Ap Chemistry Workbook 1278

ADVANCED PLACEMENT CHEMISTRY WORKBOOK AND NOTE SET

Student Name ______________________________________________________________

Table of Contents

Fundamental Review

2

Atomic Character and Structure

2

Atomic Number

2

Mass Number and Isotopes

3

Ions and Electrical Charge

5

Molecules and Molecular Compounds; Ions and Ionic Compounds

6

Quantitative Aspects of Compounds

7

Percentage Composition

7

Empirical and Molecular Formulas

7

Hydrated Compounds and their Formulas

10

Mass Ratios

11

Stoichiometry Review

12

Limiting Reactants

12

Theoretical and Percent Yield

13

Percent Error

14

Review of the Periodic Table

15

Oxidation Numbers

16

Naming Compounds and Writing Formulas

19

Introduction to Descriptive Chemistry: Solutions and Introductory Solution Chemistry

22

The Solution Process for Ionic Compounds

22

The Solution Process for Molecular Compounds

23

Electrolytic Properties of Solutions

24

General Aqueous Solubility Guidelines

22

Solution Concentration

27

Dilutions

28

Reactions in Aqueous Solution

30

Writing Equations in Aqueous Solution

32

Using the Ideal Gas Equation and Gas Stoichiometry

Periodic Table of the Elements

34

35

Atomic Radius

36

Ionization Energy

37

Electron Affinity

39

II

Oxidation-Reduction Reactions and Introductory Electrochemistry

41

Oxidation-Reduction Reactions

42

Reduction Potentials

44

Oxidizing and Reducing Agents

47

Common Oxidizing and Reducing Agents

48

Balancing Redox Reactions

50

Electrochemistry

54

Voltaic Cells

Chemical Kinetics and Reaction Mechanisms Collision Theory

54

59 60

Activation Energy

61

Collision Orientation

62

Chemical Kinetics and Rates of Reaction

63

Definition of Reaction Rate

63

Rates in of Concentration

65

Stoichiometric Differences

66

Concentration Dependence of Rate

67

Reaction Order

69

Method of Initial Rates

70

Concentration Changes over Time

72

First-Order Reactions

72

First-Order Half-Life

73

Second-Order Reactions

74

Second-Order Half-Life

75

Rate and Temperature

75

The Arrhenius Equation

75

Reaction Mechanisms

77

Catalysts

85

III

Chemical Equilibrium I: Generalized Equilibrium and Le Châtelier’s Principle

86

The Concept of Equilibrium

87

Graphic Representation of Equilibrium

87

Quantitatively Expressing Equilibrium

88

The Equilibrium Constant

89

The Magnitude of the Equilibrium Constant

91

Manipulating Equations and Kc Values

92

Heterogeneous Equilibria

93

Calculating Equilibrium Constants

94

Reaction Quotients

98

Calculating Equilibrium Concentrations

100

Le Chatelier’s Principle

101

Chemical Equilibrium II: Acid-Base Equilibrium and Solubility Equilibrium

104

The Arrhenius Acid and the Arrhenius Base

105

The Brønsted-Lowry Acid and the Brønsted-Lowry Base

105

Conjugate Acid-Base Pairs

107

Relative Strengths of Acids and Bases

108

The Autoionization of Water

110

The pH Scale

112

Measuring pH

112

Strong Acids and Strong Bases

114

Equilibrium Expressions of Strong Acids and Strong Bases

Weak Acids and Weak Bases

115

116

Equilibrium Expressions of Weak Acids

116

Calculating Ka from pH

117

Determining the Validity of Assumptions and Percent Ionization

119

Calculating pH from Ka

120

Equilibrium Expressions of Weak Bases

121

Calculating

[OH-]

of a Weak Base Solution

121

The Equilibrium Concentrations of Polyprotic Acids

123

The Relationship Between Ka and Kb

124

Acid-Base Properties of Salt Solutions/Hydrolysis

125

Acid-Base Strength and Chemical Structure

127

The Common-Ion Effect

128 IV

Determining the pH of Solutions Exhibiting the Common-Ion Effect

129

Buffered Solutions

131

Buffering Capacity

131

The pH of Buffered Solutions and the Henderson-Hasselbalch Equation

132

The pH of Buffered Solutions when Strong Acids or Strong Bases are Added

135

Acid-Base Titrations

136

Titration Curves

137

Strong Acid – Strong Base Titrations

137

Strong Acid-Weak Base or Strong Base-Weak Acid Titrations

140

Summary of Acid-Base Titration Characteristics

144

The Value of the Solubility-Product Constant, Ksp

147

Factors Affecting Solubility

150

Thermochemistry, Thermodynamics and Electrochemistry II

153

The Nature of Energy

154

Distinguishing a System from the Surroundings

155

The First Law of Thermodynamics

155

The Internal Energy

156

State Functions

159

Enthalpy

160

Reaction Enthalpies

161

Hess’s Law

162

Bond Enthalpies

164

Heats of Formation

167

Calorimetry

169

Spontaneous Processes

173

Entropy and the Second and Third Laws of Thermodynamics

173

Gibbs Free Energy

177

The Electrochemistry-Redox-Thermochemistry-Equilibrium Relationship

183

Concentration Cells

185

Electrolysis

186

Electroplating

187

Electrolysis and Corrosion

187

Quantitative Electrolysis

188

V

Principles of Chemical Bonding Ionic Bonding

189 190

Born-Haber Cycles

Covalent Bonding

191

194

Review of Typical Geometry and Bond Number

195

Polar Bonds and Polar Molecules

196

Multiple Bonds

197

Lewis Structures

198

Drawing Lewis Structures

198

Deciding Between Several Lewis Structures: Formal Charge

202

Resonance Structures

203

Exceptions to the Octet Rule

205

VSEPR Theory

207

Electron Density Geometries

209

Molecular Geometries

210

Bond Angles

218

Hybridization Theory

218

The Behavior of Solutions, Liquids and Gases; Phase Change and Intermolecular Forces

225

Intramolecular Review: Ionic, Covalent and Metallic Bonding

226

Electronegativity and Polarity

227

Intermolecular Forces

228

Ion-Dipole Forces

229

Dipole-Dipole Forces

229

London Dispersion Forces

230

Hydrogen Bonding

232

Relative Strength of Intermolecular Forces

233

Vapor Pressure and Boiling Point

234

Properties and Intermolecular Forces

235

Change of Phase

238

Heating Curves

239

Critical Temperature and Critical Pressure

242

Phase Diagrams

243

Process of Solution

245

Degree of Saturation and Solubility

246

Factors Affecting Solubility

247

VI

Colligative Properties

248

Vapor Pressure Lowering

249

Boiling Point Elevation

250

Freezing Point Depression

251

Osmotic Pressure

252

The Gas Laws and Gas Behavior

254

Boyle’s Law

254

Charles’s Law

255

Combined Gas Law

256

Avogadro’s Law

256

Gay-Lussac’s Law

257

Ideal Gas Equation

257

Variations of the Ideal Gas Equation

258

Dalton’s Law of Partial Pressures

259

Mole Fractions

261

Kinetic Molecular Theory of Gases

261

Deviations from Ideal Behavior

262

Van der Waals Equation of State

264

Diffusion and Effusion

264

Collecting Gas Over Water

265

Descriptive Chemistry

267

Introduction

268

Synthesis Reactions

269

Decomposition Reactions

270

Single-Replacement

271

Double-Replacement (Metathesis) Reactions

272

Hydrolysis Reactions

273

Outline of Aqueous Metathesis Reactions

273

Redox Reactions

274

Oxidizing and Reducing Agents

275

Electrolysis

276

Naming Common Complexes

276

Coordination Chemistry

277

Organic Reactions

278

Except for artwork, material herein © David Alan Nelson; artwork used with permission. VII

ADVANCED PLACEMENT CHEMISTRY

Fundamental Review

Students will be able to:

identify the character, location and nature of the subatomic particles in the atom

describe the relationship between the subatomic particles and atomic number, mass number and ion charge

calculate average atomic mass from natural abundances

describe the formation of molecular compounds, ionic compounds and compare the properties of these classes of compounds ompounds

determine empirical and molecular formulas from percent composition data & reaction analysis data

calculate percent composition

determine the formulas and describe the nature of hydrated compounds

discuss the relationships given by chemical equati equations

use chemical equations to determine limiting reactants and product yields in chemical reactions

name compounds using the stock system of naming

name acids, polyatomic ions and hydrated compounds

predict valence electron structure from elements’ locations on the periodic table

assign and predict oxidation states to atoms, ions and compounds

describe the solution process for molecular and ionic compounds

describe and predict the electrolytic properties of solutions based upon the character of the ssolute

predict the solubility of ionic and molecular compounds, including acids and bases

write ionization/dissociation equations for appropriate compounds, and write full molecular, ionic and net ionic equations for fo reactions in aqueous solution

calculate solution concentration, and determine volumes of solutions required based upon concentration data

calculate required amounts for and perform dilutions

predict the occurrence of metathesis reactions and write their chemical equations using appropriate phase indicators

use the ideal gas equation and perform gas stoichiometry

Part 1: ATOMIC CHARACTER AND STRUCTURE

By the time of Bohr's model of the atom, much was in fact known about the physical properties of atoms. The electron had been discovered, followed by protons and the nucleus, and finally, the neutron. This section discusses many of the properties of the atom fundamental to the study of chemistry.

Many of the points below are extended upon in the sections of the pages that follow. The figure on the right represents a "textbook" view of the atom. Realize that the atom is far more complicated than this figure, and we will add detail during our study this year.

The atom is made of protons, neutrons and electrons.

The protons and neutrons are located in a dense core, and all of the mass of the atom is located here. Electrons surround the nucleus in a region of space.

Protons have a 1+ electrical charge, electrons have a 1- electrical charge, and neutrons have no electrical charge. Because the protons are positively charged and the neutrons have no electrical charge, the overall charge on the nucleus is positive.

Figure 1. A textbook model of an atom. The darker area surrounding the nucleus represents the region occupied by electrons as they orbit.

The atomic number of an atom is the number of protons in its nucleus. The atomic number of an element is unique.

The mass number of an atom is the sum of its protons and neutrons; recall, only protons and neutrons have mass. The mass number is central in determining the average atomic mass of an element, and atoms of an element that have different numbers of neutrons are called isotopes.

An atom with equal numbers of protons and electrons is electrically neutral; each 1+ proton "cancels" the charge on a 1electron. When this is not the case the atom is more specifically called an ion.

ATOMIC NUMBER Each element has an atomic number, which is given on nearly every periodic table. The atomic number of an element is equal to the number of protons in the nucleus of an atom of the element. Atomic number defines an element. No two elements have the same atomic number, and to change the atomic number is to change the

Figure 2. The atomic number of carbon is shown as 6. This tells you that an atom of carbon has 6 protons in its nucleus. The atomic number of hydrogen is shown as 1, which tells you that an atom of hydrogen has 1 proton in its nucleus. The proton number of an atom cannot change or the identity of the atom has also changed. Atomic number is often represented with an italicized Z.

identity of an atom. See Figure 2. 2

The atomic number can provide many details when attempting to characterize the physical properties of atoms:

You will read soon about atomic mass and mass number, both of which are dependent upon the number of protons in an atom; thus, they are related to atomic number.

Your determination of other characteristics will be made easier if you that atomic number does not change for an element. Electron number and neutron number may vary, but the number of protons is constant for an element.

MASS NUMBER & ISOTOPES

Figure 3. The three atoms above show atomic mass clearly - it is the number of protons plus the number of neutrons. You can see that the hydrogen atom has a mass of 1 (one p+ plus zero n0), the carbon atom has a mass of 12 (six p+ plus six n0), and the uranium atom has a mass of 238 (ninety-two p+ plus one hundred forty-six n0). The electrons are shown, but recall that they are massless and do not contribute to atomic mass.

Protons and neutrons are the only subatomic particles that have mass, and these are located in the nucleus. Together, these particles contribute to the atomic mass of an atom. The mass number of an atom is the number of neutrons and protons in its nucleus. Figure 4 shows the atomic mass of three atoms.

Atoms of an element can have different numbers of neutrons in their nuclei. Isotopes of atoms contain different numbers of neutrons in their nuclei. Another common term used to describe a specific nucleus is nuclide. For example, the isotope carbon-12’s nucleus could be described as the carbon-12 nuclide.

The name of a particular isotope is the element name plus its mass number. The name of the isotope on the far left in Figure 4 is carbon-12, the middle isotope is carbon-13 and the isotope on the right is carbon-14. This system of naming isotopes helps distinguish between the varying Figure 4. You can see in the figure above that there are three different isotopes of carbon atoms present in the lump of coal. The atoms do not differ in atomic number (because atomic number defines the atoms as carbon atoms), but they do have different numbers of neutrons; these represent the isotopes of carbon. These isotopes have the same chemical properties, but they do have a significant difference in the physical property of mass. The atomic mass of the first isotope is 12, which is the sum of the 6 p+ and 6 n0 in its nucleus. The atomic mass of the second isotope shown is 13, the sum of 6 p+ and 7 n0. The atomic mass of the third carbon isotope is 14, the sum of 6 p+ and 8 n0.

isotopes of an element that are present in nature. A shorthand notation for naming isotopes is to indicate the mass number of the isotope to the upper-left of its symbol. This notation is shown in Figure 4. The number in the lower-left corner is the atomic number of the atom. 3

The average atomic mass of an element is the weighted-average mass of all of the naturally-occurring isotopes of an element. As an analogy, consider a class where your grade is made up of several categories. For example, perhaps tests are 70% of your grade, homework is 10% and laboratory work is 20%. You can see that your performance on tests will have a greater impact on your final grade than a few homework assignments will. In a similar manner, if there are three isotopes of an element in nature and one of those isotopes makes up 70% of all naturally-occurring isotopes, then we would expect that the average atomic mass of the element would be near the mass of this particular isotope (much in the same way that your final grade would be near your test average). The average atomic mass is often simply called the atomic mass.

Practice 1.1

Isotope

Atomic mass

Percent abundance

From the table you can see that the mass of silicon-28

silicon-28

28

92.23

contributes more than the other two naturally-occurring

silicon-29

29

4.67

silicon-30

30

3.10

isotopes to the average atomic mass of silicon, which is 28.09. We can calculate the average atomic mass by determining the weighted average of the values shown.

The weighted average – the average atomic mass – is calculated by summing the results of the following calculation for each isotope: ൬

percent abundance ൰ x isotope mass = contribution of isotope to average atomic mass 100

Determine the average atomic mass of the silicon atom based on the data above.

The average atomic mass of an element is 58.69, and two naturally-occurring isotopes of the element are known to exist. If the mass of the isotope with an abundance of 78.3% is found to be 58.4, what is the mass of the second isotope?

4

IONS AND ELECTRICAL CHARGE

Atoms that contain equal numbers of electrons and protons are electrically neutral. However, atoms can easily lose or gain electrons and form ions, which are electrically-charged chemical species. The identity of the element does not change – only the electrical charge of the species changes. Recall from first-year chemistry that atoms lose or gain electrons based upon their metal or nonmetal character, and that the number of electrons gained or lost is related to valence structure.

Metals tend to lose electrons to form positive ions called cations. For example, an atom of sodium will readily lose its outermost electron and form the sodium ion, which has a charge of 1+. The charge is due to the difference between positive charges (protons) and negative charges (electrons): an atom of sodium contains 11 protons and 11 electrons; an ion of sodium contains 11 protons and 10 electrons – this gives a charge of (+11) + (–10) = +1. Metal ions are named just as their atoms are. We represent the ion by placing its charge to the upper-right of the symbol: sodium ion is represented as Na+.

Nonmetals tend to gain electrons to form negative ions called anions. For example, an atom of nitrogen will readily gain three electrons to fill its valence shell and form the nitrogen ion, which has a charge of 3-. The charge is due to the difference between positive charges (protons) and negative charges (electrons): an atom of nitrogen contains 7 protons and 7 electrons; an ion of nitrogen contains 7 protons and 10 electrons – this gives a charge of (+7) + (–10) = –3. Nonmetal ions are named by changing their name ending to “–ide.” The nitrogen ion is called “nitride ion.” We represent the ion by placing its charge to the upper-right of the symbol: nitrogen ion is represented as N3–.

Practice 1.2

Answer the following questions about atoms, isotopes and ions.

What is the number of protons, neutrons and electrons in a neutral atom of silicon-31?

How many protons and neutrons are in an isotope of lead-204? 82; n0 = 122

What is the mass number of an isotope of carbon that contains 8 neutrons and 6 electrons?

Identify the charge and total number of electrons in the most common ion of the following elements. Write the chemical symbols for the ions you select.

calcium

oxygen

chlorine

lead

zinc

5

Part 2: MOLECULES AND MOLECULAR COMPOUNDS; IONS AND IONIC COMPOUNDS

Although we tend to use the term loosely, a molecule is a discrete collection of covalently-bonded atoms that form a single unit – thus, we should not use the term molecule to describe ionic compounds, metallic compounds or alloys. Molecules include water (H2O), hexane (C6H14) and ammonia (NH3). From these examples we might conclude the general statement:

Molecules are units of matter composed of a definite number of (usually) nonmetallic atoms bonded to one another. Unlike ionic compounds – which can be broken along their ionic bonds and

Figure 5. A molecule of histidine. The bonds in a molecule are covalent bonds.

maintain their identity – the bonds of molecules are covalent bonds that cannot be broken unless a chemical change occurs.

Ionic compounds are made of ions that are attracted by electrostatic attraction due to their opposite charges – the ions do not share electrons. Thus, because the ions of ionic compounds are not sharing electrons, we do not refer to them as molecules; instead, they are referred to as formula units.

Notice in Figure 5 that breaking the bonds in the histidine molecule changes the chemical structure – it is no longer histidine. However, because the formula of NaCl is just that – NaCl – there is no change in chemical nature if a “layer” of Na–Cl ions is removed – the sample is simply smaller (Figure 6).

It is important to note that polyatomic ions are charge-carrying covalently-bonded ions. The bonds between the atoms in the polyatomic ion are covalent, while the attraction that holds the ion as a whole to another ion is an ionic bond. Compounds that form in this manner – all of which are ionic compounds – separate upon dissolving along the ionic bonds and never along their covalent bonds within the polyatomic ion. Again, to break a covalent bond is to form a new substance.

Figure 6. A unit of NaCl. The crystal is a series of repeating formula units rather than a "molecule."

6

Part 3: QUANTITATIVE ASPECTS OF COMPOUNDS

PERCENTAGE COMPOSITION

Percentage composition indicates the percent composition by mass of the elements in a compound. It is easily determined when the formula of a compound is known: divide the mass of an element of interest by the molar mass of a compound.

Practice 1.3

What is the percent composition by mass sodium in sodium carbonate?

A 2.424 g sample of a hydrocarbon is burned in excess oxygen to produce 3.756 g water and 7.342 g carbon dioxide. Use this information to determine the percent composition carbon and hydrogen in the hydrocarbon.

EMPIRICAL AND MOLECULAR FORMULAS

Empirical formulas provide one with the relative numbers and identities of the atoms in a molecular compound. However, they do not necessarily represent the actual number of each atom – they only give a whole-number ratio of the atoms in the compound. A molecular formula, by comparison, gives the actual number and identities of the atoms in a compound – it is a multiple of the empirical formula. The empirical formula can be a molecular formula if the actual numbers of atoms in a molecule is represented by the empirical formula.

If we know the molecular formula for a compound, we can easily determine its empirical formula, but other methods are required to determine a molecular formula if we know the empirical formula.

7

Formulas are mol ratios of the atoms in the compound – thus, in order to determine the empirical formula, simply think about how to find the ratio of mol in the compound from the data provided.

Divide the percent composition

Divide the larger mol values by

Write the chemical formula

or mass of each element by its

the smallest – multiply to

using the mol ratios determined

molar mass to determine mol

eliminate decimal values if

in the previous step

necessary

You should know how to determine the empirical formula of a compound by the following methods: mass composition, percent composition and combustion analysis

Practice 1.4

A certain compound is found to be 71.40% by mass carbon, 9.59% by mass hydrogen and 19.02% by mass oxygen. Determine the empirical formula of the compound.

Convert all percents into moles by dividing by the atomic mass of the element; divide by smallest; change to whole numbers

52.14 / 12.011 © = 4.34 mol C

13.13 / 1.0079 = 13.02 mol H

34.73 / 15.9994 = 2.17 m4 / 2.17 = 2 mol C

Imagine a 14.50 g sample of a compound containing only nitrogen and oxygen was decomposed into its elements; a mass of 3.76 grams of nitrogen was collected. What is the empirical formula of the compound?

8

If we know the products of a reaction that a particular compound undergoes, then we can analyze the products to determine the composition of the original compound. This allows us to determine the original compound’s empirical formula. For example, imagine that an unknown compound containing only carbon, hydrogen and oxygen (e.g., a sugar) is burned. The products of this reaction are carbon dioxide and water. We will perform 4 steps to determine the empirical formula of the sugar: •

Determine the percent composition carbon in carbon dioxide and the percent composition hydrogen in water.

•

Use the percent composition of carbon and hydrogen to determine the mass of carbon and hydrogen in the original compound.

•

Subtract the mass of carbon and the mass of hydrogen from the mass of the original compound to obtain the mass of oxygen.

•

Solve for empirical formula.

Practice 1.5

Imagine a 0.5540 gram sample of an unknown sugar is burned. Sugars contain only carbon, hydrogen and oxygen. The combustion of the sugar produces 0.6793 g water and 1.383 g carbon dioxide. What is the empirical formula of the sugar?

Molecular formulas are whole number multiples of empirical formulas; that is, a molecular formula and empirical formula are related by the ratio of their masses – thus, in order to determine the molecular formula, simply divide the experimentally determined molar mass by the mass of the empirical formula, and multiply the empirical formula by the result.

Determine the mass of

Divide the mass of the molecular

Multiply the empirical formula

the empirical formula

formula by the mass of the

by the value of the ratio obtained

empirical formula

between the masses

9

Practice 1.6

The major ingredient in antifreeze is composed of 38.7 g carbon, 9.70 g hydrogen and 51.6 g oxygen per 100.0 grams of antifreeze. Its molar mass has been found to be 62.1 g/mol. What is the molecular formula of antifreeze?

HYDRATED COMPOUNDS AND THEIR FORMULAS

A hydrated compound is an ionic compound that contains water molecules in positions of the crystalline lattice not occupied by ions. For example, the compound CoCl2, a purple compound, is the dehydrated form of the compound CoCl2 • 6H2O, which contains six water molecules in various lattice positions. The term for a compound that is the dehydrated form of a hydrated compound is anhydrous. In the examples provided, the purple cobalt(II) chloride is called anhydrous cobalt(II) chloride, while the hydrated form is called cobalt(II) chloride hexahydrate. The naming of a hydrate involves adding the prefixed form of hydrate after the name of the ionic compound. We can determine the number of water molecules that are contained in the hydrated form by removing the water through dehydration, which involves driving off the water by heating. You can often distinguish between the hydrated and anhydrous forms of most ionic compounds because they are different colors.

Practice 1.7

A 1.023 g sample of a hydrated copper(II) sulfate salt was dehydrated and found to have a mass of 0.6540 g. What is the percent composition water of the hydrate, and what is the number of molecules of water in the hydrate’s formula? Name the compound.

10

MASS RATIOS

Mass ratios indicate the ratio of the masses of the elements in a chemical compound. They can be easily determined by determining the mass of each element and establishing the ratios between the elements. Although it may seem elementary, it is worth noting that the mass ratio of “A to B” places A in the numerator and B in the denominator!

Practice 1.8

What is the mass ratio of chromium to oxygen in the compound chromium(VI) oxide?

11

Part 4: STOICHIOMETRY REVIEW: LIMITING REACTANTS & THEORETICAL YIELDS

The following relationships can be made between the species in a stoichiometric set-up:

When considering stoichiometry questions, be sure to

Grams of Substance A Grams of Substance B

consider:

For what are you being asked to solve?

Is the chemical equation balanced?

Are there any prior problems that must be

Use molar mass of A

Use molar mass of B

solved (empirical or molecular formula, percentage composition or limiting reactant)?

Mol of Substance A

Check the units in the set-up to ensure that

Ratio from chemical equation

Mol of Substance B

all of the units except the one desired in the answer cancel.

Is there a problem that must be solved afterward (percent yield, percent error)?

Are the conversions correctly set up – for example, did you indicate “there are 6.022 x 1023

atoms in a mol” and not incorrectly

make the relationship “there are 6.022 x 1023

[One mole of any particle is 6.022 x 1023 of the particles. The particles could be atoms, molecules or ions.]

mol in an atom?”

Through the use of Avogadro’s number, you can move to atoms or molecules from mole values

If you are looking for atoms, did you correctly consider the number of atoms in the formula of interest?

Did you to correctly calculate the molar masses of diatomic molecules and polyatomic ions?

LIMITING REACTANTS

A limiting reactant is a reactant in a chemical equation that is not present in sufficient stoichiometric quantity to react with another chemical. You will know that a calculation requires a limiting reactant calculation prior to stoichiometric determinations because the question will provide the mass of more than one reactant – this clues you that you must determine which reactant will be used-up first. Alternatively, most questions that do not require a limiting reactant determination expressly state that one of the reactants is “in excess.” Determining the limiting reactant requires simply determining the stoichiometry between the quantities available, and then using the reactant of which there is limiting quantity in the problem of interest.

12

Practice 1.9

A 35.0 gram of nitrogen is placed in a sealed container with a 5.00 gram sample of hydrogen. Assuming 100% reaction, what mass of ammonia can be produced?

THEORETICAL YIELD AND PERCENT YIELD

A theoretical yield is the stoichiometric amount of product that one could calculate as expected from a chemical reaction that occurs to 100% completion. However, most reactions do not yield the theoretical yield of product for various reasons that we will discuss later. The amount of product that is actually collected during a chemical reaction is called the actual yield or laboratory yield. Like all percent calculations, the percent yield is an expression of the ratio between an actual quantity and a maximum quantity:

From laboratory →

laboratory yield

From calculation →

theoretical yield

x 100 = percent yield

Practice 1.10

A student collected a mass of 0.2850 grams of lead(II) iodide when she reacted a 0.2500 gram sample of lead(II) nitrate with an excess sample of potassium iodide according to the double-displacement reaction shown here:

Pb(NO3)2 + 2 KI → PbI2 + 2 KNO3

What is the student’s percent yield?

13

A 200.0 g sample of CH4 was burned in 35.6% yield, and the H2O product collected. What mass of H2O was collected?

The reaction between methane and a halogen is called a substitution reaction – one in which a hydrogen on methane is replaced by a halogen atom.

Write the reaction between bromine and methane.

If the reaction occurs to about 29.5%, what mass of bromine is required to prepare 500.0 g of HBr?

PERCENT ERROR

A percent error provides a measure of the closeness of a laboratory value to the accepted value of a measurement – it is not a measure of the “correctness” of an answer!

lab value: recorded lab value

|lab value – accepted value|

accepted value: value from reference work, table of data, etc.

accepted value

x 100 = percent error

Practice 1.11

What is a student’s percent error when he collects 5.55 g carbon dioxide after burning 2.50 grams methane?

14

The combustion of octane, C8H18, produces the typical products of hydrocarbon combustion.

Write the balanced chemical equation for the combustion of octane.

How many grams H2O are produced when 265.5 g C8H18 is burned in the presence of 685.0 g O2?

How many atoms of oxygen are used each time 1.00 mol octane burns?

How many molecules of carbon dioxide are produced when a 4.00 mol sample of octane burns completely?

Part 5: REVIEW OF THE PERIODIC TABLE

Electrons are responsible for the chemical behavior of atoms. And, we will see here, that it is only the electrons in the highest-occupied quantum level that are of real significance for us. Why should only a few electrons be responsible for behavior? There are several explanations that will make your understanding on this point more clear: •

In order to react with one another, atoms must lose, gain or share electrons. We should not be surprised, then, that the electrons farthest out on the atom are those that will most likely "tangle" up with the electron clouds of other atoms. Thus, these electrons are the electrons responsible for the observed chemical behavior of atoms.

•

Full quantum levels are not very reactive. All the quantum levels before an atom's outermost quantum are full. Thus, only the partially-filled outer quantum is responsible for most of the chemical behavior of atoms. This is why noble gases (Group 18 atoms) are not very reactive; the Group 18 atoms have a full quantum level.

15

As an example, let’s review oxygen and sulfur. Oxygen has electrons in two quantum levels: 1 and 2. Sulfur has electrons in three t quantum levels: 1, 2 & 3. Look at the electrons in the highest quantum level for each element, which is quantum level two for oxygen and quantum three for sulfur. Oxygen has 6 electrons in the highest quantum (2s2 and 2p4), and sulfur also has 6 electrons (3s2 and 3p4). The electrons in the highest quantum for each element are an element's valence electrons, and you can ssee ee that elements in the same column on the periodic table have the same number of valence electrons; the electrons are just in different quantum levels. Oxygen O and sulfur both have six valence electrons, and we say that their valence shells are "s2p4." •

Valence electrons are the electrons that are in an atom's highest quantum level. They are the s or p electrons of the quantum level of highest energy. Because the transition metals are filling a previous quantum's energy level, they are not part of the valence vale structure. Instead, the valence shell of most transition metals is s2. Electrons that are not valence electrons are called core electrons.

•

Valence electrons are responsible for chemical behavior, and that is why the elements in the same column on the periodic table tabl have similar chemical behavior - they have the same valence structure, which is shown in the periodic table below.

•

An important aspect of valence electron configuration to consider is called Hund’s Rule,, which can be useful in explaining the chemical behavior and physical properties of elements. Data suggests that each orbital in a sublevel (e.g., the three orbitals orbital of the ‘p’ sublevel) vel) possesses one electron before any of the orbitals possess two. This leads to a regular pattern of half-filled and filled orbitals for the elements. For example, the nitrogen atom’s three ‘p’ electrons are not paired up in the first ‘p’ orbital; rather, r each of the three orbitals possesses one electron, which gives nitrogen three half half-filled orbitals. We shall see the effect of this in later units of study.

OXIDATION NUMBERS The valence structure of elements allows us to predict their oxidation states in compounds, which are the actual charges that ions will take, or the hypothetical charge that covalently covalently-bonded atoms would take if their bonds were completely ionic. ionic The use of oxidation numbers is fundamental to understanding how compounds bond and predicting the formulas for chemical compounds. For example, in the compound NaCl, the ionic charge on sodium ion is 1 1+,, while the ionic charge on chlorine is 1–. 1 This is because sodium has lost an electron and chlorine has gained an electron when ions formed. However, think about the case of NH3, which does not involve the production of ions because the atoms share electrons. How do we assign the “charge” on each species if there are no ions present? This is where oxidation states (oxidation numbers) come in useful. They allow us to keep track of the electrons shared in a covalent bond just as easily as we can keep track of transferred electrons in an ionic bond. 16

We assign oxidation numbers to the atoms or ions in a bond according to a set of rules, which are based upon experimental evidence. Moreover, they are assigned in the order shown here:

The oxidation number of a species in elemental form is zero

The oxidation number of a monatomic ion is the charge on the ion

The oxidation number of oxygen is usually 2–; of hydrogen, usually 1+; and of the halogens, usually 1–

The sum of the oxidation numbers of a charged species is the charge on the species

The sum of the oxidation numbers of a neutral compound is zero; and negative oxidation states are assigned to more electronegative elements

You should keep in mind that oxidation numbers are a man-made concept, and their use does not mean that actual charges are on the species being discussed, although this might be the case for ions. We shall use the concept of formal charge later to better describe the location of electrons, but even that concept ignores some information about how electrons are arranged – only sophisticated quantum mechanical calculations can provide definitive information about how electrons are arranged in bonds. However, we shall use both oxidation numbers and formal charge, as these are of great value in predicting and explaining chemical reactions.

Experimental evidence suggests the following oxidation numbers are common for atoms:

Metals may take on positive oxidation states equal to their valence electron count, their ‘s’ electron count, their ‘d’ electron count or the sum of their ‘s’ and ‘d’ electron count. They may also take on states related to any ‘p’ valence electron they possess. Typically, oxidation numbers we discuss are on the order of 1+ to 4+, with a few middle transition metals taking on larger states.

Nonmetals may exhibit positive oxidation numbers equal to their valence electron count

Nonmetals may exhibit positive oxidation numbers equal to the number of ‘s’ or ‘p’ electrons they possess

Nonmetals may exhibit negative oxidation numbers equal to the number of valence electrons required to fill their valence shells (which are the typical numbers seen when the nonmetal form ions)

At any rate, keep in mind the rules on asg oxidation numbers before randomly-asg oxidation sates! And, let the periodic table be your most valuable tool!

Practice 1.12

For each species below, determine the oxidation number on all elements:

1) CaBr2

2) H2O

3) NaNO3

4) Pb(NO3)2

5) NH4OH

6) SO42-

17

7) Cr2O72-

8) NaNO2

9) SnBr4

10) P2O5

11) NCl3

12) PO43-

For the reaction below, identify the oxidation number for each element.

Cd(s) + NiO2(s) + H2O(l) → Cd(OH)2(s) + Ni(OH)2(s)

This is the reaction that occurs in nickel-cium – “nicad” – batteries to generate electricity.

Write the full electron configuration for each of the elements below, and then use this information to predict oxidation states that should be observed.

S

Na

Cr

Si

18

Part 6: NAMING COMPOUNDS USING THE STOCK SYSTEM AND WRITING CHEMICAL FORMULAS

You likely have learned that molecular compounds are named by using prefixes, that ionic compounds might be named using Roman numerals to indicate charge and that cations forming only a single charge ion cannot be named with Roman numerals. You must believe that simple binary compounds (those containing only two elements or two ions) can ALL be named using the Roman numeral system, and it will be beneficial to you to simply do this! Very few elements only take on a single oxidation state; it is easier to always indicate the state than to consider which elements do or do not exhibit variable oxidation states. •

Naming a binary compound that does not contain a polyatomic ion. 1.

Identify the two elements in the compound.

2.

Identify the oxidation state of the first element – this is the Roman numeral to be enclosed in parentheses and used after the element name. (No space goes between the element and the Roman numeral.) It would be unusual, however, to indicate the oxidation state on hydrogen.

3. •

Change the name of the second element with the “-ide” ending, and put the two names together.

Naming a binary compound that contains a polyatomic anion. 1.

Identify the two components in the compound.

2.

Identify the oxidation state of the first element – this is the Roman numeral to be enclosed in parentheses and used after the element name. (No space goes between the element and the Roman numeral.)

3. •

•

•

•

Do not change the name of the polyatomic anion and include it as the second part of the name of the compound.

Naming a binary compound that contains a polyatomic cation. 1.


2.

Name the cation directly.

3.

Change the name of the second element with the “-ide” ending, and put the two names together.

Naming a binary compound that contains two polyatomic ions. 1.


2.

Name the cation and anion directly, and put the names together.

Writing the formula of an ionic compound. 1.

Identify the ions in the compound and their charges – the Roman numeral of a monatomic ion is its charge.

2.

Select the lowest whole-number ratio of ions that provides for a neutral compound.

Writing the formula of a molecular compound. 1.

If the compound is named with the prefix system, simply use the indicated number of each atom. (Except H2O is always “water” and NH3 is always “ammonia.”)

2.

If the compound is named with the stock system, determine the symbol and oxidation state of each element. Select the ratio of atoms that provides for a neutral molecule.

19

•

Naming Acids

Acids are named according to the anion from which an acid is derived.

Anion ends in “-ide”

Corresponding acid has the prefix “hydro-” and the suffix “-ic”

Anion ends in “-ate”

Corresponding acid has the suffix “-ic” (Retain any prefixes on the anion)

Anion ends in “-ite”

Corresponding acid has the suffix “-ous” (Retain any prefixes on the anion)

Practice 1.13

Name the acids below.

HClO

HClO2

HClO3

HClO4

Write the formulas for the following acids.

phosphoric acid

phosphorus acid

Select two sulfur polyatomic ions and name their acids.

20

Practice 1.14

Name the following compounds, or write their formulas, as appropriate. Include the stock and prefix names of molecular compounds.

iron(III) nitrate

carbon(IV) oxide

nitrogen(V) oxide

sodium chloride

PF3

ammonium carbonate

CuSO4

Cu2SO4

CoCl6 • 5H2O

H2O2

sodium acetate

HCl

xenon(VI) chloride

21

Part 7: INTRODUCTION TO DESCRIPTIVE CHEMISTRY: SOLUTIONS AND INTRODUCTORY SOLUTION CHEMISTRY

Many of the reactions that we will discuss – and most of those that we will perform in the lab – will be done in aqueous solution, which is a solution made when a soluble compound is dissolved in water. Although we will not discuss solutions in detail until later in the year, there are several points that we should make now to help us understand the nature of aqueous chemistry: •

A solution is a homogeneous mixture made up of two or more substances that do not chemically combine. Instead, the substances mix uniformly in the solution.

•

The solvent in a solution is the substance present in largest quantity by volume. Usually, the solvent of interest is water.

•

Figure 7. The solution process of an ionic compound in water.

The solute is the substance that is dissolved in the solvent.

In Figure 7 above, you can see that the ionic compound (represented by the +/– spheres) is dissolving in water. The positive ion is attracted to the negative end of the polar water molecule, while the negative ion is attracted to the positive end of the water molecule. Solutions generally form when the substances of which the solutions are made are of similar nature in of polarity: that is, polar substances dissolve polar substances (or ionic substances), while nonpolar substances dissolve nonpolar substances.

THE SOLUTION PROCESS FOR IONIC COMPOUNDS

To illustrate the process of solution, we will look at the solution of NaCl in water, which is represented in the figure above.

Recall that the water molecule is polar, with a concentration of negative charge on the more electronegative oxygen atom. This leaves the hydrogen ‘ends’ with partial positive charge. The negative ion in a soluble ionic compound is strongly attracted to the positive regions of water molecules, while the positive ion is strongly attracted to the negative end of the water molecules. When the conditions are right, this attraction exceeds any attraction between the ions themselves and causes dissociation. For an ionic compound like NaCl, the process of solution occurs because the solvent-solute attractions between water and salt are greater than the lattice energy of NaCl.

An ionic compound that dissociates considerably into its component ions in solution is termed soluble. For example, the compound NaCl separates into sodium ions and chloride ions in solution – the process of solution for ionic compounds involves the breaking of chemical bonds.

22

Practice 1.15

Into what ions do the following compounds dissociate?

NH4OH

NaCH3COO

H2SO4

THE SOLUTION PROCESS FOR MOLECULAR COMPOUNDS

If a molecular compound is water-soluble, then it will dissolve in water. However, unlike ionic compounds, the individual atoms of most molecular compounds do not separate in solution. That is, the molecules of most molecular compounds stay together – they are bonded covalently and would require a much stronger “pull” from water molecules to separate into ions. A solution of methanol, CH3OH, for example, does not separate into OH- and CH3+ or any other combination in solution. A solution of methanol is molecules of water and molecules of methanol.

For example, look at the figure below, which shows a model of an alcohol (1-propanol). When this substance dissolves in water, the individual atoms that make up the molecules do not separate as they do when ionic compounds dissolve.

In a molecular compound like this alcohol, the hydrogen bonding that occurs between the molecules hold the liquid together. In the presence of water, the molecules hydrogen bond to the water as this hydrogen bond is disrupted.

Some common exceptions to this general statement about molecular compounds’ behavior in water include hydrochloric acid, which completely ionizes into hydrogen ions and chloride ions, and several chemicals that react with water to form ions, including ammonia. (Notice that the term “dissociate” is used

for

ionic

The hydrogen bond between the alcohol molecules is being disrupted by the water molecule’s attraction – a new hydrogen bond is forming between water and the alcohol.

compounds’

separation, while the term “ionize”

Figure 8. The characterization of the solution process for a molecular compound.

is used to describe molecular compounds’ separation.)

23

ELECTROLYTIC PROPERTIES OF SOLUTIONS Pure water cannot conduct electricity, but the solutions formed when compounds dissociate or ionize in solution do conduct electricity. Electrical conductivity of solutions is directly proportional to the concentration of ions in solution. Compounds that ionize or dissociate in solution – thus, causing the solution to conduct electrical current – are called electrolytes.

Strong electrolytes are ionic compounds that dissociate (or molecular compounds that ionize) nearly 100% in solution. Strong bases, strong acids and soluble salts are strong electrolytes.

Weak electrolytes are ionic compounds that dissociate (or molecular compounds that ionize) very little in solution. Weak soluble bases, weak acids and sparingly soluble salts are weak electrolytes.

The strong acids, some common weak acids and the strong bases are provided below, and a list of the guidelines for solubility of ionic compounds is provided on Page 25. You should memorize these lists, and assume that all other acids and bases are weak. Also, assume that any compound not listed as soluble is insoluble.

Common strong acids – The common strong acids are soluble and 100% dissociated or ionized in solution; they are strong electrolytes HCl

hydrochloric acid

HBr

hydrobromic acid

HI

hydroiodic acid

HNO3

nitric acid

HClO4

perchloric acid

HClO3

chloric acid

H2SO4

sulfuric acid

Some common weak acids – The common weak acids are soluble but are not significantly dissociated or ionized in solution; they are weak electrolytes HF

hydrofluoric acid

CH3COOH

acetic acid

HCN

hydrocyanic acid

HNO2

nitrous acid

H2CO3

carbonic acid

H2SO3

sulfurous acid

H3PO4

phosphoric acid

Common strong bases – The common strong bases are 100% dissociated in solution; they are strong electrolytes NaOH

sodium hydroxide

LiOH

lithium hydroxide

KOH

potassium hydroxide

Ca(OH)2

calcium hydroxide

Sr(OH)2

strontium hydroxide

Ba(OH)2

barium hydroxide

24

GENERAL AQUEOUS SOLUBILITY GUIDELINES

1.

The common inorganic acids are soluble. Low-molecular-weight organic acids, e.g., acetic acid, are soluble.

2.

All common compounds of the Group 1 metals are soluble.

3.

All common ammonium ion compounds are soluble.

4.

All common nitrates, acetates, chlorates and perchlorates are soluble.

5.

The common halide compounds are soluble except as noted: these fluoride compounds are insoluble: MgF2, CaF2, SrF2, BaF2 & PbF2; these halide compounds are insoluble: AgX, Hg2X2, PbX2.

6.

The common sulfates are soluble except barium sulfate, lead sulfates and mercury sulfates

7.

The common metal hydroxides are insoluble except for those of the Group 1 metals and Ca, Sr and Ba from Group 2.

8.

The common carbonates, phosphates and arsenates are insoluble except for those of Group 1 metals and those of ammonium.

9.

The common sulfides are insoluble except those of Group 1 metals, Group 2 metals and those of ammonium.

10. Polar compounds are generally soluble. 11. Nonpolar compounds are generally insoluble, or are soluble to very low concentrations.

Summary of solubility and solution character –

NO Soluble ionic compound or

NO Is the compound an acid?

Is the compound ammonia or molecular base?

strong base?

YES

NO

YES

Nonelectrolyte

YES

NO Strong Electrolyte

Is it a strong acid?

Weak electrolyte

YES

Strong electrolyte

25

Practice 1.16

For the following compounds:

- identify the compound as ionic or molecular - identify any as acids, bases or salts - identify any acids or bases as strong or weak - identify salts as soluble or insoluble - label all acids, bases and salts as weak or strong electrolytes - for all compounds that dissociate or ionize, write the dissociation or ionization equation

CaCl2

HNO3

C2H5OH

LiOH

HCl

C6H12O6

CH3COOH

HBr

HF

NaF

PbF2

Ca(OH)2

Mg(OH)2

26

SOLUTION CONCENTRATION

The concentration of a solution is measured by the amount of solute per given quantity of solution. We generally use the molarity of a solution to discuss concentration. Molarity is a measure of mol solute per liter of solution. The unit of molarity is shown with an italicized uppercase em, M. Spoken, molarity can be described as “mol per liter.”

molarity =

amount of solute (mole) ࢓࢕࢒ = volume of solution (liter) ࡸ

You can rearrange the molarity equation to solve for mol of solute or for liters of total solution. You must that the concentration of a solution expressed in molarity is based on the unit liter – this means that any milliliter measurements must be converted in order to calculate correct values. Additionally, the solute amount is in mol – not mass. Thus, any mass amounts must be converted to mol in order to use the concentration unit of molarity.

Practice 1.17

Answer the following questions.

How many mol of NaCl are in 750. mL of a solution that is described as 2.50 M?

What is the molarity of 2.25 L of solution in which 0.0550 mol silver nitrate is dissolved?

What mass of CaBr2 is required to make 0.650 L of 0.500 M solution?

How many mol of NaCl are in 35.0 mL of a solution that is described as 2.35 M?

How many grams of NaCl are in 35.0 mL of a solution that is described as 0.550 M?

27

It is important to note that compounds that ionize or dissociate in solution will contain a greater molarity of ions than is expressed by simply indicating the molarity of the solution. For example, consider a solution that is 1 M NaCl. This measurement tells us that there is 1 mol of sodium chloride ion each liter of solution. However, sodium chloride is an ionic compound that dissociates in solution into sodium ions and chloride ions. Thus, the total mol ions in the solution is 2 mol : 1 mol of sodium ions and 1 mol chloride ions – the solution is, therefore, 2 M in ions. This would not be observed for nonionizing molecular compounds.

Practice 1.18

How many mol species (ions or molecules) are present in 0.750 L of the following solutions? Where appropriate, identify the ions. Give the molarity of ions for the solutions that dissociate.

0.50 M NaCl

2.3 M glucose

1.2 M sulfuric acid

0.750 M (NH4)3PO4

DILUTIONS

Common laboratory dilutions are performed with stock solution, which is concentrated solution. To determine the final molarity of a dilute solution, the following equation can be used:

M1V1 = M2V2, where

M1 and V1 are the molarity and volume of the stock solution

M2 and V2 are the molarity and volume of the desired solution

The equation will provide the amount of stock solution you need to dilute to achieve a total solution volume, V2, of the molarity desired, M2.

Practice 1.19 What mL volume of stock solution is required to make 1.20 L of 0.500 M solution of HCl? Stock HCl is 12.5 M.

28

It is often that you will want to dilute a prepared volume of dilute solution to a lower molarity. In this case, do not forget to for the volume of solution already present – , the dilution equation gives the total final volume of the dilute solution.

Practice 1.20

How much water should be added to 110.0 mL of 0.5000 M HCl to make a solution of 0.0850 M?

Imagine that you have found 45.0 mL 2.00 M calcium acetate solution.

What volume of water should be added to make 0.350 M solution?

How many total mol of ions are present in the new solution?

Would it be possible to make a more concentrated solution from the original sample? Explain.

29

Part 8: REACTIONS IN AQUEOUS SOLUTIONS

Imagine the following scenario for the reaction on the right. Before this picture was taken, two solutions were prepared: 0.70 M Pb(NO3)2 and 1.4 M KI. In the sample of the first solution, Pb(NO3)2, there are two ion species present: Pb2+ and NO3–. In the sample of the second solution there are also two ion species present: K+ and I-. When the two solutions are mixed, the mixture initially contains all four species. Immediately, however, the Pb2+ ions and I- ions react with one another to bond and form PbI2, an insoluble compound. The formation of an insoluble compound is called precipitation, and the insoluble compound is called a precipitate. The remaining two ions – K+ and NO3– – do not react in solution. In this reaction the cations of each compound switched anion "partners;" i.e., lead ion is now associated with the iodide ion and the potassium ion is now associated with the nitrate nitra ion.

Why did this mixture of solutions form lead to a chemical reaction while other combinations do not? The driving force for a metathesis reaction – a reaction in which cations and anions exchange bonding partners – is the removal of ions from solution. solution Here, the lead ions and iodide ions were removed: the compound is more stable than the free ions in solution. Because the other ions, K+ and NO3–, are more stable in solution than the compound KNO3, these two ions do not form a compound in solution. However, if the water is evaporated away, then the two ions will form a compo compound und because the ions are less stable out of solution than is the compound KNO3.

Gas-forming Reactions

When one of the products of a double-displacement displacement reaction is a gas, then the reaction will occur. Most commonly, the gases that form are NH3 (when “NH4OH” decomposes into NH3 and HOH), CO2 (when “H2CO3” decomposes into CO2 and HOH), SO2 (when “H2SO3” decomposes into SO2 and HOH) or H2S (directly from the combination of hydrogen ion and sulfide ion). Gas-forming Gas reactions are commonly those that involve a solid ionic compound and an acid.

ZnS(s)

+

HCl(aq)

→

Na2CO3(aq)

+

HCl(aq)

→

H2S(g)

H2CO3(aq)

+

+

ZnCl2(aq)

NaCl(aq)

↑ H2CO3 (H2O + CO2) H2SO3 (H2O + SO2) NH4OH (H2O + NH3)

30

Molecular Compound-formation reactions

A double-replacement will occur when a non-ionizing molecular compound is formed. The molecular compound is usually water but can be others, including weak acids. Common examples of this reaction type are acid-base reactions, which are also called acid-base neutralization reactions. In these reactions an acid and a base react to form a salt and the non-ionizing molecule H2O. The general form of these double-replacement reactions is shown here:

CH3COOH(aq)

+

NaOH(aq)

→

NaCH3COO(aq)

+

HOH(l) ↑ water

Precipitation Reactions

A double-replacement precipitation reaction is one in which two soluble, dissociated compounds are mixed and one - or both - of the exchange species is an insoluble salt. You need to know what compounds are soluble and what compounds are insoluble in order to predict the results of any particular reaction attempt - this information is contained in the solubility rules. The general form of doublereplacement precipitation reactions is shown here:

AgNO3(aq)

+

KCl(aq)

→

AgCl(s)

+

KNO3(aq)

↑ precipitate

31

WRITING EQUATIONS IN AQUEOUS SOLUTION

Are both reactants soluble ionic compounds?

