+]

[CN]

[Cu(CN)42]

7.50×104

0.050

0

Change Equil

3) Since the equilibrium is so large, and the concentration for cyanide is much greater than that for Cu2+, virtually all of the copper ion is consumed. Let x represent the small amount of copper that remains.

Cu2+(aq) + 4CN(aq)

Initial

Cu(CN)42(aq)

[Cu2+]

[CN]

[Cu(CN)42]

7.50×104

0.050

0

Change (7.50×104) 4(7.50×104) (+7.50×104) 0.003 Equil

.

x

0.050

7.50×104

270

Chapter 16. Aqueous Ionic Equilibrium 4) Substitute the equilibrium expression with the concentration values from the ICE table.

[Cu(CN)4 2 ] = 1.0  1025 [Cu2+ ] [CN ]

Kf =

7.50  104 = 1.0  1025 x (0.050)

Substitute the numerical value of Kf and solve for x.

x =

7.50  104 (0.050)1.0  1025

= 1.50  10 27

[Cu2+] = 1.50 × 1027 mol L-1

Check

The copper ion concentration is very low since the formation constant is so large. The value of x was small compared to the change in copper ions (7.50×104).

.

271

Chapter 17. Gibbs Energy and Thermodynamics Student Objectives 17.1 Spontaneous and Nonspontaneous Processes  

Define spontaneous and understand the difference between spontaneous and nonspontaneous processes.

Know and understand that spontaneity and speed are not the same. Spontaneity is dictated by thermodynamics, while speed is dictated by kinetics.

17.2 Entropy and the Second Law of Thermodynamics        

Know that spontaneity is not solely dependent on the enthalpy change for a reaction or process.

Some endothermic processes, like ice melting, are spontaneous under certain conditions.

Know that when ice melts, liquid water evaporates, or NaCl dissolves in water, the arrangement of molecules or ions becomes more disordered.

Define entropy in words and know Boltzmann’s equation for entropy.

Know and understand the difference between a macrostate and a microstate.

Define and understand the second law of thermodynamics.

Understand the example of gas particles in connected flasks as an illustration of entropy change and number of microstates.

Know and rationalize the entropy changes associated with state changes.

Predict the sign of S for common processes.

17.3 Heat Transfer and Changes in the Entropy of the Surroundings    

Know and understand that the entropy change for the universe is defined in terms of the entropy change for the system and the entropy change for the surroundings.

Know that a process in which the system's entropy decreases can still be spontaneous depending on the degree to which the surroundings’ entropy changes.

Know and understand that an exothermic process increases the entropy of the surroundings and an endothermic process decreases the entropy of the surroundings.

Calculate the entropy change for the surroundings in terms of the enthalpy change for the system and the temperature.

17.5 Gibbs Energy   

Define Gibbs energy and express it mathematically in terms of the temperature, enthalpy change, and entropy change of the system.

Understand the meaning of the mathematical sign of the change in Gibbs free energy with respect to spontaneity.

Compute the change in Gibbs free energy and predict spontaneity from T, H, and S.

.

272

Chapter 17. Gibbs Energy and Thermodynamics 17.6 Entropy Changes in Chemical Reactions: Calculating ∆rS°    

Define and understand standard conditions.

Use standard molar entropies to calculate the standard entropy change for a chemical reaction or physical process.

Define and understand the third law of thermodynamics.

Understand the dependence of standard molar entropy on physical state, molar mass, atomic arrangement, and molecular complexity.

17.7 Gibbs Energy Changes in Chemical Reactions: Calculating ∆rG°      

Calculate the standard free energy change for a reaction using the temperature, the standard enthalpy change, and the standard entropy change.

Interpret the meaning of the mathematical sign for the standard free energy change.

Calculate the standard free energy change for a reaction using the standard free energies of formation of products and reactants.

Calculate the standard free energy change for a stepwise reaction.

Define reversible and irreversible reactions and processes.

Understand what is meant by “free” energy.

17.8 Gibbs Energy Changes for Nonstandard States: The Relationship between ∆rG° and ∆rG  

Calculate the free energy change for a reaction or process under nonstandard conditions.

Understand that the free energy change for a reaction or process at equilibrium is exactly zero, indicating nonspontaneity for the reaction or process in either direction.

17.9 Gibbs Energy and Equilibrium: Relating ∆rG° to the Equilibrium Constant (K)  

Calculate and interconvert between the standard free energy change and the equilibrium constant.

Use a two‐point equation to calculate the equilibrium constant for a reaction or process at two different temperatures.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

273

Chapter 17. Gibbs Energy and Thermodynamics Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

17.1 Spontaneous and Nonspontaneous Processes  Spontaneous processes  Nonspontaneous processes  Spontaneity versus speed o Kinetics o Thermodynamics



17.3 Entropy and the Second Law of Thermodynamics  Spontaneity and enthalpy o Endothermic processes: driven by increase in disorder o Examples: ice melting, water vaporizing, NaCl dissolving  Entropy o Definition: S = k ln W o Arrangements  Macrostate  Microstate o Second law of thermodynamics  Suniv > 0 o S = Sfinal  Sinitial o Examples with gas particles and connected flasks: microstates o State changes o “Places” for energy  Translation  Rotation  Vibration



  

        

.

Intro figure: Escher‐type illusion of perpetual motion Figure 17.1 Mechanical Potential Energy and Chemical Potential Figure 17.2 Thermodynamics and Kinetics Unnumbered figure: photos and molecular models of diamond and graphite

Unnumbered figure: photo and molecular models of ice melting Unnumbered figure: photo and molecular models of liquid water evaporating Unnumbered figure: photo and molecular models of NaCl dissolving in water Unnumbered figure: photo of Boltzmann’s tombstone Unnumbered figures: simple illustrations of energy and number of microstates Unnumbered figures: illustrations of gas molecules in separate but connected flasks to illustrate macrostates and microstates Unnumbered figure: illustration of energy flow for melting ice cubes Figure 17.3 Entropy and State Change Figure 17.4 “Places” for Energy Example 17.1 Predicting the Sign of Entropy Change

274

Chapter 17. Gibbs Energy and Thermodynamics Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

17.1 Spontaneous and Nonspontaneous Processes  Mechanical potential energy and chemical potential energy typically increase through nonspontaneous processes. Ask the class for additional examples of nonspontaneous processes.



A common misconception is that spontaneous processes occur at high rates.

This is often but certainly not always true, e.g., the combustion of octane at 25 oC in the absence of a spark or catalyst.

17.2 Entropy and the Second Law of Thermodynamics  Endothermic processes can occur spontaneously if the arrangement of particles changes from an orderly one to a more disorderly one. Even so, the spontaneity of these processes depends on temperature (e.g., the melting of an ice cube).  Entropy is a thermodynamic state function that increases with the number of energetically equivalent arrangements (microstates).  There are many statements of the second law of thermodynamics. The one that is particularly relevant in this section is that for any spontaneous process, the entropy of the universe increases.  Another example to consider using is the probabilities associated with flipping multiple coins.  The entropy of materials increases in transitioning from the solid phase to the liquid and then to the gas phase. There are more different ways to define the locations of the particles as a result of those phase changes, and there are more “places” for energy (vibration, rotation, translation).  Conceptual Connection 17.1 Entropy



Spontaneity is not synonymous with exothermic. Endothermic processes also can be spontaneous. Exothermic processes can be nonspontaneous.

A beginner’s view of entropy relies on the analogy to disorder, but the better picture of entropy is energy equivalence. Spontaneous processes occur such that energy is evenly distributed, e.g., thermal equilibration.



.

275

Chapter 17. Gibbs Energy and Thermodynamics Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

17.3 Heat Transfer and Changes in the Entropy of the Surroundings  Entropy change of the universe o Suniv = Ssys + Ssurr  Temperature dependence of Ssurr o Entropy: energy dispersal per unit temperature o Ssurr = Hsys/T

  

Unnumbered figures: bar graphs of entropy changes for universe, surroundings, and system Figure 17.5 Maxwell–Boltzmann Distribution of Kinetic Energy for Two Different Temperatures Example 17.2 Calculating Entropy Changes in the Surroundings

17.4 Entropy Changes for Phase Transitions  Process occurring reversibly at constant temperature  rS = rH/T

17.5 Gibbs Energy  Gibbs energy (chemical potential) o G = TSuniv o G = H  TS  Spontaneity o Can be determined with data for just the system o Spontaneous: G < 0 o Nonspontaneous: G > 0 o Effect of T on spontaneity  Four cases in Table 17.1

.

  

Figure 17.6 Gibbs Energy Table 17.1 The Effect of rH, rS, T on Spontaneity Example 17.3 Computing Gibbs Energy Changes and Predicting Spontaneity from rH and rS

276

Chapter 17. Gibbs Energy and Thermodynamics Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

17.3 Heat Transfer and Changes in the Entropy of the Surroundings  Spontaneity can be driven by either Ssys, Ssurr, or both.  Conceptual Connection 17.2 Entropy and Biological Systems

17.4 Entropy Changes for Phase Transitions  For constant pressure processes occurring reversibly the entropy can be calculated from the enthalpy of the system.

17.5 Gibbs Energy  The great advantage of using Gibbs energy is that it allows one to focus only on the enthalpy change and entropy change for the system.  Since G = H  TS, the change in free energy can be composed of positive or negative enthalpy and entropy terms. The signs and magnitudes of those terms, as well as the magnitude of T, determine whether a reaction or process is spontaneous.  Conceptual Connection 17.3 H, S, and G

.

 



A decrease in free energy corresponds to a spontaneous process.

Students occasionally forget to watch the unit consistency. H is often in kJ mol‐1, while S is often in J.(molK)‐1.

For a reaction in which G > 0, the reverse reaction is spontaneous.

277

Chapter 17. Gibbs Energy and Thermodynamics Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

17.6 Entropy Changes in Chemical Reactions: Calculating ∆rS°  Standard states o Gas: pressure of 1 bar o Liquid or solid: stable form at 1 bar (often at 25 oC) o Solution: concentration of 1 mol L‐1  Third law of thermodynamics o S = 0 for perfect crystal at 0 Kelvin  Standard molar entropies o Tables of values  Phases  Molar mass  Allotropes  Molecular complexity  Dissolution  Standard entropy change for a reaction  r S o   n p S o ( products)   nr S (reac tants) o

           

17.7 Gibbs Energy Changes in Chemical Reactions: Calculating ∆rG°  Change in Gibbs free energy from ∆rH° and ∆rS° o calculate ∆rH° from tabulated values o calculate ∆rS° from tabulated values o

Change in Gibbs free energy from ΔGf



rG° =  np fG° (products) –  nr fG° (reactants) only for reactions at 298 K Change in Gibbs energy for stepwise reaction "Free" energy o Energy available to do work o Reversible reactions o Irreversible reactions



o

Example 17.6 Estimating the Standard Change in Gibbs Energy for a Reaction at a Temperature Other than 25 oC Using  r G o = r H o  T  r S o

     

.

Example 17.5 Calculating the Standard Change in Free Energy for a Reacting using  r G o = r H o  T  r S o

 r G o = r H o  T  r S o



 



Figure 17.7 Zero Entropy Table 17.2 Standard Molar Entropy Values (So) for Selected Substances at 298 K Unnumbered table: comparative So values for H2O(l) and H2O(g) Unnumbered table: comparative So values for noble gases Unnumbered table: comparative So values for diamond and graphite Unnumbered table: comparative So values for Ar and NO gases Unnumbered table: comparative So values for CO and C2H4 gases Figure 17.8 “Places” for Energy in Gaseous NO Unnumbered table: comparative So values for NO, NO2, and N2O4 gases Unnumbered table: comparative So values for KClO3(s) and KClO3(aq) Figure 17.9 Water Before and After Dissolution of AlCl3 Example 17.4 Calculating Standard Entropy Changes (∆rS°)

Table 17.3 Standard Molar Gibbs Energies of Formation (∆fG°) for Selected Substances at 298 K Example 17.7 Calculating ∆rG° from Standard Gibbs Energies of Formation Chemistry in Your Day: Making a Nonspontaneous Process Spontaneous Example 17.8 Calculating ∆rG° for a Stepwise Reaction Figure 17.10 Gibbs Energy Figure 17.11 A Reversible Process

278

Chapter 17. Gibbs Energy and Thermodynamics Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

17.6 Entropy Changes in Chemical Reactions: Calculating ∆rS°  The third law of thermodynamics defines the entropy of a perfect crystal at 0 Kelvin as zero, suggesting that for all other cases the standard molar entropy values are positive. These values are related to phase, molar mass, structure, and molecular complexity.  Standard entropy changes are determined from the standard entropies of the products and reactants; a larger table of these values can be found in Appendix II.  The standard entropies of many aqueous ions (e.g., OH) are negative. This is because aqueous standard entropies are referenced to that of H+, defined to be 0.



The equation for calculating ∆rS° is nearly identical to that for calculating ∆rH°.

17.7 Gibbs Energy Changes in Chemical Reactions: Calculating ∆rG°  The free energy change for a reaction can be calculated through several methods.  The conditions for reversible reactions make them unrealistic; real reactions are irreversible and do not achieve the theoretical limit of available free energy.



For reactions at temperatures other than 298 K, the equation  r G o = r H o  T  r S o is



.

approximately valid. Molar enthalpy and entropy are functions of temperature, but we use the values tabulated at 298 K as a good approximation even for other temperatures assuming a small variation over that temperature range.

Calculating the free energy change using free energies of formation can be done only for T = 298 K.

279

Chapter 17. Gibbs Energy and Thermodynamics Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

17.8 Gibbs Energy Changes for Nonstandard States: The Relationship Between ∆rG° and ∆rG  Nonstandard conditions o

 r G = r G +RT  ln Q

o o o

G at standard conditions G at equilibrium G at other nonstandard conditions

 

o



17.9 Gibbs Energy and Equilibrium: Relating ∆rG° to the Equilibrium Constant (K)  Free energy change at equilibrium o G = 0 at equilibrium



 r G o =  RT  ln K



Temperature dependence of K  r H o  1   r S o o ln K=  + R T  R



o 



o

ln

K2 K1

=

 r H o  1 R

1     T2 T1 

.

Unnumbered figure: photo of a wet floor and molecular models of H2O Figure 17.12 Gibbs Energy versus Pressure for Water Example 17.9 Calculating ∆rG° Under Nonstandard Conditions

Figure 17.13 Gibbs Energy and the Equilibrium Constant Example 17.10 The Equilibrium Constant and ∆rG° Example 17.11 Using the van’t Hoff Equation to Determine the Standard Enthalpy and Entropy of Reaction Example 17.12 Using the Two‐Point Form of the van’t Hoff Equation to Predict the Equilibrium Constant at a Given Temperature

280

Chapter 17. Gibbs Energy and Thermodynamics Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

17.8 Gibbs Energy Changes for Nonstandard States: The Relationship Between ∆rG° and ∆rG  For nonstandard conditions, the free energy change can be calculated using the reaction quotient, Q.  The free energy change for a reaction at equilibrium is 0.  Conceptual Connection 17.4 Gibbs Energy Changes and Le Châtelier’s Principle



Beginning with a spontaneous reaction (G < 0), the free energy change increases until equilibrium is reached (G = 0).

17.9 Gibbs Energy and Equilibrium: Relating ∆rG° to the Equilibrium Constant (K)  The equilibrium constant for any reaction can be determined by calculating ∆rG at the temperature of interest.  The natural logarithm of the equilibrium constant is linearly related to the inverse of the temperature. A plot of ln K vs. 1/T can be used to find ∆rH° (from the slope) and ∆rS° (from the y‐intercept).



This chapter has a relatively large number of equations, and many students try to memorize all of them rather than working with a few to derive the rest.

.

281

Chapter 17. Gibbs Energy and Thermodynamics

Consider the combustion reaction for calcium:

Additional Problem for Computing Gibbs Energy Changes and Predicting Spontaneity from rH and rS (Example 17.3)

2 Ca(s) + O2(g)  2 CaO(s) H = 1270 kJ, S = 365 J K-1

a) Calculate G at 25 °C and determine whether the reaction is spontaneous. b) Determine at what temperature (if any) the reaction is spontaneous.

a) Use G = H – TS to calculate G from values of H and S (in the same units).

