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C HEMICA L P RINCIPLES THE QUEST FOR INSIGHT Sixth Edition PETER ATKINS · LORETTA JONES · LEROY LAVERMAN FFMMTTOOCC..iinndddd PPaaggee ii 1100//1188//1122 88::0077 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp SIXTH EDITION CHEMICAL PRINCIPLES T H E Q U E S T F O R I N S I G H T PETER ATKINS Oxford University LORETTA JONES University of Northern Colorado LEROY LAVERMAN University of California, Santa Barbara W. H. Freeman and Company New York FFMMTTOOCC..iinndddd PPaaggee iiii 3300//1100//1122 33::0088 PPMM uusseerr--FF440088 //UUsseerrss//uusseerr--FF440088//DDeesskkttoopp Associate Publisher: Jessica Fiorillo Library of Congress Control Number: Senior Developmental Editor: Randi Blatt Rossignol Your EPCN application for a Library of Congress control Marketing Manager: Alicia Brady number for Media and Supplements Editors: Dave Quinn and Title: “Chemical principles” Heidi Bamatter ISBN: “1429288973” Assistant Editor: Nicholas Ciani was successfully transmitted to the Library of Congress. Photo Editor: Bianca Moscatelli Senior Project Editor: Georgia Lee Hadler ISBN-13: 978-1-4292-8897-2 Full-Service Project Management: Aptara ISBN-10: 1-4292-8897-3 Cover Designer: Victoria Tomaselli International Edition International Edition ISBN-13: 978-1-4641-2467-9 Cover design: Dirk Kaufman ISBN-10: 1-4641-2467-1 Cover image: Nastco/iStockphoto.com Text Designer: Marsha Cohen © 2013, 2010, 2005, 2002 by P. W. Atkins, L. L. Jones and Illustration Coordinator: Bill Page L. E. Laverman Illustrations: Peter Atkins and Leroy Laverman with All rights reserved Network Graphics Production Manager: Paul Rohloff Printed in the United States of America Composition: Aptara Printing and Binding: RR Donnelley First printing W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 Houndmills, Basingstoke RG21 6XS, England www.whfreeman.com Macmillan Higher Education Houndmills, Basingstoke RG21 6XS, England www.macmillanhighered.com/international FFMMTTOOCC..iinndddd PPaaggee iiiiii 1100//1188//1122 88::0077 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp Contents in Brief FUNDAMENTALS F1 Introduction and Orientation, Matter and Energy, Elements and Atoms, Compounds, The Nomenclature of Compounds, Moles and Molar Masses, Determination of Chemical Formulas, Mixtures and Solutions, Chemical Equations, Aqueous Solutions and Precipitation, Acids and Bases, Redox Reactions, Reaction Stoichiometry, Limiting Reactants Chapter 1 THE QUANTUM WORLD 1 Chapter 2 QUANTUM MECHANICS IN ACTION: ATOMS 31 Chapter 3 CHEMICAL BONDS 67 MAJOR TECHNIQUE 1 • Infrared Spectroscopy 105 Chapter 4 MOLECULAR SHAPE AND STRUCTURE 107 MAJOR TECHNIQUE 2 • Ultraviolet and Visible Spectroscopy 146 Chapter 5 THE PROPERTIES OF GASES 149 Chapter 6 LIQUIDS AND SOLIDS 189 MAJOR TECHNIQUE 3 • X-Ray Diffraction 223 Chapter 7 INORGANIC MATERIALS 227 Chapter 8 THERMODYNAMICS: THE FIRST LAW 259 Chapter 9 THERMODYNAMICS: THE SECOND AND THIRD LAWS 317 Chapter 10 PHYSICAL EQUILIBRIA 367 MAJOR TECHNIQUE 4 • Chromatography 419 Chapter 11 CHEMICAL EQUILIBRIA 421 Chapter 12 ACIDS AND BASES 463 Chapter 13 AQUEOUS EQUILIBRIA 519 Chapter 14 ELECTROCHEMISTRY 561 Chapter 15 CHEMICAL KINETICS 611 MAJOR TECHNIQUE 5 • Computation 665 Chapter 16 THE ELEMENTS: THE MAIN-GROUP ELEMENTS 667 Chapter 17 THE ELEMENTS: THE d-BLOCK 725 Chapter 18 NUCLEAR CHEMISTRY 765 Chapter 19 ORGANIC CHEMISTRY I: THE HYDROCARBONS 797 MAJOR TECHNIQUE 6 • Mass Spectrometry 821 Chapter 20 ORGANIC CHEMISTRY II: POLYMERS AND BIOLOGICAL COMPOUNDS 823 MAJOR TECHNIQUE 7 • Nuclear Magnetic Resonance 854 iii cc1111CChheemmiiccaallEEqquuiilliibbrriiaa..iinndddd PPaaggee 442211 77//1177//1122 55::2266 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp 11 CHEMICAL EQUILIBRIA Why Do You Need to Know This Material? The dynamic equilibrium toward which Reactions at Equilibrium every chemical reaction tends is such an important aspect of the study of chemistry 11.1 The Reversibility of Reactions that the next four chapters