Step 1:

Is one of the reactants a non-oxidizing acid and the other a solid carbonate, hydroxide, sulfite or sulfide?

NO

Check the reactants YES

STOP. NO

No chemical reaction will occur.

YES

Is one of the potential products: Step 2:

NH3, or weak acid)

Check the products

STOP.

a molecular compound (e.g., H2O,

a decomposing compound

an insoluble gas

an insoluble ionic compound?

NO


YES 1.

Identify the decomposition products of NH4OH, H2CO3 or H2SO3, if present

Step 3:

2.

Write the equation in molecular form

Write the net ionic equation

3.

Write all soluble compounds, strong acids and strong bases in dissociated form with appropriate electrical charges

4.

Cancel all species that remain unchanged from reactant-

Stop at 2 for the full molecular equation

Stop at 3 for the full ionic equation

side to product-side

Practice 1.21

For the following proposed reactions write molecular and net ionic equations. Give a reason if no reaction should occur.

barium chloride solution is added to sodium sulfate solution

potassium chloride solution is mixed with sodium sulfate solution

32

sulfuric acid is dropped onto a solid sample of calcium carbonate

a solution of HCl is added to a sample of aqueous sodium hydroxide

potassium hydroxide and cobalt(III) nitrate are mixed together

aqueous preparations of calcium chloride and sodium phosphate are mixed

pieces of sodium acetate are dropped into hydrochloric acid

small chunks of lithium metal are added to cold water

sodium sulfide solution is dripped into lead(II) acetate solution

33

Part 9: USING THE IDEAL GAS EQUATION AND GAS STOICHIOMETRY

While we will explore the use of the ideal gas equation in more detail later in a formal unit of study, we need to use it in its fundamental form throughout the year. Thus, we will review it here and leave additional details to be added later. According the kinetic molecular theory of gases, an ideal gas behaves according to the ideal gas law: ࡼࢂ = ࢔ࡾࢀ P = the pressure of the gas (in atmospheres, kilopascals, torr or mm Hg); V = the volume of the gas (in liters when used with any value of R); T = the temperature of the gas (in kelvin when used with any value of R: °C + 273 = K); R = 0.0821 when pressure is in atm, 8.314 when pressure is in kPa Use the relationships 760 torr = 760 mm Hg = 1 atm = 101.325 kPa to convert pressure.

Using the above equation, we can easily show that 1 mol of any gas occupies 22.414 L at standard temperature and pressure, STP, which is 273 K (0°C) and 1 atm (101.325 kPa).

Moreover, the equation and the relationship above allow us to convert between mol of gas and volume of gas in order to perform stoichiometry using gases. This is essential to the course.

Practice 1.22 Convert the following pressure values to the unit requested using dimensional analysis. 1.25 atm to torr

85.0 kPa to atm

The reaction between hydrogen gas and oxygen gas produces liquid water. Write the balanced equation for the reaction

What mass of water can be produced when 2.5 L of hydrogen is burned in excess oxygen at STP?

Use the ideal gas equation to find the pressure of the water gas above when it is transferred to a container with a volume of 0.500 L at a temperature of 200°C.

34


Periodic Table of the Elements


compare the relative sizes of the atoms of elements based upon their location on the periodic table and explain why & how the radii of atoms changes across a period and down a group

identify the property of ionization energy and discuss and explain the periodic trend of ionization energy

identify the property of electron affinity and discuss and explain the periodic trend of electron affinity

describe the effects of atomic radius, ionization energy and electron affinity on the observed properties ties of atoms; e.g., formation of ions and reactivity

35

This booklet is simply a quick review of the fundamental trends of atomic radius, ionization energy, electron affinity and general atomic properties as they can be seen using the periodic table. Much more on the periodic table will be incorporated into additional units of study.

Part 1: ATOMIC RADIUS

Atomic radius is most closely aligned with the effective nuclear charge (Zeff) experienced by the electrons on an atom. The effective nuclear charge is a measure of the nuclear charge experienced by an electron. As the effective nuclear charge increases for electrons, they will be pulled closer to the nucleus. For example, the nuclear charge on nitrogen is 7+, while for oxygen it is 8+. Thus, the 2s electrons on the two atoms experience different attraction by the nucleus – the 2s electrons of oxygen are more strongly attracted, which leads to a smaller

Figure 9. The trend of atomic radius can easily be seen in this figure - the increasing number of protons left-to-right and the increasing electron shells top-to-bottom cause this trend.

radius for oxygen compared to nitrogen.

Atomic radius generally decreases across the periodic table as the number of protons increases and the effective nuclear charge increases.

Atomic radius generally increases down a group of elements as the effective nuclear charge decreases on the electrons in larger shells and the repulsion between electrons increases.

Two atomic radius measurements are important in our discussion: bonding radius and nonbonding radius.

Figure 10. In graph form, the trend of atomic radius across a period is quite apparent.

36

The radius of atoms changes when they bond because the bonded atoms’ nuclei are pulling on one another’s electrons – this pulling causes the atoms in the bond to become distorted and “squeezed” by the electrostatic attraction between the nuclei and electrons of the atoms involved.

The nonbonding radius of atoms is somewhat larger than the bonding radius because the atoms are not subjected to the pulling force of another atom’s nucleus.

Atomic radius is measured in the SI unit of picometer, pm, which is 10-12 meter, or 1/1 000 000 000 000 meter. Atoms have atomic radii ranging from about 30 pm for hydrogen to about 300 pm for larger metals in Groups 6 and 7.

Part 2: IONIZATION ENERGY

As atoms become smaller, you should be able to imagine that it would require more energy to remove electrons – the attraction is greater between the nucleus of a small atom and the atom’s electrons than between the nucleus of a larger atom and its electrons. Thus, ionization energy – the amount of energy required to remove an electron from a gaseous atom or ion – increases as atoms decrease in atomic radius and decreases as atoms become larger.

For example, the ionization of the first electron from sodium(g) is shown here:

Na(g) → Na+(g) + e-

Figure 11. The trend of ionization energy is explained by evaluating the size of the atoms. You should be able to explain several small but notable exceptions.

∆H = +496 kJ/mol

This energy is called the first ionization energy, I1, of sodium because it is the energy required to remove the most

loosely-held valence electron. The second ionization energy, I2, of sodium is the amount of energy required to remove the most looselyheld electron from the sodium ion, Na+ (i.e., the second electron). Notice the very large (almost ten-fold) increase in the ionization energy when the second electron is removed, an observation for which we will explore an explanation soon: Na+ (g) → Na2+(g) + e-

∆H= +4560 kJ/mol

37

Table 1. The removal of valence electrons requires a significantly lower energy than does the removal of non-valence electrons, as is shown in this table of the first through seventh ionization energies for the Period 3 elements.

IE

Na

Mg

Al

Si

P

S

Cl

Ar

First

496

738

578

787

1012

1000

1251

1520

Second

4562

1451

1817

1577

1903

2251

2297

2665

Third

6912

7733

2745

3231

2912

3361

3822

3931

Fourth

9543

10540

11575

4356

4956

4564

5158

5770

Fifth

13353

13630

14830

16091

6273

7013

6540

7238

Sixth

16610

17995

18376

19784

22233

8495

9458

8781

Seventh

20114

21703

23293

23783

25397

27106

11020

11995

The trends in ionization energy can be readily explained by looking at effective nuclear charge: electrons farther from the nucleus are more shielded (that is, experience a lesser nuclear charge) than closer-in electrons. This shielding is caused by the number of repulsive electrons between any given electron and the nucleus. Clearly, for example, s electrons are less shielded than are the p electrons in the same sublevel. Thus, the p electrons exhibit lower ionization energy than their same-level s electrons – and they certainly require less energy to remove than any (n – 1) electrons.

Figure 12. For the same atom, notice that the outermost electron would experience a lesser effective nuclear charge than a closer-in electron. Thus, the ionization energy of the first electron is less than that of the second or subsequent electrons.

The trend of ionization energy is influenced by the size of the atom. It seems intuitive that a smaller atom exerts a greater attraction between the nucleus and electrons than does a larger atom. Thus, the smaller atoms exhibit higher ionization energies than the larger atoms.

Although the general trend of ionization energy is an increase left-to-right across the periodic table, there are cases where a decrease in ionization energy is evidenced. For example, the small decreases noticed as we move from Group 15 to Group 16 can be explained in of Hund’s rule: the atoms of Group 16 possess filled p-orbitals, which increases the electron repulsions and makes the first ionization energy lower than is seen in the ½-filled p-orbitals of Group 15 where the electron-electron repulsion is not as great because no electrons are paired.

38

Part 3: ELECTRON AFFINITY

Ionization energy expresses the amount of energy required to remove electrons from gaseous atoms to form positivelycharged ions. It is also expected to see atoms gain electrons to fill their valence shells. A measure of the energy change associated with the formation of negatively-charged ions when gaseous atoms take electrons into their valence shells is expressed as the electron affinity of atoms. The addition of electrons to atoms is generally exothermic, which results in a negative sign on the energy change1, as shown here: Cl(g) + e- → Cl–(g) ∆H= –349 kJ/mol

Figure 13. The electron affinity of the atoms generally becomes more negative (increases) across a period.

The more negative the value of ∆H – corresponding to a larger release of energy – the greater the electron affinity of an atom; this generally means that the negative ion is more

stable relative to the free atom. As ∆H becomes less negative there is lower tendency for the negative ion to form. Indeed, a positive ∆H indicates that the addition of energy is required for the addition of an electron.

Practice 2.1

Place each set in order of increasing atomic radius:

P, Se, S, As

Na, Be, Mg

Arrange the following in increasing order of predicted first ionization energy: Ne, Na, P, Ar, K

There are at least two widespread sign conventions on ∆H for electron affinity. Recognize that the bottom-line is that the addition of an electron is favorable 39 for a nonmetal atom, while it is less favorable for a metal atom or a nonmetal ion that is already charged to a full valence.

1

Consider the ionization energies seen here:

I1 = 577.5 kJ, I2 = 1816.7 kJ, I3 = 2744.8 kJ, I4 = 11577 kJ, I5 = 14842 kJ, I6 = 18379 kJ

In what group is this atom likely located? Justify your response.

Place the following in predicted order of electron affinity from greatest electron affinity (most negative) to lowest electron affinity (most positive).

Na, Ar, Cl, Si

The Group 1 atoms have slightly negative values of electron affinity, while the Group 2 metals have slightly positive values. Explain this observation.

Why might the Group 15 elements have more positive electron affinity values than other nonmetals, even though we know they typically form negative ions?

40


Oxidation Oxidation-Reduction Reduction Reactions and Introduction to Electrochemistry


identify oxidation and reduction of chemical species; identify oxidants and reductants

describe the nature of a redox reaction

use the activity series of the metals to predict the outcome of a proposed redox reaction

balance redox reactions by the half half-reaction ion method

characterize species as oxidizers izers or reducers, and predict the redox nature of species, including predicting the products of potential oxidations and reductions

relate oxidation and reduction ability to the periodic table

describe the fundamentals of an electrochemical cell, and iden identify tify the essential characteristics of a cell

determine the cell potential for a given cell

determine reduction potentials for half half-reactions used in cells

propose and construct spontaneous electrochemical cells

use shorthand notation for describing electrochemical cells

identify the nature of batteries and how their structure relates to their function

41

Part 1: OXIDATION-REDUCTION REACTIONS

Understanding chemistry requires a complete understanding of oxidation-reduction reactions, which are reactions that occur with a transfer of electrons from one species to another. For example, look at the reaction here, which occurs in acid solution: 2 H+(aq) + Zn(s) → Zn2+(aq) + H2(g)

During a redox reaction, one species increases its oxidation number, while another species decreases its oxidation number. The species that decreases its oxidation number is said to have been “reduced,” while the species that increases its oxidation number is said to have been “oxidized.” All redox reactions have both an oxidized and reduced species.

A reduction is the result of gaining electrons – the increase in electrons provides a less positive value for the oxidation number. That is, the oxidation number is reduced. This is the reduction half-reaction.

An oxidation is the result of losing electrons – the decrease in electrons provides a more positive value for the oxidation number. That is, the oxidation number is increased. This is the oxidation half-reaction. +1

0

2 H+(aq) + Zn(s) → Zn2+(aq) + H2(g)

0

+2

In this example equation, the oxidation number of zinc has increased from zero to positive two; zinc has been oxidized, which is due to the loss of electrons. The oxidation number of hydrogen has gone from positive one to zero; hydrogen has been reduced, which is due to the gain of electrons. In this case, zinc is the reducing agent (reductant) – a substance that loses electrons, and hydrogen is the oxidizing agent (oxidant) – a substance that gains electrons. The vocabulary can be quite difficult to sort – here’s a succinct statement that uses all of these new redox :

A reducing agent loses electrons to reduce the oxidation number on another species – it is itself oxidized. An oxidizing agent gains electrons to increase the oxidation number on another species – it is itself reduced. The “oxidized” and “reduced” refer to the oxidation numbers – reduction involves a decrease, while oxidation involves an increase.

There are two particular reactions that are of interest currently: the oxidation of metals by acids and the oxidation of metals by salts. The processes that occur are as described above: one chemical species is oxidized (a pure metal), and another species is reduced (either hydrogen ion or a metal ion).

42

Oxidation of metals by acids

The oxidation of a metal by an acid results in the formation of a positive metal ion and hydrogen gas. Many of the reactions that you will perform in lab are of this type. One of the most common is the oxidation of magnesium metal by hydrochloric acid: Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) (molecular equation) Mg(s) + 2 H+ (aq) → Mg2+(aq) + H2(g)

(net ionic equation)

So, what’s happening? You now know that in a redox reaction one species is reduced while another is oxidized. In the case above, valence electrons of magnesium are being transferred to hydrogen ion, resulting in the reduction of hydrogen from +1 to 0 as H2. Magnesium metal is being oxidized from Table 2. Activity Series of Selected Elements. Elements’ reduction equations are shown in this table. Also shown are the elements’ reduction potentials. Relative to the reduction of hydrogen ion to hydrogen, a reduction potential provides a relative measure of the ease with which a particular reduction will occur.

Element

Reduction reaction

0 to +2 as it loses electrons.

Oxidation of metals by salts

The oxidation of a metal by a salt occurs Ered

Element Reduction reaction

Ered

when a more active metal is placed in a solution containing a less active metal.

Li

Li+ + e– → Li

–3.05

Fe

Fe2+ + 2e– → Fe

–0.440

K

K+ + e– → K

–2.925

Ni

Ni2+ + 2e– → Ni

–0.28

Ba

Ba2+ + 2e– → Ba

–2.90

Sn

Sn2+ + 2e– → Sn

–0.136

Ca

Ca2+ + 2e– → Ca

–2.87

Pb

Pb2+ + 2e– → Pb

–0.126

Na

Na+ + e– → Na

–2.71

H

2H+ + 2e– → H2

0.00

Mg

Mg2+ + 2e– →Mg

–2.37

Cu

Cu2+ + 2e– → Cu

+0.337

–1.66

Ag

The less active metal is reduced; the more active metal is oxidized (Fig. 14).

The discussion of “active” in the sense of metals and redox reactions fundamentally involves a discussion of the ease with which an elemental metal sample will lose electrons: a more active metal will readily lose electrons to a less active metal. That is, a more

Al

Al3+

+ 3e → Al –

Ag+

+ e → Ag –

+0.799

active metal will be oxidized by a less active metal, and a less active metal will

Mn

Mn2+ + 2e– → Mn

–1.18

Hg

Hg2+ + 2e– → Hg

+0.854

Zn

Zn2+ + 2e– → Zn

–0.763

Pt

Pt2+ + 2e– → Pt

+1.188

be reduced by a more active metal. We shall now turn our attention to discussing how we can quantitatively

Cr

Cr3+ + 3e– → Cr

–0.74

Au

Au3+ + 3e– → Au

+1.52

measure the “activity” of metals.

43

Figure 14. Zinc is more active than copper. When we place a piece of Zn in a solution of CuSO4, the zinc atoms on the metal strip will lose their electrons to the copper atoms in solution, and solid copper will deposit on the strip; the less active metal undergoes reduction, while the more active metal undergoes oxidation. Reduction, Cu2+(aq) + 2 e- → Cu(s), is occurring in the top of the figure, and oxidation, Zn(s) → Zn2+(aq) + 2 e-, is occurring at the bottom of the figure.

REDUCTION POTENTIALS, Ered

We will simplify our discussion by simply stating that there is a potential energy difference between elemental metal and its ions in solution. We can compare this potential energy difference for any metal and its ions to that of the potential energy difference between hydrogen ion and elemental hydrogen to obtain a relative potential energy difference, which is measured in volts, V. You can see in Table 2 that the potential energy difference – or simply, the potential – for 2H+ + 2e- → H2 is zero, which is a defined potential. Moreover, we can further simplify the discussion of potentials by always considering the potential for a reduction reaction, even if a reaction occurs as an oxidation. This provides us the following:

A standard reduction potential measures the electromotive force of a particular reduction relative to the reduction of hydrogen ion. A more negative reduction potential indicates that the force behind a particular reduction is less than that of another, and a more positive reduction potential indicates that the force behind a particular reduction is greater than that of another. Thus, the more positive a reduction potential, the more likely a reaction is to occur as a reduction.

Thus, if one considers a pair of reduction reactions, the reduction that has the more positive reduction potential value will occur as a reduction, while the reaction with the more negative reduction potential will occur as an oxidation; i.e., in a battle between two reductions, the reduction with a more positive reduction potential wins. This provides the basis for the activity series of the metals, which allows us to predict whether or not a reaction will occur – notice that the most active metals have large negative reduction potential values, which corresponds to the likelihood of undergoing oxidation.

44

Practice 3.1

Consider a solid piece of lead placed into a solution containing the copper(II) ion. We are interested in determining whether or not the lead will reduce copper(II) ion into elemental copper and itself be oxidized to lead(II). That is, will the following reaction occur?

Pb(s) + Cu2+(aq) → Pb2+(aq) + Cu(s)

First, determine the values of each reduction’s potential, and then compare the values of Ered. It is wise to write both reactions as reductions to ensure you look up the correct potential.

Consider a solid piece of Ag placed into a solution containing the chromium(III) ion. Will a chemical reaction occur? In this case, first write the proposed chemical reaction before beginning.

Practice 3.2

A solution contains the iron(II) ion. Will the addition of either zinc or nickel cause the removal of iron(II) ion from solution? Be sure to write all proposed reactions before beginning. Clearly show that the use of reduction potentials s your answer.

45

Practice 3.3

Use reduction potentials to determine whether each of the reactions below will occur. Write the proposed reaction for each first, and write the net ionic equation for any reaction that occurs. Justify your response in each case with reduction potentials.

1) iron is dropped in an aqueous solution of nickel nitrate

2) to a solution of lead(II) nitrate, a piece of silver metal is added

3) hydrogen gas is bubbled through a solution of copper(II) chloride

4) barium is added to cold water

5) platinum is placed in hydrochloric acid

Consider a solution containing silver ion, copper(I) ion and tin(IV) ion. Can you propose a series of solid metal additions that will selectively precipitate each ion from the solution? That is, how could you use reduction potentials to separate this mixture? (Assume that any new ions that form are removed by another method after their formation.)

46

E = Ered(reduction) – Ered(oxidation)

We conclude this statement from the observations:

If the value of E (the potential difference) between the reduction and the oxidation is positive, then the reaction will occur spontaneously; otherwise, the reaction will not occur spontaneously. Notice, importantly, that we use the reduction potential when applying this equation whether the process occurs as a reduction or an oxidation.

Part 2: OXIDIZING AND REDUCING AGENTS

The relative values of Ered can tell us whether or not species will serve as oxidizing agents or as reducing agents; we have already said that the more positive the Ered value for a half-reaction, the greater the tendency for the reactant of the half-reaction to be reduced and, therefore, to oxidize another species.

For example, recall the oxidation of zinc by hydrogen: 2 H+(aq) + Zn(s) → Zn2+(aq) + H2(g) In this reaction, zinc is oxidized (Zn → Zn2+), while hydrogen is reduced (H+ → H2). The value of Ered for the reduction of hydrogen ion to hydrogen atom is 0.00 V; the value of Ered for the reduction of zinc ion to zinc atom (the reverse of the oxidation of zinc atom) is -0.76 V. Because the value 0.00 V is more positive than the value -0.76 V, we can see that hydrogen is more likely to be reduced, which is to be an oxidizing agent.

Oxidizing Agents – strong oxidizers have more positive values for Ered.

Oxidizing agents are species that can cause the oxidation of another species; strong oxidizing agents are easily reduced. A strong oxidizer must be able to take electrons from another species – thus, the halogens, oxygen and many oxyanions (like permanganate ion, nitrate ion, dichromate ion) are strong oxidizers. They are easily reduced to their reduced forms: fluorine atom → fluorine ion, chlorine atom → chlorine ion, and dichromate ion → chromium(III) [from chromium(VI)]. Other common oxidizing agents include peroxides and some metal ions. Strong oxidizers are weak reducers.

Reducing Agents – strong reducers have less positive values for Ered.

Reducing agents are species that can easily lose electrons to other species. Among common strong reducers are the metal atoms, which can easily lose electrons to reduce another species; the metal atoms are themselves oxidized. Strong reducing agents are weak oxidizers.

Redox and the Periodic Table

It is possible to use the periodic table to qualitatively discuss the oxidation or reduction character of elements: species that can accept electrons from another species are oxidizers, as they can be reduced. Species that can provide electrons to another species are reducers, as they can be oxidized. Besides just free element, consider whether or not one of the species in a compound or ion can accept or donate electrons, too. For example, perhaps a metal ion in a compound can be reduced or oxidized to a different ion. 47

COMMON OXIDIZING AND REDUCING AGENTS •

Common oxidizing agents – these species will cause a loss of electrons in another species, and they will themselves gain electrons. Oxidizing agents undergo reduction.

•

MnO4- in acid solution

→

Mn2+

MnO2 in acid solution

→

Mn2+

MnO4– in neutral or base solution

→

MnO2(s)

Cr2O72- in acidic solution

→

Cr3+

HNO3 (conc)

→

NO2

HNO3 (dilute)

→

NO

H2SO4 (hot, conc)

→

SO2

metallic ions

→

metallous ions (lower oxidation state)

free halogens

→

halide ions

Na2O2

→

NaOH

HClO4

→

Cl–

H2O2

→

H2O

S2O82-

→

SO42-

CrO42-

→

Cr3+

Common reducing agents – these species will cause a gain of electrons in another species, and they will themselves lose electrons. Additional reducing agents will be added. Reducing agents undergo oxidation.

halide ions

→

free halogens

free metals

→

metal ions

sulfite ions or SO2

→

sulfate ions

nitrite ions

→

nitrate ions

free halogens, dilute basic solution

→

hypohalite ions

free halogens, conc basic solution

→

halate ions

metallous ions

→

metallic ions (higher oxidation state)

C2O42-

→

CO2

MnO2 in base solution

→

MnO4–

You must learn to recognize the relationships between ionization energy, electron affinity, valence structure, reduction potentials and atomic character in order to evaluate problems surrounding electrochemistry and reactions – this is not a time for memorization; rather, it is your last best chance to take the time to learn this chemistry before we continue into more challenging material. Your assignment sets will include items helping you with this. 48

Practice 3.4

Identify the following as good oxidizers or good reducers. For your selection, write the product.

(A) oxygen (B) chlorine (C) sodium (D) Cr2O72(E) iodine Rank the following from the strongest reducer to the weakest reducer without using the table of potentials. Then, look up the values to check your answer.

Al, Fe, Br

Predict whether the following species are likely better reducers or better oxidizers. Be prepared to explain your selection. For each, write the reduction or oxidation reaction. Br–

Br2

NO3–

MnO4–

Use reduction potentials to determine the value of the potential difference when the following reactions occur. Silver nitrate solution is dripped over a solid piece of copper

A piece of solid magnesium is placed into a solution of hydrochloric acid.

The following reaction occurs as written. Write the products of the reaction. Identify the oxidizing agent and the reducing agent.

MnO4– (aq) + Fe2+(aq)+ H+(aq) → ?

49

Part 3: BALANCING REDOX EQUATIONS

You are undoubtedly familiar with balancing equations to observe the conservation of mass. However, a new consideration for redox reactions is the balancing of charge, too; more complex reactions will not balance as to charge and mass quite so easily. We can balance redox reactions by the half-reaction method. The half-reaction methods involves separating the oxidation half of the reaction from the reduction half of the reaction, balancing each separately for atoms and charge, and then putting the equations back together – it is much easier than it sounds when reading the preceding sentence! The half-reaction method works because the number of electrons gained by the reduction half-reaction must equal the number of electrons lost by the oxidation half-reaction.

When balancing redox reactions, we are typically discussing reactions that are occurring in acidic or basic solutions. The character of solution is important to balance redox equations in this manner!

50

Balancing redox in acid solution

Identify the oxidation half reaction and the reduction half

Cu + NO3– → NO2 + Cu2+

reaction

Write two half-reactions: one for the oxidation and one for the reduction

Balance all atoms except O and H in both half-reactions

Balance oxygen by adding one water molecule for each oxygen atom needed

Balance hydrogen by adding H+ as needed to either reaction.

Balance charges by adding electrons to the reactions.

This will result in adding e- to the left side of one reaction and the right side of the other. Multiply through the equation(s) as needed to cancel the electrons. The electrons MUST cancel.

Add the reactions – cancel any common ions or molecules; simplify if possible.

Write complete equation.

51

Balancing redox in base solution

Identify the oxidation half reaction and the reduction half reaction

Pb(OH)42- + ClO– → PbO2 + Cl–

Write two half-reactions: one for the oxidation and one for the reduction

Balance all atoms except O and H in both half-reactions

Balance oxygen by adding one water molecule for each oxygen atom needed

Balance hydrogen by adding H+ as needed to either reaction.

Balance charges by adding electrons to the reactions. This will result in adding e- to the left side of one reaction and the right side of the other. Multiply through the equation(s) as needed to cancel the electrons. The electrons MUST cancel.

Add the reactions – cancel any common ions or molecules; simplify if possible. Add an OH- ion for each H+ ion shown. Combine them to form water. Add the same number of OH- to the other side, too.

This step is performed in basic solution only to eliminate the hydrogen ions present in the equation. Why?

Cancel common water molecules.

There will not be an excess of H+ in basic solution!

Write complete equation.

52

Practice 3.5

Balance the following redox reactions by the half-reaction method.

CN– (aq) + MnO4– (aq) ⇋ CNO– (aq) + MnO2(s) in acid solution

Cu(s) + NO3– (aq) ⇋ Cu2+(aq) + NO2(g) in acid solution

Cr(OH)3(s) + ClO– (aq) ⇋ CrO42-(aq) + Cl2(g) in base solution

53

Part 4: ELECTROCHEMISTRY

VOLTAIC CELLS

Recall that in a spontaneous redox reaction there is a potential energy difference. When the reaction occurs, energy is released, and this energy can be harnessed to perform electrical work (although, because of the Second Law, we cannot harness the maximum work possible, which we shall see when studying thermodynamics).

A voltaic cell (or galvanic cell) is a device that allows the transfer of electrons between the reactants of a redox reaction without between the reactants. The energy released as the electrons travel is harnessed to perform work.

The Structure of a Voltaic Cell

In a typical voltaic cell, two metals are placed in solutions of their ions2; e.g., zinc metal is placed in a solution of zinc ions, and copper metal is placed in solution of copper ions. Because the two metals are not in with one another, the redox reaction cannot occur until a method of allowing for the flow of electrons is provided. This can be accomplished by connecting the two metal pieces with wire. The wire provides a conduit for the transfer of electrons from one electrode to the other. An electrode is a sample of metal that is connected to another by Figure 15. A spontaneous voltaic cell.

an external circuit – here, the wire is the external circuit.

In one half-cell of a voltaic cell, a reduction reaction occurs. Here, the reduction reaction is Cu2+(aq) → Cu(s). In the other half-cell of a voltaic cell, an oxidation reaction occurs. Here, the oxidation reaction is Zn(s) → Zn2+(aq). The compartments of a voltaic cell are given names: the anode compartment and the cathode compartment.

2

For metals that take on several oxidation states, one could use a platinum or graphite electrode at the cathode to cause the reduction of the metallic ion to an ion with a lower state. These are not common to

see in AP Chemistry, but we will work a few in free-response practice in order to present a complete discussion.

54

Reduction occurs at the cathode, and oxidation occurs at the anode. Thus, electrons are traveling from the anode to the cathode in order that cations in the cathode are reduced to elemental metal. This means that at the anode elemental metal is being oxidized.

By convention, the cathode is labeled positive, while the anode is labeled negative; however, the electrode compartments are neutral. In fact, if the electrode compartments take on electrical charge, then the voltaic cell ceases to operate.

There are five essential pieces of a voltaic cell: •

A reduction half-cell; this is the cathode

•

An oxidation half-cell; this is the anode

•

A conduit for the transfer of electrons; this is the circuit

•

A method of opening and closing the circuit; this might be a switch

•

A source of positive and negative ions that move into the electrode compartments to maintain neutral solution; this is called a salt bridge, and it is filled with a salt that can provide ions to the compartments (a porous disk may be used; this allows for the transfer of ions through the disk)

The Operation of a Voltaic Cell

A) Zinc metal at the anode is losing electrons. This removes the zinc atoms from the electrode and increases the zinc ions in solution. The mass of the anode decreases as the atoms are lost to solution. The positive charge in the compartment solution is increasing as zinc cations are released.

D) A constant transfer of ions from the salt bridge into the electrode compartments occurs in order that the compartments maintain a neutral character.

B) The electrons travel through the circuit to the cathode.

C) Copper ions in the cathode compartment’s solution are gaining electrons. This removes copper ions from solution, which are deposited on the copper electrode. The mass of the cathode increases as atoms of metal are deposited. The negative charge in the compartment solution is increasing as copper cations are removed. 55

A voltaic cell will operate as long as there is a potential difference between the anode and cathode. For a spontaneous voltaic cell to operate in the desired direction, the value of the reduction potential at the cathode is more positive than the reduction potential of the anode, because the value of the cell potential must be positive. This is not unlike the single-replacement discussion on previous pages: E = Ered(reduction) – Ered(oxidation)

Ecell = Ered(reduction) – Ered(oxidation)

Ecell = Ered(cathode) – Ered(anode)

Notice that we always use the values of the reduction potentials even though one reaction occurs as an oxidation. You can determine the cathode and anode of a cell by looking at the reduction potentials for the two half-reactions – the more positive reduction potential occurs as a reduction, which means it is the cathode. You can also use the value of Ecell to determine the values of reduction potentials for reactions by rearranging the equation for Ecell.

Practice 3.6

The following reaction occurs in a voltaic cell: Zn(s) + Cu2+(aq) ⇋ Zn2+(aq) + Cu(s). What is the value of the Ecell?

A voltaic cell operates spontaneously; it is constructed of cells containing Ni(NO3)2(aq) and Ni(s) and CuNO3(aq) and Cu(s). A) Determine the half-reactions occurring in each cell, write the balanced reaction and identify the species undergoing oxidation and the species undergoing reduction. Identify the number of electrons being transferred.

B) Which reaction is occurring at the cathode? Justify your choice.

C) In which compartment is the mass of the electrode increasing? Justify your choice.

D) What is the value of the cell potential?

56

A voltaic cell operates spontaneously with Ecell = +1.46 V. The reactions in the cell are shown here:

In+(aq) → In3+(aq) + 2e– Br2(g) + 2e– → 2Br– (aq)

A) Write the balanced overall reaction, and indicate the number of mol electrons transferred.

B) Which reaction is occurring at the cathode? Which reaction is occurring at the anode? Justify each choice.

C) Using a table of reduction potentials and the value of cell emf provided in the question, determine the value of Ered for the reduction of indium(III) to indium(I).

Practice 3.7

Consider the following reaction: Cr2O72-(aq) + I– (aq) ⇋ Cr3+(aq) + I2(s)

Balance the reaction in acid solution, determine the anode and cathode half-reactions if this is used in a voltaic cell, and determine the value of Ecell. (In the case of substances such as these – i.e., not metals – the electrodes are generally made of graphite or an inert metal such as Pt. Thus, instead of deposition of metal, solution species will change.)

57

Two reduction reactions are shown here. Using the data for Ered for the reactions, determine the reactions that occur if these are used in a voltaic cell. Identify the reaction that occurs at the cathode and the reaction that occurs at the anode. Justify Justi your choice. Then, use the blank set-up up below to completely structure the operating voltaic cell. There are several things that should always be labeled when constructing a cell. Cd2+(aq) + 2e- → Cd(s) Sn2+(aq) + 2e- → Sn(s)

Later, we will see that the concentration of the reactants and the temperature of the cell. For now, we are working with standard cells, which are cells at 298 K using 1 M solutions. Shorthand Notation for Voltaic Cells We can easily represent the details of a voltaic cell by using shorthand notation to describe the cell. We write the oxidation half-cell cell on the left, separating the reactant from the product with a single bar. We show the separation of cells by using a double bar, which represents thee salt bridge, and then show the reduction half halfcell in the same manner as we represented the anode reaction.

Practice 3.8

Show the cium-tin tin cell in Practice 3.7 in shorthand notation.

Part 5: APPLICATIONS OF ELECTROCHEMISTRY: BATTERIES, CORROSION AND ELECTROLYSIS

Redox titration, batteries, atteries, corrosion and electrolysis will be eexplored xplored in a laboratory setting, and more on electrochemistry will be added when we study thermodynamics later. 58


Chemical Kinetics and Reaction Mechanisms


Explain the concept of reaction rate

Describe, predict and discuss the factors that affect rates of reaction

Explain reaction rates as they relate to the stoichiometry of chemical reactions

Discuss, determine and use rate constants

Calculate rate laws

Determine concentrations based on rate integrated rate laws and half half-life

Interpret graphical data to draw conclusions about reaction orders

Explain activation n energy and transition states

Use the Arrhenius equation and explain its significance; determine activation energy and rate constants at variable temperatures

Recognize that reactions may occur in steps, and discuss the various steps of reactions

Determinee whether or not proposed mechanisms of reaction are consistent with observed reactions

Define, identify and discuss the significance of a catalyst in a chemical reaction

59

Part 1: COLLISION THEORY

Although sugar and oxygen will react, it is not often that we see a sugar bowl go up in flames even in the presence of atmospheric atmosp oxygen. (This is a good thing.) Beyond , then, there must be more behind the occurrence of chemical reactions. A widelywidel accepted model of chemical reaction is called collision theory,, which suggests that the reactants in a chemical reaction must collide with one another in order to react. This explains several observations about chemical reactivity, including items we will wil discuss soon.

However, not all collisions lead to a chemical reaction. We know this because we do not see sugar spontaneously reacting with oxygen in the air to produce carbon dioxide and water. According to collision theory, in order for a chemical rea reaction ction to proceed more than simple collision must occur. Collision Theory includes more than a simple statement that molecules or atoms must collide to react. In addition: •

There is an energy component to collision theory

•

There is an orientation component to collision theory

Svante Arrhenius,, someone we will see again when we study acids, was the first to suggest that molecules must collide with sufficient kinetic energy; his theory was presented in 1888.

Thus, we can summarize collision theory eory as as: Chemical reactions occur when molecules or atoms collide with sufficient kinetic energy – the activation energy – and the collision occurs in a favorable orientation

60

ACTIVATION ENERGY

Look at the figure at left. It shows the energy changes that are associated with the reaction shown here: CO + NO2 → CO2 + NO

You can see that the reactant molecules possess an amount of potential energy due to their chemical composition and the interactions of the subatomic particles and atoms in the molecules. According to collision theory, an additional amount of Figure 16. Reactants must reach the transition state to react - they can reach the transition state by possessing the activation energy.

energy must be present in order to allow a reaction to take place. The reactants CO and NO2 must

possess a certain amount of kinetic energy upon collision in order for the reaction to proceed to form the activated complex, or the transition state, which is the highest-energy arrangement of atoms during a chemical reaction.

Due to the high energy of the transition state, it is very unstable and short-lived. It is at this point in a chemical reaction that the reaction will proceed to products or back to reactants. The amount of energy absorbed to reach the transition state is equal to the activation energy, Ea, of the reaction, which is the minimum amount of energy required for a reaction to occur.

As shown in the figure above, the reaction is exothermic, as the products possess less potential energy than the reactants, and the difference between the final potential energy and the initial potential energy is the enthalpy change of the reaction, ∆H. In order for the reverse reaction to occur,

Figure 17. In (A) the reaction has both the correct orientation and energy. In (B), the reactants may possess enough energy, but their orientation is incorrect.

the reactants (CO2 and NO for the reverse reaction) must possess activation energy equal to the sum of ∆H for the forward reaction and the Ea of the forward reaction.

The figure at right shows an additional example of the transition state for a chemical reaction; successful in (A), but note that the collision in (B) is not oriented properly to form products. 61

In a sample, only a certain fraction of the molecules will possess enough kinetic energy to react. You can see in the figure at right, as the

Figure 18. This graph represents the kinetic energy possessed by two samples of molecules. Many have very high kinetic energy, while many have very low kinetic energy. At the far right, we can see that a very small fraction at either temperature has enough kinetic energy to react. However, there is a greater fraction with this minimum activation energy at the higher temperature.

temperature of a sample increases, the fraction of molecules with enough kinetic energy to react increases.

The graph to the right has a special characteristic: it represents a Boltzmann distribution, which is a graph that plots probabilities of specific cases – like how many molecules possess a certain amount of energy – occurring in a large sample. Ludwig Boltzmann studied how the observation of the behavior of large samples of material could be used to predict molecular behavior; thus, molecular details are inferred from bulk behavior. Boltzmann believed in the existence of subatomic particles, which was not a widely-held belief in the late 1800s. Unable to face the criticism and attacks on his theories, Boltzmann committed suicide in 1906, just as the subatomic world became apparent trough the work of Thomson, Marsden, Geiger, Rutherford and others.

COLLISION ORIENTATION

The second statement of collision theory involves the orientation of the atoms or molecules. The figure at right shows two potential orientations of reacting species – but only one orientation is effective at producing a chemical reaction. In figure (a), the iodide ion collides with the carbon atom on CH3Br, which results in an orientation that is favorable. In figure (b), the iodide ion collides with the bromine atom in the molecule, which – even with sufficient kinetic energy – does not allow a reaction to occur. The reasons orientation might be important are many, among them include that the colliding species must have attractions at their points or that only a single orientation will allow the species to fit together.

Figure 19. The iodine atom on the left in (a) will collide in the proper orientation to allow for reaction. No reaction will occur in (b) because an iodine/bromine collision is not properly oriented.

We might assume that gases at room temperature collide about 1010 times per second, but only one in every 1013 collisions produces a reaction. : collision, proper orientation and sufficient kinetic energy are required for the reaction to occur. 62

Part 2: CHEMICAL KINETICS AND RATES OF REACTION

Chemical kinetics is the area of chemistry concerned with the rates of chemical reactions, or the speed at which chemical reactions occur. Rates of reaction can be affected by several factors, which we have previously discussed.

Practice 4.1 For each of the following: describe what is meant by the statement, provide an explanation for its effect on rates of reaction in of collision theory, and provide an example of each that s your statement.

Concentration of reactants

Temperature at which the reaction occurs

Presence of a catalyst

Surface area of reacting molecules

time, t (minutes)

Mole A

Mole B

0

1.00

0.00

10

0.75

0.25

DEFINITION OF REACTION RATE Imagine a reaction that can be represented as A → B. Because the stoichiometry is 1:1, for each molecule of A that is consumed in the reaction a molecule of B is

20

0.55

0.45

30

0.40

0.60

40

0.30

0.70

amount of substance B present, ∆mol B, divided by

50

0.22

0.78

the elapsed time, ∆t. This provides the speed of the

60

0.16

0.84

formed. If we time this reaction over 60 minutes, we might collect the data shown in the table at left. We can calculate the rate of reaction as the change in the

reaction, or the rate at which substance B is produced:

average rate =

∆ mol B ∆t

Notice that we have elected to express the rate of reaction in of the appearance of substance B. We could easily have elected to express the rate in of the disappearance of substance A:

average rate =

∆ mol A ∆t

63

Practice 4.2 Calculate the rate of reaction for the reaction shown in the table on Page 63. Like any rate, reaction rate expresses a change over time. In this case, the change could be production of substance B or consumption of substance A. Average Rate of Appearance of Product, substance B ∆t

time interval

mol B initial

mol B final

∆ mol B

Average rate, mol min-1

t = 0 min to t = 10 min t = 10 min to t = 20 min t = 20 min to t = 30 min t = 30 min to t = 40 min t = 40 min to t = 50 min t = 50 min to t = 60 min

Average Rate of Disappearance of Reactant, substance A time interval

∆t

mol A initial

mol A final

∆ mol A

Average rate, mol min-1

t = 0 min to t = 10 min t = 10 min to t = 20 min t = 20 min to t = 30 min t = 30 min to t = 40 min t = 40 min to t = 50 min t = 50 min to t = 60 min

Why is the rate of reaction the same magnitude for the appearance of B as it is for the disappearance of A over each time interval?

How is the rate changing as the product is produced?

64

RATES IN TERM OF CONCENTRATION

The reaction we looked at above gave the amount of reactant and product in of mol of each substance. However, we may prefer to use the concentration of reactants and products to express the rate of reaction. Let’s look at the reaction shown here: C4H9Cl(aq) + HOH(l) → C4H9OH + HCl time, s (seconds)

[C4H9Cl(aq)], M

0.0

0.100

50.0

0.091

100.0

0.082

150.0

0.074

200.0

0.067

300.0

0.055

Imagine a 0.10 M solution of C4H9Cl(aq), which is an organic compound called chlorobutane. If we measure the disappearance of the chlorobutane over time we might gather the data shown in the table at left.

Notice that we have measured the concentration of the reactant remaining after each time interval rather than the mol reactant remaining. Rate is a measure of change over time, and the change may be measured in various units, just as the time may; here, we have changed from measuring in minutes to measuring in seconds.

400.0

0.045

500.0

0.037

(This reaction is typical of the reactions of haloalkanes, organic compounds that possess a halogen in place of one hydrogen in an alkane-like structure. The

800.0

0.020

1000.0

0.000

halogen is replaced with –OH from water, and the result is an alcohol and an inorganic acid. The products in the reaction above are the alcohol butyl alcohol and hydrochloric acid: R-X + HOH → R-OH + HX.)

Practice 4.3

Write an expression to show how the rate of disappearance of water relates to the disappearance of chlorobutane. Also, write expressions to show the rates of production of butyl alcohol and hydrochloric acid.

Calculate the rate of change in M s-1 during the interval t = 50.0 s to t = 150.0 s and the interval t = 500.0 s to t = 1000.0 s

How does the rate of reaction change when comparing the two intervals above? Propose an explanation for this observation.

65

STOICHIOMETRIC DIFFERENCES

So far we have looked at two reactions, both of which had stoichiometric relationships of 1:1 between all species. Let’s now look at reactions where the stoichiometric ratios are not 1:1, like in the reaction between hydrogen and iodine to produce hydrogen iodide: 2 HI(g) → H2(g) + I2(g)

In the reaction above, you can see that 2 mol HI are used for each mol H2 produced and each mol I2 produced. This changes the calculation of the rate of reaction because we need to for the fact that 2 mol of HI are used in each time interval examined if we want to compare the rate of appearance of H2 or I2 to the rate of disappearance of HI.

If we express the rate of appearance of H2 as average rate =

∆[H2 ] ∆t

then you should recognize that the rate of disappearance of HI is twice this rate, or average rate =

∆[HI] ∆[Hଶ ] =2 ∙ ∆t ∆t

(An observation such as this is valid when comparing the rates of disappearance or appearance of the species in a chemical reaction with one another, and it does not change the fact that rate of disappearance of HI alone is, in fact, ∆[HI] / ∆t.)

We can conclude that generally, for a reaction of the form aA + bB → cC + dD the relative rates of change are given by: average rate=

∆[A] ∆[B] ∆[C] ∆[D] = = = a∆t b∆t c∆t d∆t

Practice 4.4 According to the equation 2 O3(g) → 3O2(g), what is the rate of the disappearance of ozone, O3, if the rate of appearance of oxygen gas is 6.0 x 10-5 M s-1?

Nitrogen dioxide is oxidized to N2O5 when burned in excess oxygen. Write the balanced chemical equation for this reaction.

Imagine that the rate of appearance of N2O5 is 4.2 x 10-7 M s-1 in a flask. What is the rate of disappearance of each reactant in the flask?

66

Part 3: CONCENTRATION DEPENDENCE OF RATE

Look at the figure at left. At the bottom of Page 65 we hinted at this phenomenon: you hopefully noticed that the rate of the chemical reactions appears to decrease as the reactants are used in the reactions. This is the case in most chemical reactions, and you can see the fact illustrated clearly if you look lo back at the reaction rates you calculated on Page 64 for the reactions whose rate data is shown in the tables. This observation is consistent with what we established earlier: the rate of chemical reaction is dependent upon the concentration of the reactants, reac among other things. Figure 20. Pay special attention to the change in rate for ∆[NO2] during the two 100 s intervals marked. This is discussed in the text at right.

More than the change in rate over time due to the concentration of reactants, however, we are

especially interested in the initial rate of a chemical reaction as a function of the concentrations of reactants. This method metho of determining rate of reaction – called alled the method of initial rates – allows us to determine how each reactant affects the overall rate of reaction. Then, we can possibly deduce the path that the reaction follows (e.g., the intermediates and transition states that form). This important information ormation gives information about the plausibility and utility of a chemical reaction in an industrial process, for example, under a set of conditions that are attainable.

One way to explore how the concentration of reactants affects the rates of reaction is to set up a variety of experiments to determine how the variation of the concentrations of reactants affects the initial reaction rate; that is, the rate of reaction at the instant of combining the reactants. You will now do this with the data we will complete in the table on the following page. Your conclusions should be made in the space provided.

Practice 4.5

In of collision theory, why hy would you expect that the rate of chemical reaction would be dependent upon the concentration of reactants?

67

Experiment Title:

The Effect of Initial Reactant Concentration on the Reaction Rate of Ion A and Ion B in Aqueous Solution to Form AB.

Experiment Purpose:

To determine the relationship between the initial rate of reaction between ammonium ion and nitrite ion when the concentrations of the ions are varied.

Experiment Number

Initial [A], M

Initial [B], M

1

0.0100

0.200

0.540 x 10-3

2

0.0200

0.200

1.08 x 10-3

3

0.0400

0.200

2.15 x 10-3

4

0.0600

0.200

3.23 x 10-3

5

0.200

0.0200

1.08 x 10-3

6

0.200

0.0400

2.16 x 10-3

7

0.200

0.0600

3.24 x 10-3

8

0.200

0.0800

4.33 x 10-3

Conclusions:

Initial Rate of Formation AB, M/s

How does the concentration of the reactants affect the initial rate of reaction? Specifically, is there a mathematical relationship that can be stated?

68

The experiment above shows: Rate ∝ [A][B] And, you hopefully recall that a proportion can be turned into an equation by multiplying through by a constant, which gives: rate = k[A][B] where k is the rate constant for the reaction. The rate constant provides a wealth of information in the study of kinetics, as we shall see soon. The dependence of the rate on the concentration of reactants is called a rate law. Taken together, the rate law and rate constant for a chemical reaction allows one to determine the rate of reaction for any set of concentrations. You can see that rearranging the equation and using known concentrations and rates can allow you to determine the rate constant. Rate constants are valid at given temperatures – the rate constant changes as reaction conditions are warmed or cooled.

Practice 4.6

Use the data in the experiments on Page 68 to determine the value of the rate constant for the reaction between Ion A and Ion B.

Use the rate constant calculated above to determine the rate of reaction when initial [A] is 0.100 M and the initial [B] is 0.100 M.

REACTION ORDER

The equation shown as average rate = k[A][B] indicates that the concentrations of both substances are raised to the first power in the rate law. This is not always the case. Several additional rate laws are shown in here:

2N2O5(g) → 2NO2(g) + O2(g)

rate law = k[N2O5]2

H2(g) + I2(g) → 2HI(g)

rate law = k[H2]0[I2]

The exponents (often shown as m and n) in a rate law are called reactant orders, and the sum of the individual reactant orders is the overall reaction order for a chemical equation, and they can only be determined experimentally. The overall reaction order for the ammonium nitrate reaction is two, while the reaction orders for the overall reactions shown above are one and two, respectively. , the overall reaction order for a chemical reaction is the sum of the orders of the reactants.