T = 273 + 25 = 298 K

G = H – TS = (1270 × 103 J) – 298 K (365 J K-1) = 1.16 × 106 J = 1.16 × 103 kJ = 1160 kJ

The reaction is spontaneous.

b) G = H – TS

Since S is negative, G will become less negative with increasing temperature. Use G = H – TS to calculate the temperature at which G changes from positive to negative (i.e., G = 0).

0 = (1270 × 103 J) – T (365 J K-1) T (365 J K-1) = 1270 × 103 J T = 1270 × 103 J / 365 J K-1 = 3479 K

The reaction will remain spontaneous until 3479 K. Above that temperature, G will be positive.

.

282

Chapter 17. Gibbs Energy and Thermodynamics Consider the reaction:

Additional Problem for Calculating the Standard Change in Gibbs Energy for a Reaction using rG° = rH° – TrS° (Example 17.5)

NO(g) + ½O2(g)  NO2(g)

Calculate rG° at 25 °C and determine whether the reaction is spontaneous.

Look up standard enthalpies and entropies Begin by looking up the standard enthalpy of formation and standard entropy for each reactant and product.

fH°(kJ mol-1)

S° (in J mol-1 K)

NO(g)

91.3

210.8

O2(g)

0

205.2

NO2(g)

33.2

240.1

Calculate rH° rH° =  np fH° (products) –  nr fH° (reactants)

Use Equation 6.15.

= 33.2 – [½(0) + 91.3] = 58.1 kJ

Calculate rS° rS° =  np S° (products) – nr S° (reactants)

Use Equation 17.16.

= 240.1 – [½(205.2) + 210.8] = 73.3 J K-1

Calculate rG° T = 273 + 25 = 298 K

Use Equation 17.17.

rG° = rH° – TrS° = 58.1 × 103 J – 298 K (73.3 JK-1) = 3.62 × 104 J = 3.62 × 101 kJ = 36.2 kJ The reaction is spontaneous at this temperature. .

283

Chapter 17. Gibbs Energy and Thermodynamics Consider the reaction: Additional Problem for Calculating rG° from Standard Gibbs Energies of Formation (Example CH3Cl(g) + Cl2(g)  CH2Cl2(l) + HCl(g) 17.7)

Use the standard free energies of formation to determine rG° for this reaction at 25°C.

Look up standard free energies of formation fG° (kJ mol-1)

Begin by looking up standard free energies of formation for each reactant and product. Remember that the standard free energy of formation of a pure element is zero.

CH3Cl(g)

60.2

Cl2(g)

0

CH2Cl2(g)

63.2

HCl(g)

95.3

Calculate rG° rG° =  np fG°(products)   nr fG° (reactants)

Use Equation 17.18.

= [63.2 + (95.3)] – [60.2 + (0)] = 98.3 kJ mol-1

The reaction is spontaneous.

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284

Chapter 17. Gibbs Energy and Thermodynamics Use the tabulated free energies of formation to calculate the equilibrium constant for the reaction at 298 K:

Additional Problem for The Equilibrium Constant and rG° (Example 17.10)

H2(g) + Cl2(g)

2HCl(g)

Look up standard free energies of formation fG° (kJ mol-1)

Begin by looking up the standard free energies of formation for each reactant and product.

H2(g)

0

Cl2(g)

0

HCl(g)

95.3

Calculate rG° rG° =  np fG° (products) –  nr fG° (reactants)

Use Equation 17.18.

= [2(95.3)] – [0 + 0] = 190.6 kJ

Calculate K Calculate K from rG° by solving equation 17.20.

rG° = RT ln K

- r G o RT 190.6 103 Jmol1

ln K  

8.314 J(mol K)  298 K  1

K  e76.9  2.50 1033

.

285

Chapter 18. Electrochemistry Student Objectives 18.1 Pulling the Plug on the Power Grid   

Know that electrical power comes from a power grid, a distribution system for electricity.

Know that fuel cell technology can replace the power source for an automobile or a home.

Know that the oxidation and reduction reactions for the combination of hydrogen and oxygen to make water can power a fuel cell.

18.2 Balancing Oxidation–Reduction Equations   

Define oxidation and reduction.

Balance aqueous redox reactions in acidic solution by using the half‐reaction method.

Balance aqueous redox reactions in basic solution by using the half‐reaction method.

18.3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions         

Understand that the components in redox reactions can be separated into two compartments or cells and that the energy released by these reactions can be used to do electrical work.

Define electrochemical cell, voltaic (galvanic) cell, electrolytic cell, half‐cell, and electrode.

Understand the rate of flowing electrons as current, and know the unit for current.

Define potential difference and know its unit.

Define electromotive force, cell potential, and standard cell potential.

Know and understand that electrochemical reactions are a combination of half‐cell reactions, oxidation occurring at the anode, and reduction occurring at the cathode.

Know and understand the components of an electrochemical cell, including the half‐cells, electrodes, and salt bridge.

Understand the direction of electron flow and the potential of a voltaic cell.

Use cell notation to represent electrochemical cells.

18.4 Standard Electrode Potentials     

Know that the standard hydrogen electrode has an electrochemical potential of 0.00 V. Understand standard potentials and use them to generate potentials for electrochemical cells.

Calculate standard potentials for electrochemical cells from standard electrode potentials of the half‐ reactions.

Calculate standard potentials and predict the direction of spontaneity for redox reactions.

Use standard potentials to predict whether metals will dissolve in acid.

18.5 Cell Potential, Gibbs Energy, and the Equilibrium Constant 

Calculate and interconvert between the standard change in Gibbs energy, G°, the standard potential for an electrochemical cell, E°, and the equilibrium constant, K.

18.6 Cell Potential and Concentration   

Use the Nernst equation to calculate the cell potential under nonstandard conditions.

Define concentration cell and understand how a concentration cell works.

Know and understand nerve cells as a biological example of a concentration cell.

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286

Chapter 18. Electrochemistry 18.7 Batteries: Using Chemistry to Generate Electricity   

Know some characteristics of dry‐cell, lead–acid storage, nickel–cadmium, nickel–metal hydride, and lithium–ion batteries.

Know the primary difference between batteries and fuel cells.

Know and understand how the hydrogen–oxygen fuel cell and the fuel‐cell breathalyzer work.

18.8 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity   

Know and understand how an electrolytic cell works.

Predict the products of some common electrolysis reactions.

Calculate and interconvert between current, time, and amount for electrolysis reactions.

18.9 Corrosion: Undesirable Redox Reactions   

Know that corrosion generally refers to the oxidation reactions of metals.

Know the steps by which iron rusts.

Know some strategies for preventing corrosion.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

287

Chapter 18. Electrochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

18.1 Pulling the Plug on the Power Grid  Fuel cells o Definition o Uses o Reaction



18.2 Balancing Oxidation–Reduction Equations  Redox reactions o Oxidation: loss of electrons, increase in oxidation number o Reduction: gain of electrons, decrease in oxidation number  Balancing redox reactions o Oxidation numbers o Separation of half‐reactions o Mass balance o Charge balance o Electron balance o Recombination of half‐reactions



18.3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions  Properties of voltaic cells o Electrical current: ampere (A)  1 A = 1 C s‐1 o Potential difference: volt (V)  1 V = 1 J C‐1  Components of voltaic cells o Half‐cells o Electrodes  Anode: oxidation  Cathode: reduction  Inert electrodes o Salt bridge  Cell notation



 



  

.

Intro figure: illustration of an MP3 player powered by a fuel cell Unnumbered figure: photo of a fuel cell capable of providing power for a house Unnumbered figure: hydrogen fuel cell bus

Example 18.1 Half‐Reaction Method of Balancing Aqueous Redox Equations in Acidic Solution Example 18.2 Balancing Redox Reactions Occurring in Basic Solution

Figure 18.1 A Spontaneous Oxidation– Reduction Reaction Figure 18.2 A Voltaic Cell Figure 18.3 An Analogy for Electrical Current Figure 18.4 Inert Platinum Electrode

288

Chapter 18. Electrochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

18.1 Pulling the Plug on the Power Grid  This example of a fuel cell is based on the reaction of hydrogen with oxygen. This reaction can be demonstrated as an explosion; the same reaction in a fuel cell releases the same amount of energy but in a controlled way.  Batteries are ubiquitous devices and should motivate interest in this subject. Ask for applications of batteries in everyone’s daily lives.



The combustion of hydrogen with oxygen occurs rapidly and often explosively. The hydrogen–oxygen fuel cell controls the kinetics but harnesses the same amount of energy.

18.2 Balancing Oxidation–Reduction Equations  Identifying the individual half‐reactions is generally straightforward.  Particularly in redox reactions, both mass and charge must be balanced.  There are other slightly different workflows for balancing redox reactions.  Working through a few examples in class helps students get the hang of balancing these reactions.



Half‐reactions that are not balanced initially with respect to mass (e.g., oxygen on one side but not the other side) are initially confusing to some students.

18.3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions  The voltage or potential between the two half‐cells drives the current.  Demonstrations of redox reactions and voltaic cells are simple and quite satisfying to students.  Electrochemical cell notation summarizes the critical components of an electrochemical cell in a compact form.



An oxidation or reduction cannot occur in the absence of its counterpart. It can appear that they are independent since each half‐ cell reaction appears to occur in an individual compartment.

The presence of inert electrodes in cell notation is initially confusing to some students.



.

289

Chapter 18. Electrochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

18.4 Standard Electrode Potentials  Standard hydrogen electrode o Reaction o Half‐cell potential  Standard electrode potentials o Measured versus reference like SHE o o o  E cell  E cathode  E anode o Table of values  Sign and value  Relative oxidizing strength o Predicting spontaneous direction o Metals dissolving in acid

           

18.5 Cell Potential, Gibbs Energy, and the Equilibrium Constant  Gibbs energy change, rGo o Faraday’s constant, F = 96,485 C/mol e o rGo = nFEo  Equilibrium constant RT 0.0592 V o ln K = log K o E = nF n

.



Figure 18.5 An Analogy for Electrode Potential Figure 18.6 The Standard Hydrogen Electrode Figure 18.7 Measuring Standard Electrode Potential Figure 18.8 Electrochemical Cell with SHE as the Anode Table 18.1 Standard Electrode Potentials at 25 oC Example 18.3 Calculating Standard Potentials for Electrochemical Cells from Standard Electrode Potentials—1 Example 18.4 Calculating Standard Potentials for Electrochemical Cells from Standard Electrode Potentials—2 Figure 18.9 Mn/Ni2+ Electrochemical Cell Example 18.5 Predicting Spontaneous Redox Reactions and Sketching Electrochemical Cells Figure 18.10 Electrochemical Cell Composed of Mg2+/Mg and Fe2+/Fe Electrodes Figure 18.10 Electrochemical Cell Composed of Mg2+/Mg and Cr3+/Cr Electrodes Unnumbered figure: photo of reaction of Zn with HCl(aq)



Unnumbered figure: relating Gibbs energy to equilibrium constant o Example 18.6 Relating rGo and Ecell



Example 18.7 Relating Ecell and K



Unnumbered figure: relating Gibbs energy to equilibrium constant with formulas

o

290

Chapter 18. Electrochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

18.4 Standard Electrode Potentials  The electrode potential of SHE has been set at zero. Other potentials are measured relatively.  The standard potential for an electrochemical cell determines the direction of spontaneity.  Conceptual Connection 18.1 Selective Oxidation  Conceptual Connection 18.2 Metals Dissolving in Acids



18.5 Cell Potential, Gibbs Energy, and the Equilibrium Constant  The standard cell potential, the standard Gibbs energy change, and the equilibrium constant can be used to predict the spontaneity of a reaction under standard conditions.  Conceptual Connection 18.3 Periodic Trends and the Direction of Spontaneity for Redox Reactions







.

Half‐reactions with negative electrode potentials can still be spontaneous depending on the potential of the other half‐ reaction.

Stronger oxidizing agents have higher electrode potentials; stronger reducing agents have more negative electrode potentials. Students sometimes forget that a strong oxidizing agent itself is reduced.

When balancing redox reactions, one must often multiply a half‐reaction by a factor such as 2 or 3 to balance the electrons. Students are tempted to multiply the standard half‐cell potential by that same factor.

Students often try to memorize all these equations and their variations, but they all can be derived easily using relationships the students already know from thermodynamics.

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Chapter 18. Electrochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

18.6 Cell Potential and Concentration  Nonstandard conditions o Reaction quotient, Q o







  

Concentration cells o Identical electrodes, different concentrations o Concentration equilibration Human nerve cells o Ion channels in membranes o Sodium–potassium pump o Membrane potential

  

18.7 Batteries: Using Chemistry to Generate Electricity  Battery types o Dry‐cell  Standard: graphite and zinc electrodes; moist MnO2 and NH4Cl  Alkaline: graphite and zinc electrodes; moist MnO2 and KOH o Lead–acid storage  Rechargeable  Six cells for total of 12 V o Other rechargeable  Nickel–cadmium  Nickel–metal hydride  Lithium–ion  Fuel cell o Definition o Example: hydrogen–oxygen cell o Example: fuel‐cell breathalyzer

       

.

Figure 18.12 Cell Potential and Concentration Example 18.8 Calculating Ecell Under Nonstandard Conditions Figure 18.12 Cu/Cu2+ Concentration Cell Example 18.9 Determining the Ksp of a Slightly Soluble Salt Using a Concentration Cell Chemistry and Medicine: Concentration Cells in Human Nerve Cells Figure 18.14 Concentration Changes in Nerve Cells Figure 18.15 Potential Changes Across the Nerve Cell Membrane

Figure 18.16 Dry‐Cell Batteries Figure 18.17 Lead–Acid Storage Battery Unnumbered figure: photo of rechargeable batteries and charger Table 18.2 Energy Density and Overcharge Tolerance of Several Rechargeable Batteries Figure 18.18 Lithium–Ion Battery Figure 18.19 Hydrogen–Oxygen Fuel Cell Chemistry in Your Day: The Lithium Battery Figure 18.20 Mesoporous Carbon

292

Chapter 18. Electrochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

18.6 Cell Potential and Concentration  The potential of a cell can be determined from the standard cell potential and the concentration ratios at nonstandard conditions, i.e., the Q expression.  Concentration cells continue to operate until the concentrations become equal. This is the same concept as thermal equilibration and heat flow.  A nerve cell is an excellent example of a concentration cell with biological relevance.  Conceptual Connection 18.4 Relating Q, K, Ecell, and E°cell

 

The Nernst equation is easily derived using the equation for G from thermodynamics.

An electrochemical cell can be established using two identical electrodes; if the concentrations are different, the cell potential will not be zero.

18.7 Batteries: Using Chemistry to Generate Electricity  Batteries are necessary for keeping electronics operating, so there may be considerable interest. A type of battery comes in a specific voltage; the size determines the amount of current available. Larger potentials represent multiple cells connected in series.  Fuel cells also generate a specific voltage.

Some devices are capable of high throughput of the chemical reactants and can generate considerable power.

.

293

Chapter 18. Electrochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

18.8 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity  Reactivity o Spontaneous: voltaic cell o Nonspontaneous: electrolytic cell  Design  Examples o Electrolysis of water o Silver plating  Predicting products o Pure molten salts o Mixtures of cations and anions o Aqueous solutions o Aqueous NaCl and overvoltage  Stoichiometry

    

18.9 Corrosion: Undesirable Redox Reactions  Corrosion definition o Oxidation of metals o Reduction of oxygen o Examples of rust  Preventing corrosion o Paint and covering o Sacrificial electrodes o Galvanizing

    

    



.

Figure 18.21 Electrolysis of Water Figure 18.22 Silver Plating Figure 18.23 Voltaic versus Electrolytic Cells Figure 18.24 Electrolysis of Molten NaCl Unnumbered figure: photo of battery‐ powered electrolysis of water Figure 18.25 Electrolysis of Aqueous NaI Figure 18.26 Electrolysis of Aqueous NaCl: The Effect of Overpotential Example 18.10 Predicting the Products of Electrolysis Reactions Figure 18.27 Electrolytic Cell for Copper Plating Example 18.11 Stoichiometry of Electrolysis

Unnumbered figure: photo of rusted ship Unnumbered figure: photo of aluminum pipe Unnumbered figure: scratch in paint Figure 18.28 Corrosion of Iron: Rusting Unnumbered figure: photo of galvanized nails Unnumbered figure: illustration of sacrificial anode

294

Chapter 18. Electrochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

18.8 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity  Electrolytic cells use electrical input from an external source to drive a nonspontaneous electrochemical reaction. The applied potential determines what kinds of reactions can occur; the current and time determine the moles of material converted.