deal with it. You need to know the composition of a reaction 11.2 Equilibrium and the Law of Mass Action mixture at equilibrium because it tells you how much product to expect. To control the 11.3 The Thermodynamic Origin yield of a reaction, you need to understand the thermodynamic basis of equilibrium of Equilibrium Constants and how the position of equilibrium is affected by conditions such as temperature and 11.4 The Extent of Reaction 11.5 The Direction of Reaction pressure. The response of equilibria to changes in conditions has considerable economic Equilibrium Calculations and biological significance: the regulation of chemical equilibrium affects the yields of 11.6 The Equilibrium Constant in products in industrial processes, and living cells struggle to avoid sinking into equilibrium. Terms of Molar Concentrations of Gases What Are the Key Ideas? Reactions proceed until the composition of a reaction 11.7 Alternative Forms of the mixture corresponds to minimum Gibbs free energy. This composition is described by Equilibrium Constant 11.8 Using Equilibrium Constants an equilibrium constant that is characteristic of the reaction. The equilibrium is dynamic The Response of Equilibria to and responds to changes in the conditions. Changes in Conditions What Do You Need to Know Already? The concepts of chemical equilibrium 11.9 Adding and Removing Reagents 11.10 Compressing a Reaction Mixture are related to those of physical equilibrium (Sections 10.1–10.3). Because chemical 11.11 Temperature and Equilibrium equilibrium depends on the thermodynamics of chemical reactions, you will need to use Impact on Materials and Biology the Gibbs free energy of reaction (Section 9.13) and standard enthalpies of formation 11.12 Catalysts and Haber’s (Section 8.17). Chemical equilibrium calculations require a thorough knowledge of molar Achievement concentration (Fundamentals, Section G), reaction stoichiometry (Fundamentals, Section 11.13 Homeostasis L), and the gas laws (Chapter 5). E arly in the twentieth century, the looming prospect of World War I created a desperate need for nitrogen compounds, because nitrates normally used for agriculture were being made into explosives. Almost all the nitrates used for fertilizers and explosives were quarried from deposits in Chile, and the limited supply could not keep up with the demand. Moreover, shipping channels were vulnerable to attack, which threatened to cut off the supply altogether. Although nitrogen is abundant in air, the methods used at the time to convert nitrogen into compounds were too costly to employ on a large scale. Any nation that could develop an economical process to “fix” atmospheric nitrogen—that is, to combine it with other elements—would have all the nitrogen compounds it needed. Scientists on both sides of the conflict were urgently seeking ways to fix nitrogen. Finally, through determination, application, and—as so often happens in research—a bit of luck, the German chemist Fritz Haber, in collaboration with the German chemical engineer Carl Bosch, found an economical way to harvest the nitrogen of the air and to provide an abundant source of compounds for both agri- 421 culture and armaments. cc1111CChheemmiiccaallEEqquuiilliibbrriiaa..iinndddd PPaaggee 442222 77//1177//1122 55::2277 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp 422 Chapter 11 Chemical Equilibria One difficulty Haber faced is that the reactions used to produce compounds from nitrogen do not go to completion, but appear to stop after only some of the reactants have been used up. At this point the mixture of reactants and products has reached chemical equilibrium, the stage in a chemical reaction when there is no further ten- dency for the composition of the reaction mixture—the concentrations or partial pressures of the reactants and products—to change. To achieve the greatest conversion of nitrogen into its compounds, Haber had to understand how a reaction approaches and eventually reaches equilibrium and then use that knowledge—as