CHAPTEResminfo.prenhall.com/.../closerlook/pdf/BLB_Ch15.pdf · 630 Chapter 15 | Chemical Equilibrium reactants equals the rate at which the reactants are formed from the products

15.1 The Concept of Equilibrium

15.2 The Equilibrium Constant

15.3 Interpreting and Working with Equilibrium Constants

15.4 Heterogeneous Equilibria

15.5 Calculating EquilibriumConstants

15.6 Applications of EquilibriumConstants

15.7 Le Châtelier’s Principle

WHAT’S AHEAD• We begin by examining the concept

of equilibrium. (Section 15.1)

• We then define the equilibriumconstant and learn how to writeequilibrium-constant expressions forhomogeneous reactions. (Section 15.2)

• We also learn how to interpret themagnitude of an equilibriumconstant and how to determine theway in which its value is affectedwhen the equation is reversed orchanged in some other fashion.(Section 15.3)

• We then learn how to writeequilibrium-constantexpressions for heterogeneousreactions. (Section 15.4)

• The value of anequilibrium constant can becalculated usingequilibriumconcentrations ofreactants andproducts. (Section 15.5)

• Equilibrium constantscan be used to predict the equilibriumconcentrations of reactantsand products and todetermine the direction inwhich a reaction must proceed inorder to achieve equilibrium.(Section 15.6)

• The chapter concludes with adiscussion of Le Châtelier’s principle,which predicts how a system atequilibrium responds to changes inconcentration, volume, pressure, andtemperature. (Section 15.7)

CH

AP

TE

R 15BROWMC15_628-667PR2 1/26/05 15:07 Page 628

TO BE IN EQUILIBRIUM IS TO BE IN A STATE OF BALANCE. A tug ofwar in which the two sides are pulling with equal

force so that the rope doesn’t move is an exampleof a static equilibrium, one in which an object

is at rest. Equilibria can also be dynamic, asillustrated in the chapter-opening photograph, which shows cars

traveling in both directions over a bridge that serves as an en-tranceway to a city. If the rate at which cars leave the city equals

the rate at which they enter, the two opposing processes are inbalance and the net number of cars in the city is constant.

We have already encountered several instances of dy-namic equilibrium. For example, the vapor above a liquid is

in equilibrium with the liquid phase. • (Section 11.5)The rate at which molecules escape from the liquid into

the gas phase equals the rate at which molecules in thegas phase strike the surface and become part of the

liquid. Similarly, in a saturated solution of sodiumchloride the solid sodium chloride is in equili-

brium with the ions dispersed in water.• (Section 13.2) The rate at which ions leave

the solid surface equals the rate at whichother ions are removed from the liquid to

become part of the solid. Both of these ex-amples involve a pair of opposing

processes. At equilibrium these oppos-ing processes are occurring at the

same rate.In this chapter we will consid-

er yet another type of dynamicequilibrium, one involving

chemical reactions. Chemicalequilibrium occurs when op-

posing reactions are proceed-ing at equal rates: The rate

at which the productsare formed from the

Chemical Equilibrium

629

TRAFFIC ENTERING AND LEAVING San Francisco over the Bay Bridge at night.

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630 Chapter 15 | Chemical Equilibrium

reactants equals the rate at which the reactants are formed from the products.As a result, concentrations cease to change, making it appear as if the reactionhas stopped. When chemical reactions occur in closed systems, the reactionswill sooner or later reach an equilibrium state—a mixture in which the concen-trations of reactants and products no longer change with time. Consequently,closed systems will always be either at equilibrium or approaching equilibri-um. How fast a reaction reaches equilibrium is a matter of kinetics.

Chemical equilibria are involved in a great many natural phenomena, andthey also play important roles in many industrial processes. In this and the nexttwo chapters, we will explore chemical equilibrium in some detail. Here wewill learn how to express the equilibrium position of a reaction in quantitativeterms, and we will study the factors that determine the relative concentrationsof reactants and products in equilibrium mixtures.

15.1 The Concept of Equilibrium

Figure 15.1 » shows a sample of solid a colorless substance, inside asealed tube resting in a beaker. When this solid is warmed until the substance isabove its boiling point (21.2°C), the gas in the sealed tube turns progressivelydarker as the colorless gas dissociates into brown gas. Eventually,even though there is still in the tube, the color stops getting darker be-cause the system reaches equilibrium. We are left with an equilibrium mixture of

and in which the concentrations of the gases no longer change astime passes. The equilibrium mixture results because the reaction is reversible.Not only can react to form but can also react to form This situation is represented by writing the equation for the reaction with adouble arrow: • (Section 4.1)

[15.1]Colorless Brown

We can analyze this equilibrium using our knowledge of kinetics. Let’s callthe decomposition of to form the forward reaction and the reactionof to re-form the reverse reaction. In this case both the forward reac-tion and the reverse reaction are elementary reactions. As we learned in Section14.6, the rate laws for elementary reactions can be written from their chemicalequations:

Forward reaction: [15.2]

Reverse reaction: [15.3]

where and are the rate constants for the forward and reverse reactions, re-spectively. At equilibrium the rate at which products are produced from reac-tants equals the rate at which reactants are produced from products:

[15.4]Forward reaction Reverse reaction

Rearranging this equation gives

[15.5]

As shown in Equation 15.5, the quotient of two constants, such as and isitself a constant. Thus, at equilibrium the ratio of the concentration terms in-volving and equals a constant. (We will consider this constant,called the equilibrium constant, in Section 15.2.) It makes no difference whetherwe start with or with or even with some mixture of the two.NO2,N2O4

NO2N2O4

kr ,kf

[NO2]2

[N2O4]=

kf

kr= a constant

kf[N2O4] = kr[NO2]2

krkf

2 NO2(g) ¡ N2O4(g) Rater = kr[NO2]2

N2O4(g) ¡ 2 NO2(g) Ratef = kf[N2O4]

N2O4NO2

NO2N2O4

N2O4(g) ∆ 2 NO2(g)

N2O4.NO2NO2,N2O4

NO2N2O4

N2O4

NO2N2O4

N2O4,

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15.1 | The Concept of Equilibrium 631

ESTABLISHING EQUILIBRIUMThe condition in which the concentrations of all reactants and products in a closed system cease

to change with time is called chemical equilibrium.

Frozen N2O4 is nearly colorless. Eventually the color stops changing as N2O4(g) and NO2(g) reach

concentrations at which they are interconverting at the same rate. The two gases are in equilibrium.

As N2O4 is warmed above its boiling point, it starts to

dissociate into brown NO2 gas.

N2O4(s) N2O4(g) 2 NO2(g) N2O4(g) 2 NO2(g)

N2O4 NO2

At equilibrium the ratio equals a specific value. Thus, there is an important con-straint on the proportions of and at equilibrium.

Once equilibrium is established, the concentrations of and nolonger change, as shown in Figure 15.2 ». Just because the composition of theequilibrium mixture remains constant with time does not mean, however, that

and stop reacting. On the contrary, the equilibrium is dynamic—some is still converting to and some is still converting to

At equilibrium, however, the two processes occur at the same rate, andso there is no net change in their amounts.

We learn several important lessons about equilibrium from this example. First,the fact that a mixture of reactants and products is formed in which concentrationsno longer change with time indicates that the reaction has reached a state of equi-librium. Second, for equilibrium to occur, neither reactants nor products can es-cape from the system. Third, at equilibrium the particular ratio of concentrationterms equals a constant. It is this third fact that we examine in the next section.

N2O4.NO2NO2,N2O4

NO2N2O4

NO2N2O4

NO2N2O4Students often incorrectly use a double-headed single arrow ( ) to denoteequilibrium reactions. Emphasize that thissymbol is reserved to show resonance andthat other uses are incorrect.

·

Á Figure 15.1 The equilibrium.N2O4(g) ∆ 2 NO2(g)

ACTIVITYChemical Equilibrium

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GIVE IT SOME THOUGHT

(a) Which quantities are equal in a dynamic equilibrium? (b) If the rate constant forthe forward reaction in Equation 15.1 is larger than the rate constant for the reversereaction, will the constant in Equation 15.5 be greater than 1 or smaller than 1?

15.2 The Equilibrium Constant

Opposing reactions naturally lead to an equilibrium, regardless of how compli-cated the reaction might be and regardless of the nature of the kinetic processesfor the forward and reverse reactions. Consider the synthesis of ammonia fromnitrogen and hydrogen:

[15.6]

This reaction is the basis for the Haber process, which, in the presence of a cata-lyst, combines and at a pressure of several hundred atmospheres and atemperature of several hundred degrees Celsius. The two gases react to formammonia under these conditions, but the reaction does not lead to completeconsumption of the and Rather, at some point the reaction appears tostop, with all three components of the reaction mixture present at the same time.

The manner in which the concentrations of and vary withtime is shown in Figure 15.3(a) ¥. Notice that an equilibrium mixture is ob-tained regardless of whether we begin with and or only with Atequilibrium the relative concentrations of and are the same, re-gardless of whether the starting mixture was a molar ratio of and orpure The equilibrium condition can be reached from either direction.


How do we know when equilibrium has been reached in a chemical reaction?