69

The reaction order can tell you quite a bit about how the variation of the concentration of reactants affects the rate of the reaction. If you double the concentration of a reactant that is first order, then you cause a proportional change in the rate of the reaction. And, if a reactant is second order, the rate change is the square of the change in reactant concentration; e.g., doubling the reactant concentration effects a four times change in rate; tripling the reactant concentration effects a nine times change in rate. This is summarized here: •

reactant is zero order: change to concentration effects no change in rate

•

reactant is first order: change to concentration effects a change in rate equal to that of the change in concentration

•

reactant is second order: change to concentration effects a squared change to rate

METHOD OF INITIAL RATES: USING INITIAL RATES TO DETERMINE RATE LAWS AND RATE CONSTANTS

Recognize that rate laws must be determined experimentally – we cannot look at a chemical equation and determine the rate law. Like we determined the rate law for the reaction in the experiment on Page 68, chemists can vary the concentrations of reactants and see how this affects the rate of the reaction; this is a method of determining a rate law.

1.

Using data provided, determine the effect of the change of concentration of the reactant(s) on the rate of the reaction; this gives the rate order(s) for the reactant(s).

2.

Write the rate law for the reaction: rate = k[A]m[B]n, where m and n are the rate order.

3.

Use the rate law to determine the rate constant by rearranging the rate law expression (substitution of data) – watch units very carefully! You can then also determine an actual rate using the rate law that has been determined.

Practice 4.7 Use the data provided to determine the rate law and rate constant for the reaction A + B → C. Trial

[A], M

[B], M

Initial Rate, M/s

1

0.100

0.100

4.0 x 10-5

2

0.100

0.200

1.6 x 10-4

3

0.200

0.200

6.4 x 10-4

Determine the rate law for the reaction. Justify your reasoning mathematically and in text.

70

What is the overall order of the reaction? Explain your answer.

Determine the rate constant for the reaction at the temperature of the trials.

What is the rate of reaction when [A] = 0.050M and [B] = 0.100M?

What concentrations should be used to obtain a rate that is twice that of the rate found in the previous item?

71

CONCENTRATION CHANGES OVER TIME •

FIRST-ORDER REACTIONS

Reactions in which the rate depends upon the concentration of a single reactant raised to the first power, as in average rate = k[A] are termed first-order reactions, and the equation

‫ ࢚࢑ – = ࢚]ۯ[ܖܔ‬+ ‫]ۯ[ܖܔ‬૙

allows us to calculate the concentration of a reactant, [A]t, after some time t has elapsed when we know the initial concentration, [A]0, and the rate constant k. The equation shown in bold above is called the first-order integrated rate law.

Practice 4.8

i) What is the concentration of an insecticide in a body of water 6 months after a spill? The initial concentration is 4.5 x 10-4 g/cm3, and the first-order rate constant for the reaction is 0.94 yr-1.

ii) How long will it take for the [insecticide] to fall to an EPA-considered safe level of 1.0 x 10-12 g/cm3?

The rate constant for the first-order reaction of the decomposition of acetaminophen, the active ingredient in Tylenol, is 0.0001103 s-1. If the concentration of 1000.0 mg of dissolved Tylenol is initially 0.00063 mg/L in a body, what is the concentration after 8.0 hours?

A notable characteristic of first-order reactants is the graph of concentration versus time. Because the first-order integrated rate law has the form y= mx + b, the graph of ln[A] versus t gives a straight line with a slope of -k and a y-intercept of ln[A]0, which could be useful in determining the order of a reaction from data or for determining a rate constant from graphical data.

Figure 21. The graph of the natural log of the concentration of a first-order reactant gives a straight line with slope equal to –k. 72

FIRST-ORDER HALF-LIFE

The half-life of a first-order reactant, t½, is the time required for the concentration of a reactant to drop to one-half its initial value. The half-life of a first order reactant is given by:

t½ =

0.693 k

[This is not a new equation – it’s just the first-order rate law simplified using [A]t = ½ and [A]0 = 1.]

Half-life calculations are important in medicine, as we will see in the exercises for this section, and we shall visit half-life again when we study nuclear chemistry. Notice that the half-life of a first order reactant does not contain a concentration variable – the half-life of a first order reactant does not depend upon the initial concentration of the

Figure 22. Notice that the half-life of the first-order reactant does not change as time or concentration change. Nuclear decay follows first order kinetics.

reactant.

We can rearrange the first-order integrated rate law to solve for k, time, or initial or final concentrations. This is, as mentioned earlier, relevant to the dosing of medicine and nuclear decay.

Practice 4.9

What is the half-life of the Tylenol in Question B on Page 72?

What amount of time has elapsed if the concentration of acetaminophen is found to be 0.0000014 mg L-1?

73

A certain isotope of nitrogen has a half-life of 455 years. Consider an ancient plant species that incorporated an amount of this isotope of nitrogen into its structure before it died. The amount of this isotope in the remains of the plant is found to be 0.0065% that of similar plants today. How long ago did the plant die?

For the item above, identify two assumptions that make this method of dating fossils and other remains valid.

SECOND-ORDER REACTIONS Reactions in which the rate depends upon the concentration of a single reactant raised to the second power, as in average rate = k[A]2, or two reactants each raised to the first-power, as in average rate = k[A][B] are termed second order reactions, and the equation ࢑࢚ =

૚ ૚ − [‫࢚]ۯ‬ [‫]ۯ‬૙

allows us to calculate the concentration of a second-order reactant A after some time t has elapsed when we know the initial concentration, [A]0, and the rate constant k. The equation shown in bold above is called the second-order integrated rate law. As with the first order reactants discussed on Page 72, a particular characteristic is notable on the graph of a second order reactant; however, in the case of a second order reactant, a straight-line graph is obtained when we plot the inverse of the concentration of the reactant versus time. In the case of a second order reactant, the slope is equal to k, and the yintercept is equal to 1/[A]0.

Figure 23. The graph of the inverse of the concentration of a second-order reactant gives a straight line with slope equal to k.

74

SECOND-ORDER HALF-LIFE

For a second-order reaction, the half-life depends upon the concentration of the reactants, and therefore changes over time. This is illustrated in Figure 24.

The complexity of dealing with second-order halflives places this concept beyond the scope of a firstyear college course; as such, we shall not explore the idea further. However, one should know at least what has been discussed to this point about secondorder half-life.

Figure 24. The half-life of the second-order reactant depicted above increases with each successive half-life. (Compare to Figure 22.)

Part 4: RATE AND TEMPERATURE

We have generally recognized that chemical reactions proceed faster as the temperature of the reactants is increased – this is apparent when you think about baking more quickly when the oven is turned to a higher temperature setting. You should realize, then, that the rate constant for a reaction that increases in rate as the temperature increases also increases.

THE ARRHENIUS EQUATION

Arrhenius noticed that the increase in the rate of reaction as temperature was increased was non-linear. Instead, he found that the increases in the rates of reactions at higher temperatures Figure 25. Taking the natural log of both sides of the Arrhenius equation allows us to use slope to find Ea. Slope = -Ea/R. Because the Ea is in joules, we use 8.315 J/mol K mol for R.

were logarithmic increases. Arrhenius found that most reactions obeyed the following equation, the Arrhenius equation:

ࢌ࢘ࢇࢉ࢚࢏࢕࢔ ࢕ࢌ ࢉ࢕࢒࢒࢏࢙࢏࢕࢔࢙ ࢝࢏࢚ࢎ ࡱࢇ = ࡭ࢋ–ࡱࢇ/ࡾࢀ

where A is a constant called the frequency factor (related to the number of collisions that will be favorably oriented for reaction), Ea is the activation energy of the reaction, R is the gas constant 8.314 J/mol K, and T is the kelvin temperature. Note carefully the discussions in the caption for Figure 25 and on Page 76. 75

Be sure to notice this:

As the value of Ea increases, the value of k decreases

As the value of T increases, the value of k increases

If we take the natural log of the Arrhenius equation, as shown below, then we obtain a convenient equation in the form y = mx + b, for which we see that the ln k = y, m = –Ea/R = slope, x = 1/T and ln A = y-intercept. (Please note that this A refers to the collision frequency, not the [A] we have used to represent concentration.) As it is, the effect of the collision frequency is insignificant in calculations and can be ignored. This equation is useful to graphically determine the activation energy of a reaction when trials are run at several temperatures.

࢒࢔ ሺࢌ࢘ࢇࢉ࢚࢏࢕࢔ ࢕ࢌ ࢓࢕࢒ࢋࢉ࢛࢒ࢋ࢙ ࢝࢏࢚ࢎ ࡱࢇ ሻ =

– ࡱࢇ ૚ ൬ ൰ ࡾ ࢀ

We can also use k measured at two temperatures and the derived equation below to determine the activation energy when the rate constant at two temperatures is known. Alternatively, knowing the activation energy and one rate constant will allow us to determine the rate constant at another temperature.

࢒࢔

࢑૚ ࡱࢇ ૚ ૚ = ൬ − ൰ ࢑૛ ࡾ ࢀ૛ ࢀ૚

Practice 4.10

What is the rate constant for a particular reaction at 430.0 K if the activation energy of the reaction is 160 kJ/mol, and the rate constant at 462.9 K is 2.52 x 10-5 s-1?

What is the activation energy for the reaction graphed in Figure 25?

76

Part 5: REACTION MECHANISMS

Generally, chemical reactions do not occur in a single step. The steps that occur during a chemical reaction as the reactants change into products are called the reaction mechanism for the reaction reaction. Reaction mechanisms attempt to show any intermediate steps that occur as a chemical reaction proceeds.

As an analogy, imagine that someone asked you what path you took to visit a relative in Boston over winter break. You might say that you left home and went north to Boston Boston. Of course, this description misses many of the smaller steps that you took on your trip. For example, a more complete description might be that you took Route 4 north out of Calvert County, got onto the Beltway at Exit 11, took the Beltway north to the I-95 95 north exit in College Park, stopped at toll booths here here-andthere, exited at rest stops, etc. This description provides a more detailed look at what occurred during the trip rather than just looking at the beginning and the end. Similarly, a reaction mechanism echanism does not look at simply the beginning and end, but instead it looks at the beginning, end and what is occurring between those two points. •

Elementary Steps

When a process that occurs in a chemical reaction involves only a single step, then the st step is called an elementary step. The isomerization of isonitrile into acetonitrile is an elementary step. Many reactions go through several elementary steps before the final products are formed. The reaction between NO(g) and ozone, O3(g), occurs in an ele elementary step if the molecules collide with sufficient energy and in the proper orientation:

NO(g) + O3(g) → NO2(g) + O2(g)

If this were not a reaction that occurred in a single, elementary step, then we would expect something to form after the NO and a O3 collide and before the NO2 and O2 are produced – but this is not the case.

The number of molecules that collide in an elementary step defines the molecularity of the step: single molecules participate in unimolecular steps, two molecules participate in bimolecular steps and three molecules participate in termolecular steps. Termolecular events are not likely, as these involve three molecules coll colliding iding in the proper orientation, all of which must possess sufficient kinetic energy. •

A Multistep Mechanism is a Series of Elementary Steps

Look at the reaction below, which shows the overall reaction of NO2(g) and CO(g), which produces NO(g) and CO2(g):

NO2 (g) + CO(g) → NO (g) + CO2(g)

77

Although we may think of this as the chemical reaction that occurs, it is, in fact, the overall reaction for a series of elementary processes that occur to effect the production of NO (g) and CO2(g):

Elementary Step 1:

NO2(g) + NO2(g) → NO3(g) + NO(g)

Elementary Step 2:

NO3(g) + CO(g) → NO2(g) + CO2(g)

In this reaction mechanism, you can see that the first process to occur is the collision of two molecules of NO2(g) to form NO(g) and NO3(g). The NO3(g) formed in this first step later collides with CO(g) to produce the products NO2(g) and CO2(g). The NO2(g) formed then continues on to react with additional NO2(g) as in Step 1. The overall reaction can be shown like this:

NO2(g) + NO2(g) + NO3(g) + CO(g) → NO3(g) + NO(g) + NO2(g) + CO2(g)

Canceling species that occur as both reactants and products gives NO2(g) + CO(g) → NO(g) + CO2(g) as the overall reaction.

The species formed in elementary steps that are used in subsequent elementary steps are called intermediates. Intermediate chemical species appear in all multistep mechanisms. Do not confuse an intermediate with the activated complexes or transition states we talked about earlier – intermediates are stable chemical species that form as chemical reactions occur; transition states are unstable “inbetween” species that exist along the path from reactants to products. We can assume that a proposed mechanism is potentially plausible if the mechanism provides for the net overall reaction; this is only the first test – we must also look at the slow step.

Practice 4.11 The conversion of ozone into oxygen appears to follow a two-step process: the first elementary step involves the decomposition of ozone into diatomic molecular oxygen and monatomic oxygen gas; the second elementary step appears to involve the reaction between ozone and the monatomic oxygen formed in the first step to produce diatomic oxygen gas.

A) Write chemical equations for each elementary step

B) Identify the molecularity of each step

C) Identify any intermediates that form

D) Write an overall chemical equation for the conversion of ozone into oxygen gas without any intermediates shown

E) What is the molecularity of the overall reaction? 78

Look at the reaction shown here:

Mo(CO)6 + P(CH3)3 → Mo(CO)5P(CH3)3 + CO

Is the following proposed reaction mechanism potentially consistent with the overall reaction? That is, could it be possible to explain the reaction shown above in the series of elementary steps shown here? Explain your response.

•

Step 1:

Mo(CO)6 → Mo(CO)5 + CO

Step 2:

Mo(CO)5 + P(CH3)3 → Mo(CO)5P(CH3)3

Rate Laws of Elementary Steps

The rate law of an elementary step can be determined by the stoichiometry of the elementary step. (This is not true for overall reactions.) Recall that a rate law describes how the concentrations of reactants affect the rate of a reaction. Thus, we can imagine that there is a relationship between the molecularity of an elementary step and its rate law. Indeed, this is the case. It is essential to note that the rate laws for reactions occurring in only a single step can be determined by their stoichiometry.

Unimolecular elementary steps have first-order rate laws

A reaction that is unimolecular depends only upon the concentration of a single reactant; thus, a unimolecular elementary step is first order in the reactant. A → products

average rate = k[A]

Bimolecular elementary steps have second-order rate laws

A reaction that is bimolecular depends upon the concentration of either of two reactants; thus; a bimolecular elementary step is second order overall, as it is first order in each reactant. A + B → products

average rate = k[A][B]

79

Practice 4.12

What is the rate law for the reaction between hydrogen gas and bromine gas if the reaction produces HBr in a single step?

In the production of two molecules of NOBr, it has been proposed that the mechanism 2 NO + Br2 → 2 NOBr might yield the product.

A) Write the equation for this reaction

B) What is the molecularity of the reaction? Justify your answer.

C) What is the rate law if the reaction follows the proposed mechanism?

D) Is this mechanism likely? Explain.

•

Rate-determining Steps and Rate Laws

Imagine the trip to Boston we spoke of earlier. Although we did not discuss it at the time, we might consider the amount of time t that the trip would take using the path we have chosen. Now imagine that at some point there is a series o off toll plazas that you must through on your trip. Plazas 1 – 4 and Plaza 6 can 1000 cars per hour, but Plaza 5 is under construction and only has 3 available booths – it can allow only 300 cars to per hour. Thus, traffic backs up at this pla plaza. za. How many cars are able to travel from Plaza 1 to Plaza 6 in a given amount of time? We can say that the rate of traffic (number of cars per unit of time) moving from Plaza 1 to Plaza 6 is limited by the rate at which traffic can move through Plaza 5 – Plaza 5 is the rate-determining plaza.

Plaza 1

Plaza 2

Plaza 3

Plaza 4

Plaza 5

Plaza 6

1000 hr-1

1000 hr-1

1000 hr-1

1000 hr-1

300 hr-1

1000 hr-1

80

The number of cars that can travel from Plaza 1 to Plaza 6 in a given amount of time is determined by the rate at which the cars can through Plaza 5, which is the “slow plaza.” Even though a later plaza – Plaza 6 – can allow 1000 per hour, this does not speed up the rate of cars ing through all 6 plazas – Plaza 5 still limits the rate of traveling the full length.

In a similar but less wordy manner, the slowest step of a multistep reaction mechanism limits the rate of a chemical reaction. If two steps are fast and a third step is slow, the overall reaction cannot proceed faster than the slowest step, which we call the ratedetermining step, or slow step. Review the equation we saw earlier for the reaction between NO and CO: NO2(g) + CO(g) → NO(g) + CO2(g) Recall that two steps were involved in this reaction: Elementary Step 1:

NO2(g) + NO2(g) → NO3(g) + NO(g)

Elementary Step 2:

NO3(g) + CO(g) → NO2(g) + CO2(g)

It has been found that the rate law is second order in NO2 and zero order in CO, rate = k[NO2]2. As such, an increase in the concentration of CO does not affect the overall rate of the reaction. Can you identify the slow step of this reaction and the fast step? The slow step is the step whose rate law matches the overall rate law, which has been determined to be k[NO2]2. Let’s accept that the following is true:

Many reactions occur in more than a single step

The reaction mechanism for a reaction proposes to show the elementary steps that occur from reactant to product

Some elementary steps are slower than others

The rate law for an elementary step can be easily determined by its stoichiometry

In order for a reaction mechanism to be plausible it must provide for the overall balanced chemical equation.

Understanding these items, our final discussion involves how we determine if a proposed mechanism is consistent with an observed rate law.

There are, then, two factors to discuss in determining whether a proposed reaction mechanism is consistent with the observed rate law:

1) Does the mechanism provide for the overall reaction? That is, do the reacting species cancel to provide for the overall reaction?

If a proposed reaction mechanism does not sum to the balanced chemical equation, then it is not a consistent reaction mechanism.

If a proposed reaction mechanism does sum to the balanced chemical equation, then it might be consistent if the second test provides for the observed rate of the overall reaction.

2) Does the reaction mechanism provide for the overall order of the reaction?

In determining the answer to this question, we look at the slow step of the mechanism. If the slow step is or can be made equivalent to the overall reaction order – and the answer to Test 1 is “yes” – then the mechanism is deemed plausible.

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Practice 4.13

As an example, look at the proposed mechanism below for the reaction between NO2 and CO to produce CO2 and NO:

NO2(g) + CO(g) → CO2(g) + NO(g)

observed rate law:

average rate = k[NO2]2

Proposed mechanism:

NO2(g) + NO2(g) → NO3(g) + NO(g)

slow

Test 1: Is the mechanism consistent with the

NO3(g) + CO(g) → NO2(g) + CO2(g)

fast

overall reaction?

Test 2: Is the rate law of the slow step consistent with the overall rate law?

As a second example, look at the proposed mechanism below for the reaction between NO and Cl2 to produce NOCl:

2NO (g) + Cl2(g) → 2NOCl (g)

observed rate law:

average rate = k[NO]2[Cl2]

Proposed mechanism:

NO (g) + Cl2(g) ⇋ NOCl2(g)

fast

Test 1: Is the mechanism consistent with the

NOCl2(g) + NO(g) → 2NOCl(g)

slow

overall reaction?

Test 2: Is the rate law of the slow step consistent with the overall rate law?

[We can substitute the reactants of an equilibrium step (or any fast step preceding a slow step, as these are always equilibrium steps) for the reactants of the subsequent slow step to apply Test 2.]

In summary:

1) Determine if the reaction mechanism is consistent with the observed balanced chemical equation.

2) If the mechanism is consistent, check the slow step to determine if it is consistent with the overall rate expression:

2a) No equilibrium step:

Directly determine the rate law for the slow step

2b) Equilibrium step:

Substitute the intermediate for its equilibrium species; determine the rate law for the new expression. Realize that a step prior to a slow step is an equilibrium step. 82

Practice 4.14

Why would a step prior to a slow step be at equilibrium?

Use the equation below to answer the following question. 2A + B → C + D

average rate = k[A][B]

Which of the following mechanisms is consistent with the observed rate law? If a mechanism is not consistent, then note why it is not. For any that are consistent, justify your selection.

(a)

(b)

A+B⇋C+M

fast equilibrium

M+A→D

slow

B⇋M

fast

M+A→C+X

slow

A+X→D

(c)

A+B⇋M

M+A→C+X X→D

Use the equation below to answer the following question.

2 NO + Br2 → 2 NOBr

average rate = k[NO]2[Br2]

Is the following mechanism consistent with the observed rate law? Justify your response.

Br2 + NO → NOBr2

slow

NOBr2 + NO → 2 NOBr

fast

83

Use the equation below to answer the following question.

H2 + 2 IBr → 2 HBr + I2

average rate = k[H2][IBr]

Which of the following mechanisms is likely given with the observed rate law? Justify all of your selections.

(a)

2 IBr + H2 → 2 HBr + I2

(b)

H2 + IBr → HI + HBr HI + IBr → HBr + I2

(c)

H2 + IBr → HBrI + H H + IBr → HBr + I HBrI → HBr + I I + I → I2

The reaction 2 NO + Cl2 → 2 NOCl has been found to observe the rate law average rate = k[NO]2[Cl2]. The following mechanism has been proposed for the reaction:

NO + Cl2 ⇋ NOCl2 NOCl2 + NO → 2 NOCl

Which step is fast and which step is slow? Justify your response.

The following mechanism has been proposed for the reaction between NO and hydrogen to form dinitrogen oxide and water: NO(g) + NO(g) ⇋ N2O2(g)

N2O2(g) + H2(g) → N2O(g) + H2O(g)

The proposed rate law is rate = k[NO]2[H2]. Justify that the mechanism is consistent with the rate law.

84

Part 6: CATALYSTS

A catalyst generally provides another mechanism for a reaction to follow. The steps of the catalyzed mechanism have lower activation energies, and thus lead to an increase in the overall rate of reaction. That is, the activation energy of the overall reaction is lowered, as shown in the figure at left. Notice that this does not change ∆H, as the initial and final energies are the same for the catalyzed or uncatalyzed reactions.

A catalyst can be easily recognized by the fact that it is present both prior to and at the Figure 26. Notice that the only change here is the activation energy of the reaction, probably owing to the fact that the catalyzed reaction follows a different mechanism. You should notice that the thermodynamics of the reaction – the enthalpy change – is not affected by a catalyst. And, be sure to notice that the activation energy of the reverse reaction is equal to the enthalpy change plus the Ea of the forward reaction.

conclusion of a chemical reaction. This is in contrast to an intermediate, which is a product in one step and a reactant in a subsequent step. Enzymes are examples of catalysts.

85


Chemical Equilibrium I: Generalized Equilibrium, the Equilibrium Expression and Le Châtelier's Principle


describe the nature and behavior of systems at equilibrium

graphically represent equilibrium and establish the kinetics kinetics-equilibrium relationship

write equilibrium expressions

predict equilibrium behavior using the magnitude of Kc

justify and perform the manipulation of equilibrium constants

calculate equilibrium constants, and use equilibrium constants to determine equilibrium concentrations

determine and use the reaction quoti quotient

predict the behavior of systems using Le Châtelier's Principle; propose changes to a system that will elicit desirable responses

86

Part 1: THE CONCEPT OF EQUILIBRIUM

Imagine the decomposition reaction of N2O4, a colorless gas, into NO2, a brown gas. The reaction equation is shown here:

N2O4(g) → 2 NO2(g)

While the equation is shown with an arrow to indicate the complete conversion of N2O4 into NO2, this is not the case Figure 27. The conversion of N2O4 to NO2 is a dynamic process that occurs in both directions.

for this reaction, and it is not the case for thousands more. Instead, the conversion of N2O4 reaches a point of dynamic equilibrium with NO2, in which the forward reaction,

N2O4 → 2NO2, occurs at the same rate as that of the reverse reaction, 2NO2 → N2O4. This is one of many cases of chemical equilibrium, in which forward and reverse processes occur at equal rates when the concentrations of reactants and products reach specific concentrations. It is chemical equilibrium that we will look at in this segment of the course.

Before we begin:

Only systems in which the reactants and products cannot be removed or added can reach equilibrium. Open systems cannot reach equilibrium, although they will continue toward it.

Reactions do not stop when equilibrium is reached, but instead they occur in both directions at equal rates.

Reactions that have reached equilibrium do not necessarily have equal concentrations of reactants and products. The system has equilibrium concentrations of reactants and products.

The importance of the statements above cannot be ignored if one is to understand chemical equilibrium.

Part 2: GRAPHIC REPRESENTATION OF EQUILIBRIUM

The graph on Page 88 makes a salient point surrounding equilibrium: the concentrations are not necessarily equal at equilibrium, and there is a definite ratio between products and reactants at a specific temperature.

Rather than going to completion – where the concentration of reactants is zero and the reaction vessel contains only product – the reaction continues until there is some concentration of both reactants and products in the reaction vessel, and the concentrations at this point are the equilibrium concentrations. It is at the point marked with the dashed line that the processes of the forward and reverse reactions are occurring at the same rate. Notice that it does not mean [N2O4] = [NO2]. At equilibrium, the production of NO2 equals the production of N2O4, and the concentrations of the reactants and products do not change (at a specific temperature – the ratio will change at other temperatures).

87

Because this reaction does not go to completion, we will use the double

arrow, ⇋, to indicate that the reaction reaches an equilibrium condition.

This is used for all reactions that reach equilibrium as opposed to going to completion. One does not use a double-ended arrow to show this. Consider the presence of this double arrow to mean that there are, in fact, two reactions occurring: N2O4 → NO2 and NO2 → N2O4.

Figure 28. At equilibrium, the concentrations of the products and reactants are not necessarily equal, but rather are present in some definite ratio.

Part 3: QUANTITATIVELY EXPRESSING EQUILIBRIUM

At equilibrium, there is some ratio between the concentrations of products and reactants. This ratio can be expressed in an equilibrium expression, to which we shall turn our attention momentarily. We will use concentration units – molarity – and partial pressure units – atm – for equilibrium expressions. First, however, we would like to conceptually derive the equilibrium expression. (There are two approaches to establishing the foundation of equilibrium: a thermodynamic approach and a kinetics approach. In that we have studied kinetics, we will use this approach now, and we will examine the thermodynamic approach at an appropriate time.) Let’s consider the

reaction we saw above: N2O4(g) ⇋ 2 NO2(g).

Practice 5.1

Write the rate law for each of the reactions here, N2O4(g) ⇋ 2 NO2(g); assume they occur in single steps..

At equilibrium, what relationship can we make between these two rate laws and their constants?

Because the rate laws at equilibrium are equivalent values (i.e., average rateforward = average ratereverse) we can show that their rates are equal using their rate expressions:

88

It is a mathematical fact that the quotient of two constants is itself a constant, and this one – the quotient of the rate constants in the rate laws above – is called the equilibrium constant. We will see more on the equilibrium constant in Part 4. For now, we can see that the concentration of reactants and products at equilibrium equals some ratio, which we label Kc or Kp.

We also note that the initial concentrations of reactants and products do not affect the ratio present at equilibrium or whether or not equilibrium will be reached. Instead, the equilibrium ratio is independent of the initial concentrations. So, although the actual concentrations present at equilibrium are affected by the initial

Figure 29. The figure shows what we saw earlier: there are in essence two reactions occurring at equilibrium, and the forward and reverse rates are equal.

concentrations, the ratio of the concentrations is not. It is very important to keep in mind that the equilibrium constant is a ratio – not an actual value of concentrations!

Finally, before continuing on to writing equilibrium expressions and their applications, we must note that equilibrium will be reached from either direction. The completion of the table below will illustrate the two previous not-to-be-missed points. You will explore these ideas in the laboratory.

Part 4: THE EQUILIBRIUM CONSTANT

Because the concentrations of reactants and products present at the beginning of the reaction do not affect the ratio of the concentrations present at equilibrium, the equilibrium expressions are much easier to write than the rate law expressions we saw in our study of chemical kinetics. (Recall that rate law expressions may or may not depend upon the initial concentrations of reactants.) Equilibrium expressions can be written using the stoichiometric coefficients in a balanced chemical equation. This is accomplished because the equilibrium ratio of reactants and products is only dependent upon the stoichiometry, not the initial concentrations or the mechanism.3 Given an equation such as the one shown here at equilibrium,

aA + bB ⇋ cC + dD the equilibrium expression for the forward reaction is written as

ࡷࢉ =

3

[۱]ࢉ [۲]ࢊ [‫[ ࢇ]ۯ‬۰]࢈

If you are interested, a more complete understanding of this is explained by the law of mass action.

89

And, because equilibrium can be reached from either direction, we could also write the expression for the reverse reaction as:

ࡷࢉ ᇱ =

[‫[ ࢇ]ۯ‬۰]࢈ [۱]ࢉ [۲]ࢊ

Then, substitution of equilibrium concentrations provides the value of the equilibrium constant. You must specify the direction for which the expression is written. You can see that the two values, Kc and Kc′, are reciprocals of one another, but they are meaningless unless the direction of the reaction is provided. Equilibrium expressions, like rate law expressions, are valid only at specific temperatures, too.

When we measure the equilibrium molar concentrations of species, we use the equilibrium constant Kc, and we often use Kp for gasphase reactions, where the equilibrium partial pressures of the gases are in atmospheres. For example, for the gas phase reaction

2A(g) + B(g) ⇋ C(g) + 3D(g), we would represent Kp as

ࡷ࢖ =

ࡼ࡯ ࡼ૜ࡰ ࡼ૛࡭ ࡼ࡮

The relationship between Kc and Kp is given as:

ࡷ࢖ = ࡷࢉ ሺࡾࢀሻ∆࢔ where ∆n is (mole product gas – mol reactant gas), R = 0.0821 L-atm/mol-K, and T is kelvin temperature

Practice 5.2

Write equilibrium expressions for the following reactions in both the forward and reverse directions. For B, show both Kc and Kp expressions. A) 2O3(g) ⇋ 3O2(g)

B) 2NO(g) + Cl2(g) ⇋ 2NOCl(g)

C) Ag+(aq) + 2NH3(aq) ⇋ Ag(NH3)2+ (aq)

90

Part 5: THE MAGNITUDE OF THE EQUILIBRIUM CONSTANT

Many industrial processes depend upon the magnitude of the equilibrium constant. If a process is not likely to go to products at certain conditions, then experiments will be done to determine how to make the reaction more favorable for the production of products. The magnitude of the equilibrium constant can allow us to determine whether a reaction is product product-favored favored or reactant-favored. reactant

The equilibrium constant tells us about the composition of an eequilibrium quilibrium mixture: its magnitude, in essence, lets us determine whether a reaction is more likely to occur in the forward direction or reverse direction. Thus, it tells us whether the equilibrium mixture m will contain significant reactant or significant product oduct at equilibrium.

side This means that the A very large equilibrium constant, Kc > 1, tells us that the reaction is favored to the right, or product side. reaction will produce significant product, and the equilibrium mixture will contain a small amount of reactant. We can say that the “equilibrium lies to the right,” or the reaction is “product “product-favored.”

A very small equilibrium constant, Kc < 1, tells us that the reaction is favored to the left, or reactant side. side This means that the reaction will not produce roduce significant product, and the equilibrium mixture will contain a large larger amount of reactant. We can say that the “equilibrium lies to the left,” or the reaction is “reactant “reactant-favored.”

, the direction from which the equilibrium constant is det determined must be specified.. It is not useful to simply say that the magnitude of Kc is 1.49 x 108. We must say that the equilibrium constant of the forward reaction is 1.49 x 108.

Practice 5.3 Determine and discuss the utility of the following reactions in producing product by the reactions shown. That is, are the reactions product-favored favored or reactant reactant-favored under the specified conditions? N2(g) + O2(g) ⇋ 2NO(g)

Kc = 1 x 10-30 at 298 K

H2(g) + I2(g) ⇋ 2HI(g)

Kp = 794 at 298 K

Why does Kc = Kp for the reactions shown above?

91

Part 6: MANIPULATING EQUATIONS AND Kc VALUES

We will consider at least three ways in which equations and their equilibrium constants can be manipulated: •

Changing a chemical equation – for example, by multiplying through by a value – increases the value of the equilibrium constant by a power equal to the change in the equation. This is easily seen by rewriting the equilibrium expression using the

new stoichiometry. For example, multiplying the equation N2(g) + O2(g) ⇋ 2 NO(g) by two yields 2 N2(g) + 2 O2(g) ⇋ 4 NO(g), which

causes the equilibrium expression to increase by the power of two, as in: Kc (N2(g) + O2(g) ⇋ 2 NO(g)) = 1 x 10-30 at 298 K, while Kc (2 N2(g) + 2 O2(g) ⇋ 4 NO(g)) = 1 x 10-60 at 298 K

•

Combining equations to arrive at an equation that is the sum of those combined requires that the individual equilibrium constants are multiplied by one another. This is easily seen by setting two expressions equal to a common species and rearranging the expressions. For example, we might be interested in the value of the equilibrium expression for the reaction 2

NOBr(g) + Cl2(g) ⇋ 2 NO(g) + 2 BrCl(g). Although we do not have this value, we do know the values of the equilibrium constants for

two reactions that sum to provide the reaction of interest: 2 NOBr(g) ⇋ 2NO(g) + Br2(g) Br2(g) + Cl2(g) ⇋ 2BrCl(g)

Kc = 0.42 Kc = 7.2

Multiplying these values of Kc gives (0.42 • 7.2) = 3.0 as the equilibrium constant for the reaction of interest. •

Reversing an equation would require using the inverse of the given Kc, as we saw in Part 4. This can easily be seen because to reverse a reaction requires the “flipping” of products and reactants.

Practice 5.4 2 A ⇋ 2 B + C (Kc,1 = 4.0) and C + D ⇋ 2 E (Kc,2 = 6.0) Show that doubling a reaction’s stoichiometry results in squaring the value of its equilibrium constant.

Show that summing the two reactions above to yield 2 A + D ⇋ 2 B + 2 E gives Kc,3 = Kc,1 • Kc,2

92

Part 7: HETEROGENEOUS EQUILIBRIA

So far we have only considered the equilibrium established between species in the same phase – e.g., aqueous or gas. However, we should also consider the equilibria that may be established when the reactants and products may not be in a single phase. For example, equilibrium is established in the following reaction between all three species: CaCO3(s) ⇋ CaO(s) + CO2(g) For reasons we shall not consider4, it is possible to write an equilibrium expression for reactions occurring with heterogeneous species without including pure solids or liquids. Thus, the equilibrium expression for the above reaction is

Kp = PCO2 The same could be said for a solvent like water. •

Pure solids and pure liquids do not appear in equilibrium expressions. However, you must not interpret this to mean that pure solids and liquids are not present at equilibrium or are not part of the chemical equation.

Practice 5.5 Determine the value of the equilibrium constant for the reaction: 2HF(aq) + C2O42-(aq) ⇋ 2F-(aq) + H2C2O4(aq) given the following data:

HF(aq) ⇋ H+(aq) + F – (aq)

H2C2O4(aq) ⇋ 2H+ + C2O42-(aq)

Kc = 6.8 x 10-4 Kc = 3.8 x 10-6

What do the values of Kc in the original equations tell us about the direction of the equilibria for the two reactions shown?

Use the values of the equilibrium constants to justify that the two acids are weak.

Is the reaction of interest likely to occur to a great extent to products? Justify your answer.

4

This stems from the law of mass action; the values are essentially 1 for these phases.

93

Write the equilibrium expressions for the following heterogeneous reactions. CO2(g) + H2(g) ⇋ CO(g) + H2O(l)

SnO2(s) + 2CO(g) ⇋ Sn(s) + 2CO2(g)

Sn(s) + 2H+(aq) ⇋ Sn2+(aq) + H2(g)

Consider for a moment the underlying concept behind the magnitude of Kc. Discuss how this might be related to the spontaneity of the reactions we saw in voltaic cells.

Part 8: CALCULATING EQUILIBRIUM CONSTANTS

There are two methods of calculating an equilibrium constant, and we will look at how to do this now. The purpose of calculating Kc might be to determine the favorability of a process at a certain temperature. •

If you know the chemical equation and the values of the concentrations or partial pressures of all species at equilibrium, then you can directly calculate the equilibrium constant. Determine the balanced equation and write the expression for the equilibrium constant. Then, substitute values.

94

Practice 5.6 Calculate the equilibrium constants Kp and Kc for the production of ammonia from its elements at 745 K. The equilibrium pressures of the species are given here: N2: 2.46 atm

H2: 7.38 atm

NH3: 0.166 atm

Calculate the equilibrium constant for the ionization of acetic acid at 298 K. The molar concentrations of the species present at equilibrium are given here: [CH3COOH]: 0.0165 M

[H+]: 0.000544 M

[CH3COO-]: 0.000544 M

When species’ equilibrium concentrations or partial pressures are unknown, the equilibrium expression and the stoichiometry of the reaction can provide a method for determining unknown values as long as some data can be obtained. Set up a table of values that you can complete for initial and equilibrium values. Once completed, the equilibrium values will provide the values to use in the equilibrium expression used to determine Kc.

1.

Tabulate the known initial and equilibrium concentrations of the species in the equilibrium constant expression. Do not include pure solid, liquid or solvent species.

2.

Calculate the change in concentrations of the species for which both initial and equilibrium concentrations are known.

3.

Use the stoichiometry of the reaction – the coefficients – to determine the changes in all species from the initial conditions to the equilibrium.

4.

Calculate the equilibrium concentrations. Use these values to calculate Kc.

Over the next few pages we shall complete several items that exemplify the material

Figure 30. Given initial data and determining product data, we can determine Kc.

discussed above. In each case, various data will be available; this should give you a good mix of problems from which you can model solutions to a wide variety of equilibrium situations.

Please do not attempt to memorize a few cases that appear to recur – rather, be sure that you can analyze various scenarios in order to be successful in solving equilibrium problems! 95

Practice 5.7 Determine the value of Kc for the equilibrium established between ammonia and water to produce the ions ammonium and hydroxide. The initial concentration in 5.0 L of water is 0.0124 M ammonia. The [OH-] at equilibrium is 0.000464 M at 298 K. What we have:

We have the equilibrium concentration of a product, and we have the initial concentration of a reactant.

What we need:

We need the equilibrium concentrations of all of the reactants and products in order to determine Kc.

What we do:

We set up a table that allows for a convenient method of determining the equilibrium concentrations of other species.

Set up a table (widely called an “ICE” chart) and insert the known values for the species. An equation may need to be written and balanced first. Tables will include the initial concentrations, the change in concentrations, and the equilibrium concentrations. NH3(aq)

⇋

H2O(l)

NH4+(aq)

OH–(aq)

Initial Change

X

Equilibrium

(Notice that we do not consider the liquid water in this heterogeneous equilibrium problem. Pure solids, liquids and solvents are not considered; here, liquid water is the solvent for ammonia.)

Determine the change values for which initial and equilibrium values are known. Here, we can calculate ∆[OH-]. NH3(aq) Initial

⇋

H2O(l)

0.0124

Change

NH4+(aq)

OH–(aq)

0.0

0.0

X 4.64 x 10-4

Equilibrium

Use the stoichiometry to determine the change and equilibrium concentrations of the other species. NH3(aq) Initial Change Equilibrium

⇋

H2O(l)

0.0124 X

NH4+(aq)

OH-(aq)

0.0

0.0 + 4.64 x 10-4 4.64 x 10-4

Use the equilibrium expression and the concentrations at equilibrium to determine Kc.

96

Determine the value of Kp for the equilibrium established between sulfur trioxide and its decomposition products, sulfur dioxide and oxygen gas. At 1000 K, a container was filled with sulfur trioxide gas to a pressure of 0.500 atm. At equilibrium, the partial pressure of sulfur trioxide is known to be 0.200 atm.

Set up a table and insert the known values for the species. An equation may need to be written and balanced first.

Determine the change values for which the initial and equilibrium values are known.

Use the stoichiometry to determine the change and equilibrium concentrations of the other species.

Use the equilibrium expression and the concentrations at equilibrium to determine Kp.

The pressure of gas PCl5 in an evacuated container was measured as 4.50 atm. After decomposition and the establishment of equilibrium, the total pressure in the container was measured as 7.00 atm. Use this information to determine the values of Kp and Kc for the decomposition of PCl5. The products are PCl3(g) and Cl2(g).

97

Part 9: REACTION QUOTIENTS

For any particular “random” mixture, the reaction may be moving toward equilibrium from reactants, moving toward equilibrium from products, or it may be at equilibrium. We may often be interested in discovering where a mixture lies on the “equilibrium continuum.” •

Determining Q, the reaction quotient

Using known values for a reaction mixture, we can determine the reaction quotient for a particular mixture. The reaction quotient is the mathematical result achieved by inserting known concentration or partial pressure values for a reaction mixture into an equilibrium expression. The reaction quotient is the ratio of products to reactants for a particular mixture of reactants and products. We can compare this ratio to the known ratio provided as Kc, as we shall see on the next page. You should already see that when the concentrations’ quotient equals the value of Kc, then the reaction is at equilibrium. (Kc is really just a special case of Q.)

Practice 5.8 What is the reaction quotient, Q, when a mixture of 2.00 mol H2(g), 1.00 mol N2(g), and 2.00 mol NH3(g) is placed in a 2.00 L container? (Hint: We cannot use mol values in a Kc or Kp expression.)

•

Using Q.

Once we have determined the reaction quotient, we can compare the value of the quotient to the equilibrium constant. The comparison of Q to Kc allows us to determine whether the mixture described will move toward the production of reactants or products, or we will see that the mixture is at equilibrium. Figure 31. The figure represents graphically the continuum from less-than-equilibrium concentrations to greater-than-equilibrium concentrations of products. It is important to note that a reaction will not remain on either side, but will instead work toward equilibrium. 98

If Q < Kc, then the mixture has a larger than equilibrium reactant concentration, and the mixture will produce additional product

If Q = Kc, then the mixture is at equilibrium

If Q > Kc, then the mixture has a larger than equilibrium product concentration, and the mixture will produce additional reactant

It might be convenient to view the nature of Q in the form below to conceptualize its meaning:

Practice 5.9

After writing the balanced equation, determine

When Q for a mixture is equal

to

Kc,

then

the reaction quotient for the following mixture

the

placed in a 0.50 L container. Then, determine

mixture is at equilibrium

whether the mixture is at equilibrium, will result

and will move in both

in the production of reactants or will result in the

directions at equal rates.

production of products. The value of the equilibrium constant is 51 at the temperature of the mixture.

0.02 mol HI(g), 0.01 mol H2(g) and 0.03 mol I2(g)

A+B

C+D

⇋

were mixed in a container. The reaction is the production of HI(g) from its elements.

When Q for a mixture is When Q for a mixture is less

greater than Kc, then the

than Kc, then the mixture is

mixture

“heavy” on reactants and

products and will move

will move toward products.

toward reactants.

is

“heavy”

on

99

Part 10: CALCULATING EQUILIBRIUM CONCENTRATIONS

In Part 8 we calculated the equilibrium constant for reactions in which we knew the equilibrium data for some species and the initial concentration for one or more species. We will now look at how to calculate the equilibrium concentrations of the species in a reaction when we know the equilibrium constant but do not know any equilibrium concentrations. Here, we are asking the question, “What will be the concentrations at equilibrium?”

We can calculate the concentrations of the species of a chemical reaction if we know the balanced chemical equation and the equilibrium constant. This provides the stoichiometry, the equilibrium expression and the ratio of products to reactants at equilibrium (i.e., Kc).

Behind the calculation of equilibrium concentrations is the observation that there is a predictable change in the concentrations based on the stoichiometry of an equation. Thus, we can utilize this to determine equilibrium concentrations. We shall see applications of this soon.

Construct a table similar in format to that used for calculating Kc. This table will include the initial concentrations (if required), the change in the concentrations as the reaction reaches equilibrium and the final equilibrium values.

Use the stoichiometry of the equation to determine the changes in concentration of the species. Because we do not know the final values at equilibrium (this is what we are trying to determine) we must use a variable for the change.

Use the initial and change concentrations to construct expressions for the equilibrium values: [initial] – [change] = [equilibrium]

Substitute the expressions obtained in the previous step into the equilibrium expression for the reaction. This may require the solving of a quadratic equation if your calculator cannot handle variables.

Practice 5.10 A 1.000-L flask is filled with 1.000 mol hydrogen and 1.000 mol iodine gas at 721 K. The value of the equilibrium constant is 50.5 at this temperature. What are the concentrations of the species at equilibrium? The reaction is shown below.

H2(g)

I2(g)

⇋

2HI(g)

Initial Change Equilibrium

100

For the equilibrium N2(g) + O2(aq)⇋ 2 NO2(g) the equilibrium constant, Kp, is 1.0 x 10-30 at 298 K. A gas cylinder is charged with NO (g) to 1.66 atm at 298 K. What are the equilibrium values of all species?

What is the value of Kc for the reaction?

Is the reaction shown above a redox reaction? Justify your response. If it is, identify the oxidation numbers on all atoms on both sides.

Part 11: LE CHÂTELIER'S PRINCIPLE

Henri-Louis Le Châtelier (pronounced “le-SHOT-lee-ay”) discovered that the disturbance of a system at equilibrium will cause the system to move again toward equilibrium. He suggested the following principle, known as Le Châtelier’s Principle:

If a system at equilibrium is disturbed by a change in temperature, pressure or reactant or product concentrations, then the system will respond by shifting its equilibrium position to counteract the effect of the disturbance.

It is important to note that a Le Châtelier response is an initial response that ceases when the disturbance has ed and the reaction again reaches equilibrium. While the system will reestablish equilibrium where Q = Kc, you should note that this does not mean the values of reactants’ and products’ concentrations remain the same – they will change, but their ratio will not change. •

Changes in Temperature

When temperature is increased, the equilibrium shifts in the direction that absorbs heat. Treat “heat/energy” as a reactant (endothermic) or product (exothermic) to consider its effect. For exothermic reactions the equilibrium shifts to form additional reactant; for endothermic reactions the equilibrium shifts to produce additional product.

101

Practice 5.11 Show the use or production of heat in an endothermic and an exothermic reaction. Provide a brief discussion as to how Le Châtelier’s Principle suggests that the equilibrium will shift to the right for endothermic reactions and to the left for exothermic reactions when a reaction mixture is heated or cooled.

•

Changes in Volume and/or Pressure When a gas phase reaction’s volume is decreased, the pressure of the equilibrium mixture must increase – thus, Le Châtelier’s Principle indicates that the equilibrium will shift to decrease the pressure to its equilibrium value. That is, the equilibrium will shift to decrease the gas molecules in the volume: the equilibrium will shift to the side of the reaction that possesses fewer molecules of gas.

When a gas phase reaction’s volume is increased, the pressure of the equilibrium mixture must decrease – thus, Le Châtelier’s Principle indicates that the equilibrium will shift to increase the pressure to its equilibrium value. That is, the equilibrium will shift to increase the gas molecules in the volume: the equilibrium will shift to the side of the reaction that possesses more molecules of gas.

Practice 5.12 Consider the equilibrium established by the reaction between N2O4(g) ⇋ 2 NO2(g). Discuss the effect of the following changes according to Le Châtelier's Principle. (Temperature is constant where pressure changes, and pressure is constant where temperature changes.)

Increase the pressure of the container at equilibrium

Decrease the volume of the container at equilibrium

Increase the temperature at equilibrium; the reaction is endothermic.

Increase the volume of the container at equilibrium

102

•

Changes in Species Concentration

The addition of a species at equilibrium will cause a system to counteract the addition of the species: more of the species will be consumed.

The removal of a species at equilibrium will cause a system to counteract the removal of the species: more of the species will be produced.

Figure 32. The addition of nitrogen gas causes a response in the reaction between nitrogen and hydrogen. The system responds by reestablishing an equilibrium so that Q = Kc.

Practice 5.13 Consider the equilibrium established by the exothermic reaction 3H2 + N2 ⇋ 2NH3. Discuss the effect of the following changes according to Le Châtelier's Principle:

Remove ammonia upon production.

Add additional nitrogen gas.

Increase the volume of the container under constant temperature and pressure

Decrease the pressure of the equilibrium mixture at constant temperature

Increase the temperature of the equilibrium mixture; the reaction is exothermic.

Add neon gas to the mixture at constant volume.

103


Chemical Equilibrium II: Acid-Base Base Equilibrium and Solubility Equilibrium


dentify and characterize Arrhenius and Brønsted Brønsted-Lowry acids and bases identify

identify and characterize conjugate acid acid-base pairs

predict the relative strengths of acids and bases

characterize the acid or base character relative to the concentrations of hydrogen ion and hydroxide ion

use the pH scale, measure pH and convert using relationships provided by Ka, Kb, Kw and pOH; determine pH

identify and describe the character of strong acids and strong bases

identify and describe the character of weak acids and weak bases

write equilibrium expressions for weak acids and weak bases; determine the pH of weak acid and weak base solutions

determine equilibrium concentrations of ions and mol molecular species in acid-base equilibria

justify the relationship between percent ionization and molarity of an acid

predict and calculate the value of the acid acid-dissociation constants for polyprotic acids

predict the acid-base-neutral neutral character of salt solut solutions; ions; use Le Châtelier's Principle to justify predictions and observations

predict acid-base base strength and character based on chemical structure; make relationships to acid-base acid character using the periodic table

predict behavior of solutions exhibiting the common-ion effect

determine the pH of solutions exhibiting the common common-ion effect

characterize and determine the pH of buffered solutions; use the Henderson Henderson-Hasselbalch Hasselbalch equation

perform titrations, make predictions as to the character of solutions during ttitration, itration, and determine pH of solutions during various points in a titration

Interpret titration curves and justify discussions about significant regions of a titration curve

use and calculate the solubility--product constant; predict the occurrence of dissolution olution and precipitation using Ksp

predict how the nature of solutions affects solubility using the concepts of chemical equilibrium

104

Part 1: THE ARRHENIUS ACID AND THE ARRHENIUS BASE

Imagine the following reactions, which occur readily in water: HCl(aq) ⇋ H+(aq) + Cl–(aq) NaOH(aq) ⇋ Na+(aq) + OH–(aq) NH3(aq) + H2O(l) ⇋ NH4

+

(aq)

+

OH–

Figure 33. Here, ammonia acts as a base by accepting hydrogen ion from water. (aq)

Leaving hydroxide ion in solution greater than before, the ammonia acts as an Arrhenius base.