we shall see in Section 11.12—to improve the yield by changing the conditions of the reaction. Like physical equilibria, all chemical equilibria are dynamic, with the forward and reverse reactions occurring at the same rate. In Chapter 10 we considered sev- eral physical processes, including vaporizing and dissolving, that reach dynamic equilibrium. This chapter shows how to apply the same ideas to chemical changes. It also shows how to use thermodynamics to describe equilibria quantitatively, which puts enormous power into our hands—the power to control the direction of And, we might add, to change the a reaction and the yield of products. We first show that a single quantity, the equi- course of history. librium constant, characterizes the composition of a reaction mixture at equilibrium. Then we explore how the equilibrium constant arises from thermodynamics and see how it is related to the free energy of the reaction. We then see how to use the equilibrium constant to calculate the equilibrium composition of a reaction mixture. Finally, we develop thermodynamic arguments to deduce how the composition of a reaction mixture at equilibrium responds to changes in the conditions under which the reaction is taking place. Reactions at Equilibrium To say that chemical equilibrium is “dynamic” means that when a reaction has reached equilibrium, the forward and reverse reactions continue to take place, but with reactants being formed as fast as they are consumed. As a result, the composition of the reaction mixture remains constant. To develop these ideas we first establish that reactions can take place in their reverse direction as soon as some products have accumulated. Then we see how to relate the equilibrium composition to the thermodynamic properties of the system. 11.1 The Reversibility of Reactions Some reactions, such as the explosive reaction of hydrogen and oxygen, appear to proceed to completion, but others seem to stop at an early stage. For example, consider the reaction that took place when Haber heated nitrogen and hydrogen The metal acts as a catalyst for this under pressure in the presence of a small amount of the metal osmium: reaction, a substance that helps to N 1g2 (cid:2)3 H 1g2 ¡ 2 NH 1g2 (A) make a reaction go faster (Section 2 2 3 15.14). Osmium is too expensive to be The reaction does produce ammonia rapidly at first, but eventually the reaction used commercially; in the industrial seems to stop (FIG. 11.1). As the graph shows, no matter how long we wait, no process, iron is used instead. more product forms. The reaction has reached equilibrium. Like phase changes, chemical reactions tend toward a dynamic equilibrium in which, although there is no net change, the forward and reverse reactions are still taking place, but at matching rates. What actually happens when the formation of ammonia appears to stop is that the rate of the reverse reaction, 2 NH 1g2 ¡ N 1g2 (cid:2)3 H 1g2 (B) 3 2 2 increases as more ammonia is formed. At equilibrium, the ammonia is decomposing as fast as it is being formed. Just as we did for phase transitions, we express this state of dynamic equilibrium by replacing the arrow in the equation with equilibrium “harpoons”: N 1g2 (cid:2)3 H 1g2 Δ 2 NH 1g2 (C) 2 2 3 cc1111CChheemmiiccaallEEqquuiilliibbrriiaa..iinndddd PPaaggee 442233 77//1177//1122 55::2277 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp 11.1 The Reversibility of Reactions 423 FIGURE 11.1 (a) In the synthesis of ammonia, the concentrations of N and H decrease with time and 2 2 that of NH increases until they finally 3 settle into values corresponding to a H2 (product) mixture in which all three are present n H2 (reactant) n and there is no further net change. o o (b) If the experiment is repeated with ntrati NH3 (product) ntrati pure ammonia, it decomposes, and e e the composition settles down into a c c n n o o mixture of ammonia, nitrogen, and c c ar ar hydrogen. (The two graphs correspond ol ol M M N2 (product) to experiments at two different