Earlier, we saw that when the reaction reachesequilibrium, a ratio based on the equilibrium concentrations of and has a constant value (Equation 15.5). A similar relationship governs the concen-trations of and at equilibrium. If we were to systematicallychange the relative amounts of the three gases in the starting mixture and thenanalyze each equilibrium mixture, we could determine the relationship amongthe equilibrium concentrations.

Chemists carried out studies of this kind on other chemical systems in thenineteenth century, before Haber’s work. In 1864, Cato Maximilian Guldberg(1836–1902) and Peter Waage (1833–1900) postulated their law of mass action,which expresses, for any reaction, the relationship between the concentrationsof the reactants and products present at equilibrium. Suppose we have the

NH3N2, H2,

NO2N2O4

N2O4(g) ∆ 2 NO2(g)

NH3.N2H23 : 1

NH3H2, N2,NH3.H2N2

NH3H2, N2,

H2.N2

H2N2

N2(g) + 3 H2(g) ∆ 2 NH3(g)

Time

(a)

Con

cent

rati

on

Con

cent

rati

on

H2

NH3N2

Equilibrium achieved

Time

(b)

H2NH3N2

Equilibrium achieved» Figure 15.3 Concentration changesapproaching equilibrium. (a) Equilibrium forthe reaction isapproached beginning with and present inthe ratio and no present.(b) Equilibrium for the same reaction isapproached beginning with only in thereaction vessel.

NH3

NH33 : 1N2H2

N2 + 3 H2 ∆ 2 NH3

Time

(a)

0

Con

cent

rati

on

Equilibriumachieved

Equilibriumachieved

(rates are equal)

Time

(b)

0

Rat

e

kf [N2O4]

kr[NO2]2

NO2

N2O4

Á Figure 15.2 Achieving chemicalequilibrium for(a) The concentration of decreases whilethe concentration of increases during thecourse of the reaction. Equilibrium is indicatedwhen the concentrations no longer change withtime. (b) The rate of disappearance of decreases with time as the concentration of decreases. At the same time, the rate of formationof also decreases with time. Equilibriumoccurs when these two rates are equal.

NO2

N2O4

N2O4

NO2

N2O4

N2O4(g) ∆ 2 NO2(g).

Recall that a catalyst only lowers theactivation energy ( ) of a reaction.Ea

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15.2 | The Equilibrium Constant 633

CHEMISTRY AT WORK | The Haber Process

W e presented a “Chemistry and Life” box in Section 14.7that discussed nitrogen fixation, the processes that con-

vert gas into ammonia, which can then be incorporatedinto living organisms. We learned that the enzyme nitrogenaseis responsible for generating most of the fixed nitrogen essen-tial for plant growth. However, the quantity of food requiredto feed the ever-increasing human population far exceeds thatprovided by nitrogen-fixing plants, so human agriculture re-quires substantial amounts of ammonia-based fertilizers thatcan be applied directly to croplands. Thus, of all the chemicalreactions that humans have learned to carry out and controlfor their own purposes, the synthesis of ammonia from hydro-gen and atmospheric nitrogen is one of the most important.

In 1912 the German chemist Fritz Haber (1868–1934) de-veloped a process for synthesizing ammonia directly from ni-trogen and hydrogen (Figure 15.4 ¥). The process issometimes called the Haber–Bosch process to also honor KarlBosch, the engineer who developed the equipment for the in-dustrial production of ammonia. The engineering needed toimplement the Haber process requires the use of temperaturesand pressures (approximately 500°C and 200 atm) that weredifficult to achieve at that time.

The Haber process provides a historically interesting ex-ample of the complex impact of chemistry on our lives. At thestart of World War I, in 1914, Germany depended on nitrate de-posits in Chile for the nitrogen-containing compounds neededto manufacture explosives. During the war the Allied naval

N2

blockade of South America cut off this supply. However, by fix-ing nitrogen from air, Germany was able to continue to produceexplosives. Experts have estimated that World War I wouldhave ended before 1918 had it not been for the Haber process.

From these unhappy beginnings as a major factor in inter-national warfare, the Haber process has become the world’sprincipal source of fixed nitrogen. The same process that pro-longed World War I has enabled scientists to manufacture fer-tilizers that have increased crop yields, thereby savingmillions of people from starvation. About 40 billion pounds ofammonia are manufactured annually in the United States,mostly by the Haber process. The ammonia can be applied di-rectly to the soil as fertilizer (Figure 15.5 Á). It can also be con-verted into ammonium salts—for example, ammoniumsulfate, or ammonium hydrogen phosphate,

—which, in turn, are used as fertilizers.Haber was a patriotic German who gave enthusiastic sup-

port to his nation’s war effort. He served as chief of Germany’sChemical Warfare Service during World War I and developedthe use of chlorine as a poison-gas weapon. Consequently, thedecision to award him the Nobel Prize for chemistry in 1918was the subject of considerable controversy and criticism. Theultimate irony, however, came in 1933 when Haber was ex-pelled from Germany because he was Jewish.Related Exercises: 15.38 and 15.63

(NH4)2HPO4

(NH4)2SO4,

Á Figure 15.5 Liquid ammonia used as fertilizer. Ammonia,produced by the Haber process, can be added directly to the soil as a fertilizer.Agricultural use is the largest single application of manufactured NH3 .

following general equilibrium equation:

[15.7]

where A, B, D, and E are the chemical species involved and a, b, d, and e aretheir coefficients in the balanced chemical equation. According to the law ofmass action, the equilibrium condition is expressed by the expression

[15.8]Kc =[D]d[E]e

[A]a[B]b

aA + bB ∆ dD + eE

N N

H H

H H

H H

H HH

N

H HH

N

�

Á Figure 15.4 The Haber process. Used to convert and tothis process, although exothermic, requires breaking the very strong

triple bond in N2 .NH3(g),

H2(g)N2(g)

Martin R. Feldman and Monica L. Tarver,“Fritz Haber,” J. Chem. Educ., Vol. 60, 1983,463–464.

For a particular reaction, the value of theequilibrium constant varies withtemperature.

ACTIVITYEquilibrium Constant

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We call this relationship the equilibrium-constant expression (or merely the equi-librium expression) for the reaction. The constant which we call theequilibrium constant, is the numerical value obtained when we substitute equi-librium concentrations into the equilibrium-constant expression. The subscript con the K indicates that concentrations expressed in molarity are used to evaluatethe constant.

In general, the numerator of the equilibrium-constant expression is theproduct of the concentrations of all substances on the product side of the equi-librium equation, each raised to a power equal to its coefficient in the balancedequation. The denominator is similarly derived from the reactant side of theequilibrium equation. (Remember, the convention is to write the substances onthe product side in the numerator and the substances on the reactant side in thedenominator.) Thus, for the Haber process, theequilibrium-constant expression is

[15.9]

Note that once we know the balanced chemical equation for an equilibrium, wecan write the equilibrium-constant expression even if we don’t know the reac-tion mechanism. The equilibrium-constant expression depends only on the stoichiom-etry of the reaction, not on its mechanism.

The value of the equilibrium constant at any given temperature does notdepend on the initial amounts of reactants and products. It also does not matterwhether other substances are present, as long as they do not react with a reac-tant or a product. The value of the equilibrium constant depends only on theparticular reaction and on the temperature.

SAMPLE EXERCISE 15.1 | Writing Equilibrium-Constant ExpressionsWrite the equilibrium expression for for the following reactions:

(a)(b)(c)

SolutionAnalyze: We are given three equations and are asked to write an equilibrium-constant expression for each.Plan: Using the law of mass action, we write each expression as a quotient havingthe product concentration terms in the numerator and the reactant concentrationterms in the denominator. Each term is raised to the power of its coefficient in the bal-anced chemical equation.

Solve: (a) (b) (c)

PRACTICE EXERCISEWrite the equilibrium-constant expression, for (a)(b)

Answers: (a) (b)

We can illustrate how the law of mass action was discovered empiricallyand demonstrate that the equilibrium constant is independent of starting con-centrations by examining some equilibrium concentrations for the gas-phasereaction between dinitrogen tetroxide and nitrogen dioxide:

[15.10]N2O4(g) ∆ 2 NO2(g) Kc =[NO2]2

[N2O4]

Kc =[CdBr4

2-]

[Cd2+][Br-]4Kc =[HI]2

[H2][I2];

Cd2+(aq) + 4 Br-(aq) ∆ CdBr4

2-(aq).H2(g) + I2(g) ∆ 2 HI(g),Kc ,

Kc =[Ag(NH3)2

+]

[Ag+][NH3]2Kc =[NOCl]2

[NO]2[Cl2],Kc =

[O2]3

[O3]2 ,

Ag+(aq) + 2 NH3(aq) ∆ Ag(NH3)2

+(aq)2 NO(g) + Cl2(g) ∆ 2 NOCl(g)2 O3(g) ∆ 3 O2(g)

Kc

Kc =[NH3]2

[N2][H2]3

N2(g) + 3 H2(g) ∆ 2 NH3(g),

Kc ,Carl W. David, “An Elementary Discussion ofChemical Equilibrium,”J. Chem. Educ.,Vol. 65, 1988, 407–409.

