Chem 21

Text

2010

Dr. van Geel

Wellesley High School

Table of Contents

Chapter 1: The Nature of Matter _________________________________________ 1 1.1 The Structure of Matter_______________________________________________________ 1 1.2 Classifying Matter ___________________________________________________________ 1

1.3 Chemical Symbols and Chemical Formulas _______________________________________ 4 1.4 Chemical and Physical Changes ________________________________________________ 5 1.5 Phases of Matter ____________________________________________________________ 7

1.6 Phase Changes______________________________________________________________ 8 1.7 Chemical Properties and Physical Properties ______________________________________ 9 1.8 Methods of Separation _______________________________________________________ 9

Chapter 2: Quantifying Chemistry _______________________________________ 11

2.1 Taking Measurements _______________________________________________________ 11 2.2 Significant Figures (Sig Figs)__________________________________________________ 13 2.3 Calculations with Sig Figs ____________________________________________________ 14

2.4 Precision vs. Accuracy _______________________________________________________ 15 2.5 Scientific Notation __________________________________________________________ 17 2.6 The Metric System__________________________________________________________ 19

2.7 Fenceposting ______________________________________________________________ 19 2.8 Converting Compound Units __________________________________________________ 22 2.9 Converting Volumes_________________________________________________________ 23 2.10 Using a Compound Unit as a Conversion Factor _________________________________ 23

2.11 Using Percentage as a Conversion Factor ______________________________________ 23 2.12 Fenceposting in Word Problems ______________________________________________ 24 2.13 Temperature _____________________________________________________________ 24

Chapter 3: Ionic Formulas and Nomenclature ______________________________ 25 3.1 Element Symbols ___________________________________________________________ 25 3.2 Molecular vs. Ionic Compounds _______________________________________________ 25 3.3 Ion Formation Rules ________________________________________________________ 26

3.4 Polyatomic Ions ____________________________________________________________ 27 3.5 Ionic Formulas _____________________________________________________________ 27 3.6 Naming Ionic Compounds____________________________________________________ 28

3.7 Naming Molecular Compounds ________________________________________________ 29

Chapter 4: The Mole __________________________________________________ 31 4.1 The Mole__________________________________________________________________ 31

4.2 Avogadro’s Number as a Conversion Factor _____________________________________ 31 4.3 Molar Mass ________________________________________________________________ 32 4.4 Using Molar Mass as a Conversion Factor _______________________________________ 33 4.5 Empirical vs. Molecular Formulas ______________________________________________ 33

4.6 Percent Composition ________________________________________________________ 35 4.7 Empirical and Molecular Formulas from Percent Composition _______________________ 35

Chapter 5: Chemical Equations _________________________________________ 38

5.1 Chemical Equations _________________________________________________________ 38 5.2 Balancing Chemical Equations ________________________________________________ 39 5.3 Stoichiometry ______________________________________________________________ 41

5.4 Limiting Reactant___________________________________________________________ 42 5.5 Theoretical Yield ___________________________________________________________ 43 5.6 The ICE Chart _____________________________________________________________ 43 5.7 Percent Yield ______________________________________________________________ 44

5.8 Classifying Reactions ________________________________________________________ 44 5.9 Predicting Products _________________________________________________________ 47 5.10 The Diatomic Seven________________________________________________________ 47

Chapter 6: Atomic Structure ___________________________________________ 49 6.1 Atomism, Joseph Proust, and John Dalton ______________________________________ 49 6.2 J. J. Thompson and the Electron ______________________________________________ 51 6.3 Robert Millikan and the Charge of the Electron___________________________________ 53

6.4 Ernest Rutherford and the Nucleus ____________________________________________ 55 6.5 The Line Spectrum__________________________________________________________ 56 6.6 Niels Bohr and Orbits________________________________________________________ 59

6.7 Werner Heisenberg and Erwin Schrödinger______________________________________ 60 6.8 The Nucleus _______________________________________________________________ 63 6.9 Atomic Number and Atomic Mass______________________________________________ 63

Chapter 7: Nuclear Chemistry __________________________________________ 65

7.1 The Strong Force ___________________________________________________________ 65 7.2 The Band of Stability ________________________________________________________ 66 7.3 Binding Energy and Mass Defect ______________________________________________ 66

7.4 Nuclear Decay _____________________________________________________________ 67 7.5 Nuclear Equations __________________________________________________________ 69

7.6 Predicting Nuclear Decay ____________________________________________________ 70 7.7 Radioactive Series __________________________________________________________ 71

7.8 Nuclear Power _____________________________________________________________ 72 7.9 Half-life ___________________________________________________________________ 75

Chapter 8: Electron Configurations ______________________________________ 78 8.1 Quantum Numbers _________________________________________________________ 78

8.2 The Energies of Orbitals _____________________________________________________ 80 8.3 Electron Configurations ______________________________________________________ 84 8.4 Condensed Configurations:___________________________________________________ 87

8.5 Special Cases: Copper and Chromium __________________________________________ 88 8.6 Electron Configuration of Ions ________________________________________________ 89 8.7 Ground State vs. Excited State ________________________________________________ 90

8.8 Orbital Shapes _____________________________________________________________ 91

Chapter 9: The Periodic Table __________________________________________ 93 9.1 Development of the Periodic Table ____________________________________________ 93 9.2 Regions of the Periodic Table _________________________________________________ 93

9.3 Effective Nuclear Charge_____________________________________________________ 95 9.4 Periodic Trends ____________________________________________________________ 97

Chapter 10: Chemical Bonds __________________________________________ 102

10.1 Chemical Bonds and Valence Electrons _______________________________________ 102 10.2 Ions and Ionic Bonds______________________________________________________ 103 10.3 Covalent Bonds __________________________________________________________ 104 10.4 Metallic Bonds ___________________________________________________________ 106

10.5 Drawing Lewis Structures __________________________________________________ 107 10.6 Lewis Structures with Resonance ____________________________________________ 110 10.7 Molecular Geometry_______________________________________________________ 110

10.8 Bond Polarity ____________________________________________________________ 114 10.9 Molecular Polarity ________________________________________________________ 115

Chapter 11: Intermolecular Forces _____________________________________ 118 11.1 Types of Intermolecular Forces _____________________________________________ 118

11.2 Strength of IMF’s _________________________________________________________ 120 11.3 IMF’s and Physical Properties _______________________________________________ 121 11.4 IMF’s and Solubility _______________________________________________________ 122

11.5 Types of Solids___________________________________________________________ 122

11.6 Specific Heat ____________________________________________________________ 126 11.7 Heats of Fusion and Vaporization____________________________________________ 127

11.8 Heating Curves___________________________________________________________ 129 11.9 Phase Diagrams __________________________________________________________ 131

Chapter 12: Solutions________________________________________________ 133 12.1 Definitions ______________________________________________________________ 133

12.2 Dissociation of Ionic Solids _________________________________________________ 134 12.3 Solubility Rules___________________________________________________________ 135 12.4 Solubility Curves _________________________________________________________ 136

12.5 Concentration____________________________________________________________ 137 12.6 Preparing Solutions _______________________________________________________ 138 12.7 Freezing Point Depression and Boiling Point Elevation ___________________________ 139

12.8 Net Ionic Equations _______________________________________________________ 141

Chapter 13: Gases __________________________________________________ 142 13.1 Pressure ________________________________________________________________ 142 13.2 Factors that Affect Pressure ________________________________________________ 142

13.3 The Ideal Gas Law________________________________________________________ 143 13.4 Ideal vs. Real Gases ______________________________________________________ 144 13.5 The Combined Gas Law ___________________________________________________ 145

13.6 Boyle’s Law and Charles’s Law ______________________________________________ 146 13.7 Standard Temperature and Pressure _________________________________________ 147 13.8 Partial Pressure (Dalton’s Law of Partial Pressures) _____________________________ 147 13.9 Graham’s Law of Diffusion _________________________________________________ 149

Chapter 14: Chemical Kinetics and Equilibrium____________________________ 152 14.1 Reaction Rates ___________________________________________________________ 152 14.2 Collision Theory __________________________________________________________ 152

14.3 Potential Energy Diagrams _________________________________________________ 153 14.4 Catalysts________________________________________________________________ 155 14.5 Reaction Mechanisms _____________________________________________________ 156 14.6 Equilibrium ______________________________________________________________ 156

14.7 The Equilibrium Constant Expression (Kc and Kp) _______________________________ 159 14.8 The Magnitude of K _______________________________________________________ 160 14.9 The Reaction Quotient, Q __________________________________________________ 161

14.10 The ICE Chart and Equilibrium Problems_____________________________________ 161 14.11 Solubility Product Constant ________________________________________________ 164

14.12 Equilibrium Shifts: Le Châtelier’s Principle____________________________________ 164

Chapter 15: Acids and Bases __________________________________________ 167

15.1 The Autoionization of Water ________________________________________________ 167 15.2 The pH Scale ____________________________________________________________ 168 15.3 Definitions ______________________________________________________________ 170 15.4 Strong vs. Weak Acids and Bases____________________________________________ 173

15.5 Problems with ICE Charts, pH, and Ka or Kb ___________________________________ 174 15.6 Indicators _______________________________________________________________ 175 15.7 Titration ________________________________________________________________ 176

15.8 Buffers _________________________________________________________________ 179 15.9 Titration Curves of Weak Acids and Polyprotic Acids ____________________________ 180

Chapter 16: Thermochemistry _________________________________________ 182

16.1 Enthalpy ________________________________________________________________ 182 16.2 Estimating ∆H with Bond Enthalpies _________________________________________ 184 16.3 Hess’s Law ______________________________________________________________ 185 16.4 Standard Enthalpy of Formation_____________________________________________ 187

16.5 Spontaneous Chemical Reactions____________________________________________ 189 16.6 Gibb’s Free Energy________________________________________________________ 190

Chapter 17: Electrochemistry__________________________________________ 193

17.1 Oxidation Numbers _______________________________________________________ 193 17.2 Oxidation-Reduction Reactions______________________________________________ 194 17.3 Balancing Redox Equations_________________________________________________ 195 17.4 Voltaic Cells _____________________________________________________________ 197

17.5 Standard Reduction Potentials ______________________________________________ 199

Chapter 18: Organic Nomenclature _____________________________________ 202 18.1 Organic Skeletal Structures_________________________________________________ 202

18.2 Naming Alkanes __________________________________________________________ 203 18.3 Naming Alkenes __________________________________________________________ 204 18.4 Naming Alkynes __________________________________________________________ 205 18.5 Naming Cyclic Alkanes_____________________________________________________ 205

18.6 Organic Functional Groups _________________________________________________ 206

1

Chapter 1: The Nature of Matter

1.1 The Structure of Matter 1.2 Classifying Matter 1.3 Chemical Symbols and Chemical Formulas 1.4 Chemical and Physical Changes 1.5 Phases of Matter 1.6 Phase Changes 1.7 Chemical Properties and Physical Properties 1.8 Methods of Separation

1.1 The Structure of Matter Chemistry is the study of matter on the submicroscopic scale. Matter is anything that has mass and occupies space—in other words, matter is physical “stuff”. Everything that you can see, touch, taste, smell (and many things you can’t) is matter: this paper, the air, your body and brain are all made of matter. But what is matter itself made of? Matter is made of atoms. We will learn about the interior structure of atoms later in the course, but for now we can visualize atoms as tiny spheres. There are over 100 different kinds of atoms: oxygen atoms, hydrogen atoms, carbon atoms, gold atoms, silver atoms, etc. While we now know that atoms can be broken down into smaller parts, splitting an atom destroys its identity as a certain substance. In other words, if a gold atom is broken into pieces, the pieces are no longer recognizable as gold. Atoms are so tiny that they can’t be seen with even the most powerful optical microscopes. Thus, when we deal with individual atoms, we are working on the submicroscopic scale. Atoms can be joined together with chemical bonds to form larger structures called molecules. Most matter is either made of individual atoms, individual molecules, or mixtures of the two. In this text, we will use the word particle to mean either an atom or a molecule. If matter is made out of particles, what is between the particles? It may be hard to picture, but the answer is empty space—a complete vacuum. 1.2 Classifying Matter All matter can be divided into two categories: pure substances and mixtures. In a pure substance, all the particles, be they atoms or molecules, are identical to all the other particles in the substance. In a mixture, there are two or more kinds of particles. So we can begin our classification scheme like this:

The simplest model of an atom is a tiny sphere.

Seven molecules, each made of two atoms.

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As we mentioned, the particles in pure substances are all identical. If there is only one kind of atom in the substance, it is called an element. For example, the substances shown on the right are both elements. Note that one is made of atoms and the other molecules, but they are still classified as elements.

In the sample to the left, all the particles (molecules) are identical, and therefore this is still a pure substance—it is not a mixture. However, because the molecules contain more than one type of atom, this is not an element; is called a compound. So now our classification tree has two more branches:

As shown at the top of the page, elements can be either individual atoms or molecules (as long as all the atoms in the molecules are the same). The first type is called an atomic element; the second type is a molecular element, because it’s made of molecules.

Pure substance

Mixture

All Matter

Element Compound

Pure substance

Mixture

All Matter

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Mixtures can also be further classified. If the substances are mixed together thoroughly, all the different kinds of particles are evenly distributed throughout the sample. If this is the case, it is a homogenous mixture. If the distribution is uneven, it is a heterogeneous mixture. This completes our classification tree:

Pure substances are rare in nature; mixtures are by far the most common form of matter. Even things we think of as very simple—such as air, tap water, steel, glass—are actually mixtures of different substances.

Element Compound

Pure substance

Mixture

All Matter

Atomic Element

Molecular Element

Homogenous Heterogeneous

Element Compound

Pure substance

Mixture

All Matter

Atomic Element

Molecular Element

Homogenous Mixture

Heterogeneous Mixture

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Summary

1.3 Chemical Symbols and Chemical Formulas Each element has its own name and chemical symbol, and these symbols are all listed on the periodic table. Many of the symbols are simply the first letter or letters of the element name:

Nitrogen = N Calcium = Ca Fluorine = F

The origins of some symbols are not so obvious; this is usually because the symbol comes from an archaic name of the element or from another language:

Sodium = Na (the Latin name is natrium) Iron = Fe (the Latin name is ferrum) Tungsten = W (the old name was wolfram)

When atoms are stuck together with chemical bonds to make a molecule, the type and number of each atom is included in a chemical formula. Water has

All Matter

Pure Substances Mixtures

Elements Compounds

Atomic Molecular

Homogenous Heterogeneous

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the chemical formula H2O. The small number 2 in the formula is called a subscript. Subscripts indicate the number of atoms of a particular element in the molecule; a molecule of water is made of two hydrogen atoms and one oxygen atom. Notice that there is no subscript after the O in H2O; if the subscript is 1, it is left out. We can also draw a water molecule like this:

H

O

H Another example: a molecule of sugar (sucrose) has the chemical formula C12H22O11, and looks like this:

O

C

CCO

C

O

C

O

CO

O

C C

O

C

O

C

O

C

C

O

O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

The molecule contains twelve carbon atoms, twenty-two hydrogen atoms, and eleven oxygen atoms—the subscripts in C12H22O11. 1.4 Chemical and Physical Changes Matter would not be very interesting unless it underwent change—a firecracker is pretty boring until it goes off. This is an example of a chemical change—a chemical change is one in which a substance is destroyed and another is created. In the case of a firecracker, gunpowder turns into a mixture of carbon dioxide, nitrogen, and various other compounds in the blink of an eye. How does this occur? How can one substance transform into another? It happens when chemical bonds are broken and/or formed. For example, nitrogen gas (N2) and oxygen gas (O2) will combine to form nitrogen monoxide (NO). Nitrogen and oxygen are both molecular elements—their atoms travel around chemically bound together in groups of two, and these bonds must be broken before a nitrogen atom can form a new bond with an oxygen atom. The whole process is diagrammed below:

A molecule of N2 and O2

Old bonds break.

New bonds form.

Two molecules of NO.

N N

O O

N N

O O

N N

O O

N N

O O

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On the submicroscopic level, a chemical change might look like this: In a physical change no new molecules are made; no chemical bonds are broken or formed. The only thing that changes is the arrangement of the molecules. For example, if we break a solid piece of ice into small pieces, we have not broken any chemical bonds—the molecules are still H2O. On the submicroscopic level, this kind of physical change might look like this: All molecules are still intact; the change is simply a large piece of the substance breaking into smaller pieces. Because in real life we can’t see particles on the submicroscopic scale, how do we know if a change is chemical or physical? The following indications give a rough guideline for distinguishing between the two.

• Physical changes tend to be reversible (water can melt and then freeze again), while chemical changes tend to be irreversible (it is impossible to “unburn” a piece of wood).

• If the change causes a change in color, it is probably a chemical change. • If the change gives off heat or light, it is probably a chemical change

(example: burning propane). • If the change absorbs heat (feels cold) it is probably a chemical change

(example: a “cold pack” for sports injuries). • If bubbles (gas) form inside a liquid, it is probably a chemical change

(exception: boiling). • If a precipitate (a solid) forms inside a liquid, it is probably a chemical

change (exception: freezing).

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1.5 Phases of Matter The phases (or states) of matter are solid, liquid, and gas. On the submicroscopic level, they look like this:

• Solids have a rigid structure and definite shape; the particles are locked in place. The particles are closely packed together, making solids the densest phase and making them not compressible.

• Liquids do not have a rigid structure; their particles can move freely past each other. Because of this, a liquid will take the shape of whatever container it is in (see below). The particles are close together, so liquids are also dense and not compressible.

• Gases have no rigid structure, so they also take the shape of their container. Gas particles also spread out to fill their container (see next page); they occupy whatever volume they are given. Because the particles are far apart, gases have low density, and they are compressible; when a gas is squeezed, its particles will get closer together.

Liquids and gases always take the shape of their containers.

Solid Liquid Gas

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No matter the size of the vessel, gas particles will distribute themselves evenly throughout. 1.6 Phase Changes When matter changes from one phase to another, it is a physical change. Depending on the circumstances, matter can change from any phase into any other. Each of these phase changes has a different name: melting, freezing, boiling, condensation, sublimation, and deposition.

Melting: Solid to Liquid

Freezing: Liquid to Solid

Boiling: Liquid to Gas

Condensation: Gas to Liquid

Sublimation: Solid to Gas

Deposition: Gas to Solid

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Why do phase changes occur? Usually phase changes happen when the substance is heated or cooled. To understand why, we need to better understand what heat is. Heat is the energy of moving particles. The hotter something is, the faster its particles are moving. In the case of solids, this means that the particles are vibrating in place (remember they are held in a rigid structure and can not move out of position). When a solid is heated, the molecules vibrate more and more until they break away from the structure. When the structure has completely collapsed, it has become a liquid. Now the particles are free to move past each other, and they move even faster as more heat is added. Eventually they move so fast that they will break free of the other particles and fly off; when all of the particles have flown apart, the substance has become a gas. When a substance is cooled, exactly the reverse process occurs. 1.7 Chemical Properties and Physical Properties Different substances have certain distinguishing characteristics, or properties. These properties can be classified as physical properties or chemical properties. A physical property is one that can be determined without changing the identity of the substance (in other words, without the substance undergoing a chemical reaction). Examples of physical properties are size, shape, density, hardness, color, smell, boiling point, melting point, etc. Chemical properties are how the substance chemically reacts with other substances. For example, a chemical property of carbon is that carbon can react with oxygen to form carbon dioxide. Conversely, we could say that a chemical property of oxygen is that it will react with carbon to form carbon dioxide. Some chemical properties have common English names; for example, flammability is a chemical property because burning is a chemical reaction. Corrosiveness is too, because corrosion is a chemical process. Stainless steel has the chemical property of not reacting with oxygen to form rust—we could say “stainless” is also a chemical property. 1.8 Methods of Separation In science and industry it is very useful to have pure substances. Unfortunately, pure substances are rare in nature; mixtures are much more common. To separate mixtures into pure substances, various techniques have been developed (many of these are centuries old).

• Filtration. Filtration separates substances based on particle size. A filter is a material that traps large particles but allows small particles to pass through; it is often used to separate solids from liquids. Using filter paper is very common; a

A filtration setup.

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circle of filter paper is folded into a cone and inserted in a funnel. The funnel is set into a flask to collect the liquid.

• Settling/Centrifuge. Settling separates based on density. If two

substances have different densities, sometimes they will separate from each other if they simply sit undisturbed. Oil and water can be separated this way. To speed up the process, a centrifuge can be used. It spins the mixture at high speed to create a force that pulls the denser material to the bottom.

• Chromatography. A drop of the mixture is placed on one end of a strip

of paper or other absorbent substance. The end of the paper is dipped into a liquid, which gradually seeps from one end of the paper to the other. As it passes through the drop of the mixture, it carries the substances along. Some substances are carried along slower than others because 1.) They don’t dissolve well in the liquid, or 2.) They stick more to the paper. The mixture is separated because some components travel farther than others.

• Distillation. Distillation separates substances based on boiling point; if two substances have different boiling points, one substance will boil and leave the other behind. The substance with the lower boiling point boils off first, and then it is cooled, condensed, and collected.

Sample of mixture (starting position)

Substance A (travels slowly)

Substance B (travels quickly)

Liquid seeps through paper from left to right.

The mixture is brought to a boil in this vessel.

The first liquid to boil away cools and condenses in this tube…

…and then drips into a collection vessel.

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Chapter 2: Quantifying Chemistry

2.1 Taking Measurements 2.2 Significant Figures (Sig Figs) 2.3 Calculations with Sig Figs 2.4 Precision vs. Accuracy 2.5 Scientific Notation 2.6 The Metric System 2.7 Fenceposting 2.8 Converting Compound Units 2.9 Converting Volumes 2.10 Using a Compound Unit as a Conversion Factor 2.11 Using Percentage as a Conversion Factor 2.12 Fenceposting in Word Problems 2.13 Temperature

2.1 Taking Measurements Precise and accurate measurements are vital to good science; it is very important to learn how to read measurement tools properly. For example, let’s say we are measuring the length of a pencil stub in centimeters, and we use the following ruler: What can we conclude about the length of the pencil? It is clearly longer than 3 cm and shorter than 4 cm, so it wouldn’t make sense to report the length as simply 3 or 4. We must estimate one more digit than the ruler provides—we use our best judgment and say that the pencil is 3.2 cm or 3.3 cm or 3.4 cm. If the ruler had finer markings, we could get a more precise measurement. What is the length of the pencil using the next ruler? We can now see that the pencil is definitely between 3.2 and 3.3 cm long, and again, we must report one more digit than the ruler provides. So we would say the length of the pencil is 3.26 cm or 3.27 cm or 3.28 cm.

1 cm 2 cm 3 cm 4 cm

1 cm 2 cm 3 cm 4 cm

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What if the length appears to be exactly on a line, like this? Always report one more digit than the ruler provides. Therefore, we would say this pencil is 3.20 cm long. That extra zero is very important! If left off, it would imply that we were using the imprecise ruler shown in the first example. In other words, the number of digits indicates how precise the measurement tool is. What about this situation? Using this ruler, the length must be reported as 3.00 cm to indicate the precision of the measurement. Remember: The measurement tool determines the number of digits. Always estimated one more digit than the tool provides.

In the lab we measure the volume of liquids with a graduated cylinder. Look closely at the liquid’s surface—it is not a straight line. This curved surface is called a meniscus, and it happens because the liquid tends to slightly creep up the sides of cylinder. Read the measurement from the bottom of the meniscus, not the top. So in this example, the volume of the liquid would be 1.56 mL, not 1.66 mL.

1 cm 2 cm 3 cm 4 cm

1 cm 2 cm 3 cm 4 cm

2 mL

1 mL

meniscus

1.5 mL

Read from the bottom of the meniscus (1.56 mL) not from the top (1.66 mL).

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2.2 Significant Figures (Sig Figs) As we just saw, the finer the gradations on the measurement tool, the more digits the measurement will have. A digit that is known with certainty is called a significant figure (sig fig for short). The value 1.5 cm has two sig figs, while the value 1.50 cm has three, implying it was measured with a finer instrument. It sometimes can be difficult to determine how many sig figs a number has. Learn the following rules: If the number has a decimal point in it: Moving from left to right, count from the first nonzero digit to the final digit all the way on the right (include all zeros on the right). If the number does not have a decimal point in it: Moving from left to right, count from the first nonzero digit to the last nonzero digit (do not include all zeros on the right). Some examples of numbers with decimal points:

counting sig figs → 1 2 3 4

9 . 6 0 8 4 sig figs

counting sig figs → 1 2 3 4 5 6 7

0 . 0 5 5 0 0 0 0 0 7 sig figs counting sig figs → 1 2 3 4 5

2 0 0 . 0 0 5 sig figs counting sig figs → 1 2 3 4

6 0 0 0 . 4 sig figs

Some examples of numbers without decimal points:

counting sig figs → 1 2 3 4

1 2 1 8 4 sig figs

counting sig figs → 1 2

5 5 0 0 0 2 sig figs counting sig figs → 1 2 3 4

2 0 0 3 0 0 4 sig figs

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A confusing situation arises when a number lacks a decimal point, but has one or more zeros on the right that are significant. Let’s say a table is 2.0 meters long—the measurement has two sig figs. What if we wanted to express the length in centimeters? If written 200 cm, the number only has one sig fig, which is less precise than the actual measurement. If we write 200., it indicates three sig figs, which is more precise than the actual measurement. (Writing 20. cm changes the value completely.)

To avoid this confusion, use scientific notation (see section 2.5). In the case above, write 2.0 × 102 cm. Another example: To indicate that the two zeros in 88,800 are not significant, write 8.88 × 104. If only the first zero is significant, write 8.880 × 104. If both zeros are significant, write 8.8800 × 104. 2.3 Calculations with Sig Figs The least precise number used in a calculation limits the precision of the calculated value. Let’s say we need to add two lengths: 9.25 cm and 6 cm. It may be tempting to do this:

9.25 +6 15.25

But can we confidently say that the .25 in this answer is correct? The 6 cm measurement was not precise enough to determine those digits; the length could be 6.24, 6.45, 6.02—there’s no way to know. The only thing known with certainty is the six, and therefore we must cut off our answer at that point:

9.25 +6 15

Pay close attention to the following rules about calculations with sig figs. For addition and subtraction: When adding or subtracting numbers of different precision, the answer is rounded off to the least precise digit. Note: this rule depends on digits, not sig figs.

89.56 ← precise to the hundredths +65.5 ← precise to the tenths 155.1 ← must be rounded to the tenths

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11500 ← precise to the hundreds - 352 ← precise to the ones 11100 ← must be rounded to the hundreds

For multiplication and division: When multiplying or dividing, the answer is rounded off so that it has the same number of sig figs as the measurement with the fewest sig figs. Note: unlike addition and subtraction, this rule depends on sig figs, not digits.

2.5688 ← 5 sig figs × 1.9 ← 2 sig figs 4.9 ← must have 2 sig figs

4 sig figs → 7644 3 sig figs → 588

= 13.0 ← must have 3 sig figs

Exact Numbers An exact number is a value known with infinite precision. Counting individual items yields an exact number; for example, the number of students in class is an exact number. We also encounter them when a number is a matter of definition. There are exactly 12 inches in a foot, and exactly 3 feet in a yard; these are defined numbers, not measured quantities. An exact number can be thought of as having an infinite number of sig figs, and so it will never be the limiting factor in a calculation. Let’s say we measure the length of a room as 15.5 feet. How many inches is that? Because there are exactly 12 inches in one foot, we calculate as follows:

15.32 ← 4 sig figs × 12 ← exact number; infinite sig figs 183.8 ← must have 4 sig figs

2.4 Precision vs. Accuracy In everyday language, the words “precision” and “accuracy” are often used interchangeably. However, they have different meanings in a scientific context. Precision is consistency; if the same object is measured multiple times, a precise instrument will give similar results each time. For example, imagine that two scales were used to weigh the same object multiple times, and the results are shown below.

Scale A Scale B 15.5 g 15.6 g 15.4 g 15.5 g

16.3 g 14.9 g 15.4 g 15.9 g

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Scale A gives results that are close to each other, while scale B’s values cover a wider range. Scale A is more consistent and is therefore more precise. Accuracy is how close the instrument, on average, gets to the true value of the object. Let’s compare two scales again; the actual mass they are weighing is 15.0 g.

Scale C Scale D 15.1 g 14.9 g 15.0 g 15.1 g

15.8 g 15.7 g 15.8 g 15.9 g

The average value from scale C is 15.0 (calculated using the proper number of sig figs), while the average value from scale D is 15.8. Because scale C’s average is closer to the true value, it is more accurate. Accuracy and precision do not always go hand in hand. A tool can be precise but not accurate or accurate but not precise. For example, if a scale measures a 15.0-g weight and gives the values:

Scale E 17.1 g 16.9 g 17.0 g 17.0 g

This scale is quite precise because its readings are very consistent, but it is not accurate because its average reading (17.0) is far from the actual value (15.0). Another scale, weighing the same 15.0-g weight, might produce:

Scale F 17.0 g 13.0 g 15.8 g 14.4 g

These readings cover a wide range; the scale is not precise. But the average reading (15.1) is close to the actual value; the scale is accurate. Measuring Precision and Accuracy One way to quantify the precision of an instrument is to use the average deviation—a measurement of variation within the data. The formula is: |measurement 1 – average| + |measurement 2 – average| … etc … number of measurements For example, let’s calculate the average deviation for the measurements from scale F. First, calculate the average of the four readings.

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(17.0 + 13.0 + 15.8 + 14.4)

4 = 15.1

Then, apply the formula.

|17.0-15.1|+|13.0-15.1|+|15.8-15.1|+|14.4-15.1| 4

= 1.35

So the average deviation for this data set is 1.35. Note that we subtract the average from the measurements. A very common mistake is to subtract the actual value instead (the formula gets confused with percent error formula, which we’ll cover next). The percent error is used to measure the accuracy of a set of data (or of a single value). The formula is:

|measured value – actual value| % error =

actual value × 100

For example, let’s calculate the percent error for the data from scale E. The average value of the data was 17.0, and the actual value was 15.0.

|17.0 – 15.0| 15.0

× 100 = 13.3%

So the percent error for these measurements is 13.3%. 2.5 Scientific Notation Chemists often deal with very large numbers and very small numbers. To avoid writing long strings of zeros (5,600,000,000,000 or 0.000000000000000454), use scientific notation. Scientific notation is written as the product of two numbers: the digit term and the exponential term. The digit term takes care of the “interesting” part of the number, while the exponential term takes care of all the zeros. For example, let’s convert the number 3,400,000,000 to scientific notation. First, take out the interesting part of the number, the digit term: 34. The digit term always has a decimal point, and the decimal point will always be directly to the right of the first digit. So in this case, our digit term is 3.4—how far did the decimal point need to move to get there?

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It started at the end of the string of zeros: 3,400,000,000. Then it moved 9 spaces to the left to get between the three and the four, so the exponent in the exponential term must be 9.

3400000000. If the number is greater than 1 the exponent will be positive.

3.4 × 109 For numbers less than 1 the exponent is negative. Let’s convert 0.00000233 to scientific notation. First, take out the digit term, 2.33. Then, count the number of jumps the decimal place made.

0.00000233 It took six jumps, so we end up with 2.33 × 10-6. To convert from scientific notation to standard notation, simply reverse the process. Let’s convert 9.45 × 106 to standard notation. The exponent is positive six, so we know the number is larger than one, which means we must move the decimal point six spaces to the right.

9.450000. And so we end up with 9,450,000. When the exponent is negative, the number is less than one, so we must move the decimal point to the left. To convert 5.4 × 10-4, move the decimal point 4 spaces to the left.

0.0005.4 And so we end up with 0.00054.

digit term exponential term

number of decimal jumps

These zeros are added on.

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How to use scientific notation on most calculators: 1. Type the digit term into the calculator. 2. Push the EE button. Do not use the “×” button. 3. Enter the exponent. Use the +/- button to change its sign. On the calculator, 2.33 × 10-6 will look like 2.33E-6. 2.6 The Metric System Scientists worldwide use the metric system for their measurements. The metric system uses base units for measuring different quantities; the most common are:

Quantity Base Unit length meter mass gram

volume liter Each base unit can be modified with a metric prefix to make larger or smaller units depending on what is being measured. It is more convenient to call the distance between Boston and New York 350 kilometers instead of 350,000 meters; it is more convenient to call the mass of a mosquito 2 milligrams instead of 0.002 grams. The prefixes are listed in the first row of the following chart; each prefix also has an abbreviation listed below it.

kilo hecto deca deci centi milli k h da d c m

×1000 ×100 ×10 ×0.1 ×0.01 ×0.001 The chart shows the relationships between the prefix and the base unit. For example,

1 km = 1000 m 1 hm = 100 m 1 dam = 10 m 1 dm = 0.1 m 1 cm = 0.01 m 1 mm = 0.001 m

2.7 Fenceposting It is frequently necessary to convert from one type of measurement unit to another. For example, if a person is 170 centimeters tall, how tall is the person in meters? There are many techniques for converting one unit to another, and

Base Units

meter gram liter

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one of the most versatile is called fenceposting (also known as the factor-label method). To use fenceposting, we must know the conversion factor that connects the starting unit to the ending unit. In this case, we need to convert from centimeters to meters, and the conversion factor between these units is:

1 m = 100 cm To use fenceposting, set up the problem like this:

170 cm m

Next, put the first unit underneath the second unit, like this:

170 cm m cm

Then, put in the numbers of the conversion factor:

170 cm 1 m 100 cm

Next, multiply all the numbers in the top row together, then all the numbers in the bottom row together, and then divide the top by the bottom:

170 cm 1 m 170 × 1 100 cm

= 100

= 1.70 m

This was a very simple example; it can get much more complicated. For example, convert 10.0 miles to meters. The following conversion factors are provided (these are all exact numbers):

1 m = 100 cm 1 mile = 5280 feet 1 inch = 2.54 cm 12 inches = 1 foot

First, analyze the problem to find the starting point and the endpoint. The starting point will always be a number given in the problem; in this case, the

starting unit with starting value ending unit

this is the conversion factor

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given information is 10 miles. The end point is what the question wants you to find; in this case, the endpoint is meters. We need to make a pathway, or unit map, that gets us from miles to meters. Each conversion factor is one step of the way. Looking at the provided conversion factors, we could make any of the following steps:

conversion factor possible steps 1. 1 m = 100 cm m → cm cm → m 2. 1 mile = 5280 feet miles → feet feet → miles 3. 1 inch = 2.54 cm inches → cm cm → inches 4. 12 inches = 1 foot inches → feet feet → inches

So what steps do we take, and in what order?

miles → ??? → ??? → ??? → ??? → ??? → meters We can convert miles to meters in four steps:

miles → feet → inches → cm → meters Now set up the fencepost:

10.0 miles ??? ??? ??? meters Then enter in the unit map in the top row:

miles feet inches cm m Then fill in the bottom row like this (see why in the next step):

miles feet inches cm m miles feet inches cm

Then add the conversion factors:

10.0 miles 5280 feet 12 inches 2.54 cm 1 m 1 mile 1 foot 1 inch 100 cm

conversion factor 2

conversion factor 3

conversion factor 4

conversion factor 1

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Now multiply everything in the top row and everything on the bottom, and then divide the top by the bottom: 10 × 5280 × 12 × 2.54 × 1 1,609,344

1 × 1 × 1 × 100 = 100

= 16,093 meters

And we’re done! There are 16,093 meters in 10 miles. 2.8 Converting Compound Units A compound unit is a combination of two other units. For example, density is defined as mass per volume; in chemistry this usually means grams per milliliter, or g/mL. To convert a compound unit to another compound unit, split the compound unit into two parts, convert each part separately, and then recombine them at the end. For example, let’s say we need to convert 300 miles/hour to meters/second. It’s important to note that 300 miles/hour means 300 miles per 1 hour, and so we can write:

300 miles 1 hour

Now we split them up and convert them separately. Some useful conversion factors are:

1 km = 0.62 miles 1 hour = 60 minutes 1 km = 1000 m 1 minute = 60 seconds

First convert 300 miles to meters:

300 miles 1 km 1000 m 0.62 mile 1 km

= 484000 m

Then, convert 1 hour to seconds:

1 hour 60 minutes 60 seconds 1 hour 1 minute

= 3600 s

Then, we put the two units back together and divide them:

484000 m 3600 s

= 134 m/s

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2.9 Converting Volumes When converting units of area or volume, there’s another twist; area units are length units squared (square feet, square centimeters, etc.) and volume units are length units cubed (cubic feet, cubic centimeters, etc.). We need to convert 4 in3 to cm3. (Remember that 1 inch = 2.54 cm.) Set it up just like before:

4 inch3 2.54 cm 1 inch

But notice that we were not asked to convert inches to centimeters, but cubic inches to cubic centimeters. So we must cube both the units and the numbers in the conversion factor:

4 inch3 2.54 cm 1 inch

Which gives us:

4 inch3 2.543 cm3 4 inch3 16.39 cm3 13 inch3

= 1 inch3

= 65.5 cm3

2.10 Using a Compound Unit as a Conversion Factor A compound unit, like density, can itself be used as a conversion factor. For example, the density of aluminum is 2.699 g/mL. This means that 1 mL of aluminum has a mass of 2.699 g; we can rewrite this as 1 mL Al = 2.699 g Al. Now we can use this as a conversion factor to convert grams Al to mL Al or mL Al to grams Al. What is the volume (in mL) of 5.00 g of Al?