Students sometimes have trouble predicting the products of electrolysis reactions in aqueous solution because they neglect the possibility of decomposing the water itself.

18.9 Corrosion: Undesirable Redox Reactions  Corrosion typically occurs when metals are converted to their metal oxides. Rust, a metal oxide of iron, does not have the desired structural properties that iron and steel have.  Corrosion is prevented by physically blocking the oxidation reaction or by providing a sacrificial electrode.  Conceptual Connection 18.5 Sacrificial Anodes

.

295

Chapter 18. Electrochemistry Additional Problem for Calculating Standard Potentials for Electrochemical Cells from Standard Electrode Potentials (Examples 18.3 and 18.4)

Use tabulated standard half-cell potentials to calculate the standard cell potential for the reaction in an electrochemical cell at 25 °C:

Zn2+(aq) + H2O2(aq)  Zn(s) + O2(g) + 2 H+(aq)

Separate into half-cells Begin by separating the reaction into oxidation and reduction half-reactions. [In this case, the zinc cations form zinc metal, a reduction reaction.]

Oxidation: H2O2(aq) + 2 e  O2(g) + 2 H+(aq)

Reduction: Zn2+(aq)  Zn(s) + 2 e-

Assign Potentials Look up the standard half-cell potentials for each half-reaction. For the oxidation, o remember to change the sign of Ered to obtain o Eox . Add the half-cell reactions together.

Ox: 2 e + H2O2(aq)  Red: Zn2+(aq)

O2(g) + 2 H+(aq)

 Zn(s) + 2 e

_________________________________________

Zn2+(aq) + H2O2(aq)  Zn(s) + O2(g) + 2 H+(aq)

o o Eox =  Ered = 0.70 V o Ered = 0.76 V

Solve Add the half-cell potentials together to obtain the overall standard potential.

o o o Ecell = Eox + Ered

= 0.70 V + (0.76 V) = 1.46 V

.

296

Chapter 18. Electrochemistry o

Use the tabulated half-cell potentials to calculate K for the oxidation of nickel by chlorine:

Additional Problem for Relating Ecell and K (Example 18.7)

Cl2(g) + Ni(s)  2 Cl-(aq) + Ni2+(aq)

Given Cl2(g) + Ni(s)  2 Cl(aq) + Ni2+(aq)

Sort You are given the redox reaction and asked to find K.

Find K

Strategize

Conceptual Plan

Use the tabulated values of the half-cell potentials o to calculate Ecell . Then use Equation 18.7 to

o o Eox , Ered

o Ecell



o o o Ecell = Eox + Ered

o

calculate K from Ecell .

o Ecell



E o cell =

Solve

Solution

Break the reaction up into oxidation and reduction half-reactions and find the standard half-cell o potentials for each. Sum the values to find Ecell .

Ox:

K

0.0592 V log K n

Ni(s) 

2 e + Ni2+(aq)

Red: Cl2(g) + 2 e  2 Cl-(aq) _________________________________ Cl2(g) + Ni(s)  2 Cl-(aq) + Ni2+(aq)

o o Eox =  Ered = +0.23 V o Ered = 1.36 V

o o o Ecell = Eox + Ered

= +0.23 V + 1.36 = +1.59 V .

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Chapter 18. Electrochemistry

0.0592 V log K n 0.0592 V 1.59 V = log K 2 2 (1.59) log K = 0.0592 K = 1053.72

o

Compute K from Ecell . The value of n (the number of moles of electrons) corresponds to the number of electrons that are cancelled in the half-reactions.

E o cell =

= 5.25  1053

The reaction is spontaneous and strongly favours the products. Chlorine is a powerful oxidizing agent.

Check

.

298

Chapter 18. Electrochemistry An electrochemical cell is based on two halfreactions:

Additional Problem for Calculating Ecell Under Nonstandard Conditions (Example 18.8)

Oxidation: Fe(s)  Fe2+(aq, 0.010 mol L-1) + 2 e

Reduction: Br2(l) + 2 e  2 Br-(aq, 1.0 mol L-1)

Compute the cell potential.

Sort

Given [Fe2+] = 0.010 mol L-1; [Br-] = 1.0 mol L-1

You are given the half-reactions for a redox reaction and the concentrations of the aqueous reactants and products. You are asked to find the cell potential.

Find Ecell

Strategize

Conceptual Plan

Use the tabulated values of the half-cell potentials o to calculate Ecell . Then use Equation 18.10 to calculate Ecell.

o o Eox , Ered



o Ecell

o Ecell , [Fe2+], [Br-]  Ecell

Ecell = E o cell 

Solve

Solution

Write the oxidation and reduction half-reactions and sum them. Find the standard half-cell potentials of o each and sum them to find Ecell .

Ox:

0.0592 V log Q n

Fe(s)  2 e + Fe2+(aq)

Red: Br2(l) + 2 e  2 Br-(aq) _____________________________________ Br2(l) + Fe(s)  2 Br-(aq) + Fe2+(aq)

o o Eox =  Ered = +0.45 V o Ered = 1.09 V

.

299

Chapter 18. Electrochemistry o o o Ecell = Eox + Ered

= +0.45 V + 1.09 = +1.54 V

o

Compute Ecell from Ecell . The value of n is the number of moles of electrons. Determine Q based on the overall balanced equation and the given concentrations of reactants and products.

0.0592 V log Q n 0.0592 V = E ocell  log [Fe2+ ] [Br  ]2 n 0.0592 = 1.54 V  log (0.010)(1.0) 2 = 1.54 V  0.0296 (  2) = 1.60 V

Ecell = E ocell 

The reaction is spontaneous. Bromine is a powerful oxidizing agent.

Check

.

300

Chapter 18. Electrochemistry

Additional Problem for Stoichiometry of Electrolysis (Example 18.11)

Aluminum can be plated out of a solution containing Al3+ according to the half-reaction: Al3+(aq) + 3 e  Al(s) What mass of aluminum (in grams) will be plated by the flow of 50 A of current 10 hours?

Sort You are given the half-reaction for the plating of aluminum, which shows the stoichiometry between the moles of aluminum and the moles of electrons. You are also given the current and time. You are asked to find the mass of aluminum that will be deposited in that time.

Given 3 mol e = 1 mol Al 50 A 10 hours

Find g Al

Strategize

Conceptual Plan

You need to find the amount of aluminum, which is related stoichiometrically to the number of electrons that have flowed through the cell.

Begin with time in hours and convert to seconds.

Given the current and time, find the number of coulombs. Then use Faraday’s constant to calculate the number of moles of electrons and the stoichiometry of the reaction to find the number of moles of aluminum. Finally, use the molar mass of aluminum to convert to the mass of aluminum.

hr  s  C  mol e-  mol Al  g Al 60 min 60 s 1 hr 1 min

Solve

Solution

Follow the conceptual plan, cancelling units to arrive at the mass of aluminum.

= 10 hr 

50 C s

1 mol e 96,485 C

1 mol Al 3 mol e 

26.98 g Al 1 mol Al

3600 s 50 C 1 mol e 1 mol Al 26.98 g Al     hr s 96,485 C 3 mol e 1 mol Al

= 168 g Al

Check

The units (g Al) are correct. The magnitude of the answer seems to make sense. (This amount is about 14 drink cans.) It takes a considerable amount of electricity to make the aluminum. In fact, aluminum is made electrolytically and is energy-intensive and expensive for this reason. This is why there is incentive to recycle aluminum cans.

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301

Chapter 19. Radioactivity and Nuclear Chemistry Student Objectives 19.1 Medical Isotopes  

Know that radioactivity, the natural emission of subatomic particles from atoms, can be used in medical applications like radiology, the practice of diagnostic imaging.

Define radioactivity.

19.2 The Discovery of Radioactivity  

Know that radioactive elements emit energy that is capable of penetrating some types of matter, including tissues of the body.

Know that radioactive substances can expose photographic plates, a phenomenon that was discovered by Becquerel and further studied by Marie and Pierre Curie.

19.3 Types of Radioactivity         

Define nuclide and recall the conventions for writing nuclear symbols.

Know that an alpha particle is the same as a helium‐4 nucleus.

Define ionizing power and penetrating power.

Know that a beta particle is the same as an electron and is emitted when a neutron changes to a proton.

Know that in beta decay, the mass number of the radioactive nucleus stays the same.

Know that a gamma particle is a high‐energy photon and is often emitted in conjunction with the other types of radiation.

Know that a positron is the antiparticle of an electron and that it is emitted when a proton changes to a neutron.

Know that electron capture occurs when an electron from an inner orbital is combined with a nuclear proton to form a nuclear neutron.

Write nuclear symbols for all the subatomic particles, and write balanced reactions for all the types of radioactive decay.

19.4 The Valley of Stability: Predicting the Type of Radioactivity    

Know that atomic nuclei are held together by the strong force.

Predict the relative stability and type of radioactive decay of isotopes based on their atomic number and N/Z ratio.

Know that the stability of an isotope depends on its absolute number of protons and neutrons.

Know that a radioactive decay series characterizes the steps by which an isotope transforms into other smaller and lighter atoms.

19.5 Detecting Radioactivity   

Know that radioactivity exposure in humans is often measured with film‐badge dosimeters, photographic film in a light‐proof case that responds to other forms of radiation besides light.

Know that the Geiger‐Müller counter is a device that can count particles emitted by radioactive nuclei.

Know that a scintillation counter is another device that detects low levels of radiation.

.

302

Chapter 19. Radioactivity and Nuclear Chemistry 19.6 The Kinetics of Radioactive Decay and Radiometric Dating       

Know that natural radioactivity is a part of our environment, unrelated to human activity, and understand that it is a consequence of the instability of all nuclei beyond atomic number 83.

Know and understand that all radioactive nuclei decay via first‐order kinetics and that the half‐life of a radioactive substance is independent of the amount of the substance.

Know that radon is a colourless, odourless gas that is emitted from the ground via the natural radioactive decay of uranium.

Know that the natural, radioactive isotope carbon‐14 decays with a known half‐life and that its concentration in dead organic material can be used to measure the age of the material through radiocarbon dating.

Calculate and predict the age of a sample based on its concentration of carbon‐14.

Know and understand that the ratio of uranium‐238 to lead‐206 within igneous rocks can be used to measure the age of the rocks.

Calculate and predict the age of a rock based on its ratio of uranium‐238 to lead‐206.

19.7 The Discovery of Fission: The Atomic Bomb and Nuclear Power    

Define nuclear fission.

Understand how the fission of an isotope like uranium‐235 can lead to a self‐amplifying chain reaction.

Know that the first nuclear weapons, developed by the United States in the 1940s, were fission weapons.

Know and understand that nuclear reactors take the energy from a small, controlled fission reaction and convert the energy to heat to produce electrical power through conventional methods, e.g., steam‐driven turbines.

19.8 Converting Mass to Energy: Mass Defect and Nuclear Binding Energy   

Know and understand why nuclear reactions release far more energy than chemical reactions.

Define mass defect and nuclear binding energy.

Calculate the mass defect and nuclear binding energy for a radioactive isotope.

19.9 Nuclear Fusion: The Power of the Sun  

Define nuclear fusion.

Know that attempts at human‐made nuclear fusion are still at experimental stages and are being pursued because fusion does not involve radioactivity or radioactive byproducts.

19.10 Nuclear Transmutation and Transuranium Elements  

Define transmutation.

Know and understand that linear accelerators and cyclotrons are devices used to transmutate elements for experimentation and to produce medically important isotopes for diagnosis and treatment of disease.

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303

Chapter 19. Radioactivity and Nuclear Chemistry 19.11 The Effects of Radiation on Life   

Know that radioactivity in the body can do damage and is classified into three groups: acute radiation damage, increased cancer risk, and genetic effects.

Know that exposure to radioactivity can come from several sources including natural radiation, diagnostic medical procedures, and consumer products.

Know and understand that the dose and time extent of a radiation exposure give rise to a variety of adverse effects, including death.

19.12 Radioactivity in Medicine and Other Applications   

Define radiotracer and recognize some examples.

Know that radioactive isotopes are used for radiotherapy in medicine.

Know that radioactivity is used in a variety of applications, including the sterilization of implanted medical devices, the destruction of bacteria and parasites that spoil foods, and the sterilization of insects that carry disease.

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304

Chapter 19. Radioactivity and Nuclear Chemistry Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

305

Chapter 19. Radioactivity and Nuclear Chemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

19.1 Medical Isotopes  Radioactivity  Medical isotopes



19.2 The Discovery of Radioactivity  Phosphorescence: glow‐in‐the‐dark  Radioactivity history o Antoine Henri‐Becquerel (1896) o Marie Sklodowska Curie o Pierre Curie



19.3 Types of Radioactivity  Alpha decay 4 o 2 protons + 2 neutrons: 2 He nucleus o High ionizing power, low penetrating power  Beta decay 0 o Electron: 1 e o Conversion of neutron to proton and electron o Medium ionizing power and penetrating power  Gamma ray emission o Photon:  o No charge or mass o Accompanies other decays o Low ionizing power, high penetrating power  Positron emission 0 o Positron: 1 e  Antiparticle of electron o Conversion of proton to neutron and positron  Electron capture o Inner‐orbital electron + proton  neutron





 

.

     

Intro figure: illustration of decaying atom, scan of infected appendix Unnumbered figure: Chalk River Laboratories

Unnumbered figure: photograph of phosphorescence Figure 19.1 The Discovery of Radioactivity Unnumbered figure: photo of Marie Curie

Unnumbered figure: photo of glow‐in‐ the‐dark watch Figure 19.2 Alpha Decay Example 19.1 Writing Nuclear Equations for Alpha Decay Figure 19.3 Beta Decay Figure 19.4 Positron Emission Example 19.2 Writing Nuclear Equations for Beta Decay, Positron Emission, and Electron Capture Table 19.1 Modes of Radioactive Decay

306

Chapter 19. Radioactivity and Nuclear Chemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

19.1 Medical Isotopes  Many students likely have personal experience with some type of radiological procedure. X‐rays are not a form of radioactivity but are high‐energy electromagnetic photons similar to gamma rays. Ask the students to suggest other radiological techniques.  Various radiological techniques collect information by penetrating tissues. Only a few involve radioactivity or nuclear chemistry.



X‐rays and gamma rays are different only by their source. X‐rays come from high‐energy electronic transitions, while gamma rays come from nuclear transitions.

19.2 The Discovery of Radioactivity  Becquerel, the Curies, and other early experimenters did not understand the power of the radioactivity that they observed by exposing photographic film. It is likely that Marie Curie shortened her life via the biological damage she sustained by conducting experiments without adequate protection.  Radium Girls: Women and Industrial Health Reform by Claudia Clark recounts women who painted radium and other radioactive isotopes on watch and clock faces. They often licked their paint brushes to maintain a "sharp" tip, thereby ingesting the radioactive material.



Radioactivity for the most part cannot be seen, but it can stimulate phosphorescence. Film and the Geiger‐Müller counter are commonly used.

19.3 Types of Radioactivity  The different types of radioactivity have different properties with respect to their detection and to their biological hazards.  Conceptual Connection 19.1 Alpha and Beta Decay



Only certain atoms and their isotopes are naturally radioactive, and each decays by a specific pathway.

.

307

Chapter 19. Radioactivity and Nuclear Chemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

19.4 The Valley of Stability: Predicting the Type of Radioactivity  Strong force  Ratio of neutrons to protons: N/Z o Valley of stability  Up to Z = 83 o N/Z too high: beta decay o N/Z too low: positron emission or electron capture  Magic numbers o Absolute numbers of protons, neutrons

 

19.5 Detecting Radioactivity  Film‐badge dosimeter  Geiger‐Müller counter  Scintillation counter

 

Figure 19.7 Film‐Badge Dosimeter Figure 19.8 Geiger‐Müller Counter

19.6 The Kinetics of Radioactive Decay and Radiometric Dating  Radioactive nuclei: first‐order decay o Rate proportional to number of nuclei o Rate constant depends on isotope 0.693 o Half‐life: t1/ 2 = k  Radon o Colourless, odourless gas o Naturally occurs from uranium in soil  Integrated rate law N o ln t = kt N0



Table 19.3 Selected Nuclides and Their Half‐ Lives Figure 19.9 The Decay of Radon‐220 Chemistry in Your Day: Uranium Isotopes and the CANDU Reactor Unnumbered figure: CANDU nuclear power plant in Qinshan Example 19.4 Radioactive Decay Kinetics Unnumbered figure: photo of Dead Sea Scrolls Unnumbered figure: photo of bristlecone pine tree Example 19.5 Radiocarbon Dating Example 19.6 Uranium/Lead Dating to Estimate the Age of a Rock



 

       

Applications o Radiocarbon dating  14C: t1/2 = 5730 yr  Dating method for organic materials < 50,000 yr o Uranium/lead dating  238U: t1/2 = 4.5 × 109 yr

.