Each of the reactions above increases the concentration of one of two ions in aqueous solution: hydrogen ion, H+, or hydroxide ion, OH–. The reactions show that the increase in the concentration may be direct or indirect:

The dissociation of a compound containing hydrogen ions or hydroxide ions directly increases the concentration of these ions in aqueous solution. This can be seen in the HCl dissociation above and in the NaOH dissociation above.

The electrostatic attraction of a hydrogen ion or hydroxide ion in solution to a compound can indirectly increase the concentration of the other ion. This can be seen in the ammonia / water reaction above. Ammonia exhibits an electrostatic attraction for hydrogen ions in solution; this indirectly increases the hydroxide concentration in solution.

A compound that increases the hydrogen ion concentration in solution is called an Arrhenius acid. A compound that increases the hydroxide ion concentration in solution is called an Arrhenius base.

A compound that increases the hydrogen ion concentration in solution need not possess the hydrogen ion; and a compound that increases hydroxide ion in solution need not possess the hydroxide ion. The only consideration for identification as an Arrhenius acid or base is whether the addition of the compound to aqueous solution will cause an increase in the concentration of either ion. The biggest limitation of the Arrhenius definitions of acids and bases is that their discussion is limited to aqueous solution.

Part 2: THE BRØNSTED-LOWRY ACID AND THE BRØNSTED-LOWRY BASE

Imagine the following reactions: HCl(g) + H2O(l) ⇋ H3O +(aq) + Cl– (aq) :NH3(g) + HCl(g) ⇋ NH4Cl(s) Each of the reactions above involves the transfer of a hydrogen ion onto another

Figure 34. In the reaction above, HCl acts as a Brønsted-Lowry acid – a proton donor – while water acts as a Brønsted-Lowry base – a proton acceptor. (Notice that HCl here also acts as an Arrhenius acid; however, water is not acting as an Arrhenius base.)

molecule. In the first case – an aqueous 105

reaction – the hydrogen ion is transferred to the water molecule to form the hydronium ion, H3O+. The second reaction occurs in gas phase between the gases ammonia and hydrogen chloride to form the solid white compound ammonium chloride. Here, a hydrogen ion is transferred onto the ammonia molecule, which Figure 35. Similar in nature to the reaction shown in Figure 34, ammonia is the Brønsted-Lowry base here, while HCl continues to acts as a Brønsted-Lowry acid.

forms the ammonium ion. This ammonium ion is attracted to the newly-formed chloride ion, which combines with ammonium to form ammonium chloride. Substances can lose

(donate) an H+ ion as it is transferred to another species. This can be seen in the

reactions above. Substances can accept an H+ ion as it is

transferred from another species. This can be seen in the reactions above.

A compound that donates a hydrogen ion to another species is called a Brønsted-Lowry acid. Because a hydrogen ion a naked proton, these species are often called “proton donors.”

A compound that accepts a hydrogen ion from another species is called a BrønstedLowry base. Because a hydrogen ion is a naked proton, these species are often called “proton acceptors.” Brønsted-Lowry bases must have an unshared electron pair.

In this manner, the Brønsted-Lowry acid-base definition is less restrictive than the Arrhenius definition: Arrhenius

Figure 36. A molecular model representation of the ammonia (Brønsted-Lowry base) and

acids and bases are limited to aqueous

water (Brønsted-Lowry acid) reaction. Brønsted-Lowry acid and bases are defined by their

solutions, while the Brønsted-Lowry

ability to donate or accept hydrogen ions. (protons).

definition includes species in gas or liquid phase, as well.

A substance that can act as both a base and an acid is called amphoteric (also amphiprotic). Generally, when the substance is reacting with something more acidic than itself it is considered a base, while it is considered an acid when reacting with a species less acidic than itself.

106

Part 3: CONJUGATE ACID-BASE PAIRS

Lowry base and acid, an Arrhenius base or acid, or a pairing that can be described using both Whether the reaction involves a Brønsted-Lowry definitions, the difference between the species present before the reaction and the species present after the reaction differ di by the presence and absence of a proton, or hydrogen ion.

Look at Figure 37 at right, whose reaction is represented here: HS– (aq) + HF(aq) ⇋ F– (aq) + H2S(aq) On the left, the acid HF can donate a proton to the species HS–, while HS– can accept a proton from the acid. So, HF – acting as a proton donor – is an acid, while HS– – acting as a

Figure 37. The conjugate acid-base base pairs can be distinguished by the presence and absence of just a hydrogen ion.

proton acceptor – is acting as a base. If we move to the right side, we see that the species can also donate or accept protons: the species H2S(aq) can donate a proton to F–, while F– can accept a proton from H2S. We can identify a simple d difference ifference between the pairs of species in the reaction above: they differ only in the presence or absence of a hydrogen ion, H+.

Species that differ in only the presence or absence of hydrogen ions are called conjugate acid-base base pairs. Every acid has a conjugate base,, which is formed when the acid loses its hydrogen ion, and every base has a conjugate acid, acid which is formed when a base accepts a hydrogen ion.

Practice 6.1

Identify the acids and basess in the reaction represented above.

Identify each conjugate onjugate acid acid-base pair.

107

Identify each acid and base as an Arrhenius acid, Arrhenius base, Brønsted-Lowry acid, Brønsted-Lowry base or by more than one of these descriptors.

Write the chemical equation for the equilibrium; i.e., Kc expression.

Practice 6.2 Identify the conjugates of each of the following species: HClO4, H2S, PH4+, HCO3–, CN-, SO42-, H2O

Part 4: RELATIVE STRENGTHS OF ACIDS AND BASES

When we discuss the strength of Arrhenius acids and bases in aqueous solution, we are discussing the extent to which an acid will donate a proton in aqueous solution or the extent to which a base will dissociate hydroxide ion. When we discuss the strength of Brønsted-Lowry acids in aqueous solution, we ae discussing the extent to which the acid donates a proton. When we discuss the strength of Brønsted-Lowry bases , we are discussing the relative ease with which the base extracts a proton from another species.

The strong acids completely transfer their acidic protons in aqueous solution, leaving mostly (~ 100%) dissociated molecules in water. Their conjugate bases have a negligible tendency to be protonated (accept protons) in solution; they are weak bases.

For example, the strong acid HCl(aq) dissociates into H+(aq) and Cl– (aq) in solution. The Cl– (aq), HCl’s conjugate base, does not exhibit a likelihood of accepting a proton to become HCl(aq) – the chloride ion is a weak conjugate base. Indeed, if this were the case, then we would not say that HCl(aq) has a likelihood of donating protons.

The weak acids only slightly transfer their acidic protons to water, leaving mostly undissociated molecules in water. Their conjugate bases have a greater tendency than the conjugate bases of strong acids to be protonated in solution; so, while the conjugate bases of weak acids are themselves also weak, they are stronger conjugates than the conjugates of strong acids. The strength of a conjugate base increases with the decreasing strength of the acid.

108

That is, as the strength of an acid decreases, the strength of its conjugate base increases. For example, the weak acid CH3COOH(l) only slightly dissociates into H+(aq) and CH3COO– (aq) in solution. CH3COO– Figure 38. The reaction between an acid HA and water.

(aq),

CH3COOH’s conjugate base,

has a tendency to become protonated to CH3COOH(l) in

solution. So, although it is a stronger conjugate than chloride ion, for example, it is still a weak base – although it is a relatively strong conjugate base.

We can think of reaction in which protons are transferred in of the ability of two bases to become protonated. This allows us to discuss the strength of acids and bases, and establish the direction of the equilibrium between the conjugate acid-base pairs. For example, take a look at the reaction between acid and water in Figure 38. We see that the conjugate base of the acid HA is the base A–, and the conjugate acid of the base H2O is the acid H3O+. We examine the extent to which the reaction occurs in one direction or another – and thus establish a measure of the strength of the acids and bases – by looking at the relative strength of the conjugate bases to accept protons. If A– is a stronger base than H2O, then A– is more likely to accept the proton and the equilibrium will lie to the left. However, if the base H2O is stronger, then the water molecule will accept the proton and the equilibrium will lie to the right.

Let’s put two actual acids in place of HX and evaluate the equilibrium; this allows us to establish the relative strengths of the acids and bases. HBr(aq) + H2O(l) ⇋ H3O+(aq) + Br– (aq) We ask the question, “Which base more strongly attracts the proton, Br– or H2O?” In solution, water more strongly attracts the proton than does bromide ion. Thus, the base Br– is weaker than the base water. The equilibrium lies in the direction of proton transfer to the stronger base, which results in the solution containing the weaker base and the weaker acid (the conjugate of the stronger base).

We conclude that in any acid-base proton transfer reaction, the equilibrium lies to the side of the reaction that contains the weaker base and weaker acid. That is, the equilibrium moves in the direction that favors the consumption of the Figure 39. Here, the stronger conjugate base is the hydroxide ion. It extracts the proton from ammonium ion (a weak acid); the equilibrium lies on the left.

stronger acid and stronger base in favor of the formation of the weaker acid and weaker base.

109

Practice 6.3

(A chart of relative strengths is shown below. We will quantitatively examine strength soon.)

For the following reaction, determine whether the equilibrium lies to the right or to the left based upon the relative strengths of the bases. H2SO4 (aq) + CO32-(aq) ⇋ HSO4– (aq) + HCO3– (aq)

For the following reaction, determine whether the equi equilibrium librium lies to the right or to the left. PO43- (aq) + H2O (l) ⇋ HPO42-(aq) + OH– (aq)

For the following reaction, determine whether the equilibrium lies to the right or to the left. NH4+ (aq) + OH– (aq) ⇋ NH3(aq) + H2O(l)

For the following reaction, determine whether the equilibrium lies to the right or to the left. HNO2 (aq) + H2O (l) ⇋ NO2– (aq) + H3O+(aq)

Acid Strength Increases HCl

H2SO4

HNO3

H3O+

HSO4–

H3PO4

HF

CH3COOH

H2CO3

H2S

NH4+

HCO3–

HPO42-

H2O

HS–

OH–

H2

Cl–

HSO4–

NO3–

H2O

SO42-

H2PO4–

F–

CH3COO–

HCO3–

HS

NH3

CO32-

PO43-

OH–

S2-

O2–

H–

Base Strength Increases

Part 5: THE AUTOIONIZATION OF WATER

Because water can act as an acid or a base, then you should expect that two water molecules can participate in a reaction such as that shown here: HOH(l) + HOH(l) ⇋ H3O+ + OH–

110

Indeed, this is observed and is called the autoionization of water. The reactions are very quick in both directions, and at room temperature only about 2 of every 109 molecules of water are ionized. This is, however, an equilibrium process, and it has an equilibrium constant expression:

Kc = [H3O+][OH–] or Kc = [H+][OH–]

This equilibrium constant expression because it involves the unique species water - has a special name, the ionproduct constant, Kw. For water, the ion-product constant is:

Kw = [H+][OH–] = 1.0 x 10-14 or Kw = [H3O+][OH–] = 1.0 x 10-14

In a neutral solution [OH–] = [H+], while in acidic solution [H+] > [OH–] and in basic solution [OH–] > [H+]. However, in Figure 40. The concentrations of hydroxide and hydrogen ions are a constant at a given temperature. At 25ºC, the value is 1.0 x 10-14. As the hydrogen ion concentration decreases, the solution becomes basic, while the solution is neutral when the ion concentrations are equal.

all solutions, the product of the ions is 1.0 x 10-14 at 25°C. Thus, in a neutral solution, the value of Kc for both ions is 1.0 x 10-7 M.

Because the product of the concentrations of hydrogen ion and hydroxide ion is always 1.0 x 10-14, then you should easily see that as the concentration of one of the ions increases, the other ion’s concentration must decrease. And, when they are equal in concentration – as in a neutral solution – the concentration of each ion is 1.0 x 10-7 M.

Practice 6.4 How does the addition of hydroxide ion affect a solution that is acidic? Be complete.

111

Part 6: THE pH SCALE

The concentrations of both hydroxide ion and hydrogen ion in even concentrated solutions are quite small. The pH scale provides a convenient way to express the very small values of the ions’ concentrations. A pH value is mathematically equal to the negative log of the concentration of pH:

– ‫[܏ܗܔ‬۶ ା ] = ‫ܘ‬۶

Figure 41 shows some pH values for common solutions. The pH scale is a logarithmic scale, so a one unit change in pH represents a ten times change in hydrogen ion concentration. For example, the pH of 1.0 M NaOH is 14, and the concentration of hydroxide ion is 1.0 mol per liter. For a solution that contains 0.1 mol hydroxide per liter, the pH changes to 13 (not one-tenth of the original pH). And, because the pH is the negative log of the concentration of hydrogen ion, an increase in the hydrogen ion concentration causes a decrease in pH. The converse is also true: a decrease in hydrogen ion concentration effects an increase in pH. You can easily estimate pH by placing a particular concentration of hydrogen ion between two known pH values; i.e., those that have y = 1.0 in y • 10–x. For example, consider the pH of a solution that has a hydrogen ion concentration of 3.26 • 10-4. We don’t know its pH exactly, but we do know that the value 3.26 • 10-4 is larger than 1.0 • 10-4 and smaller than 1.0 • 10-3. Thus, the pH will fall between 3 and 4. The calculated pH is 3.49.

Often, the pOH is used to express the concentration of OH- in solution. This value, obtained in the same manner as pH, provides a measure of the hydroxide concentration instead of

Figure 41. Some pH values for common substances.

the concentration of hydrogen ions. It is, however, useful to make the following relationship:

pH + pOH = 14.00

Part 8: MEASURING pH

We can measure pH in any of several ways, which we will do extensively in laboratory:

Acid-base indicators – an acid-base indicator is a substance that changes colors based on the concentration of hydrogen ions in the solution. Although they are useful for many applications, acid-base indicators change color over a wide range (about 2 pH units), and they only tell an observer that a pH is lower or higher than the pH at which an indicator changes. Some common indicators will be seen in a laboratory investigation soon.

pH meters – pH meters work by conducting voltage from solution across two electrodes. The inner-workings of pH meters will be taken up when we study electrochemistry. 112

Practice 6.5

Determine whether the following solutions are acidic, neutral or basic:

[H+] = 4 x 10-9 M

[OH–] = 1 x 10-7 M

[OH–] = 7.0 x 10-13 M

A solution is 0.010 M in [OH–]. What is the concentration of H+? Is the solution basic, neutral or acidic?

A solution is 2 x 10-6 M in hydronium ion. What is the molarity of the hydroxide ion?

What is the concentration of hydrogen ion in a solution whose pH is measured to be 3.76?

Estimate the pH of a solution whose hydrogen ion concentration is found to be 0.000653 M.

Estimate the pH of a solution whose hydroxide concentration is found to be 0.000653 M.

A solution of antacid has a pH of 8.99. What are the hydrogen ion and hydroxide ion concentrations of the antacid solution?

A solution has pOH = 3. What is the pH of the solution?

113

Part 9: STRONG ACIDS AND STRONG BASES

Strong acids

are 100% dissociated in solution

are strong electrolytes whose solutions conduct electricity

are generally the only source of H+ in their aqueous solutions

There are seven common strong acids – six monoprotic (one acidic proton) and one diprotic (two acidic protons). They are shown here with their common dissociation equations. Recall that the anion species on the right in each equation is the conjugate base of the acid on the left. Notice that equilibrium arrows are not used for the dissociation equations because the dissociation lies completely to the right. hydrochloric acid HCl

HCl(aq) → H+(aq) + Cl– (aq)

hydroiodic acid

HI(aq) → H+(aq) + I– (aq)

HI

hydrobromic acid HBr

HBr(aq) → H+(aq) + Br– (aq)

chloric acid

HClO3

HClO3 (aq) → H+(aq) + ClO3– (aq)

perchloric acid

HClO4

HClO4 (aq) → H+(aq) + ClO4– (aq)

nitric acid

HNO3

HNO3 (aq) → H+(aq) + NO3– (aq)

sulfuric acid

H2SO4

H2SO4 (aq) → H+(aq) + HSO4– (aq)

In a solution of a monoprotic strong acid, we can assume that the concentration of H+ is equal to the molarity of the acid. That is, if a 0.20 M solution of nitric acid is used, then the concentration of H+ is 0.20 M. However, the more complex case of the diprotic sulfuric acid will be seen soon.

Strong bases

are 100% dissociated in solution

are strong electrolytes whose solutions conduct electricity

There are seven common strong hydroxide bases – all of them hydroxides of Group 1 or Group 2 metals. The oxides of sodium and calcium are also strong bases, as are the ionic hydride compounds and the anion nitride. They are shown here with the equations of their reaction with water, which is the source of their basicity:

sodium, lithium, potassium, rubidium, cesium, calcium, barium and strontium hydroxides, MOH or M(OH)2: MOH(aq) ⇋ M+(aq) + OH– (aq)

MOH

M(OH)(aq) ⇋ M2+(aq) + 2 OH– (aq)

M(OH)2

The hydroxides provide one or two mol hydroxide ion per mol compound. See sample exercise below.

oxides of sodium and calcium, among others, Na2O or CaO: O2-(aq) + HOH (l) ⇋ 2 OH– (aq)

Additional soluble ionic metal oxides may also participate in this type of reaction with water. Each mol oxide ion will deprotonate two mol water, which provides for two mol OH– to be produced.

114

compounds containing the nitride ion, N3-: N3-(aq) + HOH (l) ⇋ NH3(aq) + 3OH–

Each mol nitride ion will deprotonate three mol water, which provides for three mol OH- to be produced.

EQUILIBRIUM CONSTANT VALUES OF STRONG ACIDS

The equilibrium constants for acids and bases have the special notations Ka and Kb, respectively, which are termed the aciddissociation constants and base-dissociation constants, respectively. The values of Ka and Kb for strong species are always Kc >> 1, which indicates that the reaction is favored toward the right, or dissociation.

Practice 6.6

What is the pH of a 0.040M solution of HNO3?

What is the pH of a 0.028M solution of NaOH?

What is the pH of a 0.0011M solution of Ca(OH)2?

What is the concentration of KOH for which the pH is 11.89? What is pOH for the solution?

What is the concentration of Ca(OH)2 for which the pH is 11.68? What is pOH for the solution?

115

Do not confuse the weak/strong with dilute/concentrated. Weak and strong refer to the extent of dissociation of acids and bases in solution, while dilute and concentrated refer to the extent of the concentration of an acid or base (or any other solute); e); that is, molarity. It is possible to have a dilute solution of a strong acid, like a 0.0005 M solution of nitric acid; it is also possible to have a concentrated solution of a weak acid, like a 9.3 M solution of acetic acid.

DILUTE AND CONCENTRATED:

STRONG AND WEAK:

used for molarity

used for the extent of dissociation

Part 10: WEAK ACIDS AND WEAK BASES

Weak acids

are slightly dissociated in solution

are weak electrolytes whose solutions conduct electricity

There are many common weak acids – generally categorized as organic acids and inorganic acids. Some are shown here with their common dissociation equations. Here we use equilibrium arrows because the reaction is at equilibrium between the acid and its conjugate base.

hydrofluoric acid HF nitrous acid

HNO2

acetic acid

CH3COOH

hydrocyanic acid HCN

HF(aq) ⇋ H3O+(aq) + F– (aq)

HNO2(aq) ⇋ H3O+(aq) + NO2– (aq)

CH3COOH (aq) ⇋ H3O+(aq) + CH3COO– (aq) HCN (aq) ⇋ H3O+(aq) + CN– (aq)

Organic acids are often recognized by the –COOH group on the end of the acid,, as in acetic acid above. above The acidic hydrogen is on this his group, not on a carbon atom.

EQUILIBRIUM EXPRESSIONS OF WEAK ACIDS The value of Ka for weak acids is always Ka < 1, which indicates that the reaction is favored toward the left, or the molecular form of the acid. Recall that this is called the acid-dissociation dissociation constant. weak ak acids. The values of the acid-dissociation acid Ka values for weak acids must be less than one because the dissociation is not favored for we constants are generally less than 10-3 for weak acids. The larger the value of Ka, the greater the extent of ionization of an acid, and the greater the extent of ionization, the greater the strength of an acid. That is, as Ka approaches one (the cut cut-off off for favoring dissociation) the strength of the acid increases. Ka values for some common acids are given in the table below below; many more are in your tables of data.. Consider the value of Ka for strong acids to be simply “large.”

116

DETERMINING THE VALIDITY OF ASSUMPTIONS and PERCENT IONIZATION When determining the pH of acid and base solutions, we will experience situations where the quadratic formula is the only manner of solving the problem. This arises when we do not know the equilibrium concentrations of the species in solution (which we begin on Page 120). To avoid this, we often assume that the dissociation of a weak acid or weak base is so small that the concentration of the acid at equilibrium is the same as it was initially. However, if more than 5% or so of the molecular form of the acid or base dissociates, then the assumption is not strictly valid. When the dissociation of an acid or base is greater than 5% at equilibrium, it is not valid to assume that [X]initial = [X]equilibrium. In these cases, one could use the quadratic formula to determine the actual concentration at equilibrium; however, in a first-year college course, it is sufficient to justify the assumption as valid by the 5% guideline and neglect the excess dissociation – just be able to explainit-away! ࢖ࢋ࢘ࢉࢋ࢔࢚ ࢏࢕࢔࢏ࢠࢇ࢚࢏࢕࢔ =

[࢏࢕࢔࢏ࢠࢋࢊ ࢉ࢕࢔ࢉࢋ࢔࢚࢘ࢇ࢚࢏࢕࢔] ࢞ ૚૙૙ [࢏࢔࢏࢚࢏ࢇ࢒ ࢉ࢕࢔ࢉࢋ࢔࢚࢘ࢇ࢚࢏࢕࢔]

Practice 6.8 Determine the percent ionization of a 0.0500 M HCN in solution at equilibrium. The pH of the solution is 5.31 at 25°C.

The equilibrium concentration of hydrogen ion in a 0.10 M solution of HF is 7.9 x 10-3 M. What is the percent ionization of HF?

A solution of H+/HCO3– has a pH of 4.39. What percent of H2CO3 ionizes at equilibrium? The concentration of carbonic acid is 0.0037 M.

Figure 42 illustrates that the percent ionization of an acid decreases as the acid’s concentration increases. Why is this true? Figure 42. Incidentally, it is worth noting that the percent ionization of weak acids decreases as the concentration of the acid increases.

117

Acid dissociation constants for some weak acids at 25°C hydrofluoric acid – 6.8 x 10-4

acetic acid – 1.8 x 10-5

phenol – 1.3 x 10-10

nitrous acid – 4.5 x 10-4

hypochlorous acid – 3.0 x 10-8

boric acid – 5.8 x 10-10

benzoic acid – 6.3 x 10-5

hydrocyanic acid – 4.9 x 10-10

hydrogen peroxide – 2.4 x 10-12

CALCULATING Ka FROM pH We can use the pH of a weak acid to calculate its acid-dissociation constant, which is the value of the equilibrium constant for the reaction of a weak acid as in: HA(aq) + H2O(l) ⇋ H3O+(aq) + A– (aq). Here, A– is the generic term for the acid’s conjugate base. The

equilibrium expression is shown here (water is not included because it is liquid): ࡷࢇ =

[۶૜ ‫۽‬ା ][‫] ିۯ‬ [۶ ା ][‫] ିۯ‬ ࢕࢘ ࡷࢇ = [۶‫]ۯ‬ [۶‫]ۯ‬

As in the equilibria discussed earlier, we need only the equilibrium concentrations in order to determine the acid-dissociation constant. , this value will tell us much about the extent of ionization of the acid. Again, we do not need to consider the value of Ka for a strong acid – the values are very large as ionization is nearly 100%.

Practice 6.7 A solution of 0.10 M formic acid was prepared and its pH was measured as 2.38. (A) Calculate the acid-dissociation constant for formic acid. (B) What percentage of the acid is ionized in this solution?

Solution: As we did earlier, we determine the equilibrium concentrations of the species in the solution. We assume that the pH is measured after the system has reached equilibrium, which allows us to calculate the concentration of hydrogen ion from the pH. Plan:

(A) Determine the concentration of hydrogen ion at equilibrium, and use this value to complete the remainder of the table. Put the calculated equilibrium values into the equilibrium expression to determine Ka. (B) We can determine the percent ionized by dividing the concentration of an ion at equilibrium by the initial concentration of acid.

HCHO2

⇋

H+

CHO2–


118

A solution of 0.020 M niacin was prepared and its pH was measured as 3.26. (A) Calculate the acid-dissociation constant for niacin. (B) What percentage of niacin is ionized in this solution?

NC5H10COOH

⇋


119

CALCULATING pH FROM Ka

We can use the acid-dissociation constant of an acid to calculate the pH of a solution of the acid.

Practice 6.9 Determine the pH of a 0.30 M solution of acetic acid, the weak acid in vinegar. Ka for acetic acid is 1.8 x 10-5. Solution:

As we did earlier, we determine the equilibrium concentrations of the species in the solution. Because we do not know the concentrations of any of the species at equilibrium (we would need pH to do this as in Practice 6.7), we need to assign variables to the change and equilibrium concentrations. Then, we solve for the value of the variable and determine the pH.

CH3COOH

⇋

H+

CH3COO–


Because the value of Ka is so small, we can typically neglect the subtraction here, which avoids the use of the quadratic formula! However, this is NOT strictly true, and you should be able to justify your use of this assumption. See Page 117.

Determine the pH of a 0.20 M solution of HCN. Ka for HCN is 4.9 x 10-10. Always check the 5% approximation; justify why you neglected it if the ionization exceeds 5% (you do not need to use the quadratic formula!). HCN

⇋

H+

CN–


120

Weak bases

slightly deprotonate water

contain a lone pair, which is responsible for the deprotonation of water

are the conjugates of acids

There are many common weak bases. A couple are shown here with their common deprotonation equations. Again, we use equilibrium arrows because the reaction is at equilibrium between the base and its conjugate acid.

ammonia

NH3

hydrosulfide ion HS–

NH3 (aq) + H2O(l) ⇋ NH4+(aq) + OH– (aq)

HS– (aq) + H2O(l) ⇋ H2S (aq) + OH– (aq)

EQUILIBRIUM EXPRESSIONS OF WEAK BASES

The value of Kb for weak bases is always Kb < 1, which indicates that the reaction is favored toward the molecular form of the base. This, similarly to its acidic counterpart, is called the base-dissociation constant.

Kb values for weak bases must be less than one because the equilibrium for hydroxide formation is not favored for weak bases. The values of the base-dissociation

Figure 43. Some common weak bases are those that contain nitrogen – its lone pair is quite attractive toward protons.

constants are generally less than 10-3 for weak bases.

The larger the value of Kb, the greater the extent of O H – production in base solution, and the greater the extent of hydroxide ion formation, the greater the strength of a base. That is, as Kb approaches one (the cut-off for favoring hydroxide ion formation) the strength of the base increases.

CALCULATING [OH–] OF A WEAK BASE SOLUTION

The determination of [OH-] is similar to that which we used for determining Ka or pH in the previous section. For the base B(aq), the equilibrium expression is: ࡷ࢈ =

[۶۰ା ][‫۽‬۶ ି ] [۰]

121

Practice 6.10

Calculate the concentration of hydroxide ion in a 0.15 M solution of ammonia. Kb for ammonia is 1.8 x 10-5.

NH3

⇋

NH4+

OH–


122

Part 11: THE EQUILIBRIUM CONCENTRATIONS OF POLYPROTIC ACIDS

Acids such as H2SO3 may ionize in successive steps: H2SO3 ⇋ H+ + HSO3–

HSO3– ⇋ H+ + SO3–

Ka1 = 1.7 x 10-2 Ka2 = 6.4 x 10-8

Performing the calculations of determining equilibrium concentrations of hydrogen ion (and thus, pH) using only the first ionization is presumptively valid when the percent ionization is less than 5%.We can assume for all acids that the second ionization is so small that its contribution to [H+] is too small to affect the pH of the solution. Thus, subsequent ionizations need not be performed at this level of chemistry. (However, this does not mean that one would not consider the extent of the first ionization to determine pH.)

When we are attempting to determine Ka for the first and successive ionizations of a polyprotic acid, simply multiply the Ka values for all ionizations as we did earlier when we combined equations. This provides the Ka value of the equation for n ionizations.

Practice 6.11 What is the value of Ka for the equilibrium H2SO3 ⇋ 2 H+ + SO32-?

Justify the reasoning for not considering the second and third ionizations of phosphoric acid when performing an equilibrium problem.

The value of Ka1 for the polyprotic acid oxalic acid, H2C2O4, is 5.9 x 10-2. What might you predict the value of Ka2 is given that the second ionization is very small? Justify your response.

The values of the acid-dissociation constants for the first, second and third ionizations of citric acid are 7.4 x 10-5, 1.7 x 10-5 and 4.0 x 10-7, respectively. What is the value of the acid-dissociation constant for the equilibrium dissociation to C6H5O73-?

123

An acid is 7.8% ionized at equilibrium. Answer the following questions.

Is the acid weak or strong? Justify your answer.

We assumed that the initial concentration of the acid, 0.05 M, is equal to the equilibrium concentration. Is this strictly valid? Explain.

If the acid is polyprotic, do we need to consider the subsequent ionizations before calculating pH? Explain.

Part 12: THE RELATIONSHIP BETWEEN Ka and Kb

Because the sum of an acid dissociation reaction and the reaction of the dissociation of its conjugate base provides for the autoionization of water – as shown below – we can make the following relationship between Ka and Kb when the two reactions discussed are related as conjugate acid-base pairs:

Ka x Kb = Kw This is a direct result of the previous idea we established in Chapter 15: the equation provided by the summation of two other equations has an equilibrium constant that is the product of the equilibrium constants for the added reactions. NH4+(aq) ⇋ NH3(aq) + H+(aq

NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH– (aq) H2O(l) ⇋ H+(aq) + OH– (aq)

The product of the Ka and Kb values for the reactions with water of a conjugate acid-base pair is the ion constant product of water, Kw. Thus, if we know that the value of Ka is very large, then we also know that the Kb value of the conjugate base must be small – the product of the two is always 1.0 x 10-14.

The product of the pKa and pKb values for the reactions with water of a conjugate acid-base pair is equal to 14.00. Why would you use this? If you need a Kb or Ka value for a species that is not listed on the acid-base dissociation tables, simply find the Ka or Kb for the conjugate base or acid, as needed, and use the relationship above. 124

Part 13: ACID-BASE PROPERTIES OF SALT SOLUTIONS/HYDROLYSIS

Recall that a salt is an ionic compound that does not contain hydrogen, hydroxide or oxide ion. Salts are one of the products of an acid-base neutralization reaction.

Soluble salts are nearly 100% ionized in solution, and their anions or cations may react with water in a hydrolysis reaction. When they do, the reaction may lead to an increase in hydrogen ion or hydroxide ion, which can influence the pH of Figure 44. Here, the carbonate ion undergoes hydrolysis to produce the hydroxide ion and the bicarbonate ion. This results in a basic solution.

the solution even though the added compound is a salt and not an acid or base.

For example, the addition of a salt containing the fluoride ion, F-, results in a reaction with water to form HF. This results in an increase in the hydroxide concentration and provides a basic solution.

F– (aq) + HOH(l) ⇋ HF(aq) + OH– (aq)

Additionally, a cation that is the conjugate acid of a weak base, as in CH3NH3+, will undergo hydrolysis to form acidic solutions: CH3NH3+(aq) + HOH(l) ⇋ CH3NH2(aq) + H3O+(aq) And, a metal cation that is not the cation of a strong base will hydrolyze to form acidic solutions. These are the most important for our study of the Lewis acids, which are chemical species that can accept electron pairs into their valence shells – the donor of the electron pair(s) is called a Lewis base. All metal cations except those of the strong bases will act as Lewis acids, and all Brønsted-Lowry bases are also Lewis bases. The only difference is that Lewis bases can donate their electrons pairs to species other than H+.

In the example shown at left, the iron ion is surrounded by 6 water molecules; recall the process of solution from our earlier studies. As the electron density from the water molecules is pulled toward the Fe-O center, the hydrogen attached to water becomes quite acidic. This character increases as the charge on the ion Figure 45. The hydrolysis of a Lewis acid – a species that undergoes hydrolysis owing to available unfilled valence shells. Generally, we are concerned with metal cations that are not those of strong bases to be Lewis acids.

increases, and an increase in Lewis acid character is noted as ionic radius decreases. 125

In order to predict the acid/base/neutral character of salt solutions, one looks at the cation and anion of the salt and then evaluates the conjugate base and acid, respectively: •

Salts derived from a strong acid and strong base do not hydrolyze – their solutions will have a pH of 7.0

•

Salts derived from a strong base and weak acid – the anion of the acid is a relatively strong conjugate base; the anion hydrolyzes to produce solutions that are basic.

•

Salts derived from a weak base and a strong acid – the cation is a relatively strong conjugate acid; the cation hydrolyzes to produce hydrogen ions and solutions that are acidic.

•

Salts derived from a weak base and a weak acid – both the cation and anion hydrolyze; the character of the resulting solution depends upon relative Ka and Kb values. [You will not likely come across these!]

Practice 6.12 For each ion in the salt solutions described below, identify the conjugate acid or conjugate base. Characterize the salt solutions as acidic, basic or neutral, and justify your selections. Rank them in order of increasing pH.

0.1 M Ba(C2H3O2)2

0.1 M NH4Cl

0.1 M NH3CH3Br

0.1 M KNO3

Predict whether the salt Na2HPO4 will form an acidic or basic solution. Justify your selection.

Calculate the pH of a 0.075 M solution of NaCH3COO(aq).

126

Part 14: ACID-BASE BEHAVIOR AND CHEMICAL STRUCTURE

Bond Strength and Polarity

The strength of an acid is related to two factors: polarity and bond strength.

Across a period, it is the polarity of the bond that drives acid strength; the more polar the bond, the stronger the acid. Thus, HF is a stronger acid than water, and water is a stronger acid than ammonia.

Within a group, the more polar the bond, the weaker the acid – this is because the bond strength increases significantly for the smaller atoms compared to the larger atoms lower in the group. Thus, HI is a stronger acid than Figure 46. See the text at right for a discussion of these trends.

HBr, and HBr is a stronger acid than HCl. HF is the only weak halide acid.

Strength of Oxoacids (Oxyacids)

An oxyacid is an acid in which the hydrogen is bonded to an oxygen atom. Thus, the variation in their strengths is due to the electronegativity of the other atom to which oxygen is bonded. As Figure 47 shows, the acid HOCl is considerably stronger than HOI because of the high electronegativity of the Cl atom compared to the I

Figure 47. See the text at left for a discussion of these

atom – the higher electronegativity pulls electron density toward the Cl

observations. The numerical values are acid-dissociation

atom, which decreases the electron density between H and O,

constants.

ultimately decreasing the strength of that bond.

For reasons similar to the previous discussion, in a family of acids like HClO, HClO2, HClO3 and HClO4 the strength of an acid increases as the number of adjacent atoms increases – the presence of electronegative oxygen atoms increases the pull of electron density away from the H―O bond, which allows the proton to dissociate Figure 48. See the text at right for a discussion of these observations. Ka values are given.

from the polyatomic ion. Note the relationship to the central atom’s oxidation state. 127

Part 15: THE COMMON-ION EFFECT

Imagine a solution of the weak acid acetic acid, HCH3COO. We expect that a small amount of acetic acid will ionize in solution to provide a solution containing acetate ion and hydrogen ions, and we expect that many of the acetic acid molecules will remain undissociated in solution, which gives the familiar equilibrium for the solution: HCH3COO (aq) ⇋ H+(aq) + CH3COO– (aq)

Ka = 1.8 x 10-5

How will the system respond if a soluble salt containing the anion, acetate ion, is added to the solution? That is, what is the effect of adding NaCH3COO, for example, to the system?

In the figure at left, we see (and could calculate) that the concentration of hydrogen ion in a 0.10 M HCH3COO solution is slightly more than 10-3 M; this gives a pH of 2.87. As acetate ion is added, we notice that the concentration of hydrogen ion decreases, which is expected according to the equilibrium expression above: the system responds – as Le Châtelier’s Principle would suggest – by decreasing the amount of acetate ion in the new solution. Looking back at the equilibrium equation should reinforce this idea for you: a disruption of the equilibrium on the right will shift the equilibrium to the left. The effect of this is to increase the pH of the solution as hydrogen ions are consumed. Figure 49. The concentration of hydrogen ion added to an acetic acid solution decreases upon the addition of a common ion. This is a direct result of Le Châtelier’s Principle.

When a strong electrolyte is added to a solution of a weak electrolyte containing

the same ion, the system responds by consuming the common ion and reestablishing equilibrium. This is called the common-ion effect. It is an application of Le Châtelier’s Principle. This is valid for any weak electrolyte solution, whether it is an acid solution, base solution or slightly soluble salt.

128

Practice 6.13

Determine which of the following compounds are strong electrolytes:

HF

NaF

FeCl3

H2SO4

AgCl

NaOH

MgBr2

HCl

Imagine a 0.05 M solution of hydrofluoric acid, HF. Write its dissociation equation in water, and discuss the effect of adding samples of the following compounds to the solution.

CaCl2

AgF

LiF

HOH

Part 16: DETERMINING THE pH OF SOLUTIONS EXHIBITING THE COMMON-ION EFFECT

Recall solving for the pH of a solution of base or acid solutions earlier: we determined the initial concentrations of all of the species, and then determined the equilibrium concentrations. Then, we substituted the values at equilibrium into the equilibrium expression for Ka or Kb, as appropriate. The final step was to solve for x, which provided the concentration of ions.

When a soluble salt containing a common ion – or effecting the production of the common ion – is introduced to a solution, we simply need to include the concentration of the common ion in our table for determining equilibrium concentrations. This should make sense from the standpoint of equilibrium: the value of an equilibrium constant is related to the ionization of a substance in solution; thus, if we already have some of the substance ionized in solution, then the value of the equilibrium concentrations will be affected by the presence of the ion.

Practice 6.14

Determine the pH of a solution that is 0.30 M in both acetic acid and sodium acetate. Ka for acetic acid is 1.8 x 10-5.

•

Write an equilibrium equation including all important species (cancel spectator ions – these do not affect results)

•

Determine the effect of any salts on the overall concentrations of important ions

•

Set up a table that includes any important equilibrium species

•

Solve for the equilibrium concentrations

•

Use the equilibrium data to address the question posed

129

CH3COOH

⇋

CH3COO–

H+


We can safely assume:

So, we can also assume:

[CH3COOH]initial ≈ [CH3COOH]equilibrium

[CH3COO–]initial ≈ [CH3COO–]equilibrium

Practice 6.15 What is the fluoride ion concentration in a 0.50 L solution containing 0.10 mol HCl and 0.20 mol HF? How would the concentration of fluoride ion change if the HCl were not present? Ka for HF is 6.8 x 10-4.

First, identify the important species present initially and at equilibrium:

Initial:

HF from the weak acid and H+ from the strong acid. HF(aq) ⇋ H+(aq) + F –(aq)

Equilibrium:

H+ from both acids and F - from the weak acid – we should expect the presence of the H+ ion from HCl will drive the equilibrium of the HF ionization to the left.

HF

⇋

H+

F–


130

Part 17: BUFFERED SOLUTIONS

We saw in the previous section that the presence of a common ion elicits a Le Châtelier response. When the common ion is the conjugate of a weak acid or weak base, a buffer solution is formed. A solution that contains a weak conjugate acid-base pair is called a buffered solution. A buffered solution resists wide change in pH because the solution contains species that act according to Le Châtelier’s Principle to counteract changes in H+ and OH-. A buffer solution contains a weak acid or weak base and a soluble salt that contains the conjugate of the acid or base.

A buffered solution containing [HX] = [X–]

After addition of OH–

After addition of H+

pH

X-

HX HX

OH–

HX

OH– + HX ⇋ H2O + X-

X-

H

+

X-

H+ + X– ⇋ HX

Figure 45. In a Le Châtelier response, the addition of acid or base to a buffer solution shifts an equilibrium to resist changes in the concentrations of those two ions; thus, changes in pH are restricted.

BUFFERING CAPACITY

Buffering capacity is an expression of how much acid or base a buffer can accept before great changes are seen in pH. As long as the concentrations of HX / X– in the buffer are much larger than the concentrations of H+ or OH– added the buffer can resist pH changes. However, when the concentration of the conjugate acid-base pair is small compared to the amount of acid or base added, then the buffering capacity of the solution is decreased. Thus, the buffering capacity depends upon the amount of the conjugate acid-base pair in the solution.

Practice 6.16 Show using equations how the buffer solution made from sodium acetate and acetic acid can resist changes in pH.

131

Part 18: THE pH OF BUFFERED SOLUTIONS & THE HENDERSON-HASSELBALCH EQUATION

The pH of a buffered solution depends upon the equilibrium concentrations of the components of the conjugate acid-base pair. Because the amount of ionized weak acid or weak base at equilibrium can generally be neglected, we simply use the initial concentrations of the components of a buffer to determine its pH.

Imagine a solution containing the weak acid HA and the soluble salt MA. The equilibrium is established as shown here (M+ is a spectator): HA(aq) ⇋ H+(aq) + A– (aq)

And, of course, the equilibrium expression is written as

‫ܭ‬௔ =

[H ା ][Aି ] [HA]

Because pH is measured as the negative log of the hydrogen ion concentration, we can take the negative log of both sides to obtain: – ‫܏ܗܔ – = ࢇࡷ ܏ܗܔ‬ Further rearrangement yields the following relationship: – ‫[ ܏ܗܔ‬۶ା ] = – ‫ ࢇࡷ ܏ܗܔ‬+ ‫܏ܗܔ‬

[۶ ା ][‫] ିۯ‬ [۶‫]ۯ‬

[‫] ିۯ‬ [‫܍ܛ܉܊ ܍ܜ܉܏ܝܒܖܗ܋‬, ‫] –ۯ‬ ‫ܘ ܚܗ‬۶ = ‫ ࢇࡷܘ‬+ ‫܏ܗܔ‬ [۶‫]ۯ‬ [‫܌ܑ܋܉ ܓ܉܍ܟ‬, ۶‫]ۯ‬

The equation above is the acid form of the Henderson-Hasselbalch equation, and we use it to determine the pH of solutions that contain a weak species and its conjugate; i.e., buffered solutions.

Notice that when the ratio of the concentration of the conjugate to acid is one, the value of the pH is equal to pKa. This observation should not be missed.

A variation of the Henderson-Hasselbalch equation is shown below; use this when the buffer is made of a weak base and its conjugate: – ‫۽[ ܏ܗܔ‬۶ ି ] = – ‫ ࢈ࡷ ܏ܗܔ‬+ ‫܏ܗܔ‬

[۶۰ା ] [‫܌ܑ܋܉ ܍ܜ܉܏ܝܒܖܗ܋‬, ۶۰ା ] ‫۽ܘ ܚܗ‬۶ = ‫ ࢈ࡷܘ‬+ ‫܏ܗܔ‬ [۰] [‫܍ܛ܉܊ ܓ܉܍ܟ‬, ۰]

Note that this solves for pOH, not pH.

You should also notice that the final term is a ratio, which allows us to use mol values in place of concentration, if desired. And, watch for salts that yield more than one ion per mol salt – this case will alter the mol and molarity of the conjugate.

132

Practice 6.17 What is the pH of a buffer solution made by adding 0.120 mol lactic acid and 0.100 mol sodium lactate to enough water to make1.0 L solution? Ka for lactic acid is 1.4 x 10-4.

What is the pH of a buffer solution made of 0.12 M benzoic acid and 0.20 M sodium benzoate? Ka for benzoic acid is 6.3 x 10-5.

What is the pH of 2.0 L buffer solution made of 0.36 mol NH4Cl and 0.20 mol NH3? Kb for NH3 = 1.8 x 10-5.

We can also calculate the amount of the conjugate acid-base pair or the ratio between the species that must be used to prepare buffers by rearranging the Henderson-Hasselbalch equation using the fact that the log of a quotient A/B is equal to log A – log B: ‫ܘ‬۶ = ‫ ࢇࡷܘ‬+ ‫]܌ܑ܋܉[܏ܗܔ –]܍ܛ܉܊ ܍ܜ܉܏ܝܒܖܗ܋[܏ܗܔ‬ It is essential that you recognize that x = log y is solved as 10x = y. 133

Practice 6.18 How many mol of sodium benzoate must be added to 500.0 mL 0.20 M benzoic acid solution to make a solution of pH 4.00?

What ratio of base to acid is required to prepare an acetic acid solution with a pH of 4.50?

What number of grams of solid sodium fluoride must be mixed with 0.100 mol solid hydrogen fluoride to prepare 0.500 L of buffer solution with a pH of 5.00?

134

Part 19: THE pH OF BUFFERED SOLUTIONS WHEN STRONG ACIDS OR STRONG BASES ARE ADDED

In order to calculate the pH of a buffer solution after the addition of strong base or strong acid using the Henderson-Hasselbalch equation, you must be very careful to note the equilibrium that is established: •

The addition of n mol strong base:

According to Le Châtelier, the addition of strong base, OH–, will cause the system to respond by u the hydroxide ion.

equilibrium shifts to the right HX(aq) + OH– (aq) ⇋ H+(aq) + X– (aq)

The weak acid will decrease in concentration by n mol, while the conjugate base will increase by n mol. •

The addition of n mol strong acid:

According to Le Châtelier, the addition of strong acid, H+, will cause the system to respond by u the hydrogen ion.

equilibrium shifts to the left HX(aq) ⇋ H+(aq) + X– (aq)

The weak acid will increase in concentration by n mol, while the conjugate base will decrease by n mol.

Practice 6.19

• Determine the pH of a 0.250 L buffer solution made with 0.060 mol acetic acid and 0.040 mol sodium acetate. • Determine the pH when 0.005 mol sodium hydroxide is added to one 250. mL sample of buffer. No change in volume occurs. • Determine the pH when 0.005 mol hydrochloric acid is added to a separate 250. mL sample of buffer. No volume change occurs.

135

Part 20: ACID-BASE TITRATIONS

•

Indicators

You have performed many titrations already. Recall that in an acid-base titration, a known molarity of acid (or base) is added to an unknown molarity of base (or acid) until the equivalence point of the titration is reached. The equivalence point is the point when stoichiometrically equivalent quantities of acid and base have reacted. An indicator can be used to determine the equivalence point. Indicators change color when specific pH values have been reached. One would choose an indicator that has an end point – the point of color change – that is close to the equivalence point of the titration. Some common indicators and the ranges over which they change color are shown at left.

Notice that most indicators have a range of usefulness over about two pH units.

136

TITRATION CURVES We can look at the titration curves of acid-base titrations, which are graphs of the pH versus the volume of added titrant. The titrant is the solution being added; i.e., the solution of known molarity. The solution to which titrant is added is the analyte. The characteristic shapes of titration curves allow us to determine the Ka and Kb values for acid and base solutions, respectively. •

Titrations

We will look at the titration curves and titration details for three types of titrations:

1.

Strong acid-strong base titrations

Strong acid-strong base titrations result in solutions that are neutral at the equivalence point. In a strong acid-strong base titration, the salt produced is the salt of a strong acid and a strong base – both of which dissociate 100% in solution; neither the anion nor cation hydrolyze.

2.

Weak acid-strong base titrations and weak base-strong acid titrations

Weak acid-strong base titrations are related to the buffer systems we discussed in the previous sections. Remaining conjugate present after the titration will hydrolyze.

3.

Polyprotic acid titrations (Curves only)

In discussing titrations, we are generally attempting to determine the pH of the mixtures of the acids and bases at points before, at and beyond the equivalence point. We can do this with equilibrium charts and/or the Henderson-Hasselbalch equation.

STRONG ACID – STRONG BASE TITRATIONS •

pH at the equivalence point

A strong acid-strong base titration will result in a neutral solution at the equivalence point •

pH before or after the equivalence point

Use the stoichiometry of the reaction and determine the excess acid or base. If the analyte is strong base,

Figure 46. The titration curve for a strong-strong titration. The pH rapidly rises when the equivalence point is reached. Any indicator that has an end point along the vertical segment would be appropriate for the titration.

then before the equivalence point there is excess base, while there is excess acid before the equivalence point if we titrate an acid. After the equivalence point, it should be apparent that the added titrant is in excess. The general idea is to determine the mol of acid/base present initially, the mol of base/acid added, and then determine the concentration of the important ion – either hydrogen ion or hydroxide ion.

Figure 46. The titration curve for a strong-strong titration. The pH rapidly rises when the equivalence point is reached. Any indicator that has an end point along the vertical segment would be appropriate for the titration. 137

We will now work through two examples of strong acid-strong base titrations. Be very careful during a titration: the molarity changes as a solution is added to another!

Practice 6.20 Determine the pH of the solution formed when the following volumes of 0.100 M NaOH are added to a 50.00 mL sample of 0.100 M HCl. Note the change in volume!