An overhead projector demonstration of theeffect of reactant concentration onequilibrium. Lee R. Summerlin and James.L. Ealy, Jr., “Equilibrium and Le Chatelier’sPrinciple,” Chemical Demonstrations, ASourcebook for Teachers, Vol. 1(Washington: American Chemical Society,1988), pp. 77–78.

Vladimir E. Fainzilberg and Stewart Karp,“Chemical Equilibrium in the GeneralChemistry Course,” J. Chem. Educ., Vol. 71,1994, 769–770.

Louise Tyson, David F. Treagust, and RobertB. Bucat, “The Complexity of Teaching andLearning Chemical Equilibrium,” J. Chem.Educ., Vol. 76, 1999, 554–558.

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15.2 | The Equilibrium Constant 635

Figure 15.1 shows the reaction proceeding to equilibrium starting with pureBecause is a dark brown gas and is colorless, the amount of

in the mixture can be determined by measuring the intensity of the browncolor of the gas mixture.

We can determine the numerical value for and verify that it is constantregardless of the starting amounts of and by performing experi-ments in which we start with several sealed tubes containing different concen-trations of and as summarized in Table 15.1 Á. The tubes are kept at100°C until no further change in the color of the gas is noted. We then analyzethe mixtures and determine the equilibrium concentrations of and as shown in Table 15.1.

To evaluate the equilibrium constant, we insert the equilibrium concen-trations into the equilibrium-constant expression. For example, using the Ex-periment 1 data, and we find

Proceeding in the same way, the values of for the other samples werecalculated, as listed in Table 15.1. Note that the value for is constant( within the limits of experimental error) even though the ini-tial concentrations vary. Furthermore, the results of Experiment 4 showthat equilibrium can be achieved beginning with rather than with

That is, equilibrium can be approached from either direction.Figure 15.6 » shows how both Experiments 3 and 4 result in the sameequilibrium mixture even though one begins with and theother with

Notice that no units are given for the values of either in Table 15.1 orin our calculation using the Experiment 1 data. It is common practice towrite equilibrium constants without units for reasons that we address laterin this section.


How does the value of in Equation 15.10 depend on the starting concentrationsof and

Equilibrium Constants in Terms of Pressure, When the reactants and products in a chemical reaction are gases, we can for-mulate the equilibrium-constant expression in terms of partial pressures in-stead of molar concentrations. When partial pressures in atmospheres are usedin the equilibrium-constant expression, we can denote the equilibrium constantas (where the subscript p stands for pressure). For the general reaction inEquation 15.7, the expression for is

[15.11]Kp =(PD)d(PE)e

(PA)a(PB)b

Kp

Kp

Kp

N2O4 ?NO2

Kc

Kc

0.0200 M N2O4.0.0400 M NO2

NO2.N2O4

Kc = 0.212,Kc

Kc

Kc =[NO2]2

[N2O4]=

(0.0172)2

0.00140= 0.211

[N2O4] = 0.00140 M,[NO2] = 0.0172 M

Kc ,

N2O4,NO2

N2O4,NO2

N2O4NO2

Kc

NO2

N2O4NO2N2O4.

TABLE 15.1 Initial and Equilibrium Concentrations of and in the Gas Phase at 100°C

Initial Initial Equilibrium Equilibrium Experiment Concentration (M) Concentration (M) Concentration (M) Concentration (M)

1 0.0 0.0200 0.00140 0.0172 0.2112 0.0 0.0300 0.00280 0.0243 0.2113 0.0 0.0400 0.00452 0.0310 0.2134 0.0200 0.0 0.00452 0.0310 0.213

Kc

NO2N2O4NO2N2O4

NO2N2O4

0

0.0100

0.0200

0.0300

0.0400[N

O2]

Time

Experiment 4

Experiment 3

Á Figure 15.6 Concentration changesapproaching equilibrium. As seen in Table15.1, the same equilibrium mixture is producedstarting with either 1.22 atm (Experiment 3)or 0.612 atm (Experiment 4).N2O4

NO2

Audrey H. Wilson, “Equilibrium: A Teaching/Learning Activity,” J. Chem. Educ., Vol. 75,1998, 1176–1177.

Jan H. Van Driel, Wobbe de Vos, and NicoVerloop, “Introducing Dynamic Equilibriumas an Explanatory Model,“ J. Chem. Educ.,Vol. 76, 1999, 559–561.

ACTIVITYEquilibriumNO2 ¬ N2O4

Penelope A. Huddle, Margie W. White, andFiona Rogers, “Simulations for TeachingChemical Equilibrium,” J. Chem. Educ.,Vol. 77, 2000, 920–926.

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where is the partial pressure of A in atmospheres, and so forth. For example,for we have


What do the symbols and represent?

For a given reaction, the numerical value of is generally different fromthe numerical value of We must therefore take care to indicate, via a sub-script c or p, which of these equilibrium constants we are using. It is possible,however, to calculate one from the other using the ideal-gas equation(Section 10.4) to convert between concentration (in molarity, M) and pressure(in atm):

[15.12]

Using the usual units, has the units of and therefore equals molari-ty, M. For substance A we therefore see that

[15.13]

When we substitute Equation 15.13 and like expressions for the other gaseouscomponents of the reaction into the expression for (Equation 15.11), we ob-tain a general expression relating and

[15.14]

The quantity is the change in the number of moles of gas in the chemicalequation for the reaction. It equals the sum of the coefficients of the gaseousproducts minus the sum of the coefficients of the gaseous reactants:

[15.15]

For example, in the reaction there are two moles of theproduct (the coefficient in the balanced equation) and one mole of the re-actant Therefore, and for this reaction.From Equation 15.14, we see that only when the same number of molesof gas appear on both sides of the balanced chemical equation, which meansthat

SAMPLE EXERCISE 15.2 | Converting Between and In the synthesis of ammonia from nitrogen and hydrogen,

at 300°C. Calculate for this reaction at this temperature.

SolutionAnalyze: We are given for a reaction and asked to calculate Plan: The relationship between and is given by Equation 15.14. To apply thatequation, we must determine by comparing the number of moles of product withthe number of moles of reactants (Equation 15.15).Solve: There are two moles of gaseous products and four moles of gaseous re-actants Therefore, (Remember that functions arealways based on products minus reactants.) The temperature, T, is 273 + 300 = 573 K.

¢¢n = 2 - 4 = -2.(1 N2 + 3 H2).(2 NH3)

¢nKpKc

Kp .Kc

KpKc = 9.60

N2(g) + 3 H2(g) ∆ 2 NH3(g)

KpKc

¢n = 0.

Kp = Kc

Kp = Kc(RT)¢n = 2 - 1 = 1,N2O4.NO2

N2O4(g) ∆ 2 NO2(g),

¢n = (moles of gaseous product) - (moles of gaseous reactant)

¢n

Kp = Kc(RT)¢n

Kc :Kp

Kp

PA =nA

V RT = [A]RT

mol>L,n>VPV = nRT, so P =

nV

RT

Kp .Kc

KpKc

Kp =(PNO2)

2

PN2O4

N2O4(g) ∆ 2 NO2(g)PA

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15.3 | Interpreting and Working with Equilibrium Constants 637

The value for the ideal-gas constant, R, is Using wetherefore have

PRACTICE EXERCISEFor the equilibrium is at 1000 K.Calculate the value for Answer: 0.335

Equilibrium Constants and UnitsThe equilibrium constant is related not only to the kinetics of a reaction but alsoto the thermodynamics of the process. (We will explore this latter connection inChapter 19.) Equilibrium constants derived from thermodynamic measure-ments are defined in terms of activities rather than concentrations or partialpressures. A full discussion of activities is beyond the scope of this book, but itis useful to explore them very briefly to understand why equilibrium constantsare usually reported without units.

The activity of any substance in an ideal mixture is the ratio of the concen-tration of the substance in molarity to a standard concentration of 1 M or, if thesubstance is a gas, the ratio of the partial pressure in atmospheres to a standardpressure of 1 atm. For example, if the concentration of a substance in an equilib-rium mixture is 0.10 M, its activity is The units of such ra-tios always cancel, and consequently, activities have no units. Furthermore, thenumerical value of the activity equals the concentration because we have divid-ed by 1. In real systems, activities are not exactly numerically equal to concen-trations. In some cases the differences are significant (See the Closer Look box inSection 13.6). However, we will ignore these differences. For pure solids andpure liquids, the situation is even simpler because the activities then merelyequal 1 (again with no units). Because activities have no units, the thermodynamicequilibrium constant derived from them also has no units. It is therefore commonpractice to write all types of equilibrium constants without units as well, a prac-tice that we adhere to in this text.


If the concentration of in an equilibrium mixture is 0.00140 M, what is itsactivity?

15.3 Interpreting and Working with Equilibrium Constants

Before doing calculations with equilibrium constants, it is valuable to understandwhat the magnitude of an equilibrium constant can tell us about the relative con-centrations of reactants and products in an equilibrium mixture. It is also useful toconsider how the magnitude of any equilibrium constant depends on how thechemical equation is expressed. We examine these ideas in this section.