Figure 19.5 Stable and Unstable Nuclei Example 19.3 Predicting the Type of Radioactive Decay Table 19.2 Number of Stable Nuclides with Even and Odd Numbers of Nucleons Figure 19.6 The Uranium‐238 Radioactive Decay Series

308

Chapter 19. Radioactivity and Nuclear Chemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

19.4 The Valley of Stability: Predicting the Type of Radioactivity  Protons are in close proximity (~1015 m) in the nucleus and repel each other. The strong force or nuclear force holds the nucleus together. The number of neutrons is important to the stability of isotopes, as demonstrated by the neutron‐to‐proton ratio.  Stable nuclides have specific patterns of numbers of neutrons and protons referred to as magic numbers. Even–even is most common, while odd–odd is rare.



19.5 Detecting Radioactivity  Radioactivity detection is simple and still makes use of a historic system: the exposure of film.  The Geiger‐Müller counter is a portable device for counting decay events.  The scintillation counter is used to measure radioactivity in low levels in liquid samples.



Radioactivity is invisible.

19.6 The Kinetics of Radioactive Decay and Radiometric Dating  The first‐order rate law, integrated rate law, and half‐life equation can be used for all radioactive decays. The rate constants for these decays vary considerably from fractions of a second to billions of years.  Radon is a radioactive gas that must be trapped to be analyzed. It is typically absorbed onto carbon or charcoal, a material with a large surface area.  Radiocarbon dating is a powerful technique but is limited to times less than 50,000 years; very short times also are not as accurately measured.  Uranium/lead dating is used for very long times and volcanic rock samples.  Conceptual Connection 19.2 Half‐Life



Radiocarbon dating can be used to determine the age of a material that contained a biological source of carbon, i.e., one that exchanged carbon with the environment. As long as carbon is being exchanged, the relative amount of carbon‐14 remains constant.



.

All nuclei above atomic number 83, regardless of N/Z, are radioactive.

The stable N/Z changes as atomic number increases. A 1:1 ratio is stable for low numbers of protons but not for relatively high numbers.

309

Chapter 19. Radioactivity and Nuclear Chemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

19.7 The Discovery of Fission: The Atomic Bomb and Nuclear Power  Uranium o 235U  Can undergo fission and create a self‐sustaining chain reaction  Only 1% of naturally occurring uranium  History o Enrico Fermi o Lise Meitner, Fritz Strassmann, Otto Hahn o Manhattan Project: Albert Einstein, J. R. Oppenheimer  Applications o Atomic weapons: Hiroshima, Nagasaki o Nuclear reactor: electricity generation



19.8 Converting Mass to Energy: Mass Defect and Nuclear Binding Energy  Einstein's equation: E = mc2  Mass defect  Nuclear binding energy  Binding energy curve o Fusion vs. fission o Predicting reactions



19.9 Nuclear Fusion: The Power of the Sun  High temperature requirement of fusion  Examples o Hydrogen bombs o Energy source of stars o Production of electricity

 

Figure 19.13 A Nuclear Fusion Reaction Figure 19.14 Tokamak Fusion Reactor

19.10 Nuclear Transmutation and Transuranium Elements  Transmutation  Facilities o Stanford Linear Accelerator o Fermilab



Unnumbered figure: photo of Joliot‐ Curies Figure 19.15 The Linear Accelerator Unnumbered figure: photos of SLAC and Fermi lab Figure 19.16 The Cyclotron

     



  

.

Unnumbered figure: photo of Lise Meitner Unnumbered figure: illustration of uranium‐235 fission Figure 19.10 A Self‐Amplifying Chain Reaction Unnumbered figure: on July 16, 1945, in the New Mexico desert, the world’s first atomic bomb Table 19.4 Selected Countries that Use Nuclear Power Figure 19.11 A Nuclear Reactor Unnumbered figure: photo of Chernobyl reactor

Example 19.7 Mass Defect and Nuclear Binding Energy Figure 19.12 Nuclear Binding Energy per Nucleon

310

Chapter 19. Radioactivity and Nuclear Chemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

19.7 The Discovery of Fission: The Atomic Bomb and Nuclear Power  Nuclear fission is not a natural phenomenon; it requires energetic neutrons bombarding target atoms.

Furthermore, the target atom is chosen so that an increase in mass is sufficient to trigger the fragmentation.  Controlled nuclear fission provides useful energy in nuclear reactors, while uncontrolled fission is used for nuclear weapons.  Nuclear reactors and the production of nuclear weapons create nuclear waste, a potential environmental problem. More than 80 percent of France’s electricity comes from nuclear power, and the French maintain a reactor capable of accommodating the waste.



Uranium‐238 does not undergo fission under normal circumstances, but it can be made to undergo fission in thermonuclear devices.

19.8 Converting Mass to Energy: Mass Defect and Nuclear Binding Energy  Unlike chemical reactions, mass is not conserved in nuclear processes.  The large energies from small mass changes make nuclear reactions far more powerful than chemical reactions.  The binding energy per nucleon vs. mass number enables prediction of atoms susceptible to fusion or fission.



The mass of a nucleus is not the simple sum of the masses of the protons and neutrons. Some of that mass is converted to energy, referred to as the binding energy.

Keeping careful track of units is key for calculations in this section.

Energy changes can be calculated per atom or per mole of atoms. It is useful to remind students that a joule is a derived SI unit.

19.9 Nuclear Fusion: The Power of the Sun  Fusion involves the lightest isotopes, typically hydrogen, deuterium, and tritium.  Fusion provides more energy per mass of fuel and does not produce radioactive byproducts.  Extremely high temperatures are required for fusion because of the electrostatic repulsion between the positively charged nuclei.



Fusion has been accomplished in thermonuclear weapons in which a fission reaction released the energy and high temperatures required to initiate the fusion reaction.

19.10 Nuclear Transmutation and Transuranium Elements  Transmutations create one element from another in devices like linear accelerators and cyclotrons. They are an important source of knowledge for specific isotopes with medical uses.  Conceptual Connection 19.3 Nuclear Transformations



Cyclotrons are not necessarily huge devices. Some are nearly small enough to fit on a tabletop.

.



311

Chapter 19. Radioactivity and Nuclear Chemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

19.11 The Effects of Radiation on Life  Types o Acute radiation damage o Increased cancer risk o Genetic defects  Measuring exposure o Becquerel (Bq) o Gray (Gy) o Sievert (Sv) o Radiation weighting factor (WR)  Sources o Natural radiation o Diagnostic medical procedures o Consumer products  Exposure effects

 

Table 19.5 Radiation Weighting Factors Table 19.6 Effects of Instantaneous Radiation Exposure

19.12 Radioactivity in Medicine and Other Applications  Diagnosis in medicine o Radiotracers o Positron emission tomography (PET)  Radiotherapy in medicine o Gamma radiation  Other applications o Kill bacteria on medical devices o Kill bacteria and parasites in food o Sterilize insects

    

Figure 19.17 A Bone Scan Table 19.7 Common Radiotracers Figure 19.18 A PET Scan Figure 19.19 Radiotherapy for Cancer Figure 19.20 Irradiation of Food

.

312

Chapter 19. Radioactivity and Nuclear Chemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

19.11 The Effects of Radiation on Life  Exposure to radiation can cause acute radiation damage (rare exposure), increased cancer risk (more common), and genetic defects.  Conceptual Connection 19.4 Radiation Exposure

19.12 Radioactivity in Medicine and Other Applications  There are substantial beneficial uses of radioactivity: diagnosis and radiotherapy in medicine, sterilization of foods, control of insects.



.

There is common misunderstanding about food irradiation. The food is treated with gamma rays and not with the radioactive materials themselves.

313

Chapter 19. Radioactivity and Nuclear Chemistry

Additional Problem for Radioactive Decay Kinetics (Example 19.4)

Fluorine-18 is a radioactive isotope that emits positrons and is used in PET scans (positron emission tomography). It has a halflife of 110 minutes. If a patient is given a 1.5 ng sample of 18F, how much of the sample in molecules persists after a day?

Sort

Given mass(F-18, initial) = 1.5 ng

You are given the initial mass of F-18 in a sample and asked to find the mass after 1 day.

t1/2 = 110 min t = 1 day

Find number of molecules(F-18, final)

Strategize

Conceptual Plan

Use the integrated rate law (Equation 19.3) to solve this problem. First determine the rate constant k using the expression for the half-life (Equation 19.1).

t1/2

The integrated rate law gives the ratio of Nt/N0, the ratio of numbers of atoms. Use the mass, the molar mass, and Avogadro’s number to find the initial number of atoms.

massinit



t1

=

2

0.693 k





N0, k, t ln

k

N0

g F-18 

1 mol F-18 6.022  1023 atoms F-18  18 g F-18 1 mol F-18

Nt

Nt =  kt N0

Solve

Solution

Follow the conceptual plan.

0.693 2 k 0.693 110min = k 0.693 k = = 6.30  10 3 min1 110 min t1

=

1.5 ng F-18  = 5.02  10

1 g F-18 1 10 ng F-18

13

.

9



1 mol F-18 6.022  1023 atoms F-18  18 g F-18 1 mol F-18

atoms F-18

314

Chapter 19. Radioactivity and Nuclear Chemistry 1 day 

Solve the integrated rate law by substituting the values. ln

24 hr 60 min  = 1440 min 1 day 1 hr

Nt =  kt N0

Nt 6.30  103 =  ( )(1440 min ) 5.02  1013 min Nt ln =  9.07 5.02  1013 Nt = e 9.07 5.02  1013 Nt = 1.15  104 (5.02  1013 ) atoms ln

= 5.78  109 atoms

Check

The units (atoms) are correct. The magnitude of the answer makes sense since it is much smaller than the starting value.

.

315

Chapter 19. Radioactivity and Nuclear Chemistry Additional Problem for Radiocarbon Dating (Example 19.5)

A papyrus scroll was buried with King Tutankhamen 3330 years ago. The half-life of carbon-14 is 5730 years.

The papyrus scroll sample had an initial decay of 15.3 disintegrations per minute per gram carbon. What is the current decay rate of the sample? (The decay rate is proportional to the amount of C-14 present.)

Sort

Given rate0 = 15.3 dis / (ming C)

You are given the initial rate of decay for the papyrus sample and its current age. You are asked to find the current decay rate.

t1/2 = 5730 years t = 3330 years

Find ratet

Strategize

Conceptual Plan

Use the expression for the half-life (Equation 19.1) to find the rate constant.

t1/2



t1

0.693 k

=

2

Use the value of the rate constant, initial rate, and age of the sample in the integrated rate law (Equation 19.4).

k



rateinit, k, t ln

ratet

rate t =  kt rate0

Solve

Solution

Follow the conceptual plan to solve the problem.

0.693 k 0.693 5730 yr = k 0.693 k = = 1.21 104 yr 1 5730 yr t1

=

rate t

 kt

2

ln Use the rate constant, time, and initial rate in the integrated rate expression.

ln

rate0 rate1 15.3

 1.21 104    3330 yr  yr  

 

rate1

 0.402 15.3 rate1  e 0.402  0.669 15.3

ln

rate t  0.669(15.3)  10.2

.

dis min  g C

dis min  g C

316

Chapter 19. Radioactivity and Nuclear Chemistry The units of the answer are correct since they are the same as the initial rate. The magnitude makes sense since the age of the sample is less than one half-life.

Greater than half of the original sample still exists (i.e., the current rate of decay is greater than half the original decay).

Check

.

317

Chapter 20. Organic Chemistry I: Structures Student Objectives 20.1 Fragrances and Odours  

Know that molecules called odourants bind with olfactory receptors in the nose, leading to the sensation of smell.

Define organic chemistry.

20.2 Carbon: Why It Is Unique   

Know that carbon has the tendency to form four covalent bonds, including the possibility of double and triple bonds.

Know the origin of the terms organic and inorganic.

Know that carbon has the tendency to catenate or bond to itself to form chain, branched, and ring structures.

20.3 Hydrocarbons   

Know that hydrocarbons contain only carbon and hydrogen and comprise alkanes, alkenes, alkynes, and aromatic compounds.

Know the generic formulas, the structures, and some examples of alkanes, alkenes, alkynes, and aromatics.

Draw and interpret drawings of hydrocarbon structures using structural formulas, condensed structural formulas, carbon skeleton formulas, ball‐and‐stick models, and space‐filling models.

20.4 Functional Groups  

Know that functional groups are specific atoms bonded to each other that can be inserted onto a hydrocarbon.

Know and identify the common functional groups: alcohol, ether, aldehyde, ketone, carboxylic acid, ester, and amine.

20.5 Constitutional Isomerism 

Know how to write constitutional isomers for hydrocarbons and for molecules containing functional groups.

20.6 Stereoisomerism I: Conformational Isomerism   

Understand that isomerism is due to rotation around a single bond.

Describe alkane conformations.

Identify ring conformations of cycloalkanes.

20.7 Stereoisomerism II: Configurational Isomerism      

Define stereoisomer, optical isomer, enantiomer, and chiral.

Identify chiral molecules.

Describe the experiment demonstrating plane‐polarized light and how an optically active sample can rotate the plane of polarization.

Establish the difference between geometric isomers and stereoisomers.

Assign E/Z configurations to alkenes.

Know how to assign absolute configurations to chiral carbons.

.

318

Chapter 20. Organic Chemistry I: Structures 20.8 Structure Determination  

Use the index of hydrogen deficiency to establish the number of pi bonds or rings in a molecule.

Know the different spectroscopic methods of structure determination: IR, NMR.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

319

Chapter 20. Organic Chemistry I: Structures Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

20.1 Fragrances and Odours  Odour and smell  Odourants  Organic chemistry o Chemistry of carbon o Foods, drugs, petroleum products, pesticides



20.2 Carbon: Why It Is Unique  Bonding o Formation of four bonds o Double and triple bonds o Catenation



Intro figure: illustration of cologne bottle, organic molecules Unnumbered figure: photo of skunk, molecular models of thiols



Unnumbered figures: Lewis structures, molecular models of methane, ethane, ethene, ethyne Unnumbered figure: Lewis structures of propane, isobutane, cyclohexane The Nature of Science: Vitalism and the Perceived Difference Between Organic and Inorganic Unnumbered figure: photo of Freidrich Wöhler

  

20.3 Hydrocarbons  Hydrocarbons: only C and H o Alkanes o Alkenes o Alkynes o Aromatics  Representations o Structural formula o Condensed structural formula o Carbon skeleton formula o Ball‐and‐stick model o Space‐filling model



Unnumbered figures: structural formula, condensed structural formula, carbon skeleton formula, ball‐and‐ stick model, space‐filling model of butane, isobutane Unnumbered figure: different representations of butane Unnumbered figures: structural formula, condensed structural formula, carbon skeleton formula, ball‐and‐ stick model, space‐filling model of propene, propyne Example 20.1 Writing Structural Formulas for Hydrocarbons Table 20.1 Alkanes, Alkenes, Alkynes Unnumbered figure: orbital overlap illustrating sigma bonds in methane Unnumbered figure: Lewis structure and valence bond model of ethane Unnumbered figure: acetylene torch Unnumbered figure: Lewis structure and valence bond model of ethyne Unnumbered figure: p orbital overlap in 1,3‐ butadiene Unnumbered figure: Lewis structure and ball‐and‐ stick model of benzene Figure 20.1 Polycyclic Aromatic Compounds

          

.

320

Chapter 20. Organic Chemistry I: Structures Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

20.1 Fragrances and Odours  The sense of smell is fascinating. One factor is sensitivity; the isopropyl and t‐butyl thiols in natural gas can be smelled at parts per billion concentration.