temperatures, and so they correspond N2 (reactant) to different equilibrium compositions.) NH3 (reactant) (a) Time (b) Time All chemical equilibria are dynamic equilibria. Although there is no net change at equilibrium, the forward and reverse reactions are still taking place. As we shall see later in the chapter, dynamic equilibria respond to changes in temperature and pressure, and the addition or removal of even a small amount of reagent can result in a change in overall composition. A reaction that is simply not taking place (such as a mixture of hydrogen and oxygen at room temperature and pressure) does not respond to small changes in the conditions and therefore is not at equilibrium. The criteria that identify a dynamic chemical equilibrium are 1. The forward reaction and its reverse are both taking place. 2. They are doing so at equal rates (so there is no net change). THINKING POINT Can you think of an experiment to show that a chemical equilibrium (a) (b) is dynamic? Think, perhaps, about using radioactive isotopes. We can use these characteristics to decide if the three systems in FIG. 11.2 are at equilibrium. All three systems appear to be unchanging. However, when methane, CH , burns with a steady flame to form carbon dioxide gas and water (Fig. 4 11.2a), the combustion reaction is not at equilibrium because the composition is not constant (reactants continue to be added and the products dissipate instead of reacting). The sample of glucose does not change (Fig. 11.2b), even if we leave it for a very long time. However, the glucose is not at equilibrium with the products of its combustion (carbon dioxide and water); it survives only because the rate of its combustion is immeasurably slow at room temperature. The gas-phase reaction (Fig. 11.2c), however, is at equilibrium because, as well as the composition being constant, additional experiments show that NO is ceaselessly forming N O and 2 2 4 that N O is decomposing into NO at the same rate. 2 4 2 FIGURE 11.2 (a) Methane burns in air with a steady flame, but, because matter is being added and removed, the reaction is not at equilibrium. (b) This sample of glucose in air is unchanging in composition, but it is not yet in equilibrium with its combustion products; the reaction proceeds far too slowly at room temperature. (c) Nitrogen dioxide (a brown gas) and dinitrogen tetroxide (a colorless volatile solid) are in equilibrium in these vessels. We can tell that they are by adding ice to change the temperature (right) and seeing the composition adjust to a new value. LAB VIDEO FIGURE 11.2 (c) cc1111CChheemmiiccaallEEqquuiilliibbrriiaa..iinndddd PPaaggee 442244 77//1177//1122 55::2277 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp 424 Chapter 11 Chemical Equilibria TABLE 11.1 Equilibrium Data and the Equilibrium Constant for the Reaction 2 SO (g) (cid:2) O (g) Δ 2 SO (g) at 1000. K 2 2 3 P (bar) P (bar) P (bar) K* SO2 O2 SO3 0.660 0.390 0.0840 0.0415 0.0380 0.220 0.00360 0.0409 0.110 0.110 0.00750 0.0423 0.950 0.880 0.180 0.0408 1.44 1.98 0.410 0.0409 *Average K: 0.0413. Chemical reactions reach a state of dynamic equilibrium in which the rates of forward and reverse reactions are equal and there is no net change in composition. 11.2 Equilibrium and the Law of Mass Action In 1864, before Haber began his work, the Norwegians Cato Guldberg (a mathema- tician) and Peter Waage (a chemist) had discovered the mathematical relation that summarizes the composition of a reaction mixture at equilibrium. As an example of their approach, look at the data in TABLE 11.1 for the reaction between SO and O : 2 2 2 SO 1g2 (cid:2)O 1g2 Δ 2 SO 1g2 (D) 2 2 3 In each of these five experiments, a mixture with different initial compositions of the three gases was prepared and allowed to reach equilibrium at 1000. K. The compositions of the equilibrium mixtures and the total pressure P were then deter- mined. At first, there seemed to be no pattern in the data. However, Guldberg and Waage noticed an extraordinary relation: the