49.00 mL NaOH

49.90 mL NaOH

50.000 mL NaOH

50.10 mL NaOH

51.00 mL NaOH

138

Determine the pH of the solution formed when the following volumes of 0.300 M HCl are added to a 50.00 mL sample of 0.100 M NaOH.

12.00 mL HCl

14.00 mL HCl

16.00 mL HCl

16.25 mL HCl

16.75 mL HCl

17.00 mL HCl

Select an indicator from the chart on Page 136 that would be useful in the titration above.

Why does no hydrolysis occur in the solution during the titration?

139

STRONG ACID-WEAK BASE OR STRONG BASE-WEAK ACID TITRATIONS

The titrations of weak acids with strong bases and the titrations of weak bases with strong acids, respectively, are more complicated than the titrations of strong bases with strong bases. This is because the ionizations are not 100% and the conjugate acids and conjugate bases will undergo hydrolysis when titration continues beyond the equivalence point. You should take the time now to examine Figure 48 on this page and sort through the significant amount of chemistry discussed in its caption.

We will discuss all of these in of the titration of a weak acid with a strong base – the titration of a weak base with a strong acid follows similar procedures. The general process for these titrations is to establish the appropriate initial equilibrium, and then use stoichiometry to determine the mol of each species present after reaction. Then, for cases where weak acid is in excess, use the Ka expression for the acid to determine the hydrogen ion concentration, or use the Kb expression if the weak base is in excess. This really comes down to two events: the stoichiometry of the reaction between the acid and the base, which is followed by the equilibrium dissociation of the excess species. If the reaction is 1:1 acid:base, then the excess species is the conjugate of the acid or base, which will undergo hydrolysis. If the titration goes beyond the equivalence point, then the strong acid or strong base is in excess, and we can use the definitions of pH and pOH after the stoichiometry calculation. Thus, for all weak-strong titrations: first, perform the stoichiometric calculation; then, perform the equilibrium calculation.

You will certainly want to recognize the time during a

Figure 48. The titration curves for weak-strong titrations are

titration of this sort when buffering occurs. During this

different in many respects compared to that of strong-strong

time, the Henderson-Hasselbalch equation can be used

titrations. A) They begin at higher acidic pH values or lower basic pH

to determine the pH of the solution.

values; B) the pH rapidly changes initially, but then levels off while noticeably changing all the while (the pH in a strong-strong changes

As mentioned before, this is not the time to try to

much less up to the equivalence point); C) At the halfway equivalence

memorize a series of steps that you simply try to follow

point, the concentration of the weak species is equal to its conjugate –

each time a problem is prsented – it is the time to apply

thus, the value of the pH is equal to pKa, which is apparent using the

the chemistry of each new situation and think about

Henderson-Hasselbalch equation; and D) from the beginning of the

what is occurring at the molecular level.

titration to the equivalence point the solution is a buffer solution. Finally, at the equivalence point, the conjugate undergoes hydrolysis. After the equivalence point the excess strong species drives the pH. 140

Practice 6.20 Determining the pH of a solution of the titration of a weak acid with strong base ending before equivalence point (excess acid)

This titration involves the reaction of a strong base with a solution of weak acid. In this case, the titration does not go to the equivalence point. For example, imagine that we titrate a 0.100 M sample of acetic acid with a 0.200 M sample of NaOH. We are asked to determine the pH of the resulting mixture after 15.0 mL of NaOH has been added to a 50.0 mL sample of the acid.

1.

Determine the appropriate equation. HCH3COO (aq) + OH– (aq) ⇋ CH3COO– (aq) + HOH(l)

2.

Determine the initial and final mol quantities of the species (except water) – this is the stoichiometry calculation. It is best to do this in of mol rather than concentration – the concentration is going to change because of the additive nature of the titration. You might want to add a row that includes the final concentrations, too.

HCH3COO

+

OH–

⇋

CH3COO–

Initial mol Change mol Final mol Mole / volume at completion [Final]

3.

Once we determine the equilibrium concentrations, we recognize the presence of weak species and its conjugate. This allows us to use the Henderson-Hasselbalch equation to solve for pH.

141

Practice 6.21 Determining the pH of a solution of the titration of a weak acid with strong base – ends after equivalence point (excess base)

This titration involves the reaction of a strong base with a solution of weak acid. In this case, the titration goes beyond the equivalence point. For example, imagine that we titrate a 0.0250 M sample of acetic acid with a 0.050 M sample of NaOH. We are asked to determine the pH of the resulting mixture after 30.0 mL of NaOH has been added to a 50.0 mL sample of the acid.

1.

Determine the appropriate equation. HCH3COO (aq) + OH– (aq) ⇋ CH3COO– (aq) + HOH(l)

2.

Determine the initial and final mol quantities of the species (except water) – this is the stoichiometry calculation.

HCH3COO

OH–

⇋

CH3COO–

Initial mol Change mol Although there will be conjugate Final mol

base, the excess strong hydroxide will drive the pH, and this species

Mole / volume at

becomes insignificant.

completion [Final]

3.

We can see that the excess OH- is the only important species – there is no additional acid in the solution. Thus, use pOH + pH = 14.00 to determine the pH of the solution.

142

Practice 6.22 Determining the pH of a solution of the titration of a weak acid with strong base – ends at equivalence point (“salt in excess”) This titration involves the reaction of a strong base with a solution of weak acid. In this case, the titration ends at the equivalence point. For example, imagine that we titrate a 0.0250M sample of acetic acid with a 0.050M sample of NaOH. We are asked to determine the pH of the resulting mixture after 25.0 mL of NaOH has been added to a 50.0 mL sample of the acid. 1.

Determine the appropriate equation for the reaction between the acid and the base. HCH3COO (aq) + OH– (aq) ⇋ CH3COO– (aq) + HOH(l)

2.

Determine the initial stoichiometric amount of conjugate base at the equivalence point – this is the stoichiometric calculation.

OH–

HCH3COO

⇋

CH3COO–

Initial mol Change mol Final mol Mole / volume after Reaction [Final]

3.

We see here that there is apparently no H+ or OH– in solution by our equilibrium calculations. Thus, what we have is the solution of a salt, NaCH3COO; i.e., the conjugate base of the analyte. Recall from our earlier discussions of hydrolysis that we can determine the pH of a salt solution by using the equilibrium expression of the weak acid from which the salt is derived. CH3COO–

+

H2 O

⇋

HCH3COO

+

OH–

Initial Change Concentration at Equilibrium

143

We will now work through several examples of each type of titration. A summary of the processes is provided here. The summary here considers that an acid is being titrated with a strong base. Adjust accordingly for the titration of a base.

SUMMARY OF ACID-BASE TITRATION CHARACTERISTICS Acid

Strong

solution –

base –

in beaker

in buret

Stoichiometric Point

Consideration at Stoichiometric Point

Process

Determine concentration strong

strong

prior to equivalence point

Excess H+ from acid

of H+ directly. Use pH = –log [H+]

Determine concentration strong

strong

after equivalence point

Excess OH– from base

of OH– directly. Use pH = 14.00 – (-log [OH–])

strong

strong

at equivalence point

[H+] = [OH–]

[H+] = [OH–] = pH 7.00

Determine concentration of H+ using weak

strong

prior to equivalence point

Weak acid is in excess

equilibrium expression for acid or Henderson-Hasselbalch

Determine concentration weak

strong

after the equivalence point

Strong base is in excess

of OH– directly. Use pH = 14.00 – (-log [OH–])

Determine the stoichiometric concentration of salt; then use the weak

strong

at the equivalence point

Salt is in excess – consider the

equilibrium expression for the

hydrolysis of the salt

hydrolysis of the salt to determine OH– directly. Use pH = 14.00 – (-log [OH–]) 144

Practice 6.23 How many mL of 0.0350 M NaOH are required to titrate the following solutions to their equivalence points? For each, suggest an appropriate indicator. (A) 40.0 mL 0.0350 M HNO3

(B) 65.0 mL 0.0620 M HBr

(C) 80.0 mL 0.0453 M HCl

A 20.0 mL sample of 0.200 M HBr is titrated with 0.200 M NaOH solution. Calculate the pH of the following solutions. (A) 20.0 mL HBr and 15.0 mL NaOH

(B) 20.0 mL HBr and 19.9 mL NaOH

145

A 50.0 mL solution of 0.150 M acetic acid is titrated with 0.150 M NaOH solution. Calculate the pH before titration and after the following volumes of NaOH have been added: (A) 25.0 mL; (B) 49.0 mL; (C) 50.0 mL; (D) 51.0 mL; (E) 75.0 mL

146

Part 21: THE VALUE OF THE SOLUBILITY-PRODUCT CONSTANT, Ksp, FOR IONIC SOLIDS IN SOLUTION

In similar fashion to other equilibria we have studied, a saturated solution of an ionic compound establishes equilibrium with undissolved solid in the solution. It is very similar in nature to the original chemical equilibrium expressions we saw earlier.

Before we discuss the equilibrium established, be sure to distinguish between solubility and the solubility product constant, Ksp. It’s just the same difference between the concentration of species at equilibrium and the value of K for an equilibrium: one refers to the actual values of molarity, while the other refers to the ratio of the molarities of all species.

Solubility refers to the quantity of a solute that dissolves in solution to make a saturated solution. Solubility might be expressed as mol/L or g/L.

The solubility-product constant, Ksp, is the equilibrium constant for the equilibrium between a dissolved solid and its saturated solution. As seen with other K values, a large value for Ksp indicates that a larger amount of solute will dissolve; smaller values indicate that little solute will dissolve. The value of Ksp for strong electrolytic salts is simply “very large.”

The equilibrium expression for Ksp is identical to the Kc values of other equilibria: it is the product of the ionized species raised to a power equal to their stoichiometric coefficients. For example, the equilibrium expression for Ag2CrO4 is shown here: Ksp = [Ag+]2[CrO42-] = 1.2 x 10-12

Practice 6.24

Write the equilibrium expression for the equilibrium established when silver chromate, Ag2CrO4, is dissolved in water to saturation.

Determine the concentration of each ion in a saturated solution of silver chromate.

Notice that when data is substituted we must double the concentration of silver ion due to its stoichiometry! 147

Ksp values can be used to determine solubility. It is important to note, however, that Ksp is not a concentration – it is the solubilityproduct constant; there is a relationship between the two quantities.

solubility of compound in g • L-1

solubility of compound in mol • L-1

molar concentration of ions

Ksp

Practice 6.25

What is the solubility of silver chromate in grams per liter based on the data above?

We can use the Ksp values for solutions to determine whether or not a precipitate will form when two solutions that contain an ion of the potential precipitate salt are mixed. This is very similar in nature to determining Q earlier – here Q is the product of the concentrations of the two ions. And, in similar fashion:

If Q < Ksp, then the solid dissolves until Q = Ksp

If Q = Ksp, then the mixture is at dynamic equilibrium – the solution is saturated. The addition of a small amount of either ion will result in precipitation.

If Q > Ksp, then precipitation occurs until Q = Ksp

It is also possible to selectively precipitate an ion from a solution containing two or more ions by adjusting the concentration of the precipitating ion so that it first just exceeds the concentration required to precipitate a first ion, and then increasing the concentration of the precipitating ion so that its concentration exceeds the concentration required to precipitate the second ion. This is shown in the second example below.

148

Practice 6.26 Will a precipitate form when 0.10 L of 3.0 x 10-3 M Pb(NO3)2 solution is added to 0.400 L of 5.0 x 10-3 M solution Na2SO4? Ksp for PbSO4 = 1.6 x 10-8.

Which salt will precipitate first in a solution that is 1.0 x 10-2 M Ag+ and 2.0 x 10-2 M Pb2+ when Cl- is added to the solution? Ksp for silver chloride is 1.8 x 10-10; Ksp for lead(II) chloride is 1.6 x 10-5. What concentration of chloride ion is needed to precipitate each metal ion?

From the data obtained above for AgCl and PbCl2, how can you selectively precipitate the ions from solution?

149

Part 23: FACTORS AFFECTING SOLUBILITY

•

The Common-Ion Effect

According to Le Châtelier’s Principle, the solubility of a solute decreases in the presence of a salt containing the cation or anion of the solute. For example, look at Figure 49 at left. In pure water, the molar solubility of magnesium fluoride is on the order of 10-4 M. However, the addition of the fluoride ion in the form of the salt sodium fluoride drives the solubility down considerably. Writing the equilibrium established will allow you to easily see this: MgF2(s) ⇋ Mg2+(aq) + 2 F– (aq) You can see that a Le Châtelier response is expected Figure 49. The molar solubility decreases significantly owing to the common ion effect.

when the fluoride ion is added, which results in the precipitation of magnesium fluoride. This is the result of Q > Ksp.

•

Solubility and pH

Ionic species that are only moderately or slightly soluble in water can be made more soluble in acid if their anions are basic. For example, the ionic solid Mg(OH)2 is only slightly soluble in solutions of pH 10.0 or so. However, when the pH is lowered to 9.0, the solubility of the compound increases over 1000 times. The solubility of ionic compounds containing a basic anion increases with decreasing pH – the basic anion reacts with the increasing H+. As the basic anion reacts, it is removed from solution. Thus, according to Le Châtelier's Principle, the salt will continue to dissolve to counteract the decrease in the anion. Again, the equilibrium equation makes Figure 50. The pH of a solution can affect solubility. This is a result of the equilibrium position of the dissolving process and Le Châtelier’s principle.

this apparent: Ni(OH)2(s) ⇋ Ni2+(aq) + 2 OH– (aq)

150

•

Solubility and Chemical Reaction

Here, the solubility of the silver chloride increase because the silver ion complexes with ammonia (for example, as Ag(NH3)2+). Again, in a typical Le Châtelier response, additional AgCl dissolves to replace the silver ion that has complexed. (Note that the initial silver ion that is present is very small as AgCl has limited solubility – however, only a small amount need be present to allow complex formation.)

Ag+(aq) + NH3(aq) ⇋ Ag(NH3)2+(s) Figure 51. Because ammonia and silver form a complex, the solubility of silver chloride increases as the concentration of ammonia increases. Again, this can be viewed in light of Le Châtelier’s principle.

Practice 6.27 Determine the solubility of CaF2 in a solution containing 0.010M Ca(NO3)2. Ksp for CaF2 = 3.9 x 10-11

CaF2 Initial

--

Change

--

Equilibrium

--

⇋

Ca2+

2 F–

151

The solubility-product constant of Mn(OH)2 is 1.6 x 10-13 at 25°C. Determine the molar solubility of the compound in water and in a solution buffered at pH 9.0.

152


Thermochemistry, Thermodynamics and Electrochemistry II


Identify and establish relationships between work, energy and heat

Recognize kinetic energy and potential energy and their conversions, and discuss and describe thermal and internal energy

Recognize and use the units of energy joule and calorie

Identify and characterize a system and surroundings when discussing the transfer of energy; recognize and describe state functions; characterize endothermic and exothermic processes

Characterize and apply the first law of thermodynamics to processes

Define, applyy and characterize enthalpy

Justify and apply Hess’s Law; determine the enthalpies of formation of various compounds

Discuss and perform calculations using calorimetry

Characterize and predict spontaneity; relate spontaneity to equilibrium conditions

Characterize, apply, calculate and predict entropy changes; establish and predict microscopic character from bulk observations

Calculate free-energy energy changes; characterize the relationship between free energy and temperature, and predict spontaneity based on the free energy difference determined when evaluating the enthalpy/entropy relationship

Quantitatively relate emf of a redox reaction to free energy change; use the Nernst equation to establish the relationship between concentration and cell emf; relat relate the Nernst equation to Le Châtelier's Principle

Calculate changes in reactants/products during electrolysis and concentration cells. Quantitatively describe oxidationoxidation reduction reactions in of amperage, voltage and time

153

Part 1: THE NATURE OF ENERGY

Molecules possess energy because of the arrangement of the atoms in the molecule. Because forces are working in a molecule (including electrostatic attractions between positive charges and negative charges), a molecule possesses potential energy, energy or energy that can be released ased when the arrangement of the atoms changes.

Energy “stored” in molecules due to their arrangement is called chemical energy. When the arrangement of atoms changes, the energy of a molecule changes – new and different forces exist. Thus, when your body uses sugar to make new molecules, there are changes in energy as Figure 52. As the temperature of the samples decreases from left left-to-right in the figure, the thermal energy of the sample decreases.

the sugar molecule’s bonds break and new bonds form in new substances. All chemical reactions involve a change in energy. Molecules can also

possess energy due to their temperature (kinetic kinetic energy energy), which is discussed as thermal energy.. The molecules in the far-left far figure at left possess more thermal energy than han the molecules in the far far-right figure.

The Units of Energy

The SI unit of energy is the joule, J. A joule is officially defined as the amount of energy required to move a 1 kilogram mass through a distance of 1 meter at an acceleration of 1 m s-2. For a more practical understanding, it takes about one joule of energy to lift a 1 kg Figure 53. As the products are made, energy is transferred as work by moving the piston or as heat to the product gas molecules and container walls.

object to a distance of 10 cm above ground, and the amount of energy required to raise the temperature of 1 gram of water from 287.5 K to 288.5 K is equal to 4.184 J. (This amount of energy is also one thermodynamic calorie – a food calorie is actually 1000 thermodynamic calories, or one kilocalorie. T Thus, the energy of one candy bar – about 300 food calories or 300000 thermodynamic calories – is a little over one-and-a-quarter quarter million J.)

Energy Transfer

Energy can be transferred in two ways: as work or as heat.

Consider the system at the right, in which gases are present in a closed cylinder, and the piston has the pressure of the atmosphere acting on it. When the gases react in an exothermic reaction, there is an increase in temperature inside the cylinder, the gases expand, and their expansion ca causes uses the piston to move against the atmosphere. Because B the energy is ultimately causing the motion o of an object against a force,, then we say work is being done: done Work is done when energy causes an object to move against a force.

154

Energy is also transferred as heat. Heat is the energy transferred to a colder object from a warmer object. Heat is a term that describes this transfer of energy – heat does not exist without the transfer of energy from a warmer object (one with greater temperature) to a cooler object (one with lower temperature). Heat is the transfer of energy due to differences in the temperature of two objects in with one another; that is, energy moving from a hotter body to a colder body is the transfer of heat. Thus, we do not say that an object “possesses” heat – we say that the object possesses “thermal energy” and that it transfers that energy as heat. Heat transfer ceases when two objects in reach the same temperature.

Part 2: DISTINGUISHING A SYSTEM FROM THE SURROUNDINGS

When we discuss a particular reaction – perhaps the combustion of octane in an engine’s cylinder, for example – we focus our discussion on two parts of the universe: the system and the surroundings. The system is the part of the universe we are examining, which might be the chemical reaction of the combustion of octane in the case above. The cylinder, pistons and everything beyond the reaction are the surroundings. We may, however, also be concerned with the behavior of the piston when exposed to the gases, in which case the piston also becomes part of the system.

Systems may be open systems, which are systems that can exchange energy and mass with the surroundings (like an open test tube in which a gas-forming reaction has occurred), or systems may be closed systems, which are systems that can exchange energy but not mass with the surroundings (like a closed test tube in which a reaction has occurred).

We would want to distinguish between system and surroundings when discussing energy changes. If a reaction occurs and heats up the air around the reaction, then we say that the system has lost energy (as heat), while we would say that the surroundings have gained energy. Notice that the discussion of which part lost energy and which part gained energy is easily discussed when we separate the system from the surroundings.

Part 3: THE FIRST LAW OF THERMODYNAMICS

In the reaction above showing the heating of the gases in the cylinder we discussed several energy changes, which are summarized here:

► chemical energy in fuel (potential energy) ► transferred as heat energy (also produced light) ► converted into thermal energy of molecules (kinetic energy) ► transferred into energy as work to move piston

Although many energy conversions and transfers occurred, there is no energy for which we cannot : all energy has been transferred as heat or work or converted into another form. This observation is the First Law of Thermodynamics: Energy cannot be created or destroyed in a chemical process or physical change. If energy is lost by a system, then an equal amount of energy is gained by the surroundings. The First Law of Thermodynamics is the Conservation of Energy.

155

Part 4: THE INTERNAL ENERGY

In order to discuss the First Law of Thermodynamics and the energy changes associated with chemical reactions, we must consider all of the energy a system possesses. There are a few we have already seen:

potential energy – the energy possessed by a sample due to its composition or position (interactions of atoms and molecules)

kinetic energy – the energy possessed by a sample due to its motion (including the electrons and nucleus); also the energy related

thermal energy – the energy possessed owing to its temperature (related, of course, to the kinetic energy)

to the temperature of the sample

All of the energy that is possessed by a system is collectively called the internal energy of the system. In the example above with the piston and cylinder, the internal energy of the gas molecules is increasing due to increasing kinetic energy as heat is absorbed. Thus, according to the first law, the internal energy (i.e., the chemical energy or potential energy) of the fuel is decreasing. The internal energy of the gas molecules subsequently decreases as work is done on the piston.

The internal energy of a system can be expressed mathematically as:

∆ࡱ = ࡱࢌ࢏࢔ࢇ࢒ − ࡱ࢏࢔࢏࢚࢏ࢇ࢒ where Einitial is the energy of the system originally, and Efinal is the energy of the system at some later time

Several points must be considered here:

The actual values of Efinal and Einitial cannot be determined – it is only ∆E with which we are concerned to apply the First Law

A system that has lost energy will have a negative value for ∆E, indicating that the initial energy is greater than the final energy (Figure 55a)

A system that has gained energy will have a positive value for ∆E, indicating that

In a chemical reaction, the final condition is the energy of the products, while the

the final energy is greater than the initial energy (Figure 55b)

initial condition is the energy of the reactants (which, recall, we don’t know – we only know ∆E).

Endothermic processes involve the absorption of energy into a system and have positive ∆E values, while exothermic processes involve the release of energy out of a

Figure 54. You can see that the internal energy of the reactants (hydrogen and oxygen gases) is higher than the internal energy of the product (liquid water). When the reaction occurs ∆E is negative as energy is lost by the system. If we change water back to hydrogen and oxygen then the sign of ∆E is positive as the internal energy increases. According to the First Law, the energy is not “gone;” rather, it is transferred as heat, light and sound from the system to the surroundings.

system and have negative ∆E values.

156

The internal energy of a system changes as heat is lost or absorbed by the system or as work is done by or on the system. We can thus relate ∆E to a mathematical relationship involving work and heat:

∆E = q + w where q is the heat added or absorbed by a system, and w is the work done on or by a system

Here, we must consider the signs of work and heat:

When heat is transferred into a system the sign of q is positive for the system, and Efinal = Eintial + q

When heat is transferred out of a system the sign of q is negative for the system, and Efinal = Eintial - q

When work is done on a system the sign of w is positive for the system, and Efinal = Eintial + w

When work is done by a system the sign of w is negative for the system, and Efinal = Eintial – w

Figure 55. In (a), the internal energy of the system in its final state is less than it was originally – energy has been lost as heat or work. In (b), the internal energy of the system in its final state is greater than it was originally – energy has been gained as heat or work. In (a) the process is exothermic (∆E < 0), while in (b) the process is endothermic (∆E > 0).

Of course, then, when both work and the transfer of energy as heat are occurring, then the sign of ∆E is dependent upon the sum of the two based on their individual magnitudes:

The sign conventions of q and w

q > 0: heat is transferred into the system from the

∆E due and the signs of q and w when q > 0 and w > 0: ∆E is > 0

surroundings q < 0: heat is transferred out of the system into the

when q < 0 and w > 0: ∆E depends on magnitudes of q and w

surroundings w > 0: work is done on the system

when q < 0 and w > 0: ∆E depends on magnitudes of q and w

w < 0: work is done by the system

when q < 0 and w < 0: ∆E is < 0

157

Practice 7.1

Imagine the following scenario. Energy nergy from the reaction has been transferred as heat to the gases inside the cylinder.

If the amount of heat absorbed by the gases is 1150 J and the gases do 450 J of work to move the piston, what is ∆E for the system ystem (the cylinder and piston)?

Has the thermal energy of tthe gases increased or decreased? Explain.

Is the reaction below the cylinder exothermic or endothermic? Explain.

What is the sign and magnitude of q for the combustion system? What is the sign of q for the cylinder system?

What is the magnitude of w for the combustion system? What is the sign and magnitude of w for the cylinder system?

For the figure at right, discuss apparent transfers of energy as heat and work; identify the signs of q and w.

158

Part 5: STATE FUNCTIONS

You may see your textbook for additional examples, but it is quite easy to understand that a state function is a property that is determined only by the final and initial conditions of the property. State functions do not depend upon the route by which the change in the property occurred.

If you travel from Owings, Maryland to Toledo, Ohio you will have traveled from 32 feet above sea level to 435 feet above sea level. It does not matter if you travel by way of the Pennsylvania turnpike or by way of I-66 through western Maryland – in either case your altitude change from Owings to Toledo will be 403 feet. Thus, altitude change is a state function – how you got to the final condition (435 feet) is not considered. However, travel distance is NOT a state function – it takes an additional 103 miles to travel by way of I-66 versus traveling the Pennsylvania Turnpike.

Another example of a state function is phase. Water at 400 K is a gas, while at 294 K water is a liquid. It does not matter if Figure 56. An example showing that heat and work are not state functions is illustrated by the battery. Regardless of how the energy is transferred, the value of ∆E for the system is a state function – it does not matter how the energy change occurs for the overall change in

the water is brought to 43 K and then heated to 615 K before settling at 294 K: the phase at 294 K does not depend on what phase the water has been in at some other times – the phase at 294 K only depends upon the final temperature.

internal energy. In (a), the energy is lost from the battery as heat, while in (b) the energy is lost from the battery as heat and work. However, this does not affect the value of ∆E for the system, which is the sum of any energy changes for the system.

Note, however, that the components of a change can be state functions even though their sum is not. As an example, consider ∆E, which is the sum of q and w. Although the total change (∆E) is not dependent upon whether the change was

effected by q or w, the values of q or w depends on how the energy change is occurring – thus, q and w are not state functions; their values depend upon the route of the energy transfer (is it heat or work?)

159

Part 6: ENTHALPY

The thermal energy of a system is a measure of the molecular motion of the system. For example, the thermal energy of liquid water is greater than the thermal energy of ice – liquid water has greater molecular motion. Thus, a hotter object has a greater thermal energy than a cooler one, and the temperature of an object with greater thermal energy is higher than the temperature of an object with lower thermal energy. “Heat” refers to the transfer of thermal energy – “hot” is a relative term to describe temperature differences. We do not say that a hot bowl of soup has “a lot of heat,” but we can say that a hot bowl of soup has “a lot of thermal energy;” and we can say that a bowl of soup is “transferring a lot of heat to the surroundings.” When the thermal energy of a system increases or decreases, we know that energy has been transferred into or out of the system as heat, respectively. This transfer of heat can be measured as a quantity called enthalpy, denoted by H. Although we cannot measure the enthalpy of a system, we can measure the change in enthalpy, ∆H, which is a measure of the amount of heat transferred into or out of a system during a process or reaction. Although the value of ∆E is the sum of the work and heat exchanges occurring for a system, compared to the value of q the value of w is quite small. Thus, we can represent the change in enthalpy – very importantly a state function - mathematically as:

∆H = Hfinal – Hinital = q

Practice 7.2 Indicate the sign of ∆H for each of the following processes:

an ice cube melts

a sample of butane burns

You are holding a beaker in your hand in which a reaction is occurring. The beaker feels cold. Is the reaction endothermic or exothermic? Explain.

Figure 57. Above, heat gained by a system indicates that the ∆H is positive (> 0) and the reaction or process is endothermic. (An equal amount of heat is lost by the surroundings.) Below, heat lost by the system to the surroundings indicates an exothermic reaction or process, and ∆H is < 0. Respectively, the figures represent an increase and decrease in thermal energy and internal energy.

160

Part 7: REACTION ENTHALPIES

All chemical reactions occur with changes in energy: heat is either released or absorbed.

Chemical reactions that occur with a release of energy have negative values of ∆H.

Chemical reactions that occur with absorption of energy have positive values of ∆H.

The energy that is released of absorbed as heat as a reaction occurs is called the enthalpy of reaction, and is abbreviated ∆Hrxn.

∆Hrxn = Hproducts – Hreactants When a reaction occurs, several things happen:

The bonds of reactants must break (bond dissociation – an endothermic process), and the bonds of products must form (bond formation – an exothermic process)

Energy is transferred out of the system and into the surroundings or is absorbed by the system from the surroundings:

If the amount of energy absorbed by bond dissociation is less than the amount released by bond formation, then the reaction is exothermic and ∆H is negative. The bond strengths of the products are greater than the bond strengths of the reactants.

If the amount of energy absorbed by bond dissociation is greater than the amount released by bond formation, then the reaction is endothermic and ∆H is positive. The bond strengths of the reactants are greater than the bond strengths of the products. ∆H of the reverse of a chemical reaction is equal in magnitude but opposite in sign to the forward reaction, as shown in Figure 58. Recall that the presence of a catalyst does not change the value of ∆H. The amount of energy required to cause a forward reaction is the activation energy, Ea; the amount of energy to cause the reverse of that reaction – i.e., Ea(reverse) – is the sum of ∆H and Ea(forward).

An example of a reaction with a very large enthalpy change is the decomposition reaction of nitroglycerin, C3H5N3O9. The bonds in this compound are relatively weak, and so little energy is required to break them. Indeed, a small jolt is all that is required to begin the decomposition. The decomposition of nitroglycerin produces the following gases according to this unbalanced equation:

Figure 58. The enthalpy change associated with the combustion of one mol of methane is –890 kJ, and for the reverse reaction, the enthalpy change is +890 kJ. This is a direct result of enthalpy’s character as a state function.

C3H5N3O9 → N2 + CO2 + H2O + O2

∆Hrxn ≈ –3500 kJ/mol

The bonds formed in the products are very strong. The formation of these strong bonds releases a great amount of energy when the products form. Thus, the reaction is explosive. The enthalpy change is very large and great amounts of energy are released into the surroundings, which provides a negative value for ∆H. (In fact, a

hallmark of reactions with large negative enthalpy changes is the formation of strong bonds in the products and weak bonds in the reactants – i.e., bond dissociation is small positive, while bond formation is large negative.) 161

Part 8: HESS’S LAW

Recall that enthalpy change is a state function: it does not matter the route that A follows to become B; for example – the value of ∆H for the conversion of A to B is the same whether it occurs simply as A → B or as A → C → D + E → B.

Imagine that you have burned a small sample of methane and produced carbon dioxide and gas water according to the equation below: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) We could express the thermochemical equation, which includes ∆H, as:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

∆H = –890 kJ/reaction

The reaction above is really two processes: the combustion of methane to produce the products carbon dioxide and gas water and the subsequent condensation of the water into liquid water:

Step 1:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)


Step 2:

2H2O(g) → 2H2O(l)


Summation

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)


When the reaction occurs, a total of 802 kJ/reaction is released as the methane burns, and a smaller amount of heat, 88 kJ/reaction, is released when the gas water condenses. The total amount of heat released as both processes occur is –802 kJ/reaction + –88 kJ/reaction = –890 kJ/reaction, or the sum of both processes. This is an application of what we call Hess’s Law: Hess’s Law tells us that the total enthalpy change of an overall process or reaction is equal to the sum of the enthalpy changes for a series of individual steps that effects the same result.

A word of caution...When working thermochemical problems based upon balanced chemical equations, it becomes important to watch the units on calculated results. For example, in the item above, a total energy change of –88 kJ is associated with the process 2H2O(g) → 2H2O(l). This value is the amount of energy released when 2 mol of water condenses – thus, we write that the enthalpy change for the process as written is -88 kJ/reaction – the reaction as written and displayed. We could alternatively write –44 kJ/mol H2O if we would like, but to use it in this particular Hess’s law calculation we still need to double the value. You are advised to work in the units of kJ/ reaction, but you must write the units and the balanced equation for this to have meaning!

162

Enthalpy change is a state function, and, as such, it is not dependent upon the path of the change, but rather only on the initial and final conditions. For any process, it does not matter whether we consider ∆H for several processes separately and sum them or consider ∆H for the overall reaction singly: the total change in enthalpy is the same. This is illustrated in Figure 59 at left. In the illustration, Hess’s Law shows that ∆H1 = ∆H2 + ∆H3.

We use Hess’s Law because we may not know the total enthalpy change for many reactions (some cannot be determined directly, for example). However, if we know a series of reactions that will provide us with the overall reaction of interest, then we can determine the enthalpy change for the reaction of interest. The two or more individual reactions that must be combined to determine the overall enthalpy change of the reaction of interest can be summed arithmetically. Figure 59. Via the combustion of methane, the formation of carbon dioxide and water releases 890 kJ/mol CH4. An alternative pathway is the formation of CO + H2O and O2, which also reacts to form carbon dioxide and water. If we call the direct combustion to CO2 and H2O ∆H1, we call the reaction to CO, H2O and O2 ∆H2, and the reaction of CO, H2O and O2 to CO2 and H2O ∆H3, then the sum of ∆H2 and ∆H3 = ∆H1 according to Hess’s law.

This is not at all unlike the summation of equations we used when determining whether or not a kinetics mechanism was consistent with an overall chemical reaction. The coefficients of the equations used must be equal – that is, multiply individual equations as needed to ensure cancellation of species; and, each reaction must be written in the correct order as far as reactants and products are concerned: change ∆H’s sign as needed. That is, if you need the reverse process, then simply reverse it and to change the sign of ∆H.

Practice 7.3

Calculate the enthalpy change associated with the combustion of nitrogen to produce NO2(g). Of course, always begin by writing the reaction of interest. Use the enthalpies of reaction shown here to answer this question.

N2(g) + O2(g) → 2 NO(g)

∆H = +180.5 kJ/reaction

2 NO(g) + O2(g) → 2 NO2(g)

∆H = –114.1 kJ/reaction

163

It would be common to determine the amount of energy released or absorbed as a mass of a compound reacts with another. In these cases, we determine the enthalpy change per reaction, and then we convert using dimensional analysis to the amount and unit of interest. What is the magnitude of ∆H associated with the combustion of 110.0 grams of nitrogen to produce NO2(g)?

You may need to reverse a reaction to make it useful. Determine the enthalpy of reaction for the production of methane from C(s) and hydrogen gas.

Enthalpies of reaction for the following reactions are known.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)


2H2(g) + O2(g) → 2H2O(l)


C(s) + O2(g) → CO2(g)


Part 9: BOND ENTHALPIES

Bond enthalpies are the energy changes associated with the breaking and formation of chemical bonds. For example, the bond enthalpy of the H-O bond is approximately 463 kJ/mol bonds – this tells us that to break O-H bonds requires 463 kJ/mol bonds and that the formation of the H-O bond releases approximately 463 kJ/mol bonds. A direct application of Hess’s Law allows us to estimate Hrxn when the bond enthalpies of the reactants and products are known. We can estimate the enthalpy change associated with a chemical reaction by simply considering the amount of energy absorbed when the bonds break in the reactants and the amount of energy that is released when the bonds form in the products. Note sign conventions when using this method of determining the enthalpy change of a reaction – negative signs are used for bond formation (as this corresponds to a release of energy), while positive signs are used for bond breaking (as this corresponds to the absorption of energy). Thus, simply assign bond breaking a positive sign and bond formation a negative sign; then, sum the result. Here, our usual products minus reactants will not work. Bond enthalpies are given for gas substances, so we may often have to also consider the enthalpy of phase change when determining overall enthalpy changes. See Figure 60 on Page 165.

∆ࡴ࢘࢞࢔ = ሺ࢈࢕࢔ࢊ ࢋ࢔࢚ࢎࢇ࢒࢖࢏ࢋ࢙ ࢕ࢌ ࢘ࢋࢇࢉ࢚ࢇ࢔࢚࢙ሻ + ሺ– ࢈࢕࢔ࢊ ࢋ࢔࢚ࢎࢇ࢒࢖࢏ࢋ࢙ ࢕ࢌ ࢖࢘࢕ࢊ࢛ࢉ࢚࢙ሻ 164

Practice 7.4

Consider the formation of CO2(g) and H2O(g) via the combustion of CH4(g) and O2(g). Use bond enthalpies to estimate the value of ∆H.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

It is recommended that you begin by sketching the molecules so that you recall their bonding patterns – this helps to ensure all bonds are considered. The reaction of interest in our example can be represented as

H H

C

H

+

H

O

O

O

O

→

O

C

O

+

H

O

H

H

O

H

Here, in order to determine the enthalpy change for the reaction, we determine the enthalpy change for each bonding event, and then sum the results:

Notice

the

determination

slight and

difference that

for

between b the

Hess’s

this law

determination earlier. There are two major reasons why bond enthalpies provide only an estimate of the enthalpy change for a chemical reaction, which are discussed on the next page.

Figure 60. The figure at left graphically represents the use of bond enthalpies in determining ∆H. In the reaction, the bonds of methane break and the bonds of Cl2 break, both of which are endothermic processes. Then, the bonds in H3CCl and HCl form, which are exothermic processes. The difference is the value of ∆Hrxn.

165

Bond enthalpies are really averages of the bond enthalpies of many cases. For example, the first C-H bond that breaks in methane requires less energy than the second or subsequent C-H bond-breaking events. Why? After the first hydrogen atom is removed, the remnant structure localizes electrostatic attraction throughout a lesser number of bonds, which leads to an increase in their strength; thus, later bond-breaking requires slightly more energy than the first.

The mean bond enthalpies only consider the bond between the two atoms of interest without regard to atoms to which the atoms of interest are bonded. This is similar in nature to the strength of oxyacid discussion in earlier units. For example, consider a molecule H-O-F and a molecule H-O-I. The electronegative character of the fluorine atom draws electron density from the H-O bond toward F, while the lesser electronegative iodine atom does not exhibit this behavior. Thus, the H-O bond enthalpy in H-O-F is less than that of the H-O bond in H-O-I; however, we use the same average bond enthalpy for both molecules – indeed, we use this average bond enthalpy for all H-O bonds.

Bond enthalpies are measures of bond strength, and we should evaluate their relative values here to provide a complete discussion of bond enthalpies and their use.

Double bonds are stronger than single bonds, and triple bonds are stronger than double bonds. Their strengths, however, do not increase in a linear manner. The bond enthalpy of the N-N bond is 160 kJ/mol, that of the N=N bond is 418 kJ/mol, and that of the N≡N bond is 941 kJ/mol. While we might predict the double-bond bond enthalpy to be twice that of the single-bond bond enthalpy, we see that it is almost 2.7 times greater – and the triple-bond bond-enthalpy is almost 6 times greater than the single-bond enthalpy. This non-linear increase in enthalpy as bond number increases is due to the shorter length of multiple bonds, which leads to an increase in electrostatic attraction and bond-strengthening.

Polar bonds (bonds in which two covalently-bonded atoms have significantly different electronegativity values) are stronger than nonpolar bonds, as the electron density (thus, the bond itself) is strongly held by one of the atoms in the bond. For example, the bond enthalpy of the N-N bond (a nonpolar bond) is 160 kJ/mol, while the bond enthalpy of the N-O bond is 200 kJ/mol and that of the N-F bond increases to almost 300 kJ/mol.

Practice 7.5 Determine the reaction enthalpy per mol octane when burned to produce carbon dioxide and water. Use mean bond enthalpies.

166

Use mean bond enthalpies to determine the amount of heat released when a 5.00 g sample of propyl alcohol (propanol) undergoes complete combustion.

Part 10: HEATS OF FORMATION

Consider again the combustion of methane to produce carbon dioxide and water. According to Hess’s law – because enthalpy is a state function – we can consider several pathways to determine the enthalpy change for the combustion. One of those pathways can simply be the formation of the compounds from their elements, called the heat of formation, ∆Hformation or ∆Hf. For example, the formation of water gas from its elements releases 240 kJ/mol of water formed. Recalling that the reverse process of a reaction has an enthalpy change that is opposite that of the forward reaction, we can conclude that the dissociation of water releases 240 kJ/mol water – it does not matter if the water formed owing to the combustion of methane or the reaction between elemental oxygen and elemental hydrogen, the result is always that ∆H for the formation of water is –240 kJ/mol water. We can use heats of formation – the enthalpy change associated with the formation of one mole of a substance from its elements – to determine the heats of reaction for various reactions. This is yet another method of estimating the energy changes associated with chemical reactions. Keep in mind, however, the following limitations and considerations when using ∆Hf values:

A heat of formation value is for the formation of one mole of a compound from its elements in their standard states, which are their states at 1 atm of pressure at 298 K. This means that if you attempt to compare bond enthalpy determinations and heats of formation determinations, you must consider the phases of the species very carefully (recall that bond enthalpies are provided for species in gas phase).

The standard heat of formation for any element in its standard state is zero. This should make sense from the standpoint that no change is required to make oxygen gas from oxygen gas, for example.

We take the sum of the standard molar heats of formation, ∆Hf°, of the reactants and subtract these from the sum of the heats of formation of the products. The subtraction eliminates the worry of sign conventions when using heats of formation. (The degree symbol denotes standard thermodynamic conditions of 1 atm, 298 K and, where relevant, 1 M solutions.)

∆ࡴ°࢘࢞࢔ = Σ ࢔ࡴ°ࢌ ሺ‫ܛܜ܋ܝ܌ܗܚܘ‬ሻ − Σ ࢓ࡴ°ࢌ ሺ‫ܛܜܖ܉ܜ܋܉܍ܚ‬ሻ

n and m are the stoichiometric coefficients of the products and reactants in the balanced chemical equation

Hf° is the standard molar heat of formation of the substances 167

Practice 7.6 Using heats of formation and guided by the figure at right – depicting the combustion of propane – determine the enthalpy change for the reaction. The steps are outlined as:

For the same reaction, determine the enthalpy change using bond enthalpies.

The reaction enthalpy for the decomposition of 1 mol of calcium carbonate is 178.1 kJ.

How much energy must be added to the system to decompose 23.5 grams of calcium carbonate?

What is the heat of formation of CaCO3? The heats of formation of CaO and CO2 are –635.5 635.5 kJ and –393.5 kJ, respectively.

168

Part 11: CALORIMETRY

In order to determine ∆H for a chemical reaction, we have to provide a method for determining the enthalpy change associated with the reaction. This can be done using a process called calorimetry, which is the process by which the heat flow of a reaction or process is measured as a change in temperature. Often, the reaction occurs in with water and ∆T for water is used to determine ∆H, as illustrated in the figure below (a “coffee-cup” calorimeter). A few vocabulary are needed to understand the process of calorimetry.

Heat capacity: The heat capacity of a substance is the amount of heat required to raise the temperature of a sample of a material by 1 K (or 1°C). The units of heat capacity can vary depending upon the amount of substance for which the heat capacity is given.

Molar heat capacity: The heat capacity of one mol of a substance, expressed as J/mol ∙ K.

Specific heat capacity: The heat capacity of 1 gram of a substance, expressed as J/g ∙ K. This is also called the specific heat of a substance. It is ‘specific’ because it is the heat capacity of a specific amount – one gram.

Practice 7.7 Which of the following samples would take the greatest amount of time to heat from 298 K to 310 K if a constant amount of heat is transferred to equal mass samples? Explain your answer.

nitrogen gas (specific heat = 1.04 J/g K)

iron solid (specific heat = 0.45 J/g K)

liquid water (specific heat = 4.18 J/g K)

calcium carbonate solid (specific heat = 0.82 J/g K)

A general equation is used to determine the amount of energy that is required to heat a sample of a substance, or, alternatively, how much heat is released as a sample of a substance cools:

q = mc∆ ∆T

m is the amount of the sample in the same unit as the heat capacity

c is the heat capacity or specific heat capacity of the substance

∆T is the change in temperature (a one degree Celsius change is equal to a one K change in temperature).

We can use the concept of heat capacity to approximate the heat of reaction for a chemical reaction occurring in water. The temperature change of water in which a

Figure 61. A coffee-cup calorimeter. See text for discussion of its use.

chemical reaction takes place is due to the heat released or absorbed by the chemical reaction: 169

∆H = –q = –[m · 4.184 J/g·K · ∆T] Notice that ∆H equals the negative of the quantity mc∆T; this is because a positive value for q means that the water absorbed energy as the reaction released energy (exothermic reactions have negative ∆Hs); and a negative value for q means that the water lost energy to the reaction (endothermic reactions have positive ∆Hs).

Practice 7.8

In a coffee-cup calorimeter, a 25.0 g sample of water underwent a temperature change of 37.5ºC as 2.5 g of a substance formed. If the molar mass of the substance is 55.0 grams, what is the value of ∆H per mol of substance?

We can also use the concept of heat capacity to determine the amount of heat absorbed (or released) by a substance in with another as the first transfers energy to the other: the amount of energy gained by one substance is equal to the amount of energy transferred from the other substance (but opposite in sign).

Imagine a whirlpool that is made with 400.0 kg of granite walls (c = 0.82 J/g · K). At night, as the water temperature falls below that of the rock bed, the rock bed begins to transfer heat to the cooler water – by morning the rocks have decreased their temperature by 22.5 K. Through what temperature change did the water undergo overnight assuming it lost no heat that was transferred to it? The whirlpool holds 350. kg water.

170

It would be relatively inconvenient to measure the enthalpy change for combustion reactions in a coffee-cup calorimeter, as the combustion of the substance is quite difficult in water; you can confirm this by attempting to light a match under water – be prepared to be disappointed. Bomb calorimeters are used to determine the enthalpy changes associated with combustion reactions. The process is essentially the same: measure the change in temperature of water to determine the enthalpy change; however, some of the heat is transferred to the calorimeter itself – this must be considered to determine the enthalpy change. A bomb calorimeter is shown in the figure at right. When a bomb calorimeter is used to determine ∆H for a chemical reaction, then we solve for heat of reaction using:

∆H = –∆T

“” is the heat capacity of the calorimeter – often called the calorimeter constant – which can be determined by running a reaction for which ∆H is known and measuring the temperature change the contents of the calorimeter undergoes:

= ∆H/∆ ∆T

Figure 62. In a bomb calorimeter, the reaction is held within a small chamber, and the water is heated indirectly as heat es through the chamber walls.

One uses the calorimeter constant in all calculations for a specific bomb calorimeter.

Use the following specific heat values to address the examples on this page.

Substance

Specific Heat (J/g-K)

Substance

Specific Heat (J/g-K)

N2(g)

1.04

H2O(l)

4.18

Al(s)

0.90

CH4(g)

2.20

Fe(s)

0.45

CO2(g)

0.84

Hg(l)

0.14

CaCO3(s)

0.82

171

Practice 7.9

Determine the amount of energy required to raise the temperature of 50 g of water from 301 K to 365 K. No phase change occurs.

Determine the amount of energy required to raise the temperature of 50.0 g of iron from 301 K to 600 K. No phase change occurs.

Determine the molar heat capacity of iron.

Determine the heat capacity (calorimeter constant) of a bomb calorimeter that undergoes a change in temperature of 2.54°C when a reaction releases 20.6 kJ to the calorimeter.

A bomb calorimeter’s temperature changes by 0.89°C when a 0.54 gram sample of metal is burned. The calorimeter’s heat capacity is 5.50 kJ/°C. What is the enthalpy change for the sample?

If the molar mass of the sample is 65.25 g/mol, what is ∆Hcomb for one mol of the metal? (∆Hcomb is often used when ∆Hrxn is that of a combustion)

172

Part 12: SPONTANEOUS PROCESSES

Processes that occur without any outside intervention are termed spontaneous processes. Processes that are spontaneous in one direction are not spontaneous in the reverse direction direction. Note that spontaneity does not address the speed of a process – only its occurrence relative ive to direction and outside intervention are addressed addressed. We might be tempted to assume that processes for which ∆H < 0 (i.e., exothermic) are spontaneous processes, while those for which ∆H > 0 (i.e., endothermic) are nonspontaneous. However, consider th the following:

ice melts when placed on a table at 39 39°C, and when placed in a freezer at –10°C, C, a sample of liquid water freezes

a piece of metal at 500°F F cools when placed in room temperature water, and the water warms

These processes are spontaneous, but the melting of ice is endothermic for the ice, and the water warming when hot iron is placed into it is also endothermic. Thus, we cannot assume that only processes that release heat are spontaneous processes. In fact, according acco to the first law, there is an attendant gain of energy in the surroundings when a system loses energy, and there is an attendant loss los of energy from the surroundings when a system gains energy. What, then, provides for a spontaneous process?

Figure 63. Spontaneity can clearly depend upon temperature, as illustrated here. For temperatures greater than 273 K, the process of melting of ice is spontaneous, while for temperatures less than 273 K, the spontaneous process is the freezing of liquid water.

Part 13: ENTROPY AND THE SECOND & THIRD LAWS OF THERMODYNAMICS

It is hopefully conceivable to you that two 1.0 L flasks – one holding 1.0 L of gas and the other empty – will spontaneously fill completely when a valve connecting them is opened. It is not conceivable to you that if the flasks were originally holding 0.5 5 L each that at some later time one flask would hold all of the 1.0 L of gas. Even though the process does not transfer heat or perform work, the movement of all of the gas into one flask from two is not a spontaneous process. Thus, there must be more to spontaneity than heat and work.