The Magnitude of Equilibrium ConstantsEquilibrium constants can vary from very large to very small. The magnitude ofthe constant provides us with important information about the composition ofan equilibrium mixture. For example, consider the reaction of carbon monoxide

N2O4

0.10 M>1 M = 0.10.

Kp .4.08 * 10-32 SO3(g) ∆ 2 SO2(g) + O2(g), Kc

Kp = Kc(RT)¢n = (9.60)(0.0821 * 573)-2 =(9.60)

(0.0821 * 573)2 = 4.34 * 10-3

Kc = 9.60,0.0821 L-atm>mol-K.

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gas and chlorine gas at 100°C to form phosgene a toxic gas used in themanufacture of certain polymers and insecticides:

In order for the equilibrium constant to be so large, the numerator of theequilibrium-constant expression must be much larger than the denominator.Thus, the equilibrium concentration of must be much greater than that ofCO or and in fact this is just what we find experimentally. We say that thisequilibrium lies to the right (that is, toward the product side). Likewise, a verysmall equilibrium constant indicates that the equilibrium mixture contains most-ly reactants. We then say that the equilibrium lies to the left. In general,

If Equilibrium lies to the right; products predominate.

If Equilibrium lies to the left; reactants predominate.

These situations are summarized in Figure 15.7 «.

SAMPLE EXERCISE 15.3 | Interpreting the Magnitude of an EquilibriumConstant

The reaction of with to form NO might be considered a means of “fixing” nitrogen:

The value for the equilibrium constant for this reaction at 25°C is Describe the feasibility of fixing nitrogen by forming NO at 25°C.

SolutionAnalyze: We are asked to comment on the utility of a reaction based on the magni-tude of its equilibrium constant.Plan: We consider the magnitude of the equilibrium constant to determine whetherthis reaction is feasible for the production of NO.Solve: Because is so small, very little NO will form at 25°C. The equilibrium lies tothe left, favoring the reactants. Consequently, this reaction is an extremely poorchoice for nitrogen fixation, at least at 25°C.

PRACTICE EXERCISEFor the reaction at 298 K and at 700 K.Is the formation of HI favored more at the higher or lower temperature?Answer: The formation of product, HI, is favored at the lower temperature because

is larger at the lower temperature.

The Direction of the Chemical Equation and KBecause an equilibrium can be approached from either direction, the directionin which we write the chemical equation for an equilibrium is arbitrary. For ex-ample, we have seen that we can represent the equilibrium as

[15.16]

We could equally well consider this same equilibrium in terms of the reversereaction:

The equilibrium expression is then

[15.17]Kc =[N2O4]

[NO2]2 =1

0.212= 4.72 (at 100°C)

2 NO2(g) ∆ N2O4(g)

N2O4(g) ∆ 2 NO2(g) Kc =[NO2]2

[N2O4]= 0.212 (at 100°C)

N2O4 - NO2

Kp

Kp = 54H2(g) + I2(g) ∆ 2 HI(g), Kp = 794

Kc

Kc = 1 * 10-30.

N2(g) + O2(g) ∆ 2 NO(g)

O2N2

K V 1:

K W 1:

Cl2 ,COCl2

CO(g) + Cl2(g) ∆ COCl2(g) Kc =[COCl2]

[CO][Cl2]= 4.56 * 109

(COCl2),

Reactants Products

Reactants

(a) K 1

(b) K 1

Products

Á Figure 15.7 K and the composition ofthe equilibrium mixture. The equilibriumexpression has products in the numerator andreactants in the denominator. (a) When there are more products than reactants atequilibrium, and the equilibrium is said to lie tothe right. (b) When there are morereactants than products at equilibrium, and theequilibrium is said to lie to the left.

K V 1,

K W 1,

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15.3 | Interpreting and Working with Equilibrium Constants 639

Equation 15.17 is just the reciprocal of the equilibrium-constant expressionin Equation 15.16. The equilibrium-constant expression for a reaction written in onedirection is the reciprocal of the one for the reaction written in the reverse direction.Consequently, the numerical value of the equilibrium constant for the reac-tion written in one direction is the reciprocal of that for the reverse reaction.Both expressions are equally valid, but it is meaningless to say that the equi-librium constant for the equilibrium between and is 0.212 or 4.72unless we indicate how the equilibrium reaction is written and also specifythe temperature.

N2O4NO2

SAMPLE EXERCISE 15.4 | Evaluating an Equilibrium Constant When an Equation Is Reversed(a) Write the equilibrium-constant expression for for the following reaction:

(b) Using information in Sample Exercise 15.3, determine the value of this equilibrium constant at 25°C.

SolutionAnalyze: We are asked to write the equilibrium-constant expression for a reaction in terms of concentrations and to determinethe value of Plan: As before, we write the equilibrium-constant expression as a quotient of products over reactants, each raised to a powerequal to its coefficient in the balanced equation. We can determine the value of the equilibrium constant by relating theequilibrium-constant expression we write for this equation to the equilibrium-constant expression in Sample Exercise 15.3, wherewe found N2(g) + O2(g) ∆ 2 NO(g), Kc = 1 * 10-30.

Kc .

2 NO(g) ∆ N2(g) + O2(g)

Kc

Solve: (a) Writing products over re-actants, we have Kc =

[N2][O2]

[NO]2

(b) The reaction is just the reverse ofthe one given in Sample Exercise15.3. Thus, both the equilibrium-constant expression and the numeri-cal value of the equilibrium constantare the reciprocals of those for thereaction in Sample Exercise 15.3.

Kc =[N2][O2]

[NO]2 =1

1 * 10-30 = 1 * 1030

Comment: Regardless of the way we express the equilibrium among NO, and at 25°C it lies on the side that favors and

PRACTICE EXERCISEFor the formation of from and at 300°C. What is the value of for the reverse reaction?Answer: 2.30 * 102

KpN2(g) + 3 H2(g) ∆ 2 NH3(g), Kp = 4.34 * 10-3H2,N2NH3

O2.N2O2,N2,

Relating Chemical Equations and Equilibrium ConstantsJust as the equilibrium constants of forward and reverse reactions are recipro-cals of each other, the equilibrium constants of reactions associated in otherways are also related. For example, if we were to multiply our original

equilibrium by 2, we would have

The equilibrium-constant expression, for this equation is

which is simply the square of the equilibrium-constant expression for the originalequation, given in Equation 15.10. Because the new equilibrium-constant expres-sion equals the original expression squared, the new equilibrium constant equalsthe original constant squared: in this case (at 100°C).0.2122 = 0.0449

Kc =[NO2]4

[N2O4]2

Kc ,

2 N2O4(g) ∆ 4 NO2(g)

N2O4 - NO2

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SAMPLE EXERCISE 15.5 | Combining Equilibrium ExpressionsGiven the following information,

determine the value of for the reaction

SolutionAnalyze: We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine theequilibrium constant for a third equation, which is related to the first two.Plan: We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equa-tions in order to come up with the steps that will add to give us the desired equation.

2 HF(aq) + C2O4

2-(aq) ∆ 2 F-(aq) + H2C2O4(aq)

Kc

H2C2O4(aq) ∆ 2 H+(aq) + C2O4

2-(aq) Kc = 3.8 * 10-6

HF(aq) ∆ H+(aq) + F-(aq) Kc = 6.8 * 10-4


How does the magnitude of the equilibrium constant for the reaction change if the equilibrium is written

Sometimes, as in problems that utilize Hess’s law (Section 5.6), we must useequations made up of two or more steps in the overall process. We obtain thenet equation by adding the individual equations and canceling identical terms.Consider the following two reactions, their equilibrium-constant expressions,and their equilibrium constants at 100°C:

The net sum of these two equations is

and the equilibrium-constant expression for the net equation is the product ofthe expressions for the individual steps:

Because the equilibrium-constant expression for the net equation is the productof two equilibrium-constant expressions, the equilibrium constant for the netequation is the product of the two individual equilibrium constants:

To summarize:

1. The equilibrium constant of a reaction in the reverse direction is the inverseof the equilibrium constant of the reaction in the forward direction.

2. The equilibrium constant of a reaction that has been multiplied by a numberis the equilibrium constant raised to a power equal to that number.

3. The equilibrium constant for a net reaction made up of two or more steps isthe product of the equilibrium constants for the individual steps.

Kc = 0.014 * 7.2 = 0.10.

Kc =[NO]2[BrCl]2

[NOBr]2[Cl]2=

[NO]2[Br2]

[NOBr]2 *[BrCl]2

[Br2][Cl2]

2 NOBr(g) + Cl2(g) ∆ 2 NO(g) + 2 BrCl(g)

Br2(g) + Cl2(g) ∆ 2 BrCl(g) Kc =[BrCl]2

[Br2][Cl2]= 7.2

2 NOBr(g) ∆ 2 NO(g) + Br2(g) Kc =[NO]2[Br2]

[NOBr]2 = 0.014

6 HI(g) ∆ 3 H2(g) + 3 I2(g)?H2(g) + I2(g)2 HI(g) ∆Kp

Solve: If we multiply the first equa-tion by 2 and make the correspond-ing change to its equilibrium constant(raising to the power 2), we get 2 HF(aq) ∆ 2 H+(aq) + 2 F-(aq) Kc = (6.8 * 10-4)2 = 4.6 * 10-7

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15.4 | Heterogeneous Equilibria 641

Reversing the second equation andagain making the correspondingchange to its equilibrium constant(taking the reciprocal) gives

2 H+(aq) + C2O4

2-(aq) ∆ H2C2O4(aq) Kc =1

3.8 * 10-6 = 2.6 * 105

Now we have two equations thatsum to give the net equation, andwe can multiply the individual values to get the desired equilibri-um constant.