20.2 Carbon: Why It Is Unique  The history of chemistry gives the background for the term organic chemistry.  The tendencies of carbon to catenate and form four bonds lead to large numbers of organic compounds, ~20 million characterized.



Other elements (e.g., silicon) catenate, but not to the extent of carbon.

20.3 Hydrocarbons  Hydrocarbons provide a good way to illustrate the idea of structural isomerism.  Encourage students to make their own models of hydrocarbons. Toothpicks, coffee stirrers, paper, tape, marshmallows, and ingenuity are all one needs for models.  Browser plug‐ins for Chime, Rasmol, JMol, and other related visualization programs are free. Websites have libraries of files of molecules.



Free rotation about single bonds can sometimes mislead students into thinking that slightly different representations of the same molecule are different molecules altogether. Making molecular models and rotating about the bonds helps alleviate this confusion.

Reminding students of the octet rule will help them draw better structures for organic compounds.



.

321

Chapter 20. Organic Chemistry I: Structures Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

20.4 Functional Groups  Specific collections or groups of atoms  Examples o Alcohol o Ether o Aldehyde o Ketone o Carboxylic acid o Ester o Amine

               

20.5 Constitutional Isomerism  Constitutional isomers in different classes of compounds



.

Table 20.2 Properties of Methane and Methanol Unnumbered figure: carbon tetrachloride, zero dipole moment Unnumbered figure: Teflon tape Table 20.3 Bonds in Organic Halides Unnumbered figure: quaternary ammonium salt Unnumbered figure: fresh fish Figure 20.2 The Structure of Aniline Unnumbered figure: Lewis structure and space‐ filling model of ethanol Unnumbered figure: Petro‐Canada gas price Unnumbered figure: primary, secondary, and tertiary alcohols Table 20.4 Solubilities of Alcohols Unnumbered figure: space‐filling model of dimethyl ether Figure 20.3 Charge Density Plots of the Carbonyl Group Unnumbered figure: Lewis structure and space‐ filling model of methanal Unnumbered figure: bananas (pentyl ethanoate) Table 20.4 Acyl Functional Groups

Unnumbered figure of propan‐1‐ol and propan‐2‐ ol

322

Chapter 20. Organic Chemistry I: Structures Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

20.4 Functional Groups  Emphasize the importance of functional groups in giving organic compounds special physical and chemical properties.



Students often have a hard time differentiating between certain functional groups such as aldehydes and ketones.

20.5 Constitutional Isomerism  Emphasize that structural isomers have the same molecular formula but a different carbon atom skeleton  Emphasize the possibility of structural isomerism among molecules with different functional groups  Conceptual Connection 20.1 Constitutional Isomers

.

323

Chapter 20. Organic Chemistry I: Structures Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

20.6 Stereoisomerism I: Conformational Isomerism  Conformational isomerism—rotation about a single bond o Newman projection formulas o Eclipsed and staggered conformations o Gauche conformation o Steric strain  Ring conformations o Angle strain

   

20.7 Stereoisomerism II: Configurational Isomerism  Cis‐trans isomerism in alkenes  Chirality o Enantiomers o Optical activity o Chemical behaviour in chiral enviroment o Absolute configurations

Unnumbered figure: three‐dimensional drawing of carbon atom Figure 20.5 Energy versus Rotation Angle for Ethane Figure 20.6 Energies of Butane Conformations Unnumbered figure: chair conformation of cyclohexane

          

20.8 Structure Determination  Index of hydrogen deficiency  Spectroscopic methods of structure determination o IR spectroscopy o NMR spectroscopy  Chemical shift  Proton NMR  Carbon NMR

     

.

Unnumbered figure: space‐filling models of cis and trans dichloroethane Table 20.6 Physical Properties of cis‐ and trans‐1,2‐Dicloroethene Example 20.2 Assigning the Configurations of Alkenes Figure 20.7: Mirror Images Figure 20.8 Rotation of Plane‐Polarized Light Unnumbered figure: sodium rubidium d‐ tartrate Unnumbered figure: illustration of tetrahedral molecule and its mirror image Unnumbered figures: illustration of enantiomers in chiral environments Figure 20.9 Absolute Configurations of Butan‐2‐ol Enantiomers Example 20.3 Assigning R,S Configurations Chemistry and Medicine: Anesthetics and Alcohol Example 20.4 Index of Hydrogen Deficiency Table 20.7 Infrared Absorptions of Selected Bonds Figure 20.10 Infrared Absorption Spectrum of Butan‐2‐one Unnumbered figure: tetramethylsilane Figure 20.11 13C‐NMR Spectrum of Cyclohexanone Example 20.5 Determining the Structure of an Unknown Organic Compound 324

Chapter 20. Organic Chemistry I: Structures Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

20.6 Stereoisomerism I: Conformational Isomerism  Allow students to visualize the single bond rotation and the fact that no single bonds break during this rotation  In the chair conformation of cyclohexane, introduce the concept of the axial up and axial down position, for better visualization.

It is also useful in the case of the equatorial positions.  Conceptual Connection 20.2 Staggered Conformations in Rings



Students often have a hard time understanding the difference between the conformers obtained as a result of a carbon– carbon single bond rotation and a constitutional isomer.

20.7 Stereoisomerism II: Configurational Isomerism  Emphasize the difference between configuration and confirmation from a three‐dimensional perspective.  Conceptual Connection 20.3 Chirality



Emphasize that assigning absolute configuration to chiral carbon is a way to track chirality on paper and that R and S configurations to not relate directly to d and l configurations.

20.8 Structure Determination  Emphasize the importance of the index of hydrogen deficiency in determining possible structures for organic compounds.



.

325

Chapter 20. Organic Chemistry I: Structures

Name the alkane:

Additional Problem for Naming Alkanes

Longest Chain Count the longest continuous chain of carbons. There are six carbons. The path straight across only includes five carbons in a row.

Substituents or Groups This compound has two substituents, both of which are named methyl.

Numbering Try numbering from the end at right. That would give “3,5-dimethyl.”

Numbering from the left gives smaller numbers: “2,4-dimethyl” General Form Substituent number(s) + substituent name(s) + base name

2,4-dimethylhexane

This molecule can also be drawn as

.

326

Chapter 20. Organic Chemistry I: Structures

Name the alkene:

Additional Problem for Naming Alkenes and Alkynes

Longest Chain Count the longest continuous chain of carbons that contains the double bond. There are six carbons. The path straight across only includes five carbons in a row.

Counting from the left, the double bond is between the third and fourth; counting from the right gives the same outcome.

Substituents or Groups This compound has two substituents, both of which are named methyl.

Numbering Try numbering from the end at right. That would give “3,5-dimethyl.”

Numbering from the left gives smaller numbers: “2,4dimethyl.”

General Form Substituent number(s) + substituent name(s) + base name

2,4-dimethyl-3-hexene

.

327

Chapter 21. Organic Chemistry II: Reactions Student Objectives 21.1 Discovering New Drugs 

Understand the idea of multistep synthesis.

21.2 Organic Acids and Bases      

Know how to write the expression for Ka for organic acids.

Know how to calculate pKa for organic acids.

Know how to write the conjugate bases of organic acids.

Know the different factors affecting acidity: inductive effect, hybridization, resonance.

Familiarize yourself with mechanisms in organic chemistry from an acid–base point of view and the curved arrows involved in movement of electrons in these mechanisms.

Understand acids and bases and their use as chemical reagents.

21.3 Oxidation and Reduction  

Know the rules for assigning oxidation states to atoms in organic molecules based on electronegativity of atoms connected to the atom to be determined.

Understand organic redox reactions and the transformation of functional groups using redox reactions.

21.4 Nucleophilic Substitution Reactions at Saturated Carbon      

Know the definition of nucleophilic and electrophilic species.

Know the meaning of the leaving group and substrate.

Know and understand the steps and reaction energy diagram of the SN1 mechanism.

Know and understand the steps and reaction energy diagram of the SN2 mechanism.

Know the factors affecting nucleophilic reactions: nucleophile type, the leaving group type, and substrate structure.

Understand carbocation stability.

21.5 Elimination Reactions     

Know the different types of elimination reactions: dehydration and dehydrohalogenation.

Understand the concept of 1,2 elimination.

Understand Zaitsev’s rule for product formation in elimination reactions.

Know and understand the steps and reaction energy diagram of the E1 mechanism.

Know and understand the steps and reaction energy diagram of the E2 mechanism.

21.6 Electrophilic Additions to Alkenes  

Understand the mechanism of hydrohalogenation as an example of electrophilic addition to alkenes.

Understand other addition reactions and be able to predict their product.

21.7 Nucleophilic Additions to Aldehydes and Ketones   

Understand nucleophilic addition of alcohols to aldehydes and ketones, as well as the formation of acetals and ketals.

Understand the mechanism of acetal and ketal formation.

Understand organomettalic reagents and the Grignard reaction. .

328

Chapter 21. Organic Chemistry II: Reactions 21.8 Nucleophilic Substitutions to Acyl Compounds    

Understand the general mechanism of nucleophilic substitution of acyl compounds and know how to apply it in the case of Fischer esterification and hydrolysis of esters.

Know the steps of the esterification mechanism.

Understand the relative reactivity within the carboxylic acid derivative series.

Understand acyl group interconversion.

21.9 Electropholic Aromatic Substitution   

Understand the preference of electrophilic substitution versus addition for aromatic compounds.

Know the steps of the general mechanism of electrophilic aromatic substitution.

Know common electrophilic aromatic reactions and their products.

21.10 Polymerization  

Define monomers and polymers.

Know the difference between step growth polymerization and addition polymerization.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

.

329

Chapter 21. Organic Chemistry II: Reactions Lecture Outline Terms, Concepts, Relationships, Skills 21.1 Discovering New Drugs  Multistep synthesis of organic compounds and their importance

21.2 Organic Acids and Bases  Relative acidity of organic acids  Inductive effects and acidity  Resonance effects on acidity o Acidic hydrogens on hydrocarbons  Mechanisms in organic chemistry o Curved arrows o Acids and bases as reagents

Figures, Tables, and Solved Examples 

Intro figure: montelukast sodium (Singulair) chemical structure and pills, child with an inhaling mask

 

Table 21.1 pKa Values for Selected Acids Unnumbered figure: inductive effect in hydrogen cyanide Unnumbered figure: ethyne and its pKa value Unnumbered figure: acetic acid and trifluoroacetic acid and their inductive effects Figure 21.1 Electrostatic Potential Surfaces Unnumbered figure: delocalization in phenoxide ion and benzene molecule Unnumbered figure: carbanion and pentan‐2,4‐dione Example 21.1 Mechanisms for Acid–Base Reactions Example 21.2 Selecting an Appropriate Acid–Base Reagent Unnumbered figure: selected Brønsted–Lowry acids and bases in order of decreasing basicity and acidity

       

21.3 Oxidation and Reduction  Hydrogenation as redox reaction  Interconversion of functional groups through redox reactions  Choosing reagents for redox reactions

 

Example 21.3 Oxidation States in Organic Molecules Table 21.2 Reagents for Selected Oxidations and Reductions Example 21.4 Choosing Reagents for Oxidation and Reductions Chemistry in Your Day: Hydrogen and the Oil Sands

 

.

330

Chapter 21. Organic Chemistry II: Reactions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

21.1 Discovering New Drugs  Identifying certain functional groups in common drugs might help students understand the significance of drug discoveries.

21.2 Organic Acids and Bases  Relate the strength of organic acids to the magnitude of their equilibrium constant.  Use curved arrows, electron flow, to write both organic and inorganic acid–base reactions.



In writing acid–base reactions, students tend to have the arrow indicate that the acidic hydrogen “departs” instead of the arrow indicating that the acidic hydrogen is being “picked up” by the base.

21.3 Oxidation and Reduction  Review the rules of assigning oxidation states for inorganic ions and elements and emphasize the difference of assigning oxidation states to atoms in organic molecules.  Conceptual Connection 21.1 Oxidation and Reduction



Establish a connection between the oxidation state of carbon and the different functional groups.

.

331

Chapter 21. Organic Chemistry II: Reactions Lecture Outline Terms, Concepts, Relationships, Skills 21.4 Nucleophilic Substitution Reactions at Saturated Carbon  Identification of substrate, nucleophile, electrophile, and leaving group  SN1 mechanism  SN2 mechanism  Factors affecting nucleophilic substitutions o Carbocation stability o Steric hindrance in substrate

Figures, Tables, and Solved Examples 



Unnumbered figure: partial charges in a polar C—X bond Figure 21.2 Energy Profile for SN1 mechanism Unnumbered figure: the two steps of the SN1 mechanism Unnumbered figure: SN1 mechanism on 2‐chloro‐2‐ methylpropane Unnumbered figure: SN1mechanism on a chiral carbon Unnumbered figure: SN2 mechanism Unnumbered figure: SN2 mechanism on bromomethane Figure 21.3 Energy Profile for SN2 Mechanism Unnumbered figure: wind inverting an umbrella Table 21.3 Selected Nucleophiles and Their Relative Effectiveness Unnumbered figure: increasing carbocation stability Unnumbered figure: steric hindrance in bromoethane, 2‐bromopropane, 2‐bromo‐2‐ methylpropane Example 21.5 Predicting Nucleophilic Substitutions

  

Unnumbered figure: position 1,2 for elimination Unnumbered figure: stepwise E1 mechanism Unnumbered figure: E2 mechanism

          

21.5 Elimination Reactions  1,2 eliminations o Dehydration and dehydrohalogenation o Zaitsev’s rule  E1 mechanism  E2 mechanism  Substitution vs. elimination o Strong bases favour elimination

.

332

Chapter 21. Organic Chemistry II: Reactions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

21.4 Nucleophilic Substitution Reactions at Saturated Carbon  Emphasize that through nucleophilic substitution a variety of different functional groups can be introduced in molecules.  Emphasize the fact that the carbon atom undergoing these reactions has to be sp3 hybridized.  Emphasize that inversion of configuration can happen only if the carbon atom is chiral.  Conceptual Connection 21.2 Leaving Groups and Base Strength



Students have a hard time differentiating between nucleophiles and bases.

21.5 Elimination Reactions  Emphasize the formation of a double bond, and therefore the formation of a new class of compounds.  The strength of different bases has to be discussed in detail for students to appreciate the full extent of the reaction.  Size of molecule and base strength is also an important concept for differentiation of products.  Conceptual Connection 21.3 Can a Strong Acid React Like a Base?



Base strength can be a challenging concept to students.

Acid and conjugate base pairs should be reviewed.



.

333

Chapter 21. Organic Chemistry II: Reactions Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Example

21.6. Electrophilic Additions to Alkenes  Hydrohalogenation mechanism  Other addition reactions



21.7 Nucleophilic Additions to Aldehydes and Ketones  Addition of alcohols o Hemiacetal and hemiketals  The Grignard reaction



21.8 Nucleophilic Substitutions of Acyl Compounds  Fischer esterification  Synthesis of different acyl compounds by interconversions



Unnumbered figure: stepwise mechanism of hydrohalogenation Table 21.4 Selected Additions to Alkenes



Unnumbered figure: partial charges in the carbon–magnesium covalent bond Unnumbered figure: mechanism of the Grignard addition to a carbonyl group Example 21.6 Grignard Reactions

 

Unnumbered figure: general mechanism of nucleophilic substitution of an acyl group Unnumbered figure: equilibrium between esterification and hydrolysis Unnumbered figure: mechanism of Fischer esterification Unnumbered figure: reactivity ranking of carboxylic acid derivatives Table 21.5 Reagents for Common Interconversions of Acyl Groups Example 21.7 Acyl Group Interconversions

    

.

334

Chapter 21. Organic Chemistry II: Reactions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

21.6. Electrophilic Additions to Alkenes  Emphasize the formation of two new sigma bonds in the molecule as a result of the addition reaction.  Emphasize the formation of new functional groups as a result of the reaction.  Conceptual Connection 21.4 H—X as an Electrophile  Conceptual Connection 21.5 Mechanism of Hydration 21.7 Nucleophilic Additions to Aldehydes and Ketones  Emphasize the difference between electrophilic and nucleophilic additions.  Emphasize the formation of new classes of compounds: acetals and hemiacetals.  Explain the formation of the cyclic form of glucose.  Emphasize the importance of the Grignard reaction as a carbon chain extension reaction.



Clarify for students the ring formation of monosaccharides.