value of the quantity 1P (cid:2)P(cid:4)22 K(cid:3) SO3 1P (cid:2)P(cid:4)221P (cid:2)P(cid:4)2 SO2 O2 was nearly the same for every experiment, regardless of the initial compositions. Here, P is the equilibrium partial pressure of gas J and P(cid:4) (cid:3) 1 bar, the standard pressure. J Note that K is unitless, because the units of P are canceled by the units of P(cid:4) in each J term. From now on, though, to keep the notation simple, we shall write simply 1P 22 K(cid:3) SO3 1P 22P SO2 O2 with each P understood to be the numerical value of the pressure in bar. J Within experimental error, Guldberg and Waage obtained the same value of K whatever the initial composition of the reaction mixture. This remarkable result shows that K is characteristic of the composition of the reaction mixture at equilibrium at a given temperature. It is known as the equilibrium constant for the reaction. The law of mass action summarizes this result: at equilibrium, the composition of the reaction mixture can be expressed in terms of an equilibrium constant, where, for any reaction between gases that can be treated as ideal, the equilibrium constant is the value of the product of the partial pressures of products divided by the product of the partial pressures of reactants each raised to a power corresponding to its coef- ficient in the balanced chemical equation. Thus, if we are interested in the reaction a A1g2 (cid:2)b B1g2 Δ c C1g2 (cid:2)d D1g2 (E) and if all the gases are treated as ideal, we write 1P 2c1P 2d K(cid:3) C D (1)* 1P 2a1P 2b A B with the numerical values of the partial pressures (in bar) at equilibrium. cc1111CChheemmiiccaallEEqquuiilliibbrriiaa..iinndddd PPaaggee 442255 77//1177//1122 55::2277 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp 11.2 Equilibrium and the Law of Mass Action 425 EXAMPLE 11.1 Writing the expression for an equilibrium constant The yearly production of ammonia is of the order of 1.3 (cid:7) 109 t, making this commod- ity one of the top 10 chemicals produced globally. If you were studying the synthesis of ammonia, you would need to work with the equilibrium constant for the reaction. Write the equilibrium constant for the ammonia synthesis reaction, reaction C. PLAN Write the equilibrium constant with the partial pressure of the product in the numerator, raised to a power equal to its coeffi cient in the balanced equation. Do the same for the reactants, but place their partial pressures in the denominator: SOLVE From K (cid:3) (P )c(P )d/(P )a(P )b, C D A B (cid:2) 1P 22 K(cid:3) NH3 P 1P 23 2 N2 H2 K = 3 As already explained, each P in this expression should be interpreted as P/P(cid:4) (that is, J J as the numerical value of the pressure in bar). Self-Test 11.1A Write the expression for the equilibrium constant for the reaction 4 NH 1g2 (cid:2)5 O 1g2 Δ 4 NO1g2 (cid:2)6 H O1g2. 3 2 2 [Answer: K (cid:3) 1P 241P 26(cid:2)1P 241P 25] NO H2O NH3 O2 Self-Test 11.1B Write the expression for the equilibrium constant for 2 H S(g) (cid:2) 2 3 O 1g2 Δ 2 SO 1g2 (cid:2)2 H O1g2. 2 2 2 Related Exercises 11.3, 11.4 We use a different measure of concentration when writing expressions for the equilibrium constants of reactions that involve species other than gases. Thus, for a species J that forms an ideal liquid or solid solution, the partial pressure in the expression for K is replaced by the molar concentration [J] relative to the standard molarity, c(cid:4) (cid:3) 1 mol(cid:3)L(cid:5)1. Although K should be written in terms of the dimension- less ratio [J]/c(cid:4), it is common practice to write K in terms of [J] alone and to inter- pret each [J] as the molar concentration with the units “mol(cid:6)L(cid:5)1” struck out. It has been found empirically, and is justified by thermodynamics, that pure liquids or solids should not appear in K. So, even though CaCO (s) and CaO(s) occur in the 3 equilibrium CaCO 1s2 Δ CaO1s2 (cid:2)CO 1g2 (F) 3 2 neither appears in the equilibrium constant, which is K (cid:3) P (cid:2)P(cid:4) (or, more simply, K (cid:3) P ). CO2 CO2 We can summarize these empirical rules by introducing the concept of the activity, a, of a substance J: J Substance Activity Simplifi ed form ideal gas a (cid:3) P/P(cid:4) a (cid:3) P J J J J solute in a dilute solution a (cid:3) [J]/c(cid:4) a (cid:3) [J] J J pure solid or liquid a (cid:3) 1 a (cid:3) 1 J J Note that all activities are pure numbers and thus are unitless. When using the simplified form of the equilibrium constant, the activity is the numerical value of the pressure in bar or the numerical value of the molarity in moles per liter. cc1111CChheemmiiccaallEEqquuiilliibbrriiaa..iinndddd PPaaggee 442266 77//1177//1122 99::3355 PPMM uusseerr--FF339933 //UUsseerrss//uusseerr--FF339933//DDeesskkttoopp 426 Chapter 11 Chemical Equilibria At this stage, activities are simply quantities introduced to facilitate writing the expression for K. In advanced work, activities are used to take into account devia- tions from ideal behavior and to accommodate real gases and real solutions, where intermolecular interactions are important. In those cases the activity of a substance may differ significantly from the calculated pressure or concentration. However, for the low-pressure gases and dilute solutions that concern us here, we can ignore the interactions between molecules and regard activities as providing a simple, uni- form way of writing general expressions for equilibrium constants and other quantities that we shall meet shortly. We have to remember, though, that because we are assuming ideal behavior when interpreting the activities, the resulting equilibrium expressions are approximations. THINKING POINT How might intermolecular and interionic interactions affect the value of an activity? The use of activities allows us to write a general expression for the equilibrium constant for any reaction: In Eq. 2a the subscript “r” in n r indicates that the values of n are 1activities of products2nr r K(cid:2) e f (2a) unitless stoichiometric coeffi cients 1activities of reactants2nr equilibrium (see Section 11.3). More specifically, for a generalized version of reaction E, with phases not specified: 1a 2c1a 2d a A1g2 (cid:3)b B1g2 Δ c C1g2 (cid:3)d D1g2 K(cid:2) C D (2b)* 1a 2a1a 2b A B Because activities are unitless, so too is K. A Note on Good Practice: In some cases you will see an equilibrium constant denoted K to remind you that it is expressed in terms of partial pressures. P However, the subscript P is unnecessary because, by definition, equilibrium constants for gas-phase reactions are expressed in terms of partial pressures. ■ Chemical equilibria with reactants and products that are all in the same phase are called homogeneous equilibria. Equilibria C, D, and E are homogeneous. Equi- libria in systems having more than one phase are called heterogeneous equilibria. Equilibrium F is heterogeneous; so too is the equilibrium between water vapor and liquid water in a closed system, H O1l2 Δ H O1g2 (G) 2 2 In this reaction, there is a gas phase and a liquid phase. Likewise, the equilibrium between a solid and its saturated solution is heterogeneous: Ca1OH2 1s2 Δ Ca2(cid:3)1aq2 (cid:3)2 OH(cid:4)1aq2 (H) 2 The equilibrium constants for heterogeneous reactions are also given by the general expression in Eq. 2; all we have to remember is that the activity of a pure solid or liquid is 1. For instance, for the calcium hydroxide equilibrium (reaction H), K(cid:2) aCa2(cid:3)1aOH(cid:4)22 (cid:2)[Ca2(cid:3)][OH(cid:4)]2 1aC4a21O4H322 1 for a pure solid because the calcium hydroxide is a pure solid. The calcium hydroxide must be present for the equilibrium to exist, but it does not appear in the expression for the equilibrium constant. Remember that each [J] represents the molar concentration of J with the units struck out. Similarly, in the equilibrium between solid nickel, gaseous carbon monoxide, and gaseous nickel carbonyl used in the purification of nickel, Ni1s2 (cid:3)4 CO1g2 Δ Ni1CO2 1g2 K(cid:2) aNi1CO24 (cid:2) PNi1CO24 4 a 1a 24 1P 24 {Ni CO CO 1