Kc

2 HF(aq) + C2O4

2-(aq) ∆ 2 F-(aq) + H2C2O4(aq) Kc = (4.6 * 10-7)(2.6 * 105) = 0.12

2 H+(aq) + C2O4

2-(aq) ∆ H2C2O4(aq) Kc = 2.5 * 105

2 HF(aq) ∆ 2 H+(aq) + 2 F-(aq) Kc = 4.6 * 10-7

PRACTICE EXERCISEGiven that, at 700 K, for the reaction and for the reaction

determine the value of for the reaction at 700 K.

Answer:(54.0)3

1.04 * 10-4 = 1.51 * 109

2 NH3(g) + 3 I2(g) ∆ 6 HI(g) + N2(g)Kp3 H2(g) ∆ 2 NH3(g),N2(g) +Kp = 1.04 * 10-4H2(g) + I2(g) ∆ 2 HI(g)Kp = 54.0

15.4 Heterogeneous Equilibria

Many equilibria, such as the hydrogen-nitrogen-ammonia system, involve sub-stances all in the same phase. Such equilibria are called homogeneous equilib-ria. In other cases the substances in equilibrium are in different phases, givingrise to heterogeneous equilibria. As an example, consider the equilibrium thatoccurs when solid lead(II) chloride dissolves in water to form a saturat-ed solution:

[15.18]

This system consists of a solid in equilibrium with two aqueous species. If wewrite the equilibrium-constant expression for this process, we encounter aproblem we have not encountered previously: How do we express the concen-tration of a solid substance? Although it is possible to express the concentrationof a solid in terms of moles per unit volume, it is unnecessary to do so in writ-ing equilibrium-constant expressions. Whenever a pure solid or a pure liquid is in-volved in a heterogeneous equilibrium, its concentration is not included in theequilibrium-constant expression for the reaction. Thus, the equilibrium-constant ex-pression for Equation 15.18 is

[15.19]

Even though does not appear in the equilibrium-constant expression,it must be present for equilibrium to occur.

The fact that pure solids and pure liquids are excluded from equilibrium-constant expressions can be explained in two ways. First, the concentration of apure solid or liquid has a constant value. If the mass of a solid is doubled, its vol-ume also doubles. Thus, its concentration, which relates to the ratio of mass tovolume, stays the same. Because equilibrium-constant expressions include termsonly for reactants and products whose concentrations can change during a chem-ical reaction, the concentrations of pure solids and pure liquids are omitted.

The omission of pure solids and liquids from equilibrium-constant ex-pressions can also be rationalized in a second way. Recall from the last sectionthat what is substituted into a thermodynamic equilibrium expression is theactivity of each substance, which is a ratio of the concentration to a referencevalue. For a pure substance, the reference value is the concentration of thepure substance itself, so that the activity of any pure solid or liquid is alwayssimply 1.

PbCl2(s)

Kc = [Pb2+][Cl-]2

PbCl2(s) ∆ Pb2+(aq) + 2 Cl-(aq)

(PbCl2)

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(a)

CaO CaOCaCO3

(b)

CaCO3

CO2 (g)CO2 (g) CO2 (g)CO2 (g)

» Figure 15.8 A heterogeneousequilibrium. The equilibrium involving CaO, and is a heterogeneous equilibrium. The equilibrium pressure of is the same in thetwo bell jars as long as the two systems are atthe same temperature, even though the relativeamounts of pure and CaO differ greatly.The equilibrium-constant expression for thereaction is Kp = PCO2

.

CaCO3

CO2

CO2

CaCO3 ,



Write the equilibrium-constant expression for the evaporation of water,, in terms of partial pressures,

As a further example of a heterogeneous reaction, consider the decomposi-tion of calcium carbonate:

Omitting the concentrations of solids from the equilibrium-constant expressiongives

These equations tell us that at a given temperature, an equilibrium among CaO, and will always lead to the same partial pressure of as long as allthree components are present. As shown in Figure 15.8 Á, we would have the samepressure of regardless of the relative amounts of CaO and

When a solvent is involved as a reactant or product in an equilibrium, itsconcentration is also excluded from the equilibrium-constant expression, pro-vided the concentrations of reactants and products are low, so that the solventis essentially a pure substance. Applying this guideline to an equilibrium in-volving water as a solvent,

[15.20]

gives an equilibrium-constant expression in which is excluded:

[15.21]Kc =[OH-][HCO3

-]

[CO3

2-]

[H2O]

H2O(l) + CO3

2-(aq) ∆ OH-(aq) + HCO3

-(aq)

CaCO3.CO2

CO2CO2

CaCO3,

Kc = [CO2] and Kp = PCO2

CaCO3(s) ∆ CaO(s) + CO2(g)

Kp .H2O(l) ∆ H2O(g)

SAMPLE EXERCISE 15.6 | Writing Equilibrium-Constant Expressions for Heterogeneous ReactionsWrite the equilibrium-constant expression for for each of the following reactions:

(a)(b)

SolutionAnalyze: We are given two chemical equations, both for heterogeneous equilibria, and asked to write the correspondingequilibrium-constant expressions.Plan: We use the law of mass action, remembering to omit any pure solids, pure liquids, and solvents from the expressions.

SnO2(s) + 2 CO(g) ∆ Sn(s) + 2 CO2(g)CO2(g) + H2(g) ∆ CO(g) + H2O(l)

Kc

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15.5 | Calculating Equilibrium Constants 643

Solve: (a) The equilibrium-constantexpression is Kc =

[CO][CO2][H2]

Because appears in the reaction as a pure liquid, its concentration does not appear in the equilibrium-constant expression.H2O

(b) The equilibrium-constant ex-pression is Kc =

[CO2]2

[CO]2

Because and Sn are both pure solids, their concentrations do not appear in the equilibrium-constant expression.

PRACTICE EXERCISEWrite the following equilibrium-constant expressions:

(a) for (b) for

Answers: (a) (b) Kp =(PH2)

4

(PH2O)4Kc =[Cr3+]

[Ag+]3 ,

3 Fe(s) + 4 H2O(g) ∆ Fe3O4(s) + 4 H2(g)Kp

Cr(s) + 3 Ag+(aq) ∆ Cr3+(aq) + 3 Ag(s)Kc

SnO2

SAMPLE EXERCISE 15.7 | Analyzing a Heterogeneous EquilibriumEach of the following mixtures was placed in a closed container and allowed to stand.Which is capable of attaining the equilibrium (a) pure (b) CaO and a pressure greater than the value of (c) some

and a pressure greater than the value of (d) and CaO?

SolutionAnalyze: We are asked which of several combinations of species can establish anequilibrium between calcium carbonate and its decomposition products, calciumoxide and carbon dioxide.Plan: In order for equilibrium to be achieved, it must be possible for both the for-ward process and the reverse process to occur. In order for the forward process tooccur, there must be some calcium carbonate present. In order for the reverse processto occur, there must be both calcium oxide and carbon dioxide. In both cases, eitherthe necessary compounds may be present initially, or they may be formed by reactionof the other species.Solve: Equilibrium can be reached in all cases except (c) as long as sufficient quanti-ties of solids are present. (a) simply decomposes, forming CaO(s) and until the equilibrium pressure of is attained. There must be enough however, to allow the pressure to reach equilibrium. (b) continues to com-bine with CaO until the partial pressure of the decreases to the equilibriumvalue. (c) There is no CaO present, so equilibrium can’t be attained because there isno way the pressure can decrease to its equilibrium value (which would requiresome of the to react with CaO). (d) The situation is essentially the same as in (a):

decomposes until equilibrium is attained. The presence of CaO initiallymakes no difference.

PRACTICE EXERCISEWhen added to in a closed container, which one of the followingsubstances— —will allow equilibrium to be established in thereaction Answer: only

15.5 Calculating Equilibrium Constants

One of the first tasks confronting Haber when he approached the problem of am-monia synthesis was finding the magnitude of the equilibrium constant for thesynthesis of at various temperatures. If the value of K for Equation 15.6 wasvery small, the amount of in an equilibrium mixture would be small relativeto the amounts of and That is, if the equilibrium lies too far to the left, itwould be impossible to develop a satisfactory synthesis process for ammonia.

H2.N2

NH3

NH3

H2(g)3 Fe(s) + 4 H2O(g) ∆ Fe3O4(s) + 4 H2(g)?