Figure 64. For the gases here, spontaneity is the movement of the gases from one full flask to equal distribu-

Consider for a moment the possible arrangements (spatial arrangement, motion

tion in both

and KE)) of one mol of water molecules at one moment under a specific set of

flasks. No work

thermodynamic conditions (P, V and T), and consider how this might ch change in

is done or heat

the next moment. Then, consider all of the other possible arrangements for these

transferred.

6.022 x 1023 molecules. Wow – this is an impossibly large number of arrangements! 173

For simplicity, consider the possible motions of just one molecule of water at a specific set of conditions, as shown in the figure at left. If we call the number of possible arrangements (which we cannot possibly count or even fathom) W, then we can define entropy as a function of W:

S = k ln W where k is Boltzmann’s constant: 1.38 x 10-23 J/K

Using this relationship, entropy is a measure of how many arrangements are possible for a given state. And, the entropy change, ∆S, is given by k ln Wfinal – k ln Winitial.

Now, then, we can simplify our Figure 65. For just one molecule of water, there are many manners of movement in both vibration and rotation for the molecule as a whole. And, these just represent a few of the possible spatial arrangements of one molecule – there are also consideration for position relative to other molecules, temperature differences and other thermodynamic characteristics.

discussion to this: the possible arrangements (positions and energies) of the particles of a sample of matter increase as the volume increases, the temperature increases or the number of

molecules increases. Thus, as the volume, temperature or number of atoms/molecules in a sample increases, the entropy increases, ∆S > 0. Entropy, S, is a measure of the possible arrangements in which a system may be found. We can now state the Second Law of Thermodynamics as: The total entropy of a closed system is continually increasing (and our closed system is the universe).

So, we now come back to our original discussion, which surrounded the additional aspect of what makes a process spontaneous: A spontaneous process is one that causes an increase in entropy for the universe; that is, a process for which the relationship between the surroundings’s entropy change and the system’s entropy change gives

∆Ssystem + ∆Ssurroundings = ∆Suniverse > 0 Be careful to note that a process can occur with ∆Ssystem< 0 and be spontaneous as long as ∆Ssurroundings is > 0 and larger in magnitude. Typically, the following processes are spontaneous, as they lead to an increase in entropy:

The formation of gases from liquids or solids

The formation of solutions or liquids from solids

An increase in molecules of product gas compared to

Figure 66. As an example of a spontaneous process, consider the dissolution of KCl in water. The possible arrangements of the ions and molecules are greater in the solution than in the solid phase. This, however, also illustrates why we do not use the term ”disorder” to define entropy – no one would suggest that the relatively orderly arrangement of the ions and water molecules is random.

molecules of reactant gas 174

Practice 7.10 Predict whether each of the following provides a negative entropy change or positive entropy change at constant temperature. You are advised to write reactions for the processes indicating the state when considering entropy changes. water melting silver ions and chloride ions reacting to form AgCl(s) iron and oxygen forming iron(III) oxide nitrogen gas reacts with oxygen gas to form nitrogen(II) oxide gas.

Which in each pair has greater entropy? Why? 1 mol NaCl(s) or 1 mol HCl(g) at 298 K

2 mol HCl(g) or 1 mol HCl(g) at 298 K

1 mol HCl(g) or 1 mol Ar(g) at 298 K

1 mol HCl(g) in a 2 L volume or in a 1 L volume

1 mol HCl(g) at 400 K or at 700 K

The change in entropy is given by

∆S = Sfinal – Sinitial When considering entropy changes, it is important to note all of the changes in order that occur as the process proceeds. For example, as we saw above, although the dissolution of potassium chloride in water increases the disorder of the solid,

Figure 67. The entropy of a substance increases as it changes phase from solid to liquid to gas.

it increases the ordering of the water molecules as they arrange themselves around the ions of K+ and Cl–. However, the change in entropy for the separation of KCl is so large that it exceeds the decrease in entropy in the ordering of water molecules. We can generally make predictions about the favorability of chemical reactions using the following summaries: •

If energy and matter become more dispersed in a reaction, then the reaction is definitely favored to occur spontaneously.

•

If only energy or matter is dispersed, then quantitative data is needed to determine the favorability of spontaneity.

•

If neither energy nor matter is dispersed, then the process is reactant-favored and will not occur spontaneously. 175

When comparing the same or similar substances, the entropy of gases is much greater than the entropy of their liquids. Solids have the lowest values of entropy; the Third Law of Thermodynamics sets the entropy of a pure crystalline solid at zero K as zero entropy. Standard molar entropy values are often provided for substances in various states in chemistry textbooks. Your textbook includes many in the Appendix. Standard molar entropy values, denoted S°, are the values of the entropy for one mol of a substance at 298 K. The units of entropy are expressed as J/K mol. Again, the standard molar entropy of any pure crystalline solid is taken as zero at 0 K.

Making some assumptions that we will not discuss, we can calculate the entropy change for the surroundings at constant temperature, which can be used to determine ∆Suniverse, if needed, according to the equation on Page 174:

―ࢗ = ∆ࡿ‫ܛ܏ܖܑ܌ܖܝܗܚܚܝܛ‬ ࢀ We concern ourselves now with the determination of the change in entropy for two processes: the expansion or contraction of gas, and chemical reactions. The entropy change associated with the expansion or contraction of a gas can be determined using:

∆S = nR ln(V2/V1) We can also quite easily calculate the entropy changes associated with chemical reactions:

∆ࡿ°࢘࢞࢔ = Σ ࢔ࡿ° ሺ‫ܛܜ܋ܝ܌ܗܚܘ‬ሻ − Σ ࢓ࡿ° ሺ‫ܛܜܖ܉ܜ܋܉܍ܚ‬ሻ

n and m are the stoichiometric coefficients of the products and reactants in the balanced chemical equation

S° is the standard molar entropy of the substances

Be sure to note that molar entropies are given in joules per mol rather than kilojoules!

Practice 7.11 Calculate ∆Ssys and ∆Ssurr associated with the vaporization of one mol of benzene, C6H6, at its boiling point 80.1°C. ∆Hvap for benzene is 30.9 kJ/mol.

The normal boiling point of ethanol is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. Calculate ∆Ssys and ∆Ssurr for the condensation of 68.3 g of ethanol gas at 1 atm at 78.3°C.

176

Practice 7.12 Predict and then calculate ∆S° for the synthesis of ammonia from its gaseous elements at 298 K. Calculate ∆Ssurr for the reaction, as well. Then, determine whether the process is spontaneous at 298 K (i.e., is ∆Suniv >0?). Which entropy change drives the spontaneity?

Al2O3(s) + 3H2(g) → 2 Al(s) + 3H2O(g) Calculate the value of ∆H° and ∆S°universe for the reaction above. Use this information to discuss the spontaneity of the reaction at 298 K.

Part 14: GIBBS FREE ENERGY

The spontaneity of a reaction is related to two thermodynamic concepts: the change in enthalpy (∆H) and the change in entropy (∆S). Josiah Willard Gibbs proposed the following relationship between these quantities, whose difference he termed the free energy:

∆Gibbs free energy, ∆G = change in enthalpy – (temperature x change in entropy): ∆G = ∆H – T∆ ∆Ssystem The change in Gibbs free energy, ∆G, tells us the following about the spontaneity of a reaction: 177

If ∆G is negative, then the reaction is spontaneous in the forward direction

If ∆G is positive, then the reaction is spontaneous in the reverse direction (although work can be added to make it spontaneous in the forward direction)

If ∆G is zero, then the reaction is at equilibrium

In any spontaneous process, free energy decreases until it reaches a minimum value, at which time a state of equilibrium exists, which allows us to make relationships between Q, Kc and G, which we shall do in depth shortly. At least for Figure 68. When Q < K, you will recall that the reaction is moving in the “forward” direction as written. As the reaction progresses the free energy is decreasing until it reaches equilibrium, at which time the free energy is zero. This is the case for all processes. A similar situation is observed if we approach the equilibrium condition from the “product” side while Q > K.

now, see Figure 68 for a relationship between Q and free energy.

We can also calculate the standard free energy change of formation, whose equation is very similar to the standard

enthalpy of formation equation we used with enthalpy and entropy earlier. The standard free energy change is the change in the free energy of a system at reference (standard) states. The standard free energy of formation for an element is zero, analogous to the enthalpy of formation for a pure element. ∆ࡳ°࢘࢞࢔ = Σ ࢔ࡳ°ࢌ ሺ‫ܛܜ܋ܝ܌ܗܚܘ‬ሻ − Σ ࢓ࡳ°ࢌ ሺ‫ܛܜܖ܉ܜ܋܉܍ܚ‬ሻ The change in standard Gibbs free energy, ∆G°, tells us the following about the spontaneity of a reaction:

If ∆G° is negative, then the reaction is spontaneous in the forward direction

If ∆G° is positive, then the reaction is spontaneous in the reverse direction (although work can be added to make it spontaneous in the forward direction)

Why is free energy “free?” Without much discussion, ∆G is the maximum useful work that can done by a system on the surroundings. ∆G° will tell us whether a particular mixture of reactants and products at standard conditions will react spontaneously, and it will tell us the amount of work that must be added to a system in order to make it occur spontaneously if it will otherwise not at standard conditions.

178

Practice 7.13 Determine ∆G° for the reaction of nitrogen gas and hydrogen gas to produce ammonia gas at 298 K. For a container charged with hydrogen and nitrogen at 298 K, is the process of the formation of NH3(g) spontaneous? Justify your response.

Calculate the value of ∆H° for the combustion of C3H8(g) (the water product is liquid) at 298 K.

Predict whether the magnitude of the free energy of formation, ∆G°, is more negative or less negative than ∆H°. Justify your choice.

Calculate the value of the free energy of formation.

179

Using the values of ∆G° and ∆H°, determine the value of ∆S°. Confirm this with a standard molar entropy calculation.

Use the value of the free energy change to determine whether the reaction between P4(g) and Cl2(g) at 298 K is spontaneous.

Consider for a moment the melting of ice water: the sign on ∆H is positive (∆H > 0) and the sign on ∆S is positive (∆S > 0). So, the process is spontaneous (∆G < 0) as long as the magnitude of the entropy term (-T∆S) is greater than the magnitude of the enthalpy term. When the have equal magnitudes then a system is at equilibrium and the value of ∆G = 0. We can make some predictions about the spontaneity of processes at certain temperatures based upon the signs of ∆H (the enthalpy term) and ∆S (the entropy term):

∆H

∆S

–T∆S

∆G

Reaction character

Example

180

Practice 7.14

The Haber process for ammonia production is a common industrial process. Compared to 298 K, predict how the free energy of formation changes if the temperature is raised to 773 K. Then determine the value of the free energy of formation for the reaction at both temperatures. Does the spontaneity change? Which term determines the spontaneity – the enthalpy term or the entropy term?

Without too much fanfare – but not to minimize their utility – we conclude our free energy discussions with several applications of the Gibbs free energy.

At nonstandard conditions (that is, the concentrations of solids and liquids is not one, the pressure of gases is not 1 atm and solutions do not have molarities of 1M):

∆ࡳ = ∆ࡳ° + ࡾࢀ ‫ࡽ ܖܔ‬

where Q is the reaction quotient with which you are familiar, and R must be expressed in of energy: 8.314 J/mol K Because the equilibrium constant K and Q are equal at equilibrium and ∆G = 0 at equilibrium, then we can derive the following useful expression:

∆ࡳ° = – ࡾࢀ ‫ࡷ ܖܔ‬ Solving the above equation for K gives:

ࡷ = ࢋି∆ࡳ /ࡾࢀ °

181

Practice 7.15

Consider the boiling of CCl4.

Write the equation that represents the boiling of CCl4(l) at 1 atm at its boiling point.

What is the value of ∆G° for the boiling of the equilibrium shown above? Do not perform any calculations, but justify your answer.

Determine the boiling point of CCl4 using thermodynamic data.

Calculate the free-energy change at 298 K of a mixture of nitrogen gas and hydrogen gas to produce ammonia gas that contains 1.0 atm N2, 3.0 atm H2 and 0.50 atm NH3. Discuss briefly whether the answer is consistent with a Le Châtelier’s principle approach.

Determine the value of the equilibrium constant for the production of ammonia from its elements at 298 K.

182

Part 15: THE ELECTROCHEMISTRY-REDOX-THERMOCHEMISTRY-EQUILIBRIUM RELATIONSHIP

We have seen that redox reactions can occur spontaneously in electrochemical cells or otherwise. It should not, then, come as a surprise that there is a relationship between the occurrence of redox reactions and the Gibbs free energy. Using the number of electrons transferred in a reaction, n, Faraday’s constant, F, and the emf of the reaction under standard conditions (1.0-molar solutions at 298 K at1 atm), E°, we get

∆ࡳ° = – ࢔Fࡱ° Faraday’s constant is 96 485 C mol-1, or the electrical charge on one mol of electrons (an alternative representation is 96 485 J/V mol, which expresses the constant in of energy and volts rather than electrical charge). And, given the relationship between ∆G°and K, we can see

࢔ࡱ° ∆ࡳ = – ࡾࢀ ‫࢔ – = ࡷ ܖܔ‬Fࡱ and ‫= ࡷ ܏ܗܔ‬ ૙. ૙૞ૢ૛ °

°

which gives us a powerful tool to relate the spontaneity of a reaction in of emf and in of Gibbs free energy, and make the relationship between cell emf and the equilibrium constant. Notice that when the cell emf = 0, the reaction is at equilibrium, and, consistent with the information in Part 14, ∆G = 0. It is at this time that we would indicate that a battery or voltaic cell is “dead.” Recall the relationship from earlier that ∆G = ∆G° + RT ln Q. Substituting and solving for E gives

ࡱࢉࢋ࢒࢒ = ࡱ°ࢉࢋ࢒࢒ –

ࡾࢀ ‫ࡽ ܖܔ‬ ࢔ࡲ

which is the Nernst equation, a valuable equation that allows us to determine the value of E at various concentrations. The Nernst equation is most often expressed in of base 10 log rather than natural log, which gives this alternative form:

ࡱࢉࢋ࢒࢒ = ࡱ°ࢉࢋ࢒࢒ –

૙. ૙૞ૢ૛ ‫ࡽ ܏ܗܔ‬ ࢔

We will see in the exercises and free-response practice sets that as the concentrations of reactants increases relative to the concentration of products, the cell emf increases, and the converse is true when the products’ concentrations is increased relative to the reactants’ concentrations. At any rate, a Le Châtelier response is expected – the Nernst equation simply provides a quantitative approach to the change.

183

Practice 7.16

Calculate the emf at 298 K generated by the cell constructed of potassium dichromate and potassium iodide solutions in which inert platinum electrodes are immersed when the concentrations of the species are [Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I–] = 1.0 M, and [Cr3+] = 1.0 × 10-5 M. First, however, use the concentrations to predict whether the value of the emf is less than or greater than that in a standard cell.

The reaction that occurs is shown here: Cr2O72-(aq) + H+(aq) + I–(aq) → Cr3+(aq) + I2(s) + H2O(l). Balance the equation first.

184

CONCENTRATION CELLS For the voltaic cells that we have previously discussed, the half half-cell solutions and their corresponding electrodes have been of different substances. For example, we have often used copper/copper(II) solutions at the anode while using silver/silver(I) solutions at the cathode. In a concentration cell, however, the electrodes are of the same metal, and the solutions are the same, also. A concentration cell using the reduction and oxidation of nickel(II) is shown at left. In the cell, the [Ni2+] = 1.00 × 10-3 M in the left half-cell and the [Ni2+] in the right half-cell is 1.00 M.. A calculation of E°cell gives zero, as E°reduction – E°oxidation equals zero because the reduction potentials in each cell are the same (-0.28 0.28 V). However, the cell operates spontaneously. Why? According to the Nernst equation:

ࡱࢉࢋ࢒࢒ = ࡱ°ࢉࢋ࢒࢒ –

. cd1 ?@ ^

Figure 69. A concentration cell is a voltaic cell in which both cells contain the same electrode solutions. Clearly, you should be able to deduce that the higher concentration half-cell cell decreases in concentration while the lower concentration half-cell half increases in ion concentration.

as long as the value of Q < 1, then a concentration cell will exhibit a voltage while working toward equilibrium; and, the reaction that occurs is the transfer of electrons from the less concentrated half half-cell to the more concentrated half-cell. cell. In this case, the reaction that occurs is Ni2+(conc) → Ni2+(dilute). Recall, as long as Q > 1 or Q < 1, then a reaction is not at equilibrium and will move toward equilibrium (from (fro the reactants when Q < 1 and from the products when Q > 1). When Q = 1, the cell will cease operation, and we can see from earlier discussions that E and ∆G are zero. In the human body, concentration differences result in electrical discharges that cause nerve impulses to travel and that cause cau the heart to beat.

Practice 7.17 Determine the value of the concentration cell depicted in the figure above. The standard reduction potential of the Ni2+ + 2e- → Ni(s) reaction is –0.28 V.

A concentration cell is constructed with Zn2+(aq)/Zn(s) half cells. One half-cell has [Zn2+] = 1.35 M and the other has [Zn2+] = 3.75 × 10-4 M.

Identify the anode reaction and the cathode reaction.

What is the cell emf?

185

ELECTROLYSIS

Electrolytic cells are quite different from the voltaic cells we have discussed. In an electrolytic cell, the electrodes are placed in the same solution, and the electrodes are not made of the solution’s metal (unless electroplating is desired); rather, they th are inert and do not participate in the reaction;; graphite and platinum are common.

In the he example shown in the figure at left, the molten salt NaCl(l) is undergoing electrolysis to yield chlorine gas and sodium metal. During electrolysis of molten salts (as long as the voltage source is oriented correctly) the result is elemental metal and elemental e nonmetal. This is not the same process that occurs during electrolysis of aqueous solutions. The process of electrolysis is Figure 70. During electrolysis with inert electrodes, we might be using a molten salt or an aqueous solution. The text describes the chemical reactions occurring.

important industry, as the active metals do not occur in elemental form – instead, they are found as salts that can be decomposed deco in this manner.

For aqueous solutions,, the electrolysis products will depend upon the ease with which the species in aqueous solution can undergo oxidation and reduction compared to the ease with which water undergoes oxidation or reduction: •

At the cathode, active metals will not undergo reduction in aqueous solution and will instead remain in ion form as water undergoes reduction to H2(g) and OH-(aq). Generally, the metals that will not undergo reduction are sodium, lithium, potassium, magnesium, m, aluminum and the other representative active metals. Essentially all transition metals of interest will undergo reduction in an electrolytic aqueous solution.

•

At the anode, the active nonmetal ions Cl–, Br–, and I– will undergo oxidation to Cl2, Br2 and I2, respectively. Other nonmetals and polyatomic ions (which are typically good oxidizers) will not undergo oxidation – in these cases, the products at the anode are the oxidation products of water, O2(g) and H+(aq).

Practice 7.18 Determine the products roducts of the following electrolysis reactions – inert electrodes are immersed in all reactant solutions and liquids. the electrolysis of molten magnesium iodide

the electrolysis of aqueous magnesium iodide

the electrolysis of aqueous copper(II copper(II) sulfate

the electrolysis of aqueous copper(II) bromide 186

ELECTROLYSIS USING NON-INERT ELECTRODES – ELECTROPLATING

In contrast to electrolysis using inert electrodes, the process of electroplating is a unique case of electrolysis. In the process, a solution containing the metal that will plate-out has into it two electrodes immersed: one the metal of the solution and the other the object to be plated. When a voltage source is applied, the metal at the cathode is reduced (like all cathode reactions), while the ions in the solution are oxidized at the anode (like all anode reactions). The reduced atoms then plate-out onto the cathode electrode, which can be a variety of materials.

In metallurgy, the purification of metals is accomplished by using impure samples of metals as the anode and pure samples of the metal as the cathode. Then, the two samples are immersed in a solution of the metal, and the metal plates-out onto the pure metal electrode. As with many chemical discoveries, the process of electrolytic purification certainly impacts the world in which we live – read more about the Hall-Héroult process by following the link on the website.

Figure 71. In electroplating, an electrode of the metal that is to plate is immersed in its solution. A second electrode is immersed, onto which the metal will plate. (Here, the nickel is plating onto the steel cathode.)

Note that in both electroplating processes above – as in all cell reactions – the mass of the anode decreases and the mass of the cathode increases. Here, however, owing to the requirement for the applied voltage, the process can be quite interesting if the voltage source is attached backwards!

ELECTROLYSIS AND CORROSION

The corrosion of metals is an interesting electrochemical process. In the process of corrosion, oxygen acts as a strong oxidizer for metal atoms that are exposed to it. For iron (Ered = –0.44 V) the process of oxidation by oxygen (Ered = 1.23 V) is easily accomplished as one portion of a metal sample acts as the anode and another portion of the metal acts as a cathode – i.e., the electrons transfer through the metal itself; this is why metals corrode and insulators do not. Later, the oxidized ions react with additional oxygen and water to form iron(III) oxide – rust – on the surface of the metal. Clearly, as the iron is removed at the anode, the metal disintegrates.

Figure 72. Oxygen oxidizes many metals, which leads to corrosion.

187

One manner of preventing corrosion is to treat metals that easily corrode with a metal that undergoes oxidation even more easily – i.e., one with a less positive reduction potential. This is termed cathodic protection, and it involves coating a metal – usually iron – with the more easily oxidized metal. Then, the more easily oxidized metal undergoes oxidation rather than the metal from which a protected piece of equipment is made. The metal coating is called a sacrificial anode, as the anode is preferentially disintegrated over the more valuable metal. The specific process of cathodic protection of iron by zinc is called galvanizing. Within your hot water tank at home there is a sacrificial anode present that prevents the corrosion of several important parts within the tank – when any sacrificial anode is consumed completely, then any metal that can be oxidized becomes an anode if other conditions are met (e.g., electrons can transfer, etc.)

Figure 73. Zinc will be preferentially oxidized here. See text.

QUANTITATIVE ELECTROLYSIS One aspect of electrolysis is to determine the amount of substance that will plate-out as current is ed through an electrolytic cell. This is possible by considering electrons the reactants in an electrolysis reaction and noting that one mol of electrons has a charge of 96 500 coulomb (C), which is one Faraday, F. One coulomb is the charge that es a point in a circuit in one second when the current is one ampere (A); thus, the total charge that es a point is the amperage, I, times the time in seconds during which the current is applied, and the amperage is easily obtained by rearrangement: ࡯ = ‫ܛ܌ܖܗ܋܍ܛ • ܍܏܉ܚ܍ܘܕ܉‬ ࡵ=

ࢗ ࢚

Of interest to us is to use the relationship between current, charge and time to determine the masses of reactants or products consumed or produced, respectively, in an electrolytic cell.

Practice 7.19 Calculate the number of grams of aluminum produced in 1.00 h when molten AlCl3 is electrolyzed under a current of 10.0 A.

The solution involves several steps: determine the reaction; determine the charge ed into the cell; determine the mol electrons transferred; relate mol electrons to mol Al; convert mol Al to mass Al

188


Principles of Chemical Bonding Students will be able to:

Describe the nature of ionic bonding; predict the occurrence of ionic bonding; perform qualitative predictions of the lattice energy based on ionic size and ionic charge

Describe the nature of covalent bond bonding; predict the occurrence of covalent bonding

Predict and characterize polar bonds based on electronegativity differences

Draw Lewis structures and calculate formal charge; predict and justify exceptions to the “octet rule”

Predict the occurrence of resonance, and justify the concept in of structural observations

Determine the electronic and molecular geometries of molecules

Use electronic and molecular geometries to explain substances’ behaviors

Use electron configurations to explain observed m molecular properties

Use the concept of hybrid orbitals to construct viable explanations of observed patterns of bonding

189

Part 1: IONIC BONDING

Metals tend to have relatively low ionization energies, while nonmetals tend to have relatively large electron affinities. Thus, metals tend to lose electrons readily, while nonmetals tend to gain electrons readily. When metals lose electrons, a positive cation is formed, and the formation of a negative anion results when a nonmetal gains electrons. The attraction between the ions – as seen in Figure 74 – results in an ionic bond, which is a bond between charged species. The species can be nonmetals, metals or polyatomic ions. The force behind an ionic bond is the electrostatic attraction between the charged Figure 74. (a) The formation of an ionic bond occurs when a metal loses

species; thus, only species of opposite charge will come together to form an ionic bond.

electrons, a nonmetal gains electrons, and (b) ions are attracted by electrostatic attraction. The bond is relatively weak compared to a covalent bond, and it can often be disrupted by attractions between ions

In Figure 74a, sodium loses a single valence electron, which is transferred to an atom of chlorine. This gives

and water molecules in aqueous solution.

both ions a full valence shell, or octet. The transfer of electrons also alters the charges exhibited by the ions: sodium is now charged +1, while chlorine is now charged –1. These opposing charges cause electrostatic attraction between the ions and lead to the formation of the ionic bond seen in the figure. The relative numbers of ions involved in an ionic bond depend upon the charges of the ions, which we will see soon.

The reaction equation for the above figure is shown here: Na(s) + ½ Cl2(g) ⇋ NaCl(s)

∆Hrxn is about –411 kJ/mol

The formation of an ionic bond involves several energy changes, including those already mentioned: the ionization energy of the metal and the electron affinity of the nonmetal. But, there are more energy changes associated with the process, as well. Let’s examine the formation of NaCl(s) for a moment to see this in detail.

The process of ionic bond formation can be pieced together using a Born-Haber cycle – an application of Hess’s law, really – named after the scientists Max Born and Fritz Haber. We look at each energy change that must occur in order to effect the final attraction of the gaseous ions of the metal and the nonmetal using the formation of one mol of the compound of interest. For the formation of NaCl several steps occur; see Figure 75 on Page 191.

190

BORN-HABER CYCLES

1.

Vaporization of the metal atoms

2.

•

Na(s) → Na(g)

•

∆Hfusion + ∆Hvaporization

•

endothermic

Separation of bonds in the nonmetal

3.

4.

•

Cl2(g) → 2Cl(g)

•

bond enthalpy

•

endothermic

Ionization of the metal atom •

Na(g) → Na+(g)

•

ionization energy

•

endothermic

Ionization of the nonmetal atom

5.

•

Cl(g) → Cl–(g)

•

electron affinity

•

exothermic

Bond formation •

Na+(g) + Cl–(g) → NaCl(s)

•

lattice energy

•

exothermic

Figure 75. A Born-Haber cycle. See text for details. (Other steps are possible, depending upon the states of the species entering the cycle, and all steps are not required. For example, the vaporization of chlorine is required during the cycle, but this would not be the case for sulfur when forming Na2S.)

The sum of the steps gives the overall enthalpy change associated with the formation of an ionic bond; in this case, the formation of NaCl. We note, however, that if we only look at the steps involved up to the attraction between the ions – i.e., the steps involving electron transfer and phase change – the process is large endothermic (above, steps 1 through 4 require 376 kJ/mol). Thus, the attraction between the ions must be very large exothermic in order for the overall process to be exothermic. Indeed, this is the case. The energy change associated with the attraction between ions is called the lattice energy. The lattice energy is the amount of energy required to separate a mol of an ionic compound into its gaseous state – it is equal in magnitude but opposite in sign to the energy released when ions in gas phase form an ionic bond. For NaCl, we see that the lattice energy is +788 kJ/mol. The lattice energy is a function of two variables: the distance between the centers of the ions (thus, the size of the Figure 76. The lattice energy is a function of the charges on the ions and their radii. Because the size difference is so small for ions, the general trend in lattice energy is due to the charges on the ions.

ions), and the charges on the ions. This relationship is given by a form of Coulomb’s law:

‫࢑ = ܡ܏ܚ܍ܖ܍ ܍܋ܑܜܜ܉ܔ‬

ࡽ૚ ࡽ૛ ࢘

where k is the constant 8.99 J-m / C2, r is the distance between the ions’ centers and Q is the charges on the ions. 191

Because the distance factor, d, does not vary much between ions, the major component that drives the lattice energy is the charges on the ions, Q1 and Q2; although for ions with the same charge we do consider the distance factor.

When atoms lose electrons, the negative-negative repulsions that existed between outer electrons are lost and the effective nuclear charge on the remaining electrons increases. Thus, when an atom loses electrons to form a cation, the cation is smaller than the atom from which it formed.

When atoms gain electrons, the negative-negative repulsions are increased between outer electrons. Additionally, the effective nuclear charge on the outer shell of electrons is generally diminished. Thus, when an atom gains electrons to form an anion, the anion is larger than the atom from which it formed. Figure 77. The sizes of ions compared to parent atoms. Notice that cations are always smaller than their parent atoms, while anions are larger than parent atoms.

The preceding statements lead us to conclude that the ions of metals are smaller than their parent atoms, and the ions of nonmetals are larger than their parent atoms. From this – for common ions – we can conclude that the size of ions increases down the periodic table, the size of ions decreases left-to-right for metals and the size of ions slightly decreases left-to-right for nonmetals. Note that a large increase is seen in ionic radius when crossing the metal-nonmetal boundary.

In a series of isoelectronic species, the species with the largest atomic number will have the smallest radius due to the fact that it has the largest effective nuclear charge. An isoelectronic series is one in which atoms and ions in series possess the same number of electrons. This should be apparent by looking at the isoelectronic series of ions Na+, Mg2+, Al3+, O2- and F-. All of these ions possess 10 electrons, and the nuclear charges are shown here: Na (11+), Mg (12+), Al (13+), O (8+) and F (9+). You can hopefully see that the greatest nuclear charge on aluminum gives it the smallest ionic radius, while the lower charge on oxygen provides it the largest radius.

192

Practice 8.1

For which of the following pairs of ions would the attraction be greatest, and for which would the attraction be the smallest? Explain your choices.

NaF, CsI or CaO

AgCl, CuO and CrN?

For which of the pairs of ions would the lattice energy be greatest, and for which would the lattice energy be the smallest? Explain your choice.

LiBr, KF, NaCl

Outline a Born-Haber cycle for the formation of MgO from its elements in their standard states. Do not attempt any calculations, but include the sign of the enthalpy change for each step.

193

Which of the following atoms will increase in size upon ionization, and which will decrease in size? Rb

Cl

N

Ca

Zn

S

1+ Identify 5 ions that are isoelectronic with P3-:

Including N3-, arrange the ions you selected above in order of decreasing radius:

Part 2: COVALENT BONDING

We saw in Part 1 that ionic bonding involves the transfer of electrons. This is possible because the metals have relatively low ionization energies, while the nonmetals have relatively large electron affinities. However, what about the case when all of the atoms involved in bonding have high ionization energies? It seems plausible that none of the atoms in this case will easily release electrons to form ions. Indeed, this is the case when nonmetals bond to other nonmetals. Because the nonmetals have relatively high electronegativity values, high ionization energies and fairly large negative electron affinities, they share electrons instead of transfer electrons when involved in bonding with one another.

Look at the Figure 78. The figure represents the attractions found between the nuclei and electrons clouds of adjacent atoms, the repulsions found between the nuclei of adjacent atoms, and the repulsions found between the electron clouds of adjacent atoms. The balance between attractions and repulsions may lead to a covalent bond, which is a bond in which the atoms share electrons to achieve a full octet or valence shell (with hydrogen needing only to complete a duet). This bonding is characteristic of the nonmetals when bonded to other nonmetals. The covalent bond is formed when the attractions are significant enough that they are not overcome by the repulsions under the conditions of the bond, as shown in (b) at right.

Temporarily, to determine the number of covalent bonds, we look at the valence

Figure 78. Attractions exist between electrons and protons, while repulsions are present between adjacent nuclei and adjacent electron clouds. (b) At a certain distance, the attractions are such that they overcome the repulsions and a covalent bond is formed.

configuration of the atoms of interest. You know that oxygen has a valence configuration of s2p4, which means that it needs only 2 electrons to fill its valence shell. Hydrogen has a valence shell of s1, which requires only the addition of a single electron to complete its shell. However, neither atom is likely to give up electrons to form positively-charged ions because their ionization energies are too high. Furthermore, they 194

both have relatively high electronegativity values, which means that they will strongly attract the electrons of atoms to which they are bonded. Notice that by sharing two electrons – one from each hydrogen atom – the oxygen atom in water achieves the “appearance” of a full valence shell: s2p4 + 2 e- = s2p6. Similarly, each hydrogen shares a valence electron from oxygen and achieves a full valence shell of s1 + 1 e- = s2. Generally, a nonmetal will form as many bonds with other nonmetals as required to form an octet, or a duet in the case of hydrogen. Figure 79 shows the number of bonds predicted for fluorine, oxygen, nitrogen and carbon, respectively. Relate this to their valence structure.

Figure 79. Notice that the number of covalent bonds is equal to the number of "empty" valence positions on an atom. You will hopefully recall the basic molecular geometries of tetrahedral, pyramidal, bent and linear.

REVIEW OF TYPICAL GEOMETRY AND BOND NUMBERS Complete the following table for nonmetals in typical covalently-bonded compounds – many more will be added!

Group

Valence Structure

Number of Covalent Bonds

Typical Geometry

Bonding Pairs of Electrons

Lone Pairs of Electrons

Representative Formula

15 16 17 18

195

POLAR BONDS AND POLAR MOLECULES When the atoms in a covalent bond have unequal electronegativity values, the atoms form a polar covalent bond. In a polar covalent bond the electrons shared in the bond are not shared equally, as represented by the figure at right, which attempts to represent the greatest electron density by the relative size and shading of the ends of the molecules. A nonpolar bond – equal distribution of electron density – is seen in H2, while a polar bond – unequal distribution of electron density – is represented by HCl. A polar bond results from the unequal sharing of electrons between the atoms in a bond. When the electronegativity difference (∆EN) is large enough, the electrons shift toward the more electronegative atom often enough that a measurable charge density is located on the more electronegative atom. (Note that when ∆EN is great enough – as in the difference between the EN values of sodium and chlorine – the electrons do not tend to shift between the atoms but are rather transferred to the more electronegative element; this results in an ionic bond.) No bond is truly 100% nonpolar, and no bond is truly 100% ionic, which leaves us to conclude that the continuum of bonding types is defined by arbitrary ∆EN; indeed, this is the case. You might even think of an ionic bond as the most polar covalent bond you can imagine, or a polar covalent bond as “not as ionic” as a true ionic bond. This provides a range of polarity to bonds based upon ∆EN that have become widely-accepted “breaking points:” •

Nonpolar: the electrons in a bond are shared equally – electronegativity difference between atoms is zero or very small, on the order of 0.0 to 0.5

•

Polar covalent: the electrons are often pulled toward one atom more than the other in the bond – electronegativity difference between atoms is intermediate, on the order of greater than 0.5 to 2.0

Ionic: the complete transfer of electrons has occurred – electronegativity

Figure 80. In the top figure representing H2, the electrons in the bond are shared equally, as the electronegativity difference between the atoms in the bond is zero. In the lower figure, the electronegativity difference is such that the electrons are shared unequally, which results in a polar bond in which the electrons are “pulled” toward the more electronegative chlorine atom.

difference between the atoms is large, on the order of greater than 2.0

There are several conventions used to indicate the polar nature of bonds when writing structural formulas for molecules, some of which are shown here. In the top structure at left, the partial negative charge is indicated by the Greek letter delta followed by a negative sign, while the partial positive end is indicated similarly with a positive sign. The partial character of this notation arises because there is no real charge on the molecule – only an induced charge as the electrons move about. In the bottom notation, the direction of the electron density is indicated by the arrowhead, while the partial positive charge is suggested by the “+” sign on the tail of the arrow. This is probably much less common to use than the delta notation of the previous example. Figure 81. We can represent the polar ends of a molecule with the Greek delta or with arrows.

Polar molecules arise when the overall distribution of electron density is asymmetrical (unequally distributed) around the central atom in a molecule, which occurs when the character of the atoms (electronegativity) causes the electron density to shift toward one region of a molecule. Polar molecules should be easily recognizable as those molecules where electron density is asymmetrical or unequally distributed. 196

MULTIPLE BONDS Several atoms may bond to another with more than a single pair of electrons. For example, the bonding in the nitrogen molecule is a triple bond, which can be represented as N≡ N. This bond involves the sharing of six electrons between each nitrogen atom, and a lone pair of electrons is present on each atom. Carbon and nitrogen will form triple bonds, and carbon, nitrogen, oxygen, silicon and sulfur will form double bonds. Make a few notes about multiple bonds: •

A multiple bond is not directly proportional in strength to a single bond. For example, in O=C=O the double bonds are not twice as strong single bonds. Rather, one of the bonds is stronger than a single bond, and the other bond is only slightly stronger than the first.

•

A multiple bond decreases the distance between atoms involved in the bond. For example, the distance between N―N is approximately 1.3 times greater than the distance between nitrogen atoms when they are bonded as N≡N. However, the tripling or doubling of bonds does not cut the distance to one-third or one-half the single bond length.

Practice 8.2 Which of the following bonds is most polar? (Recall that electronegativity increases as the radii of atoms decrease.) B―Cl or C―Cl?

P―F or P―Cl?

Select any four pairs of atoms that form polar covalent bonds and sketch them with the delta notation described above.

Sketch the molecule ethanol and the molecule 1,2-dichloropentane. Indicate the polarity of the molecule using the delta notation.

Use multiple bonds to predict the bonding in a molecule containing one each of carbon, nitrogen, oxygen and hydrogen. Are isomers of your sketch possible within the context of typical bonding patterns? Explain.

197

Part 3: LEWIS STRUCTURES

Lewis structures are diagrams with which you are already familiar. They are diagrams drawn to represent the predicted bonding of atoms in a compound and the location of the electrons on the atoms. They will be very useful in understanding the details of molecular geometry and the behavior of molecular compounds.

A Lewis structure is a figure that represents the bonding of atoms in a compound and shows the arrangement of electrons on a molecule or ion.

DRAWING A LEWIS STRUCTURE

In order to successfully draw Lewis structures you must be able to arrange the shared electrons in a molecule between atoms in order to provide each atom an octet (or a duet for hydrogen). Then, apply lone pairs of electrons on molecules so that all atoms have an octet (or a duet for hydrogen).

While there are many ways to plan and organize the process for drawing Lewis structures, most structures can easily be drawn based upon the bonding patterns of the nonmetals with which you are already familiar.

i.

Determine the number of available valence electrons. Subtract electrons for cations, and add electrons for anions.

ii.

Determine the central atom on the molecule. When given a formula, it is often the case that the central atom is given first. Otherwise, choose the atom that forms the greatest number of bonds as a starting point.

iii.

Form bonds between the atoms in order to achieve complete bonding of all atoms.

iv.

Place electrons on all outlying atoms in order to form octets for these atoms.

v.

Place any additional electrons around the central atom, even if doing so results in more than an octet.

vi.

If the central atom still does not achieve an octet, and it is expected to do so, move lone pairs from the outlying atoms as bond(s) between the central atom and the outlying atoms. In doing so, be sure to evaluate if other lone pairs will yield equivalent structures; if this is the case, then resonance describes the molecule or ion; otherwise, use the obtained structure by itself.

Without much fanfare, we shall begin to draw Lewis structures and discuss their utility.

198

Practice 8.3 A Lewis structure for a simple compound: PCl3

We first total the valence electrons from each atom appearing in the formula.

The second step is to arrange the atoms in a reasonable manner to suggest their orientation to one another. It is often wise to start with the atom that makes the greatest number of bonds as the central atom.

Using two electrons each, place bonds between the atoms so that all atoms have the predicted number of bonds. Use double and triple bonds where necessary. Here, we have used 6 of the available 26 e-.

Place electrons around the outside atoms to complete their octets. This uses a total of 18 e-, leaving 26 – 6 – 18 = 2 e- remaining.

Finally, place the remaining electrons on the central atom, even if this results in more than an octet for the atom. Count the electrons to ensure you have used them all. Here, we have used 26 e-.

199

Practice 8.4 A Lewis structure for an ion: SO42-

We first total the valence electrons from each atom appearing in the formula. Add electrons for anions, subtract electrons for cations.

The second step is to arrange the atoms in a reasonable manner to suggest their orientation to one another. It is often wise to start with the atom that makes the greatest number of bonds as the central atom.

Using two electrons each, place bonds between the atoms so that all atoms have the predicted number of bonds. Use double and triple bonds where necessary. Here, we have used 8 of the available 26 e-.

Place electrons around the outside atoms to complete their octets. This uses a total of 24 e-, leaving 32 – 8 – 24 = 0 e- remaining.

Finally, we show ion in brackets with the charge shown to the upper-right of the brackets.

200

Practice 8.5 Draw Lewis structures for the following molecules and ions: NH3, CH2Cl2, HCN, BrO3–, NO3–, CH3OH, NH4+ and H2CO. Identify those that are polar and be able to explain why you so indicated.

201

DECIDING BETWEEN SEVERAL POSSIBLE LEWIS STRUCTURES: FORMAL CHARGE

When several Lewis structures can be drawn that all obey the octet rule, we need a method of deciding which structure is most reasonable. We can do this by calculating the formal charge of each atom in the Lewis structure, which is the charge an atom in a molecule would have if all of the atoms had the same electronegativity.

In order to assign formal charge, we assign the electrons to atoms in the molecule, and then we look at the difference between the number of electrons assigned to the atom and the number of valence electrons the non-bonded atom possesses. First, assign the electrons in the Lewis structure: •

All nonbonding electrons are assigned to the atoms on which they appear

•

Half of the bonding electrons are assigned to each atom in the bond

•

Then, subtract the number of assigned electrons from the number of valence electrons on the non-bonded atom.

The formal charge calculation that provides formal charges closest to zero and in which the most electronegative element possesses negative formal charge generally identifies the most likely Lewis structure. This is not an actual assignment of charge on the atoms – electronegativity differences and other considerations determine the charge distribution on molecules. This is simply a convenient method of determining the most reasonable Lewis structure!

Practice 8.6

Determine the formal charge on the atoms in the following possible Lewis structures for NCO–. Choose the more reasonable Lewis structure based upon this determination.

Draw 2 possible Lewis structures for CO2, and use formal charge to determine the most likely structure.

202

RESONANCE STRUCTURES

We often see molecules whose Lewis structures cannot be adequately described by a single Lewis structure. For example, consider the nitrate ion, for which three possible structures are shown here:

Notice that the double bond could be placed on any of the three N―O bonding regions.

Practice 8.7 What would we observe about the bonding regions of the nitrate ion if one of these structures alone represented the structure of the nitrate ion?

The problem with the structures for nitrate ion above is that the bonds are dissimilar, which is not observed for nitrate ion – it has been experimentally shown that the bonds are all of equal length. In the cases where more than one Lewis structure is possible, we must accept that more than one Lewis structure exists and draw the Lewis structure to be an average of all of the possibilities. See step (iv) on Page 198 to decide when resonance is required. We then draw the structure as resonance structures, which is a series of structures that represents the observed properties of the molecule or ion. For nitrate ion, we draw:

Figure 82. When we attempt to draw the ion with an octet, we notice that a pair of electrons from any one of the oxygen’s gives the same structure. Thus, resonance exists for the ion. The correct structure includes all three representations.

203

Practice 8.8 Draw the possible Lewis structures for the ozone molecule showing that resonance plays a role in describing its structure.

It is of importance to note that resonance structures are NOT equivalent to several structures whose bonding pairs are “flipflopping,” but are instead truly two or more structures that must be used together to describe a molecule or ion. Additionally, the bond lengths in molecules displaying resonance are shorter than those displaying “pure” single or “pure” double bonds. In the example above showing the nitrate ion, three structures adequately describe the nitrate ion, and each of them shows a different placement of the double bond. We would expect that the bond length in the nitrate ion is not equal to that of a typical N-O or N=O bond, but is rather an intermediate bond length between the two. We describe the bonds by dividing the number of bonds by the number of bonding regions; e.g., 4/3 bonds, or one-and-a-third bonds. In this example, the bonds are shorter and stronger than single bonds, but longer and weaker than pure double bonds.

Practice 8.9 Which is predicted to have the shorter bond lengths, SO3 or SO32-? Using concepts of chemical bonding, explain your answer.

How do the bond lengths in NO3– compare to the bond lengths in CO32-? Justify your response using Lewis structures.

204

An important molecule that exhibits resonance is the benzene molecule, C6H6. It is shown below with the elements in place and in a common form in which C atoms are represented by the intersection of two lines; iitt is often drawn with a circle to represent the delocalization of the electrons in the bonding ding.

EXCEPTIONS TO THE OCTET RULE

You should become familiar with three common exceptions to the octet rule, and recognize this as another example of the difficulties associated with painting all chemical behavior with a broad brush.

Molecules with an odd number of electrons

In previous structures, the number of available electrons has been even. However, there are cases of bonding where the number of available electrons is odd. It is, quite frankly, simply not possible to draw a stru structure cture in which all atoms have an octet. Do not try to do so.

Practice 8.10 Draw Lewis structures for the NO and NO2 molecules.

205

Molecules with atoms that have less than an octet.

These molecules most often involve boron (3 bonds; 6 electrons) and beryllium (2 bonds; 4 electrons).

Practice 8.11 Draw three possible Lewis structures for BF3: one with all single bonds on boron with no lone pairs, one with lone pairs on boron, and one with a double-bond between a fluorine atom and boron. Then, use formal charge to select the most reasonable structure.

Notice that one of these structures above requires fluorine and boron to be double-bonded, and one requires that fluorine have only two lone pairs. This is not consistent with the fact that fluorine has a very high electronegativity. Thus, we expect that the structure in which

Figure 83. Molecules that donate both bonding electrons

boron does not achieve an octet is the most important structure. Using

in a single bond pair with molecules that accept both

the concept of formal charge helps us recognize the importance of the

electrons to form a coordinate covalent bond. These

single-bond-only structure.

species are often the Lewis acids and Lewis bases.

As an aside, it is important to note that boron compounds especially form coordinate covalent bonds, which are bonds that are created when both bonding electrons are provided by the same atom. We see this most often with boron, as its empty p-orbitals are available to be occupied by lone pairs of electrons on atoms and molecules. In fact, one of the most common compounds of this type to see is the H3N:BX3 coordinate compound; see Figure 83.

Molecules with atoms that have more than an octet.

These molecules are the most common of the exceptions to see. For example, look at the only possible Lewis structure for PCl5 at right. In this molecule, there must be ten electrons placed around the phosphorus atom. This is possible because the period 3 and later atoms have empty d-orbitals that can be used in bonding (however, recent studies suggest our explanation may be flawed). The period 2 atoms cannot exhibit these expanded valence shells because they do not have available d-orbitals. Thus, while we see PCl5, we will not see NCl5. Other common expanded valence shell molecules and ions include SF4, AsF6–, ICl4–. It is worth noting again that similar structures like NF5 and OF4 have not been observed due to their lack of “extra” bonding orbitals found in the d-sublevel. In addition to the consideration of available d-orbitals, there are two additional considerations to think about when deciding if an expanded valence shell is possible: 206

•

Expanded valence shells are more likely as the central atom becomes large. This is seen for the large atoms, including iodine and xenon.

•

Often, molecules with central atoms that contain expanded valence shells will be surrounded by very electronegative atoms like oxygen, fluorine and chlorine.

And, lastly, consider that if there are several possible structures, you should accept the Lewis structure that provides for an octet if it is possible to do so. Formal charge will help you see that an expanded structure is unlikely versus an octet structure.

Practice 8.12 Draw two possible Lewis structures for the phosphate ion – one with an expanded octet – and then use formal charge to dismiss one of them as unlikely.

Expanded octets can also arise when the lone pairs of electrons are the cause. Attempt a Lewis structure for XeF5.

Circle the atoms from the list that follows that could exhibit expanded valence shells:

C

N

P

Br

Xe

F

O

I

Cl

S

According to current theory, what is the single-most important consideration for predicting the possibility of an expanded octet?

Part 4: VSEPR THEORY

VSEPR – valence-shell electron pair repulsion – theory is a topic with which you are somewhat familiar from first-year chemistry. The bottom-line of the theory is that electron pairs, whether bonding pairs or non-bonding pairs, will repel one another and can be used to predict the molecular geometry of molecules and ions: The most likely arrangement of electron pairs on molecules and 207

ions provides for the maximum distance between bonding and non non-bonding pairs of electrons.

It is important to note that there are two simultaneous discussions occurring when describing the arrangement of electron pairs and molecular geometry:

•

Electron pair geometry discusses the arrangement of the regions of electron density (bonding regions and lone pair regions) around an atom

•

Molecular geometry discusses the arrangement of bonding regions of electron density in a molecule or ion

•

Electron geometry gives rise to the molecular geometry.

Predicting electronic and molecular geometries is a fairly straightforward process if you can easily draw Lewis structures.

Draw the Lewis structure for the molecule or ion.

Arrange all pairs of electrons around the central atom in such a way as to minimize repulsions between the th bonding pairs/nonbonding pairs. Describe the electronic geometry in of the arrangement of regions of electron density. density

Describe the three-dimensional dimensional molecular geometry in light of the electronic geometry in of the arrangement of bonding regions of electron density – double- and triple-bonds are considered a single region of electron density for discussions of molecular geometry.

To clarify our objective before we begin, let’s look at the molecule ammonia, NH3.

Figure 84. We represent the shape of a molecule by first drawing the Lewis structure, then evaluating the electron geometry, and finally evaluating molecular geometry.

