Chemical Principles: The Quest for Insight (Part 2)

571 Pages·121.675 MB·English

by  Atkins P., Jones L., Laverman L.

C
HEMICA L
P
RINCIPLES
THE QUEST FOR INSIGHT
Sixth Edition
PETER ATKINS · LORETTA JONES · LEROY LAVERMAN
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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
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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
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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 nitro-
gen. 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.
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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 pres-
sures 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 equilib-
rium. 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 compo-
sition 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 compo-
sition 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 composi-
tion 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 decompos-
ing 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
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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 meth-
ane, 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)
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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 equilib-
rium 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.
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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.
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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 quan-
tities 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 gen-
eral 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 pres-
ent 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