O2(g)H2(g), H2O(g),Fe3O4(s)

CaCO3

CO2

CO2

CO2

CO2CO2

CaCO3,CO2

CO2(g)CaCO3

CaCO3Kp ,CO2CaCO3

Kp ,CO2CaCO3,CaCO3(s) ∆ CaO(s) + CO2(g):

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Haber and his coworkers therefore evaluated the equilibrium constants forthis reaction at various temperatures. The method they employed is analogousto that described in constructing Table 15.1: They started with various mixturesof and allowed the mixtures to achieve equilibrium at a specifictemperature, and measured the concentrations of all three gases at equilibrium.Because the equilibrium concentrations of all products and reactants wereknown, the equilibrium constant could be calculated directly from theequilibrium-constant expression.

SAMPLE EXERCISE 15.8 | Calculating K When All Equilibrium ConcentrationsAre Known

A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibri-um at 472°C. The equilibrium mixture of gases was analyzed and found to contain7.38 atm 2.46 atm and 0.166 atm From these data, calculate the equilib-rium constant for the reaction

SolutionAnalyze: We are given a balanced equation and equilibrium partial pressures andare asked to calculate the value of the equilibrium constant.Plan: Using the balanced equation, we write the equilibrium-constant expression.We then substitute the equilibrium partial pressures into the expression and solvefor

Solve:

PRACTICE EXERCISEAn aqueous solution of acetic acid is found to have the following equilibriumconcentrations at 25°C: and

Calculate the equilibrium constant for the ioniza-tion of acetic acid at 25°C. The reaction is

Answer:

We often don’t know the equilibrium concentrations of all chemical speciesin an equilibrium mixture. If we know the equilibrium concentration of at leastone species, however, we can generally use the stoichiometry of the reaction todeduce the equilibrium concentrations of the others. The following steps out-line the procedure we use to do this:

1. Tabulate all the known initial and equilibrium concentrations of the speciesthat appear in the equilibrium-constant expression.

2. For those species for which both the initial and equilibrium concentrationsare known, calculate the change in concentration that occurs as the systemreaches equilibrium.

3. Use the stoichiometry of the reaction (that is, use the coefficients in the bal-anced chemical equation) to calculate the changes in concentration for allthe other species in the equilibrium.

4. From the initial concentrations and the changes in concentration, calculatethe equilibrium concentrations. These are used to evaluate the equilibriumconstant.

1.79 * 10-5

HC2H3O2(aq) ∆ H+(aq) + C2H3O2

-(aq)

Kc[C2H3O2

-] = 5.44 * 10-4 M.[HC2H3O2] = 1.65 * 10-2 M; [H+] = 5.44 * 10-4 M;

Kp =(PNH3)

2

PN2(PH2)3 =

(0.166)2

(2.46)(7.38)3 = 2.79 * 10-5

Kp .

N2(g) + 3 H2(g) ∆ 2 NH3(g)

Kp

NH3.N2,H2,

NH3,N2, H2,ACTIVITYUsing an Equilibrium Table

Steven F. Watkins, “Applying the ReactionTable Method for Chemical ReactionProblems (Stoichiometry and Equilibrium),”J. Chem. Educ., Vol. 80, 2003, 658–661.

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15.5 | Calculating Equilibrium Constants 645

SAMPLE EXERCISE 15.9 | Calculating K from Initial and Equilibrium ConcentrationsA closed system initially containing and at 448°C is allowed to reach equilibrium.Analysis of the equilibrium mixture shows that the concentration of HI is Calculate at 448°C for the reac-tion taking place, which is

SolutionAnalyze: We are given the initial concentrations of and and the equilibrium concentration of HI. We are asked to calculatethe equilibrium constant for the reaction Plan: We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calcu-late the equilibrium constant.

H2(g) + I2(g) ∆ 2 HI(g).Kc

I2H2

H2(g) + I2(g) ∆ 2 HI(g)

Kc1.87 * 10-3 M.2.000 * 10-3 M I21.000 * 10-3 M H2

Solve: First, we tabulate the initialand equilibrium concentrations ofas many species as we can. We alsoprovide space in our table for listingthe changes in concentrations. Asshown, it is convenient to use thechemical equation as the headingfor the table.

Initial 0 M

Change

Equilibrium 1.87 * 10-3 M

2.000 * 10-3 M1.000 * 10-3 M

2 HI(g)∆I2(g)+H2(g)

Comment: The same method can be applied to gaseous equilibrium problems, in which case partial pressures are used as tableentries in place of molar concentrations.

PRACTICE EXERCISESulfur trioxide decomposes at high temperature in a sealed container: Initially, the vessel ischarged at 1000 K with at a partial pressure of 0.500 atm. At equilibrium the partial pressure is 0.200 atm. Calculate thevalue of at 1000 K.Answer: 0.338

Kp

SO3SO3(g)2 SO3(g) ∆ 2 SO2(g) + O2(g).

Second, we calculate the change inconcentration of HI, which is thedifference between the equilibriumvalues and the initial values: Change in [HI] = 1.87 * 10-3 M - 0 = 1.87 * 10-3 M

Third, we use the coefficients in thebalanced equation to relate thechange in [HI] to the changes in and [I2]:

[H2]

a1.87 * 10-3 mol HI

Lb ¢ 1 mol I2

2 mol HI≤ = 0.935 * 10-3

mol I2

L

a1.87 * 10-3 mol HI

Lb ¢1 mol H2

2 mol HI≤ = 0.935 * 10-3

mol H2

L

Fourth, we calculate the equilibriumconcentrations of and usingthe initial concentrations and thechanges. The equilibrium concentra-tion equals the initial concentrationminus that consumed:

I2 ,H2

[I2] = 2.000 * 10-3 M - 0.935 * 10-3 M = 1.065 * 10-3 M

[H2] = 1.00 * 10-3 M - 0.935 * 10-3 M = 0.065 * 10-3 M

The completed table now looks likethis (with equilibrium concentra-tions in blue for emphasis): Initial 0 M

Change

Equilibrium 1.87 * 10-3 M1.065 * 10-3 M0.065 * 10-3 M

+1.87 * 10-3 M-0.935 * 10-3 M-0.935 * 10-3 M

2.000 * 10-3 M1.000 * 10-3 M

2 HI(g)∆I2(g)+H2(g)

Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed.

Finally, now that we know the equi-librium concentration of each reac-tant and product, we can use theequilibrium-constant expression tocalculate the equilibrium constant.

Kc =[HI]2

[H2][I2]=

(1.87 * 10-3)2

(0.065 * 10-3)(1.065 * 10-3)= 51

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Reactionformsproducts

Reactionformsreactants

Equilib-rium

K

Q

K Q K

Q

Á Figure 15.9 Predicting the direction ofa reaction by comparing Q and K. The relativemagnitudes of the reaction quotient Q and theequilibrium constant K indicate how the reactionmixture changes as it moves toward equilibrium.If Q is smaller than K, the reaction proceeds fromleft to right until When thereaction is at equilibrium and has no tendency tochange. If Q is larger than K, the reactionproceeds from right to left until Q = K.

Q = K,Q = K.


15.6 Applications of Equilibrium Constants

We have seen that the magnitude of K indicates the extent to which a reaction willproceed. If K is very large, the reaction will tend to proceed far to the right; if K isvery small (that is, much less than 1), the equilibrium mixture will contain mainlyreactants. The equilibrium constant also allows us to (1) predict the direction inwhich a reaction mixture will proceed to achieve equilibrium and (2) calculate theconcentrations of reactants and products when equilibrium has been reached.

Predicting the Direction of ReactionFor the formation of from and (Equation 15.6), at 472°C.Suppose we place a mixture of 2.00 mol of 1.00 mol of and 2.00 mol of

in a 1.00-L container at 472°C. How will the mixture react to reach equilib-rium? Will and react to form more or will decompose to form

and To answer this question, we can substitute the starting concentrations of

and into the equilibrium-constant expression and compare itsvalue to the equilibrium constant:

To reach equilibrium, the quotient will need to decrease fromthe starting value of 0.500 to the equilibrium value of 0.105. This change canhappen only if the concentration of decreases and the concentrations of and increase. Thus, the reaction proceeds toward equilibrium by forming

and from that is, the reaction proceeds from right to left.The approach we have illustrated can be formalized by defining a quantity

called the reaction quotient. The reaction quotient, Q, is a number obtained by substi-tuting starting reactant and product concentrations or partial pressures into anequilibrium-constant expression. Thus, in our example, when we substituted the start-ing concentrations into the equilibrium-constant expression, we obtained

When we substitute the starting partial pressures into the equilibrium-constant expression, we will label the reaction quotient as .

To determine the direction in which the reaction will proceed to achieveequilibrium, we compare the values of and or and Three possiblesituations arise:

• The reaction quotient will equal the equilibrium constant only if thesystem is already at equilibrium.

• The concentration of products is too large and that of reactants toosmall. Thus, substances on the right side of the chemical equation will reactto form substances on the left; the reaction moves from right to left in ap-proaching equilibrium.

• The concentration of products is too small and that of reactants toolarge. Thus, the reaction will achieve equilibrium by forming more prod-ucts; it moves from left to right.

These relationships are summarized in Figure 15.9 «.

Q 6 K:

Q 7 K:

Q = K:

Kp .QpKcQc

Qp

Qc = 0.500.