Wee are attempting to do two things here: identify the three dimensional nature of the shape of the molecule in of the regions re of electron density, and we are attempting to identify the three dimensional nature of the shape of the molecule in of the regions of bonding electron density. In this case, the Lewis structure is fairly straightforward. Then, we arrange four regions regio of electron density in such a way that they form the four corners of a tetrahedron, which gives tetrahedral electronic geometry. geome Then, the molecular geometry considers only those three bonding regions of density, which here yields trigonal pyramidal geometry. 208

ELECTRON DENSITY GEOMETRIES There are essentially five electronic geometries to discuss; let’s look at the common ele electron ctron geometries, from which the molecular geometries can be derived.

209

Practice 8.13 Provide an example molecule or ion for each of the electronic geometries.

MOLECULAR GEOMETRIES

Once we identify the electronic geometry of a molecule – that is, the total regions of electron density surrounding the central atom – we consider the character of the regions of electron density as bonding or non-bonding to consider the molecular geometry of the molecule, which is the arrangement of the bonded atoms in a molecule.

Again, the idea is that the pairs of electrons on a molecule will arrange themselves to maximize the distance between pairs. For the total number of pairs of electrons n around a central atom, there are generally n-1 number of molecular geometries.

We use the following designation to represent the bonding/lone pair character of atoms in a molecule:

A indicates the central atom

B indicates bonded atoms

E indicates lone pairs of electrons

210

Electron Densities: 2 (linear electronic geometry)

Molecular Geometries: 1 (linear)

Designation: AB2

Number of Regions of Electron

Electron Geometry

Bonding

Nonbonding

Regions

Regions

2

0

Molecular Geometry

Example Compound and Ion

Density

2 linear

linear

Electron Densities: 3 (trigonal trigonal planar electronic geometry geometry)

Molecular Geometries: 2

three regions of density with three bonding regions – trigonal planar molecular geometry

AB3

three regions of density with two bonding regions – bent molecular geometry

AB2E


Electron Geometry

Bonding

Nonbonding

Regions

Regions

3

0

Molecular Geometry

Example Compounds and Ions

Density

3

trigonal planar

trigonal planar

2

1

bent

211

Electron Densities: 4 (tetrahedral electronic geometry)


four regions of density with four bonding regions – tetrahedral molecular geometry

AB4

four regions of density with three bonding regions – trigonal pyramidal molecular geometry

AB3E

four regions of density with two bonding regions – bent molecular geometry

AB2E2


Electron Geometry

Bonding

Nonbonding

Regions

Regions

4

0

Molecular Geometry


Density

4

tetrahedral

tetrahedral

3

1

trigonal pyramidal

2

2

bent

212

Electron Densities: 5 (trigonal bipyramidal electronic geometry)


five regions of density with five bonding regions – trigonal bipyramidal molecular geometry

AB5

five regions of density with four bonding regions – seesaw molecular geometry

AB4E

five regions of density with three bonding regions – tee-shaped molecular geometry

AB3E2

five regions of density with two bo bonding regions – linear molecular geometry

AB2E3


Electron Geometry

Bonding

Nonbonding

Regions

Regions

5

0

Molecular Geometry


Density

5

trigonal bipyramidal

trigonal bipyramidal

4

1

see-saw

3

2

tee-shaped

2

3

linear

213

Electron Densities: 6 (octahedral electronic geometry)

Molecular Geometries: 3 to consider

six regions of density with six bonding regions –octahedral molecular geometry

AB6

six regions of density with five bonding regions – square pyramidal molecular geometry

AB5E

six regions of density with four bonding regions – square planar molecular geometry

AB4E2


Electron Geometry

Bonding

Nonbonding

Regions

Regions

6

0

Molecular Geometry


Density

6

octahedral

octahedral

5

1

square pyramidal

4

2

square planar

214

Regions of Electron

Electronic

Density

Geometry

Bonding Pairs

Non-bonding or

Molecular

Example Molecule

Lone Pairs

Geometry

or Ion

2

3

4

215

Regions of Electron Density

Electronic Geometry

Bonding Pairs

Non-bonding or Lone Pairs

Molecular Geometry

Example Molecule or Ion

5

6

216

Practice 8.14 For the molecules or ions below, provide the Lewis structure, electronic geometry and molecular geometry.

BrF4

PCl5

NH3

BF3

H2S

SnCl3-

CO32-

ClF3

CH4

Cl2CO

217

We will often discuss the geometry around several central atoms, an example of which is represented below for the molecule H3CCOOH.

BOND ANGLES Nonbonding pairs of electrons and double- and triple-bonds take up more space than do single bonding pairs of electrons. Thus, when a double- or triple-bond is involved in bonding or a molecule possesses a lone pair of electrons (or more), then the bonding angles tend to decrease between bonding pairs as the nonbonding pairs or multiple bonding pairs of electrons repel more than we see

Figure 85. For tetrahedral structures, the predicted bond angle is 109.5º. However, lone pairs push on bonding regions to decrease this angle.

with single-bonding pairs.

Decrease the predicted angles between bonding pairs when there are multiple bonds or lone pairs on a molecule. The effect of multiple bonds and lone pairs is to cause a decrease in bond angles due to the larger space occupied by lone pairs and multiple bonds.

HYBRIDIZATION THEORY In discussing the various molecular geometries, we have considered that all of the bonds have been identical and that the number of bonds is equivalent to the number of empty valence shell orbital positions (excepting the special case of expanded octets) – however, a close examination of the beryllium atom and its bonding sheds some doubt on the straightforward approach we have used to this point. Consider the electronic structure of the neutral beryllium atom, which is shown below. If we look at this structure, we really do not see any reason for the atom to bond twice in of our current discussion of bonding (i.e., empty orbitals). The 1s and the 2s are both occupied to full, and the 2p filling to share electrons would suggest 6 bonds for beryllium – a situation that is not observed. In short, the beryllium atom should not form covalent bonds if valence shell positions are the only consideration. Thus, there must be more to bonding than the aforementioned straightforward approach. Hybridization theory helps explain the observed formation of covalent bonds.

218

In the case of Be, consider that a 2s electron is promoted to a 2p orbital. This would give the configuration shown at right. Using this idea, we can see how beryllium might form two bonds – an electron-sharing event with the 2s and another with the 2p. However, this still poses a problem because the bonds should be different – i.e., different lengths, angles, energies – if this is the case. And, the bonds have been observed to be the same in a molecule such as BeF2, for example.

So, let’s consider that the two half-filled orbitals now blend to form a new orbital that is intermediate in nature to an s and a p, and we’ll call it an sp orbital. Now we appear to have satisfied our need to form two equivalent hybrid orbitals that Be uses to form covalent bonds.

The above example shows the essence of hybridization theory, which attempts to extend upon the ideas of valence shell bonding theory to explain the bonding of atoms and the observations regarding bond character. Let’s look at a second example together using carbon. First, however, the following statements about hybridization theory should be made:

When pure orbitals do not adequately explain the bonding found in molecules, we turn to hybridization theory to better represent the bonding

The number of hybrid orbitals is equal to the number of orbitals involved in the promotion events. For example, the promotion of one electron from an s in Be to a p in Be gave two hybrid orbitals.

The notation for hybrid orbitals formed is to indicate the orbital letter and the number used to form the complete set of hybrid orbitals. For example, above we used a single s and a single p, which results in two hybrid orbitals termed sp orbitals.

Hybrid orbitals are equivalent to one another, except that for a set of hybrid orbitals, the spatial orientation of each is different.

Practice 8.15 For the molecule methane, the four bonds must be equivalent according to the studies of the molecule and its bonds.

In the boxes below, use arrows to indicate the valence shell structure of carbon.

219

Using only the number of empty valence positions and their character (that is, two bonds from an empty orbital and two bonds from half-empty orbitals) suggest that the four bonds in CH4 cannot be equivalent. So, we promote one of the s electrons to the empty p orbital.

Still, here, however, the four bonds cannot be equivalent because the three p orbital bonds would be different than the single s orbital bond. Thus, we form four hybrid orbitals by combining the single 2s and the three 2p orbitals.

We now have four equivalent orbitals that carbon uses for bonding.

What is the notation used for these hybrid orbitals?

Practice 8.16 For the following molecules and ions, show that hybridization can be used to describe the bonding in the molecule or ion.

A Lewis structure is almost essential in order to ensure that lone pairs of electrons are considered in the hybridization. Lone pairs appear as filled orbitals after promotion.

BF3

NH3

220

BrF3

CO2

H2O

Hopefully you see the following as you review the examples we completed: the number of hybrid orbital is equal to the number of regions of electron density on the central atom of a molecule. Thus,

two regions of electron density arise from sp hybridization using an s-orbital and a p-orbital

three regions of electron density arise from sp2 hybridization using one s-orbital and two p-orbitals

four regions of electron density arise from sp3 hybridization using one s-orbital and three p-orbitals

five regions of electron density arise from s sp3d hybridization using one s-orbital, three p-orbitals and one d-orbital

six regions of electron density arise from sp3d2 hybridization using one s-orbital, three p-orbitals and two d-orbitals

Practice 8.17

For each molecule or ion: indicate the hybridization, electronic geometry and molecular geometry

NH4+

221

SF4

PF5

ClF5

XeF4

PF6–

222

Especially when considering carbon, oxygen and nitrogen, we speak of the overlapping or head-on nature of the orbitals in a molecule containing these elements. A head-on bonding between the orbitals of two atoms is termed a sigma, σ, bond. To see the nature of a sigma bond, let’s examine the molecule water. Water exhibits sp3 hybridization, which involves hybridization of one s-orbital and all three p-orbitals. Two of these hybrid orbitals contain lone pairs, while two interact in a head-on bond formation with the orbital of hydrogen atom. This head-on interaction results in the formation of a sigma bond.

Figure 86. All four orbitals in the second quantum level are hybridized in oxygen. To form water, two hybridized orbitals bond head-on with hydrogen atoms. This results in a sigma bond.

In the case of ethyne, HC2H, however, we do not see all of the orbital undergo hybridization – carbon in ethyne is sp hybridized (two regions of electron density). Thus, two p-orbitals are unhybridized. Unhybridized

orbitals form bonds by overlapping rather than head-on bonding, which is the termed a pi bond. Because pi bonds results from overlapping orbitals, they do not allow for rotation around the atoms involved – thus, their structures are more rigid. In our eyes, it is the breaking of pi bonds by light that cause molecular structure changes in the rhodopsin to allow for the

Figure 87. In ethyne, there are two unhybridized orbitals on each carbon and two hybridized orbitals. Hybridized orbitals bond head-on (sigma bond), while unhybridized orbitals bond by overlap.

translation of visible light into color.

Because all bonds involve at least one region of electron density, then all bonds contain at least one sigma bond (head-on bond). So, we can draw the following conclusions:

a single bond is a sigma bond

a double bond is one sigma bond and one pi bond

a triple bond is one sigma bond and two pi bonds

Practice 8.18 Draw the structures of propane, propyne and propanol. For each: indicate the hybridization on each carbon, on any oxygen atoms present, and clearly indicate the number of sigma bonds and pi bonds in each molecule.

223

Sketch the molecule acetonitrile, H3CCN. Describe the hybridization around the carbon atom, and indicate the total number of pi bonds and sigma bonds in the molecule.

For benzene, C6H6, what do you predict about the total number of sigma bonds and pi bonds?

224


The Behavior of Solutions, Liquids and Gases; Gases Phase Change and Intermolecular Forces Students will be able to:

Identify the intermolecular forces of attraction and repulsion that exist between molecules and ions in compounds

Describe the relationships between the observed behavior of substances and mixtures and the intermolecular forces of attraction present resent in the substance or mixture

Discuss how size, composition and molecular shape affect the observed behavior of substances and mixtures

Define and explain viscosity and surface tension, and establish the relationships between these physical properties and intermolecular forces

Predict the occurrence and explain the behavior of hydrogen bonding

Recognize the enthalpy changes that accompany phase changes, and interpret or produce a phase diagram for a substance

Define and discuss vapor pressure

Describe the behavior of solids based on their crystalline structures, and characterize solids based on their molecular structure

Characterize the solution process and predict the occurrence of dissolution

Justify Henry’s law on the molecular level; perform calculations using Henry’s Law

Express the concentration of solutions in various units

Characterize the behavior of solutions versus pure solvents; explain the observed behaviors of solutions versus pure pur solvents; perform calculations to determine the behavior of solutions versus pure solvents

Use the gas laws and explain the behavior of gases in of the gas laws

Explain ideal gas behavior and use the ideal gas equation to determine the characteristics characterist of the behavior of gas samples

Use the variations of the ideal gas law to determine quantities

Use and explain the concept mole fraction and Dalton’s Law of Partial Pressures to explain the observed behaviors of gases

Explain the behavior of gases in t of the kinetic molecular theory of gases

Discuss and explain the effects of the limitations of the KMT and the ideal gas equation

Express the observed variances of the behavior of gases in of the van der Waals equation

Use effusion and diffusion to determine and explain some properties of gases

225

Part 1: INTRAMOLECULAR REVIEW – COVALENT, IONIC AND METALLIC BONDING

You will recall that there are several bonding patterns found between the atoms in a substance, and these patterns of bonding affect the behavior of a substance. Specifically, three bonding patterns are explored with respect to these intramolecular (literally, “within molecules”) forces of attraction. Intramolecular forces of attraction are the bonds of covalent bonding and the significant electrostatic attractions between ions. Covalent bonding occurs between atoms of nonmetals as they share valence electrons to fill their valence shells. (Rarely this pattern will be observed between nonmetals and metals, as is the case with titanium dioxide, and more often, between beryllium and nonmetals, which always bond covalently.)

Figure 89. An ionic compound is made of

Ionic bonding occurs between metal atoms

positively- and negatively-charged ions that are electrostatically attracted.

and nonmetal atoms as the metal loses electrons and the nonmetal gains electrons.

Recall that the metal is termed a cation, and the nonmetal is termed an anion. The electrostatic attraction between the ions is the force that holds an ionic compound together. The ions bond in ratios that provide neutral compounds. Figure 88. A covalently-bonded

Above, Figure 89 is an example of

molecule.

an ionic compound. As expected, the sodium has lost electrons to

form a positively-charged cation, and the nonmetal chlorine has gained an electron to form the negatively-charged anion. Metallic bonding occurs when metals lose their electrons and form positive ions – these ions are held together by the electrostatic force that exists between the cations and the free electrons. A figure of metallic bonding is shown in Figure 3. The spheres in the center of the rings represent positivelycharged cations that have lost their valence electrons. The electrons “float” freely between the cations and act as an electrostatic glue to hold the metal sample together. The significant result of the pattern of metallic bonding is that metals have the properties of malleability, electrical conductivity and

Figure 90. Metal atoms easily lose their valence

ductility – all properties owing to the ability of the cations to move freely and

electrons, which then exist sandwiched between

slide past one another. This is not the case in ionic bonding, in which case the

ions of pure metal. This explains the malleability of

ions are held strongly in place and to move them cleaves the crystal lattice.

metals and their electrical conductivity.

Again, the forces of attraction discussed above are all between the atoms and ions of a substance; the intermolecular forces to which we will now turn our attention are between the molecules of a substance. However, we will review polarity of molecular compounds first, as this is essential to understanding how molecules are attracted. 226

Part 2: ELECTRONEGATIVITY AND POLARITY

Recall that nonmetal atoms with different electronegativity values will bond to form a polar covalent bond. In a polar covalent bond the electrons shared in the bond are not shared equally, as represented by the diagram at left, which suggests the electrons being pulled more strongly by the more electronegative oxygen atom.

A bond in which the electrons are attracted to one atom more than another creates a dipole, or a bond that has two “poles:” one that appears to be more negative than another. We then say that the bond is partially-negative on the more electronegative atom and partially positive on the less electronegative atom – the direction of the dipole is conventionally discussed in of the direction of electron flow; thus, the direction of the bond dipole in the O-H bond is toward the oxygen.

Polar molecules arise when the overall distribution of electrons is asymmetrical or there is otherwise an unequal distribution of electrons around the central atom. Additionally, the Figure 91. Water is polar because the high electronegativity of oxygen leads to asymmetrical or unequal distribution of electron density.

presence of a lone pair induces polarity. (Ionic compounds can be thought of as the most extremely polar molecules – the electrons are induced to move so completely in one direction that they are transferred.)

Of course, while molecules can be polar or nonpolar, we can also consider that regions of molecules are such. For example, look at the molecule shown in Figure 92 below. In the vitamin A molecule (a), the hydrocarbon region on the left of the molecule is nonpolar, owing to the relative equal distribution of electron density. On the right of the molecule however, the -OH group imparts a small region of polar character. The vitamin C molecule (b) is much more polar, with four regions of asymmetrically-arranged electron density distributed around the molecule.

It is molecules’ ionic/polar character that affects the attractions that exist between molecules of substances and mixtures. We will see that substances’ physical properties, including such properties as melting point, phase at room temperature, boiling point and solubility are the result of intermolecular forces of attraction, which are a direct result themselves of the

Figure 92. Like regions of smaller molecules, the regions of large molecules, too, can have varying

orientation of the atoms and

degrees of polar character. The interactions between molecules depend upon this character, and,

their intramolecular forces

indeed, affect their solubility, melting point and other physical properties.

of attraction (i.e., bonding). 227

Practice 9.1 Rank the intramolecular forces in order of increasing strength.

What forces are disrupted when new substances form: intramolecular or intermolecular?

What forces are disrupted when a substance melts or boils: intramolecular or intermolecular?

What forces are disrupted when one substance dissolves in another: intramolecular or intermolecular?

Mark the molecules (i) – (v) at left as polar or nonpolar.

Part 3: INTERMOLECULAR FORCES

We have seen how the atoms of compounds or the ions of compounds are held together by intramolecular forces. Covalent bonds, metallic bonds and ionic bonds are the three types of intramolecular forces. Again, the behavior a molecule exhibits owes a lot to the patterns of intramolecular bonding present in the molecule. Whether or not the molecule will conduct electricity in solution is an example of a property of a molecule that is dependent upon the bonding pattern the molecule possesses.

But beyond the attraction within a molecule, there are also attractions between molecules of the same compound and molecules of different compounds. Figure 93. The covalent bond that holds the molecule together is a strong intramolecular force (a chemical bond), while the intermolecular attraction is what attracts one molecule to another. The rupture of intramolecular forces, recall, results in the formation of a new substance, while the disruption of intermolecular forces orces at most causes a change of phase.

This concept is reinforced in the figure at left: both the intramolecular ecular covalent bond and the intermolecular attraction are shown. The degree of these attractions explains many of the observed properties of substances.

Intermolecular forces of attraction vary in strength, and they are affected by the structure, mass and size of the interacting molecules and ions. At any rate, however, intermolecular forces are at their greatest 15% as strong as covalent or ionic bonds. bonds

228

ION-DIPOLE FORCES

Ion-dipole forces exist between the ions of a compound and a polar molecule to which the ion is attracted by electrostatic attraction. The most common occurrence of this is a dissolved solute ion in the polar molecules of a solvent. For example, water is a polar molecule in which ions such as Na+ and Cl– can be easily dissolved.

In Figure 94, you can see that the polar water molecules exert attraction for the negative ions at the water molecules’ partial positive ends (the hydrogen

Figure 94. Ion-dipole forces are responsible for the dissolution process of ionic salts in water or other polar solvents. Solution occurs when the ion-dipole attractions exceed those of the intramolecular ionic bonds attractions. Notice that an ionic solid may easily have its intramolecular electrostatic attractions disrupted by water molecules – this reinforces the idea that ionic bonds are much weaker than covalent bonds.

ends). These attractions between the polar water molecule and the ions of the solute are the ion-dipole attractions. Similar attractions exist between the partial negative end of the water molecules and the positive sodium ion.

Because there is attraction in this scenario, polar solvents will dissolve polar solutes, while nonpolar solutes will not readily dissolve in polar solvents. Recall that the extent to which an ionic solid will dissolve is related to the lattice energy of the ionic compound, the enthalpy change of the solution process and entropy.

DIPOLE-DIPOLE FORCES

Dipole-dipole forces are the forces important when two polar molecules interact. For example, the most common dipole-dipole interaction with which you are familiar is that between the polar solvent water and a polar solute like ethanol, which is shown in Figure 95. The polar water molecule is attracted to the polar region of the ethanol molecule owing to the electrostatic attractions between two dipoles – notice the orientation of the partially-positive region on water to the partially-negative region on ethanol. A dipole-dipole attraction exists when the partial positive end of a polar molecule is attracted to the partial negative end of an adjacent molecule. Several dipole-dipole arrangements are possible for even just two molecules that interact, as illustrated in Figure 96 on Page 230. Figure 95. When two polar molecules near one another, their opposite polar ends attract, which results in a relatively strong intermolecular attraction.

Overall, the attractive interactions in dipole-dipole attracted substances lasts longer than the repulsive interactions that exist when like poles are oriented toward one another. Thus, the overall attraction in a 229

substance exhibiting dipole-dipole forces is an attractive force. In order to be realized, the molecules experiencing dipole-dipole interactions must be close to one another; in fact, they must be much nearer one another than the species involved in ion-dipole interactions because the dipole-dipole force is much weaker than the ion-dipole force (as evidenced by the fact that the ion-dipole attraction disrupts the dipole-dipole force in solutions of soluble salts).

Dipole-dipole forces are affected by the mass, size and character of the molecules and of the atoms in molecules. •

Molecules that can interact more closely at their polar regions will exhibit greater dipole-dipole interactions. Thus, molecules that are smaller with similar polarities experience the dipole-dipole force more strongly.

•

Dipole-dipole interactions are affected by the relative sizes of polar regions – as the polar region becomes larger, then the dipole-dipole interactions increase.

LONDON DISPERSION FORCES

All substances can be turned into liquids, which suggests that their molecules must exhibit

Figure 96. Dipole-dipole

some attraction for adjacent molecules. However, nonpolar and non-ion samples would not

attractions are present between

be expected to exhibit ion-dipole forces or dipole-dipole forces – so there must be some

polar molecules.

other force that causes attractions between molecules that do not exhibit dipoles or ionic charge. Fritz London, a German-American scientist, studied this phenomenon and arrived at the idea of induced dipoles, which are also called London dispersion forces or van der Waals forces.

To understand London forces, consider two atoms of helium. The electrons of the helium atom are in constant motion. We should expect that the electrons are on average equally distributed around the two atoms. However, the electrons may, at times, be situated in such a way that the nucleus is exposed as a partial positive charge on one atom while the electrons provide a partial negative charge on an adjacent atom (Figure 97a). This results in an induced dipole on the atom, and an attraction between adjacent atoms: in the figure, the instantaneous partial-negative charge of the right-hand atom results in an attraction between the nucleus of atom 1 and the electron density of atom 2 (Figure 97b). As shown in the figure, the induced dipole gives a moment of electrostatic attraction between the atoms that causes a brief and weak attraction. However, as the electrons are moving Figure 97. Although atoms are by nature nonpolar, the movement of electrons and the attractions by

rapidly, the attraction is

nuclei induce temporary polar character, or induced dipoles. This causes a short-lived attraction

very short-lived and very

between atoms and molecules that otherwise should not exhibit significant attractions.

weak. 230

The exhibition of induced dipoles varies with the character of atoms. Large atoms that have electrons far from the nucleus are more likely to experience induced dipoles through polarizability, which is the result of the shift of electrons for two reasons: the large atom itself does not exert a great attraction for its outer electrons due to shielding, so they are easily influenced by adjacent atoms’ nuclei. Thus, large molecular weight atoms are more likely to experience polarizability and significant London forces. For example, the atoms of fluorine are not significantly polarizable, as the atom is quite small; however, the atoms of iodine are quite polarizable, as the influence of iodine’s nucleus is not as great on its own electrons. (A case study for the relationship between melting point and polarizability can be seen in the halogens: as the polarizability increases – resulting in more significant London forces – the melting point increases. This is evidenced by the phase at which the halogens exist at room temperature.) Molecular shape affects the strength of the London forces: the more London force interactions between molecules, then the greater the London force attraction. For example, in Figure 98 are shown two isomers of pentane, C5H12. Notice that the straight-chain isomer (top) has more length over which London dispersion forces may exist, and thus has different properties than the more “condensed” neopentane isomer shown below it. When evaluating the contributions intermolecular attractions for two substances, generally consider the size and shape of the molecules, along with any polar character: when the molecules are close in size and shape, the London forces are approximately equal; any polarity is the decisive factor.

Practice 9.2 Which force is decisive in driving the observed physical properties related to intermolecular forces for each of the compounds here?

CH3CN

CH3I

For which substances are dipole-dipole attractions decisive: Br2, HCl, HBr

Figure 98. London forces increase when the number of “points-of-” increases between two molecules. Thus, n-pentane has significantly more LDF attractions than neopentane, and the melting points are quite different because of this.

and N2?

For which substance is London force the most significant force of attraction: Br2, HCl, HBr and N2?

Of the four substances above, which would be expected to have the highest and lowest melting points? Explain.

231

HYDROGEN BONDING

Hydrogen bonding is not a bond at all, but is rather an attraction of much greater strength than London forces or other dipole-dipole attractions.. A hydrogen bond occurs when an electronegative atom such as N, O or F is bonded to hydrogen, which virtually exposes hydrogen as a proton because of the electronegativity of the bonded atom N, O or F. The hydrogen atom is significantly attracted to a lone pair of electrons on an atom in an adjacent molecule

Figure 99. Due to the electronegativity of oxygen, the electrons of

(Figure 99). The lone pair is also usually found on N, O or

the hydrogen atoms are strongly pulled away from the hydrogen

F. Thus, a hydrogen bond is really a unique dipole-dipole

atoms. This results in a significant dipole, which results in

interaction that occurs under the aforementioned

hydrogen bonding for water and other molecules.

conditions.

The hydrogen bond is responsible for many biologicallyimportant functions, including the proper folding of proteins and the attractions that hold DNA together. Hydrogenbonding is also responsible for the unique behavior of water when it freezes. Liquid water is more dense than solid water, which is the result of the distance from one another that water molecules situate in the frozen state due to hydrogen bonding – they are farther from one another in the solid state than in the liquid state.

It is important to note that not all cases of intermolecular attractions involving hydrogen are hydrogen-bonding. In order for hydrogen-bonding to occur, the hydrogen atom must be attached to an electronegative atom such as N, O or F. As shown in Figure 100, the intermolecular force between H and O is not a case of hydrogen bonding, but rather a case of dipole-dipole interaction. Hydrogen bonding occurs only when a hydrogen atom is bonded directly to an atom N, O or F in the same molecule and the interaction occurs between the O, N or F of an adjacent molecule and the hydrogen atom so

Figure 100. Be sure to note the difference between bonds in which hydrogen participates and the specific case of intermolecular attraction called the hydrogen bond. Moreover, all intermolecular forces involving hydrogen are not cases of hydrogen bonding.

bonded.

A summary of the intermolecular attractions is shown below. Please be sure to review it and organize your thoughts surrounding how:

intermolecular forces and intramolecular forces are different

the strengths of the intermolecular forces compare

polarity and other structural considerations affect how molecules are attracted to one another

232

Are ions involved? YES

NO Are polar molecules involved?

Are polar molecules and ions involved? YES

NO

NO YES

Is hydrogen bonded to nitrogen, oxygen or fluorine? NO

IONIC BONDING

YES ION-DIPOLE INTERACTIONS

LONDON FORCES ONLY HYDROGEN BONDING

DIPOLE-DIPOLE FORCES

Part 4: RELATIVE STRENGTH OF INTERMOLECULAR ATTRACTIONS

The strength of intermolecular attractions is based on many factors. The force that causes all of the attractions is electrostatic force, and its strength can be predicted based upon the magnitude of the apparent charges that are attracted to one another. For example, an ion-dipole attraction between a three-plus ion and water is stronger than the ion-dipole attraction between a plus-one ion and water; similarly, the attraction between long, chainlike molecules will be greater than the attraction between small, spherical molecules. The attractions are shown in order of generally decreasing strength below. At any rate, the strongest intermolecular attraction is at most 15% as strong as chemical bonding. And, the strength drops significantly as one moves across the spectrum; i.e., the strength of LDFs is orders of magnitudes less than the strength of H-bonding, for example.

Ion-Dipole Forces

Hydrogenbonding

Dipole-Dipole

London

Forces

Dispersion Forces

(Do not take the ranking above to indicate that the most significant force that drives the behavior of a substance is due to the strongest force present – the behavior is due to the cumulative effect of all interactions, of which any one of the above could be more significant than others that are also present in a sample.)

233

Part 5: VAPOR PRESSURE AND BOILING POINT

Before we discuss the properties based upon intermolecular forces, assuring an understanding of the concepts of vapor pressure and boiling is essential.

In Figure (a) below, you can see that above the surface of a liquid there are a certain number of molecules that escape the liquid. In Figure (b) – where the container has been capped to allow equilibrium to be established – we can measure the equilibrium vapor pressure. This represents the vapor pressure of a substance: the pressure that is exerted by the gaseous phase above the surface of the substance’s liquid or solid phase at a certain temperature.

Vapor pressure is a factor of the Figure 101. Vapor pressure is a measure of the pressure exerted upward by the molecules

atmospheric pressure pushing

at the surface of a liquid or solid and the atmosphere pushing down on it.

down on the substance, the temperature and the intermolecular forces of the

substance. When the vapor pressure of a substance exceeds the atmospheric pressure exerted downward, then the substance begins to boil. The normal boiling point of a substance is the temperature at which the vapor pressure of a liquid is large enough to overcome normal atmospheric pressure, one atmosphere. For example, at sea level the normal boiling point of water is 373 K. At 373 K the pressure exerted by water vapor is great enough to overcome the intermolecular attractions of the liquid phase and the force of the atmosphere. Thus, the molecules in gas phase can completely escape the liquid. At high elevations, the boiling point of water is less, such as in Denver, Colorado.

234

Part 6: PROPERTIES AND INTERMOLECULAR ATTRACTIONS

The properties of substances are in large measure due to their intermolecular forces. Among the properties of interest: boiling point (affecting phase at room temperature), surface tension, viscosity and vapor pressure. Boiling point. One should be able to predict that the boiling points of substances are based upon the attractions between the molecules of the substances. For example, in order for water to boil, enough energy must be added so that the molecules possess enough kinetic energy to overcome the pressure of the atmosphere on the molecules, the gravity force pulling molecules downward and overcoming the strong hydrogen bond attractions between the molecules.

As the intermolecular forces of attraction increase between the molecules of a substance, the boiling point of the substance increases, which is illustrated in Figures 102 and 103.

Figure 102. The boiling point of water is significantly higher than nonpolar CH4 due to the hydrogen-bonding present in water. Notice, too, the position of polar H2S: less than that of water, but greater than that of CH4. H2S exhibits only dipole-dipole interactions, and methane only London forces. It is important to note the effect of molecular

Molecular weight also affects boiling point. The graph here shows how the molecular weight and intermolecular forces affect boiling point. It is worth noting in light of the discussion of boiling point versus molecular weight that given similarlymassed compounds, the intermolecular forces drive the boiling point.

weight on boiling point, as well.

Vapor pressure. The intermolecular forces of attraction in a substance affect the vapor pressure of a substance, as should be apparent in that the boiling points are affected. As the intermolecular forces of attraction strengthen, the amount of energy required to escape the liquid is greater, which leads to a lower vapor pressure.

As the intermolecular forces of attraction increase for a substance, the vapor pressure of the substance decreases. That is, all other things being equal, nonpolar substances will have higher vapor pressures and ionic solids will have the lowest vapor pressures. Relate this to boiling points.

As an aside, a substance that evaporates quickly – i.e., low intermolecular attractions – has a high vapor pressure, and it is described by the term volatile.

Figure 103. The vapor pressure, too, is a function of intermolecular forces of attraction. 235

Surface tension. Surface tension is a measure of the energy required to disrupt the intermolecular attractions that occur at the surface of a liquid. At the surface, a liquid’s molecules experience “pulls” of intermolecular forces inward. The inward force can cause a liquid to appear as if it has “skin” layer that must be broken in order to access the liquid.

As the intermolecular forces of attraction increase, the surface tension of liquids increases. For example, water has a much higher surface tension than hexane (C6H6) – thus, bugs that can walk on water might not be able to walk on hexane!

Viscosity. Viscosity is a measure of how easily molecules of a liquid can move with respect to one another. Liquids with low viscosity can flow quite easily, while liquids with high viscosity cannot flow as easily. Viscosity is related to intermolecular attractions, temperature and the shapes of molecules. (We define viscosity as the resistance to flow.)

Molecules that exhibit little intermolecular attractions can flow more easily than liquids that exhibit a high degree of intermolecular attraction. For example, a liquid with only London forces of attraction will flow more easily than a liquid that exhibits strong dipole-dipole interactions.

Others things being equal, a liquid that is larger or that is composed of molecules that can become entangled will be more viscous than a small-molecule liquid or one that is simple in structure. For example, we would expect that mercury is less viscous than oil because mercury is a simple substance that will not become entangled; oil, on the other hand, is long and sinewy and will become entangled.

Generally, liquids’ viscosities decrease as the temperature of the liquid increases. This is due to the increased ability of the faster-moving molecules to disrupt the intermolecular forces of attraction.

Practice 9.3 For the four halogens F2, Cl2, Br2 and I2, rank them as to:

polarizability

boiling points

For the four noble gases He, Ne, Ar and Kr, rank them as to:

polarizability

boiling points

For the four compounds here, identify the most significant intermolecular force and other less significant forces, and then rank them as to the properties that follow.

CH4, CH3OH, HCOCH3, CH3CH2OH

236

boiling point

surface tension of their liquids

Select the molecule with the higher boiling point, and justify your response.

Select the molecule from below that should be predicted to show the greater intermolecular attractions. Justify your response. (The oxygen atom of each alcohol is indicated with an arrow.)

Select the molecule from the two shown below that should be predicted to have the higher vapor pressure. Justify your response.

The boiling points of the isomers propanol and ethyl methyl ether are 97.2°C and 10.8°C, C, respectively. Sketch Lewis structures of the molecules, and d discuss iscuss their differences in of intermolecular forces, molecular weight and related concepts. Prepare this separate from this note set in a more formal manner than the briefer statements above.

237

Part 7: CHANGE OF PHASE

Changes of phase, which are represented above, are always accompanied by energy changes. When energy is added to a substance it may disrupt the intermolecular attractions and cause a phase change; it is also true that the removal of energy from a sample can increase the intermolecular attractions and cause a phase change. A phase change occurs when the intermolecular attractions between molecules are disrupted, or a phase change occurs when the intermolecular attractions increase. As we saw in our discussion of bond enthalpies, the disruption of forces of attractions (bonds or intermolecular) requires that energy be added, while the formation of attractions (bonds or intermolecular) requires that energy be released from a system.

We are concerned with six phase changes through which substances go:

Solid → liquid phase change: melting (reverse of freezing)

The process of melting is called fusion, and the energy needed to cause melting is called the heat of fusion, or the enthalpy of fusion, and it is represented by ∆Hfusion, which can be read as the “enthalpy change of fusion” or “heat of fusion.” Like enthalpy changes we have already seen, we will express the heat of fusion in the unit kJ/mol; for freezing, ∆Hfusion has a negative sign because energy is released from the system, while for melting, ∆Hfusion has a positive sign. The intermolecular attractions that provide a solid its structure are disrupted when a substance melts.

Liquid → gas phase change: vaporization (reverse of condensation)

The process of vaporization is the change of state from liquid to gas phase. The amount of energy required for this to occur is called the enthalpy of vaporization or the heat of vaporization, and it can be represented in kJ/mol as ∆Hvaporization. The heat of vaporization for a substance is much higher than a substance’s heat of fusion because vaporization requires total separation of intermolecular attractions, whereas the phase change to liquid from solid or solid to liquid requires only the temporal breaking and reforming of intermolecular attractions. (This significant change in energy explains why steam burns are so damaging.)

Solid → gas phase change: sublimation (reverse of deposition)

The process of sublimation occurs when a substance’s molecules undergo a change of state from solid to gas without existing in the liquid phase. This process, too, requires energy, which we term the enthalpy of sublimation, ∆Hsublimation. The energy change for the process is the sum of the enthalpy of fusion and the enthalpy of vaporization, as Hess’s Law would predict.

Figure 104. Phase represents the allowance of molecular motion in a sample; as intermolecular forces increase, the temperature at which phase change occurs increases.

238

Practice 9.4 Identify the following as endothermic (absorbs energy) or exothermic (releases energy). Provide an example of deposition and sublimation.

freezing

melting

condensation

vaporization

sublimation

Example:

deposition

Example:

A great amount of energy is required to turn a cup of liquid water into a cup of steam. If you touch steam, it burns even more than scalding hot liquid water. In of the energy that is released when the steam s your skin, why does steam burn more than hot liquid water?

We have already seen the enthalpy change associated with the heating of a phase (in contrast to the new discussions above surrounding phase change). This was earlier given as the specific heat of a substance, and it varies considerably for each phase. For example, the specific heat of liquid water is 4.184 J/g • K, that of solid water is 2.092 J/g • K, and that of gas water is 1.841 J/g • K. Do not forget that specific heats are per gram of a substance in joules, while the phase enthalpy changes are in kJ per mol.

HEATING CURVES

Heating curves reflect the temperature changes associated with the heating of a substance as a function of the heat added (Figure 16). Of course, a reverse graph would provide a cooling curve showing decreasing temperatures and a release of energy.

At Point A the water exists in solid phase, and the heat added from Point A to Point B is disrupting the hydrogen bonds holding the crystalline structure of solid water together. The specific heat of ice is used to determine

Figure 105. A heating curve for 100.0 g of water.

the amount of energy required to effect this temperature change. 239

At Point B the ice is undergoing melting, and the

enthalpy of fusion, ∆Hfus, can be used to determine the amount of energy

required to melt the sample of ice. Notice the temperature does not change as melting occurs.

The melting of ice is complete at Point C, and it is from here to Point D that the liquid phase of water is heating. The specific heat of liquid water is used to determine the amount of energy required to heat the sample from zero Celsius to 100 Celsius (or 273 K to 373 K).

When the liquid begins to vaporize at Point D, notice that the temperature does not increase until vaporization is complete. The amount of energy required to vaporize the sample is the heat of vaporization, ∆Hvap. The complete vaporization of this sample is complete at Point E. At Point E the water vapor (gas) phase of water begins to heat. The specific heat of water vapor is used to determine the amount of energy required to heat the gas phase.

A summary of the energy changes for the heating curve shown above is shown here. Note that for a cooling curve, all of the reverse processes would release energy equal in magnitude to those shown but would be opposite in sign.

Point or line

Process or change occurring...

Point A

Sample begins as solid ice at -25 °C.

A→B

Solid heats from -25 °C to 0 °C.

Point B

Solid begins to melt (fusion).

B→C

Solid is melting. No temperature change occurs until melting is complete.

Point C

Melting is complete. Liquid water begins to heat.

C→D

Liquid water heats from 0 °C to 100 °C.

Point D

Liquid water begins to vaporize.

D→E

Liquid is vaporizing. No temperature change occurs until vaporization is complete.

Point E

Vaporization is complete. Water vapor (gas) begins to heat.

E→F

Gas phase heats from 100 °C to 125 °C.

Amount of energy required to cause change... --Specific heat of solid water J/g • K heat of fusion kJ/mol

---

--Specific heat of liquid water J/g • K heat of vaporization kJ/mol

---

--Specific heat of water vapor J/g • K

240

In order to determine the amount of energy required to raise a sample from Temperature X to Temperature Y, you must consider if there are any phase changes in the problem, and, as needed:

Raise the temperature of the solid to its melting point (specific heat of solid)

Melt the sample of solid (heat of fusion)

Raise the temperature of the liquid to its vaporization point (specific heat of liquid)

Vaporize the sample (heat of vaporization)

Raise the temperature of the vapor to the final temperature (specific heat of vapor)

Some example problems follow.

Practice 9.5 How much energy is released as a 25.6 g sample of water cools from 48°C to room temperature, which is about 25°C?

How much energy is absorbed to raise the temperature of a 250.0 g sample of solid water from –45.0°C to –5.00°C?

Calculate the amount of energy released when a 16.5 g sample of water vapor is cooled from 400. K to 285 K.

Calculate the enthalpy change, ∆H, when a 0.500 mol sample of ice water stored at –15.0°C is changed into water vapor and heated to 105°C.

241

When Freon-12, CCl2F2, vaporizes it can take the heat away from a sample of liquid water, which cools the water. How many grams of Freon-12 are required to change the temperature of a 146.5 mol sample of liquid water by 40.0°C? The heat of vaporization of Freon-12 is 34.9 kJ/mol.

What is the heat of fusion of methanol, CH3OH if 4.01 kJ of heat is released when 25.23 g of methanol freezes?

Part 8: CRITICAL TEMPERATURE AND PRESSURE

The liquefaction of gases can occur by two routes:

1.

Decrease the temperature so that the intermolecular attractions increase sufficiently to cause the gas-liquid phase change, or

2.

increase the pressure on the gas to a pressure sufficient to increase intermolecular attractions, which will cause the gasliquid phase change

For example, carbon dioxide exists as a gas at normal atmospheric pressure at room temperature (1 atm / 298 K). The intermolecular attractions between nonpolar CO2 molecules are not great enough to overcome the kinetic energy of the molecules and cause the liquid or solid phase. However, if one increases the pressure of carbon dioxide at 298 K to 75 times the pressure at sea level (75 atm), then the carbon dioxide interactions (London dispersion forces) are great enough to cause the liquid phase to form. Similarly, if one would decrease the temperature of the gas sample at 1 atm to less than 190 K or so, the gas would change phase, too (gas → solid). However, above certain temperatures, it is not possible to achieve the liquid phase, and instead the gas phase simply becomes more dense.

Practice 9.6 Why does it seem plausible that there could be a temperature at which a gas cannot be liquefied?

242

The temperature above which a gas can no longer be liquefied – because the intermolecular attractions will not overcome the kinetic energy of the molecules – is called the critical temperature of a substance. The pressure at which a substance must be held to cause liquefaction at the critical temperature is called the critical pressure.

Practice 9.7

Would you expect that substances with weak intermolecular forces (like only limited London forces) or substances with stronger intermolecular forces (like hydrogen bonding) have lower critical temperatures? Explain.

A sampling of critical temperatures and pressures is shown here.

Substance (IMF)

Critical Temperature

Critical Pressure

ammonia (dipole)

405.6 K

111.5 atm

argon (London)

150.9 K

48 atm

carbon dioxide (London)

304.3 K

73.0 atm

water (H-bonding)

647.6 K

217.7 atm

hydrogen sulfide (dipole)

373.5 K

88.9 atm

Part 9: PHASE DIAGRAMS

A phase diagram is a graphical representation of the phase of matter that is stable for a substance at a given set of pressure and temperature variables. The following are important considerations:

The boundary shown between two phases is a dynamic equilibrium boundary – the substance is transitioning back-andforth between the two phases at equal rates.

Because the liquid phase cannot exist beyond the critical temperature, the liquid-gas equilibrium curve ends at the critical temperature.

Points not on a curve represent points at which only a single phase is present.

A phase diagram and its details is shown on Page 244.

243

Figure 106. The phase diagram shows how variations of pressure and temperature influence the stable phase for a substance.

Practice 9.8 Evaluating a phase diagram.

the curve A – B

the curve A – C

the curve A – D

triple point

critical point

the slope of A – D

points along 1 atm

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Part 10: THE PROCESS OF SOLUTION

Much of the chemistry we have studied has been in aqueous solution, which means that the chemicals are dissolved in water. In the case of aqueous solutions, the solvent is water and the solute is the dissolved species. To illustrate the process of solution and its favorability, we will look at the solution of NaCl in water. Three interactions must be considered when describing the process of solution: the interactions between solute particles, the interactions between solvent particles and the interactions between solute and solvent particles. Disrupting the attractive interactions of solutesolute interactions and solvent-solvent interactions requires the addition of energy: the disruption of attractive forces requires energy. We will refer to the enthalpy change that occurs to disrupt these attractions as ∆H1 and ∆H2, respectively. The third interaction is that between the separated solute particles and the separated solvent particles: the attraction between solvent and solute particles releases energy. We will call this enthalpy change ∆H3. Thus, the overall enthalpy change can be given as:

∆Hsolution = ∆H1 + ∆H2 + ∆H3 where ∆H3 is negative; ∆H1 and ∆H2 are positive

Figure 107. The attractions that exist between the charged ions of NaCl and the water molecules’ dipoles is great enough that the solid NaCl dissolves in water. Note that this is the enthalpy change only, and it alone does not provide for a discussion of spontaneity – one must also consider the entropy change associated with the solution process. When the difference between ∆Hsolution - T∆S < 0, then the process is spontaneous. Recall that the water molecule is polar, with a concentration of negative charge on the more electronegative oxygen atom. This leaves the hydrogen ‘ends’ with partial positive charge. The negative ion in an ionic compound is strongly attracted to the positive regions of water molecules, while the positive ion is strongly attracted to the negative end of the water molecules. When the conditions are right, this attraction exceeds any attraction between the ions themselves and causes dissociation. For an ionic compound like NaCl, the process of solution occurs because the solvent-solute attractions between water and the ions are greater than the lattice energy of NaCl.

Figure 108. Sepaation of the solvent and solute particles is endothermic. Then, three cases exist for ∆H3: ∆H3 < ∆H1 + ∆H2 (endothermic solution process); ∆H3 = ∆H1 + ∆H2 (no enthalpy change for the solution process); or ∆H3 > ∆H1 + ∆H2 (exothermic solution process). 245

The energy changes associated with solution explain why nonpolar substances do not readily dissolve in water or other polar substances. In order to dissolve, the nonpolar molecules would need to exert a greater attraction on the water molecules than the water ter molecules exert on one another – this is not likely in light of the intermolecular attractions that exist between nonpolar molecules. For similar reasons, it is not likely that the nonpolar molecules in octane are likely to exert a greater attraction for or the ions in NaCl than the ions exert for one another; thus, ionic compounds are not likely to dissolve in nonpolar liquids.

The dissolutions discussed here are physical processes that do not cause new substances to form. Additional dissolutions, like that of nickel in acid, are chemical processes that change the substances.

Figure 109. The text describes the three stages of solution formation shown here. that an entropy consideration is important in determining spontaneity.

Part 11: SATURATED SOLUTIONS AND SOLUBILITY

Solutions may be unsaturated, saturated or supersaturated.. Only a certain amount of solute can be dissolved in a given amount of solvent. When this limit is reached, no additional solute can be dissolved in the solution. The amount of solute that can be dissolved is related to many factors, including the size of the solvent and solute particles, the attractions between the particles par and the temperature of the solution. A solution that holds less than the maximum amount of solute it can is said to be unsaturated. A mass of about 36 grams of NaCl can be dissolved in 100 mL of water at 273 K, which makes a solution with a molarity of about 6.16M. A NaCl solution containing less than 36 grams per 100 mL is unsaturate unsaturated. A solution that holds a mass of solute equal to the maximum it can hold is said to be saturated saturated. Saturated solutions are in a state of dynamic equilibrium, where:

solute + solvent ⇋ solution. crystallization

dissolution

The rates of dissolution and crystallization at saturation are equal. The amount of solute required to form a saturated solution solut is called the solubility of the solute. The solubility of NaCl is about 36 grams per 100 mL water. We could use the value of Ksp to determine the characteristic saturation of compounds at specific temperatures. Supersaturated solutions can be formed by creating an environment at which the solute has a greater solubility (e.g., ( raising the temperature), and then causing a change iin n the environment that decreases the solubility of the solute (e.g., ( lowering the temperature). Supersaturated solutions are very unstable, and the addition of even a very small crystal of the solute will cause the crystallization of the excess solute.

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Part 12: FACTORS AFFECTING SOLUBILITY

Solute-solvent interactions

As the solvent and solute exhibit greater attraction for one another – like two polar substances – the solubility of the solute increase. For example, a molecule that is polar would exhibit a greater solubility in water than would hexane, a nonpolar hydrocarbon.

For molecules with nonpolar and polar regions (like the alcohols), solubility in polar solvents (like water) decreases as the alcohol’s nonpolar region grows larger. A small alcohol, like CH3OH, is infinitely miscible in water, while the large alcohol hexanol, C6H13OH, is only soluble to 0.0058 mol / 100 g water.

Nonpolar gases, including oxygen and carbon dioxide, exhibit low solubilities in water. However, if chemical reaction occurs – as is the case with chlorine gas – the solubility of the gas increases, often on the order of 102 – 103 times larger (according to Le Châtelier’s Principle and concepts of equilibrium).

Pressure

Although limited to discussing the solubility of gas in water, an increase in pressure generally increases solubility. This can easily be related to the equilibrium established as the pressure increases: greater pressure increases the number of solute particles entering the solution.

Figure 110. At the lower pressure of (a), fewer molecules of gas will enter the liquid phase. As the pressure is increased (b), more gas molecules will be dissolved

The relationship between the solubility of a gas and the pressure of the solution

in solution. Thus, soda is bottled under pressure to increase the CO2 content – as

can be given by Henry’s Law:

soon as it is opened to 1 atm of pressure, Sg = kPg

the CO2 solubility decreases.

where Sg is the solubility of the gas (usually in molarity), Pg is the partial pressure of the gas over the solution and k is a unique constant for each solution called Henry’s Law constant.

An application of Henry’s Law is the bottling of soda pop under great pressure: More carbon dioxide gas can be held in the solution under about 4 atm pressure than can be held at atmospheric pressure. The solubility of CO2 drops from about 0.12 mol per liter at 4 atm to about 0.000093 mol per liter after opened to 1 atm. Thus, pop goes “flat” after being opened (see also the temperature discussion). See Figure 110.