21.8 Nucleophilic Substitutions of Acyl Compounds  Emphasize the difference in reactivity of the these types of carbonyl compound (substitution vs. addition).  Conceptual Connection 21.6 The Reaction Between an Acyl Chloride and NH3



Students have a hard time distinguishing between compounds undergoing nucleophilic substitution and nucleophilic addition.

.

335

Chapter 21. Organic Chemistry II: Reactions Lecture Outline Terms, Concepts, Relationships, Skills 21.9 Electrophilic Aromatic Substitutions  Nitration of benzene

Figures, Tables, and Solved Example 

Unnumbered figure: general mechanism of electrophilic substitution Unnumbered figure: nitration mechanism Table 21.6 Common Electrophilic Aromatic Substitution Reactions

 

21.10 Polymerization  Monomers vs. polymers  Step‐growth polymers  Addition polymers

   

Figure 21.4 Poly(ethylene trephthalate) Chemistry in Your Day: Kevlar Figure 21.6 Poly(vinyl chloride) Unnumbered figure: ethylene and polyethylene, chloroethane and poly(vinyl chloride)

.

336

Chapter 21. Organic Chemistry II: Reactions Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

21.9 Electrophilic Aromatic Substitutions  Emphasize why aromatic compounds undergo electrophilic substitution rather than electrophilic addition.  Conceptual Connection 21.7 Resonance Structures 21.10 Polymerization  Emphasize the importance of polymers in everyday life.

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Chapter 22. Biochemistry Student Objectives 22.1 Diabetes and the Synthesis of Human Insulin   

Know that diabetes is a disease caused by the absence or insufficiency of insulin, a protein synthesized in the pancreas that helps regulate cellular and blood glucose levels.

Know that genetic engineering has allowed modified bacteria to produce human insulin, allowing it to be harvested for human use.

Know that biochemistry is the chemistry of living organisms, often involving macromolecules such as lipids, carbohydrates, proteins, and nucleic acids.

22.2 Lipids        

Give a general description of lipids with respect to structure and function.

Know that fatty acids can be saturated (all C–C bonds) or unsaturated (at least one C=C bond).

Know and understand how the tri‐alcohol glycerol combines with fatty acids to form triglycerides.

Know and understand that fats are solids and oils are liquids at room temperature, a consequence of the structures of the fatty acids in triglycerides.

Know that natural geometric isomers of unsaturated fatty acids are typically cis, that trans forms are introduced when hydrogenation is used to control potential rancidity, and that trans fatty acids are not considered healthy by nutrition experts.

Give a general description of phospholipids with respect to structure and function.

Draw and recognize representations of phospholipids and their role in lipid bilayers.

Recognize the structure of cholesterol and know that it serves as a structural component of cell membranes and as a precursor to other steroids in the body.

22.3 Carbohydrates   

Know that carbohydrates are polyhydroxy aldehydes and ketones that are nutritional sugars and starches and that serve as energy sources for the body.

Recognize the general structures of monosaccharides like glucose and fructose.

Know and understand how monosaccharides like glucose and fructose form rings that can combine to make more complex structures such as disaccharides (e.g., sucrose) or polysaccharides (e.g., cellulose, starch, glycogen).

22.4 Proteins and Amino Acids   

Know and understand that proteins are macromolecules that serve as structural components, enzymes, and hormones and have additional roles in transport, storage, protection, and mobility.

Know that there are 20 common amino acids and that 10 essential ones must be consumed since the body cannot produce them from other molecules.

Know the general structure of amino acids and write reactions for the formation of peptide bonds.

22.5 Protein Structure      

Know that proteins are characterized as fibrous or globular.

Know that protein structure consists of four levels: primary, secondary, tertiary, and quaternary.

Know that the primary structure of a protein gives the sequence of amino acids.

Recognize the two common components of secondary structure, ‐helices and ‐pleated sheets, and know that hydrogen bonding enables the formation of both.

Understand the different intermolecular interactions that maintain tertiary structure.

Know that multimeric proteins exhibit quaternary structure. .

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Chapter 22. Biochemistry 22.6 Nucleic Acids: Blueprints for Proteins   

Understand the basic structure of nucleic acids: sugars, bases, and phosphate links.

Know and understand how the nucleotides are covalently combined to make a single strand and how two complementary strands form a double helix through hydrogen bonding.

Define and understand codon, gene, and chromosome.

22.7 DNA Replication, the Double Helix, and Protein Synthesis  

Know and understand the process of DNA replication.

Know how proteins are synthesized from an RNA copy of DNA and understand how the codons direct the ribosome to attach particular amino acids to each other to make the specific protein sequence.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 22. Biochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

22.1 Diabetes and the Synthesis of Human Insulin  Diabetes o Insulin o Biotechnology: genetic engineering  Biochemistry

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22.2 Lipids  Fatty acid structure o Even number of carbons, C4 to C20 o Chains  Saturated  Unsaturated  cis  trans  Glycerides o Triglycerides o Phospholipids  Phosphatidylcholine  Lipid bilayers  Cholesterol o Membrane structure o Biosynthetic precursor to testosterone, estradiol

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

22.3 Carbohydrates  General structure: (CH2O)n  Monosaccharides o Carbonyl: aldehyde or ketone o OH groups o Cyclic form  Disaccharides o Glycosidic linkage o Sucrose (from glucose and fructose)  Polysaccharides o Cellulose o Starch o Glycogen

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         .

Intro figure: illustration of spiral timeline, structures of biomolecules Unnumbered figure: Frederick Banting

Unnumbered figures: Lewis structures, space‐filling models of myristic acid, oleic acid Table 22.1 Fatty Acids Figure 22.1 The Effect of Unsaturation Figure 22.2 The Formation of Tristearin Unnumbered figure: photo of lard package Unnumbered figure: Lewis structure, space‐ filling model of tristearin, triolein Chemistry and Medicine: Dietary Fat: The Good, the Bad, and the Ugly Figure 22.3 Cis and Trans Fats Figure 22.4 Phosphatidylcholine Figure 22.5 Schematic for Phospholipid or Glycolipid Figure 22.6 Lipid Bilayer Unnumbered figure: Lewis structure of cholesterol, testosterone, ‐estradiol Unnumbered figure: Lewis structure, space‐ filling model of glucose, fructose Figure 22.7 Fischer Projection of Carbon Atom with Four Different Substituents Example 22.1 Absolute Configurations of Carbohydrates Figure 22.8 Intramolecular Reaction of Glucose to Form a Ring Unnumbered figure: Lewis structures of ring forms of glucose, fructose, galactose Figure 22.9 Formation of a Glycosidic Linkage Figure 22.10 Hydrolysis of Disaccharides Unnumbered figures: Lewis structures of cellulose, starch Figure 22.11 Hydrolysis of Polysaccharides 340

Chapter 22. Biochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

22.1 Diabetes and the Synthesis of Human Insulin  There are several examples of essential proteins that are missing and can be replaced.

Diabetes affects about 11 percent of the population. Other examples include human growth hormone and thyroid hormone.  Biotechnology or genetic engineering is commonly used to make important proteins and biomedical molecules.



Not all missing or damaged proteins can be replaced. Insulin is an example of one that can be.

22.2 Lipids  Fatty acids and glycerol are the building blocks of triglycerides, the components of nutritional fats and oils. The space‐filling models give a good sense of the shapes.

Melting points give a sense of how well the molecules can pack together.  Trans fats and the trans fatty acids from which they are made appear often in the nutrition literature and in the media. Trans fats do occur in nature but in small quantities.  The interactions of phospholipids with water make the membranes containing phospholipids even more stable.



The structural properties of fats and oils come primarily from the chain length and degree of unsaturation of the fatty acids.

22.3 Carbohydrates  When compared with the lipids, carbohydrate structures are more complex, including issues such as the nature of the monosaccharide, ring structures, and the glycosidic linkage between any two monosaccharides in more complex molecules.  The microscopic and macroscopic properties of the more complex sugars arise from the nature of the monosaccharides and the glycosidic linkages between them.



Monosaccharides occur primarily in the cyclic form in water.

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Chapter 22. Biochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

22.4 Proteins and Amino Acids  Classes o Enzymes o Hormones o Structure, transport, storage, contraction, protection  Amino acids o General structure o 20 common amino acids o 10 essential amino acids  Cannot be biosynthesized

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22.5 Protein Structure  Structural categories o Fibrous (linear)  Example: collagen o Globular  Example: hemoglobin  Levels o Primary: amino acid sequence o Secondary: ‐helices, ‐pleated sheets o Tertiary: intramolecular interactions o Quaternary: subunits

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     

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22.6 Nucleic Acids: Blueprints for Proteins  Basic nucleotide structure o Phosphate o Sugar: ribose or deoxyribose o Base: purine or pyrimidine  Genetic code o Nucleotide o Codon o Gene o Chromosome

      

.

Table 22.2 Protein Functions Unnumbered figure: Lewis structures of amino acids Unnumbered figure: alanine Unnumbered figure: illustration of different types of amino acid side chains Unnumbered figure: zwitterion Table 22.3 Common Amino Acids Unnumbered figure: peptide bond Example 22.2 Peptide Bonds

Figure 22.12 Fibrous and Globular Proteins Figure 22.13 Levels of Protein Structure Chemistry and Medicine: The Essential Amino Acids Table 22.4 Essential Amino Acids Figure 22.14 Primary Structure of Egg‐White Lysozyme Unnumbered figure: photo of normal and sickled red blood cells Figure 22.15 The ‐Helix Structure Figure 22.16 The ‐Pleated Sheet Structure Figure 22.17 Interactions Within Proteins

Figure 22.18 DNA Structure Unnumbered figure: Lewis structures of DNA and RNA sugars Unnumbered figure: DNA and RNA bases Figure 22.19 Base Pairing in DNA Figure 22.20 Short Strand of DNA Figure 22.21 Genetic Structure Figure 22.22 Chromosomes

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Chapter 22. Biochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

22.4 Proteins and Amino Acids  The combination of amino acids to form polypeptides is an example of a condensation reaction. The acid portion of one amino acid combines with the amine portion of another, yielding a peptide bond and a water molecule.  The 20 common amino acids can be placed in a few categories according to their R group: alkyl groups, alcohols and thiols, carboxylic acids, amines and amides, and aromatic rings.  Conceptual Connection 22.1 Peptides



The essential amino acids are ones that cannot be biosynthesized by humans and therefore must be consumed (eaten).

22.5 Protein Structure  Protein structural categories arise from different functions.

The fibrous examples like collagen serve as structural elements in muscles for example. Globular proteins are often more water soluble and serve as enzymes or to support other functions besides structure.  The levels of protein structure arise from the size, complex structure, and functions of many proteins. The intramolecular interactions often arise from the side chains of the amino acids.  Sickle‐cell anemia is due to a protein defect in which a single amino acid is replaced with one that is less functionalized. The ability of the red blood cell to carry oxygen is reduced.



Protein structure involves intramolecular interaction between amino acid components that extends far beyond the mere order of the amino acids in a protein.

22.6 Nucleic Acids: Blueprints for Proteins  The structures of RNA and DNA are relatively simple considering their importance. For example, only four bases are used to code for a host of proteins. The simple structure, though, enables duplication and repair.  Conceptual Connection 22.2 The Genetic Code



Base‐pairing in DNA occurs through specific hydrogen bonding interactions. Base‐ pair mismatches do not have those proper hydrogen bonding interactions.

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343

Chapter 22. Biochemistry Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

22.7 DNA Replication, the Double Helix, and Protein Synthesis  Replication o DNA double helix structure o Complementary single strands o Base‐pair match  Protein synthesis o RNA copies of DNA o Codons o Ribosomes

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Figure 22.23 Watson and Crick Unnumbered figure: complementary strains Figure 22.24 DNA Double Helix Figure 22.25 DNA Replication Figure 22.26 Protein Synthesis Chemistry and Medicine: The Human Genome Project

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Chapter 22. Biochemistry Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

22.7 DNA Replication, the Double Helix, and Protein Synthesis  The Human Genome Project has provided the identity and order of bases for all genetic material in humans.

Now the effort focuses on identifying which chromosome each gene resides in and the mechanisms by which the massive amounts of information are organized.  Protein synthesis makes use of three amino‐acid codons.

As each codon is read by the ribosome, an amino acid is added to the polypeptide.

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345

Chapter 22. Biochemistry Examine the structure of deoxyribose, part of the backbone in DNA. Does deoxyribose exhibit optical isomerism? If so, which carbon atoms are chiral?

Additional Problem for Carbohydrates and Optical Isomerism

Chiral Centres Examine each carbon to see which have four different atoms or groups attached.

Carbons 3 and 4 are chiral centres. Carbon 1 is the carbonyl C=O; carbon 2 and carbon 5 have two H attached apiece.

With two chiral centres, deoxyribose exhibits optical isomerism.

Optical Isomerism

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346

Chapter 22. Biochemistry Additional Problem for Peptide Bonds (Example 22.2)

Show the tripeptide that results from linking the amino acids glycine, valine, and serine, in that order. Label the N-terminus and C-terminus ends of the tripeptide.

Peptide Bonds A peptide or amide bond forms when the hydroxyl –OH from the carboxylic acid of one amino acid is removed, together with the hydrogen from the amino group of the adjacent amino acid.

Water is removed each time a peptide bond is formed. They appear in the boxes.

Tripeptide The resulting tripeptide is:

H2N – glycine – valine – serine – CO2H

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347

Chapter 23. Chemistry of the Nonmetals Student Objectives 23.1 Insulated Nanowires  

Know that carbon forms nanotubes and that boron nitride, a material that is isoelectronic with carbon, also forms nanotubes.

Know that carbon nanotubes are electrical conductors, while boron nitride nanotubes are electrical insulators.

23.2 The Main‐Group Elements: Bonding and Properties 

Recall the periodic trends for radius, the general properties of metals and nonmetals, and the major divisions of the periodic chart.

23.3 Silicates: The Most Abundant Matter in Earth’s Crust      

Know that silicon and oxygen together constitute the majority of the Earth’s crust.

Define silicate.

Know that quartz and glass have the empirical formula SiO2 but consists of tetrahedral SiO4 structural units.

Know that aluminosilicates form with Al atoms in Si sites, giving the formula unit AlO2.

Know some general characteristics of the different silicate structures.

Predict the structure of a given silicate.

23.4 Boron and Its Remarkable Structures    

Know that the different allotropes of elemental boron, B12, are based on an icosahedral structure.

Know that boron reacts with halogens to form trihalides.

Know that boron reacts with oxygen to form network covalent solids.

Define and describe the different borane structures: closo, nido, and arachno.

23.5 Carbon, Carbides, and Carbonates       

Describe the common crystalline allotropes of carbon: graphite and diamond.

Know that coal is an amorphous form of carbon that forms under pressure from the decomposition of ancient organic material and consists of three main types: lignite, bituminous, and anthracite.

Know some general characteristics of coke, charcoal, activated carbon, soot, and carbon black.

Know that carbon also forms fullerenes and nanotubes.

Know that carbon forms ionic carbides, covalent carbides, and metallic carbides when combined with different elements.

Know that carbon combines with oxygen to form carbon dioxide and carbon monoxide.

Know that carbon dioxide and water combine to form the carbonates: carbonic acid, bicarbonate, and carbonate.

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Chapter 23. Chemistry of the Nonmetals 23.6 Nitrogen and Phosphorus: Essential Elements for Life        

Recall that elemental nitrogen is a diatomic gas that comprises 78 percent of Earth’s atmosphere.

Know some general characteristics of the main allotropes of phosphorus: white phosphorus, red phosphorus, and black phosphorus.

Know the common nitrogen hydrides: ammonia, hydrazine, and hydrogen azide.

Know the common nitrogen oxides: nitrogen oxide, nitrogen dioxide, dinitrogen trioxide, nitrous oxide, and dinitrogen tetraoxide.

Know the oxyanions and oxyacids of nitrogen: nitrate, nitric acid, and nitrite.

Know that phosphorus forms phosphine (PH3) and the phosphine halides PX3 and PX5.

Know that phosphorus reacts with oxygen to form P4O6 and P4O10.

Know that phosphoric acid and the phosphates are among the most important and useful phosphorus compounds.

23.7 Oxygen   

Recall that most elemental oxygen is a diatomic gas that comprises 21 percent of Earth’s atmosphere.