NH3 ;H2N2

H2

N2NH3

[NH3]2>[N2][H2]3

[NH3]2

[N2][H2]3 =(2.00)2

(1.00)(2.00)3 = 0.500 whereas Kc = 0.105

NH3N2, H2,

H2 ?N2

NH3NH3,H2N2

NH3

N2,H2,Kc = 0.105H2N2NH3

SAMPLE EXERCISE 15.10 | Predicting the Direction of Approach to EquilibriumAt 448°C the equilibrium constant for the reaction

is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448°C if we start with mol of HI,of and of in a 2.00-L container.I23.0 * 10-2 molH2,1.0 * 10-2 mol

2.0 * 10-2

H2(g) + I2(g) ∆ 2 HI(g)

Kc

Students commonly have troublediscriminating between Q and . The valueof Q equals that of only when the systemis at equilibrium. However, they arecalculated in the same way: usesequilibrium concentrations whereas Q usesnonequilibrium concentrations (usuallyinitial concentrations).

Kc

Kc

Kc

The techniques presented on the nextseveral pages require students to be able tomanipulate algebraic equations. Having aproper math background is more importanthere than in any other section of the text.

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15.6 | Applications of Equilibrium Constants 647

Solve: The initial concentrations are

[I2] = 3.0 * 10-2 mol>2.00 L = 1.5 * 10-2 M

[H2] = 1.0 * 10-2 mol>2.00 L = 5.0 * 10-3 M

[HI] = 2.0 * 10-2 mol>2.00 L = 1.0 * 10-2 M

The reaction quotient is thereforeQc =

[HI]2

[H2][I2]=

(1.0 * 10-2)2

(5.0 * 10-3)(1.5 * 10-2)= 1.3

Because the concentration of HI must increase and the concentrations of and must decrease to reach equilibrium;the reaction will proceed from left to right as it moves toward equilibrium.

I2H2Qc 6 Kc ,

SolutionAnalyze: We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which directionthe reaction must proceed to achieve equilibrium.Plan: We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting con-centrations into the equilibrium-constant expression to calculate the reaction quotient, Comparing the magnitudes of theequilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed.

Qc .

PRACTICE EXERCISEAt 1000 K the value of for the reaction is 0.338. Calculate the value for and predict the di-rection in which the reaction will proceed toward equilibrium if the initial partial pressures are

Answer: and so the reaction will proceed from right to left, forming more SO3.Qp = 16; Qp 7 Kp ,PO2 = 2.5 atm.

PSO2 = 0.41 atm;PSO3 = 0.16 atm;Qp ,2 SO3(g) ∆ 2 SO2(g) + O2(g)Kp

SAMPLE EXERCISE 15.11 | Calculating Equilibrium ConcentrationsFor the Haber process, at 500°C. In an equilibrium mixture of the three gases at500°C, the partial pressure of is 0.928 atm and that of is 0.432 atm. What is the partial pressure of in this equilibriummixture?

SolutionAnalyze: We are given an equilibrium constant, and the equilibrium partial pressures of two of the three substances in theequation ( and ), and we are asked to calculate the equilibrium partial pressure for the third substance Plan: We can set equal to the equilibrium-constant expression and substitute in the partial pressures that are known. Then wecan solve for the only unknown in the equation.

Kp

(NH3).H2N2

Kp ,

NH3N2H2

Kp = 1.45 * 10-5N2(g) + 3 H2(g) ∆ 2 NH3(g),

Calculating Equilibrium ConcentrationsChemists frequently need to calculate the amounts of reactants and productspresent at equilibrium. Our approach in solving problems of this type is similarto the one we used for evaluating equilibrium constants: We tabulate the initialconcentrations or partial pressures, the changes therein, and the final equilibri-um concentrations or partial pressures. Usually we end up using theequilibrium-constant expression to derive an equation that must be solved foran unknown quantity, as demonstrated in Sample Exercise 15.11.

Solve: We tabulate the equilibriumpressures as follows: Equilibrium pressure (atm) 0.432 0.928 x

N2(g) + 3 H2(g) ∆ 2 NH3(g)

Because we do not know the equilib-rium pressure of we representit with a variable, x. At equilibriumthe pressures must satisfy theequilibrium-constant expression:

NH3,

Kp =(PNH3)

2

PN2(PH2)3 =

x2

(0.432)(0.928)3 = 1.45 * 10-5

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In many situations we will know the value of the equilibrium constant andthe initial amounts of all species. We must then solve for the equilibriumamounts. Solving this type of problem usually entails treating the change inconcentration as a variable. The stoichiometry of the reaction gives us the rela-tionship between the changes in the amounts of all the reactants and products,as illustrated in Sample Exercise 15.12.

SAMPLE EXERCISE 15.12 | Calculating Equilibrium Concentrations from Initial ConcentrationsA 1.000-L flask is filled with 1.000 mol of and 2.000 mol of at 448°C. The value of the equilibrium constant for the reaction

at 448°C is 50.5. What are the equilibrium concentrations of and HI in moles per liter?H2, I2 ,

H2(g) + I2(g) ∆ 2 HI(g)

KcI2H2

Solve: First, we note the initialconcentrations of and in the1.000-L flask:

I2H2

Initial 1.000 M 2.000 M 0 M

Change

Equilibrium

2 HI(g)∆I2(g)+H2(g)

and [I2] = 2.000 M[H2] = 1.000 M

Third, we use the stoichiometry ofthe reaction to determine the changesin concentration that occur as the re-action proceeds to equilibrium. Theconcentrations of and will de-crease as equilibrium is established,and that of HI will increase. Let’srepresent the change in concentra-tion of by the variable x. The bal-anced chemical equation tells us therelationship between the changes inthe concentrations of the three gases:

H2

I2H2For each x mol of that reacts, x mol of are consumed and 2x mol of HI areproduced:

Initial 1.000 M 2.000 M 0 M

Change

Equilibrium

+2x-x-x

2 HI(g)∆I2(g)+H2(g)

I2H2

Fourth, we use the initial concentra-tions and the changes in concen-trations, as dictated by stoichiometry,to express the equilibrium concentra-tions. With all our entries, our tablenow looks like this:

Initial 1.000 M 2.000 M 0 M

Change

Equilibrium 2x M12.000 - x2 M11.000 - x2 M+2x-x-x

2 HI(g)∆I2(g)+H2(g)

SolutionAnalyze: We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container andare asked to calculate the equilibrium concentrations of all species.Plan: In this case we are not given any of the equilibrium concentrations. We must develop some relationships that relate the ini-tial concentrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.9,where we calculated an equilibrium constant using initial concentrations.

Comment: We can always checkour answer by using it to recalculatethe value of the equilibrium constant:

Kp =(2.24 * 10-3)2

(0.432)(0.928)3 = 1.45 * 10-5

PRACTICE EXERCISEAt 500 K the reaction has In an equilibrium mixture at 500 K, the partial pressure of

is 0.860 atm and that of is 0.350 atm. What is the partial pressure of in the equilibrium mixture?Answer: 1.22 atm

Cl2PCl3PCl5

Kp = 0.497.PCl5(g) ∆ PCl3(g) + Cl2(g)

Second, we construct a table in whichwe tabulate the initial concentrations:

When solving for x, especially when usingthe quadratic formula, it is imperative tocheck if the answer obtained is reasonable.

We now rearrange the equation tosolve for x:

x = 35.01 * 10-6 = 2.24 * 10-3 atm = PNH3

x2 = (1.45 * 10-5)(0.432)(0.928)3 = 5.01 * 10-6

For the equation , thesolution x is given by the quadratic formula;see Appendix A.

0 = a x

2 + bx + c

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12.98.8

20.8

16.926.018.9

37.8

32.242.9

35.927.4

48.854.947.8

38.7

80

60

40

20

0200

300Total pressure (atm)

Temperature (�C)

400 400

450

500

600

500

Percent NH3at equilibrium 60.6

15.7 | Le Châtelier’s Principle 649

Fifth, we substitute the equilibriumconcentrations into the equilibrium-constant expression and solve forthe single unknown, x:

Kc =[HI]2

[H2][I2]=

12x2211.000 - x212.000 - x2 = 50.5

If you have an equation-solving cal-culator, you can solve this equationdirectly for x. If not, expand this ex-pression to obtain a quadratic equa-tion in x: 46.5x2 - 151.5x + 101.0 = 0

4x2 = 50.51x2 - 3.000x + 2.0002

Solving the quadratic equation(Appendix A.3) leads to two solu-tions for x:

x =-1-151.52 ; 31-151.522 - 4146.521101.02

2146.52 = 2.323 or 0.935

When we substitute intothe expressions for the equilibriumconcentrations, we find negative con-centrations of and Because anegative concentration is not chemi-cally meaningful, we reject this solu-tion. We then use to findthe equilibrium concentrations:

x = 0.935

I2 .H2

x = 2.323

[HI] = 2x = 1.870 M

[I2] = 2.000 - x = 1.065 M

[H2] = 1.000 - x = 0.065 M

Check: We can check our solutionby putting these numbers into theequilibrium-constant expression:

Kc =[HI]2

[H2][I2]=

11.8702210.065211.0652 = 51

Comment: Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions will not be chemicallymeaningful and should be rejected.