5.00 g Al 1 mL Al 2.699 g Al

= 1.85 mL

2.11 Using Percentage as a Conversion Factor Percentage is defined as parts per hundred. For example, “2% milk” is 2% fat by mass. This means that if we have 100 g of milk, we have 2 g of fat. (We could use any mass unit; 100 kg of milk contains 2 kg fat.) This can be used as a conversion factor: 100 g milk = 2 g fat. How much fat (in kg) is contained in 1.5 kg of milk?

1.5 kg milk 2 kg fat 100 kg milk

= 0.03 kg fat

3

This step uses density as a conversion factor.

This step uses percentage as a conversion factor.

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2.12 Fenceposting in Word Problems The versatility of fenceposting becomes apparent when it is used to solve word problems. The key is to extract conversion factors from the language of the problem. Example: A small factory can produce 20 toy cars in one workday. Each car requires four wheels. The factory buys wheels in cartons of 100. Each carton costs $50. There are 240 workdays per year. How much money will the factory spend on wheels in two years?

First, pull out all the conversion factors in the problem:

20 cars = 1 workday 1 car = 4 wheels 1 carton = 100 wheels 1 carton = $50 1 year = 240 workdays

Then make the unit map:

2 years → workdays → cars → wheels → cartons → dollars

Finally, set up the fencepost:

2 years 240 workdays 20 cars 4 wheels 1 carton $50 1 year 1 workday 1 car 100 wheels 1 carton

= $19,200

2.13 Temperature Scientists usually use the Celsius and Kelvin temperature scales. In Celsius, 0 degrees is defined as the freezing point of water, and 100 degrees is the boiling point of water. A degree in the Kelvin system is the same size as a degree in the Celsius system, but the reference points are different. Zero degrees Kelvin is defined as absolute zero: the lowest possible temperature (there are no negative numbers on the Kelvin scale). In Kelvin, water freezes at 273 and boils at 373. To convert between Celsius and Kelvin:

K = C + 273 or C = K - 273

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Chapter 3: Ionic Formulas and Nomenclature

3.1 Element Symbols 3.2 Molecular vs. Ionic Compounds 3.3 Ion Formation Rules 3.4 Polyatomic Ions 3.5 Ionic Formulas 3.6 Naming Ionic Compounds 3.7 Naming Molecular Compounds

3.1 Element Symbols It saves time to know the element symbols on the periodic table. You do not need to memorize them all; we will concentrate on some of the more commonly encountered elements. Memorize the following 42 symbols:

Symbol Name Symbol Name H Hydrogen Hg Mercury Li Lithium B Boron Na Sodium Al Aluminum K Potassium C Carbon Pu Plutonium Si Silicon Ti Titanium U Uranium Be Beryllium Sn Tin Mg Magnesium Pb Lead Ca Calcium N Nitrogen Xe Xenon P Phosphorus Ba Barium As Arsenic Cr Chromium O Oxygen Mn Manganese S Sulfur Co Cobalt Kr Krypton Fe Iron F Fluorine Ni Nickel Cl Chlorine Zn Zinc Br Bromine Pt Platinum I Iodine Cu Copper He Helium Ag Silver Ne Neon Au Gold Ar Argon

3.2 Molecular vs. Ionic Compounds The elements on the periodic table are classified as metals and nonmetals (there are other, finer classifications which we will discuss in chapter 9). The dividing line between metals and nonmetals is a zigzag line that starts at aluminum and ends at polonium:

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H He

Li Be B C N O F Ne

Na Mg Al Si P S Cl Ar

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Lr Rf Ha Sg Ns Hs Mt

Elements to the left of this line are classified as metals; elements to the right are classified as nonmetals. There are two types of compounds: molecular and ionic. Molecular compounds are made from two nonmetals, for example: CO2, N2O5, XeF6. Ionic compounds are made from a metal and a nonmetal, for example: NaCl, MgO, CaF2. The two types of compounds also differ in the type of bond that holds them together; molecular compounds use covalent bonds, in which atoms in the compound share electrons, and ionic compounds use ionic bonds, in which the atoms stick together because one is positively charged and the other is negatively charged. (We will study these bonds in greater detail in chapter 10.) 3.3 Ion Formation Rules When a metal and a nonmetal combine to form an ionic compound, the metal atom gets a positive charge, and the nonmetal atom gets a negative charge. (When an atom has an electric charge, it is called an ion. Positive ions are called cations; negative ions are called anions.) Ions are written with their charges as superscripts: Al+3, F-, S-2. Different atoms form ions of different charges, generally according to their column on the periodic table. Know the following ion formation rules: Cations

• All elements in column 1 will form ions with a +1 charge (Na+, K+). • All elements in column 2 will form ions with a +2 charge (Mg+2, Ca+2). • Aluminum will form an ion with a +3 charge (Al+3). • Silver will form an ion with a +1 charge (Ag+). • Cadmium and Zinc will form ions with a +2 charge (Zn+2, Cd+2).

For other elements not listed above, the charge of the ion can vary. For example, iron (Fe) can form both Fe+2 and Fe+3 ions, but sodium (Na, a column 1 element) can only form Na+.

This zigzag line divides metals and nonmetals on the periodic table.

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Anions • All elements in column 17 will form ions with a -1 charge (F-, Cl-). • Sulfur, oxygen, and selenium will form ions with a -2 charge (O-2, S-2,

Se-2). • Nitrogen and phosphorus will form ions with a -3 charge (N-3, P-3).

3.4 Polyatomic Ions As the name suggests, a polyatomic ion is made from multiple atoms. This cluster of atoms behaves as a single particle with a single charge. Here are some common polyatomic ions:

NO3- nitrate CO3

-2 carbonate SO4

-2 sulfate HCO3- bicarbonate

PO4-3 phosphate OH- hydroxide

NH4+ ammonium C2H3O2

- acetate 3.5 Ionic Formulas When cations and anions combine to form an ionic compound, the resulting compound is always neutral—it has zero charge. Thus, the cations and anions must combine in a certain ratio so that their charges cancel each other out. For example, when Na+ and Cl- combine, they must do so in a 1-to-1 ratio, because one sodium ion requires one chlorine ion to balance out the charges:

Na+ Cl- The charges add up to zero, so the formula is NaCl. When Mg+2 and Cl- combine, each Mg+2 requires two Cl- ions to cancel out the +2 charge: Mg+2 Cl- Cl- The charges add up to zero, so the formula is MgCl2. What about K+ and N-3? We need three K+ to balance one N-3: K+ K+ K+ N-3 The formula is K3N. It’s a little trickier when the charges are +2 and -3 or -2 and +3, for example, Mg+2 and P-3: Mg+2 Mg+2 Mg+2 P-3 P-3 The formula is Mg3P2. Note: In all ionic compounds, the cation (positive ion) is always listed first in the formula. Polyatomic ions work just the same way. However, if the polyatomic ion is doubled or tripled in the formula, put parenthesis around it with a subscript

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outside the parentheses to indicate the number of polyatomic ions. For example, if Al+3 and NO3

- are combined, we need three NO3- to balance one Al+3:

Al+3 NO3

- NO3- NO3

- The formula is Al(NO3)3. If NH4

+ and CO3-2 are combined:

NH4

+ NH4+

CO3-2 The formula is (NH4)2CO3.

3.6 Naming Ionic Compounds To name ionic compounds, use the following procedure.

1. Name the cation. 2. If the charge on the cation can vary (in other words, if it is not one of the

cations listed in section 3.3), write the charge in parentheses as a Roman numeral.

3. Drop the last syllable of the anion and replace it with “–ide.” If the anion is a polyatomic ion, simply name the polyatomic ion without changing the last syllable.

For example, let’s name MgCl2.

1. Name the cation.

magnesium

2. If the charge on the cation can vary, write the charge in parentheses as a Roman numeral.

Magnesium is in column 2, and therefore its charge cannot vary, so we do not write a roman numeral.

3. Drop the last syllable of the anion and replace it with “–ide.” If the anion

is a polyatomic ion, simply name the polyatomic ion without changing the last syllable.

magnesium chloride

Now let’s try FeCO3.

1. Name the cation.

iron

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2. If the charge on the cation can vary, write the charge in parentheses as a Roman numeral.

The charge on iron can vary, so we must indicate its charge with a Roman numeral. CO3 is a polyatomic ion with a charge of -2, so the iron must have a charge of +2 to balance it out. So we write: iron(II)

3. Drop the last syllable of the anion and replace it with “–ide.” If the anion

is a polyatomic ion, simply name the polyatomic ion without changing the last syllable.

iron(II) carbonate

3.7 Naming Molecular Compounds Use the following procedure.

1. Name the elements in the order they appear in the formula. 2. Drop the last syllable of the final element and replace it with “–ide.” 3. Add Greek numerical prefixes to the name of each element to tell how

many atoms of the element are in the compound. Exception: If the first element is singular, it doesn’t get a prefix.

The Greek prefixes for 1 through 8 are: mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octo-. Let’s name S4N2, following the rules step by step.

1. Name the elements in the order they appear in the formula.

sulfur nitrogen

2. Drop the last syllable of the final element and replace it with “–ide.”

sulfur nitride

3. Add Greek numerical prefixes to the name of each element to tell how many atoms of the element are in the compound. Exception: If the first element is singular, it doesn’t get a prefix.

Tetrasulfur dinitride

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Now let’s try CO:

1. Name the elements in the order they appear in the formula.

carbon oxygen 2. Drop the last syllable of the final element and replace it with “–ide.”

carbon oxide

3. Add Greek numerical prefixes to the name of each element to tell how

many atoms of the element are in the compound. Exception: If the first element is singular, it doesn’t get a prefix.

carbon monoxide

Notice that due to the exception in rule 3, this is not monocarbon monoxide.

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Chapter 4: The Mole

4.1 The Mole 4.2 Avogadro’s Number as a Conversion Factor 4.3 Molar Mass 4.4 Using Molar Mass as a Conversion Factor 4.5 Empirical vs. Molecular Formulas 4.6 Percent Composition 4.7 Empirical and Molecular Formulas from Percent Composition

4.1 The Mole One sip of water contains approximately 1,000,000,000,000,000,000,000,000 (one septillion) molecules of water. Because quantities that we use every day are unimaginably huge on the particulate level, chemists have devised a more manageable unit of measurement. A similar strategy is used when buying eggs; it is more convenient to ask for two dozen eggs than twenty-four. Because we commonly deal with trillions and trillions of molecules, the “chemist’s dozen” needs to contain a lot more than twelve. If we converted the molecules of water in one sip to dozens of molecules, we would get: 1,000,000,000,000,000,000,000,000

12 = 83,300,000,000,000,000,000,000

dozen Which isn’t much of an improvement. The chemist’s dozen needs to be large enough to bring astronomical quantities of molecules down to earth; in fact, a chemist’s dozen contains 602,000,000,000,000,000,000,000 particles, or more concisely, 6.02 × 1023 particles. This number is called Avogadro’s number (in honor of Lorenzo Romano Amedeo Carlo Avogadro, a famous scientist of the 1800’s), and instead of calling it a chemist’s dozen, it is called a mole (abbreviated as “mol”). There are, therefore…

1,000,000,000,000,000,000,000,000 6.02 × 1023 = 1.66 moles of water

…in one sip. 4.2 Avogadro’s Number as a Conversion Factor If we know the number of particles, we can calculate the number of moles, and vice versa. To do so, use the following as a conversion factor: 1 mole of particles = 6.02 × 1023 particles A sample of helium contains 5.40 × 1025 atoms. How many moles is this?

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5.40 × 1025 atom He 1 mol He 6.02 ×1023 atoms He = 89.7 mol He

How many molecules are in 3.50 moles of water?

3.50 mol water 6.02 ×1023 molecules water 1 mol water

= 2.11 × 1024 molecules

4.3 Molar Mass Different atoms weigh different amounts. The average atom of oxygen weighs 2.65 × 10-23 g, while the average hydrogen atom weighs 1.68 × 10-24 g. Because each oxygen atom weighs about 16 times more than each hydrogen atom, one mole of oxygen atoms will weigh about 16 times more than one mole of hydrogen atoms—just as a dozen bowling balls weighs more than a dozen tennis balls. The mass of one mole of an element is called the molar mass of that element. The molar mass of each element is listed on the periodic table:

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Co

Cobalt 58.93

Molecules also are counted using moles; one mole of CO2 contains 6.02 × 1023 molecules of CO2. Like elements, compounds also have molar mass—the mass of one mole of molecules. The molar mass of a compound is calculated by adding up the molar masses of the elements in the compound. For example, the molar mass of glucose (C6H12O6), is found as follows: The periodic table provides the molar masses of carbon, hydrogen, and oxygen: C: 12.01 g/mol H: 1.01 g/mol O: 16.00 g/mol Multiply the molar masses by the subscripts in the chemical formula, and add up the results: 12.01 × 6 = 72.06 1.01 × 12 = 12.12 16.00 × 6 = 96.00 180.18 g/mol

One mole of cobalt (6.02 × 1023 atoms of cobalt) has a mass of 58.93 grams.

Avogadro’s number is used as a conversion factor.

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4.4 Using Molar Mass as a Conversion Factor It is often very important in chemistry to know the number of particles (moles) in a sample of a substance. Unfortunately, there is no laboratory instrument that can directly measure the number of particles contained in a sample. We must weigh the sample and then use the molar mass to calculate the number of moles. We can do this with fenceposting, using molar mass as a conversion factor. For glucose, with a molar mass of 180.18 g/mol, this can be expressed mathematically as:

180.18 g glucose = 1 mole glucose So, if we have 10.0 g glucose, how many moles do we have? Using fenceposting:

10.0 g glucose 1 mol glucose 180.18 g glucose

= 0.0555 mol glucose

We can also convert moles to grams. What is the mass of 4.20 mol glucose?

4.20 mol glucose 180.18 g glucose 1 mol glucose

= 757 g glucose

Because we will convert between grams and moles very frequently, setting up fenceposts can take up too much time. A handy way to convert between the two is by using the triangle shown at right. You can solve for any value on the triangle using the other two values. For example, to find the number of moles, divide the grams by the molar mass (notice that grams are positioned above molar mass, like a numerator over a denominator). To find the grams, multiply the molar mass by the number of moles (the molar mass is right next to the moles, as if written “molar mass × moles”. To find the molar mass, divide the grams by the moles. 4.5 Empirical vs. Molecular Formulas There are two versions of the chemical formula: empirical and molecular. The molecular formula is the one we have learned already. It is the most complete description of the molecule—the number of every atom in the molecule is given by the subscripts. For example, a molecule of glucose has 6 carbons, 12 hydrogens, and 6 oxygens, and so the molecular formula is C6H12O6. However, sometimes the exact structure of a molecule is not known, and the only information given by experiments is the ratio of each element to each other. An analysis of glucose

The molar mass is used as a conversion factor.

grams

molar mass

moles

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might only reveal that there are equal amounts of carbon and oxygen, with double that amount of hydrogen. This is still useful information, and it can be expressed in the empirical formula, in which the subscripts express the ratio of each element within the molecule: CH2O is the empirical formula for glucose. In other words, the empirical formula is a simplified version of the molecular formula, in which the ratios of the elements are expressed in the lowest possible whole number subscripts.

Molecular Formula Empirical Formula C5H10 CH2 C6H14 C3H7

C4H10O3 C4H10O3 Notice in the last example that the empirical formula and molecular formulas are the same; the molecular formula could not be reduced any further. Converting from the empirical formula to the molecular formula is possible if you know the molar mass of the actual molecule. Imagine that our lab analysis has revealed that the empirical formula of a compound is CH2, and that the molar mass of the compound is about 56 g/mol. What is the molecular formula of the compound? We know that the molecular formula must be a multiple of the empirical formula. The possibilities are therefore:

CH2 C2H4 C3H6 C4H8 C5H10 C6H12 etc. We need to find the one that has the right molar mass:

CH2 C2H4 C3H6 C4H8 C5H10 C6H12 etc. 14.03 g/mol 28.06 g/mol 42.09 g/mol 56.12 g/mol 70.15 g/mol 84.18 g/mol

C4H8 is the closest match, so it must be the molecular formula. Instead of laboriously calculating a sequence of molar masses, we can also calculate the multiple by dividing the experimental molar mass by the empirical molar mass, and then multiply the empirical formula by the result. In this case,

experimental molar mass → 56 empirical molar mass → 14.03

≈ 4 ← multiple

CH2 × 4 = C4H8

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4.6 Percent Composition Some chemical analyses won’t even produce the empirical formula of a compound—they produce the percent composition. The percent composition is the percentage, by mass, of each element in the compound. For example, in the molecule C4H8, the molar mass is 56.12 g/mol. The four carbon atoms in this molecular together contribute 4 × 12.01 = 48.04 g/mol to the total, while the eight hydrogen atoms together contribute 8 × 1.01 = 8.08 g/mol to the total. Thus, the percent composition of carbon is:

48.04 56.12

× 100 = 85.6 %

And the percent composition of hydrogen is:

8.08 56.12

× 100 = 14.4 %

4.7 Empirical and Molecular Formulas from Percent Composition If the chemical analysis provides the percent composition, we can convert it to an empirical formula and then use the molar mass to get the molecular formula. Use the procedure outlined in the following example. An unknown compound is found to be 27.5% carbon, 24.4% oxygen, 42.7% nitrogen, and 5.4% hydrogen by mass. The molar mass of the compound is approximately 262 g/mol. Find the empirical and molecular formulas. The first step is to pretend that there is 100 g of the substance (because the data is in percent, the actual amount doesn’t matter). Then figure out how many grams of each element are in this imaginary sample, by multiplying the 100 g by the percentages (in decimal form):

C: 100 g × 0.275 = 27.5 g O: 100 g × 0.244 = 24.4 g N: 100 g × 0.427 = 42.7 g H: 100 g × 0.054 = 5.4 g

Next, convert each of these masses to moles by dividing each mass by the molar mass of the element:

C: 27.5/12.01 = 2.29 mol O: 24.4/16.00 = 1.53 mol N: 42.7/14.00 = 3.05 mol H: 5.4/1.01 = 5.3 mol

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Next, divide each value by the lowest value, which is 1.53 in this case:

C: 2.29/1.53 = 1.50 mol O: 1.53/1.53 = 1.00 mol N: 3.05/1.53 = 1.99 mol H: 5.3/1.53 = 3.5 mol

With luck, these will all be very close to whole numbers, and we’re done—just write the whole numbers as subscripts of the empirical formula. If they’re not whole numbers, they should be very close to a particular fraction, such as 1/2, 1/4, 3/4, 1/3 or 2/3. (As decimals, these are 0.5, 0.25, 0.75, 0.33, and 0.66, respectively.) If one of the results is a decimal, multiply the formula by a small integer to get whole numbers. For example:

0.00 – don’t multiply 0.25 – multiply by 4 0.50 – multiply by 2

0.75 – multiply by 4 0.33 – multiply by 3 0.66 – multiply by 3

In our problem, the numbers end up like this:

C: 1.50 mol = 1.5 mol O: 1 mol = 1 mol N: 1.99 mol ≈ 2 mol H: 3.5 mol = 3.5 mol

Because we have 1.5 and 3.5, multiply everything by 2 to get:

C: 1.5 x 2 = 3 mol O: 1 x 2 = 2 mol N: 2 x 2 = 4 mol H: 3.5 x 2 = 7 mol

These numbers are the subscripts, so the empirical formula is C3O2N4H7. To help remember the procedure for finding the empirical formula, here’s a rhyme:

Percent to mass Mass to moles Divide by smallest Multiply ‘til whole!

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Next, find the molecular formula as we did in section 4.5. First, find the molar mass of the empirical formula, C3O2N4H7.

3 × 12.01 = 36.03 2 × 16.00 = 32.00 4 × 14.00 = 56.00 7 × 1.01 = 7.07

131.1 g/mol Knowing that the molar mass of the compound is 262, we need to multiply the empirical formula by 2 to get the molecular formula:

C3O2N4H7 × 2 = C6O4N8H14

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Chapter 5: Chemical Equations

5.1 Chemical Equations 5.2 Balancing Chemical Equations 5.3 Stoichiometry 5.4 Limiting Reactant 5.5 Theoretical Yield 5.6 The ICE Chart 5.7 Percent Yield 5.8 Classifying Reactions 5.9 Predicting Products 5.10 The Diatomic Seven

5.1 Chemical Equations A chemical equation is a written description of a chemical reaction. For example, when the methane gas in a Bunsen burner burns, the reaction is written:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) The chemicals to the left of the arrow, CH4 and O2, are called the reactants; they are the starting point of the reaction. The arrow itself can be read as “yields” or “forms”; it represents the progress of the reaction, the process of chemical transformation. The CO2 and H2O to the right of the arrow are called the products: what the reaction creates. The letters in parentheses after the chemical formulas are called state symbols. They represent the phase of matter of the corresponding chemical: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous, which means the chemical is dissolved in water. The numbers to the left of the chemical symbols are called coefficients. They indicate the number of molecules that takes part in the reaction. (If the coefficient is one, it is omitted.) In the reaction above, the coefficients tell us that 1 molecule of CH4 reacts with 2 molecules of O2 to produce 1 molecule of CO2 and 2 molecules of H2O. Remember that chemists usually deal with moles instead of individual molecules, so it is more practical to read the equation as: 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O. Sometimes there are words, symbols, or chemicals written above the arrow; these indicate various conditions under which the reaction occurs. For example, a temperature may be written above the arrow to indicate that the reaction should be performed at that temperature. If a chemical symbol or formula is written above the arrow, that substance acts as a catalyst for the reaction (a catalyst is a substance that speeds up a chemical reaction but is not a reactant or a product). The symbol Δ (delta) indicates that the reactants should be

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heated to help them react. Sometimes the word “electricity” is written above the arrow, meaning that electricity must flow through the reactants to make them react. 5.2 Balancing Chemical Equations A chemical reaction obeys the Law of Conservation of Mass, which states that matter can neither be created nor destroyed. In other words, all the atoms on the reactant side of the arrow must be accounted for on the product side, no more, no less. Chemical reactions only rearrange the atoms by breaking old bonds and forming new ones. A common chemistry problem is to balance chemical equations to make sure that all atoms on the reactant side match the atoms on the product side. When doing this, the subscripts in the chemical formulas never change; the only numbers that may change are the coefficients. For example, take this unbalanced equation:

N2 + H2 → NH3 To begin with, there are two N’s on the left, but only one on the right. So to balance the N’s, we give NH3 a coefficient of 2:

N2 + H2 → 2NH3 Now the N’s are balanced, but the H’s are not. We have two H’s on the left and 6 H’s on the right (two molecules of NH3, and each molecule contains 3 H’s). So H2 needs a coefficient of 3:

N2 + 3H2 → 2NH3 Sometimes balancing is easier using simple molecular models to keep track of the atoms. Let’s balance the following equation using this method.

C3H8 + O2 → H2O + CO2 The simple molecular models look like this (basically chemical formulas with each atom represented—no subscripts):

CCCHHHHHHHH OO → HHO COO

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This makes it easier to count the number of each atom. There are eight H’s on the left and only two on the right. So add three more HHO’s to balance the H’s:

CCCHHHHHHHH OO → HHO COO HHO HHO HHO Now the H’s balance. There are three C’s on the left, but only one on the right, so add two more COO’s to balance the C’s:

CCCHHHHHHHH OO → HHO COO HHO COO HHO COO HHO Now we just need to balance the O’s. There are two on the left and ten on the right. So add four more OO’s to balance them out.

CCCHHHHHHHH OO → HHO COO OO HHO COO OO HHO COO OO HHO OO Now count up the numbers of each molecule, and use the results as the coefficients in the equation.

C3H8 + 5O2 → 4H2O + 3CO2 Balancing is basically trial and error, and experience is the best teacher. There are a couple of general guidelines:

1. Balance elements last. For example, when balancing the following equation, it is much easier to balance oxygen last.

C2H2 + O2 → H2O + CO2

2. If the same polyatomic ion appears on both sides of the equation, balance

the polyatomic ion as a unit, not as individual elements. For example, the nitrate ion (NO3

-) in the following reaction can be balanced simply by adding the coefficient 2 to NaNO3.

NaNO3 + CaCl2 → NaCl + Ca(NO3)2

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5.3 Stoichiometry Stoichiometry is the mathematics of the molar ratios in a chemical reaction; stoichiometry’s core is the coefficients of the reaction. The coefficients express the ratios of particles (moles) between the chemicals in the equation. Remember, the coefficients DO NOT express ratios of mass, only moles!

2C3H6 + 9O2 → 6CO2 + 6H2O For example, the above equation says that if 2 mol C3H6 react, 9 mol O2 will also react, 6 mol CO2 will form, and 6 mol H2O will form. Likewise, the equation tells us that if we want to form 6 mol of H2O, we need to combine 2 mol C3H6 and 9 mol O2. All these ratios can be used as conversion factors in fenceposting problems. Example: If 55.4 moles of O2 react, how many moles of CO2 will form?

55.4 mol O2 6 mol CO2 9 mol O2

= 36.9 mol CO2

Example: How many moles of O2 will react with 43.5 mol C3H6?

43.5 mol C3H6 9 mol O2 2 mol C3H6

= 196 mol O2

Example: To produce 140 mol H2O, how many moles of C3H6 must react?

140 mol H2O 2 mol C3H6 6 mol H2O

= 46.7 mol C3H6

If the problem deals with units of mass, we need to add mole-to-mass conversions (using molar mass values): Example: If 65.9 g of O2 react, how many grams of CO2 will be formed?

65.9 g O2 1 mol O2 6 mol CO2 44 g CO2 32 g O2 9 mol O2 1 mol CO2

= 60.4 g CO2

The 6:9 ratio from the coefficients are used as a conversion factor.

Conversions between moles and grams are made using the molar mass.

The coefficient ratio is still the core of the problem.

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Example: To produce 99.9 g H2O, how many grams of C3H6 must react?

99.9 g H2O 1 mol H2O 2 mol C3H6 42 g C3H6 18 g H2O 6 mol H2O 1 mol C3H6

= 77.7 g C3H6

5.4 Limiting Reactant In real life, it is rare that the reactants are combined in a perfect molar ratio. Usually, one of the reactants is used up before the others—this is called the limiting reactant (also called the limiting reagent). Consider the reaction below:

H2 + O2 → H2O2 In the reaction, 1 mol H2 combines with 1 mol O2. If, in a real experiment, we combined exactly 1 mol H2 and 1 mol O2, both will be completely used up. However, if we combined 1 mol H2 and 3 mol O2, we will use up all the H2, 1 mole of the O2, and 2 mol O2 will be left over. The H2 is the limiting reactant, while O2 is said to be in excess. Identifying the limiting reagent is not always this easy. One way to figure out the limiting reagent is to compare the actual amounts of reactants to the ideal amounts indicated by the equation. For example: If 16.6 g C3H6 and 47.1 g O2 react in the equation below, which is the limiting reactant?

2C3H6 + 9O2 → 6H2O + 6CO2 First, pick one of the reactants (it doesn’t matter which one), and ask the question: for this reactant to be used up, how much of the other reactant is needed? Answer the question with a fencepost:

16.6 g C3H6 1 mol C3H6 9 mol O2 32 g O2 42 g C3H6 2 mol C3H6 1 mol O2

= 56.9 g O2

So, if 56.9 g of O2 are needed, but the problem stated there was only 47.1 g, O2 must be the limiting reagent. If we had chosen to do the fencepost starting with O2 instead of C3H6, it would be:

47.1 g O2 1 mol O2 2 mol C3H6 42 g C3H6 32 g O2 9 mol O2 1 mol C3H6

= 13.7 g C3H6

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According to this calculation, 13.7 g C3H6 are needed to use up the O2, and since the problems gives us 16.6 g C3H6, we know that it will not be used up—the conclusion is, again, that O2 must be the limiting reagent. 5.5 Theoretical Yield The limiting reagent controls (limits) the amount of products that the reaction can make; once one of the reactants is gone, the reaction stops and no more product is made. The amount of product made when the limiting reagent is gone is called the theoretical yield; it is the maximum amount of product that can be produced by the given amounts of reactants. The theoretical yield is calculated using a fencepost, which must start with the amount of the limiting reagent. Example: If 16.6 g C3H6 and 47.1 g O2 react in the equation below, what is the theoretical yield of H2O?

2C3H6 + 9O2 → 6H2O + 6CO2 In the previous section, we determined that O2 was the limiting reagent. So, we fencepost starting with 47.1 g O2, and solve for the amount of H2O formed:

47.1 g O2 1 mol O2 6 mol H2O 18 g H2O 32 g O2 9 mol O2 1 mol H2O

= 17.7 g H2O

5.6 The ICE Chart A useful technique for keeping track of all the moles in a reaction is the ICE chart. “ICE” stands for “initial, change, end.” Let’s use the reaction:

2Mg + O2 → 2MgO If 12 mol Mg and 5 mol O2 are combined and allowed to react, how many moles of all species are present when the reaction is over? First, we need to determine the limiting reagent as we did in section 5.4; in this problem, the limiting reagent is O2. The ICE chart begins with the initial (starting) amounts of everything. Note that before the reaction has occurred, the amount of MgO is zero:

2Mg + O2 → 2MgO Initial 12 5 0

Because O2 is the limiting reagent, we know that it will be used up completely. Therefore, the amount of O2 will decrease by 5 moles. According to the

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coefficients in the equation, if one mole of O2 reacts, 2 mol of Mg also react. Therefore, if the 5 mol O2 react, 10 mol Mg also reacts. Likewise, the coefficients also tell us that if 5 mol O2 react, then 10 mol MgO are formed. This information goes into the “change” line of the ICE chart:

2Mg + O2 → 2MgO Initial 12 5 0

Change -10 -5 +10 Then, some simple subtraction to calculate the ending amounts:

2Mg + O2 → 2MgO Initial 12 5 0

Change -10 -5 +10 End 2 0 10

The ICE chart tells us that the O2 is used up completely, there are 2 moles of Mg left over, and 10 mol MgO will be formed. 5.7 Percent Yield The theoretical yield is rarely achieved in actual practice; chemical reactions hardly ever produce as much as they should. The amount that is actually produced is only some percent of what could be produced, and this is called the percent yield. It is calculated as follows:

actual amount of product theoretical yield

× 100 = % yield

In our example, if the theoretical yield of a reaction was 43.2 g CO2, but only 35.5 g were actually produced, our % yield would be:

35.5 g CO2 43.2 g CO2

× 100 = 82.2%

5.8 Classifying Reactions Although there are many different types of chemical reactions, there are five common types:

• Synthesis • Decomposition • Single replacement • Double replacement • Combustion

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Synthesis reactions create a single new substance from its component parts. They follow the general pattern of:

A + B → AB Some examples: 2H2 + O2 → 2H2O 2Na + Cl2 → 2NaCl H2O + CO2 → H2CO3 We can think of synthesis reactions as “a couple going on a date”; A and B start out single, then form a happy couple. Decomposition reactions are just the opposite of synthesis reactions: they break down a single substance into its component parts. They follow the general pattern of:

AB → A + B Some examples: 2H2O → 2H2 + O2 2NaCl → 2Na + Cl2 H2CO3 → H2O + CO2 We can think of decomposition reactions as “a couple breaking up”; AB starts out as a happy couple, and then they split up. Single replacement reactions start with an element and a compound. In the reaction, the element trades places with one of the elements in the compound. They follow the general pattern:

A + BC → B + AC Notice how A has taken B’s place in the compound, and B is now in elemental form. Some examples:

K + NaCl → Na + KCl Ca + MgBr2 → Mg + CaBr2 3Na + AlCl3 → 3NaCl + Al

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In the last example, notice that when Na replaced Al, it formed NaCl, and NOT NaCl3. NaCl3 is not a valid compound because the charges on the ions do not balance out. The products of the reaction must never violate the formula rules! We can think of single replacement reaction as “someone cuts in at the dance”; BC is a couple, and then A comes along and cuts in, kicking out B and forming a new couple, AC. Double replacement reactions start with two compounds. In the reaction, two elements change places. They follow the general pattern:

AB + CD → CB + AD A has replaced C, and C has replaced A. Some examples:

NaBr + KCl → NaCl + KBr MgCO3 + CaCl2 → MgCl2 + CaCO3 2AlCl3 + 3MgO → Al2O3 + 3MgCl2

In the second example, notice that the polyatomic ion CO3

-2 remains intact and behaves as a unit. And again, all the product formulas must be valid. We can think of double replacement as “a double date gone horribly wrong”; A and B start out as a couple, C and D start out as a couple, but by the end of the date they’ve switched partners. Combustion reactions are the reactions of burning. We will focus on reactions that involve burning common fuels: compounds containing carbon, hydrogen, and oxygen. They follow the general pattern:

CxHyOz + O2 → CO2 + H2O The subscripts in the hydrocarbon x, y, and z can be any number; there is a huge number of possibilities. Oxygen (O2) will always be a reactant in combustion reactions (nothing can burn without oxygen). No matter what the fuel is, the products of combustion reactions will always be CO2 and H2O. Some examples:

CH4 + O2 → CO2 + 2H2O 4C2H5O + 13O2 → 8CO2 + 10H2O C6H12O6 + 6O2 → 6CO2 + 6H2O

Just to extend our dating metaphor, we can think of combustion reactions as “fiery love.”