Know the three oxide types: oxide, peroxide, and superoxide.

Know some general characteristics of another common allotrope of oxygen, ozone (O3).

23.8 Sulfur: A Dangerous but Useful Element      

Know that elemental sulfur is found in a few natural deposits believed to have been formed by anaerobic bacteria and is extracted by the Frasch process.

Know some general characteristics of dihydrogen sulfide.

Assign oxidation states to sulfur in redox reactions.

Know that sulfur dioxide forms naturally from volcanic activity and forms as a pollutant from coal and oil combustion as well as metal extraction.

Know that sulfuric acid is the most abundantly produced chemical in the world.

Know some general uses of sulfuric acid.

23.9 Halogens: Reactive Elements with High Electronegativity    

Determine the oxidation states of halogens in redox reactions.

Know that halogens are not found in their elemental forms because of their high electronegativities.

Know some general characteristics of fluorine, hydrofluoric acid, and chlorine.

Know some general characteristics of interhalogen compounds and halogen oxides.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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349

Chapter 23. Chemistry of the Nonmetals Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

23.1 Insulated Nanowires  Carbon nanotubes o Structure o Electrical conductivity  Boron nitride nanotubes o Structure o Electrical insulation

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23.2 The Main‐Group Elements: Bonding and Properties  Atomic radius  Electronegativity  Metals, nonmetals, metalloids

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 

23.3 Silicates: The Most Abundant Matter in Earth’s Crust  Quartz and glass o Formula unit: SiO2 o Structure and shape: SiO4 tetrahedra  Aluminosilicates o AlO2 in place of SiO2  Silicate units, chains, sheets o Orthosilicate: SiO44 o Pyrosilicate: Si2O76 o Pyroxene  Repeating unit: Si2O64  Formula unit: SiO32 o Amphibole  Repeating and formula unit: Si4O116 o Phyllosilicate  Repeating unit: Si4O104  Formula unit: Si2O52  Predicting structures

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Intro figure: illustration of carbon nanotubes Table 23.1 Properties of BN and C Figure 23.1 Boron Nitride Nanotube

Unnumbered figure: periodic tables p‐ block elements

Figure 23.2 Major Elements in Earth’s Crust Figure 23.3 SiO4 Tetrahedron Figure 23.4 Structure of Quartz Example 23.1 Determining the Composition of an Aluminosilicate Figure 23.5 Pyrosilicate Structure Figure 23.6 Pyroxene Structure Unnumbered figure: photo of asbestos fibres Unnumbered figure: photo of mica sheets Figure 23.7 Amphibole Structure Figure 23.8 Phyllosilicate Structure Table 23.2 Types of Silicate Structures Example 23.2 Composition and Charge Balance of Silicates Examples 23.3 and 23.4 Predicting Silicate Structures

350

Chapter 23. Chemistry of the Nonmetals Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

23.1 Insulated Nanowires  Nanotubes are modern examples of the descriptive chemistry of carbon, boron, and nitrogen. The microscopic structures are nearly the same, but the macroscopic properties (e.g., electrical conductivity) are different.

23.2 The Main‐Group Elements: Bonding and Properties  Review the periodic properties of the p‐block elements: electron configurations, electronegativity, radius, electron affinity, and metallic character.

23.3 Silicates: The Most Abundant Matter in Earth’s Crust  The silicates form a variety of structures, though most of them are based on a tetrahedral shape. This recognizable green pyramid is shown in all of the ball‐and‐stick models.



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Silicates, like SiO2, do not exist as discrete molecules but instead are network covalent solids.

351

Chapter 23. Chemistry of the Nonmetals Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

23.4 Boron and Its Remarkable Structures  Elemental boron: B12 icosahedron  Boron halides, BX3  Boron oxygen compounds o Trigonal BO3 in hexagonal shapes  Boron hydrides o closo‐boranes o nido‐boranes o arachno‐boranes

   

23.5 Carbon, Carbides, and Carbonates  Crystalline carbon forms o Graphite: planar, hexagonal o Diamond: tetrahedral  Coal o Lignite o Bituminous o Anthracite  Other types: coke, charcoal, activated carbon, soot, carbon black  Specialized structures: fullerenes, nanotubes  Carbides o Ionic o Covalent o Metallic  Carbon oxides o Carbon dioxide o Carbon monoxide  Carbonates o Carbonic acid o Washing soda o Baking soda o Baking powder

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Figure 23.9 B12 Icosahedron Figure 23.10 B2O3 Structure Figure 23.11 closo‐Borane Structures Figure 23.12 nido‐ and arachno‐Boranes

Figure 23.13 Graphite Structure Figure 23.14 Diamond Structure Table 22.3 Approximate Composition of the Main Types of Coal Unnumbered figure: photo of tires Figure 23.15 C60 and a Geodesic Dome Figure 23.16 Carbon Nanotubes Figure 23.17 The Nanocar Figure 23.18 Calcium Carbide Structure Unnumbered figure: micrograph of steel Unnumbered figure: photo of Alka‐Seltzer tablets in water

352

Chapter 23. Chemistry of the Nonmetals Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

23.4 Boron and Its Remarkable Structures  Boron predominantly forms trigonal structures for boron halide and boron oxide compounds but cage structures for hydride compounds.

23.5 Carbon, Carbides, and Carbonates  The crystalline allotropes of carbon are well known; other forms of carbon (e.g., coal, charcoal) are widely used. The newer forms, fullerenes and nanotubes, have been discovered and developed in the last quarter century.  The carbides, particularly the covalent and metallic carbides, are important commercially because of their hardness.

Carbide tips are added to saw blades made of steel to make them sharper and cut longer.  The combustion products of carbon, CO2 and CO, are well known and highly relevant to daily life.  Students are always fascinated by the fullerenes and carbon nanotubes.  Conceptual Connection 23.1 Phase Changes and Pressure  Conceptual Connection 23.2 Carbonate Solubility

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 

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Graphite has a high melting point because of the strong bonds within the sheets, but it is slippery because of the weak interactions between graphite sheets.

Large amounts of carbon are sequestered in ocean water, in shells of marine organisms, and in rocks.

Fullerenes, particularly C60 and C70, are naturally occurring.

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Chapter 23. Chemistry of the Nonmetals Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

23.6 Nitrogen and Phosphorus: Essential Elements for Life  Elemental forms o Nitrogen: N2 (g), 78% of Earth’s atmosphere o Phosphorus  White: P4 tetrahedra  Red: P4 chains  Black: layered like graphite  Nitrogen compounds o Hydrides: ammonia, hydrazine, hydrogen azide o Oxides: nitrogen monoxide, nitrogen dioxide, dinitrogen trioxide, dinitrogen tetraxoxide, nitrous oxide o Nitric acid, nitrate, nitrite  Phosphorus compounds o Hydride: phosphine o Halides: PX3, PX5, POX3 o Oxides: P4O6, P4O10 o Phosphoric acid, phosphates

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23.7 Oxygen  Elemental oxygen o O2 o Ozone, O3  Oxide types o Oxide: O2 o Peroxide: O22 o Superoxide: O2  Ozone o Strong oxidizing agent o Disinfectant o Ozone layer

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Figure 23.19 White Phosphorus Figure 23.20 Red Phosphorus Table 23.4 Oxidation States of Various Nitrogen Compounds Figure 23.21 Hydrazine and Hydrogen Peroxide Unnumbered figure: vehicle crash test Figure 23.22 Tetraphosphorus Hexaoxide and Decaoxide, P4O6 and P4O10 Table 23.5 Uses of Phosphates in the Food Industry

Unnumbered figure: photo of ocean vents Table 23.6 Types of Oxides

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Chapter 23. Chemistry of the Nonmetals Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

23.6 Nitrogen and Phosphorus: Essential Elements for Life  A large amount of nitrogen occurs as the elemental gas N2 in the atmosphere. The nitrogen oxides also are gases.  The most common ions containing nitrogen are nitrate and nitrite. Nitric acid is a strong mineral acid used commercially in the fertilizer industry.  Ammonia is also a gas. Its natural supply comes from bacteria that are capable of “fixing” nitrogen: cleaving the triple bond in N 2.  The allotropes of elemental phosphorus include the P4 tetrahedron of white phosphorus, the chain forming interconnecting tetrahedra in red phosphorus, and the graphite‐like structure of black phosphorus.  Phosphoric acid is used to remove scale (i.e., rust) in the steel industry, in the production of fertilizers, and as a flavour agent and preservative in soft drinks. Phosphates are widely used in detergents and in the food industry.



23.7 Oxygen  The atmospheric component gases O2 and O3 are well known, but large quantities of the most common element in Earth’s crust are found as water and oxides of many other elements.  Chemical cycles of bacterial life near some volcanic vents in the ocean are based not on oxygen but on sulfur.

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

.

Both extremes of nitrogen oxidation states occur in molecules that are important as fertilizers: the 3 oxidation state of N in ammonia and the +5 oxidation state of N in nitrate.

Phosphorus can form molecules with an expanded octet because of low‐lying d‐ orbitals.

Ozone in the upper atmosphere and as a disinfectant is useful, but ground‐level ozone is a pollutant.

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Chapter 23. Chemistry of the Nonmetals Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

23.8 Sulfur: A Dangerous but Useful Element  Elemental form o Isolation by Frasch process o Oxidation of recycled H2S  Hydrogen sulfide o Toxicity, odour o Product of bacteria and rotting o Different properties from those of H2O  Sulfur dioxide o Natural volcanic source o Combustion product of coal and petroleum  Sulfuric acid o Most abundant chemical produced in the world o Common uses

  

23.9 Halogens: Reactive Elements with High Electronegativity  Elemental forms o Not naturally occurring because of high electronegativity o F2 from oxidation of F o Cl2 from electrolysis  Interhalogen compounds o ABn for n = 1, 3, 5, 7 o A = large, B = small (B = F for n = 5, 7)  Halogen oxides o Often explosive o Chlorine oxides o Often strong oxidizing agents

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  

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Unnumbered figure: photo of molten sulfur Figure 23.23 Quenching Liquid Sulfur Figure 23.24 Balzac Natural Gas Plant, near Calgary, Alberta Example 23.5 Balancing of and Assigning Oxidation States to Sulfur Reactions Table 23.7 Common Metal Sulfides Figure 23.25 Dehydration of Sucrose

Table 23.8 Selected Properties of the Halogens Example 23.6 Determining the Oxidation State of Halogens Table 23.9 Comparison of Halogen X—X Bond Energy Example 23.7 Identifying Changes in Oxidation States

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Chapter 23. Chemistry of the Nonmetals Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

23.8 Sulfur: A Dangerous but Useful Element  Elemental sulfur is isolated directly by the Frasch process or by oxidation of H2S that is recycled from other processes. Hydrogen sulfide is toxic, so it must be removed or recycled anyway.  Hydrogen sulfide has the characteristic smell of rotting eggs. It can be smelled at about 20 ppb, but the sense of smell is desensitized after exposure.  Sulfur dioxide is an environmental pollutant that arises from combustion of sulfur‐ containing coal and petroleum products.

Large quantities of sulfur dioxide are also produced from volcanic activity.



Hydrogen sulfide looks as if it should resemble water based on their formulas.

However, nearly everything about these two compounds, including bond angle, toxicity, and physical properties, is different.

23.9 Halogens: Reactive Elements with High Electronegativity  Elemental halogens can be formed by electrolysis of the halide salts.  Halogen oxides are often explosive, and most of the known ones of chlorine are strong oxidizing agents.



Like the group 1 and 2 metals, none of the elemental forms of the halogens occur naturally because of their high reactivity.

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Chapter 23. Chemistry of the Nonmetals

Predict the silicate structure for the mineral talc Mg3Si4O10(OH)2 that serves as the basis for talcum powder, and show that the formula is charge neutral.

Additional Problem for Predicting Silicate Structures (Examples 23.3 and 23.4)

Si to O Ratio Determine the ratio of Si to O in the formula.

Si:O ratio is 4:10 or 2:5.

The other two oxygens occur as hydroxides and are not part of the silicate structures. (That situation is indicated by the OH in the parentheses.)

Match Ratio to List Match the Si:O ratio to the type of silicate in Table 23.2.

The 2:5 ratio is indicated for phyllosilicate sheets with a formula unit of Si2O52.

Anion Charge Si2O52: 2 @ 2 = 4

Determine the total anion charge.

OH: 2 @ 1 = 2 Total anionic charge = 6

Cation Charge Mg2+: 3 @ +2 = +6

Determine the total cation charge and show that it matches the total anion charge.

Total cationic charge = +6

The anionic and cationic charges balance, so the formula is charge neutral.

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Chapter 23. Chemistry of the Nonmetals Calculate the oxidation state for Cl in the following compounds:

Additional Problem for Determining the Oxidation State of Halogens in Compounds (Example 23.6)

a) Cl2O6

b) ClF3

c) HClO4

Oxidation states:

Solution a

oxygen: 3 @ 2 = 6 chlorine: 2 @ ?? = +6

Chlorine must be +3 oxidation state.

Oxidation states:

Solution b

fluorine: 3 @ 1 = 3 chlorine: 1 @ ?? = +3

Chlorine must be +3 oxidation state.

Oxidation states:

Solution c

hydrogen: 1 @ +1 = +1 oxygen: 4 @ 2 = 8 chlorine: 1 @ ?? = +7

Chlorine must be +7 oxidation state.

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Chapter 24. Metals and Metallurgy Student Objectives 24.1 Vanadium: A Problem and an Opportunity   

Know that petroleum contains trace components other than the hydrocarbons that are the target fuel.

Know that petroleum contains vanadium compounds as contaminants at levels that suggest that they were part of some extinct species’ biological systems.

Know that metallurgy includes all processes associated with mining, separating, and refining metals to their pure forms and then combining metals into alloys that have different properties

24.2 The General Properties and Natural Distribution of Metals   

Know the common properties of metal, such as conductors of electricity and heat, malleability, and ductility.

Know that metals make up the crust and core of the Earth.

Know that minerals are homogeneous, naturally occurring, crystalline inorganic solids. An ore is a rock that contains a high concentration of a specific mineral.

24.3 Metallurgical Processes     

Know that extractive metallurgy involves separating the metal‐containing components and nonmetal‐containing components. Crushed ore can be separated by air in an industrial cyclone or by solutions of water, oil and detergents.

Know that in pyrometallurgy, heat is used to extract metal from a mineral.

Know that hydrometallurgy uses an aqueous solution to extract metals from their ores.

Know that electrometallurgy uses electrolysis to convert a metal salt into the metal.

Know that powder metallurgy uses micron‐sized metal high pressure and heat to form the desired compound at temperatures below melting point but high enough for the particles to fuse.

24.4 Metal Structures and Alloys      

Know that structures of metals can be described as the closest packing of spheres into crystal types: face‐centred cubic, body‐centred cubic, and hexagonal closest packed.

Know that an alloy is a metallic material that contains more than one type of element in solid solutions or in specific ratios.

Know that in substitution alloys, the two metals form a miscible solid solution.

Know that solid compositions that melt at temperatures lower or higher than either of the two pure metals are common.

Know that some alloys exhibit phase regions with limited solubility of one of the components and various mixtures can have different crystal structures.

Know that interstitial alloys form when a second metal can fit in the hole made by the other metal atoms in closest‐packed crystals.

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Chapter 24. Metals and Metallurgy 24.5 Sources, Properties, and Products of Some of the 3d Transition Metals 

     

Know that titanium is the fourth most abundant metal on Earth; it occurs mainly as oxides, is isolated by conversion of the oxides to intermediate TiCl4 and then reduction, is a strong and light metal used in propellers and jet engine parts, and has its largest use in the form of TiO2, a white powder used in paints.

Know that chromium compounds are brightly coloured, the oxide ore is reduced to form the meta,; an alloy with steel called stainless steel resists rusting, and the ions chromate and dichromate are strong oxidizing agents used with organic molecules.

Know that manganese occurs as mineral ores, is reduced to the metal by Al or Na, is used to alloy with steel, as permanganate is used as an oxidizing agent, and as MnO2 is an additive to glass used to decolourize it.

Know that cobalt is found in ores of other metals and sulfide minerals, is ferromagnetic and important for the production of magnets, is an additive to steel, as salts it gives brilliant blue colour to pigments and inks, and is part of vitamin B12.