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Because as the temperatures of solutions increase the kinetic energy of particles increases, more opportunities for the breaking and reforming of intermolecular attractions exist. As such, the solubility of solids generally increases with increasing temperature.

Because as the temperatures of solutions increase the kinetic energy of particles increases, gases’ motion causes molecules to be “pushed” out of solution. As such, the solubility of gases generally decreases with increasing temperature.

Figure 111. As intermolecular forces are broken and reform in warmer solution, the solubility of solids generally increases as temperature increases. The opposite trend is generally observed for gases.

Part 13: COLLIGATIVE PROPERTIES

The presence of a solute in a solvent often imparts new physical properties to the mixture that the solvent does not have when pure. For example, pure water freezes at 0°C, while its freezing point can be made as low as –10°C to –20°C by adding diethylene glycol, which is the chemical in antifreeze. Water’s boiling point is also brought much higher upon the addition of antifreeze, often to over 150°C, which allows the mixture to keep the engine cool even at very high temperatures. This behavior is characteristic of colligative properties, which are properties affected by the concentration of solute particles within a solvent; the identity of the solute particles does not, however, affect colligative properties. For example, adding a quantity of any solute would affect the boiling point of water.

There are two new units of concentration that we use in the determination of the values of colligative properties for solutions.

The molal concentration (molality), m, of a solution is the quotient of the mol solute divided by the kilogram of solvent. It is similar in design to the molarity of a solution, except that the summative volume is not used, but instead the mass of the solvent is important. Notice that the density or mass of the solvent must be known to determine molality. ࢓࢕࢒ࢇ࢒࢏࢚࢟, ࢓ =

࢓࢕࢒ ࢙࢕࢒࢛࢚ࢋ ࢑࢏࢒࢕ࢍ࢘ࢇ࢓ ࢙࢕࢒࢜ࢋ࢔࢚ 248

The mole fraction, X, is the ratio of the mole of a component of interest to the total mole present in a solution. Because it is a ratio, a mole fraction has not units. ࢓࢕࢒ ࢌ࢘ࢇࢉ࢚࢏࢕࢔, ࢄ =

࢓࢕࢒ ࢉ࢕࢓࢖࢕࢔ࢋ࢔࢚ ࢚࢕࢚ࢇ࢒ ࢓࢕࢒ ࢕ࢌ ࢇ࢒࢒ ࢉ࢕࢓࢖࢕࢔ࢋ࢔࢚࢙

Practice 9.9 Calculate the molality and molarity of a solution made of 4.35 g C6H12O6 in 25.0 mL of water. Assume the volume of the solution is that of the solvent.

Calculate the mol fraction of glucose for the solution.

We will consider four colligative properties: boiling point elevation, freezing-point depression, vapor pressure lowering and osmotic pressure.

VAPOR PRESSURE LOWERING: The vapor pressure of a solution is lower than the vapor pressure exhibited by the pure solvent.

For nonelectrolyte solutions, we can easily provide the equation for Raoult’s Law, which establishes the relationship between the vapor pressure of a pure solvent, the mole fraction of a solute and the vapor pressure of the solution. According to Raoult’s law, the addition of a nonvolatile, nonelectrolyte solute lowers the vapor pressure of a solution compared to the vapor pressure of the solvent alone.

PA = XAP°A

PA is the vapor pressure of the solution

XA is the mole fraction of solvent

P°A is the vapor pressure of the pure solvent

Figure 112. In the pure solvent (a), the vapor pressure has a greater value than in a solution containing a nonvolatile nonelectrolyte (b). 249

The vapor pressure lowering can be attributed to the enhanced intermolecular forces that exist in the solution compared to the solvent alone. In order to form a solution, the solute-solvent particles must have an attraction. This increased attraction causes a greater “pull” downward for the molecules within the mixture, resulting in a lower vapor pressure.

Solutions that follow Raoult’s law are termed ideal solutions. Ideal solutions are more closely approximated when the solventsolute interactions are similar, the concentration is low and molecular size is small. As significant intermolecular forces dominate (such as very weak or very strong), solutions begin to behave less ideal. It is important to note that if the solute is volatile or an electrolyte, then Raoult’s law does not adequately address the behavior of the solution. We shall always only consider nonvolatile and nonelectrolyte solutes when considering the applications of vapor pressure.

Practice 9.10

Determine the vapor pressure of a solution of 50.0 mL glycerin and 500.0 mL water. The density of glycerin is 1.26 g/mL. Water has a vapor pressure of 23.8 Torr at 25.0°C.

Which solution should behave most ideally: benzene dissolved in water or methanol dissolved in water? Explain.

BOILING POINT ELEVATION: the boiling point of a solution is higher than the boiling point of the pure solvent.

As discussed in the introduction, the temperature at which a solvent boils increases when solute particles are dissolved in the solvent. Water-antifreeze mixtures are probably the most common example. Because boiling is related to the vapor pressure above a substance’s liquid – and vapor pressure is lowered for solutions versus the pure solvent (that is, more energy is required to cause the vapor pressure to exceed atmospheric pressure) – the temperature at which a solution boils is always higher than the pure solvent alone according to:

∆ࢀ࢈ = ࢏ࡷ࢈ ࢓

∆Tb is the change in boiling point (note that this is the change in boiling point, not the new boiling point)

Kb is the molal boiling-point-elevation constant, which is unique for each solvent; expressed in °C/m. (Distinguish this from the base dissociation constant Kb.)

m is the molal concentration of the solute

i is the van’t Hoff factor of the solute 250

The van’t Hoff factor is the number of discrete particles into which a solute dissociates in solution. For a nonelectrolyte, the van’t Hoff factor is one, while for an electrolyte such as the soluble salt NaCl the van’t Hoff factor is two5. For weak electrolytes (weak acids and weak bases), the van’t Hoff factor is between one and two. The colligative properties are based upon the number of solute particles, so the van’t Hoff factor is important in determining the values of boiling point elevation and freezing point depression. (We only take under consideration nonelectrolytes when applying vapor pressure lowering.) FREEZING POINT DEPRESSION: The freezing point of a solution is lower than the freezing point of the pure solvent. As discussed in the introduction, the temperature at which a solvent freezes decreases when solute particles are dissolved in the solvent. Water-antifreeze mixtures are probably the most common example. Another common example is seen when salt or other ionic compounds are put on roadways to decrease the freezing point of snow and ice on the road. When the solute particles dissolve in the ice and snow the freezing point of water decreases, which prevents the water-salt solution from re-freezing – the temperature at which a solution freezes is always lower than the pure solvent alone according to:

∆ࢀࢌ = ࢏ࡷࢌ ࢓

∆Tf is the change in freezing point (note that this is the change in freezing point, not the new freezing point)

Kf is the molal freezing-point-depression constant, which is unique for each solvent; expressed in °C/m.

m is the molal concentration of the solute

i is the van’t Hoff factor of the solute

Practice 9.11 Rank the following in order of their increasing van’t Hoff factor: CaCl2, NaCl, HCl, HC2H3O2, C12H22O11

A 4.025 mol sample of ethylene glycol is added to 750 g of pure water to be added to a radiator. What is the boiling point of this new mixture compared to that of pure water? The value of Kb for water is 0.51°C/m.

What mass of ethylene glycol is required to increase the boiling point of the water in the radiator to 120.0°C?

5

Because of intermolecular attractions, the effective van’t Hoff factor is never realized. That is, the dissociation of NaCl may be taken to be

100%, but the ions are strongly attracted within aqueous solution so as to not all behave as discrete particles. This does not affect our predictive abilities, and neither does it significantly affect calculated results.

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Arrange the following in order of predicted freezing points, with lowest freezing point first – all are aqueous solutions. 0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m HC2H3O2, 0.10 m C12H22O11

An interesting and often-used aspect of colligative properties is their utility in determining molar mass – much in the same manner as that of volumetric analysis. Use the freezing point depression or another property to determine the molality of the solution, and then convert this into mol solute to determine its molar mass. A 0.5246 gram sample of a soluble nonelectrolyte was dissolved in 10.0012 g of lauric acid, for which the value of Kf is 3.90°C/m. The freezing point of lauric acid was lowered by 1.68°C. Determine the molar mass of the nonelectrolyte.

Osmosis occurs between two solutions separated by a semipermeable membrane, which is a membrane through which some molecules can and others cannot. For two solutions on either side of a membrane, solvent particles through in both directions, but always in a net direction across the membrane to the more concentrated side. For example, a primary method of transport in the human body is the through osmosis, which does not require energy expenditure (active transport does). The osmotic pressure, π , is the pressure required to stop osmosis from occurring, and is described in Figure 113 at left in some detail. Two solutions having the same osmotic pressure are isotonic to one another. When a solution Figure 113. Osmosis occurs at the semipermeable membrane as water es from the region of lower solute concentration outside the immersed chamber into the region of greater solute concentration inside the immersed chamber. As the solution level rises in the tube, the pressure at the membrane increases. Eventually, the pressure reaches a point where water can no longer enter the chamber even though a concentration gradient still exists. This is the osmotic pressure of the solution.

of higher solute concentration is separated from one with lower solute concentration, the solution of higher concentration is said to be hypertonic to the other – the solution of lower solute concentration is termed hypotonic to the first. 252

OSMOTIC PRESSURE: The osmotic pressure of a solution is affected by the molar concentration of solute particles.

π = ࡹࡾࢀ ‫ܚܗ‬

࢔ࡾࢀ ࢂ

π is the osmotic pressure of the solution

V is the total volume of solution; n is the number of mol solute; R is the ideal gas constant 0.0821 L atm / mol K; T is the kelvin temperature of the solution; M is the molarity of the solution

Practice 9.12

The average osmotic pressure of blood is 7.70 atm at 25.0°C. What concentration of glucose in an IV bag will be isotonic with blood?

How many grams of glucose should be dissolved in each 250. mL IV bag?

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Part 14: THE GAS LAWS AND GAS BEHAVIOR

You will recall from first-year chemistry that there are several equations that can quantitatively explain the behavior of gases under a variety of conditions of pressure, volume and temperature. We shall see the equations and the effects of changing the conditions under which gases are held on the behavior of gases in this section.

BOYLE’S LAW

Robert Boyle (1627 – 1691)

The pressure exerted on a gas is inversely proportional to the volume of a gas: when the temperature and amount of a sample of gas is held constant, then a change in the pressure exerted on the gas causes a change in the volume of the gas; V ∝ 1/P. The relationship is inverse, which means that:

as the pressure exerted on a gas is increased the volume of the gas decreases, and

as the pressure exerted on a gas is decreased the volume of the gas increases

This relationship is called Boyle’s Law, and it can be represented mathematically as:

P1V1 = P2V2 Figure 114. Notice the inverse proportion provided by Boyle’s law when the temperature and amount of gas is held constant.

In Figure 114, you can see that as the pressure exerted on the gas is decreased, the volume increases. Moreover, as the pressure is halved and then halved again (from 4.0 atm to 2.0

atm to 1.0 atm) the volume increases by the same factor (from V to 2V to 4V). It is convenient to learn how to recognize the changes that will occur without performing calculations.

Practice 9.13 A weather balloon expands to 568 L when the pressure on the balloon is decreased as it rises into the atmosphere. If the pressure at launch on the ground is 100.4 kPa, what is the volume of the balloon prior to launch? The balloon will settle at an altitude where the pressure is 30.5 kPa. Neglect any changes in temperature.

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CHARLES’S LAW

Jacques Charles (1746 – 1823)

The volume of a gas is directly proportional to the temperature of a gas: when the pressure and amount of a sample of gas is held constant, then a change in the temperature of a gas causes a change in the volume of the gas; V ∝ T. The relationship is direct, which means that:

as the temperature of a gas is increased the volume of the gas increases

as the temperature of a gas is decreased the volume of the gas decreases

When considering the kelvin temperature scale, the relationship is 1:1 – a doubling of the temperature results in the doubling of the volume, for example. This relationship is called Charles’s Law, and it can be represented mathematically as:

Figure 114. A direct relationship exists between the volume of a gas and its temperature. When plotted against kelvin temperature, the relationship is 1:1, indicated by the dashed lines.

ࢂ૚ ࢂ૛ = ࢀ૚ ࢀ૛ Notice that the graph of Charles’s Law is a straight line in that the relationship is direct between Kelvin temperature and volume. This provides an interesting discussion that the volume of a gas is “zero” at absolute zero, or 0 K. It is very important to recall that all calculations with gases must be performed using the Kelvin temperature scale, which is rooted in the absolute molecular motion of particles of matter. The Celsius and Fahrenheit scales are arbitrary scales based upon the freezing and boiling points of water.

Practice 9.14 A cylinder of gas is marked “Contents Under Pressure – Do Not Heat or Incinerate.” Use Charles’s Law to explain why this warning is on the container.

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COMBINED GAS LAW

The gas laws discussed above can be combined into a single equation, which can then be used for various situations that require the description of gas behavior at certain conditions:

ࡼ૚ ࢂ૚ ࢀ૛ = ࡼ૛ ࢂ૛ ࢀ૚ Practice 9.15 Determine the pressure of a neon light that contains a 1.45 L volume of gas at STP when the light is turned on and the temperature rises to just over 327°C.

AVOGADRO’S LAW

Amadeo

Avogadro (1776 – 1856)

The volume of a gas is directly proportional to the number of mol of gas present: when the pressure and temperature of a sample of gas are held constant, then a change in the amount of a gas causes a change in the volume of the gas; V ∝ n. The relationship is direct, which means that:

as the amount of a gas is increased the

Figure 115. For any ideal gas, the volume of one mol is 22.4 L at 1

volume of the gas increases

atm of pressure at 0°C – the identity of the gas is not important.

as the amount of a gas is decreased the

However, we will see that not all gases behave ideally, and that some

volume of the gas decreases

gases are more ideal than others.

This relationship is called Avogadro’s Law, and it can be represented mathematically as:

ࢂ૚ ࢂ૛ = ࢔૚ ࢔૛ Familiar statements resulting from Avogadro’s Law include:

Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

The standard molar volume of an ideal gas is 22.4 L at STP, which is 273 K and 101.325 kPa

Notice that neither of the preceding statements includes a discussion of the identity of the gas – only the number of mol is referenced.

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Figure 116. Variations in Avogadro’s law can be seen in this graphical representation. However, the variations are small enough that we neglect them in common calculations. We do note the explanations for these variations later.

GAY-LUSSAC’S LAW

We have many times over used the concept associated with Gay-Lussac’s law without asg a name to it. Gay-Lussac’s law states that at a given temperature and pressure volumes of gases combine in whole-number ratios.

Figure 117. According to Gay-Lussac’s law – really, a variation of Avogadro’s law – two volumes of hydrogen and one volume of oxygen form two volumes of steam because 2 H2 + O2 → 2 H2O

This was the forerunner, really, of Avogadro’s law, and it can be useful when chemical reactions occur at constant temperature and pressure: the resulting volume of gas produced from gas reactants is proportional to the stoichiometry of the chemical equation.

IDEAL GAS EQUATION

Gases that behave exactly as we would expect according to the relationships discussed above are called ideal gases, while gases that deviate from this behavior are called real gases. Gases that behave ideally can be described according to the ideal gas equation, which is derived from the laws discussed above:

Boyle’s Law:

V ∝ 1/P

or V = 1/P • k

Charles’s Law:

V∝T

or V = T • k

Avogadro’s Law:

V∝n

or V = n • k

This gives: V = knT / P or PV = knT. Using R to represent the constant k gives:

PV = nRT

R is the ideal gas constant – use the appropriate constant based on the units of pressure

P = kPa: R = 8.314 kPa L / mol K; or, when P = atm: R = 0.0821 atm L / mol K

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Practice 9.16 The value of the gas constant R is not a secret. Simply use the value of standard temperature, standard pressure in the desired unit, one mol of gas and 22.414 L for T, P, n and V to solve for any desired R value corresponding to a particular pressure. For example, the standard pressure in mm Hg is 760 mm Hg. Determine the value of R when pressure is given in mm Hg.

On each of the cylinders below, sketch the relative position at which the piston would appear at the conditions below each cylinder.

VARIATIONS OF THE IDEAL GAS EQUATION

Because the number of mol, n, is equal to the mass of a sample divided by its molar mass, we can easily see how this expression can be substituted into the ideal gas equation to yield:

ࡼࢂ =

ሺ࢓ࢇ࢙࢙ሻࡾࢀ ࡹ

where m is the mass of a sample of gas and M is the molar mass of the gas. Similarly, the expression D = mass √ volume can be substituted for mass/V in the above variation of the ideal gas law:

ࡼ=

ሺ࢓ࢇ࢙࢙ሻࡾࢀ ࡰࡾࢀ = ࢂࡹ ࡹ

Using either of these variations allows us to determine the molar mass, sample size or density of a gas when other information is known. As with other calculations, watch that units match when solving!

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Practice 9.17 A sample of an unknown gas occupies a volume of 3.52 L at STP. The density of the gas is found to be 1.78 g/L. What is the identity of the elemental gas?

Be careful to consider the diatomic nature of some gases when performing calculations such as these. What mass of nitrogen gas occupies a volume of 3.55 L at STP?

DALTON’S LAW OF PARTIAL PRESSURES

The total pressure exerted by a sample of gas is equal to the sum of the individual pressures of the various gases that make up the sample. This should make sense to you because the ideal gas equation does not address the composition of any gas; rather it only includes a term for the number of moles of gas. The contribution of each gas is based upon its mol value, as we should expect from the ideal gas equation: P1 = n1 (RT/V) → P1/n1 = (RT/V) P2 = n2 (RT/V) → P2/n2 = (RT/V) P3 = n3 (RT/V) → P3/n3 = (RT/V)

The expressions above show that the pressure

Figure 118. For the gases separately (a) and (b), the pressures can be

exerted by each gas is due to the number of mol of

obtained from the ideal gas equation. When added to one another (c), the

each gas. Thus, the sum of the individual pressures

total pressure is equal to the gases’ individual pressures.

would be the total pressure exerted by all of the gases collectively: ࡾࢀ

ࡼ࢚࢕࢚ࢇ࢒ = ሺ࢔૚ + ࢔૛ + ࢔૜ + ⋯ ሻ ቀ ࢂ ቁ = ࡼ૚ + ࡼ૛ + ࡼ૜ + ⋯

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Practice 9.18 A sample containing 0.32 mol of N2, 0.14 mol O2, 0.04 mol Ar and 0.33 mol Ne is held in a container with a volume of 2.0 L. At a temperature of 200.°C, C, what is the pressure exerted by each gas, and what is the total pressure exerted?

A 40.00 L container has a pressure of 325.6 kPa at a temperatur temperature e of 800 K. The container holds 4 gases: 35.5 g each of the first 3 halogens, and an unknown mass of the last halogen, I2. What mass of I2 is present in the container?

If the container above has a rated burst pressure of 12.00 atm, at what temperature will it burst?

A 4.98 g sample of ammonia gas was placed in a 1.00 L container at 500. K where it decom decomposed posed into its product gases. What mass of each species is present at some later time when the total pressure in the container was measured m as 14.0 atm?

Assuming that the pipes of the assembly below have essentially zero volume, what is the total pressure in the container when both valves are opened and the gas allowed to move throughout the assembly?

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MOLE FRACTIONS

We have already used mol fractions, and we revisit them here to see that they are convenient to use with gas behavior problems because the contribution to a total pressure of a gas system can be easily determined.

Practice 9.19 The amount of nitrogen in a gas mixture is 3.2 mol. If the total sample of the gas mixture exerts 129.6 kPa and contains 14.5 mol gas, what is the mole fraction of nitrogen and its contribution to the total pressure?

KINETIC MOLECULAR THEORY OF GASES

Rudolf Clausius (1822 – 1888)

When we discuss the behavior of gases, we make several assumptions about their behavior:

Gas particles are in constant, random motion. Attractive forces and repulsive forces between gas particles are

negligible. The average kinetic energy of a gas sample does not change with

time and collisions so long as the temperature of the sample does not change – i.e., the collisions are perfectly elastic, which means kinetic

Figure 119. The temperature of a sample of gas is

energy is conserved.

directly proportional to the average kinetic energy of a

The average kinetic energy of a sample of gas is directly proportional

to the absolute (kelvin) temperature of the sample. The volume occupied by the gas particles in a sample of gas is

essentially zero relative to the volume of the gas sample.

sample: ½mv2, µ, or root-mean-square speed, rms. However, this does not mean that all of the particles have the same speed. A fraction will have a greater speed, while a fraction will have a lesser speed.

As suggested by the kinetic molecular theory, the behavior of gases can be related to the average kinetic energy of the molecules of gases, which is measured as the temperature of the sample. Additionally, the average speed of the molecules of a gas sample is also related to the average kinetic energy of the molecules, although many molecules will be moving more slowly than the average speed, and many molecules will be moving more quickly than the average speed (Figure 119).

261

An important note to make using Figure 119 is the root-mean-square-speed, or rms. This is the speed that a gas particle possessing the average kinetic energy will have according to ½mv2. We note that the rms speed is greater for a sample at a greater temperature, and for two samples at the same temperature, the rms speed of the heavier molecules is lower (Figure 120). Figure 120. The average speed of molecules at the same temperature is inversely proportional to the molecular mass of the molecules of gas.

We will use this information in discussing effusion and diffusion later.

Practice 9.20 Consider the following statements and explain them in of molecular motion and the kinetic molecular theory of gases.

The pressure of a gas sample decreases when its temperature is decreased at constant volume

The pressure of a gas sample decreases when its volume is increased at constant temperature

DEVIATIONS FROM IDEAL BEHAVIOR

The behavior of gases cannot be exactly determined from the ideal gas equation for most gases – the assumptions of the kinetic molecular theory are valid only for real gases, which most gases are not. An ideal gas is one that most closely behaves as the kinetic molecular theory suggests, while a real gas is one that exhibits characteristics that cause behavior that deviates from what kinetic molecular theory predicts.

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Practice 9.21 What would contribute to the real behavior of a gas? That is, what characteristics would make a gas behave less ideally?

Would a gas be more ideal or real at high temperatures? Explain briefly.

Would a gas be more ideal or real at low pressures? Explain briefly.

Consider two samples of a gas: one is held under very low temperatures in a small volume at high pressure. The other is held under a very high temperature in a large volume at low pressure. From which gas could we predict more ideal behavior? Explain briefly.

Figure 121. When a gas exhibits ideal behavior, the value PV/RT will equal one. These figures may help address the items on this page.

263

VAN DER WAALS EQUATION OF STATE

Johannes van der Waals (1837 – 1923)

The van der Waals equation is an equation that can be used to correct for some of the item we have discussed that make gases behave less ideally. Specifically, you can use the van der Waals equation to correct for molecular volume and molecular attraction.

ቆࡼ +

࢔૛ ࢇ ቇ ሺࢂ − ࢔࢈ሻ = ࢔ࡾࢀ ࢂ૛

The constants a and b are experimentally determined for gases. You will not likely be asked to use the van der Waals equation to determine a specific value, but you will be expected to predict the effect on behavior when values for a and b are provided. The constants a and b generally increase as molecules exhibit greater intermolecular attractions and become larger, respectively.

Practice 9.22 The constant a corrects for variations in ideal behavior due to attractions. How does the value of a for water gas compare to the value of a for carbon dioxide gas? Explain.

The constant b corrects for variations in ideal behavior due to volume. How does the value of b for water gas compare to the value of b for carbon dioxide gas? Explain.

DIFFUSION AND EFFUSION

Diffusion here represents the movement of gas particles through another substance, such as air. The idea is quite simple: lighter molecules move faster, and heavier molecules move more slowly. We can determine the rate of diffusion – really, the rms speed – using

ࣆ࢘࢓࢙

૜ࡾࢀ =ඨ ࡹ

Figure 122. Diffusion (a) and effusion (b) illustrated. Diffusion occurs where perfume fills the air in a room; effusion is exemplified by a balloon that goes flat. 264

Effusion is the movement of molecules through a small hole in a container. Graham’s Law of Effusion tells us that the ratio of the rate of effusion of Gas 1 to the rate of effusion of Gas 2 is equal to the square root of the ratio of the molar mass of Gas 2 to the molar mass of Gas 1 – again, heavier gases move more slowly. Graham’s Law can be used to determine the molar mass of a gas sample when its rate of effusion compared to another gas is determined.

࢘ࢇ࢚ࢋ૚ ࡹ૛ =ඨ ࢘ࢇ࢚ࢋ૛ ࡹ૚ Practice 9.23 Calculate the rms speed of nitrogen molecules, N2(g) at 298 K. (Like other thermodynamic calculations, the value of R requires joules; and, because the unit joule is defined in of the kilogram, we need to express molar mass in of kg mol-1.)

A gas effuses at a rate 2.4 times that of nitrogen. What is the molar mass of the gas?

COLLECTING GAS OVER WATER

For any substance, there is an amount of pressure exerted by the substance above the surface of the substance owing to a number of molecules that escape against the pressure of the atmosphere. This pressure is quite significant for liquids, and it is less significant for solids. A common lab practice is to collect gas over water to determine the quantity produced, and if we collect a gas by displacing water, then the total pressure of the gas above the

Figure 123. When we collect gas over water (a), the gas in the collection volume consists of both the gas made during the reaction and the water gas present due to the vapor pressure of water (b). Thus, to determine the mol of gas collected (using the ideal gas equation), we must equalize the pressure inside the collection container so that it equals Patm and subtract the vapor pressure of water from this value.

water will be the sum of the vapor pressure of the water and the pressure of the gas. In order to determine the number of mol of gas that have been collected, then, we must correct for the pressure of water vapor in our sample. You can use a data table of vapor pressures of water at specific temperatures and atmospheric pressures to determine the correction for PH2O. 265

ࡼ࢚࢕࢚ࢇ࢒ = ࡼࢍࢇ࢙ + ࡼ࢝ࢇ࢚ࢋ࢘

Practice 9.24

Imagine that you have collected a sample of oxygen gas over water by displacing the water. You collected 0.250 L of gas over the water at 26°C at sea level. How many mol of oxygen was collected? How many grams of potassium chlorate must be decomposed to collect this much oxygen?

The End

266


Descriptive Chemistry

267

INTRODUCTION

A major section of the Advanced Placement Chemistry exam is the reaction section of the Free-Response session – the notorious Question 4. Provided three descriptions of chemical reactions, Question 4 asks you to write chemical formulas for the given reactants and the products of the reaction. In addition, a follow-up question is asked that requires fundamental knowledge of chemical processes. The instructions are provided here (College Board, Chemistry Course Description): 4. For each of the following three reactions, in part (i) write a balanced equation for the reaction and in part (ii) answer the question about the reaction. In part (i), coefficients should be in of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work, but only equations that are written in the answer boxes provided will be graded.

The follow-up question surrounds applications of the reaction, the chemistry of the reaction or myriad other topics related to the chemical reaction. Points will be awarded for the correct formulas of the reactants, correct formulas for products, and correct answers to the follow-up question. Note that you can earn points for any one part of the item without earning points (or even answering) other parts of the item. For example, you might know that the products of a hydrocarbon combustion are water and carbon dioxide, but you may not know the formula for the reactant 2,4-dimethyl hexane.

The reaction limits for Question 4 are those commonly discussed in a first-year college course laboratory or in the context of the curriculum – as in oxidation reduction reactions, simple organic reactions, single- and double-replacement reactions, acid-base reactions (including hydrolysis), synthesis reactions and decomposition reactions. Notable inclusions are those reactions of copper with the oxidizing acid nitric acid, and the reactions that result in decomposing products like carbonic acid, sulfurous acid and ammonium hydroxide. You will not be expected to know more than cursory organic chemistry reactions, and you will not typically be required to know complex ion formation.

Reactions must be written in net-ionic form

Reactions must be balanced as to mass and charge

Reactions do not have to be represented with phase symbols

Reactions always occur – there are no ‘no reaction’ options

You will have a table of reduction potentials available to you on the exam.

For clarity, the reactions on the following pages are not balanced, and the phases of the species are not shown unless it is germane to the reaction. Moreover, the net ionic equations are not shown so as to present a more complete picture from which you can learn.

268

SYNTHESIS REACTIONS

Inorganic reactions classified as synthesis reactions are those that produce a single product from two or more reactants.

Metals and non-oxygen nonmetals form salts: Na + Cl2 → NaCl Fe + Br2 → FeBr2 (limited bromine) Fe + Br2 → FeBr3 (excess bromine) Metals and oxygen form metal oxides: Li + O2 → Li2O

for Group 1 metals and silver

Fe + O2 → FeO (limited oxygen) Fe + O2 → Fe2O3 (excess oxygen)

Use the common oxidation numbers of 1+

Use the common oxidation numbers of 2+ for Group 2 metals and zinc

Nonmetals and oxygen form nonmetal oxides:

aluminum

N2 + O2 → NxOy C + O2 → CO (limited oxygen) Reactions of

Increase the oxidation number of a transition metal if the nonmetal is present

C + O2 → CO2 (excess oxygen)

in excess

Elements and

Use the common oxidation number of 3+ for

Nonmetals form molecular compounds:

Increase the oxidation number of a less

Reactions

P + Cl2 → PCl3 (limited chlorine)

electronegative nonmetal (if possible) with

of Oxides

P + Cl2 → PCl5 (excess chlorine)

halogens or oxygen if the halogen or oxygen is present in excess

[including water] Soluble metal oxides and water form hydroxides:

under the hydrolysis of salts section

Na2O + H2O → NaOH CaO + H2O → Ca(OH)2

Other reactions with water are shown

Note that reactions of water are not the same as reactions in water

Nonmetal oxides and water form ternary acids:

should be written dissociated

N2O5 + H2O → HNO3 CO2 + H2O → H2CO3

Most products formed here are soluble and

The oxidation states of the elements do not change in the formation of acids or salts

Nonmetal oxides and metal oxides form salts: CaO + SO3 → CaSO4 N2O5 + Na2O → NaNO3

269

DECOMPOSITION REACTIONS

Decomposition reactions are suggested by the presence of a single reactant, which will decompose into elements or compounds.

Binary compounds decompose into their elements: NaCl → Na + Cl2 HgO → Hg + O2 Carbonates decompose into metal oxides and carbon dioxide: MgCO3 → MgO + CO2 Chlorates decompose into metal chlorides and oxygen gas: Mg(ClO3)2 → MgCl2 + O2

Group 1 carbonates do not decompose into metal oxides and carbon dioxide

Alkali and alkaline nitrates decompose into nitrites and

hydroxides (i.e., strong bases) –

oxygen gas:

dissociate rather than decompose

Mg(NO3)2 → Mg(NO2)2 + O2 General Decomposition

All hydroxides here are solid – aqueous

Ammonium salts may also follow

Metal hydroxides may decompose into metal oxides and

various other schemes – these will be

water:

addressed individually Ca(OH)2 → CaO + H2O

Reactions

Most decomposition reactions require heat or light

Some ammonium salts decompose to lose ammonia: (NH4)2SO4 → NH3 + H2SO4

Electrolysis can be used to decompose molten salts (i.e., melted), but electrolysis in aqueous solution

Ammonium salts containing nitrate or dichromate decompose to produce an oxide, water and nitrogen gas –

follows other patterns of reactivity, which are discussed on Page 276.

the N or Cr undergoes reduction: (NH4)2Cr2O7 → Cr2O3 + H2O + N2 Water and hydrogen peroxide decompose as shown: H2O2 → H2O + O2 H2O → H2 + O2 Ternary acids decompose into nonmetal oxides and water: H2CO3 → CO2 + H2O

270

SINGLE-REPLACEMENT REACTIONS

Single-replacement reactions are a form of redox reactions in which a more active metal replaces a less active metal or hydrogen or a more active nonmetal replaces a less active nonmetal.

More-active metals replace less-active metals in aqueous solution: Mg + FeCl3 → Fe + MgCl2 Pb + Cu(NO3)2 → Pb(NO3)2 + Cu The active metals Li, K, Na and Ca replace hydrogen from cold water, steam and non-oxidizing acids and

At the Advanced Placement level, one should use reduction potentials to

release hydrogen gas:

determine the occurrence of single-

Li + H2O → LiOH + H2

replacement reactions, but do not lose sight General Single-

The active metals Mg, Al, Mn, Zn, Cr and Fe replace

of periodic table location to determine

Replacement

hydrogen from steam and non-oxidizing acids and

activity

Reactions

release hydrogen gas:

For transition metals, select a common oxidation state when forming the new

Al + H2O → Al(OH)3 + H2

compound The active metals Cd, Co, Ni, Sn and Pb replace hydrogen from non-oxidizing acids and release

The common oxidizing acid is HNO3, which does not participate as shown

hydrogen gas: Ni + HBr → NiBr2 + H2 Active nonmetals replace less-active nonmetals from their compounds in aqueous solution: Cl2 + KI → KCl + I2

271

DOUBLE-REPLACEMENT (METATHESIS) REACTIONS

Metathesis reactions occur owing to the removal of ions from solution. Metathesis reactions are never redox reactions. (See outline on Page 274.)

Two soluble salts’ aqueous solutions may produce an insoluble compound (a precipitate): NaCl + AgNO3 → NaNO3 + AgCl(s) A solid sulfite or carbonate react with non-oxidizing acids to produce a decomposing compound that

common exceptions to solubility to

results in the formation of a molecular oxide, water

correctly identify the precipitate. Look for

and a salt: Precipitation and

Na2SO3(s) + HCl → NaCl + H2SO3 → NaCl + H2O + SO2

Gas-forming

Na2CO3(s) + HCl → NaCl + H2CO3 → NaCl + H2O + CO2

Reactions

It will be especially important to learn the

silver, barium, lead cations, and look for phosphate, carbonate and hydroxide anions

You must learn the three decomposing compounds – H2SO3, H2CO3 and NH4OH – and the direct formation of H2S [See the

Ammonium salts react with soluble hydroxides to

note about these compounds under

form ammonia, water and a salt:

Hydrolysis Reactions on Page 273]

NH4Cl + NaOH → NaCl + NH4OH → NaCl + NH3 + H2O Sulfides react with non-oxidizing acids to form hydrogen sulfide gas and a salt: Na2S (s) + HCl → NaCl + H2S

Salts that contain the conjugate of a weak acid will react with acids to form the weak acid: Molecular

form in these reactions are water and the

Ca(CH3COO)2 + HCl → CH3COOH + CaCl2

Compoundformation Reactions

The common molecular compounds that

weak acids Acids and bases form a salt and water [see also

Watch for quantities to be given in acid base reactions with polyprotic acids – react only

below]:

the correct number of hydrogen ions

HBr + KOH → HOH + KBr

Anhydrides of acid and bases first react with water to form their respective acid or base, and then the Acid-Base Neutralization Reactions

acid or base reacts: SO2 + Ca(OH)2 → i) SO2 + H2O → H2SO3 ii) H2SO3 + Ca(OH)2 → CaSO3 + HOH

Acid anhydrides are the nonmetal oxides that form acids upon reaction with water, and the basic anhydrides are the metal oxides that form bases upon reaction with water [See Reactions of Oxides on Page 269]

272

HYDROLYSIS REACTIONS

Conjugate bases of weak acids react with water to

form the weak acid – they hydrolyze: NaCH3COO + HOH → CH3COOH + NaOH

The decomposition reactions of H2SO3, H2CO3 and NH4OH we saw upon their formation in precipitation reactions does

Hydrolysis

Ammonium ions will undergo hydrolysis to form

NOT occur during hydrolysis owing to their

ammonium hydroxide:

very low concentrations

NH4Cl + HOH → HCl + NH4OH

The metal ions of strong bases do not act as Lewis acids

Many metal cations act as Lewis acids in water to

form metal hydroxides and a strong acid:

Do not hydrolyze the strong acids’ conjugates

AlCl3 + HOH → Al(OH)3 + HCl

OUTLINE OF AQUEOUS METATHESIS REACTIONS

Step 1: Check the reactants

Are both reactants soluble ionic compounds?

Is one of the reactants a nonoxidizing acid and the other a solid carbonate, hydroxide, sulfite or sulfide?

NO

YES

STOP.

NO


YES

Is one of the potential products: Step 2:

a molecular compound (e.g., H2O, NH3, or weak acid)

Check the products

a decomposing compound

an insoluble gas

an insoluble ionic compound?

STOP.

NO


YES

Step 3: Write the net ionic equation

6.

Identify the decomposition products of NH4OH, H2CO3 or

7.

Write the equation in molecular form

8.

Write all soluble compounds, strong acids and strong bases

H2SO3, if present

Stop at 2 for the full molecular equation

in dissociated form with appropriate electrical charges 9.

Cancel all species that remain unchanged from reactantside to product-side

Stop at 3 for the full ionic equation

10. Write the reaction equation showing all species that remain

273

REDOX REACTIONS

Many of the above reactions are redox reactions, but there are also additional reactions that are arbitrarily classified here for convenience. A table of common reducing agents and oxidizing agents is given on Page 275.

MnO4- undergoes reduction: MnO4– → Mn2+ [acid solution] MnO4– → MnO2 [base or neutral solution]

This is not an exhaustive representation – these are simply some of the more common oxyanions that you will encounter

Cr2O72- undergoes reduction in acid solution: Oxyanion

Cr2O72- → Cr3+

Reactions

– one species undergoes reduction and one species undergoes

IO3- undergoes reduction in acid solution:

oxidation; look at the other species in the reaction (often a metal) and try to deduce

IO3– → I2

the chemistry in of that species C2O42- undergoes oxidation to carbon dioxide: C2O42- → CO2 ([via CO32-]

Copper with concentrated sulfuric acid: Cu + H2SO4 → CuSO4 + SO2 + H2O Atypical Copper Reactions

Other Considerations

Copper with nitric acid:

Note atypical reactions of copper: these are not ordinary single-replacement, as the

Cu + HNO3 → Cu(NO3)2 + NO + H2O [dilute acid]

products are not the salt and hydrogen gas

Cu + HNO3 → Cu(NO3)2 + NO2 + H2O [conc acid]

as in those on Page 271.

See the table on Page 276 for more information on the reduction or oxidation of individual species

Practicing these and understanding the concept of electron transfer will cause this list to become engrained in your brain!

274

COMMON OXIDIZING AND REDUCING AGENTS •

Common oxidizing agents – these species will cause a loss of electrons in another species, and they will themselves gain electrons. Oxidizing agents undergo reduction.

•

MnO4- in acid solution

→

Mn2+

MnO2 in acid solution

→

Mn2+

MnO4– in neutral or base solution

→

MnO2(s)

Cr2O72- in acidic solution

→

Cr3+

HNO3 (conc)

→

NO2

HNO3 (dilute)

→

NO

H2SO4 (hot, conc)

→

SO2

metallic ions

→

metallous ions (lower oxidation state)

free halogens

→

halide ions

Na2O2

→

NaOH

HClO4

→

Cl–

H2O2

→

H2O

S2O82-

→

SO42-

CrO42-

→

Cr3+

Common reducing agents – these species will cause a gain of electrons in another species, and they will themselves lose electrons. Additional reducing agents will be added. Reducing agents undergo oxidation.

halide ions

→

free halogens

free metals

→

metal ions

sulfite ions or SO2

→

sulfate ions

nitrite ions

→

nitrate ions

free halogens, dilute basic solution

→

hypohalite ions

free halogens, conc basic solution

→

halate ions

metallous ions

→

metallic ions (higher oxidation state)

C2O42-

→

CO2

MnO2 in base solution

→

MnO4–

We will add species to this list throughout the course.

275

ELECTROLYSIS

Electrolysis involves the age of an electrical current into a molten salt (which is a simple decomposition), or the age of electrical current into an aqueous solution of a salt (which is not simple decomposition).

Cathode reactions (reduction):

For metals that are easily reduced (i.e., lessactive metals):

M+(aq) → M(s) occurs

These are among the most difficult reactions to address

For metals that are not easily reduced (i.e.,

more-active metals)

The result of an electrolysis of an aqueous salt solution is either the reduction of water

HOH(l) → H2(g) + OH–(aq) occurs

at the cathode or the oxidation of water at the anode – or – the reduction of a metal at Electrolysis Reactions

the cathode or the oxidation of a nonmetal at the anode: the occurrence is based upon

Anode reactions (oxidation):

the battle for reduction at the cathode

For nonmetals that are easily oxidized (i.e.,

between the metal ion and water and the

less-active nonmetals)

battle at the anode for the oxidation of the

X– (aq) → X(v) occurs

anion or water

For nonmetals that are not easily oxidized (i.e., more-active nonmetals)

The symbol (v) indicates variable phase for the oxidized nonmetal

HOH(l) → O2(g) + H+(aq) occurs

NAMING COMMON COMPLEXES

Complex ions are named using the general format here. If no cation or anion is present (that is, only the complex ion is named, then drop ‘cation +’ or ‘+ anion,’ as appropriate, below)

cation + prefix for number of ligands + name of the ligand + name of the metal in complex ion (-ate ending)(Roman numeral) or prefix for number of ligands + name of the ligand + name of the metal in complex ion(Roman numeral) + anion

Examples –

[Cu(OH)4]2- is dihydroxocuprate(II)

Latin names for complex anions:

copper (cuprate), gold (aurate), silver (argentate), tin (stannate), iron (ferrate), lead (plumbate) –

Na2[Fe(CN)4] is sodium tetracyanoferrate (II)

others use element name with -ate suffix

[Zn(NH3)4](NO3)2 is tetramminezinc nitrate 276

COORDINATION CHEMISTRY

Complex ions – although more rare on the exam than other reactions – can be readily determined using a few simple rules. They are important in the qualitative analysis of ion solutions (i.e., determining the presence of ions), so we shall review them as part of descriptive chemistry.

Lewis acids react with excess cyanide ion to form cyano complexes:

Ag+(aq) + CN– (aq) → [Ag(CN)2] – (aq) Fe3+

+ CN (aq) → [Fe(CN)6 –

The common metals ions that form complex ions or coordination complexes are the Lewis acids iron ion, cobalt ion, nickel ion,

]3-(aq)

chromium ion, copper ion, zinc ion, silver Lewis acids react with excess hydroxide ion to form hydroxo complexes: Zn2+(aq)

ion and aluminum ion

+ OH (aq) → [Zn(OH)4 –

]2-(aq)

NH3, (ammine complexes), cyanide ion, CN–,

Al3+ + OH– (aq) → [Al(OH)6]3-(aq)

(cyano complexes), hydroxide ion, OH–, (hydroxo complexes), thiocyanate ion, SCN–,

Lewis acids react with excess thiocyanate ion to form thiocyanato complexes: Ag+(aq) + SCN– (aq) → [Ag(SCN)2] – (aq)

(thiocyanato complexes)

(called the coordination number)

Lewis acids react with excess ammonia ion to form Complex Ion Formation and

Ag+(aq) + NH3 (aq) → [Ag(NH3)2]+(aq)

Dissolution

Cu2+ + NH3(aq) → [Cu(NH3)4]2+(aq)

The number of ligands is best predicted as two times the charge on the metal ion

Fe3+ + SCN– (aq) → [Fe(SCN)6]3-(aq)

ammine complexes:

The most common ligands are ammonia,

Treatment of a complex ion with acid often results in its dissolution of the complex

The reactions shown at left are those that should cover about 90% of the complexes presented on the AP Exam. However, additional complexes may form with:

Solid metal hydroxides of Lewis acids react with •

halo- ligands F–, Br–, Cl–, I–

•

carbonato ligand CO32-

•

hydrido ligand H–

•

nitrato/nitrito ligands NO3–, NO2–

Addition of dilute acid to a complex typically

•

oxalato ligand [ox] C2O42-

releases the ligand as its ion and releases the metal

•

oxo ligand O2-

•

sulfato SO42-

•

thiosulfato S2O32-

•

ethylenediamine (CH2)2(NH2)2

•

methylamine CH3NH2

•

dimethylamine (CH3)2NH2

•

aquo H2O

ammonia to produce ammine complexes and hydroxide ion: Cu(OH)2(s) + NH3(aq) → [Cu(NH3)4]2+(aq) + OH– (aq)

as its ion or a precipitate if the acid contains an anion with which the metal ion is insoluble: [Cd(NH3)4]2+(aq) + HNO3 → Cd2+ + NH4+ [Ag(CN)2] – (aq) + HCl(aq) → AgCl(s) + CN–

Lewis acids and Lewis bases react to form a coordinate covalent compound: BF3 + NH3 → H3N-BF3

277

ORGANIC REACTIONS

Simple organic reactions are common on the AP Chemistry exam.

Combustion produces carbon dioxide and water, carbon monoxide and water, or carbon and water: CH4 + O2 → CO2 + H2O [excess oxygen] CH4 + O2 → CO + H2O [limited oxygen] CH4 + O2 → C + H2O [very limited oxygen] Halogenation or hydrogenation occurs in alkenes and alkynes: H2C=CH2 + Br2 → BrH2C-CH2Br H2C=CH2 + H2 → H3C-CH3 HC≡CH + Cl2 → ClHC=CHCl [limited chlorine] General Organic

Practice will be essential for the organic reactions

HC≡CH + Cl2 → Cl2HC–CHCl2 [excess chlorine]

Note the use of limited or excess reactants – the presence of a catalyst is almost always

Reactions

required Substitution reactions result in no change in bonding, only replacement of an atom to which

Additional notes will be made as needed for organic reactions

carbon is bonded; an inorganic acid forms: Cl2 + CH4 → CH3Cl + HCl [excess halogen results in di-, tri- or tetra-substitution] Esterification results when an organic acid and an alcohol react – the acid loses its –OH group and the alcohol loses the alcohol group’s hydrogen; water forms: CH3COOH + CH3OH → CH3COOCH3

278

ORGANIC NAMING

There are essentially nine organic classes with which you should be familiar. Organic compounds are those compounds of carbon (except the carbonates and carbon dioxide, which are inorganic compounds).

The basic hydrocarbons: alkanes, alkenes and alkynes •

Contain only carbon and hydrogen

•

React with oxygen to produce typical combustion products

•

Are identified by their –ane (saturated hydrocarbons – all single bonds), –ene (double bonds between carbons), or –yne (triple bonds between carbons) endings

The alcohols •

Contain the –OH function

•

React with oxygen to produce typical combustion products

•

Are identified by their –ol ending

•

React with carboxylic acids to form esters

The ethers •

Contain the –O– function

•

Are identified by their –oxy component

The carboxylic acids •

Contain the –COOH function

•

Are identified by their –oic ending

•

React with alcohols to form esters

The esters •

Contain carbon, hydrogen and the –CO–O function

•

Are identified by their –oate ending

•

Form when alcohols and carboxylic acids react in a condensation reaction

The aldehydes and ketones •

Contain the –CO function in a middle (ketone) or end (aldehyde) position

•

Are identified by their –one or -al ending

The amines and amides •

Amines: R – NH2 (ammonia derivatives)

•

Amides: R – CO – NH2 (carboxylic-amine derivatives)

•

Are identified by their amine or amide ending

Common groups appended to stems: –CH3 (methyl), –CH2CH3 (ethyl), –CH2CH2CH3 (propyl), –CH2CH2CH2CH3 (butyl)

Halogens are often added before the stem as chloro–, bromo–, fluoro– and iodo–.

279

Naming organic compounds begins with naming the basic alkanes, alkenes and alkynes. Please use the notes sections to add additional considerations as we progress. We will consider naming of alkanes, alkenes, alkynes, alcohols, carboxylic acids, esters, aldehydes and ketones. We will reserve the naming of ethers, amines and amides for later.

Naming Alkanes

Identify the longest chain of carbon atoms

Name the alkane:

n = 1 → methane

n = 6 → hexane

n = 2 → ethane

n = 7 → heptane

n = 3 → propane

n = 8 → octane

n = 4 → butane

n = 9 → nonane

n = 5 → pentane

n = 10 → decane

If there are substituents (appended organic groups of carbon and hydrogen), then:

name the longest chain and give it the name of the basic alkane

name the substituents using the group names on the bottom of Page 279

indicate the position of the substituents by numbering the carbons to give the substituents the lowest possible numbers

indicate the presence of more than one substituent of the same kind using a Greek prefix and separating the positions with commas

where more than one substituent type is present, name the substituents as above and place them in alphabetical order; use hyphens to separate the “pieces”

Naming Alkenes (double bond present)

Identify the longest chain of carbon atoms that contains a double bond, and name this using the alkane stem; give the double bond(s) the lowest numbers, and change the ending to -ene

If there are substituents, name these as in alkanes while still giving the alkene position(s) the lowest number(s)

Naming Alkynes (triple bond present)

Identify the longest chain of carbon atoms and name this using the alkane stem; give the triple bond(s) the lowest numbers, and change the ending to -yne

If there are substituents, name these as in alkanes while still giving the alkyne position(s) the lowest number(s)

Naming Alcohols

Identify the longest chain of carbon atoms containing the function, and give the carbon with the alcohol function the lowest number, and then change the stem of the corresponding alkane by dropping the –e and adding –ol

Naming Carboxylic Acids

If there are substituents, name these as before while always giving the alcohol position(s) the lowest numbers

Drop the –e from the name of the corresponding alkane and append the name with –oic acid

Naming Esters

Name the alcohol part of the compound with an –yl ending, and then name the acid part of the compound including the –COO carbon with an -oate ending.

Naming Aldehydes and Ketones

Name the longest string of carbon atoms that contains the function, and append –one or –al to the name. 280

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