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5.9 Predicting Products The type of reaction can also be determined using only the reactants.

Reaction Type Reactant Pattern Synthesis A + B →

Decomposition AB → Single replacement A + BC → Double replacement AB + CD →

Combustion CxHyOz + O2 → Once we classify the reaction from the reactants, we can predict the products. For example, let’s say the reactants are:

K + Ca(NO3)2 → This fits the pattern of a single replacement reaction. The K replaces the Ca, and so the complete reaction will be:

K + Ca(NO3)2 → KNO3 + Ca Notice that the compound formed (KNO3) has balanced charges; K forms a +1 ion, and NO3

- is -1. A common mistake is to maintain the number of NO3-‘s on

the product side, making the compound K(NO3)2. This probably arises out of a desire to balance the equation, but make sure to write valid product compounds first, and then balance the equation:

2K + Ca(NO3)2 → 2KNO3 + Ca 5.10 The Diatomic Seven Seven elements are diatomic (the atoms are bound together in groups of two) in their most common form: H2, N2, O2, F2, Cl2, Br2, and I2. To remember “the diatomic seven”, notice that their symbols make the shape of a 7 on the periodic table (except for hydrogen):

H He

Li Be B C N O F Ne

Na Mg Al Si P S Cl Ar

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Lr Rf Ha Sg Ns Hs Mt

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Remember the diatomic seven when predicting products. For example, when predicting the products of the following reaction:

CO → This is a decomposition reaction, and the products are elemental carbon and oxygen. It may be tempting to write:

CO → C + O But oxygen is diatomic, so it must be:

2CO → 2C + O2

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Chapter 6: Atomic Structure

6.1 Atomism, Joseph Proust, and John Dalton 6.2 J. J. Thompson and the Electron 6.3 Robert Millikan and the Charge of the Electron 6.4 Ernest Rutherford and the Nucleus 6.5 The Line Spectrum 6.6 Niels Bohr and Orbits 6.7 Werner Heisenberg and Erwin Schrödinger 6.8 The Nucleus 6.9 Atomic Number and Atomic Mass

6.1 Atomism, Joseph Proust, and John Dalton In the Western world, atomism—the idea that matter is made out of atoms—was first proposed in ancient Greece, and is commonly credited to Democritus (ca. 460—ca. 370 BC). He pictured atoms as tiny, indivisible spheres, much as we have done so far in this course. The idea did not gain widespread acceptance for many centuries, until rigorous experimentation provided strong supporting evidence. In this chapter, we will learn about some of the most important scientists and experiments that supported atomism and led to the development of the modern model of the atom. French scientist Joseph Proust is credited with discovering the Law of Definite Proportions, which can be paraphrased: Any given compound always contains the same ratio of elements. For example, water is always composed of 1 oxygen atom and 2 hydrogen atoms (which works out to 8/9 oxygen and 1/9 hydrogen by mass); there is no other “version” of water with a different ratio. This consistency can be explained if oxygen and hydrogen are made out of atoms, and when these atoms combine to form water, they do so in a certain ratio. In contrast, imagine what would happen if matter were not made of particles; imagine instead that each element was like clay: there’s hydrogen clay, oxygen clay, carbon clay, etc. Two blobs of element clay could conceivably combine in any possible ratio of amounts. Because Proust’s experiments demonstrated that this was not the case, atomism began to look more likely.

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English scientist John Dalton strengthened the case further by demonstrating the Law of Multiple Proportions. Let’s say we have 12 grams of carbon (1 mole), and we combine it with oxygen to make the compound carbon monoxide (CO); it requires 16 grams of oxygen (1 mole) to react with all the carbon—we can’t use up all the carbon with less than this amount. Then let’s say we start again with the same 12 grams of carbon, but this time we combine it with oxygen to make carbon dioxide (CO2); this time it requires 32 grams (2 moles) of oxygen. The amount of oxygen in the second compound is exactly double the amount of oxygen in the first. If we start with 12 g carbon, we can’t make a compound with 20 or 24 g of oxygen; as long as the starting amount of carbon remains 12 g, oxygen must combine with it in multiples of 16 g. Once again, this finding only makes sense if matter is made out of atoms. In the first compound, each atom of carbon combines with one atom of oxygen to form one molecule of carbon monoxide (CO). In the second compound, each atom of carbon combines with two atoms of oxygen to form one molecule of carbon dioxide (CO2). So a given amount of carbon will only combine with a certain amount of oxygen, or exactly double that amount of oxygen, or exactly triple, etc. The Law of

H H

O

H H

O

H H

O

H H

O

O

O

O

H

H

H

If matter were made of atoms, it is easy to explain why a certain compound always has a consistent ratio of elements.

In the “clay” model, water could be formed by any amount of hydrogen and oxygen. Proust’s research disproved this.

C + O → C O

C + O O → O C O

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Multiple Proportions is: If two elements can form two or more different compounds, a fixed amount of the first element will combine with whole number multiples of the second element. Put another way, we can make XY or XY2 or XY3, but not XY1.5 or XY2.7. With this evidence in hand, Dalton and Proust revived atomism. They envisioned atoms as tiny, unbreakable spheres, and they believed that atoms were the smallest increment of matter. 6.2 J. J. Thompson and the Electron Late in the 1800’s, J. J. Thomson revolutionized this view of atoms. Much of his research involved a cathode ray tube (CRT), which he set up as follows:

A cathode ray tube consists of a glass tube with a bulb at one end. The bulb is painted with phosphorescent paint. A power source provides an electric charge to two metal plates inside the tube: the cathode (negative) and the anode (positive). There is another set of charged metal plates inside the bulb. When the power is turned on, a cathode ray shoots from the cathode, through a hole in the anode, between the plates in the bulb, and strikes the phosphorescent paint. The paint glows where the ray strikes. Previous experiments had shown that the cathode ray is not deflected by the charged plates inside the bulb (the ray followed path b in the diagram). Because beams of light are also not deflected by electric charge, many people assumed that a cathode ray was a kind of light beam. However, Thomson found that these experiments were flawed because the tube contained trace amounts of gas. When he was able to completely evacuate the tube, he found that the

+ plate

— plate

— +

power source

cathode (-) anode (+)

phosphorescent paint

a b c

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charged electric plates in the bulb actually would deflect the cathode ray toward the positive plate (path a in the diagram). He thus concluded that the cathode ray carried a negative charge (because opposites attract), and moreover, that the cathode ray was actually a stream of particles because only matter was capable of holding an electric charge. Further experiments showed that a stream of hydrogen ions (H+) would follow path c; because they have a positive charge, they are deflected in the opposite direction of the negative cathode ray particles. Notice also that they are not deflected as far off center as the cathode ray; Thomson realized that cathode ray particles must be lighter than hydrogen ions because they are more easily deflected (it’s easier to knock a tennis ball off course than a heavy bowling ball). This discovery was remarkable because Thomson had discovered a particle that was even smaller than hydrogen, which is the smallest atom! In other words, he had shown that the smallest bit of matter was not the atom, that there was something about 1000 times smaller; he named this new particle the electron. Thomson assumed that the atoms in the cathode plate must contain the electrons, and he now pictured the atom as a positively charged sphere with electrons imbedded in it. This is known as the plum pudding model of the atom (a plum pudding consists of dough studded throughout with pieces of fruit). Because the negative electrons are balanced out by the positive sphere, overall the atom is electrically neutral.

+

e- e-

e- e-

e-

e-

e- e-

e- + +

+ +

+

+ +

+

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6.3 Robert Millikan and the Charge of the Electron Early in the 1900’s, Robert Millikan conducted his famous oil drop experiment with the following apparatus:

Millikan’s apparatus contained two electrically charged plates; the top plate was given a positive charge and the bottom plate a negative charge. An oil sprayer produced a fine mist of tiny oil drops, and some of these drops would fall through a hole in the top plate into the space between the plates. X-rays shot into this space would knock electrons off of air molecules, and bunches of these electrons would stick to the oil drops giving them various amounts of negative charge—a drop with more electrons had a greater negative charge than one with fewer electrons. By viewing the drops through the microscope, Millikan could determine the size of each drop and therefore its mass. Knowing the mass, he could calculate the force of gravity pulling down on the drop. An electrical force counteracted the force of gravity; the negatively charged drop was attracted to the positive plate above it and repelled by the negative plate below it. Millikan could adjust the charges on the plates so that the upward

X-rays

negative plate

positive plate

oil sprayer

oil droplets

microscope

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electrical force exactly equaled the downward pull of gravity—and the drop would hang in midair.

Upward force depends on: • the negative charge on the drop (unknown) • amount of charge on plates (known)

Downward force depends on: • mass of drop (known) • force of gravity (known)

Because Millikan knew that the upward force equaled the downward force, and he knew the charge on the plates, the mass of the drop, and the force of gravity, he was able to calculate the charge on any given oil droplet. His great discovery was this: the magnitude of every charge he measured was a multiple of the number 1.592 × 10-19 coulomb (a coulomb is a unit of electric charge). Millikan thus deduced that this must be the charge of a single electron. (Millikan’s value was slightly off; later experiments have shown the charge of an electron to be 1.602 × 10-19 coulombs.) When discussing subatomic particles, it is more convenient to use atomic units instead of coulombs to denote charge. In atomic units, the charge of an electron is simply -1.

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6.4 Ernest Rutherford and the Nucleus Around the same time as Millikan, Ernest Rutherford was doing research with alpha particles—particles that had a positive charge and were much heavier than electrons. His apparatus is diagrammed below. A source of alpha particles shoots a stream of the particles at an extremely thin piece of gold foil. The gold foil is surrounded by a fluorescent screen that glows wherever an alpha particle strikes it.

Based on the plum pudding model of the atom, Rutherford expected the alpha particles to pass directly through the gold foil and strike the screen straight on (path A). Rutherford reasoned that electrons, because they were so light, would not be able to knock the heavy alpha particles off course. He also thought that the positively charged sphere would have little effect on the pathway because the positive charge was spread throughout the entire volume of the atom; it was too diffuse to have much effect on a heavy, fast-moving alpha particle.

The results were astonishing. Most of the alpha particles did, in fact, travel as expected. However a small number of them were slightly deflected by the foil (path B), and about 1 in 8000 actually bounced back off the foil as if hitting a brick wall (path C).

alpha particle emitter

gold foil

path C path B

path A

fluorescent screen

+

e- e-

e- e-

e-

e-

e- e-

e- + +

+ +

+

+ +

+

Rutherford expected alpha particles to pass directly through gold atoms without being deflected.

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Rutherford knew that anything able to deflect an alpha particle must be heavy. He also knew that it must be small because the collisions were so rare. Instead of the positively charged, diffuse sphere in the plum pudding model, he discovered that the mass of an atom was highly concentrated into a small central core; he named this core the nucleus.

The new model of the atom consisted of a small, dense, positively charged nucleus surrounded by a cloud of electrons—the rest of Rutherford’s “new” atom was empty space. 6.5 The Line Spectrum One problem with Rutherford’s model was that it could not explain line spectra—the characteristic colors of light that atoms emit when they absorb and then release energy. Let’s see how this works. The electrical attraction between a negatively charged electron and a positively charged nucleus is similar to the gravitational attraction between a brick and the earth. Lifting a brick off the ground requires an input of energy; if the brick remains held up in the air, the brick retains that energy. Only when the brick is released and it falls back to earth is the energy released. We can say that the higher up the brick is held, the more energy it possesses. Electrons are similar; it requires an input of energy to pull an electron further away from its nucleus. An electron far from its nucleus has high energy, and that energy will be released when the electron “falls” closer to the nucleus.

You cannot physically lift an electron like a brick; electrons absorb energy in other ways: heating the substance, running an electric current through a substance, and shining a light on a substance can all push electrons to higher energy levels.

Because almost all of the mass of the atom is concentrated in the small nucleus, a small number of alpha particles will hit it directly and be deflected.

+

e-

e-

e- e-

e- e-

e-

e-

e-

e-

e-

+

e-

e- high-energy brick

low-energy brick

high-energy electron

low-energy electron

energy input energy released energy input energy released

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No matter how the electron absorbs energy, when it falls back toward the nucleus it will release its energy in the form of an electromagnetic (EM) wave. The energy of EM waves varies greatly, and the amount of energy in a wave depends on the frequency of the wave. If the frequency is high, the wave oscillates back and forth very quickly, and it will have a short wavelength.

High energy wave: If the frequency is low, the wave oscillates back and forth slowly, and it will have a long wavelength. Low energy wave: The entire range of EM wave energy is called the electromagnetic spectrum. Different regions of the spectrum have different names. The farther an electron falls, the higher the energy of the EM wave it gives off. In the example shown here, the electron on the left falls farther and emits a high-energy wave (X-ray), while the electron on the right falls a shorter distance and emits a low-energy wave (visible light with a blue color).

short wavelength

long wavelength

High energy High frequency Short wavelength

Low energy Low frequency Long wavelength

radio waves

microwaves infrared waves

visible light (ROYGBV)

UV rays

X-rays gamma rays

+

e-

e-

e-

e-

X-ray blue light

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If Rutherford’s model was correct—if electrons floated randomly in a cloud around the nucleus—then an atom should be able to give off a continuous spectrum of EM waves. An electron could gain any amount of energy, and it could fall toward the nucleus any distance, thereby giving off any amount of energy. But reality was quite different. Each element gave off a limited number of very specific wavelengths. The collection of wavelengths emitted by a certain element is called a line spectrum because when the emitted energy is viewed with a spectroscope (a prism-like device that separates each color of light), each different

wavelength of EM wave appears as a glowing line. For example, hydrogen’s line spectrum looks like this:

Wavelength EM wave 94 nm UV ray 95 nm UV ray 97 nm UV ray 103 nm UV ray 122 nm UV ray 410 nm Visible light (violet) 434 nm Visible light (violet) 486 nm Visible light (blue) 656 nm Visible light (red) 1094 nm Infrared 1282 nm Infrared 1875 nm Infrared 2624 nm Infrared 4050 nm Infrared 7456 nm Infrared

+

e- e- e-

e-

e-

e-

e-

e-

e-

e-

e-

In Rutherford’s model, electrons could fall from any conceivable distance, giving off every possible EM wave.

0 nm 1000 nm 500 nm 2000 nm 4000 nm 8000 nm

UV visible light infrared

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n = 1

n = 2

3

+

4 5

6

6.6 Niels Bohr and Orbits Shortly after Rutherford’s discovery, Niels Bohr theorized that electrons must be constrained to certain energy levels; the energy of electrons was quantized. Only certain distances from the nucleus (energy levels) were allowed. Bohr pictured electrons traveling around a nucleus in concentric circular paths called orbits (his model is sometimes called the planetary model). When an electron absorbs energy, it jumps up to a higher orbit; when it loses energy, it falls to a lower orbit. Because the distance between orbits is fixed, only certain amounts of energy can be released; thus an atom can only give off the wavelengths in its line spectrum, not a continuous range of wavelengths.

Note: the distances between orbits are not to scale.

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The hydrogen atom shown on the previous page has six orbits, and each orbit is assigned a quantum number 1 through 6. (The quantum numbers are represented by the letter n.) Every possible energy drop is represented by an arrow. The farther the electron falls, the more energy (EM with a shorter wavelength) is released:

Electron Jump Wavelength EM wave 6 → 1 94 nm UV ray 5 → 1 95 nm UV ray 4 → 1 97 nm UV ray 3 → 1 103 nm UV ray 2 → 1 122 nm UV ray 6 → 2 410 nm Visible light (violet) 5 → 2 434 nm Visible light (violet) 4 → 2 486 nm Visible light (blue) 3 → 2 656 nm Visible light (red) 6 → 3 1094 nm Infrared 5 → 3 1282 nm Infrared 4 → 3 1875 nm Infrared 5 → 4 2624 nm Infrared 6 → 4 4050 nm Infrared 6 → 5 7456 nm Infrared

Bohr’s model used a mathematical formula (developed by Swedish physicist Johannes Rydberg many years before) that accurately predicted the amount of energy for each orbital. Brilliant though the model was, it had a major drawback: it could only predict the energy levels for atoms containing only one electron. It also could not account for subsequent discoveries about the nature of matter, which are discussed in the next section. 6.7 Werner Heisenberg and Erwin Schrödinger The Uncertainty Principle Werner Heisenberg, a student and colleague of Bohr’s, developed one of the seminal theories of modern physics: the uncertainty principle. The details and mathematics of Heisenberg’s work are (far) beyond the scope of this text. However, for our purposes, the uncertainty principle states that it is impossible to know the exact position of an electron; Heisenberg showed that the best we can do is define a region of space near the nucleus where a certain electron is most likely to be found. Bohr’s image of electrons traveling in predictable circular orbits around the nucleus was inaccurate; according to the uncertainty principle, electrons cannot be so neatly pinpointed. In contrast to Bohr’s orbit, the region of space in which a certain electron is most often found is called an orbital.

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In the orbital shown above, the darker regions are where the electron spends most of its time—we say the region has high electron density. (Note that this does not mean there are a lot of electrons packed close together in this area; it means that there is a high probability of finding a certain electron in this area.) Wave/Particle Duality Subatomic particles behave very differently than everyday objects. Under certain circumstances, subatomic particles behave less like solid particles and more like waves; this confusing characteristic is known as wave/particle duality. (When an electromagnetic wave behaves as a particle, we refer to it as a photon—a particle of light.) Waves can be described with equations (just as a sine wave is described by y = sin(x)), and in 1925 Erwin Schrödinger developed a mathematical description of an electron wave called a wave function. Viewing an electron as a wave can help us understand a couple of aspects of electron “behavior”—how an electron can be described as not a particle in space but as a probability distribution, and why different electrons have different levels of energy. Let’s use a vibrating slinky as an analogy; imagine two people holding the ends of a slinky and shaking it so that the slinky forms a stable wave with one peak: It requires a certain amount of energy to generate this wave, and if the people shake faster (making a wave with higher energy), they will reach another point where the wave is stable, has two peaks, and looks like this:

+ +

With Bohr’s orbit, the electron is constrained to a definite circular path.

An orbital is a region of space where the electron is most often found.

e-

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If they shake faster still, they will eventually create a stable, higher-energy wave with three peaks: As more energy is put into the slinky, other stable waveforms will occur; the number of peaks will increase by one each time. If you try this yourself, you will feel that these three waves have a certain innate stability. As you change the rate of shaking to move from one stable wave to the next, the wave collapses and the slinky just jiggles randomly. There are certain set frequencies (energies) that create a stable wave; the slinky only “allows” certain levels of energy. This is an example of energy quantization, as we saw with the orbits in the Bohr model. Electron waves are similarly quantized; while electrons don’t travel in neat circular orbits, they still can only possess certain amounts of energy. The height of the wave is directly related to the electron density. For example, in the first wave, there is one peak in the very middle of the wave—this means that the most likely place to find an electron is halfway between the two people. At the very ends of the slinky, where it does not move up and down, the probability of finding an electron is zero (such a region is called a node). With the second wave, there are two areas of high electron density and three nodes:

high probability zero probability

zero probability

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Schrödinger’s orbitals will be discussed at greater length in chapter 9. 6.8 The Nucleus We now know that the nucleus contains two kinds of subatomic particles: protons and neutrons. The protons give the nucleus its positive charge; each proton has a charge of +1 atomic units. Neutrons are neutral (0 charge). Both protons and neutrons have a mass of 1 atomic mass unit, abbreviated as “u” (1 u = 1.66 × 10-27 kg). The characteristics of the three subatomic particles we’ll study in this class are summarized below.

Particle Charge (atomic units) Mass (u) proton +1 1 neutron 0 1 electron -1 1/1823 ≈ 0

6.9 Atomic Number and Atomic Mass An atom’s atomic number is the number of protons in its nucleus. Each element has a unique atomic number, and the elements on the periodic table are listed from lowest atomic number (hydrogen, 1) to highest (newly created elements have atomic numbers above 110). Because each element has a different atomic number, the identity of the element can be determined by the atomic number. Any atom with 6 protons is a carbon atom; any atom with 64 protons is a gadolinium atom. The mass number of an atom is the number of protons in the nucleus plus the number of neutrons in the nucleus. (Because the mass of an electron is so small compared to the other particles, it can been ignored when calculating the mass.) The number of neutrons in atoms of the same element can vary; while all carbon atoms must have 6 protons, some carbon atoms have 6 neutrons and others have 7 neutrons. (These are the two versions of carbon found in nature; carbon atoms with other numbers of neutrons can be created under artificial conditions.) The varying number of neutrons means that different carbon atoms will have different mass numbers: 12 or 13. These varieties of atoms of the same element are called isotopes. The two natural isotopes of carbon can be written as C-12

high probability

zero probability

zero probability

zero probability

high probability

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and C-13, or as 12C and 13C, or as 12C6 and 13C6 (the subscript in the latter case is the atomic number). Because there are two naturally occurring isotopes of carbon, the atomic mass listed on the periodic table is a combination of the two. It is not, however, a simple average of 12 and 13 (which would be 12.5). The mass listed on the periodic table (12.01) takes into account the abundance of each isotope; it is called a weighted average. C-12 is by far the most common isotope, making up 98.93% of all natural carbon. C-13 makes up the remaining 1.07%. The weighted average is calculated by multiplying the masses of each isotope by the abundances (percentages) and adding the results together:

atomic mass = (12.0 × 0.9893) + (13.0 × 0.0107) = 12.01 u Because the atomic masses listed on the periodic table are weighted toward the most abundance isotope, it is a good bet that the integer closest to the listed mass is the most common isotope. For example, the listed mass of iron is 55.85, and the most common isotope of iron is Fe-56. The number of protons and electrons determines the overall electrical charge on an atom. Protons have a +1 charge and electrons have a -1 charge; to calculate the overall charge, subtract the number of electrons from the number of protons. For example, if an atom of sodium has 11 protons and 10 electrons, the charge is +1. If an atom of phosphorus has 15 protons and 18 electrons, the charge is -3.

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Chapter 7: Nuclear Chemistry

7.1 The Strong Force 7.2 Neutron Glue 7.3 Binding Energy and Mass Defect 7.4 Nuclear Decay 7.5 Nuclear Equations 7.6 Predicting Nuclear Decay 7.7 Radioactive Series 7.8 Nuclear Power 7.9 Half-life

7.1 The Strong Force Because each proton in the nucleus carries a +1 charge, there are tremendous repelling forces constantly trying to force the protons apart—it is like a highly compressed spring just waiting for a chance to uncoil. So what holds the protons together? The strong force keeps the nucleus together. It is one of the four Fundamental Forces of Nature. (The other three are gravity, electromagnetism, and the weak force.) We experience two of these fundamental forces, gravity and electromagnetism, in our daily lives; we feel gravity pull us toward the earth and know that magnets stick to the refrigerator. But the strong force and weak force are completely outside of our experience because they are only capable of acting over very short distances. In fact, the strong force only works on objects that are within 1.5 x 10-15 m of each other—about the diameter of a nucleus. Because the strong force has a very limited range, any nucleus that is too big will be unstable; its diameter will be larger than the range of the strong force.

range of strong force

range of strong force

The nucleus on the left is stable because it is small; its diameter is smaller than the range of the strong force. The larger nucleus on the right exceeds the range of the strong force and is therefore less stable.

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7.2 The Band of Stability Neutrons are attracted to protons and other neutrons by the strong force, and so they help to hold a nucleus together. Neutrons thus act as a “glue,” and with the right amount of neutrons, a nucleus can contain a lot of protons. The glue analogy is not perfect, however, because more neutrons only add stability up to a point. For reasons we won’t delve into, it turns out that the ratio of protons to neutrons is what matters most. For small nuclei, the most stable ratio of neutrons to protons is about 1 to 1. For larger nuclei, the stable ratio tends toward 1.5 to 1. The line on the graph at right represents a 1 to 1 ratio. The dots represent stable nuclei. As the graph shows, stable nuclei lie close to the line. Larger stable nuclei lie above the line. It turns out that the most stable nuclei are isotopes of nickel and iron. Because they are the most stable, we can say that smaller nuclei would “like” to get bigger, and bigger nuclei would “like” to get smaller. 7.3 Binding Energy and Mass Defect Imagine we have 6 separate, individual protons and 6 separate, individual neutrons. They are not clustered together to form a nucleus; they are too far apart for the strong force to hold them together. A free proton has a mass of 1.007825 u, so six protons = 6 × 1.007825 = 6.04695 u. A free neutron has a mass of 1.008665 u, so six neutrons = 6 × 1.008665 = 6.05199 u. Therefore, all twelve particles would have a total mass of 12.09894 u. Now imagine that the protons and neutrons are pushed together close enough that the strong force sticks them all together to form a carbon nucleus (C-12). Because this nuclear formation is a more stable arrangement than having all the particles separate, energy is released by the particles. (Energy is always given off when something reaches a more stable state—recall the falling brick analogy in the previous chapter.) This energy is called the binding energy, and the more stable the

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resulting nucleus, the more energy is given off. Because nickel and iron are the most stable of all the elements, they also have the greatest binding energies.

Where does this energy come from? Energy cannot be created from nothing, and in this case the energy actually is created from the particles themselves. Some of the matter in the particles is converted into energy—this means that the particles themselves actually become lighter; they lose mass. The amount of matter lost is called the mass defect. So the mass of the carbon nucleus we just formed is actually less than the 12.09894 u of all the separate particles; the mass of a carbon nucleus is 12.00000 u; the mass defect of carbon is 0.09894 u, or approximately 0.82%. The more stable the nucleus, the greater the binding energy, and the greater the mass defect. 7.4 Nuclear Decay What happens to unstable nuclei? An unstable nucleus will undergo nuclear (radioactive) decay; a process that produces radiation (high-speed subatomic particles or EM waves). The five different decay processes are discussed below. The first type of nuclear decay is alpha decay. The nucleus ejects an alpha particle, which is a cluster of 2 protons and 2 neutrons. This cluster also happens to be identical to a helium (He-4) nucleus. Therefore, an alpha particle can be represented either by the Greek letter alpha (α) or the helium symbol:

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4α2 or 4He2 The second type of nuclear decay is beta decay. The nucleus ejects a beta particle, which is simply an electron. (This is a newly formed electron, not one of the electrons in the cloud surrounding the nucleus.) A beta particle is represented by the Greek letter beta (β) or e, for electron:

0β-1 or 0e-1 The third type is gamma emission. Gamma rays are a type of high-energy electromagnetic wave that is generated by an internal rearrangement of particles in the nucleus; as the particles move from an unstable arrangement to a more stable one, they release EM energy. Gamma rays have no mass and no charge, and are represented by the Greek letter gamma (γ):

0γ0 or more often, just γ

These three types of radiation differ in their penetrating power; alpha particles penetrate the least, and gamma rays penetrate the most. For example, a piece of ordinary paper will stop alpha particles, a piece of wood can stop beta particles, and lead or concrete is needed to stop gamma rays. The fourth type is electron capture; this produces x-rays. The nuclear reaction of electron capture is described in the next section. Finally, there is positron emission. A positron is a positively charged electron. Positrons are sometime referred to as “beta positive” radiation and are represented by the symbols:

0β+1 or 0e+1

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7.5 Nuclear Equations Like chemical reactions, nuclear reactions can be described by equations. Nuclear equations must balance in terms of both mass and charge. For example, when an atom of polonium-210 undergoes alpha decay, the equation is:

210Po84 → 206Pb82 + 4He2 The masses balance because 210 = 206 + 4. The charges balance because 84 = 82 + 2. [Remember that the atomic number (the subscript in an isotope’s symbol) is the number of protons, and is therefore equivalent to the charge on the nucleus (because each proton has a +1 charge). For example, the charge of a 14N7 nucleus is +7.] When C-13 undergoes beta decay, the equation is:

13C6 → 13N7 + 0e-1 The masses balance because 13 = 13 + 0. The charges balance because 6 = 7 + (-1). Notice that the mass of the atom has not changed, and the charge of the atom has increased. Because the beta particle has a -1 charge, the charge of the nitrogen must go up; a proton must have been produced along with the beta particle. But where did it come from? If we look carefully, we see that C-13 has 6 protons and 7 neutrons. N-13 has 7 protons and 6 neutrons. The transformation has created a new proton and a new beta particle, and we’ve lost a neutron; the neutron actually has changed into a proton and an electron! In gamma emission, the equation looks like this:

230Th90* → 230Th90 + 0γ0 Because a gamma ray has no mass and no charge, the mass and charge numbers on the thorium atom don’t change. The asterisk (*) indicates that the protons and neutrons in the thorium nucleus are in an unstable, high-energy arrangement, and as the particles return to a stable state, that excess energy is released as a gamma ray.

mass reactants

mass products

charge reactants

charge products

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In electron capture, one of the innermost electrons in the electron cloud is captured by the nucleus. This electron combines with a proton to create a neutron. The nuclear equation looks like this:

26Al13 + 0e-1 → 26Mg12 When this inner electron is captured, it leaves an empty spot in a low-level orbital. A higher-energy electron will fall into its place, and this transition releases an electromagnetic wave. Because this is a large drop in energy, the wave is high-energy wave: an x-ray. In positron emission, a proton turns into a neutron and a positron, as follows:

11C6 → 11B5 + 0e+1 Many other nuclear reactions are possible that are not classified as nuclear decay. For example:

2 12C6 → 20Ne10 + 4He2

235U92 + 1n0 → 94Sr38 + 140Xe54 + 2 1n0 7.6 Predicting Nuclear Decay As mentioned before, there are two reasons a nucleus can be unstable. If the nucleus is so large (over 84 protons) that it exceeds the boundaries of the strong force, the strong force cannot hold it together. Also, if the ratio of neutrons to protons is not correct, it will be unstable. Generally speaking, an unstable nucleus tends to undergo the type of decay that will correct the cause of the instability. For example, if a nucleus is too large, it will try to rid itself of excess protons and neutrons by emitting alpha particles. If the N:P ratio is too high (too many neutrons, too few protons) it will undergo beta decay, which destroys a neutron, creates a proton, and lowers the ratio. If the N:P ratio is too low (too few neutrons, too many protons), it will undergo either electron capture or positron emission, both of which destroy a proton, create a neutron, and increase the ratio. In other words, when the ratio is off, the nucleus tends to decay toward the band of stability, as shown in the following graph.

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7.7 Radioactive Series For many unstable nuclei, stability will not be attained after one nuclear decay process. An unstable nucleus often will decay into another unstable nucleus, which decays into another unstable nucleus, etc.; a series of nuclear transformations follows until a stable nucleus is reached. This is called a radioactive series or decay chain. A decay chain is shown at right; it starts with 249Cf98 at the top right and ends with 205Tl81 at the bottom left. Each time the line moves down and to the left, alpha decay has occurred. Each time the line moves directly to the right, beta decay has occurred.

Decay moves in this direction.

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7.8 Nuclear Power Nuclear Weapons When a neutron strikes a uranium-235 nucleus, the following nuclear reaction occurs:

1n0 + 235U92 → 3 1n0 + 90Rb37 + 143Cs55 This is called a fission reaction: a reaction in which a nucleus splits into smaller nuclei. Notice that this reaction also produces three more neutrons. If those neutrons go on to strike other uranium nuclei, those nuclei split and produce more neutrons, which split more nuclei, which makes more neutrons, which split more nuclei, and so on—a fission chain reaction occurs.

The lines in this figure represent the paths of neutrons. Each branching point is where a uranium nucleus has split, producing three more neutrons. Each time a uranium nucleus splits a tremendous amount of energy is released. If a large amount of U-235 undergoes a fission chain reaction all at once, the result is a nuclear explosion. This kind of sustained chain reaction is only possible if the piece of uranium is large enough. If the piece of uranium is too small, the neutrons end up flying out into empty space instead of colliding with nuclei inside the piece of uranium. In other words, if the chunk of uranium is too small, the chain reaction doesn’t happen, and no explosion occurs—it is a subcritical mass of uranium. If it is just large enough to sustain the chain reaction, it is called a critical mass. If it is larger than the critical mass, it is a supercritical mass.

1 neutron

n

n

n

n

235U

90Rb

143Cs

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A nuclear bomb starts with two or more subcritical masses of uranium. As soon as they are brought together, they form a supercritical mass, which explodes.

subcritical supercritical subcritical

For the bomb to work, the subcritical masses must be propelled together extremely rapidly. This is done with a conventional (non-nuclear) explosive. The bombs the United States dropped on Japan in World War II were this type of bomb (they are called “atomic bombs” or “A-bombs”). Modern nuclear weapons use nuclear fusion instead of nuclear fission. Nuclear fusion happens when two small nuclei come together to form a larger one, as in the following reaction:

1H1 + 3H1 → 4He2 This process also releases an enormous quantity of energy. In fact, fusion releases much more energy than fission, and fusion bombs (called “hydrogen bombs” or “H-bombs”) are thousands of times more powerful than fission bombs. For two nuclei to get close enough to fuse, they must be moving extraordinarily fast. Remember: Heat is the energy of moving molecules; the hotter something is, the faster its molecules are moving. So fusion reactions require very high temperatures to occur. Generating this amount of heat requires a fission

Supercritical Mass

Subcritical Mass

ka- blooie!

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reaction! So a fusion bomb is actually two bombs in one. A small fission bomb explodes first, which heats up a quantity of hydrogen enough to make it fuse. Fusion reactions also power the stars; our sun is an enormous fusion reactor. Small atoms such as hydrogen and helium fuse inside the sun, making larger atoms and releasing huge quantities of heat and light. Nuclear Reactors A fission reaction can be slowed down to a consistent, stable, rate; it doesn’t have to happen all at once in an explosion. This is what happens in a nuclear reactor; a controlled fission reaction generates electricity. The uranium is shaped into fuel rods. Each fuel rod has some fission going on inside it, sending neutrons to neighboring rods, causing their atoms to undergo fission, as shown at right.