Know that copper can be found in its elemental form and thus has been used as a metal for many years; it has high conductivity, is used to manufacture pipes and metal surfaces on roofs, and is formed into bronze and brass.

Know that nickel is used as an alloy with steel to form stainless steel.

Know that zinc comes from ores, it forms alloys, and it is used as an anticorrosion coating.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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Chapter 24. Metals and Metallurgy Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

24.1 Vanadium: A Problem and an Opportunity  Petroleum contaminants o Vanadium compounds o Ancient organism biology based on vanadium  Metallurgy



Intro figure: pieces of vanadium metal

24.2 The General Properties and Natural Distribution of Metals  Properties o Opaque o Conductors of heat and electricity o Malleable o Ductile  Minerals and ores





Table 24.1 Thermal Conductivity and Electrical Resistivity of Several Metals Unnumbered figure: salt production by desalinization Unnumbered figure: vein of gold metal

24.3 Metallurgical Processes  Isolation steps o Mine o Separate o Refine  Extractive metallurgy o Pyrometallurgy Calcination Roasting Smelting  Flux  Slag o Hydrometallurgy  Leaching  Example: gold o Electrometallurgy  Aluminum: Hall process  Copper o Powder metallurgy

    

Figure 24.1 Separation by Air Figure 24.2 Separation by Using a Solution Figure 24.3 The Hall Process Figure 24.4 Copper Electrolysis Cell Unnumbered figure: different metal products

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Chapter 24. Metals and Metallurgy Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

24.1 Vanadium: A Problem and an Opportunity  Use vanadium as an example to understand the pros and cons of certain sources and uses of different metals.



Vanadium may only be present in trace amounts, but the process of converting petroleum to fuel products or burning the fuels that contain it may concentrate the samples.

24.2 The General Properties and Natural Distribution of Metals  The distribution of metals in nature requires an understanding of the structure and composition of ores and minerals.



Most metals do not occur naturally in their elemental form but must undergo reduction of the cationic form from ores that have been concentrated.

24.3 Metallurgical Processes  Metallurgical processes have been developed over time to make the isolation of metals more efficient.  The technique used for a particular metal is likely to depend on the nature and quantity of an elemental metal or its ore.



Metallurgy takes advantage of basic chemical principles to concentrate an element or mineral or to reduce metal cations to the elemental form.

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Chapter 24. Metals and Metallurgy Lecture Outline Terms, Concepts, Relationships, Skills 24.4 Metal Structures and Alloys  Closest packing of spheres o Face‐centred cubic o Body‐centred cubic o Hexagonal closest packed  Alloys o Substitutional  Miscible solid solutions  Alloys with limited solubility  Phase diagrams o Interstitial  Octahedral holes  Tetrahedral holes

Figures, Tables, and Solved Examples  

Table 24.2 The Crystal Structures of the 3d Elements Unnumbered figure: face‐centred structure of crystallized copper Unnumbered figure: structure of nickel copper alloy Figure 24.5 Cu–Ni Phase Diagram Figure 24.6 Cr–V Phase Diagram Examples 24.1and 24.2 Determining Alloy Compositions from a Phase Diagram Figure 24.7 Cr–Ni Phase Diagram Examples 24.3 and 24.4 Alloy Compositions in a Solid Solution with Limited Solubility Figure 24.8 Octahedral Holes in Closest Packed Crystals Figure 24.9 A Different View of an Octahedral Hole Figure 24.10 A Tetrahedral Hole in a Closest Packed Crystal Table 24.3 Formulas of Several Interstitial Alloys

         

24.5 Sources, Properties, and Products of Some of the 3d Transition Metals  Titanium o Sources (ores, minerals), uses  Chromium o Sources (ores, minerals), uses, alloys  Manganese o Sources (ores, minerals), uses, alloys  Cobalt o Sources (ores, minerals), uses, alloys  Nickel o Sources (ores, minerals), uses, alloys  Zinc o Sources (ores, minerals), uses, alloys



Table 24.4 The Colours of Various Chromium Compounds Figure 24.11 Chromium Compounds Unnumbered figure: tetrahedral arrangements of chromate and dichromate ions Unnumbered figure: ignition of ammonium dichromate Unnumbered figure: varied applications of copper metal Unnumbered figure : zinc phosphate adhering to steel

    

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Chapter 24. Metals and Metallurgy Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

24.4 Metal Structures and Alloys  Knowledge of crystal lattices takes on a practical approach. Substitutional alloys require knowledge of the size of a potential component and melting information as a function of the mole percent of the components. Interstitial alloys can take advantage of the holes in the closest packed crystals.  Conceptual Connection 24.1 Interstitial Alloys



The formulation of alloys takes advantage of knowledge of both microscopic properties and macroscopic properties.

24.5 Sources Properties and Products of Some of the 3d Transition Metals  Seven metals are compared and contrasted in terms of isolation, practical applications, and their compounds.



The descriptive chemistry of seven metals is outlined: the isolation of the metal depends on the primary source while the practical uses depend on the microscopic properties of the metal.

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Chapter 25. Transition Metals and Coordination Compounds Student Objectives 25.1 The Colours of Rubies and Emeralds  Know that the colours of gemstones are attributable to electronic transitions between the d‐orbitals of transition elements.

25.2 Electron Configurations of Transition Metals  Recall that the transition metals occupy the d block of the periodic table.  Recall how to write electron configurations and determine oxidation states for d‐block elements.  Recall trends in radius, ionization energy, and electronegativity for d‐block elements.

25.3 Coordination Compounds     

Define complex ion, ligand, coordination compound, and coordinate covalent bond.

Recognize monodentate, bidentate, and polydentate ligands.

Define chelate and chelating agent.

Recognize the common geometries of complex ions.

Write names and formulas for coordination compounds.

25.4 Structure and Isomerization    

Differentiate between structural isomers and stereoisomers.

Recognize and identify coordination isomers and linkage isomers.

Identify and draw geometric isomers.

Recognize and draw optical isomers.

25.5 Bonding in Coordination Compounds  State the hybridization of the metal centre using valence bond theory for particular geometries.  Explain the crystal field theory of d‐orbital splitting in octahedral and tetrahedral complexes.  Understand the relationship between crystal field splitting and the colours of transition metal complexes.  Know the difference between weak‐field, moderate‐field, and strong‐field ligands.  Know the relationship between crystal field splitting and the magnetic properties of transition metal complexes.  Understand the difference between high‐spin and low‐spin in transition metal complexes.  Predict the number of unpaired electrons in an octahedral complex.

25.6 Applications of Coordination Compounds  Know that EDTA is used as a chelating agent to bind toxic heavy metals.  Know the role of transition metals in biomolecules such as hemoglobin, cytochrome c, chlorophyll, and carbonic anhydrase.

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Chapter 25. Transition Metals and Coordination Compounds Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls

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367

Chapter 25. Transition Metals and Coordination Compounds Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

25.1 The Colours of Rubies and Emeralds  Crystal field  Cr3+ o In Al2O3 (ruby, red) o In Be3Al2(SiO3)6 (emerald, green) 2+  Fe o In garnet (red) o In peridot (yellow‐green)  Cu2+ o In turquoise



Intro figure: illustration of ruby colour and the shape of the d z 2 orbital



Unnumbered figure: photos of ruby, emerald, garnet, peridot, and turquoise

25.2 Electron Configurations of Transition Metals  Electron configurations  Periodic properties o Atomic radius o First ionization energy o Electronegativity  Oxidation states



25.3 Coordination Compounds  Coordination complex o Complex ion plus counterions  Common ligands o Monodentate: water, ammonia, chloride, carbon monoxide, thiocyanate o Bidentate: oxalate, ethylenediamine o Polydentate: ethylenediamine tetraacetate  Geometries and coordination number o Linear o Square planar o Tetrahedral o Octahedral  Names o Ligands and prefixes o Metal centre o Names of metals in anionic complexes

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         

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Table 25.1 First‐Row Transition Metal Orbital Occupancy Figure 25.1 First‐Row Transition Metal Oxidation States

Unnumbered figure: molecular model of [Co(NH3)6]Cl3 Unnumbered figure: Lewis acid–base reaction Unnumbered figure: bidentate ligand Table 25.2 Common Ligands Figure 25.5 Bidentate and Polydentate Ligands Coordinated to Co(III) Table 25.3 Common Geometries of Complex Ions Unnumbered table: Guidelines for Naming Complex Ions Unnumbered figure: EDTA Table 25.4 Names and Formulas of Common Ligands Table 25.5 Names of Common Metals when Found in Anionic Complex Ions Examples 25.1 and 25.2 Naming Coordination Compounds 368

Chapter 25. Transition Metals and Coordination Compounds Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

25.1 The Colours of Rubies and Emeralds  The colours of gemstones, which are caused by transition metals in different mineral environments, are an introduction to crystal field theory, coordination compounds, and complex ions.

25.2 Electron Configuration of Transition Metals  The electron configurations of transition metals follow the rules and guidelines set forth previously in Section 8.3.  Oxidation states are likely to be somewhat confusing because of the variety of values some elements can assume. In general, the first‐row transition metals can have oxidation states of +2 and +3. Elements in the centre of the row have more possible states than the ones on the ends. For example, Sc is only +3, Zn is only +2, but Mn can be any state from +2 to +7.



25.3 Coordination Compounds  Students may be confused about the organization of metal cations, ligands, and additional properties for complex ions. Table 25.2 lists the common ligands, and Table 25.3 lists the shapes and coordination numbers.

Likewise, one can often predict the geometry given the formula of a complex ion.  Students may struggle with predicting the formulas of complex ions. For example, what does Fe2+ form when reacted with ammonia?  Naming complex ions follows some basic guidelines. Often, coordination chemistry relies more on structures than on names.





.

Electron configurations of the first‐row transition metals fill the 4s before the 3d unless all the 3d orbitals can be filled evenly (e.g., Cr and Cu).

When forming cations of transition metals, remove electrons from the 4s before the 3d.

In complexes, all of the valence electrons on the metal are found in the d orbitals. For this reason, complexes of Ir+ are isoelectronic with complexes of Pt2+ and Au3+.

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Chapter 25. Transition Metals and Coordination Compounds Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

25.4 Structure and Isomerization  Structural isomers o Coordination isomers o Linkage isomers  Stereoisomers o Geometric isomers  Cis–trans  Fac–mer o Optical isomers

      

25.5 Bonding in Coordination Compounds  Valence bond theory o Geometry o Hybridization o Orbitals  Crystal field theory o d orbital orientation o Octahedral field  Splitting pattern  Strong field  Weak field  Colours and crystal field splitting energy  Spectrochemical series  Magnetism  High‐spin  Low‐spin o Tetrahedral field  Splitting pattern o Square planar field  Splitting pattern

         

.

Table 25.6 Ligands Capable of Linkage Isomerization Figure 25.3 Linkage Isomers Figure 25.4 Cis–trans Isomerism Figure 25.5 Fac–Mer Isomerism in Co(NH3)3Cl3 Examples 25.3 and 25.4 Identifying and Drawing Stereoisomers Figure 25.6 Stereoisomerism in [Co(en)3]3+ Example 25.5 Recognizing and Drawing Enantiomers

Figure 25.7 Relative Positions of d Orbitals and Ligands in an Octahedral Complex Figure 25.8 d Orbital Splitting in an Octahedral Field Figure 25.9 Colours of Complex Ions Figure 25.10 The Colour Wheel Figure 25.11 The Colour and Absorption Spectrum of [Ti(H2O)6]3+ Example 25.6 Calculating Crystal Field Splitting Energy Unnumbered figures: orbital diagrams of low‐ and high‐spin Fe2+ complexes Examples 25.7 and 25.8 High‐ and Low‐ Spin Octahedral Complexes Figure 25.12 Splitting of d Orbitals by a Tetrahedral Ligand Geometry Figure 25.13 Splitting of d Orbitals by a Square Planar Ligand Geometry

370

Chapter 25. Transition Metals and Coordination Compounds Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

25.4 Structure and Isomerization  Some ligands can be attached in two distinct ways, giving rise to linkage isomers.  Models are likely to be helpful for seeing or visualizing the arrangement of the ligands in octahedral structures.  Fac and mer require three ligands of the same kind in an octahedron.  Visualizing optical isomerism in octahedral complexes is much more difficult than in tetrahedral complexes. Models are essential for the beginning student.



Tetrahedral complexes cannot exhibit cis– trans isomerism, and square planar complexes cannot exhibit optical isomerism.

25.5 Bonding in Coordination Compounds  The crystal field model is completely electrostatic and views all ligands as merely point negative charges. This model does not explain the spectrochemical series at all.  Strong‐ and weak‐field effects of ligands can be confirmed by the number of unpaired electrons in octahedral complexes.  Demonstrations of colour (especially cobalt complexes) and magnetism (diamagnetism, paramagnetism, and number of unpaired electrons) are easy and highly illustrative.  Conceptual Connection 25.1 Weak‐ and Strong‐Field Ligands



Students often will simply memorize the splitting patterns for octahedral, tetrahedral, and square planar complexes. These patterns can be simply deduced by considering the orientation of the d orbitals and the positions of the incoming ligands.

Students often confuse the terms weak‐ and strong‐field with high‐ and low‐spin.



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Chapter 25. Transition Metals and Coordination Compounds Lecture Outline Terms, Concepts, Relationships, Skills

Figures, Tables, and Solved Examples

25.6 Applications of Coordination Compounds  Chelating agents: EDTA  Chemical analysis: SCN  Colouring agents  Biomolecules o Hemoglobin and cytochrome c: iron in porphyrin ligand o Chlorophyll: magnesium in porphyrin o Carbonic anhydrase: zinc  Drugs and therapeutic agents o Cisplatin: cis‐[Pt(NH3)2Cl2]

       

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Figure 25.14 Chemical Analysis with SCN Table 25.7 Transition Metals and Some of Their Functions in the Human Body Unnumbered figure: structures of cytochrome c, heme Figure 25.15 Porphyrin Structures Figure 25.16 Hemoglobin Figure 25.17 Chlorophyll Figure 25.18 Carbonic Anhydrase Figure 25.19 Cisplatin

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Chapter 25. Transition Metals and Coordination Compounds Teaching Tips Suggestions and Examples

Misconceptions and Pitfalls

25.6 Applications of Coordination Compounds  The use of chelating agents to remove lead and mercury is a documented treatment when a case of heavy metal poisoning is known or strongly suspected.  With an unsubstantiated belief that mercury‐ containing preservatives found in a few antiviral vaccines cause autism, some children are being subjected to EDTA treatment. What are some of the potential dangers or hazards of chelation therapy?

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Chapter 25. Transition Metals and Coordination Compounds

Write the ground state electron configuration for Nb and Nb3+.

Additional Problem for Writing Electron Configurations for Transition Metals

Identify the noble gas Identify the noble gas that precedes the element and put it in square brackets.

Nb is atomic number 41 in period 5 and Group 15.

Complete noble gas shell = [Kr]

Count the periods Count down the periods to determine the outer principal quantum level—this is the quantum level for the s orbital. Subtract one to obtain the quantum level for the d orbital.

Nb is in the fifth period:

[Kr] 5s4d

Count the electrons Count across the row to see how many electrons are in the neutral atom and fill the orbitals accordingly.

Nb has five more electrons than [Kr]:

[Kr] 5s25d3 For an ion, remove the required number of electrons, first from the s and then from the d orbitals.

Nb3+ has lost three electrons relative to Nb:

[Kr] 5s04d2 or [Kr] 4d2

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Chapter 25. Transition Metals and Coordination Compounds How many unpaired electrons would you expect for the complex ion [Cu(NH3)6]2+?

Additional Problem for High- and Low-Spin Octahedral Complexes (Examples 25.7 and 25.8)

Charge and number of d electrons Begin by determining the charge and number of d electrons on the metal.

Complex ion = +2 NH3: 6 @ +0 = 0 Cu: 1 @ +2 = +2  Cu2+

Cu: [Ar] 4s13d10 Cu2+: [Ar] 4s03d9  Cu2+ has d9 configuration.

Ligands in spectrochemical series NH3 is a strong-field ligand, so  will be relatively large.

Look at the spectrochemical series to determine whether the ligand is a strong-field or weak-field ligand.

High-spin or low-spin?

Decide if the complex will be high- or low-spin and draw the electron configuration.

Strong-field ligands yield low-spin configurations.

Unpaired electrons Count the unpaired electrons.

This configuration has one unpaired electron.

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