If nothing intervened, the fuel rods would overheat and cause a meltdown. To slow down the rate of reaction, control rods are lowered between the fuel rods. The control rods are made out of a material that can absorb neutrons. When the control rods are between the fuel rods, the neutrons of one fuel rod can’t cause as much fission in a neighboring fuel rod, and the whole reaction slows down. A schematic of a basic nuclear reactor is on the following page. The fuel rods and the control rods make up the reactor core. When the reactor is

active, the fission chain reaction heats a liquid coolant surrounding the rods. The hot coolant is pumped through pipes to boil water in the steam generator; the pressurized steam flows through the steam line to the turbine, which powers the generator. The steam is then cooled and condensed by water from the cooling tower, and then pumped back to the steam generator. So a nuclear power plant is nothing more than a very sophisticated steam engine!

Uranium fuel rods exchange neutrons, inducing fission in neighboring fuel rods.

The white control rods block the exchange of neutrons, slowing down the chain reaction and preventing overheating.

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7.9 Half-life While there is no way to predict exactly when any particular unstable nucleus will decay, a large collection of nuclei will decay in a predictable fashion—every radioactive isotope has a unique half-life (t1/2): the time it takes for half of the sample to decay. If an isotope decays very quickly, its half-life is short; if it decays slowly, its half-life is long. The range of half-lives is quite large:

Isotope Half-life U-238 4.5 × 109 years C-14 5715 years Sr-90 28.8 years Sc-46 83.8 days Cu-62 9.67 minutes Mt-226 3.4 milliseconds

The following graph shows the decay of 1000 g of a radioactive isotope. Notice that the amount of isotope decreases by half with every half-life.

reactor core steam generator

control rods

coolant fuel rods turbine/

generator

cooling tower

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A simple way to solve half-life problems is to use a chart set up like this: # of half-lives time passed amount remaining Example: A certain isotope has a half-life of 2 days. If you start with 400 g, how much of the isotope will be left after 8 days? The “# of half-lives” always starts at zero, and increase by one as we move across: # of half-lives 0 1 2 3 4 5 time passed amount remaining “Time passed” also always starts at zero and jumps up by one half-life in each box: # of half-lives 0 1 2 3 4 5 time passed 0 2 4 6 8 10 amount remaining The “amount remaining” starts with 400 g in this case, and decreases by half in each box. # of half-lives 0 1 2 3 4 5 time passed 0 2 4 6 8 10 amount remaining 400 200 100 50 25 From the chart, we can see that when eight days have passed, 25 g of the isotope remain. Example: You started with 2000 mg of a radioactive isotope 45 hours ago; you now have 250 mg. What is the half-life of the isotope? In this case, we know that the sample has decreased by halves from 2000 to 250 mg, and we can plug that into our chart: # of half-lives time passed amount remaining 2000 1000 500 250 By filling in the half-lives row, we see that three half-lives have passed:

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# of half-lives 0 1 2 3 time passed amount remaining 2000 1000 500 250 The problem tells us that 45 hours have passed, so we can fill in the end of the “time passed” row, as well as put zero at the beginning: # of half-lives 0 1 2 3 time passed 0 45 amount remaining 2000 1000 500 250 The chart tells us that 3 half-lives took 45 hours; therefore the half-life must be 45/3 = 15 hours.

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Chapter 8: Electron Configurations

8.1 Quantum Numbers 8.2 The Energies of Orbitals 8.3 Electron Configurations 8.4 Condensed Configurations: 8.5 Special Cases: Copper and Chromium 8.6 Electron Configuration of Ions 8.7 Ground State vs. Excited State 8.8 Orbital Shapes

8.1 Quantum Numbers Bohr’s model of the atom confined electrons to concentric circular orbits around the nucleus; he assigned quantum numbers to the orbits, starting with 1 at the inside and increasing by one with each orbit outward. When Schrödinger replaced circular orbits with regions of space called orbitals, he retained the idea of quantum numbers; his system of equations actually assigns every electron in an atom a set of four quantum numbers. Each electron’s set of quantum numbers is unique to it—no other electron in the atom can have the same four numbers (this is called the Pauli Exclusion Principle). The first quantum number in Schrödinger’s system is designated n, and it describes the “shell” that an electron is in; these shells are analogous to Bohr’s orbits. The possible values of n are 1, 2, 3, 4, and on, theoretically, to infinity. The higher the n, the higher the energy of the shell. Each shell is split into subshells, which are designated with an l quantum number, with the possible values 0, 1, 2, 3, etc. The higher the energy of the shell (the higher the n number), the more subshells it can contain:

Shell n = 1 contains 1 subshell (with l value = 0) Shell n = 2 contains 2 subshells (with l values = 0, 1) Shell n = 3 contains 3 subshells (with l values = 0, 1, 2) Shell n = 4 contains 4 subshells (with l values = 0, 1, 2, 3)

To make matters more complicated, the subshells are also designated by letters; a subshell with l = 0 is called an s subshell. When l = 1, it is a p subshell. When l = 2, it is a d subshell. When l = 3, it is an f subshell. The subshells also have different energy levels; from low to high they are s, p, d, and f (larger shells will also contain g, h, i, etc., but these are not commonly encountered). Chemists don’t usually refer to subshells by their numbers—they use the letters:

Shell n = 1 contains 1 subshell, called 1s Shell n = 2 contains 2 subshells, called 2s and 2p Shell n = 3 contains 3 subshells, called 3s, 3p, and 3d Shell n = 4 contains 4 subshells, called 4s, 4p, 4d, and 4f

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It gets worse. Each subshell is divided into orbitals, which have yet another quantum number, called ml, which has the possible values of positive and negative integers surrounding zero (…-2, -1, 0, +1, +2…). Every s subshell contains 1 orbital (ml = 0). Every p subshell contains 3 orbitals (ml = -1, 0, +1). Every d subshell contains 5 orbitals (ml = -2, -1, 0, +1, +2). Every f subshell contains 7 orbitals (ml = -3, -2, -1, 0, +1, +2, +3). Within a given subshell, the orbitals all have the same energy.

subshell s p d f # orbitals 1 3 5 7

And finally, each orbital can contain two electrons. Each electron has another quantum number assigned to it, designated ms. The only possible values for ms are +1/2 and -1/2, which are called spin up and spin down, respectively. In shorthand, they are represented by ↑ and ↓.

subshell s p d f # orbitals 1 3 5 7 e- capacity 2 6 10 14

So, every electron in an atom has a unique set of quantum numbers: n, l, ml, and ms. All of the preceding information is summarized in the following table:

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n l ml ms +1/2 1 s (l = 0) 0 -1/2 +1/2 s (l = 0) 0 -1/2 +1/2 -1 -1/2 +1/2 0 -1/2 +1/2

2 p (l = 1)

+1 -1/2 +1/2 s (l = 0) 0 -1/2 +1/2 -1 -1/2 +1/2 0 -1/2 +1/2

p (l = 1) +1 -1/2

+1/2 -2 -1/2 +1/2 -1 -1/2 +1/2 0 -1/2 +1/2 +1 -1/2 +1/2

3

d (l = 2)

+2 -1/2 +1/2 s (l = 0) 0 -1/2 +1/2 -1 -1/2 +1/2 0 -1/2 +1/2

p (l = 1) +1 -1/2

+1/2 -2 -1/2 +1/2 -1 -1/2 +1/2 0 -1/2 +1/2 +1 -1/2 +1/2

d (l = 2)

+2 -1/2 +1/2 -3 -1/2 +1/2 -2 -1/2 +1/2 -1 -1/2 +1/2 0 -1/2 +1/2 +1 -1/2 +1/2 +2 -1/2 +1/2

4

f (l = 3)

+3 -1/2

8.2 The Energies of Orbitals Life would certainly be easier if the energies of all these shells and subshells and orbitals were organized as follows (each box represents an orbital, and each arrow represents an electron):

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g ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ f ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ n = 5 p ↑↓ ↑↓ ↑↓ s ↑↓

f ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ n = 4 p ↑↓ ↑↓ ↑↓ s ↑↓

d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ n = 3 p ↑↓ ↑↓ ↑↓ s ↑↓

p ↑↓ ↑↓ ↑↓ n = 2 s ↑↓

n = 1 s ↑↓ But alas, reality is more complicated. The energies of these orbitals are ordered like this:

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Which is horrible indeed.

7p ↑↓ ↑↓ ↑↓ n = 6 6d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 5f ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 7s ↑↓ 6p ↑↓ ↑↓ ↑↓ 5d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ n = 5 4f ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 6s ↑↓ 5p ↑↓ ↑↓ ↑↓ 4d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 5s ↑↓ n = 4 4p ↑↓ ↑↓ ↑↓ 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4s ↑↓ n = 3 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ n = 2 2s ↑↓ n = 1 1s ↑↓

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Fortunately, the periodic table itself is organized according to energy level, as shown below. 1s 1s

2s 2p

3s

3p

4s 3d 4p

5s 4d 5p

6s 5d 6p

7s 6d 7p

4f

5f

The periodic table is divided into “blocks” according to the subshells. Groups 1 and 2 are the s-block, 3-12 are the d-block, 13-18 are the p-block, and the f-block is usually shown separately below the rest of the table, just to save space. If the f-block were in the proper position in the table, it would look like this: 1s 1s

2s 2p

3s 3p

4s 3d 4p

5s

4d 5p

6s 4f 5d 6p

7s 5f 6d 7p

If we read the lines of the periodic table like a page of text (left to right, top to bottom), the shells and subshells all fall into place. The order we end up with is this:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p It is much easier to remember the blocks on the periodic table than memorize this sequence.

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8.3 Electron Configurations Different atoms have different numbers of electrons. With such a large number of orbitals to occupy, how does an electron know where to go? Fortunately, the placement of electrons is not random. A complete set of empty subshells and orbitals is shown below.

7p 6d 5f 7s 6p 5d 4f 6s 5p 4d 5s 4p 3d 4s 3p 3s 2p 2s 1s

A neutral xenon atom has 54 electrons; what orbitals will they occupy?

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7p 6d 5f 7s 6p 5d 4f 6s 5p ↑↓ ↑↓ ↑↓ 4d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 5s ↑↓ 4p ↑↓ ↑↓ ↑↓ 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4s ↑↓ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

The first rule of placing electrons is to fill the orbitals from the bottom up. This is called the Aufbau Principle (Aufbau is German for “build up”). In other words, electrons will occupy the lowest possible energy levels. A useful analogy is to think of electrons as college students moving into a dorm. Each subshell is a floor of the dorm, and each orbital is a room that can hold two roommates (electrons) each. The first students to arrive will move into the rooms on the first floor to avoid hauling their stuff up the stairs. Once the first floor is full, the next students have no choice but to move in to the rooms on the second floor. The students will always move into the lowest available floor to avoid carrying their stuff up more stairs. The locations of the electrons is called the electron configuration. Instead of drawing a big diagram, the quick way of indicating the electron configuration for xenon is like this:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 The superscripts indicate the number of electrons contained in each subshell.

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What about arsenic? It has 33 electrons, and the configuration looks like this (to save space the empty high-energy orbitals are not shown):

4p ↑ ↑ ↑ 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4s ↑↓ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

Notice that the electron configuration is not:

4p ↑↓ ↑ 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4s ↑↓ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

The dormitory analogy continues: Every student wants to have a single; if possible, she will avoid having a roommate. So if there are three students on the 4p floor and three available rooms (orbitals), they will occupy the rooms individually instead of pairing up. The reason for this is that electrons are all negatively charged, so they repel and try to get as far away from each other as possible. This tendency is called Hund’s Rule. The electron configuration of arsenic is:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 The electrons only pair up when they have to. For example, sulfur (16 electrons):

3p ↑↓ ↑ ↑ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

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The electron configuration of an atom greatly influences its properties. The magnetic properties of an atom, for example, depend on how many unpaired electrons it has. Look at the electron configurations on the previous pages. Sulfur has two unpaired electrons in the 3p subshell. Arsenic has three unpaired electrons in the 4p subshell, and xenon has no unpaired electrons. Atoms with no unpaired electrons are called diamagnetic, and they have a very weak response to a magnetic field. Atoms with unpaired electrons are called paramagnetic, and they have a somewhat stronger response to a magnetic field. 8.4 Condensed Configurations: The electron configurations for Ne, Na, Al and Cl are:

Ne: 1s2 2s2 2p6

Na: 1s2 2s2 2p6 3s1 Al: 1s2 2s2 2p6 3s2 3p1 Cl: 1s2 2s2 2p6 3s2 3p5

Notice that Na, Al and Cl all contain the same electron configuration as Ne (1s2 2s2 2p6), with more subshells added on top. To speed things up, we can write [Ne] instead of 1s2 2s2 2p6. So the electron configurations of Na, Al, and Cl are:

Na: [Ne] 3s1 Al: [Ne] 3s2 3p1 Cl: [Ne] 3s2 3p5

These are called condensed or short form electron configurations. Only the elements in group 18 can be put in brackets to make a condensed configuration. Some more examples of condensed configurations:

O: [He] 2s2 2p4 Ca: [Ar] 4s2 I: [Kr] 5s2 4d10 5p5 Pb: [Xe] 6s2 4f14 5d10 6p2

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8.5 Special Cases: Copper and Chromium It turns out that having an exactly full subshell or an exactly half-full subshell is an especially stable (low energy) configuration. This can make the electron configuration for certain atoms different from the expected configuration according to the Aufbau principle. For example, we would expect copper (29 electrons) to have the configuration:

3d ↑↓ ↑↓ ↑↓ ↑↓ ↑ 4s ↑↓ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6 3s2 3p6 4s2 3d9

But in fact, it is more stable for it to fill the 3d subshell by taking one electron from the 4s subshell:

3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4s ↑ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6 3s2 3p6 4s1 3d10

Likewise, with chromium (24 electrons), we would expect:

3d ↑ ↑ ↑ ↑ 4s ↑↓ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6 3s2 3p6 4s2 3d4

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But again, it is more stable to fill the 3d subshell halfway by taking an electron from 4s, giving the configuration:

3d ↑ ↑ ↑ ↑ ↑ 4s ↑ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6 3s2 3p6 4s1 3d5

8.6 Electron Configuration of Ions Negative ions have more than the usual number of electrons, and we write their electron configurations the standard way. For example, the O-2 ion has 2 extra electrons for a total of 10, and the electron configuration is:

2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6

Positive ions have fewer than the usual number of electrons, but their configurations are not written the standard way. The rule here is that the electrons are always taken away from the highest numbered shell, not the highest energy subshell. For example, a neutral nickel atom has the configuration:

3d ↑↓ ↑↓ ↑↓ ↑ ↑ 4s ↑↓ 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6 3s2 3p6 4s2 3d8

When it loses two electrons to make a Ni+2 ion, the two electrons come from the 4s subshell, not the 3d subshell, because the highest number shell is 4, not 3:

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3d ↑↓ ↑↓ ↑↓ ↑ ↑ 4s 3p ↑↓ ↑↓ ↑↓ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6 3s2 3p6 3d8

Notice that the empty 4s subshell is not listed in the electron configuration. 8.7 Ground State vs. Excited State Most of the time, the electrons occupy the lowest possible energy levels, according to the Aufbau Principle; this is called being in the ground state. As we learned in the previous chapter, when electrons absorb energy from an external source (heat, electricity, light) they are “promoted” to higher energy levels; this is called being in an excited state. For example, the ground state electron configuration of phosphorus (15 electrons) is:

3d 4s 3p ↑ ↑ ↑ 3s ↑↓ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓

1s2 2s2 2p6 3s2 3p3

But when the electrons absorb energy, they are bumped up to higher energy levels, and the exited state configuration may look like this:

3d or ↑ ↑ 4s ↑ 3p ↑ ↑ ↑ 3s ↑↓ ↑↓ 2p ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 2s ↑↓ ↑↓ 1s ↑↓ ↑↓

1s2 2s2 2p6 3s2 3p2 4s1 1s2 2s2 2p6 3s2 3p1 3d2

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8.8 Orbital Shapes Remember that orbitals are regions of space in which a certain electron is found most of the time. The size and shape of those regions of space depend on the subshell. Orbitals in s subshells (usually referred to as “s orbitals”) have the shape of a sphere:

y

z

x

P orbitals look like dumbbells. There are three orbitals in a p subshell, and they all have the same shape. They differ only in their orientation:

y

z

x

y

z

x

y

z

x

D orbitals vary in both shape and orientation; four of them look like cloverleaves, and one looks like a dumbbell with a donut around the middle. The five types of d orbitals are shown below:

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

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All orbitals are centered around the atom’s nucleus, so they overlap each other a lot. While parts of multiple orbitals may occupy the same space, the electrons in those orbitals still have different energy. The following diagram shows overlapping 2s and 2p orbitals.

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Chapter 9: The Periodic Table

9.1 Development of the Periodic Table 9.2 Regions of the Periodic Table 9.3 Effective Nuclear Charge 9.4 Periodic Trends

9.1 Development of the Periodic Table Much of the research that led to the modern periodic table was presented by Russian scientist Dmitri Mendeleev in 1869-70. Mendeleev listed the elements in order of increasing atomic mass, and then organized the elements into groups (columns). The elements in a certain group shared similar chemical properties, such as the ratio in which they combine with other elements. (For example, all the elements in group 1 combine with chlorine in a 1-to-1 ratio: HCl, LiCl, NaCl, etc. The elements in group 2 combine with chlorine in a 1-to-2 ratio: BeCl2, MgCl2, CaCl2, etc.) This repetition of properties as one marches through the list of elements is called the Periodic Law. At the time, certain elements (such as germanium and gallium) had not yet been discovered, and critics pointed to the conspicuous gaps in Mendeleev’s table. So confident was Mendeleev that he not only predicted that the missing elements would be discovered (they were), but he also predicted (correctly) various properties of these theoretical elements such as atomic mass, atomic radius, and density. His ordering of the elements by atomic mass still left a few inconsistencies (potassium belongs in the first period despite being slightly lighter than argon, for example), but these were ironed out when Henry Moseley reordered the elements by atomic number in 1913. 9.2 Regions of the Periodic Table The elements on the periodic table can be divided into groups based on shared properties. The first three categories we will discuss are metals, nonmetals, and metalloids. Metals The elements classified as metals are highlighted below.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 H He

2 Li Be B C N O F Ne

3 Na Mg Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

7 Fr Ra Lr Rf Ha Sg Ns Hs Mt

La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

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In general, metals have the following physical properties: • Malleable (can be bent and molded into different shapes) • Ductile (can be pounded into thin sheets and drawn into wires) • Luster (can be polished to a shiny surface) • Good heat conduction • Good electrical conduction • High density • High melting points (all are solid at room temperature except mercury)

Nonmetals The elements classified as nonmetals are highlighted below.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 H He

2 Li Be B C N O F Ne

3 Na Mg Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

7 Fr Ra Lr Rf Ha Sg Ns Hs Mt

La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

In general, nonmetals have the opposite physical properties of metals:

• Brittle (not malleable or ductile) • Dull surface (no luster) • Poor heat conduction • Poor electrical conduction • Low density • Low melting points (many are gases at room temperature)

Metalloids The elements classified as metalloids are highlighted below.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 H He

2 Li Be B C N O F Ne

3 Na Mg Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

7 Fr Ra Lr Rf Ha Sg Ns Hs Mt

La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

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Metalloids have physical properties intermediate between metals and nonmetals. Silicon (a semiconductor of electricity) is especially useful for making computer chips. Other Categories and Terminology The 18 vertical columns of the periodic table are referred to as groups or families; just as you may share characteristics with your family members, elements in the same family often have similar chemical properties. The 7 horizontal rows of the periodic table are called periods. Other categories are:

• Alkali metals (group 1) • Alkaline earth metals (group 2) • Transition metals (groups 3-12) • Inner transition metals (a.k.a. rare earth metals)

a. Lanthanides b. Actinides

• Halogens (group 17) • Noble gases (group 18)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 H He

2 Li Be B C N O F Ne

3 Na Mg Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

7 Fr Ra Lr Rf Ha Sg Ns Hs Mt

La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

9.3 Effective Nuclear Charge Because the nucleus has a positive charge and electrons have negative charge, there is an attraction between them. But not all electrons in an atom are attracted to the nucleus equally—some electrons “feel” the pull of the nucleus more than others. How much charge an electron “feels” is called the effective nuclear charge (Zeff). Two things influence Zeff. The first is how much positive charge is in the nucleus—the number of protons. The more protons, the more the nucleus will attract an electron. The number of protons, or atomic number, is called Z:

alkali metals alkaline earth metals halogens noble gases

transition metals

lanthanides actinides

inner transition metals

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Z = number of protons in the nucleus (atomic number)

The bigger the Z, the bigger the Zeff.

Zeff is diminished by anything that gets between an electron and its nucleus. If an electron is in the valence shell (the shell with the highest n value), all the electrons in the lower shells (lower n values) are between that electron and the nucleus: that valence electron “feels” less charge from the nucleus. The inner electrons are called core electrons or shielding electrons, and are represented by an “S”. Every shielding electron cancels out the charge of one proton.

S = number of shielding electrons (electrons in inner n levels)

The bigger the S, the smaller the Zeff. Mathematically speaking: Zeff = Z – S What is the Zeff for an electron in the valance shell of calcium? Look at the electron configuration:

1s2 2s2 2p6 3s2 3p6 4s2

The shielding electrons are in shells 1, 2, and 3. Shell 1 contains 2 electrons, shell 2 contains 8 electrons, and shell 3 contains 8, giving a total of 18 (S = 18). Cacium has 20 protons (Z = 20). So the Zeff of calcium is:

Zeff = 20 – 18 = 2 What is the Zeff for Gallium? It has 31 protons, so Z = 31. The electron configuration is:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 The outermost shell is n = 4. The inner shells are 1, 2, and 3, and altogether they contain 28 electrons. So S = 28, and

Zeff = 31 – 28 = 3

Z = +20

n = 1

n = 2

n = 3

n = 4

# e = 2

# e = 8

# e = 2

# e = 8

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9.4 Periodic Trends Effective Nuclear Charge Below is an abridged periodic table; it has only the “main group elements”—all the elements except the transition and inner transition metals. The Zeff values are shown for each element.

1 2

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8

1 2 tran

sitio

n m

etal

s

3 4 5 6 7 8

1 2

Notice that Zeff does not change moving vertically within a group. Zeff increases moving from left to right across a period. We can represent this trend with a single arrow, as follows:

Atomic Radius The atomic radius is the distance from the nucleus to the valence electrons measured in Angstroms (1 Å = 1 x 10-10 m). The atomic radii are:

0.79 0.49

2.05 1.4 1.17 0.91 0.75 0.65 0.57 0.51

2.23 1.72 1.62 1.44 1.23 1.09 0.97 0.88

2.77 2.23 1.81 1.52 1.33 1.22 1.12 1.03

2.98 2.45 2 1.72 1.53 1.42 1.32 1.24

3.34 2.76 tran

sitio

n m

etal

s

2.08 1.81 1.63 1.53 1.43 1.34

2.7 2.23

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In general, atomic radii increase moving from top to bottom within a group. The reason for this is that the number of electron shells increases. Moving down group one, the electron configurations are:

H: 1s1 Li: 1s2 2s1 Na: 1s2 2s2 2p6 3s1 K: 1s2 2s2 2p6 3s2 3p6 4s1 Rb: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 Cs: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s1 Fr: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s1

Notice that each element adds another shell moving down the group. The more shells of electrons, the larger the radius. Moving from left to right across a period, atomic radii tend to decrease. Let’s look at the electron configurations of the elements in period 2:

1s2 2s1 1s2 2s2 1s2 2s2 2p1 1s2 2s2 2p2 1s2 2s2 2p3 1s2 2s2 2p4 1s2 2s2 2p5 1s2 2s2 2p6 As we move across, electrons are simply filling up the 2nd shell; the number of shells does not increase or decrease, so this can’t account for the trend. Remember, however, that Zeff increases from left to right; the pull on the valence electrons increases, pulling them closer to the nucleus and decreasing the radius.

The overall trend in atomic radii is shown below. The smallest atoms are in the upper right; the largest in the lower left.

Z = +3

n = 1

n = 2

# e = 2

# e = 1

Z = +6

n = 1

n = 2

# e = 2

# e = 4

Z = +10

n = 1

n = 2

# e = 2

# e = 8

Lithium, carbon, and neon atoms are shown above. Their values of Zeff are, respectively, 1, 4, and 8. A higher Zeff pulls the outer electrons closer to the nucleus, resulting in a smaller atomic radius.

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When an atom becomes a negative ion by gaining an electron, the radius increases. The additional electron increases the repulsive forces in the outer shell, which then expands:

The opposite happens when an atom loses an electron to become a positive ion—the loss of a valence electron decreases repulsive forces in the valence shell, and the radius decreases.

Z = +9

n = 1

n = 2

A neutral fluorine atom has 7 electrons in its valence shell. If it gains an electron to become an F- ion, the increased repulsion between electrons in the valence shell causes the shell to expand; the radius increases.

Z = +9

n = 1

n = 2

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Ionization Energy The amount of energy needed to remove a valence electron from an atom is called the ionization energy (measured in kJ/mol). This is a measure of how strongly the nucleus holds on to the outer electrons.

1312.0 2372.3

520.2 899.4 800.6 1086.4 1420.3 1313.9 1681.0 2080.6

495.8 737.7 577.6 786.4 1011.7 999.6 1251.1 1520.5

418.8 589.8 578.8 762.1 947.0 940.9 1139.9 1360.7

403.0 549.5 558.3 708.6 833.7 869.2 1008.4 1170.4

375.7 508.1 tran

sitio

n m

etal

s

595.4 722.9 710.6 821.0 920 1047.8

380 514.6

Moving from top to bottom within a group, ionization energy decreases—it is easier to remove electrons from larger atoms nearer the bottom of the table. Zeff does not change moving down a group, so it cannot be responsible for this vertical trend in ionization energy. The trend is due to the increase in radius moving down a group. In a small atom, the valence electrons are close to the nucleus, and therefore it is harder to pull them away (high ionization energy). In a large atom, the valence electrons are farther from the nucleus, so it is easier to separate them (low ionization energy). The attraction between a valence electron and a nucleus is similar to the attraction between two magnets. If two magnets are very close together, it is hard to pull them apart. If the same two magnets start out a small distance from each other, it is much easier to separate them.

+ -

+ -

Close together: harder to separate, high ionization energy Farther apart: easier to separate, low ionization energy

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Ionization energy tends to increase from left to right across a period. Not only does atomic radius decrease across a period, but Zeff increases as well. Both of these contribute to higher ionization energies toward the right side of the periodic table. Overall, the trend is ionization energy is:

Electronegativity As we will discuss in the next chapter, chemical bonds occur when atoms share valence electrons. The sharing is not always equal; some atoms “hog” shared electrons more than others. Electronegativity is a measure of how much an atom “hogs” electrons when it is bound to other atoms. The range of electronegativity values is 0.00 to 4.00, and the values for the main group elements are shown below:

2.2

0.98 1.57 2.04 2.55 3.04 3.44 3.98

0.93 1.31 1.31 1.9 2.19 2.58 3.16

0.82 1 1.81 2.01 2.16 2.55 2.96

0.82 0.95 1.78 1.96 2.05 2.1 2.66

0.79 0.89 tran

sitio

n m

etal

s

2.04 2.33 2.02 2 2.2

0.7 0.9

Just like ionization energy, electronegativity tends to decrease moving down a group and increase moving to the right across a period. The reasons for the trends are also the same: the increasing radius moving down a group, and the decreasing radius/increasing Zeff moving across a period.

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Chapter 10: Chemical Bonds

10.1 Chemical Bonds and Valence Electrons 10.2 Ions and Ionic Bonds 10.3 Covalent Bonds 10.4 Metallic Bonds 10.5 Drawing Lewis Structures 10.6 Lewis Structures with Resonance 10.7 Molecular Geometry 10.8 Bond Polarity 10.9 Molecular Polarity

10.1 Chemical Bonds and Valence Electrons Imagine you broke yourself down into individual elements. If you weighed 70 kg (154 pounds) you would end up with the following:

Element Weight (lb) Element Weight (lb) Oxygen 100.4 Phosphorus 1.5 Carbon 28.0 Potassium 0.5

Hydrogen 15.7 Sulfur 0.4 Nitrogen 4.8 Sodium 0.2 Calcium 2.3 Magnesium 0.1

If you were to sell your elements at chemical-catalogue prices, you’d be worth about $400. But if you collected some of the atoms and made a functioning kidney out of them, you could sell it on the black market for at least $1000. And in their current, walking/talking/thinking arrangement, your friends and family would value them as priceless (presumably). What makes a pile of elements worth so little, but an organ or a human being so valuable? The value lies not in the elements themselves, but in how they are arranged; the organization of matter is what we deem precious. What holds atoms together in such complex and beautiful ways? Chemical bonds. Chemical bonds occur when the valence electrons of adjacent atoms interact. Remember that valence electrons are the electrons in the shell with the highest n value. For example,

C: 1s2 2s2 2p2 K: 1s2 2s2 2p6 3s2 3p6 4s2

Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

has 4 valence electrons. has 2 valence electrons. has 7 valence electrons.

The shortcut (which only works for neutral atoms, not ions) for finding the number of valence electrons is to look at the group number. Atoms in group 1 have 1 valence electron, atoms in group 2 have 2, group 13 have 3, group 14 have 4, group 15 have 5, group 16 have 6, group 17 have 7, and group 18 have

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8. (Most of the transition metals, groups 3-12, have 2 valence electrons because moving from left to right, the electrons fill up the d subshell of the second-highest shell.) It is also useful to picture the valence electrons with a Lewis dot structure. A Lewis dot diagram consists of the element symbol with the valence electrons represented by dots surrounding it:

Na Mg Al Si

P S Cl Ar

The Lewis dot diagrams for the elements in period 3 are shown above. Notice that the first four electrons are drawn on the four sides of the element symbol; electrons are only paired when electrons 5-8 are drawn. So the following is incorrect:

Al To understand why atoms bond together, we need to understand what “motivates” an atom. (Of course, atoms don’t actually have thoughts and feelings, but it is helpful to imagine that they do.) The dream of every atom is to feel complete, and atoms accomplish this by having eight electrons in the valence shell. This “full octet” is very stable, and atoms undergo chemical reactions to obtain it. Noble gases already have a full octet, and therefore rarely react with other atoms. 10.2 Ions and Ionic Bonds One way an atom can obtain a full octet is by becoming an ion—by gaining or losing electrons. Atoms in group 17 have 7 valence electrons, so they form a full octet by gaining one more (they become ions with a charge of -1). Atoms in group 16 start with 6 and want two more (they form ions with a charge of -2). Group 15 elements start with 5, and want to gain three (they form -3 ions). On the other side of the periodic table, elements behave differently. Group 1 atoms only have one valence electron, and the easiest way for them to get a full valence shell is to lose that electron. For example, sodium has one valence electron in shell 3: 1s2 2s2 2p6 3s1.

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When sodium loses its valence electron it becomes Na+, with the electron configuration 1s2 2s2 2p6, which has a full valence octet in shell 2. Likewise, group 2 elements will lose two electrons to become +2 ions, and group 13 elements tend to lose three to become +3 ions. Carbon and the other elements in group 14 are right in the middle with 4 valence electrons, and the don’t usually gain or lose electrons at all; we’ll see how they manage to fill their octets in the next section. Generally speaking, metals lose electrons to become stable and nonmetals gain electrons to become stable. Thus, when metals encounter nonmetals, the metal will transfer its unwanted electrons to the nonmetals, which eagerly snap them up:

Na Cl Na Cl

Once the two ions are formed, they stick together because opposite charges attract: this is an ionic bond. The resulting compound is NaCl. What about magnesium and chlorine? Mg has two unwanted electrons, but chlorine only needs one. So magnesium requires two atoms of chlorine, one for each electron:

Cl Mg Cl Cl Mg+2 Cl

And so the resulting compound is MgCl2. 10.3 Covalent Bonds When two nonmetals encounter each other they can bond without forming ions. Instead of transferring electrons from one atom to another, they share valence electrons. When two chlorine atoms come in contact, they share a pair of electrons like this (the electrons of different atoms have different shading):

Cl Cl Cl Cl

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Because only one pair of electrons is shared, this is called a single bond. Single bonds (one pair of electrons) are also represented by a single line, as shown in the figure on the right. What about oxygen? Each oxygen atom has six valence electrons, and when two oxygen atoms share electrons, they must share two pairs of electrons to be “happy”, making a double bond:

O O O O

Nitrogen atoms have five valence electrons, share three pairs of electrons, and make a triple bond:

N N NN In the molecules above, there are two types of electron pairs: pairs that are shared between atoms pairs that are not. These are called bonding pairs and lone pairs, respectively. Because valence electrons are shared in covalent bonds, there is some overlap between electron clouds. The more overlap between the clouds, the more stable the bond. But there is another factor which limits how much overlap there can be: the repulsion of the positively-charged nuclei. The following is a potential energy diagram. The x-axis is the distance between the nuclei, and the y-axis is the potential energy of the two atoms. Remember: low energy corresponds to stability and high energy corresponds to instability.

+ +

+ +

+ +

+ +

A D

C

B

Distance between nuclei

Pote

ntia

l ene

rgy

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When the atoms are far apart, as in position A, they have no interaction with each other. Having no interaction with each other is the baseline; it is neither high nor low in energy; the stability is neutral. As the atoms approach each other in position B, their electron clouds start to overlap and sharing of electrons begins. Each atom is somewhat satisfied by this position as their octets are starting to be filled. The more the electron clouds overlap, the more sharing occurs, and the more stable the situation becomes. However, as the clouds overlap more and more the nuclei also get closer and closer together, and the repulsion between the nuclei gets stronger and stronger. At position D, when the nuclei are very close the potential energy is quite high; the arrangement is unstable. The atoms must strike a balance between overlapping clouds and repelling nuclei; the ideal distance is at the low point of the graph at position C. 10.4 Metallic Bonds In a metal, all of the valence electrons of the atoms are shared throughout the material. This is called metallic bonding. On the submicroscopic level, a piece of metal looks like a “sea” of loose valence electrons with metal cations floating around in it: This loose metallic structure is responsible for giving metals many of their unique properties (conductivity of heat and electricity, malleability, ductility).

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e- e

-

e-

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10.5 Drawing Lewis Structures A Lewis structure is a picture of a molecule that clearly shows how all the atoms in the molecule are bound together. It shows all covalent bonds (single, double, and triple) and all valence electrons. Use the following procedure for drawing Lewis structures.

1. Add up the valence electrons for all the atoms in the molecule. If the molecule has a charge, add or subtract electrons to account for the charge. For example, if the molecule has a +1 charge, subtract one electron from the total; if it has a -2 charge, add two electrons to the total.

2. Divide by two to get the number of electron pairs. 3. Arrange atoms in a tentative arrangement according to the following

guidelines: a. C is usually the central atom; C tends to bond to itself, forming

chains of C atoms. b. H is always a terminal atom (an atom around the outside of the

molecule). c. O rarely bonds to itself. d. If C is absent, the atom present in the least quantity is usually the

central atom. e. Most structures are symmetrical. f. Pay attention to how the formula is written—it often gives clues

about how the atoms are arranged. 4. Connect the atoms with single bonds. 5. Add pairs to terminal atoms to fill their octets (hydrogen will not have an

octet—its valence shell is full once it has two electrons: 1s2). Once their octets are filled, add any remaining pairs to the central atom. Carbon rarely has a lone pair.

6. If the central atom has less than an octet, transfer lone pair electrons from terminal atoms to create double or triple bonds between the central atom and the terminal atoms to complete the central atom’s octet.

a. B is satisfied with six valence electrons and Be is satisfied with four. b. F only makes single bonds.

For example, let’s draw the Lewis structure of CCl2O. Step 1: Add up all the valence electrons. The C atom has 4; each chlorine has 7; the O atom has 6. The total is 4 + 2(7) + 6 = 24. Step 2: Divide this number by 2 to get the number of electron pairs. 24/2 = 12.

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Step 3: According to rule 3a, the carbon atom will be in the center of the structure, and the other atoms will surround it. So we can draw:

C

O

Cl Cl Step 4: Connect the atoms with single bonds.

C

O

Cl Cl Step 5: After we add pairs of electrons to the terminal atoms to complete their octets, we have used up all of the 12 pairs we started with: three pairs are the single bonds, and there are nine lone pairs on the terminal atoms.

C

O

Cl Cl

Step 6: Carbon does not yet have an octet; the three bonding pairs surrounding it only consist of only 6 electrons total. So we need to transfer a lone pair from one of the terminal atoms to make a double bond. If we make a double bond with one of the chlorine atoms, we get:

C

O

Cl Cl

If we make a double bond with the oxygen, we get:

C

O

Cl Cl

Notice that the second structure is symmetrical; the first one is not. Therefore, the second one is the preferred Lewis structure.

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Another example: CH3CHCH2 Step 1: Adding up all the valence electrons, we get 3(4) + 6(1) = 18. Step 2: Divide by 2 two get 9 electron pairs. Step 3: Guidelines 3a and 3b tell us that the C atoms will form a chain, and the H atoms will be surrounding it. Guideline 3f says that the chemical formula can give clues about how to arrange the atoms; in this case it suggests that three H’s will be bound to the first C, one H to the middle C, and two H’s to the third C.

C C

H

H

H

H

C

H

H

Step 4: When all these are connected with single bonds, it looks like this:

C C

H

H

H

H

C

H

H

Step 5: We have used up 8 of our 9 pairs of electrons. Normally, we would put excess pairs on one of the terminal atoms to try to fill its octet; however, all the H’s are already satisfied with two electrons each. We would then usually put this lone pair on one of the C’s, but that is discouraged by this rule. Let’s move on to the next step and see if this works itself out. Step 6: The middle and third carbon do not have octets. There are no lone pairs on terminal atoms to transfer to make a double bond. But we do have a pair of electrons left over from step 5 which we couldn’t place. If we use that extra pair of electrons to make a double bond between the middle and third carbons, everything works out:

C C

H

H

H

H

C

H

H

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10.6 Lewis Structures with Resonance There are two possible ways of writing the Lewis structure of SO2:

O

S

O O

S

O

These two Lewis structures are equally valid; they differ only in the position of the double bond. This is an example of resonance. We can think of resonance as the double bond rapidly switching from one place to another, and this is depicted with a double arrow like this:

O

S

O O

S

O

This is not a perfect picture of reality, however. The double bond doesn’t actually flip from one side to the other. The bonds are actually exactly the same; each bond is somewhere between a single bond and a double bond—we can think of it as a “one-and-a-half bond.” The compound that we drew in the previous section can also be written in two different ways, both valid Lewis structures, and it may be tempting to think that resonance will occur:

C C

F

H

H

H

C

H

H

CC

H

H

H

H

C

F

H

Don’t be fooled! The problem here is that not only is the position of the double bond changing, but the position of an atom is too: a hydrogen has switched position from one side to the other. Remember, electrons and bonds can move, but atoms cannot. Therefore there is only one valid Lewis structure for this compound. But which is it? If the formula is written C3H5F, it wouldn’t be clear. To specify which structure is meant, the formulas could be written H2FCCH=CH2 and HFC=CHCH3, respectively. 10.7 Molecular Geometry Lewis structures do not accurately depict the three-dimensional shapes of molecules: the molecular geometry. The arrangement of the electron pairs (the electron geometry) provides the basic shape of the molecule; think of the electron geometry as the underlying skeleton to which atoms are attached. The term electron domain refers to a

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bond (single, double, or triple) or a lone pair. Single, double, and triple bonds each count as one domain. Electron domains are negatively charged, and therefore they repel one another. They are all connected to the central atom, so they arrange themselves so that they are as far away from each other as possible. This positioning of electron domains based on mutual repulsion is known as valence shell electron pair repulsion theory, or VSEPR (pronounced “vesper”). For example, if there are two electron domains around the central atom, they will repel each other until they are on directly opposites side of the central atom: X This electron geometry is called linear. The domains are 180° apart. If there are three electrons domains, they arrange themselves like this:

X

This electron geometry is called trigonal-planar. The domains are 120° apart. If there are four electrons domains, they arrange themselves like this:

X

This electron geometry is called tetrahedral. The domains are 109.5° apart. The dashed line is meant to indicate that it is receding into the distance, while the wedge-shaped line is projecting out towards the viewer. Once actual atoms and lone pairs are put on the electron domains, we can describe the molecular geometry. The molecular geometry is determined by the arrangement of the atoms, not the electron domains. If the electron geometry is trigonal-planar, and there are no lone pairs, then the molecular geometry is also called trigonal-planar:

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B

H H

H

If the electron geometry is trigonal-planar, and there is one lone pair, then the molecular geometry is called bent. To emphasize that lone pairs occupy space just like bonding pairs and participate in electron-domain repulsion, lone pairs on the central atom are shown as teardrop-shaped clouds. (Lone pairs on terminal atoms are drawn in the usual way, because we are not concerned with their geometries.)

S

O O

Notice how the bent shape refers only to the position of the atoms, not including the lone pair. If the electron geometry is tetrahedral, and there are no lone pairs, then the molecular geometry is also tetrahedral.

H

C

H

H

H

If the electron geometry is tetrahedral, and one of the domains is a lone pair, then the molecular geometry is trigonal-pyramidal (“triangle-shaped pyramid”).

N

HH

H

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If the electron geometry is tetrahedral, and two of the domains are lone pairs, then the molecular geometry is bent.

O

H

H

The following chart summarizes this information.

Number of Lone Pairs on Central Atom Electron

Geometry 0 lone pairs 1 lone pair 2 lone pairs

Linear X X X

Molecular Geometry: Linear

Trigonal Planar

X

X

X

X

Molecular Geometry: Trigonal Planar

X

X X

Molecular Geometry: Bent

Tetrahedral X

X

X

X

X

Molecular Geometry: Tetrahedral

X

XX

X

Molecular Geometry: Trigonal Pyramidal

X

X

X

Molecular Geometry: Bent

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10.8 Bond Polarity We can imagine that in every chemical bond, the atoms at either end are pulling on the pair(s) of bonding electrons. Atoms with higher electronegativity can pull harder than atoms with lower electronegativity. The range of electronegativity values is 0.00 – 4.00, and the values for the main group elements are shown below.

H 2.2

He

Li 0.98

Be 1.57

B 2.04

C 2.55

N 3.04

O 3.44

F 3.98

Ne

Na 0.93

Mg 1.31

Al 1.31

Si 1.9

P 2.19

S 2.58

Cl 3.16

Ar

K 0.82

Ca 1.0

Ga 1.81

Ge 2.01

As 2.16

Se 2.55

Br 2.96

Kr

Rb 0.82

Sr 0.95

In 1.78

Sn 1.96

Sb 2.05

Te 2.1

I 2.66

Xe

Cs 0.79

Ba 0.89 Tr

ansi

tion

Met

als

Tl 2.04

Pb 2.33

Bi 2.02

Po 2.0

At 2.2

Rn

Fr 0.7

Ra 0.9

If the tug-of-war is very lopsided (one atom is much more electronegative than the other), then the stronger atom pulls an electron away from the weaker one completely, and we have an ionic bond. The more electronegative atom gains the electron and becomes a negative ion, and the less electronegative atom loses the electron and becomes a positive ion. If one atom is a little bit stronger than the other (but not strong enough to win completely), the bonding electrons are held closer to the more electronegative atom. Because chlorine is more electronegative than hydrogen, the bonding pair is pulled closer to the chlorine:

ClH This type of bond is called polar covalent. Because the electrons are closer to the chlorine atom, the chlorine atom has a partially negative charge (not a full -1 charge, because it can’t completely pull the electrons away from the hydrogen). Likewise, the hydrogen has a partially positive charge because the electrons are far away from it. Polar covalent bonds thus have one partially negative end and one partially positive end. If the battle is close to a tie (both atoms have about the same electronegativity), the bonding electrons remain about the same distance from each atom. In the

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following picture, both hydrogen and carbon pull about equally on the bonding pair.

CH This type of bond is called non-polar covalent. Using the values from the table we can quantify how lopsided the battle is. For example, when sodium and chlorine bond, the difference in their electronegativities is 3.16 – 0.93 = 2.23; this large difference indicates an ionic bond. When hydrogen bonds to carbon, the difference is 2.55 – 2.2 = 0.35, and the bond is non-polar covalent. When hydrogen bonds to fluorine, the difference is 3.44 – 2.55 = 0.89, which indicates a polar covalent bond. Where are the cutoffs between the bond types? There are shades of grey between types, but for convenience we will use the following classifications:

EN difference Classification 0 to 0.5 Non-polar covalent 0.5 to 1.7 Polar covalent Over 1.7 Ionic

10.9 Molecular Polarity For reasons that will be clear a bit later, it is very useful to visualize the polarity of all the bonds in a molecule. Once the polarities are determined using the electronegativity values, the polarities can be represented with arrows pointing toward the more electronegative atoms in the bonds. For example, in NH3 each N-H bond is polar-covalent, and N is more electronegative than H:

N

HH

H

The back end of the arrow has a plus sign, indicating that the bond is more positive on that side. The other way to indicate polarity is the lowercase Greek letter delta (δ):

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N

HH

H

“δ+” indicates the atom has a partial positive charge; “δ-” indicates the atom has a partial negative charge. All of the bond polarities in a molecule combine to form the overall molecular polarity of the molecule. For example, let’s look at the CO2 molecule. The oxygen is more electronegative than the carbon, so the bond polarities are:

C OO

The two arrows point in exactly opposite directions—because they oppose each other perfectly, they cancel each other out and the molecule, overall, is not polar. Let’s look at SO2.

S

O O

Notice that this time, the polarity arrows cancel each other out from left to right, but not up and down—they both point somewhat downward. So if we combine the two arrows to get the molecular polarity, it is represented by an arrow pointing directly downward:

S

O O

δ+

δ+

δ- δ+

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Let’s contrast this with BCl3, in which the bonds are polar towards the chlorine atoms:

B

Cl

Cl Cl

In this case, the arrows end up canceling each other out, so the molecule is nonpolar. Look at the following molecules and their bond polarities.

Cl

C

ClCl

Cl

N

HH

H

H

O

H

And now look at their molecular polarities.

Cl

C

ClCl

Cl

N

HH

H

H

O

H

In CCl4, the bond polarities cancel each other out completely making the molecule nonpolar. NH3 and H2O are polar because the bond polarities do not cancel out.

nonpolar

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Chapter 11: Intermolecular Forces

11.1 Types of Intermolecular Forces 11.2 Strength of IMF’s 11.3 IMF’s and Physical Properties 11.4 IMF’s and Solubility 11.5 Types of Solids 11.6 Specific Heat 11.7 Heats of Fusion and Vaporization 11.8 Heating Curves 11.9 Phase Diagrams

11.1 Types of Intermolecular Forces We’ve seen that atoms can stick together with chemical bonds to form molecules. But what holds molecules together? Why do a bunch of water molecules stay together in a drop instead of just gradually drifting away from each other? The forces that hold molecules together are called intermolecular forces (“inter” means between, as in international = between countries). These IMF’s are not chemical bonds like ionic, covalent, and metallic bonds—while weaker than chemical bonds, they still play a critical role in determining the properties of a substance. All IMF’s are due to the attraction between positive and negative charges, and they all involve molecular polarity in some form. In the context of IMF’s, polar molecules are called dipoles. As we saw in the previous chapter, dipoles have a region of partial positive charge and a region of partial negative charge. We’ll picture them like this:

δ+ δ- The intermolecular forces are classified according to the types of molecules interacting. The first type of IMF is called ion-dipole, because it is between an ion and a polar molecule:

δ+ δ- + The partially negative side of the dipole is attracted to the positive ion. A negative ion would be attracted to the positive side. The next type is dipole-dipole:

δ+ δ- δ+ δ- The partially negative side of one dipole is attracted to the partially positive side of a neighboring dipole.

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Next: dipole-induced dipole. Polar and non-polar molecules still have some attraction between them. When one side of a polar molecule approaches a non-polar molecule, the electrons in the non-polar molecule shift away: δ+ δ- In the picture above, the negative side of the polar molecule repels the electrons in the molecule on the right. Because most of the electrons are now on the right side of this molecule, it has now become polar temporarily; it is called an induced dipole:

δ+ δ- δ+ δ-

The last kind of IMF is induced dipole-induced dipole. Electrons move randomly around the nuclei of the atoms in a molecule. Due to random chance, sometimes the electrons will be unevenly distributed throughout the molecule, and the molecule will spontaneously, temporarily become polar. In that instant of polarity, it can induce a dipole in a neighboring molecule. This induced dipole can induce another dipole in another neighbor, and so on, until the molecules look like this: δ+ δ- δ+ δ- δ+ δ- δ+ δ- δ- δ+ δ- δ+ δ- δ+ δ- δ+ δ+ δ- δ+ δ- δ+ δ- δ+ δ- So the induced dipole-induced dipole IMF happens because electron clouds are constantly “sloshing” back and forth creating temporary dipoles and inducing dipoles in neighboring molecules. Because all molecules have electrons, all molecules have this kind of IMF (a polar molecule will have sloshing electrons, too). This type of IMF is also called London dispersion forces.

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11.2 Strength of IMF’s The four types of IMF’s have different strengths. We’ve discussed them in order of strongest to weakest: ion-dipole, dipole-dipole, dipole-induced dipole, and induced dipole-induced dipole. There is also tremendous variation within a type of IMF. Not all dipole-dipole attractions are equally strong, for example. Because all the IMF’s are due to positive/negative attraction, the strength varies with the amount of polarization of the molecules.

• The strength of ion-dipole depends on the charge of the ion and the strength of the dipole.

• The strength of dipole-dipole, and dipole-induced dipole depends on the strength of the dipole.

• The strength of induced dipole-induced dipole depends on the number of electrons: more electrons sloshing around means stronger temporary dipoles. In other words, the larger the molecule, the stronger the induced dipole-induced dipole IMF.

Sometimes molecules are so polar, and the dipole-dipole interactions are so strong, that we give the resulting IMF another name: hydrogen bonding. Remember this is not a chemical bond; though it has “bond” in the name, it is still just an IMF. Hydrogen bonding occurs only under certain circumstances: when hydrogen is covalently bound to nitrogen, oxygen, or fluorine. These three elements are extremely electronegative, so when they bond to H, the resulting bond is highly polar; the H will be partially positive and the N, O, or F will be partially negative. The high degree of polarization makes the IMF quite strong. The following shows a number of molecules engaging in hydrogen bonding. The dotted lines represent hydrogen bonds.

O

H H F

H

NH

H

H

O

H

C

H

H

H

O

H

H

H

H

NH

Notice that the hydrogens bonded to the carbon atom do NOT participate in hydrogen bonding, because the C-H bond is not polar enough.

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11.3 IMF’s and Physical Properties IMF’s play a huge role in determining the physical properties of a substance. We will examine their effects on boiling point, melting point, volatility, surface tension, and viscosity. Melting Point and Boiling Point The three states of matter look like this:

As the solid heats up, the molecules vibrate in place faster and faster. IMF’s try to hold them in place, but as they absorb heat energy they eventually move fast enough to overcome the IMF’s. When this happens, the rigid structure of the solid falls apart—it melts—and the substance becomes a liquid. The specific temperature at which the heat energy overcomes the IMF’s is called the melting point. If the IMF’s are very strong, a higher temperature is needed to overcome them. The stronger the IMF’s, the higher the melting point. Now that the material is a liquid, it can flow—the molecules have enough energy to move past each other. Intermolecular forces still hold them together, but they can’t lock them into the rigid structure of the solid. If even more heat is added, the molecules will move faster and faster until they gain enough energy to fly apart. This is the process of boiling, and it also happens at a certain temperature: the boiling point. A liquid with strong IMF’s resists boiling until a lot of energy is put into the substance: the stronger the IMF’s, the higher the boiling point. So there is always a battle between heat and IMF’s—more heat causes the molecules to move more, but IMF’s try to hold the molecules in place. Volatility Evaporation is similar to boiling, but doesn’t happen at a specific temperature; it happens to some degree at all temperatures (it happens faster at high temperatures and slower at low temperatures). The ease of evaporation is called the volatility, and it too is determined by the strength of the IMF’s. If the IMF’s are weak, it is easy for molecules to escape; if the IMF’s are strong, it is hard for the molecules to escape. You may have noticed that certain liquids dry up (evaporate) faster than others; rubbing alcohol (isopropanol) dries up much faster than water, while vegetable oil hardly seems to evaporate at all.

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This is also due to the strength of the IMF’s—the stronger the IMF’s, the lower the volatility. Surface Tension The surface of a liquid behaves like a skin that is somewhat difficult to puncture. Even objects that are denser than the liquid can float on top of it; a steel needle can sit on the surface of water (if it is pushed through the surface, it sinks). This property of liquids is called surface tension. The stronger the IMF’s, the stronger the surface tension. Viscosity As we discussed above, when a liquid flows, its molecules move past each other. The ease of flow is called the viscosity; a fluid with low viscosity flows easily, while a highly viscous one is thick and gooey. The stronger the IMF’s, the higher the viscosity. 11.4 IMF’s and Solubility The solubility of a substance is also heavily influenced by IMF’s. When something dissolves, its molecules evenly disperse throughout the liquid, forming a homogenous mixture. Imagine we start out with a chunk of sugar submerged in water. Water is a polar substance and sugar is a polar substance, and therefore the water and sugar will have dipole-dipole IMF’s between them. Because there is this strong attraction between them, they mix together easily. Now imagine submerging a chunk of candle wax in the water instead of the sugar. Candle wax is non-polar, so the only type of interaction between the water molecules and the wax is dipole-induced dipole. Now, this is not as strong as dipole-dipole, but it is still an IMF, so we might still expect the wax to dissolve. But it doesn’t dissolve at all. We can think about it like this: The water has a choice; it can mix with the wax and have dipole-induced dipole interactions, or it can stick to itself only and have stronger dipole-dipole interactions. Put another way, it is more stable for the water to stick to itself than to stick to the non-polar wax. Polar and non-polar substances don’t mix well. Generally speaking, polar substances dissolve well in other polar substances, and non-polar substances dissolve well in non-polar substances. In other words: like dissolves like. 11.5 Types of Solids Solids can be classified into four different categories depending on how their atoms/molecules are held together. The way the solid holds together influences its melting point, boiling point, solubility, and conductivity.

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Melting point/Boiling point: The stronger the forces holding the particles together, the higher the melting and boiling points will be. Solubility: Some solids are made of charged or polar particles; these solids usually dissolve well in polar substances such as water. Solids made of non-polar particles usually dissolve well in non-polar substances such as oil or hexane. Electrical conductivity: The flow of electricity is the movement of charged particles (either electrons or entire ions). Therefore, conductive substances must have “loose” charged particles. If the electrons or ions are held tightly in place, the substance will not be a good conductor.

Ionic solids Ionic solids are composed of ions (either mono- or polyatomic). There are no distinct molecules in ionic solids; rather, the ions are arranged in a repeating structure called a crystal lattice, shown at right. The lattice energy is the amount of energy needed to separate all the ions in one mole of an ionic substance. The stronger the ionic bonds, the higher the lattice energy. MP/BP: Because ionic bonds are very strong, ionic substances tend to melt and boil at very high temperatures. Solubility: The ions dissolve well in polar solvents but poorly in non-polar solvents. Conductivity: In the solid phase, ionic solids are poor conductors; the ions are not freely mobile. However, when in the liquid phase or dissolved, they conduct very well—once the ions are free to move, they can conduct. A note on hydrates: Many ionic compounds will trap molecules of water within their crystal lattice—such a compound is said to be hydrated. This results in the compound having a higher molar mass than the anhydrous form. For example, CoCl2 combines absorbs water in a 1:6 ratio; one mole of CoCl2 will hold six moles of water. The chemical formula of this hydrate is CoCl2•6H2O, and the molar mass of the compound is the molar mass of CoCl2 plus six times

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the molar mass of water: 129.83 + 6(18.02) = 237.95 g/mol. To name a hydrated solid, include a numerical prefix in front of the word hydrate:

Formula Name CoCl2•6H2O cobalt(II) chloride hexahydrate Ca(NO3)2•3H2O calcium nitrate trihydrate Na2SO4•10H2O sodium sulfate decahydrate

When weighing out a certain mass of an ionic substance in the lab, be sure to note of the solid is a hydrate or not; if you don’t take the mass of the extra water into account, you will end up with too few moles of the compound. Molecular solids Molecular solids are composed of individual molecules held together with IMF’s. MP/BP: Because IMF’s are weaker than chemical bonds, molecular solids tend to melt and boil at relatively low temperatures (many molecular substances are liquids or gases at room temperature). Solubility: If the molecules are polar, the substance will dissolve in polar solvents. If the molecules are non-polar, the substance will dissolve in non-polar solvents. Conductivity: Most electrons in molecular solids are participating in chemical bonds or are held tightly to the atoms. There are no ions either, so molecular solids are poor conductors whether solid, liquid, or dissolved.

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Network covalent solids Network covalent solids consist of atoms held together with covalent bonds. There are no individual molecules; the entire substance can be thought of as one large molecule. Common examples include diamond (left) and graphite (right). Both are pure carbon, but the atoms are connected in different ways—this greatly changes their properties (and their prices!).

MP/BP: Covalent bonds are strong, and therefore network covalent solids melt/boil at very high temperatures. Solubility: It is very difficult to separate the atoms in these solids; they dissolve poorly in both polar and non-polar solvents. Conductivity: Electrons are tightly bound up in most of these solids; they are poor conductors (graphite is an exception). Metallic solids For the structure of metallic solids, review section 10.4. MP/BP: The melting and boiling points of metals vary a lot, but most are high. Solubility: Metals do not dissolve well in either polar or non-polar solvents. Conductivity: Due to the freely mobile electrons, metals are excellent conductors.

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C C

C

CC

C C

C

C

C

C C

C C

C C

C

C

C

CC

C

C

C

C

C

C C

C C

C

C

CC

C

C

C

C

C

C C

CC

C C

C

C C

C

CC

C C

C

C

C

C C

C C

C C

C

C

C

CC

C

C

C

C

C

C C

C C

C

C

CC

C

C

C

C

C

C C

CC

C C

C

C C

C

CC

C C

C

C

C

C C

C C

C C

C

C

C

CC

C

C

C

C

C

C C

C C

C

C

CC

C

C

C

C

C

C C

CC

C C

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Summary:

Solubility

Solid MP/ BP polar

solvent

non-polar

solvent

Conductivity

Ionic High yes no Solid-poor Liquid/dissolved-good

Molecular Low Like dissolves like poor

Network covalent High no no poor

Metallic High no no good

11.6 Specific Heat Imagine running a metal spoon and a wooden spoon under hot water for the same amount of time, so that they both absorb the same amount of heat energy from the water. The metal spoon will reach a much higher temperature than the wooden one. The same amount of energy will change the temperatures of different substances by different amounts. How easy it is to raise the temperature of a substance is called the specific heat capacity or specific heat. If it takes a small amount of energy to change the temperature, the substance has a low specific heat. If it takes a large amount of energy to change the temperature, the substance has a high specific heat. Specific heat is represented by the variable c. The definition of specific heat is: the amount of energy (Joules) needed to raise one gram of a substance by one degree Kelvin. Thus, the units for c are J/g-K. The mass of a substance also influences how easily it changes temperature. A large pot of water on the stove needs much more energy to heat up than a small pot of water, even though they both have the same specific heat. We represent the mass (in grams) with the variable m. So far we haven’t mentioned how far we want to change the temperature. If we only want to increase it a few degrees, we don’t need much heat. But to make a big increase in temperature, we need more. The size of the temperature change is represented by the variable ∆T (delta T). So, there are three things that influence the amount of heat needed to raised the temperature: the specific heat (c), the mass of the sample (m), and the increase

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in temperature (∆T). The amount of heat (in joules) is represented by the variable q, and so we can write the formula:

q = mc∆T Example: The specific heat of water is 4.18 J/g-K. If you have 40 g of water, how much energy do you need to raise its temperature from 15 °C to 40 °C?

From the question, we know:

m = 40 c = 4.18 ∆T = 40 – 15 = 25

Enter this into our equation, and get:

q = mc∆T q = 40(4.18)(25) = 4180 J Example: 50 g of a certain substance absorbs 2000 J, and the temperature increases from 35 °C to 95 °C. What is the specific heat of the substance?

From the question, we know: m = 50 q = 2000 ∆T = 95 – 35 = 60

Enter this into our equation, and get: q = mc∆T 2000 = 50(c)(60) c = 0.667 J/g-K 11.7 Heats of Fusion and Vaporization Melting or boiling a substance also requires energy. The amount of energy depends on the substance; it may take only a small amount of energy to melt one substance, but a lot of energy to melt another. For our purposes, the amount of heat needed to melt something does not include the energy needed to heat it up to its melting point; it is only the heat put into the substance from when it starts to melt to when it has melted completely. The amount of heat needed to melt one mole of a substance is called the molar heat of fusion, represented by the variable ∆Hfus. The amount of heat needed

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to boil one mole of a substance is called the molar heat of vaporization, represented by the variable ∆Hvap. Thus, the units for ∆H are kJ/mol. As with changing the temperature, the more of a substance we have, the more energy we need to melt it or boil it. When dealing with phase changes, we don’t use mass, we use moles, represented by the variable n. Again the amount of heat is represented by q, which, in this situation, is in kilojoules (kJ), not joules. So the equation is this:

q = n∆Hfus or vap Example: How much energy is needed to melt 20 g of ice? The heat of fusion of water is 6.02 kJ/mol.

First, we need to know how many moles we are dealing with. So, convert 20 g to moles using the molar mass of water (18 g/mol).

n = 20/18 = 1.11 mol

And then plug into our equation:

q = n∆Hfus

q = 1.11(6.02) = 6.69 kJ

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11.8 Heating Curves Imagine heating a substance with a constant rate of input of energy. The substance begins as a solid which heats up, then it melts, then the liquid continues to heat up, then it boils, and then the gas continues to heat up. If the temperature is plotted vs. the energy input, you end up with a heating curve:

This is the heating curve for 100 grams of an imaginary substance. It starts out at 0 K, and the temperature rises as heat is added. The temperature increases from 0 to 140, and this requires 4 kJ of energy. The temperature then remains constant while the substance melts. This is because all the energy during this period is used up by the melting process; the energy goes towards overcoming the IMF’s that hold the particles together. During this flat part of the graph (a plateau) both the solid and liquid phases are present. At the very beginning of the plateau, there is only solid. As more energy is added, the solid becomes liquid and by the end of the plateau, melting is complete. The melting process requires a total of 6 kJ of energy. Notice we only count the energy put in during the plateau; we don’t include the first 4 kJ needed to raised the temperature to the melting point. After melting is complete, the temperature of the liquid then increases until it reaches the boiling point (this requires 6 kJ). When boiling begins, the temperature plateus again; the temperature can only continue to increase once all the liquid has turned to gas. Boiling the substance requires 16 kJ.

0 5 10 15 20 25 30 35 40

400

300

200

100

0

S

S+L

L

L+G

G

Energy input (kJ)

Tem

pera

ture

(K)

melting

boiling

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A heating curve provides a lot of information about a substance. For example, the temperatures of the plateaus are the melting and boiling points. By looking at the slopes of the graph, the specific heats of each phase of the substance can be calculated. Let’s say we need to calculate the specific heat of the liquid phase. On the graph, we can see that the liquid phase increases in temperature from 140 to 190; therefore the ∆T value is 280 – 140 = 140 K. During this period, the substance absorbed 6 kJ, or 6000 J, or energy. Remember that we only focus on the liquid segment of the graph, and not on anything that came before! While it took 4 kJ to heat the solid from 0 to 140, and 6 more to melt it, the energy put in during the liquid segment of the graph is only 6 kJ. So, we know that:

q = 6000 m = 100 (that’s what the heating curve specified) ∆T = 140

And plug into the equation: q = mc∆T 6000 = 100(c)(140) c = 0.429 J/g-K The length of the plateaus can be used to calculate the ∆Hfus or ∆Hvap values. First, we must know how many moles of the substance are being heated. The heating curve is for 100 g, and we must convert this value to moles using the molar mass, which is 20 g/mol:

n = 100/20 = 5 mol The graph shows that it required 14 kJ to boil the substance. Again, remember that we only focus on the segment of the graph where melting occurs, not anything that came before. Now we can plug into the equation:

q = n∆Hvap 14 = 5∆Hvap ∆Hvap = 2.8 kJ/mol

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11.9 Phase Diagrams The following is a phase diagram. It shows how pressure and temperature influence the phase of a substance.

The solid lines on the diagram represent phase boundaries. The phase boundaries delineate the three phases: solid, liquid, and gas. From the diagram, we can determine the phase of the substance at any combination of temperature and pressure. For example, when the temperature is 5 °C and the pressure 0.9 atm, the substance is a solid. At 60 °C and 0.2 atm, it is a gas. The phase boundary between liquid and gas terminates at the critical point. When the pressure and temperature are above this point, the substance exists as a strange phase of matter called a supercritical fluid, which has the properties of a liquid (it can dissolve other substances) and a gas (it diffuses well). The triple point is where the three phase boundaries all meet. At this particular temperature and pressure, all three phases coexist. A phase changes occurs any time one of the phase boundaries is crossed. We are most familiar with changing phase by heating or cooling a substance, but phase changes also can occur with pressure changes. For example, it is possible to make this substance boil (go from liquid to gas) by decreasing the pressure.

1.00

17 56 Temperature (°C)

Pres

sure

(at

m) solid

liquid

gas

super-critical fluid

melting

freezing

deposition

sublimation condensation

boiling critical point

triple point

normal melting point

normal boiling point

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The diagram demonstrates that the melting and boiling points change depending on the pressure. As pressure increases, both the melting and boiling points increase. Because of this dependence on pressure, it is useful to define a standard pressure at which to measure melting and boiling points—1 atm. The melting point at 1 atm is called the normal melting point; the boiling point at 1 atm is called the normal boiling point. The general shape of the above phase diagram is typical for most substances. Water’s phase diagram is unique in that the slope of the solid/liquid phase boundary slopes to the left instead of the right: As a result, water’s melting point decreases as pressure increases, the opposite of most substances. Thus, ice will melt if it is put under high enough pressure.

Temperature (°C)

Pres

sure

(at

m)

L

S

G

Ice will melt if put under enough pressure, even when the temperature stays the same.

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Chapter 12: Solutions

12.1 Definitions 12.2 Dissociation of Ionic Solids 12.3 Solubility Rules 12.4 Solubility Curves 12.5 Concentration 12.6 Preparing Solutions 12.7 Freezing Point Depression and Boiling Point Elevation 12.8 Net Ionic Equations

12.1 Definitions A solution is a homogenous mixture of a solute dissolved in a solvent. The solvent is always present in greater quantity than the solute. For example, in a mixture of 20% methanol and 80% water, the water is the solvent and the methanol is the solute. If the percentages are reversed, the methanol is the solvent and the water is the solute. Both the solute and the solvent can be in any phase of matter:

• Saline solution is a solution of solid NaCl in liquid water. • Soda is a solution of CO2 gas in liquid water. • Alcoholic beverages are solutions of liquid ethanol in liquid water. • Air is a solution of oxygen gas in nitrogen gas. • Yellow gold is a solution of solid copper in solid gold (solutions of two or

more metals are also called alloys). • Old dental fillings were made from liquid mercury dissolved in solid silver. • Hydrogen gas can be dissolved in solid platinum. • Humid air is a solution of liquid water in nitrogen gas. • Iodine sublimating into air creates a solution of solid iodine in nitrogen

gas. In colloids, the molecules of the solute are clumped together; they are not floating in the solvent as individual molecules. Therefore, colloids are cloudy; solutions are clear. Like solutions, the solute in a colloid will not settle to the bottom of the container over time. Examples of colloids are fog (water droplets in air), milk (fat and protein in water), and smoke (carbon compounds in air). A suspension is similar to a colloid, but the clumps of solute are larger than in colloids. Suspensions are also cloudy, but they will separate if allowed to sit for a period of time. Examples of suspensions are mud (dirt suspended in water), blood (blood cells suspended in plasma) and paint (pigment particles suspended in water or oil).

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Summary

Particle size Appearance Settling?

Solution Atoms, ions, or molecules Clear No

Colloid Medium clumps (<100 nm diameter) Cloudy No

Suspension Large clumps (>100 nm diameter) Cloudy Yes

12.2 Dissociation of Ionic Solids When an ionic substance dissolves in water, the ions dissociate from one another: they split apart and float throughout the solution. After they dissociate, water molecules surround the ions and stick to them through ion-dipole IMF’s. The layer of water molecules surrounding the ions is called the hydration shell.

H

O

H

H

OH

H

O

H

HO

H H

OH

H

O

H

H

O

HH

O

HH

O

H

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12.3 Solubility Rules Some ionic compounds dissolve well in water (they are soluble); others don’t (they are insoluble). Use the following general rules to determine if an ionic compound is soluble in water:

1. Compounds containing a group 1 cation or ammonium (NH4+) are soluble.

2. Compounds containing nitrates (NO3-) or acetates (C2H3O2

-) are soluble.

3. Compounds containing halogen anions (other than F-) are soluble, except when paired with Ag, Hg(I), and Pb.

4. Sulfate (SO4-2) compounds are soluble, except those with Ba, Sr, Ca, Pb,

Ag, and Hg(I).

5. Carbonates (CO3-2), hydroxides (OH-), oxides (O-2), silicates (SiO3

-2), and phosphates (PO4

-3) are insoluble, except for group 1 cations and ammonium.

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12.4 Solubility Curves While compounds can generally be classified as soluble or insoluble, the solubility of a solute depends partly on the temperature of the solvent; it is easy to dissolve sugar in hot tea and much harder in iced tea. The solubility of solids and liquids tends to increase with higher temperatures, while the solubility of gases tends to decrease. The relationship between temperature and solubility is shown in a solubility curve, shown below. All solubility curves must specify the amount of solvent; the standard amount is 100 g. Let’s look at the curve for KNO3. The graph tells us that at 10 °C, 100 g of water can dissolve a maximum of 22 g of KNO3. At 60 °C, 100 g of water can dissolve 106 g of KNO3. If a certain solution is holding the maximum amount of solute at

KNO3 Pb(NO3)2

Al2(SO4)3

HgCl2 NH3

HCl

NaCl

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a given temperature, we say the solution is saturated. If it is holding less, it is unsaturated; for example, if 100 g of water at 60 °C had 30 g of KNO3 dissolved in it, it would be classified as unsaturated. Sometimes a solution can hold more than the maximum amount determined by the graph. Such a solution is called supersaturated, and it is unstable. If a supersaturated solution is disturbed by stirring, shaking, or dropping something into it, the excess solute quickly precipitates out until the solution is merely saturated. 12.5 Concentration Molarity and Molality If a solution contains a large proportion of solute, it is concentrated. If it contains a small amount, it is dilute. Two of the most common ways to measure concentration are molarity (M) and molality (m). The two measurements are used for different applications.

moles solute molarity (M) =

liters of solution For example, if 2 liters of a solution contains 5 moles of NaCl, that solution has a molarity of 5 mol/2 L = 2.5 M. We say that it is a 2.5 molar solution.

moles solute molality (m) =

kg of solvent For example, if 3 moles of sugar are dissolved in 4 kg of water, the solution has a molality of 3 mol/4 kg = 0.75 m. We say that it is a 0.75 molal solution. Notice two key differences between molarity and molality—molarity used liters of solution, while molality uses kilograms of solvent. Fenceposting with Concentration Because molarity is a compound unit that includes moles and volume, it can be used as a conversion factor in fenceposting problems. Example: 5.5 L of 0.3 M KCl contains ________ g of KCl

5.5 L solution 0.3 mol KCl 74.6 g 1 L solution 1 mol KCl

= 123.1 g KCl

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Example: If 1.2 L of 0.75 M CaCO3 solution reacts, how many grams of water will be produced?

CaCO3 + 2HCl → CO2 + CaCl2 + H2O 1.2 L solution 0.75 mol CaCO3 1 mol H2O 18 g H2O 1 L solution 1 mol CaCO3 1 mol H2O

= 16.2 g H2O

12.6 Preparing Solutions By Mixing Solute and Solvent A common task in the chemistry lab is to prepare a certain volume of a solution with a particular concentration. For example, to make 2 L of 0.5 M NaOH, first calculate how many grams of NaOH are needed:

2 L solution 0.5 mol NaOH 40 g 1 L solution 1 mol NaOH

= 40 g NaOH

Don’t be tempted to measure out 2 L of water, add the 40 g of NaOH, and stir. The problem with this is that the NaOH itself has volume, and when this combines with the 2 L of water, the total volume of the solution will be greater than 2 L; the molarity will be too low. To avoid this, add the NaOH to the container first, and then add water up to the 2 L mark on the container. The easiest way to do this is to use a volumetric flask like the one shown here; it has a large bulb on the bottom for easy swirling, and it tapers to a narrow neck with a mark for accurate volume control. By Diluting a Concentrated Solution It is also very common to keep chemicals stored as concentrated stock solutions, and then dilute them to the desired concentration. For example, hydrochloric acid (HCl) is usually sold as a 12 M solution, but this is too concentrated for most applications. Let’s say we need to make 2 L of 0.5 M HCl using the concentrated stock solution (12 M). We must take a small amount of the concentrated solution and mix it with water, and end up with exactly the right concentration and exactly the right volume. Remember that the definition of molarity is moles/liter of solution:

M = n/V We can rearrange this to solve for the number of moles

n = MV.

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The key to solving the problem is this: the small amount of concentrated HCl contains exactly as many moles of HCl as the dilute solution. Mathematically speaking,

ncon = ndil or MconVcon = MdilVdil We want 2 L of dilute solution (Vdil = 2). The dilute solution needs to be 0.5 M (Mdil = 0.5). The concentrated solution is 12 M (Mcon = 12). Plug into the equation to find the volume of concentrated solution needed:

MconVcon = MdilVdil 12Vcon = 0.5(2) Vcon = 1/12 = 0.0833 L

So to make the solution, mix together 0.0833 L of concentrated solution, and enough water (1.9167 L) to make 2 liters of solution. Safety note: When diluting concentrated acid, the acid should always be added to water already in the container. It is unsafe to add water to concentrated acid; it can make the solution heat up, boil, and splash out of the container. “Do as you oughtta; add acid to watta.” 12.7 Freezing Point Depression and Boiling Point Elevation Pure water freezes at 0 °C and boils at 100 °C (at 1 atm). But if water has something dissolved in it, the freezing point decreases and the boiling point increases. The complete explanation for this phenomenon is complicated, but in a sense, the solute interferes with the processes of boiling and freezing. Because of this interference, a higher temperature must be reached for boiling to occur and a lower temperature must be reached for freezing to occur. The temperature change is greater the more concentrated the solution. For freezing, this is expressed mathematically as:

∆Tf = Kfm where ∆Tf is the change in the freezing point, m is the molality of the solution, and Kf is the freezing point depression constant (for water, Kf = -1.86 °C/m). For boiling, the equation is very similar, but it uses the boiling point elevation constant, Kb:

∆Tb = Kbm

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For water the Kb is 0.51 °C/m. Note that Kf’s are always negative, and Kb’s are always positive. Example: A solution of water has a molality of 0.8 m. What are the freezing point and boiling point of the solution?

∆Tf = Kfm = -1.86 × 0.8 = -1.49 °C So the new freezing point is 0 – 1.49 = -1.49 °C

∆Tb = Kbm = 0.51 × 0.8 = 0.41 °C

So the new boiling point will be 100 + 0.41 = 100.41.

The classic application of this is antifreeze (ethylene glycol) dissolved in the radiator water of cars. It prevents the water from freezing in winter, yet it also prevents the water from boiling over when the engine is generating a lot of heat. We also put salt on roads to prevent them from freezing; the salt dissolves in the moisture on the road and lowers the freezing point. When an ionic substance dissolves in water, it dissociates into its component ions. Therefore, “NaCl(aq)” actually means “Na+(aq) + Cl-(aq)”. This has a large impact on freezing point depression and boiling point elevation; if the solute dissociates, the molality of the solution is multiplied. For example, when BaCl2 dissolves in water, it dissociates into three ions (Ba+, Cl-, Cl-). Therefore, in a solution of 1 m BaCl2, the overall molality is 3 m. A solution of 1 m Al2(SO4)3 has an overall molality of 5 m. This is taken into account in the equations by adding the variable “i" or van’t Hoff factor—it is the number of ions the solute breaks up into. It changes the equations as follows:

∆Tf = iKfm ∆Tb = iKbm Example: An aqueous solution of AlCl3 has molality of 2 m. What is the freezing point of the solution?

AlCl3 consists of four ions, so i =4:

∆Tf = iKfm = 4 × -1.86 × 2 = -14.9 °C So the new freezing point is 0 – 14.9 = -14.9 °C

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12.8 Net Ionic Equations Let’s look at a double replacement reaction between two aqueous ionic substances:

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) If we remember that all aqueous ionic substances are actually separate ions, we can rewrite the equation like this: Ba+2(aq) + 2Cl-(aq) + 2Na+(aq) + SO4

-2(aq) → BaSO4(s) + 2Na+(aq) + 2Cl-(aq) Notice that nothing actually happens to the sodium ions and the chloride ions; they have not changed from one side of the arrow to the other. They did not participate in any chemical reaction. They are called spectator ions, because all they did was “watch” the reaction from the sidelines. The real chemical reaction was between the barium ions and the sulfate ions, which joined together to form a precipitate of barium sulfate. If we remove the spectator ions from the equation, we are left with the net ionic equation—a pared-down version of the full equation that only focuses on the real reaction:

Ba+2(aq) + SO4-2(aq) → BaSO4(s)

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Chapter 13: Gases

13.1 Pressure 13.2 Factors that Affect Pressure 13.3 The Ideal Gas Law 13.4 Ideal vs. Real Gases 13.5 The Combined Gas Law 13.6 Boyle’s Law and Charles’s Law 13.7 Standard Temperature and Pressure 13.8 Partial Pressure (Dalton’s Law of Partial Pressures) 13.9 Graham’s Law of Diffusion

13.1 Pressure Gas molecules are constantly colliding with the sides of their container, and the combined force of the collisions exerts a pressure on the container. Pressure is defined as force per unit of area (P = F/A). Imagine someone is pushing on you with her palm with a certain amount of force. Now imagine she pushes on you with the same force, only with the tip of her finger. The finger transmits the same amount of force to a much smaller area, and therefore exerts more pressure than the palm (and therefore it hurts more). Some units of pressure are atmospheres (atm), torr (named after 15th-century scientist Evangelista Torricelli), and mmHg (millimeters of mercury). 1 atm is approximately the air pressure at sea level. They can be interconverted as follows:

1 atm = 760 torr = 760 mmHg 13.2 Factors that Affect Pressure Moles of Gas Imagine two identical containers, A and B. B contains more moles of gas than A, so B will have a higher pressure than A—the underlying reason for this goes back to molecular collisions and the definition of pressure. The internal surface area of the container is the same, but more gas molecules collide with the walls in container B; the number of collisions in greater, so the force is greater, so the pressure is greater. Mathematically, the relationship between pressure and moles is: P = k1n

n is the number of moles k1 is a constant

P

n

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Temperature Particles move fast when the temperature is high and slow when the temperature is low. The faster the particles move, the harder they hit the walls of their container, and therefore the pressure is higher:

P = k2T

T is the temperature of the gas in degrees Kelvin (K) k2 is a constant Volume of the Container If we decrease the volume of a container, the gas molecules are crammed into a smaller space, and so they collide with the walls more frequently. Thus, decreasing the volume results in an increase in pressure, and increasing the volume causes a decrease in pressure. So volume and pressure are inversely proportional:

k3 P = V

V is the volume of the gas in liters (L)

k3 is a constant 13.3 The Ideal Gas Law Combining the above three equations into one produces:

k1k2k3nT P =

V The three constants can be merged into the term R, the ideal gas constant. It has a value of 0.0821 L⋅atm/mol⋅K. So the equation becomes:

RnT P =

V This is usually rearranged and written as PV = nRT: the Ideal Gas Law. Example: A 4-L container holds 0.4 mol H2 at 25 °C. What is the pressure? The temperature must be in Kelvin, so convert: 25 + 273 = 298 K PV = nRT P(4) = 0.4(0.0821)(298) P = 2.45 atm

P

T

P

V

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The Ideal Gas Law can also become part of a stoichiometry problem. Example: A 2-L container holds F2 and NaCl. The temperature is 300 K, and the pressure is 1.5 atm. If all the F2 reacts, how many grams of NaF will be made?

F2(g) + 2NaCl(aq) → Cl2(g) + 2NaF(aq) The first step is to use PV=nRT to find the number of moles of F2:

PV=nRT 1.5(2) = n(0.0821)(300) n = 0.122 mol F2

We can then use fenceposting to go from moles of F2 to grams of NaF.

13.4 Ideal vs. Real Gases The Ideal Gas Law makes a couple of assumptions that are not completely valid. According to the Ideal Gas Law, if a gas is compressed enough, its volume will decrease to zero. This would only be possible if the gas particles themselves had no volume, and this is not true. As a real gas is compressed, the volume decreases and the particles get closer and closer together, but the volume never reaches zero because the particles take up a tiny bit of space. Second, the Ideal Gas Law assumes that there are no attractive forces (no IMF’s) between gas particles; this also is not true. The attraction between particles in real gases slightly decreases the pressure. Whenever two particles bounce off each other, they don’t move quite as fast as they did before the collision; the attraction between the particles slows them down a bit as they move apart. Thus, they collide with the walls at a lower speed. These differences do make a difference in how real gases behave, but to keep things simpler we will assume that all gases behave in an ideal fashion.

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13.5 The Combined Gas Law It is sometimes useful to predict the pressure (or temperature, or volume) of a gas when the conditions of the gas have changed. For example, if we have 5 L of gas at 1 atm and 760 K, and then we increase the volume to 7 L while simultaneously decreasing the temperature to 500 K, what will the new pressure be? Notice that in this situation the number of moles of gas hasn’t changed, and remember that the Ideal Gas Law can be applied to both of the conditions:

Condition 1 P1 = 1 V1 = 5 T1 = 760 P1V1 = nRT1

Condition 2 P2 = ??? V2 = 7 T2 = 500 P2V2 = nRT2

We can rearrange each version of the Ideal Gas Law like this:

P1V1 P2V2 n = RT1

n = RT2

Because the moles (n) are the same before the change and after:

P1V1 P2V2 RT1

= RT2

And the R’s cancel out, leaving the Combined Gas Law:

P1V1 P2V2 T1

= T2

Then, plugging in the given values, and solving for P2:

1(5) P2(7) 760

= 500

P2 = 0.470 atm The units for this equation are the same as for the Ideal Gas Law: P in atm, V in L, and T in K. When doing problems with this equation, if the problem doesn’t mention one of the variables, assume that the variable is the same from condition 1 to condition 2—eliminate it from the equation.

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Ideal Gas Law or Combined Gas Law? Sometimes it may be hard to know which equation to use to solve a problem. If the problem mentions the number of moles of gas, chances are good the Ideal Gas Law is the right choice. If moles are not mentioned, the Combined Gas Law is most likely needed.

Example: A helium balloon with a volume of 55 mL and a temperature of 300 K is cooled down to 280 K. What is the final volume of the balloon? (Moles are not mention; use the Combined Gas Law.) Example: There are 2.5 moles of CO2 in a tank with a volume of 3 Liters at a temperature of 320 K. What is the pressure of the gas? (The problem gives a mole value; use the Ideal Gas Law.)

13.6 Boyle’s Law and Charles’s Law Two relationships contained within the combined gas law are named after their discoverers. The relationship between pressure and volume is called Boyle’s Law, and the relationship between temperature and volume is called Charles’s Law. Boyle’s law assumes that the temperature of the gas remains constant. In this case, the temperature terms cancels out in the combined gas law, leaving:

P1V1 P2V2 T1

= T2

P1V1 = P2V2

Notice that the larger V2 gets, the smaller P2 must be, and vice-versa; the smaller V2 gets, the larger P2 must be. Boyle’s Law is this inverse relationship (which was discussed in section 13.2). Charles’s law assumes that the pressure of the gas remains constant. Now the pressure terms cancel out in the combined gas law:

P1V1 P2V2 T1

= T2

P

V

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V1 V2 T1

= T2

When T2 is large, V2 must also be large; when T2 is small, V2 must also be small. Thus Charles’s Law is this direct relationship between temperature and volume: 13.7 Standard Temperature and Pressure Chemists often use the phrase “at STP”. This for consistency—if someone says “one liter of gas” without specifying the temperature and pressure, he could be talking about any number of moles of that gas. So there is a specific pressure and temperature called Standard Temperature and Pressure (STP), which is defined as follows:

STP: P = 1 atm T = 0 °C (273 K) Fact to memorize: At STP, 1 mole of gas has a volume of 22.4 L. 13.8 Partial Pressure (Dalton’s Law of Partial Pressures) Suppose a container holds a 50/50 mixture of gases A and B, and the pressure inside the container is 1 atm. Both of the gases are bouncing off the walls, and so both contribute to the pressure. Because there are equal numbers of molecules A and B, the pressures that each of them applies to the container are equal. The total pressure (PT) is 1 atm, so each gas exerts a pressure of 0.5 atm. In other words:

PT = PA + PB PA and PB are called partial pressures. Partial pressure is defined as the amount of pressure a gas would exert on its container if it were the only gas in the container. The total pressure is the sum of the partial pressures of all the gases in the container:

PT = PA + PB + PC + … Suppose a container held a mixture that was 50% O2, 30% H2, and 20% Ne. Another way of putting this is that the mole fraction of O2 is 0.5, the mole fraction of H2 is 0.3, and the mole fraction of Ne is 0.2. Mole fractions are

V

T

148

abbreviated as “X”; the mole fraction of O2 is represented as XO2. The partial pressure of a gas is equal to the total pressure times the mole fraction of the gas:

PO2 = PT × XO2 This simply means: because O2 makes up 30% of the total gas, it exerts 30% of the total pressure. Partial pressures can be used in the Ideal Gas Law: Example: A 3-L container holds a mixture of gases. The total pressure inside the vessel is 2 atm, and the temperature is 315 K. The mole fraction of oxygen is 0.35. How many moles of oxygen are in the container?

First calculate the PO2 = PT × XO2 = 2 × 0.35 = 0.70 atm Then plug this value into PV=nRT to find the moles of O2 (0.0812 mol).

We could also solve this problem by using PV=nRT to find the total moles, then multiply the total moles by the mole fraction.

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Collecting Gases Over Water Chemists often collect gases over water. In this technique, a gas is produced by a chemical reaction, and the gas is bubbled into a long glass tube filled with water (called a eudiometer). The eudiometer is graduated so the volume of gas can be measured. While this is a convenient way to collect a gas, it has a disadvantage; water evaporates into the top of the eudiometer, creating a mixture of the gas product and water vapor. Let’s say we collect 0.01 L of O2 in the eudiometer, and would like to use PV=nRT to calculate the number of moles of O2. Although the total pressure inside the eudiometer is 1 atm (assuming we’re close to sea level), part of that pressure is due to O2 and the rest is due to the water vapor. In other words:

PT = PO2 + PH2O The PH2O is called the vapor pressure of water. The vapor pressure of water depends on the temperature; the higher the temperature, the more water will evaporate, and the greater the vapor pressure will be. A table of vapor pressure values is shown here. According to the table, at room temperature (25 °C), the vapor pressure of water is 0.313 atm. So from the equation above:

PT = PO2 + PH2O 1 = PO2 + 0.0313 PO2 = 0.9687 atm

This is the value we should plug into PV = nRT to find the moles of O2. 13.9 Graham’s Law of Diffusion Diffusion is the movement of particles from an area of high concentration to an area of low concentration. For example, if we put a drop of food coloring into a cup of water, the molecules of food coloring will gradually spread out until the whole cup has an even concentration; the dye molecules start out all close together in the drop (concentrated) while the rest of the water is clear (dilute). The molecules diffuse from the high concentration to the low concentration until the concentration is the same throughout the solution.

Temp (°C)

Vapor Pressure

(atm)

Temp (°C)

Vapor Pressure

(atm) 1 0.0065 21 0.0246

2 0.0070 22 0.0261

3 0.0075 23 0.0277

4 0.0080 24 0.0294

5 0.0086 25 0.0313

6 0.0092 26 0.0332

7 0.0099 27 0.0351

8 0.0106 28 0.0373

9 0.0113 29 0.0395

10 0.0121 30 0.0418

11 0.0129 31 0.0443

12 0.0138 32 0.0469

13 0.0148 33 0.0496

14 0.0158 34 0.0525

15 0.0168 35 0.0555

16 0.0180 36 0.0586

17 0.0191 37 0.0619

18 0.0203 38 0.0653

19 0.0217 39 0.0690

20 0.0231 40 0.0727

gas produced

gas collected

eudiometer

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Gases can also diffuse through space. When someone opens a bottle of perfume, the perfume molecules diffuse through the air to reach everyone’s nostrils. At equal temperatures, lighter molecules will diffuse faster than heavier ones. Remember that temperature is a measurement of the kinetic energy of particles. If the temperatures are equal, the average kinetic energies of the particles are also equal. The kinetic energy of a particle is determined by the equation:

K.E. = ½mv2 where m is the mass and v is the velocity (rate of

diffusion) If the kinetic energies (temperature) of two particles, A and B, are the same, then

½mAvA2 = ½mBvB

2 If the kinetic energies are equal, and if A is lighter than B, then A must have a greater velocity than B. In other words, the molar mass of a gas determines how fast it can diffuse: lighter gases diffuse faster than heavier gases at the same temperature. To compare the rates of diffusion of two gases, rearrange the equation above to get Graham’s Law of Diffusion:

½mAvA2 = ½mBvB

2 mAvA

2 = mBvB2

vA

2 mB vB

2 =

mA vA mB vB

= mA

So, if we know the rate of diffusion of one gas, we can calculate the rate of diffusion of another gas using the molar masses. Example: Neon gas diffuses through a tube at 1 m/min. At what speed will Argon gas diffuse through the same tube?

vNe mAr vAr

= mNe

1 40 vAr

= 20

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1 vAr

= 1.41

vAr = 0.707 m/min

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Chapter 14: Chemical Kinetics and Equilibrium

14.1 Reaction Rates 14.2 Collision Theory 14.3 Potential Energy Diagrams 14.4 Catalysts 14.5 Reaction Mechanisms 14.6 Equilibrium 14.7 The Equilibrium Constant Expression (Kc and Kp) 14.8 The Magnitude of K 14.9 The Reaction Quotient, Q 14.10 The ICE Chart and Equilibrium Problems 14.11 Solubility Product Constant 14.12 Equilibrium Shifts: Le Châtelier’s Principle

14.1 Reaction Rates Some chemical reactions happen faster than others. Trinitrotoluene (TNT) detonates in a fraction of a second; the iron in a car muffler takes years to rust through. The study of the speed of chemical reactions is called chemical kinetics. What do we mean by the speed of a reaction? A rate always involves a change over time; the speed of a car is measured in distance over time: miles per hour (miles/hr). The speed of light is measured in meters per second (m/s). In a chemical reaction, we measure how fast the reactants are used up, or how fast the products are produced. In the case of iron rusting, the reaction is:

4Fe(s) + 3O2(g) → 2Fe2O3(s) To determine the rate of this reaction, we could measure how fast iron or oxygen is used up or how fast rust (Fe2O3) is created. The units would be moles per unit time. A slow reaction could be measured in moles per year or moles per hour. If a reaction happens quickly, it could be moles per second. 14.2 Collision Theory For two molecules to react, they must come in contact with each other—they must collide. But not all collisions are productive; the molecules must collide with the correct orientation. Let’s look at the reaction:

O2(g) + N2(g) → 2NO(g) If the molecules collide like this, nothing happens: If the molecules collide like this, nothing happens:

O O N N

O

O

N

N

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Only with the correct alignment will the reaction happen: With this orientation, the N’s and the O’s are lined up with each other, and it is possible for N–O bonds to form. To increase the reaction rate, we can increase the number of collisions. This is accomplished by packing more molecules into a smaller space; in other words, by increasing the concentration. When the molecules are packed together, they bump into each other more frequently—increasing the concentration increases the reaction rate. Another simple way to speed up a reaction is to increase the surface area of the reactants. For example, a large chunk of baking soda will react slowly with vinegar, because the two substances can only react where they make contact with each other—at the surface. If the baking soda is broken into small particles (powdered), the vinegar will make contact with a lot more of the baking soda and the reaction happens faster. 14.3 Potential Energy Diagrams The molecules must also collide with enough energy to break their old bonds. In the reaction above, the O–O bond and the N–N bond must break; after those bonds break, two N–O bonds form. The reactant molecules O2 and N2 are very stable, and remember that high stability corresponds to low energy. The product molecules (NO) are also stable, and also have low energy. But in the transition between reactants and products, in the instant that old bonds break and new bonds form, there is a high-energy, unstable state called an activated complex. In our reaction, it might look something like this:

O

O N

N

baking soda

slow fast

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O

O N

N O

O N

N O

O N

N We can graph the energy of the chemical species as the reaction progresses on a potential energy diagram: As the diagram shows, the reactants require a certain amount of energy to reach the activated complex—this amount of energy is called the activation energy (Ea) of the reaction. If the two molecules do not collide with enough energy to overcome the activation energy and make it to the activated complex, then no products are formed; the collision is not productive. Notice also that the products have less potential energy than the reactants. This means that the reaction gives off energy to the surroundings; such a reaction is called exothermic.

reactants

activated complex

products

reactants activated complex products

Ea

∆H

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If the products end up at a higher energy than the reactants, the reaction takes in energy from the surroundings and is called endothermic. The difference in energy between the products and reactants is called the enthalpy of reaction, and is represented by ∆H. (We encountered two specialized ∆H values in chapter 11: ∆Hfus and ∆Hvap, and we will learn more about ∆H in chapter 16). So the graph can tell us two things: the activation energy and the ∆H value. How do we give the reactants enough energy to overcome Ea? Heat them up! When the reactants are at a higher temperature, more of them will have enough energy to overcome the activation energy—thus increasing the temperature increases the reaction rate. 14.4 Catalysts Another way to increase the reaction rate is to add a catalyst to the reaction. A catalyst works by stabilizing the activated complex, thereby lowering the activation energy:

Ea

∆H

lower activation energy due to catalyst

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With a lower Ea, more reactants can make it over the hump, and the reaction goes faster. 14.5 Reaction Mechanisms Most chemical reactions actually occur in a sequence of reactions instead of all at once. These intermediate reactions are called elementary steps. Consider the following reaction:

CO + NO2 → CO2 + NO This reaction actually takes place in two elementary steps:

2NO2 → NO3 + NO CO + NO3 → NO2 + CO2

The first elementary step produces NO3 and NO, while the second combines the NO3 with CO to make NO2 and CO2. Notice that NO3 is produced in one step and immediately consumed by the next. We therefore call NO3 an intermediate—it is produced and consumed in the course of a reaction, but it does not appear in the overall equation. The elementary steps in a mechanism never happen at the same rate. In the example above, the first step happens slowly and the second quickly. The slowest step is called the rate-determining step because it limits the rate of the overall reaction. It’s similar to a production line in a factory; the slowest worker determines the overall rate of production. No matter how fast the other workers are, the slowest person determines the rate. 14.6 Equilibrium So far we have assumed that chemical reactions can only move from left to right—from reactants to products. However, many chemical reactions are reversible. For example, N2 and H2 can react together to form NH3:

N2 + 3H2 → 2NH3 And NH3 can also decompose back into N2 and H2:

2NH3 → N2 + 3H2 Instead of writing both equations, we can write:

N2 + 3H2 ⇄ 2NH3

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The double arrow doesn’t just mean that both the forward and the reverse reactions can happen; it means the forward and the reverse reactions are happening at the same time. How is this possible? Let’s look at the simplest reaction possible, where reactants (R) turn into products (P), and vice-versa:

R(aq) ⇄ P(aq) Imagine the reaction starts out with a certain amount of reactant and zero product. The concentration of the reactant is high at first, while the concentration of the product is zero. The reaction proceeds forward (to the right), and product starts to form. As the reactants get used up, the forward reaction slows down because the concentration of the reactants decreases; remember that according to collision theory, lowering the concentration decreases the reaction rate:

As more and more product forms, the product concentration increases, and so the reverse reaction speeds up:

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Eventually, the rate of the forward reaction will slow down enough, and the reverse reaction will speed up enough, so that the two reactions will happen at the same rate:

This state is called equilibrium. At equilibrium, the reaction appears to have stopped, because the concentrations of reactants and products are constant. They are constant not because the reactions have stopped, but because the forward and reverse reactions are going at the same rate. At equilibrium, P is always being formed by the forward reaction, and it is always being destroyed by the reverse reaction. Because the rates of formation and destruction are equal, the concentration of P remains constant. The same is true for R; the forward reaction destroys it, the reverse reaction creates it, but its concentration is constant because it is formed and destroyed at the same rate. In the following graph, we see that the concentration of reactant decreases in the beginning and then remains constant once equilibrium has been established. Likewise, the concentration of product increases initally and then stabilizes. Note that once equilibrium is reached, the concentrations are not equal.

159

At equilibrium: • The concentrations of all species are constant. • The concentrations of the reactants and the products are hardly ever

equal. • Both the forward and the reverse reaction continue; they do not stop. • The rate of the forward reaction equals the rate of the reverse reaction.

14.7 The Equilibrium Constant Expression (Kc and Kp) The state of equilibrium can be reached from many different starting concentrations of reactants and products. Let’s start the following reaction with only reactants: 1 M of R, and 0 M of P.

R(aq) ⇄ P(aq) The reaction then proceeds until equilibrium is reached. At equilibrium, the concentrations are 0.8 M R and 0.2 M P. Now let’s say we start with the concentrations reversed: 0 M R and 1 M P. The reaction proceeds again until it reaches equilibrium. At equilibrium, we find again: 0.8 M R and 0.2 M P. The reaction can begin with any concentrations of R and P. Here’s what happens when many different starting concentrations are tried:

Starting Concentrations

Equilibrium Concentrations

R P R P 1 0 0.8 0.2 0 1 0.8 0.2

1.5 1 2 0.5 500 20 416 104 333 45 302.4 75.6

Notice that no matter what the starting concentrations are, at equilibrium the ratio of P to R is always the same: it is always 1:4 or 0.25. This ratio is called the equilibrium constant, or Kc. It is a ratio of the concentrations of the products to the ratio of reactants. It doesn’t matter how much product or reactant we start with; at equilibrium, the ratio of the products to the reactants will always equal Kc.

[P] Kc =

[R]

160

The brackets mean molarity; “[P]” means “the molarity of P”. This formula for Kc only applies to the reaction P ⇄ R. To generalize the formula for Kc to any reaction, let’s take the generic reaction:

aA + bB ⇄ cC + dD (The lowercase letters are the coefficients.) For this reaction,

[C]c[D]d Kc =

[A]a[B]b Notice that the coefficients become superscripts in the equation. For example, take the reaction:

Ca(OH)2(aq) ⇄ Ca+2(aq) + 2OH-(aq)

[Ca+2][OH-]2 Kc =

[Ca(OH)2] If the chemicals are in the gas phase, use partial pressure values instead of concentration. Instead of Kc, we use Kp for gases. For example:

3H2(g) + N2(g) ⇄ 2NH3(g)

PNH32

Kp = PH2

3PN2 Liquids and solids are never included in an equilibrium constant expression, because they do not affect the rates of chemical reactions. For example:

H2CO3(aq) ⇄ H2O(l) + CO2(aq)

[CO2] Kc = [H2CO3]

Notice that the liquid water is not included in the equilibrium expression. 14.8 The Magnitude of K There values of K have a huge range: K can be lower than 1 × 10-20 and higher than 1 × 1020. What does the magnitude mean? Remember that the K is a ratio of products to reactants at equilibrium. So if K is very high, the concentration of products is much higher than the concentration of reactants; we say the

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equilibrium “lies to the right” or “favors the products”. If K is very low, the concentration of reactants is much higher than the concentration of products; we say the equilibrium “lies to the left” or “favors the reactants”. 14.9 The Reaction Quotient, Q When a chemical reaction is not at equilibrium, it naturally reacts until equilibrium is reached. With too much product, the reaction will “shift to the left” to reach equilibrium. With too much reactant, the reaction will “shift to the right” to reach equilibrium. It’s not always easy to determine which way a certain combination of products and reactants will shift; it depends on their specific concentrations and the value of K. Fortunately, it’s easy to predict mathematically with the reaction quotient, or Q. Q has the same formula as K, only the concentrations plugged in are the actual, current concentrations, not the concentrations that will exist at equilibrium. For example:

HC2H3O2(aq) ⇄ H+(aq) + C2H3O2-(aq) Kc = 1.8 × 10-5

Let’s say we make a solution so that:

[HC2H3O2] = 0.5 M [H+] = [C2H3O2-] = 0.3 M

Which direction will this reaction go, if these are the starting concentrations? First, calculate Qc:

[H+][C2H3O2-] 0.3(0.3)

Qc = [HC2H3O2]

= 0.5

= 0.18

Because this value is higher than the Kc value, we conclude that the current concentration of the products is too high; thus the reaction will shift to the left to attain equilibrium. The rules are:

If Q > K, reaction shifts to the left. If Q < K, reaction shifts to the right. If Q = K, reaction is already at equilibrium; no shift occurs.

14.10 The ICE Chart and Equilibrium Problems We first encountered the ICE chart in chapter 5. It is a powerful tool for solving problems using K values. For this type of problem “ICE” stands for “initial, change, equilibrium” instead of “initial, change, end”. Let’s say we make a 1 M HF solution. The HF then dissociates according to the following reaction:

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HF(aq) ⇄ H+(aq) + F-(aq) Kc = 6.8 × 10-4 Now we need to calculate the concentrations of all species at equilibrium. Before the reaction starts, the initial concentrations of H+ and F- are zero, so the chart begins:

HF(aq) ⇄ H+(aq) + F-(aq) initial 1 0 0

To reach equilibrium, the reaction must proceed to the right; a certain amount of HF will dissociate forming H+ and F-, but we don’t know exactly how much (yet). So we’ll call this amount “x”. The concentration of HF will decrease by x, and the concentration of H+ and F- will both increase by x. Therefore we add to our chart:

HF(aq) ⇄ H+(aq) + F-(aq) initial 1 0 0

change -x +x +x The equilibrium concentrations must then be:

HF(aq) ⇄ H+(aq) + F-(aq) initial 1 0 0

change -x +x +x equil. 1 – x x X

So at equilibrium:

[HF] = 1 – x [H+] = [F-] = x We can then plug these values into the Kc expression to solve for x:

[H+][F-] x(x) x2 Kc = 6.8 × 10-4 =

[HF] =

1 – x =

1 – x

x2 6.8 × 10-4 = 1 – x

6.8 × 10-4 - 6.8 × 10-4(x) = x2

x2 + 6.8 × 10-4(x) - 6.8 × 10-4 = 0

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What we have here is a quadratic equation, which is no fun to solve. If we do solve it, it yields:

x = 0.0257 M 1 – x = 0.9743 M However, if we backtrack a few steps, we can make a useful approximation that simplifies the calculation considerably. Notice that the value of x turned out to be very small. We could have predicted this because the Kc value of this equation is very low; we know that only a small quantity of the HF will dissociate. In other words, the value of x will be very small compared to the initial concentration of HF. Therefore, 1 – x will be approximately 1. So:

x2 6.8 × 10-4 = 1 – x

becomes

x2 6.8 × 10-4 ≈ 1

And we can solve for x and 1 – x:

x ≈ 0.0261 1 – x ≈ 0.9379 Compare this to the value of x found by solving the quadratic equation, and the % error is only 1.6%. Sometimes the same type of problem is done in reverse; the equilibrium concentrations are given in the problem, and we must find the K value. For example, let’s say a 0.250 M solution of butyric acid reacts as follows:

HC4H7O2(aq) + H2O(l) ⇄ H3O+(aq) + C4H7O2-(aq)

At equilibrium, it is found that [H3O+] = 1.9 × 10-3 M. Calculate Kc for this reaction. Here’s what we know at the outset:

HC4H7O2(aq) + H2O(l) ⇄ H3O+(aq) + C4H7O2-(aq)

initial 0.250 0 0 change equil. 1.9×10-3

164

The stoichiometry of the equation tells us that the amount of H3O+ formed must be equal to the amount of C4H7O2

- formed. And we know that this is also the amount of HC4H7O2 that was consumed in the reaction. Notice we ignore the water; because it is a liquid, it will not appear in our equilibrium expression.

HC4H7O2(aq) + H2O(l) ⇄ H3O+(aq) + C4H7O2-(aq)

initial 0.250 0 0 change -1.9×10-3 +1.9×10-3 +1.9×10-3 equil. 0.2481 1.9×10-3 1.9×10-3

Now that we have the equilibrium concentrations, we can calculate the Kc:

[H3O+][C4H7O2-] 1.9×10-3(1.9×10-3)

Kc = [HC4H7O2

-] =

0.2481 = 1.46 × 10-5

14.11 Solubility Product Constant Every reaction that reaches equilibrium has a K expression and K value, but for certain types of reactions the K value is given a special name. The equilibrium equations for weak acids and bases have Ka and Kb values, respectively (see next chapter). The K expression for a reaction that describes the dissolving of a substance is called the Ksp (the solubility product constant). This reaction will reach equilibrium only in a saturated solution, when some of the solid remains undissolved. For example, in a saturated solution of MgF2, the following equilibrium exists:

MgF2(s) ⇄ Mg+2(aq) + 2F-(aq)

Ksp = [Mg+2][F-]2 = 5.16 × 10-11 Remember, solids do not appear in equilibrium expressions. 14.12 Equilibrium Shifts: Le Châtelier’s Principle We have already seen in our discussion of Q that when the concentration of products is too large, the reaction shifts to the left, and when the concentration of reactants is too large, the reaction shifts to the right. When a reaction is not at equilibrium, it shifts right or left to reestablish equilibrium. There are a number of ways in which equilibrium can be disturbed: Adding or Removing a Product or Reactant The following reaction starts out at equilibrium:

A(aq) ⇄ B(aq)

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If we change the concentration of the product or the reactant, the equation will no longer be at equilibrium. If we add more A, for example, the reaction must shift to the right to regain the equilibrium ratio. If we add B, the reaction shifts left. If we remove some A, the reaction shifts left. If we remove B, the reaction shifts right. If the reaction includes solids or liquids, adding or subtracting them will NOT shift the equilibrium because they do not appear in the equilibrium expression. Change in Temperature As discussed in section 14.3, exothermic reactions release energy, and endothermic reactions absorb energy. Because exothermic reactions give off heat, we can consider heat a product of the reaction, and we can even include it in the chemical equation for convenience:

A(aq) ⇄ B(aq) + heat If the reaction is endothermic, we consider heat a reactant:

heat + A(aq) ⇄ B(aq) It is now possible to predict the direction of shift when the reaction is heated or cooled:

• If an exothermic reaction is heated, the reaction shifts left to counteract the addition of heat.

• If an exothermic reaction is cooled, the reaction shifts right to counteract the removal of heat.

• If an endothermic reaction is heated, the reaction shifts right to counteract the addition of heat.

• If an endothermic reaction is cooled, the reaction shifts left to counteract the removal of heat.

Change in Pressure Pressure only affects reactions involving gases; it has no effect on other reactions. When pressure on the system increases, the reaction will shift to counteract the increase in pressure; it does this by shifting toward the side of the reaction with fewer moles of gas. For example:

A(g) ⇄ B(g) + 2C(g) In this reaction, there is one mole of gas on the left and three moles of gas on the right. If the pressure increases, the reaction will try to decrease the pressure by decreasing the amount of gas—it will shift to the left. Conversely, if the

166

pressure decreases, the reaction will try to increase the pressure by increasing the amount of gas—it will shift to the right. Catalysts As we saw in section 14.4, catalysts lower the activation energy of a reaction and thereby speed it up. At equilibrium, both the forward and reverse reactions are happening at the same rate. Adding a catalyst speeds up both reactions, and therefore the equilibrium balance is maintained. Catalysts do not cause shifts in equilibrium.

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Chapter 15: Acids and Bases

15.1 The Autoionization of Water 15.2 The pH Scale 15.3 Definitions 15.4 Strong vs. Weak Acids and Bases 15.5 Problems with ICE Charts, pH, and Ka or Kb 15.6 Indicators 15.7 Titration 15.8 Buffers 15.9 Titration Curves of Weak Acids and Polyprotic Acids

15.1 The Autoionization of Water Water is always undergoing a reversible autoionization, where H2O molecules split into H+ and OH- ions. Water maintains the following equilibrium:

H2O(l) ⇄ H+(aq) + OH-(aq) Kw = 1.0 × 10-14 Kw is the equilibrium constant of this reaction; it is called the ionization constant of water. The equilibrium expression is:

Kw = 1.0 × 10-14 = [H+][OH-] No matter what other chemicals are mixed with the water, this equation always holds true! In pure water, what are the concentrations of H+ and OH-? We need to use an ICE chart to figure this out:

H2O(l) ⇄ H+(aq) + OH-(aq) I N/A 0 0 C -x +x +x E N/A-x x x

Kw = 1.0 × 10-14 = [H+][OH-] = x(x) = x2 x = 1 × 10-7

M So in pure water, [H+] = [OH-] = 1 × 10-7

M If something is added to the water that changes the H+ concentration, the OH- must also change to maintain the equilibrium. For example, if the H+ concentration increases to 0.0001 M, what will the OH- concentration be? Use the formula for Kw:

Kw = 1.0 × 10-14 = [H+][OH-]

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1.0 × 10-14 = (0.0001)[OH-] [OH-] = 1 × 10-10

As you can see, the OH- concentration has decreased in response to the increased H+ concentration. Conversely, increasing the OH- concentration decreases the H+ concentration. If we make a solution where the OH- concentration is 1 × 10-5 M, what will the H+ concentration be?

Kw = 1.0 × 10-14 = [H+][OH-] 1.0 × 10-14 = [H+](1 × 10-5) [H+] = 1 × 10-9

What does any of this have to do with acids and bases? If a solution has more H+ than OH-, we it call it acidic. If it has more OH- than H+, we call it basic. If the concentrations are equal, we call it neutral. 15.2 The pH Scale To describe the acidity or basicity of a solution, we could simply give the H+ concentration or OH- concentration. But the range of possible H+ and OH- concentrations is huge; they can be well over 1 M and far less than 1 × 10-14 M. Because of this large range, chemists avoid using molarity; they’ve developed another way of expressing the concentration of H+ and OH-: the pH scale. The pH of a solution is the negative logarithm of the H+ concentration:

pH = -log[H+] For example, if the H+ concentration of a solution is 2.2 × 10-4, the pH is:

pH = -log[H+] = -log(2.2 × 10-4) = 3.66 The pH scale is a bit counterintuitive: Acidic solutions (high H+ concentration) have a low pH. Basic solutions (low H+ concentration) have high pH. If we know the pH, we can work in reverse to calculate the H+ concentration. For example, if a solution has a pH of 8.4, what is the H+ concentration?

pH = -log[H+] 8.4 = -log[H+] -8.4 = log[H+] [H+] = 10-8.4 = 3.98 × 10-9 M

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Although it’s not used very much, the pOH of a solution is defined as the negative logarithm of the OH- concentration:

pOH = -log[OH-] The following table contains a range of H+ and OH- values. Notice that because [H+][OH-] = 1.0 × 10-14, when H+ is high, OH- is low. As a result of this, when pH is high, pOH is low, and vice versa.

[H+] pH pOH [OH-]

1 (1 × 100) 0 14 0.00000000000001 (1 × 10-14)

0.1 (1 × 10-1) 1 13 0.0000000000001 (1 × 10-13)

0.01 (1 × 10-2) 2 12 0.000000000001 (1 × 10-12)

0.001 (1 × 10-3) 3 11 0.00000000001 (1 × 10-11)

0.0001 (1 × 10-4) 4 10 0.0000000001 (1 × 10-10)

0.00001 (1 × 10-5) 5 9 0.000000001 (1 × 10-9)

0.000001 (1 × 10-6) 6 Aci

dic

8 0.00000001 (1 × 10-8)

0.0000001 (1 × 10-7) 7 neutral 7 0.0000001 (1 × 10-7)

0.00000001 (1 × 10-8) 8 6 0.000001 (1 × 10-6)

0.000000001 (1 × 10-9) 9 5 0.00001 (1 × 10-5)

0.0000000001 (1 × 10-10) 10 4 0.0001 (1 × 10-4)

0.00000000001 (1 × 10-11) 11 3 0.001 (1 × 10-3)

0.000000000001 (1 × 10-12) 12 2 0.01 (1 × 10-2)

0.0000000000001 (1 × 10-13) 13 1 0.1 (1 × 10-1)

0.00000000000001 (1 × 10-14) 14

Basi

c

0 1 (1 × 100)

Another useful equation can be derived from [H+][OH-] = 1.0 × 10-14:

[H+][OH-] = log([H+][OH-]) =

log[H+] + log[OH-] = -log[H+] + -log[OH-] =

pH + pOH =

1.0 × 10-14

log(1.0 × 10-14) -14 14 14

Thus, it is easy to calculate the pOH if the pH is known, and vice-versa. Because the H+ concentration in pure water is 1 × 10-7, the pH of pure water is:

pH = -log[H+] = -log(1 × 10-7) = 7 Therefore, pH = 7 is neutral. Anything with a pH under 7 is acidic; anything with a pH over 7 is basic.

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15.3 Definitions Alchemists and chemists have known acids and bases for centuries. Not surprisingly, the precise definitions of “acid” and “base” have changed over time. In the 1880’s, Swedish chemist Svante Arrhenius gave us the following definitions, which are still useful today: Arrhenius Acid: anything that dissociates to release H+ ions in water:

HCl(g) → H+(aq) + Cl-(aq) Note: After the H+ ions are released, they are carried by water in the form of hydronium ions: H3O+. The reaction that forms the hydronium ions is: H+(aq) + H2O(l) → H3O+(aq) For this reason, H+(aq) and H3O+(aq) are often used interchangeably. Arrhenius Base: anything that dissociates to release OH- ions in water:

NaOH(s) → Na+(aq) + OH-(aq) How can we recognize an Arrhenius acid or an Arrhenius base? Acids usually have the H’s written in the front of the formula: HCl, H2SO4, HBr, etc. Bases are ionic compounds containing hydroxide ions: NaOH, KOH, Ca(OH)2, etc. Hydrogen ions and hydroxide ions are very reactive; both can cause chemical burns. However, acids and bases neutralize each other. The H+ ions combine with the OH- ions to form harmless water:

H+(aq) + OH-(aq) → H2O(l) Another way of looking at neutralization is to see the acids and bases react with each other directly:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + H2O(l)

2H3PO4(aq) + 3Mg(OH)2(aq) → Mg3(PO4)2(aq) + 3H2O(l) In all of these reactions, water is formed along with a salt (in chemistry, the term “salt” means an ionic compound). The next, more general way of defining acids and bases, was proposed by Johannes Brønsted and Thomas Lowry in 1923:

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Brønsted-Lowry Acid: something that donates H+ ions. Brønsted-Lowry Base: something that accepts H+ ions. [Because an H+ ion is nothing more than a proton (the hydrogen atom consists of one proton orbited by one electron), Brønsted-Lowry acids are “proton donors” and bases are “proton acceptors.”]

H2SO4(aq) + NH3(aq) → HSO4-(aq) + NH4

+(aq) In the above reaction, H2SO4 donates a proton to NH3: H2SO4 is acting as a Brønsted-Lowry acid and NH3 is acting as a Brønsted-Lowry base. H2O(l) + CO3

-2(aq) → OH-(aq) + HCO3-(aq)

In the above reaction, H2O donates a proton to CO3

-2: H2O is acting as a Brønsted-Lowry acid and CO3

-2 is acting as a Brønsted-Lowry base. Note the differences between this definition and Arrhenius’s. First, the Brønsted-Lowry definition does not specify water as a solvent; the acids and bases could be in any phase of matter. Second, while a Brønsted-Lowry acid still must have hydrogen in it, a Brønsted-Lowry base need not have hydroxide, it must simply accept a proton from an acid. Brønsted-Lowry acids and bases come in conjugate acid-base pairs. For example:

HCl(aq) + C2H3O2-(aq) → Cl-(aq) + HC2H3O2(aq)

In this equation, HCl donates a proton (acid) and the acetate ion accepts it (base). If we reverse this reaction we get:

Cl-(aq) + HC2H3O2(aq) → HCl(aq) + C2H3O2-(aq)

In this case, the HC2H3O2 is the proton donor (acid), and the Cl- is the proton acceptor (base). So we have:

HCl = conjugate acid of Cl- Cl- = conjugate base of HCl HC2H3O2 = conjugate acid of C2H3O2

- C2H3O2- = conjugate base of HC2H3O2

These are conjugate acid-base pairs. We can tell them apart because, while they look similar, the acid always has the proton (H), while the base lacks it.

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Sometimes a molecule is capable of either donating or accepting a proton; such a molecule is called amphoteric. The bicarbonate ion is an example; in the first reaction, it donates a proton to water, and in the second equation, it accepts a proton from water:

HCO3

- + H2O → CO3-2 + H3O+

HCO3

- + H2O → H2CO3 + OH- Notice that in the above equations, water is also acting as an amphoteric molecule. When a Brønsted-Lowry acid is added to water, the water accepts a proton:

HF(aq) + H2O(l) ⇄ H3O+(aq) + F-(aq) When a Brønsted-Lowry base is added to water, the water donates a proton:

NH3(aq) + H2O(l) ⇄ NH4+(aq) + OH-(aq)

This is how a Brønsted-Lowry base generates OH- ions without containing them itself (like an Arrhenius base). It is important to know how acids and bases behave when they dissolve in water, as they are usually found in aqueous solution. Acids increase the H+ concentration by dissociating, releasing H+ ions into solution:

HA(aq) → H+(aq) + A-(aq) Bases increase the OH- concentration in one of two ways. Compounds containing hydroxide ions dissociate:

BOH(aq) → B+(aq) + OH-(aq) Other bases accept H+ from water molecules, creating OH-:

B(aq) + H2O(l) → BH+(aq) + OH-(aq)

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15.4 Strong vs. Weak Acids and Bases The H+ (H3O+) produced by acids is the reactive part of an acidic solution; it’s what burns things, what reacts with other chemicals. If one acid can make more H+ than another, we say it is a stronger acid. Consider two acids, HX and HY. HX is a strong acid, and HY is a weak acid. When HX dissolves in water, the following reaction occurs:

HX(aq) → H+(aq) + X-(aq) The HX dissociates completely. By the stoichiometry of this reaction, 1 mole of HX will produce 1 mole of H+ ions. But when HY dissolves in water, it does not dissociate completely; an equilibrium is established, and only a small amount of H+ is formed:

HY(aq) ⇄ H+(aq) + Y-(aq) Ka = 1.5 × 10-4 Therefore, 1 mole of HY only produces a small amount of H+. (We will figure out the exact amount using an ICE chart later in the chapter.) Because strong acids dissociate completely, they produce the maximum possible amount of H+. Because weak acids don’t dissociate completely, they reach an equilibrium and produce small amounts of H+. The strength of a weak acid (the degree of dissociation) is represented by its Ka value (which is simply the Kc of dissociation reaction); the larger the Ka, the stronger the acid. There are seven common strong acids to remember; assume any other acid is a weak acid. HCl, hydrochloric acid HBr, hydrobromic acid HI, hydroiodic acid HNO3, nitric acid H2SO4, sulfuric acid HClO3, chloric acid HClO4, perchloric acid The strong bases are the hydroxides of alkaline metals and the alkaline earth metals, excluding Be(OH)2 and Mg(OH)2. These are named according to the standard ionic compound rules: sodium hydroxide, calcium hydroxide, etc. Weak bases have a Kb value, which is the Kc of the base accepting a proton from water: B(aq) + H2O(l) ⇄ HB+(aq) + OH-(aq). The larger the Kb, the stronger the base. Please remember that “strong” and “weak” refer only to the degree of ionization, not the concentration of a solution. We can have a solution of a strong acid with a low concentration (0.001 M HCl) and a solution of a weak acid with a high concentration (6 M HC2H3O2).

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15.5 Problems with ICE Charts, pH, and Ka or Kb Example: Calculate the pH of a 0.8 M solution of a weak acid with Ka = 3 × 10-6.

First, write the chemical equation for the dissociation of the acid, and use an ICE chart:

HA(aq) ⇄ H+(aq) + A-(aq) I 0.8 0 0 C -x +x +x E 0.8-x x x

Plug the equilibrium values into the Ka expression:

Ka = 3 × 10-6 = x2/(0.8 – x)

Because the Ka is small, x will be very small compared to 0.8, and therefore the equation can be simplified to:

3 × 10-6 = x2/0.8

Solve for x = 1.55 × 10-3 pH = 2.81

Example: A 0.5 M solution of a weak acid has a pH of 3.5. Calculate the Ka of the acid.

First, calculate the H+ concentration from the pH: 10-3.5 = 3.16 × 10-4 M Then fill in an ICE chart:

HA(aq) ⇄ H+(aq) + A-(aq) I 0.5 0 0 C 0.5 - 3.16 × 10-4 +3.16 × 10-4 +3.16 × 10-4 E ~0.5 3.16 × 10-4 3.16 × 10-4

Plug the equilibrium values into the Ka expression:

Ka = (3.16 × 10-4)2/0.5 = 2.0 × 10-7

Example: A solution of a weak acid with Ka = 8 × 10-8 has a pH of 3.4. What is the concentration of the solution?

First, calculate the H+ concentration = 10-3.4 = 3.98 × 10-4 M

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Plug the known values into an ICE chart:

HA(aq) ⇄ H+(aq) + A-(aq) I x 0 0 C x - 3.98 × 10-4 +3.98 × 10-4 +3.98 × 10-4 E ~x 3.98 × 10-4 3.98 × 10-4

Plug the equilibrium values into the Ka expression:

Ka = 8 × 10-8 = (3.98 × 10-4)2/x x = 1.98 M

15.6 Indicators One way to approximate a solution’s pH is to use indicators. An indicator is a substance that changes color depending on the pH. Indicators are weak acids that undergo the reaction:

HInd(aq) ⇄ H+(aq) + Ind-(aq) (red) (colorless)

The HInd molecule has a different color than the Ind- ion. For example, HInd may be red, while Ind- is colorless. If the reaction is shifted to the left, the solution has a high concentration of HInd, and the solution appears red. If the solution is shifted to the right, there will be a high concentration of Ind-, and the solution is clear. This reaction shifts, according to Le Châtelier’s principle, with the addition of acid or base. For example, if the pH is low (high concentration of H+), the reaction shifts to the left, and the solution turns red. If the pH is high (low concentration of H+), the reaction shifts to the right, and the solution is clear.

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Different indicators change color at different pH levels:

Indicator Color

changes at pH of:

Color on the Acid

Side

Color on the Base

Side Thymol Blue 2 red Yellow Methyl yellow 3.5 red Yellow Bromphenol blue 3.8 yellow blue-violet Alizarin sodium sulfonate 4.5 yellow Violet Bromcresol green 4.8 yellow Blue Methyl red 5.3 red Yellow Bromcresol purple 6 yellow Purple Phenol red 7.2 yellow Red Thymol blue 8.8 yellow Blue Phenolphthalein 9 colorless Red Thymolphthalein 10 colorless Blue Alizarin yellow 11 yellow Lilac Trinitrobenzoic acid 12.7 colorless orange-red

(Thymol blue is listed twice because it changes color twice over the pH range. Under pH 2, it is red; between 2 and 8.8 it is yellow; above 8.8 it is blue.) We can zero in on a pH by testing with multiple indicators. For example, if a solution is tested with methyl red and it turns red, the pH must be less than 5.3. If we then test it with bromcresol green, and it turns blue, the pH is above 4.8. This narrows down the pH to somewhere between 4.8 and 5.3. Indicators are a cheap but imprecise way to measure pH. A pH meter is more accurate and easier to use—stick a probe in the solution and a digital readout gives the pH. 15.7 Titration Titration is an analytical technique for determining the concentration of a solution. One of the most common types of titration is an acid-base titration. Let’s say that we need to determine the concentration of a solution of HCl. First, we would set up an apparatus as shown at right. The ring stand holds a buret: a graduated tube from which liquid can be dispensed by opening the valve at the bottom. A certain volume (let’s say 80 mL) of unknown HCl is in the Erlenmeyer flask, and the buret holds a basic solution of known concentration; in this case, we’ll use 1 M NaOH. When the valve is opened, the NaOH drips into the HCl and the following neutralization reaction occurs:

NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

buret valve

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If we monitored the pH of the solution in the Erlenmeyer flask as more and more NaOH was added, we would get a titration curve:

Before any NaOH has been added, the pH is about 0.5. As more NaOH is added, it gradually neutralizes the HCl and the pH slowly rises. The pH rises rapidly around the equivalence point—when exactly enough NaOH has been added to neutralize all of the HCl. As excess NaOH is added beyond this point, the pH rises further. In our example, the equivalence point is reached when 24 mL (0.024 L) of NaOH have been added. Because the concentration of the NaOH was 1 M, we calculate that this contained 0.024 mol of NaOH. Looking back at the neutralization reaction, we see that the NaOH and HCl react in a 1-to-1 ratio. We therefore know that the sample of HCl also must have contained 0.024 moles of HCl. Because the original HCl solution had a volume of 80 mL (0.08 L), we can then calculate that its concentration was 0.024 moles/0.08 L = 0.3 M, and the problem is solved. Instead of monitoring the pH and generating a graph to find the equivalence point, it is much more common to add a pH indicator to the Erlenmeyer flask that changes color when the equivalence point is reached. As soon as the solution changes color, we know we’ve reached the equivalence point, and we determine how much NaOH has been added to HCl. Example: You wish to discover the concentration of an HCl solution using titration. You place 100 mL of the HCl in an Erlenmeyer flask and add a drop of phenolphthalein (an indicator). You titrate the acid using 1 M NaOH. The solution turns pink after 25 mL of NaOH have been added. What was the concentration of the acid?

equivalence point

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Remember that NaOH reacts with the HCl as follows:

NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

First, calculate the amount of NaOH added:

25 mL = 0.025 L 0.025 L × 1 M = 0.025 mol NaOH

Because the NaOH and the HCl react in a 1-to-1 ratio (see equation above), we know that the 100 mL of HCl contained 0.025 mol of HCl. Now find the concentration of the HCl:

0.025 mol/0.100 L = 0.25 M HCl

To solve titration problems quickly, we can derive a simple equation. The key to a titration is this: at the equivalence point, the moles of OH- added equals the moles of H+ in the starting solution:

moles OH- added = moles H+ in the starting solution The moles can be expressed as the volume of the solution times the concentration of the solution, or MV:

MBVB = MAVA

Where MB is the molarity of the base, VB is the volume of base added, MA is the molarity of the acid, and VA is the volume of the acid.

The situation is slightly more complicated when dealing with a polyprotic acid (an acid with multiple protons, like H2SO4 and H3PO4) or with a base with multiple hydroxide ions (like Ca(OH)2). If the acid is diprotic, it has 2 moles of H+ in every mole of acid. Therefore the moles of H+ is two times the moles of acid. Likewise, Ca(OH)2 has two moles of OH- for every mole of base. To generalize the formula, we must take these facts into account:

nBMBVB = nAMAVA

Where nB is the number of hydroxide ions (OH-) in the base and nA is the number of protons (H+) on the acid.

Example: You wish to discover the concentration of an H3PO4 solution using titration. You place 100 mL of the H3PO4 in an Erlenmeyer flask and add a drop of phenolphthalein. Using a buret, you titrate the acid using 1 M Ca(OH)2. The

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acid solution turns pink after 40 mL of Ca(OH)2 have been added. What was the concentration of the acid?

Using our equation:

nBMBVB = nAMAVA 2(1)(0.040) = 3(MA)(0.100)

MA = 0.267 M 15.8 Buffers In chemical and biological systems, it is often useful to maintain a solution at a certain pH level. For example, human blood has a pH of approximately 7.4, and a change of 1 pH in either direction is deadly. The blood keeps the pH at the proper level using a buffer: a solution that contains roughly equal quantities of a weak acid and its conjugate base. For example, the buffer system in blood is H2CO3/HCO3

-:

H2CO3 ⇄ H+ + HCO3-

When a person exercises vigorously, her muscles release lactic acid, which adds H+ to her bloodstream. Without a buffer, this added acid would catastrophically decrease the pH of her blood. However, the extra H+ combines with the HCO3

-, and the reaction shifts to the left (according to Le Châtelier’s principle). The H+ concentration is hardly affected. Likewise, if something basic is added to the bloodstream, the H+ concentration decreases, and the equilibrium shifts to the right to compensate.

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15.9 Titration Curves of Weak Acids and Polyprotic Acids When a weak acid is titrated, the titration curve differs from a strong acid in subtle ways. The first difference is seen in the beginning of the weak acid’s curve; there is a slight increase in the pH right at the beginning, whereas the strong acid’s curve is flat. The second difference is that the equivalence point of a strong acid always occurs at a pH of 7; the equivalence points of weak acids can occur at a wide range of pH values.

The titration curve for a polyprotic acid (diprotic) is shown below. Because the acid has two protons, the curve has two equivalence points. The protons are neutralized in two steps, not at the same time. For example, when H2CO3 is titrated, the first proton reacts with the OH- in the following reaction:

H2CO3 + NaOH → HCO3- + H2O

Once this first reaction is complete (the first equivalence point is reached), then the base reacts with the HCO3

- until the second equivalence point is reached:

HCO3- + OH- → CO3

-2 + H2O

weak acid

strong acid

The weak acid’s curve increases right at the beginning.

The weak acid’s equivalence point is not at pH 7.

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The first equivalence point

The second equivalence point

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Chapter 16: Thermochemistry

16.1 Enthalpy 16.2 Estimating ∆H with Bond Enthalpies 16.3 Hess’s Law 16.4 Standard Enthalpy of Formation 16.5 Spontaneous Chemical Reactions 16.6 Gibb’s Free Energy

16.1 Enthalpy Every chemical system possesses a certain amount of potential energy called enthalpy (represented by H). Enthalpy is measured in joules or kilojoules. If a chemical or physical change occurs, the enthalpy of the system also changes, and energy is either released to the surroundings or absorbed from the surroundings. This enthalpy change is represented by ∆H, which is called the heat of reaction or enthalpy of reaction. As we discussed in chapter 14, if the products of a reaction have lower enthalpy than the reactants, the system releases energy to the surroundings, and we call the reaction exothermic. Because the system has lost energy, the ∆H value will be negative. Exothermic reactions will warm up their containers. The physical changes of condensation and freezing are also exothermic; as the particles lose kinetic energy (by slowing down), the energy is absorbed by the surroundings. This is why steam causes bad burns; if steam comes into contact with your hand it condenses, releasing a lot of heat, which is then absorbed by your hand. If the products of a reaction have higher enthalpy than the reactants, the system absorbs energy from the surroundings, and we call the reaction endothermic. Because the enthalpy of the system has increased, the ∆H value will be positive. Endothermic reactions will cool down their containers by pulling energy away from it. A cold pack for first aid contains an endothermic chemical reaction; it draws heat away from a sprained ankle to reduce swelling. Melting, boiling, and evaporation are endothermic processes; our sweat draws heat from our bodies as it evaporates, and chocolate feels cool in the mouth as it melts.

Exothermic Energy released to surroundings

Reaction vessel gets warmer

Products have less enthalpy than reactants

∆H is negative

Endothermic Energy absorbed from surroundings

Reaction vessel cools down

Products have more enthalpy than reactants

∆H is positive

Determining ∆H with a Calorimeter Many reactions we do in lab take place in aqueous solution. If the solution heats up as the reaction progresses, we know that it is an exothermic reaction; if the

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solution cools, we know it is endothermic. But how can we quantify the energy change? If we perform the reaction in an insulated reaction vessel (called a calorimeter), we can measure the temperature change of the solution without gaining or losing heat to the surroundings. Then, we can use q = mc∆T to calculate the amount of heat generated or absorbed by the reaction. Let’s say 100 mL of 1.00 M HCl and 100 mL of 1.00 NaOH are mixed together in a calorimeter, and the following reaction occurs:

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Both solutions start at 21.1 °C. After the reaction has occurred, the temperature of the mixture is 27.8. Because the temperature rose, we know this reaction is exothermic. The total volume is 200 mL, and because it is mostly water (with density 1 g/mL), the mass is 200 g. The water absorbed the heat given off by the reaction, and we can calculate this amount of heat using: q = mc∆T = 200(4.18)(6.7) = 5600 J From the information given, we can also calculate the ∆H of the reaction. ∆H is always expressed in kilojoules per mole (kJ/mol), so we first convert the 5600 J to 5.6 kJ. In this problem we have 0.1 mol of both reactants, so divide 5.6 by 0.1:

∆H = 5.6/0.1 = 56 kJ/mol Remember: we know that this reaction is exothermic, and therefore the ∆H must be negative! So our answer must be: -56 kJ/mol. Calculating the Heat Given Off/Absorbed Using ∆H Another common type of problem asks for the amount of heat given off by a certain reaction, given the ∆H value. Example: Calculate the amount of heat, in kJ, released when 50 g of methane (CH4) burns. The ∆H for the combustion of methane is -802 kJ/mol.

The ∆H value tells us that burning 1 mol of methane releases 802 kJ of energy. First, we convert our mass to moles: 50 g CH4 = 3.125 mol CH4. Then, multiply this by the ∆H to find the amount of heat:

3.125(-802) = -2506 kJ

(The negative sign indicates the reaction is exothermic.)

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16.2 Estimating ∆H with Bond Enthalpies The approximate ∆H of a reaction can also be calculated using bond enthalpy (also known as bond energy) values. Bond enthalpy is the energy required to break a chemical bond (measured in kJ/mol); the stronger the bond, the more energy is needed. Conversely, when a bond forms, the bond energy is released. The ∆H of a reaction thus depends on the number and strength of bonds broken and the number and strength of bonds formed. Breaking old bonds uses up a certain amount of energy; forming new bonds releases a certain amount of energy. The ∆H can be calculated by finding the difference between these two amounts. In these calculations, the variable D represents bond enthalpies; DH-H represents the bond enthalpy of a hydrogen-to-hydrogen bond, DO=O represent the bond enthalpy of an oxygen-to-oxygen double bond. Example: Find the ∆H for the combustion of methane using the following:

Bond Bond Enthalpy (kJ/mol) C-H 413 C-C 347 O=O 498 C=O 805 H-O 464

To solve, we need to know exactly what bonds break and what bonds form. So first write a balanced equation:

CH4 + 2O2 → CO2 + 2H2O Next, drawing Lewis structures to picture all the bonds (lone pairs have been omitted for clarity):

H

CH H

H

O O

O O

O C O

O

H H

O

H H

+ +

Then calculate the amount of energy needed to break all the bonds in the reactants (four C-H bonds, and two O=O bonds):

4DC-H + 2DO=O = 4(413) + 2(498) = 2648 kJ/mol

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Then calculate the amount of energy released when the bonds of the products form (four C=O bonds, and four H-O bonds)

2DC=O + 4DH-O = 2(805) + 4(464) = 3466 kJ/mol So, 2648 kJ/mol were used to break the old bonds, and 3466 kJ/mol were released when the new bonds form. So overall: ∆H = 2648 – 3466 = -768 kJ/mol The general equation for this type of calculation is: ∆H = ΣD(bonds broken) - ΣD(bonds formed) 16.3 Hess’s Law The following three facts are very handy:

1. If a reaction is reversed, the sign (+/-) of the ∆H switches. For example, take the reaction:

A + 2B → AB2 ∆H = -150 kJ/mol

If this reaction is reversed, the ∆H value of the new equation is now positive:

AB2 → A + 2B ∆H = +150 kJ/mol

2. If the coefficients of a reaction are multiplied or divided, the ∆H undergoes the same multiplication or division. So:

A + 2B → AB2 ∆H = -150 kJ/mol

If the entire equation is doubled: 2A + 4B → 2AB2 ∆H = 2(-150) = -300

If the entire equation is halved:

1/2A + B → 1/2AB2 ∆H = -150/2 = -75 kJ/mol

3. If two chemical equations are added together to make a new equation, the ∆H of the new equation is the sum of the two ∆H’s of the added reactions. This fact is called Hess’s Law.

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A + 2B → AB2 ∆H = -150 kJ/mol AB2 → B + AB ∆H = +400 kJ/mol

Adding these two reactions together, we get the reaction: A + 2B + AB2 → AB2 + B + AB

We can then cancel out the species that appear on both sides (notice that we cross out the B on the right, and decrease the coefficient of the 2B on the left by one):

A + 2B + AB2 → AB2 + B + AB

Leaving:

A + B → AB ∆H = -150 + 400 = +250 The reason these facts are useful is this: to determine the ∆H of a reaction without doing any kind of calorimetry experiment, manipulate and add other equations with known ∆H values to get the needed equation. Example: Find the ∆H of the following reaction:

1/2N2(g) + O2(g) → NO2(g) Use the following information:

1. N2(g) + O2(g) → 2NO(g) ∆H = +180.5 2. NO2(g) → NO(g) + 1/2O2(g) ∆H = +57.07

N2 must appear on the left side of the target equation, and equation 1 has it there already. But the coefficient for N2 in equation 1 is 1, and in the final equation it must be 1/2. So, we need to divide equation 1 in half, and the ∆H is also halved:

1/2N2(g) + 1/2O2(g) → NO(g) ∆H = +180.5/2 = +90.25

On the right side of the final equation, we see NO2. NO2 is on the left side of equation 2, so we need to reverse equation 2, and change the sign of the ∆H:

NO(g) + 1/2O2(g) → NO2(g) ∆H = -57.07

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Adding these two new equations and their ∆H values yields the target equation (we must cross out the species that appear on both sides of the arrow combine the 1/2O2’s on the left side):

1/2N2(g) + 1/2O2(g) + NO(g) + 1/2O2(g) → NO(g) + NO2(g) 1/2N2(g) + O2(g) → NO2(g) ∆H = +90.25 – 57.07 = +33.18

16.4 Standard Enthalpy of Formation When a chemical reaction describes the formation of one mole of a compound from its constituent elements in their standard states, the ΔH of the reaction is called the standard enthalpy of formation (ΔH°f) for that compound. Standard state is the most stable state at a pressure of 1 atm and temperature of 25 °C (don’t confuse this with STP). For liquid water, the equation would be:

H2(g) + 1/2O2(g) → H2O(l) ΔH°f = -285.8 kJ/mol The equation fits all of the criteria: One mole of water is formed. The hydrogen and oxygen are in standard states (they are in the gas phase, and they are diatomic). The following equations do not have all the species in standard states…

N(g) + 1/2O2(l) → NO(g) …because nitrogen should be diatomic, and oxygen should be a gas, not a liquid, at 1 atm and 25 °C.

C(diamond) + O2(g) → CO2(g) …because at 1 atm and 25 °C, the most stable form of carbon is graphite, not diamond. Important fact: by definition, the ΔH°f of an element in its standard state is zero. Why do we care about ΔH°f? Thousands of ΔH°f values have been determined and recorded. These ΔH°f values are very useful because it provides yet another way to determine the ΔH of a reaction without doing it experimentally. Here’s how it works. Take the equation for the combustion of methane:

CH4(g) + 2O2(g) → 2H2O(l) + CO2(g)

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If we write the equations for the formation of everything in this equation and look up the ∆Hf values, we get:

CH4(g) C(s) + 2H2(g) → CH4(g) -103.85 kJ/mol O2(g) N/A (element) 0 kJ/mol H2O(l) H2(g) + 1/2O2(g) → H2O(l) -285.8 kJ/mol CO2(g) C(s) + O2(g) → CO2(g) -393.5 kJ/mol

Once we have these equations and values, we will always be able to use Hess’s Law to find the ∆H of the overall reaction.

CH4(g) → C(s) + 2H2(g) 2H2(g) + O2(g) → 2H2O(l)

C(s) + O2(g) → CO2(g) CH4(g) + 2H2(g) + O2(g) + C(s) + O2(g) → C(s) + 2H2(g) + 2H2O(l) + CO2(g)

CH4(g) + 2O2(g) → 2H2O(l) + CO2(g) Notice that: The first equation is the ΔH°f of CH4(g), only reversed. The second is double the ΔH°f of H2O(l). The third is the ΔH°f of CO2(g). And so the ∆H of the reaction is:

∆H = [ΔH°f of CO2(g)] + 2[ΔH°f of H2O(l)] – [ΔH°f of CH4(g)] = -393.5 + 2(-285) – (-103.85) = -859.65 kJ/mol

Because all the ΔH°f values can be looked up in a table, we can determine the ΔH of the combustion reaction without burning any CH4. This technique always works, and is generalized by the formula:

ΔH°rxn = ΣnΔHf

°(products) - ΣmΔH f°(reactants)

It may look complicated, but all it means is this: To figure out the ΔH of any equation, add up the ΔH°f values of the products and multiply them by their coefficients, then subtract all ΔH°f values of the reactants multiplied by their coefficients. Example: Calculate the ∆H of the following reaction:

Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)

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Use the following information:

Substance ΔH°f (kJ/mol) H2O(l) -285.83 Ca(OH)2(aq) -1002.82

Applying the equation, we get:

ΔH = [1(-1002.82) + 1(0)] – [1(0) + 2(-285.83)] = -431.16 kJ/mol

16.5 Spontaneous Chemical Reactions From our everyday experience, we know that some things happen without anybody making them happen; if someone drops a book, it falls to the floor without anyone pushing it to the floor. A hot cup of coffee cools off automatically, eventually reaching room temperature. These processes involve a loss of energy, and reversing them requires an input of energy; it requires energy to lift the book from the floor, and it requires energy to reheat the coffee. Processes that lose energy (exothermic processes) tend to happen spontaneously, while processes that use up energy (endothermic processes) tend not to. In other words, the sign (+/-) of ∆H for a reaction determines, partly, the spontaneity of the reaction. There is another factor besides ∆H that influences the spontaneity of a process, as illustrated by the famous nursery rhyme:

Humpty Dumpty sat on a wall. Humpty Dumpty had a great fall. All the king’s horses and all the king’s men Couldn’t put Humpty together again.

Clearly, the process of breaking the egg is pretty easy; reversing the process is extremely difficult. Why? An egg is a very specific, very organized collection of matter. Disrupting this order is easy; restoring it is almost impossible. There is a natural tendency towards disorder—this is the other factor besides ∆H that influences the spontaneity of a process. When a drop of food coloring is placed in water, the molecules of food coloring disperse throughout the liquid. The

coefficient for Ca(OH)2(aq)

∆H°f for Ca(OH)2(aq)

∆H°f for H2(g)

coefficient for H2(g)

∆H°f for Ca(s)

∆H°f for H2O(l)

coefficient for Ca(s)

coefficient for H2O(l)

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reverse process will never happen, or is exceedingly unlikely to; once the molecules have spread out, they will not spontaneously organize themselves again into a concentrated drop. The dispersed state is more favorable than the concentrated state. The amount of disorder in a system is called entropy. Processes that increase entropy are more likely to be spontaneous than processes which decrease it. These changes in entropy are represented by ∆S—the counterpart to ∆H, for energy. A positive value of ∆S indicates an increase in disorder, while a negative value indicates a decrease in disorder. The different states of matter have different amounts of entropy. Solids, with all their particles locked in place, are highly ordered and have the lowest entropy. Gases, with particles flying around all over the place, have the highest. Liquids and aqueous states are somewhere in the middle. So depending on the states of matter in a reaction, we can roughly predict whether entropy will increase or decrease. For example:

Na2CO3(s) → Na2O(s) + CO2(g) There are no moles of gas on the reactant side and 1 mole on the product side. So there is a net gain in moles of gas, and therefore the entropy of the system increases (∆S of this reaction is positive). In the reaction:

N2(g) + 3H2(g) → 2NH3(g) There are four moles of gas on the reactant side (1 mole of N2 and 3 moles of H2) and two moles of gas on the product side. There is a net decrease in moles of gas, so the entropy of the system decreases (∆S of the reaction is negative). In the reaction:

NaCl(s) → Na+(aq) + Cl-(aq) The reaction starts with a solid (low entropy) and ends with two moles of aqueous ions (higher entropy); there is a net gain in entropy (∆S is positive). 16.6 Gibb’s Free Energy ∆H and ∆S can be combined into one equation that tells us about the overall spontaneity of a reaction:

∆G = ∆H – T∆S

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This is called the Gibb’s free energy equation. Ignore the T in the equation for now, and let’s concentrate on ∆H and ∆S. If both ∆H and ∆S favor spontaneity—that is, when ∆H is negative and ∆S is positive—the reaction must be spontaneous and the sign of ∆G must be negative. Conversely, if both ∆H and ∆S favor non-spontaneity—that is, when ∆H is positive and ∆S is negative—the reaction must be non-spontaneous and the sign of ∆G must be positive. Now let’s get back to that T in the equation, which stands for temperature (in Kelvin). For some reactions, the spontaneity depends on temperature. For example, let’s take the melting of ice. Melting is an endothermic process; it requires an input of heat, and this does not favor spontaneity. However, the liquid state is much more disordered than the solid state, and so melting the ice would mean an increase in entropy—this favors spontaneity. So the ∆H and ∆S are in conflict with each other. Using actual values for the reaction:

H2O(s) → H2O(l) ∆H = 6.010 kJ/mol ∆S = 0.022 kJ/mol-K

∆G = ∆H – T∆S ∆G = 6.010 – T(0.022)

If the temperature is high (350 K), ∆G ends up negative, and the ice melts:

∆G = 6.010 – 350(0.022) = -1.690 kJ/mol And when the temperature is low (200 K), ∆G ends up positive, and the reaction does not occur; the ice remains solid:

∆G = 6.010 – 200(0.022) = 1.610 kJ/mol What temperature marks the border between a non-spontaneous process and a spontaneous one? The switching point between positive and negative ∆G is zero, so let’s plug that into the equation and solve for the temperature:

0 = 6.010 – T(0.022) T = 273 K

Convert this value to Celsius, and get 0 °C: the melting point of ice. In other words, the reason that the melting point of ice is 0 °C is that below this temperature the reaction is not spontaneous, but above this temperature, it is spontaneous. The following chart sums up the relationships between ∆H, ∆S, and ∆G:

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∆S is + ∆S is –

∆H is + ∆G depends on temperature

∆G must be +; reaction is non-

spontaneous

∆H is – ∆G must be –;

reaction is spontaneous

∆G depends on temperature

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Chapter 17: Electrochemistry

17.1 Oxidation Numbers 17.2 Oxidation-Reduction Reactions 17.3 Balancing Redox Equations 17.4 Voltaic Cells 17.5 Standard Reduction Potentials

17.1 Oxidation Numbers An electrical current is the movement of charged particles (usually electrons). While all chemical reactions involve electrons, some chemical reactions involve a transfer of electrons from one atom to another, and thereby generate an electrical current. One way to keep track of electrons moving from atom to atom is to use oxidation numbers. If a neutral atom gains electrons, it becomes a negative ion; if it loses electrons, it becomes a positive ion; the charge of an ion tells us whether the atom has gained or lost electrons. In molecular compounds atoms share electrons, and it isn’t immediately obvious which to which atom a given electron “belongs”. To generally indicate how many electrons an atom has gained or lost, we use the oxidation number (or oxidation state), which is defined as the number of electrons an atom has lost, gained, or shared. We can assign an oxidation number to any atom using the following rules:

1. The oxidation number of an element is zero. 2. The oxidation numbers of a molecule add up to the total charge of the

molecule. 3. In compounds, the oxidation number of group 1 metals is +1 and the

oxidation number of group 2 metals is +2. 4. In compounds, fluorine’s oxidation number is -1. 5. In compounds, hydrogen’s oxidation number is +1. 6. In compounds, oxygen’s oxidation number is -2. 7. In compounds, group 17 elements’ oxidation number is -1, group 16

elements’ oxidation number is -2, and group 15 elements’ oxidation number -3.

Sometimes these rules will conflict. When this happens, follow the rule higher up the list. Example: Assign the oxidation numbers to the atoms in Na2O2.

Rule 2 says that the sum of the oxidation numbers must be zero. Rule three says sodium is +1. Rule six says oxygen is -2. If we follow rules three and six, the sum of the oxidation numbers is -2, which violates rule two. We must either ignore rule three and make sodium +2, or ignore

194

rule six and make oxygen -1. Rule three is higher on the list, so it wins: sodium is +1 and oxygen is -1.

Example: Assign the oxidation numbers to the atoms in SF2.

Rule two says that the sum of the oxidation numbers must be zero. Rule four says fluorine is -1. Rule seven says sulfur is -2. If we follow rules four and seven, the sum of the oxidation numbers is -4, which violates rule two. We must obey rules two and four and violate rule seven; F is -1 and S must be +2.

17.2 Oxidation-Reduction Reactions If the oxidation numbers of atoms change during the course of a reaction, the reaction is classified as an oxidation-reduction reaction (or redox reaction, for short). For example:

Fe2O3 + 3CO → 2Fe + 3CO2 In Fe2O3 on the reactant side, Fe is +3. As elemental Fe on the product side, it is 0. The iron has been reduced—its oxidation number has decreased. In CO on the reactant side, C is +2. In CO2 on the product side, C is +4. The carbon has been oxidized—its oxidation number has increased. We also say that the iron was reduced by the CO, and therefore the CO is termed the reducing agent. Likewise, we say that the carbon was oxidized by the Fe2O3, and therefore Fe2O3 is termed the oxidizing agent. Oxidation and reduction go hand-in-hand; when something is oxidized it loses electrons, and something else must be reduced and accept those electrons. Thus, every redox reaction can be broken into two half-reactions; an oxidation half-reaction and a reduction half-reaction. For example, take the simple redox reaction:

Fe+2 + Cu → Cu+2 + Fe The copper loses two elections to become oxidized, so the oxidation half-reaction is:

Cu → Cu+2 + 2e-

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The two electrons lost by the copper are accepted by the iron in the reduction half-reaction:

2e- + Fe+2 → Fe Sometimes the half-reactions are not so obvious. For example:

KIO3 + H2SO3 → KI + H2SO4 (This reaction isn’t balanced, but don’t worry—we’ll get to that in the next section.) To split this into half-reactions, look for two similar molecules on either side of the equation. That gives us:

KIO3 → KI and H2SO3 → H2SO4 In the first half-reaction, the iodine is reduced from a +5 to -1; therefore this is the reduction half-reaction. In the second half-reaction, the sulfur is oxidized from a +4 to +6; therefore this is the oxidation half-reaction. Again, don’t worry about the fact that the oxygens don’t balance. These are just the skeletons of the complete half-reactions; we will learn how to balance them in the next section. 17.3 Balancing Redox Equations Redox equations can be balanced in acidic or basic solution; the problem will specify which one. There is a specific procedure to follow; let’s do an example. The starting equation which needs to be balanced in acidic solution is:

MnO4- + C2O4

-2 → Mn+2 + CO2 1. Write the skeletons of the oxidation and reduction half-reactions, as we did

above. The skeletons contain the compounds that are oxidized and reduced; there will be O’s and H’s that don’t balance yet.

MnO4

- → Mn+2 C2O4-2 → CO2

2. Add coefficients to balance all elements other than hydrogen and oxygen.

MnO4- → Mn+2 C2O4

-2 → 2CO2

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3. Balance the oxygen by adding H2O to the side that needs more O’s.

MnO4- → Mn+2 + 4H2O C2O4

-2 → 2CO2 4. Balance the hydrogen atoms by adding H+ ions to the side that needs more

H’s.

8H+ + MnO4- → Mn+2 + 4H2O C2O4

-2 → 2CO2 5. Balance the charge by adding electrons, e-.

5e- + 8H+ + MnO4- → Mn+2 + 4H2O C2O4

-2 → 2CO2 + 2e- 6. If the electrons lost in one half-reaction is not equal to the electrons gained in

the other half-reaction, multiply one or both of the half-reactions by an integer that will make them equal.

2 × (5e- + 8H+ + MnO4

- → Mn+2 + 4H2O) 5 × (C2O4-2 → 2CO2 + 2e-)

10e- + 16H+ + 2MnO4

- → 2Mn+2 + 8H2O 5C2O4-2 → 10CO2 + 10e-

7. Recombine the half-reactions and cancel out redundant species on both sides

of the arrow. The electrons should always cancel out completely.

10e- + 16H+ + 2MnO4- + 5C2O4

-2 → 2Mn+2 + 8H2O + 10CO2 + 10e- If the reactions happen in basic solution, complete steps 1-4 above, then continue with the following procedure: 5. Wherever there are H+ ions, add the same number of OH- ions. Add the

same number of OH- ions to the other side to keep the atoms and charges balanced.

8OH- + 8H+ + MnO4

- → Mn+2 + 4H2O + 8OH- C2O4-2 → 2CO2

6. Combine the H+ ions and OH- ions that are on the same side of the equation

to form water.

8H2O + MnO4- → Mn+2 + 4H2O + 8OH- C2O4

-2 → 2CO2

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7. Cancel or combine the H2O molecules.

8H2O + MnO4- → Mn+2 + 4H2O + 8OH- C2O4

-2 → 2CO2

4H2O + MnO4- → Mn+2 + 8OH- C2O4

-2 → 2CO2 8. Balance the charge by adding electrons, e-.

5e- + 4H2O + MnO4- → Mn+2 + 8OH- C2O4

-2 → 2CO2 + 2e- 9. If the electrons lost in one half-reaction is not equal to the electrons gained in

the other half-reaction, multiply one or both of the half-reactions by an integer that will make them equal.

2 × (5e- + 4H2O + MnO4

- → Mn+2 + 8OH-) 5 × (C2O4-2 → 2CO2 + 2e-)

10e- + 8H2O + 2MnO4

- → 2Mn+2 + 16OH- 5C2O4-2 → 10CO2 + 10e-

10. Recombine the half-reactions and cancel out redundant species on both sides

of the arrow. The electrons should always cancel out completely.

10e- + 8H2O + 2MnO4- + 5C2O4

-2 → 2Mn+2 + 16OH- + 10CO2 + 10e-

8H2O + 2MnO4- + 5C2O4

-2 → 2Mn+2 + 16OH- + 10CO2 17.4 Voltaic Cells When a redox reaction is split into half-reactions, the oxidation half-reaction gives off electrons, and the reduction half-reaction accepts electrons. For example:

Zn(s) + Cu+2(aq) → Zn+2(aq) + Cu(s) Splitting this into half-reactions and balancing the charges with electrons, we get:

Zn(s) → Zn+2(aq) + 2e- and Cu+2(aq) + 2e- → Cu(s) When the zinc is oxidized, it releases two electrons. Copper ions accept those electrons and form solid copper. Overall, this reaction doesn’t do much good; zinc turns into zinc ions and copper ions turn into copper metal. But what if we physically separate the two half-reactions from each other by putting them in separate beakers?

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With this arrangement, the electrons given off by the zinc can’t reach the Cu+2. There needs to be a wire connecting the two half-reactions to allow the electrons to move from one side to the other: What are electrons moving through a wire? Electricity! We’ve created a battery (technical term: voltaic cell). Our voltaic cell needs one more thing to function. As it is now, when electrons move from left to right, a charge imbalance will build up; the left side will become more and more positive by losing electrons and the right side will become more and more negative by gaining electrons. This charge buildup will stop the flow of electrons; as negative charge builds up on the right, it will repel incoming electrons. As positive charge builds up on the left, it will pull back on outgoing electrons. The cell needs a salt bridge—a piece of tubing that contains a dissolved ionic compound, like NaNO3. When positive charge builds up on the left, NO3

- ions flow into the beaker to neutralize the charge. When negative charge builds up on the right, Na+ ions flow into that beaker to neutralize the charge.

Zn → Zn+2 Cu+2 → Cu

Zn → Zn+2 Cu+2 → Cu

2e-

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Some terminology: Each beaker is called a half-cell. The pieces of metal are called electrodes. The electrode where oxidation occurs (Zn in this case) is called the anode. The electrode where reduction occurs (Cu in this case) is called the cathode. 17.5 Standard Reduction Potentials Some substances have a greater tendency to be reduced than others; this tendency is measured in volts and is called the standard reduction potential (abbreviated E°red). In a voltaic cell using two different metals, which one will be reduced and which one will be oxidized? Which direction will the electrons flow? A table of standard reduction potentials allows comparison of the metals’ tendencies to be reduced. A voltaic cell essentially pits two half-reactions against each other; the half-reaction with the higher E°red “wins”, and it proceeds as it is written in the table—as a reduction reaction. The half-reaction with the lower E°red “loses” and is forced to go in reverse—as an oxidation reaction. Example: A voltaic cell is set up with the following two half-reactions. What half-reaction will occur at the anode and which at the cathode?

Pb+2(aq) + 2e- → Pb(s) E°red = -0.13 V Mg+2(aq) + 2e- → Mg(s) E°red = -2.37 V

The first reaction has the higher E°red, and so it occurs as it is written:

Pb+2(aq) + 2e- → Pb(s)

Zn → Zn+2 Cu+2 → Cu

2e-

Na

+ →

NO

3 - →

NaNO3(aq)

200

The lead will thus be the cathode, and the magnesium will be the anode. The magnesium reaction will be an oxidation reaction, and must be reversed:

Mg(s) → Mg+2(aq) + 2e-

The greater the difference in E°red, the more voltage the cell generates. The voltage generated by a cell is the difference in E°red values. The formula for overall cell voltage is:

E°cell = E°red(cathode) - E°red(anode)

For the Pb/Mg cell in the previous problem:

E°cell = -0.13 – (-2.37) = +2.24 V

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Standard Reduction Potentials in Water at 25°C Reduction Reaction E°red F2 + 2e- → 2F- +2.87 Co+3 + e- → Co+2 +1.8 PbO2 + 4H+ + SO4

-2 + 2e- → PbSO4 + 2H2O +1.69 MnO4

- + 8H+ + 5e- → Mn+2 + 4H2O +1.49 PbO2 + 4H+ + 2e- → Pb+2 + 2H2O +1.46 Cl2 + 2e- → 2Cl- +1.36 Cr2O7

-2 + 14H+ + 6e- → 2Cr+3 + 7H2O +1.33 O2 + 4H+ + 4e- → 2H2O +1.23 Br2 + 2e- → 2Br- +1.07 NO3

- + 4H+ + 3e- → NO + 2H2O +0.96 Hg+2 + 2e- → Hg +0.85 Ag+ + e- → Ag +0.8 Fe+3 + e- → Fe+2 +0.77 I2 + 2e- → 2I- +0.54 Cu+ + e- → Cu +0.52 NiO(OH) + H2O + e- → Ni(OH)2 + OH- +0.491 MnO2 + H2O + e- → MnO(OH) + OH- +0.421 O2 + 2H2O + 4e- → 4OH- +0.40 Fe(CN)6

-3 + e- → Fe(CN)6-4 +0.36

Cu+2 + 2e- → Cu +0.34 Cu+2 + e- → Cu+ +0.15 Sn+4 + 2e- → Sn+2 +0.15 2H+ + 2e- → H2 0 Fe+3 + 3e- → Fe -0.04 Pb+2 + 2e- → Pb -0.13 Sn+2 + 2e- → Sn -0.14 Ni+2 + 2e- → Ni -0.25 Co+2 + 2e- → Co -0.29 PbSO4 + 2e- → Pb + SO4

-2 -0.359 PbI2 + 2e- → Pb + 2I- -0.365 Cr+3 + e- → Cr+2 -0.4 Cd+2 + 2e- → Cd -0.4 Fe+2 + 2e- → Fe -0.41 Cr+3 + 3e- → Cr -0.74 Zn+2 + 2e- → Zn -0.76 Cd(OH)2 + 2e- → Cd + 2OH- -0.809 2H2O + 2e- → H2 + 2OH- -0.83 Zn(OH)2 + 2e- → Zn + 2OH- -1.129 V+2 + 2e- → V -1.18 Mn+2 + 2e- → Mn -1.18 Al+3 + 3e- → Al -1.66 Al(OH)3 + 3e- → Al + 3OH- -2.31 Mg+2 + 2e- → Mg -2.37

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Chapter 18: Organic Nomenclature

18.1 Organic Skeletal Structures 18.2 Naming Alkanes 18.3 Naming Alkenes 18.4 Naming Alkynes 18.5 Naming Cyclic Alkanes 18.6 Organic Functional Groups

18.1 Organic Skeletal Structures All organisms are carbon-based; all biological molecules have an underlying structure made of linked carbon atoms. In chemistry, the term organic has thus come to refer to any compound containing carbon, even artificially synthesized compounds. To save time and space, organic molecules are often written as skeletal structures—a kind of chemical shorthand. Each line in a skeletal structure represents a carbon-to-carbon covalent bond. Hydrogen atoms are not shown in the skeletal structure, nor are carbon-to-hydrogen bonds. If there are other elements besides carbon and hydrogen in the molecule, their symbols are included in the skeletal structure. When hydrogen is bound to a non-carbon atom, the “H” is written directly next to the atom, without a line to indicate a bond.

Lewis Structure Skeletal Structure

C C

H

H

H

H

H

H

H C C C C C H

H

H

H

H

H

H

H

H

H

H

H C

H

H

C C C

H

H

H

H

H

C C C

H

H

H

H

C

C

C

O

H

H

H

H

H

H

O

N C C O H

H

H

H

H

H

H

H2N

OH

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18.2 Naming Alkanes An alkane is an organic molecule that only has single bonds. Every compound has a main chain of carbon atoms; the main chain is always the longest chain present in the molecule. For example, there are three possible main chains in the following molecule—the first contains 6 carbons, the second contains eight, and the third contains nine. Thus, the nine-member chain is the main chain of this molecule.

The number of carbons in the main chain determines the numerical prefix of the overall molecule:

1 carbon: meth- 6 carbons: hex- 2 carbons: eth- 7 carbons: hept- 3 carbons: prop- 8 carbons: oct- 4 carbons: but- 9 carbons: non- 5 carbons: pent- 10 carbons: dec-

To name a single-chain alkane, simply attach the numerical prefix to the suffix “-ane”. For example:

propane pentane octane

Branches off the main chain use the same numerical prefixes, but have the suffix “-yl”. Here are the first five side chain names:

1 carbon: methyl 2 carbons: ethyl 3 carbons: propyl 4 carbons: butyl 5 carbons: pentyl

Side chains can attach to any carbon in the main chain. The particular carbon is indicated with a number. For example:

3-methyloctane 4-ethyloctane

The main chain is shown here in bold.

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Notice that we have numbered the carbon atoms in the main chain from left to right. This is because the side branch is closest to the left hand side of the main chain. If the side group were closer to the other side, we would number from right to left. For example:

2-methyloctane

If the main chain has two or more of the same side group, another numerical prefix (di-, tri-, tetra-, etc.) is added to the side group, and multiple numbers are written in front of the name to indicate the position of the side groups. For example:

3,5-dimethyloctane 3,4-diethyloctane

If there are two different side chains, each is listed, with the position number, in alphabetical order. For example:

4-ethyl-3-methyloctane 4-ethyl-3-methyl-5-propyloctane

4-ethyl-3,3-dimethyl-5-propylnonane 4,4-diethyl-3,3,7-trimethyl-5-propylnonane 18.3 Naming Alkenes Alkenes are compounds that contain one or more double bonds; they all have the suffix “-ene” instead of “-ane”. The position of the double bond is indicated with a number, similar to how a side chain is numbered. For example:

1-pentene 2-octene

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If there are multiple double bonds, a numerical prefix is added directly before the “ene”. For example:

2,6-octadiene 2,4,7-decatriene 18.4 Naming Alkynes Alkynes contain triple bonds and are named just like alkenes, only using the suffix “-yne”. For example:

3-hexyne 1,3,5-heptatriyne

18.5 Naming Cyclic Alkanes These ring structures have the prefix “cyclo-” followed by the name of the alkane. For example:

cyclopentane cyclooctane

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18.6 Organic Functional Groups The following is a list of functional groups, so called because their presence on a carbon chain dramatically alters the chemical behavior of the molecule. R, R’, and R’’ represent generic hydrocarbon chains.

Name Structure Example

Alcohol

R OH

OH

Ether

R O R'

O

Aldehyde

R C

H

O

C

O

H

Ketone

R

C

R'

O

C

O

Carboxylic Acid

R C

OH

O

C

O

OH

Ester

R C

O

O

R'

C

O

O

Amine

R NH2

or

R NH

R' or

R N

R'

R''

NH2

NH

N

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Another common structure found in organic molecules is the phenyl group; a ring of six carbon atoms with three resonating double bonds:

Instead of drawing the two resonance structures, the following drawing is often used:

208

Index of Important Terms

A absolute zero, 24 abundance, 64 accuracy, 16 acidic, 168 actinides, 95 activated complex, 153 activation energy, 154 alkali metals, 95 alkaline earth metals, 95 alkane, 203 alkene, 204 alkyne, 205 alloys, 133 alpha decay, 67 amphoterism, 172 anhydrous, 123 anions, 26 anode, 199 aqueous, 38 Arrhenius acid and base, 170 atomic element, 2 atomic mass, 64 atomic number, 63 atomic radius, 97 atomic units, 54 atomism, 49 atoms, 1 Aufbau Principle, 85 autoionization, 167 average deviation, 16 Avogadro’s number, 31

B balancing equations, 39 band of stability, 66 base units, 19 basic, 168 bent, 112, 113 beta decay, 68 binding energy, 66 Bohr, Niels, 59 boiling, 8 boiling point, 121 boiling point elevation, 139 boiling point elevation constant, 139 bond enthalpy (energy), 184 bond polarity, 114 bonding pair, 105 Boyle’s Law, 146 Brønsted-Lowry acids and bases, 171 buffer, 179 buret, 176

C calorimeter, 183 catalyst, 38, 155 cathode, 199 cathode ray tube, 51 cations, 26 Celsius, 24 centrifuge, 10 chain reaction, 72 charge, 64 Charles’s Law, 146 chemical bond, 102 chemical bonds, 1 chemical change, 5 chemical equation, 38 chemical formula, 4 chemical kinetics, 152 chemical properties, 9 chemical symbol, 4 Chemistry, 1 chromatography, 10 coefficients, 38 collision theory, 152 colloid, 133 Combined Gas Law, 145 combustion, 46 compound, 2 compound unit, 22 concentration, 137 condensation, 8 condensed configurations, 87 conjugate acid-base pairs, 171 control rods, 74 conversion factor, 20 coolant, 74 cooling tower, 74 core electrons, 96 covalent bond, 26, 104 critical mass, 72 critical point, 131 crystal lattice, 123 current, 193 cyclic alkane, 205

D Dalton, John, 50 Dalton’s Law of Partial Pressures, 147 d-block, 83 decay chain, 71 decomposition, 45 Democritus, 49 deposition, 8 diamagnetism, 87

209

diatomic seven, 47 diffusion, 149 digit term, 17 dipole-dipole, 118 dipole-induced dipole, 119 dipoles, 118 dissociation, 134 distillation, 10 double bond, 105 double replacement, 46

E effective nuclear charge, 95 electrodes, 199 electromagnetic spectrum, 57 electromagnetic wave, 57 electron capture, 68 electron configurations, 84 electron density, 61 electron domain, 110 electron geometry, 110 electronegativity, 101 element, 2 elementary steps, 156 empirical formula, 34 endothermic, 155, 182 enthalpy, 182 enthalpy of reaction, 155, 182 entropy, 190 equilibrium, 156 equilibrium constant expression, 159 equivalence point, 177 eudiometer, 149 exact number, 15 excited state, 90 exothermic, 154, 182 exponential term, 17

F factor-label method, 20 families, 95 f-block, 83 fenceposting, 20 filtration, 9 fission, 72 freezing, 8 freezing point depression, 139 freezing point depression constant, 139 frequency, 57 fuel rods, 74 functional groups, 206 fusion, 73

G gamma emission, 68 gas, 7

generator, 74 Gibb’s free energy, 190 Graham’s Law of Diffusion, 149 ground state, 90 groups, 95

H half-cell, 199 half-life, 75 half-reactions, 194 halogens, 95 heat, 9 heat of fusion, 127 heat of reaction, 182 heat of vaporization, 128 heating curve, 129 Heisenberg, Werner, 60 Hess’s Law, 185 heterogeneous, 3 homogenous, 3 Hund’s Rule, 86 hydrates, 123 hydration shell, 134 hydrogen bonding, 120 hydronium ions, 170

I ICE chart, 43 ideal gas constant, 143 Ideal Gas Law, 143 indicators, 175 induced dipole, 119 induced dipole-induced dipole, 119 inner transition metals, 95 insoluble, 135 intermediate, 156 intermolecular forces (IMF’s), 118 ion, 26 ion formation rules, 26 ion-dipole, 118 ionic bond, 26, 114 ionic bonds, 103 ionic compounds, 26 ionic formula, 27 ionic solids, 123 ionization constant of water, 167 ionization energy, 100 isotopes, 63

K Kelvin, 24

L lanthanides, 95

210

lattice energy, 123 Law of Conservation of Mass, 39 Law of Definite Proportions, 49 Law of Multiple Proportions, 50 Le Châtelier’s Principle, 164 Lewis dot structure, 103 Lewis structures, drawing, 107 limiting reactant (reagent), 42 line spectrum, 56, 58 linear, 111 liquid, 7 London dispersion forces, 119 lone pair, 105

M mass defect, 67 mass number, 63 matter, 1 meltdown, 74 melting, 8 melting point, 121 Mendeleev, Dmitri, 93 meniscus, 12 metallic bonds, 106 metallic solids, 125 metalloids, 94 metals, 25, 93 metric prefix, 19 metric system, 19 Millikan, Robert, 53 mixtures, 1 molal, 137 molality, 137 molar, 137 molar mass, 32 molarity, 137 mole, 31 mole fraction, 147 molecular compounds, 26 molecular element, 2 molecular formula, 33 molecular geometry, 110, 111 molecular polarity, 115 molecular solids, 124 molecules, 1 Moseley, Henry, 93

N net ionic equation, 141 network covalent solids, 125 neutral, 168 neutralization, 170 noble gases, 95 node, 62 nonmetals, 25, 94 non-polar covalent, 115 normal boiling point, 132

normal melting point, 132 nuclear decay, 67 nuclear equations, 69 nuclear power, 72 nuclear reactor, 74 nuclear weapons, 72 nucleus, 56, 63

O oil drop experiment, 53 orbital, 60 orbitals, 79 orbits, 59 organic compound, 202 oxidation number (oxidation state), 193 oxidation-reduction reaction, 194 oxidization, 194 oxidizing agent, 194

P paramagnetism, 87 partial pressure, 147 particles, 1 Pauli Exclusion Principle, 78 p-block, 83 percent composition, 35 percent error, 17 percent yield, 44 percentage, 23 Periodic Law, 93 periodic table, 93 periodic trends, 97 periods, 95 pH, 168 phase boundaries, 131 phase changes, 8 phase diagram, 131 phases of matter, 7 phenyl group, 207 photon, 61 physical change, 6 physical property, 9 planetary model, 59 plum pudding model, 52 polar covalent, 114 polyatomic ion, 27 polyprotic acid, 178 positron emission, 68 potential energy diagram, 105, 153 precipitate, 6 precision, 15 predicting products, 47 pressure, 142 products, 38 properties, 9 Proust, Joseph, 49 pure substances, 1

211

Q quantization, 59 quantum number, 60 quantum numbers, 78

R radiation, 67 radioactive series, 71 rate-determining step, 156 reactants, 38 reaction mechanisms, 156 reaction quotient, 161 reaction rates, 152 reaction types, 44 reactor core, 74 real gases, 144 redox reaction, 194 reducing agent, 194 reduction, 194 resonance, 110 Rutherford, Ernest, 55

S salt, 170 salt bridge, 198 saturated, 137 s-block, 83 scientific notation, 17 semiconductor, 95 settling, 10 shielding electrons, 96 significant figures (sig fig), 13 single bond, 105 single replacement, 45 skeletal structures, 202 solid, 7 solubility curve, 136 solubility product constant, 164 solubility rules, 135 soluble, 135 solute, 133 solution, 133 solutions, preparing, 138 solvent, 133 specific heat (capacity), 126 spectator ions, 141 spectroscope, 58 spin, 79 spontaneity, 189 standard enthalpy of formation, 187 standard reduction potential, 199 standard state, 187 standard temperature and pressure (STP), 147 state symbols, 38

states of matter, 7 steam generator, 74 steam line, 74 stoichiometry, 41 strong force, 65 strong vs. weak acids and bases, 173 subcritical mass, 72 sublimation, 8 submicroscopic scale, 1 subscript, 5 subshells, 78 supercritical fluid, 131 supercritical mass, 72 supersaturated, 137 surface tension, 122 suspension, 133 synthesis, 45

T tetrahedral, 111, 112 theoretical yield, 43 Thomson, J. J., 51 titration, 176 titration curve, 177 transition metals, 95 trigonal-planar, 111 trigonal-pyramidal, 112 triple bond, 105 triple point, 131 turbine, 74

U uncertainty principle, 60 unit map, 21 unsaturated, 137

V valence shell, 96 valence shell electron pair repulsion theory, 111 van’t Hoff factor, 140 vapor pressure, 149 viscosity, 122 volatility, 121 voltaic cells, 197 volts, 199 VSEPR, 111

W wave function, 61 wave/particle duality, 61 wavelength, 57 weighted average, 64