PRACTICE EXERCISEFor the equilibrium the equilibrium constant has the value 0.497 at 500 K. A gas cylinder at500 K is charged with at an initial pressure of 1.66 atm. What are the equilibrium pressures of and at thistemperature?Answer: PPCl5 = 0.967 atm; PPCl3 = PCl2 = 0.693 atm

Cl2PCl5 , PCl3 ,PCl5(g)KpPCl5(g) ∆ PCl3(g) + Cl2(g),

15.7 Le Châtelier’s Principle

In developing his process for making ammonia from and Haber soughtthe factors that might be varied to increase the yield of Using the valuesof the equilibrium constant at various temperatures, he calculated the equilibri-um amounts of formed under a variety of conditions. Some of his resultsare shown in Figure 15.10 ». Notice that the percent of present at equilib-rium decreases with increasing temperature and increases with increasingpressure. We can understand these effects in terms of a principle firstput forward by Henri-Louis Le Châtelier* (1850–1936), a French in-dustrial chemist. Le Châtelier’s principle can be stated asfollows: If a system at equilibrium is disturbed by achange in temperature, pressure, or the concentra-tion of one of the components, the system will shiftits equilibrium position so as to counteract the ef-fect of the disturbance.

In this section we will use Le Châtelier’sprinciple to make qualitative predictions abouthow a system at equilibrium responds to variouschanges in external conditions. We will consider three

NH3

NH3

NH3.H2,N2

¥ Figure 15.10 Effect of temperature andpressure on the percentage of in anequilibrium mixture of and Eachmixture was produced by starting with a molarmixture of and The yield of is greatestat the lowest temperature and at the highestpressure.

NH3N2 .H2

3 : 1NH3 .N2 , H2 ,

NH3

* Pronounced “le-SHOT-lee-ay.”

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ways that a chemical equilibrium can be disturbed: (1) adding or removing a re-actant or product, (2) changing the pressure by changing the volume, and(3) changing the temperature.

Change in Reactant or Product ConcentrationsA system at equilibrium is in a dynamic state; the forward and reverseprocesses are occurring at equal rates, and the system is in a state of balance.Altering the conditions of the system may disturb the state of balance. If thisoccurs, the equilibrium shifts until a new state of balance is attained. LeChâtelier’s principle states that the shift will be in the direction that mini-mizes or reduces the effect of the change. Therefore, if a chemical system is atequilibrium and we add a substance (either a reactant or a product), the reaction willshift so as to reestablish equilibrium by consuming part of the added substance. Con-versely, removing a substance will cause the reaction to move in the direction thatforms more of that substance.

As an example, consider an equilibrium mixture of and

Adding would cause the system to shift so as to reduce the newly increasedconcentration of This change can occur only by consuming and simul-taneously consuming to form more This situation is illustrated inFigure 15.11 ¥. Adding more to the equilibrium mixture would likewisecause the direction of the reaction to shift toward forming more Remov-ing would also cause a shift toward producing more whereas adding

to the system at equilibrium would cause the concentrations to shift in thedirection that reduces the newly increased concentration; that is, some ofthe added ammonia would decompose to form and

In the Haber reaction, therefore, removing from an equilibrium mix-ture of and causes the reaction to shift from left to right to formmore If the can be removed continuously, the yield of can beincreased dramatically. In the industrial production of ammonia, the iscontinuously removed by selectively liquefying it; the boiling point of

is much higher than that of and Theliquid is removed, and the and are recycled to form more asdiagrammed in Figure 15.12 ». By continuously removing the product, the re-action is driven essentially to completion.

NH3,H2N2NH3

H2 (-253°C).N2 (-196°C)NH3 (-33°C)

NH3

NH3NH3NH3.NH3N2, H2,

NH3

H2.N2

NH3

NH3

NH3,NH3

NH3.N2

NH3.N2

H2H2.H2

N2(g) + 3 H2(g) ∆ 2 NH3(g)

NH3 :N2, H2,

Part

ial p

ress

ure

Time

N2

NH3

H2

Initialequilibrium

Equilibriumreestablished

H2 added at this time

» Figure 15.11 Effect of addingto an equilibrium mixture of

and When is added, aportion of the reacts with to form

thereby establishing anew equilibrium position that has thesame equilibrium constant. The resultsshown are in accordance withLe Châtelier’s principle.

NH3 ,N2H2

H2NH3 .N2 , H2 ,H2

John Olmsted III, “Amounts Tables as aDiagnostic Tool for Flawed StoichiometricReasoning,” J. Chem. Educ., Vol. 76, 1999,52–54.

E. Weltin, “Calculating EquilibriumConcentrations by Iteration: Recycle YourApproximations,” J. Chem. Educ., Vol. 72,1995, 36–38.

Consider changing the concentration orpressure of a substance in a reaction atequilibrium as creating stress. The removalof something creates a “deficit stress,” andadding something creates an “excess stress.”Reactions will always respond to stress byreducing the deficit (making more of whatwas removed) or by reducing the excess(consuming some of what was added). Thisdescription works regardless of which side ofthe reaction is changed.

Michael Sutton, “Man of Principle,”Chemistry in Britain, June 2000, 43–44.

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Expandinggases cool

RecycledunreactedN2 and H2

Liquid NH3

Refrigeratedunit

Heatexchanger

Heatexchanger

Pump to circulateand compress gases

Heatingcoil

Catalyst(460–550�C)

Preheatedfeed gasesNH3 outlet

N2, H2inlet

« Figure 15.12 Schematic diagramsummarizing the industrial production ofammonia. Incoming and gases are heatedto approximately 500°C and passed over acatalyst. The resultant gas mixture is allowed toexpand and cool, causing to liquefy.Unreacted and gases are recycled.H2N2

NH3

H2N2


What happens to the equilibrium if (a) isadded to the system, (b) NO is removed?

Effects of Volume and Pressure ChangesIf a system is at equilibrium and its volume is decreased, thereby increasing itstotal pressure, Le Châtelier’s principle indicates that the system will respond byshifting its equilibrium position to reduce the pressure. A system can reduce itspressure by reducing the total number of gas molecules (fewer molecules of gasexert a lower pressure). Thus, at constant temperature, reducing the volume of agaseous equilibrium mixture causes the system to shift in the direction that reduces thenumber of moles of gas. Conversely, increasing the volume causes a shift in the di-rection that produces more gas molecules.

For example, let’s again consider the equilibrium What happens if the total pressure of an equilibrium mixture is increased bydecreasing the volume as shown in the sequential photos in Figure 15.13 »? Ac-cording to Le Châtelier’s principle, we expect the equilibrium to shift to theside that reduces the total number of moles of gas, which is the reactant side inthis case. We therefore expect the equilibrium to shift to the left, so that isconverted into as equilibrium is reestablished. In Figure 15.13(a) andFigure 15.13(b), compressing the gas mixture initially causes the color to dark-en as the concentration of increases. The color then fades as equilibrium isreestablished, Figure 15.13(c). The color fades because the pressure increasecauses the equilibrium to shift in favor of colorless N2O4.

NO2

N2O4

NO2

N2O4(g) ∆ 2 NO2(g).

O22 NO(g) + O2(g) ∆ 2 NO2(g)

William R. Smith and Ronald W. Missen,“Chemical Equilibrium and PolynomialEquations: Beware of Roots,” J. Chem. Educ.,Vol. 66, 1989, 489–490.

Richard S. Treptow, “Le Châtelier’sPrinciple,” J. Chem. Educ., Vol. 57, 1980, 417–420.

Lee R. Summerlin and James L. Ealy, Jr.,“Effect of Concentration on Equilibrium:Cobalt Complex,” Chemical Demonstrations,A Sourcebook for Teachers, Vol. 1(Washington: American Chemical Society,1988), pp. 81–82.

MOVIENitrogen Dioxide and Dinitrogen Tetroxide

ANIMATIONSEquilibrium, Temperature

Dependence of EquilibriumNO2 ¬ N2O4

BROWMC15_628-667PR2 1/26/05 15:08 Page 651


NO2

N2O4

The volume and hence the pressure are changed by moving the plunger.

Compression of the mixture temporarily increases the concentration

of NO2.

An equilibrium mixture of brown NO2(g) (red) and

colorless N2O4(g) (gray) held in a gas-tight syringe.

When the mixture reestablishes equilibrium, the color is as light as that at the beginning because the

formation of N2O4(g) is favored by the pressure increase.

LE CHATELIER’S PRINCIPLE^

If a system at equilibrium is disturbed by a change in temperature, pressure,or the concentration of one of the components, the system will shift itsequilibrium position so as to counteract the effect of the disturbance.

Á Figure 15.13 Effect of pressure on anequilibrium.


What happens to the equilibrium if the volume ofthe system is increased?

For the reaction there are four moleculesof reactant consumed for every two molecules of product produced. Conse-quently, an increase in pressure (decrease in volume) causes a shift toward theside with fewer gas molecules, which leads to the formation of more , asindicated in Figure 15.10. In the case of the reaction the number of molecules of gaseous products (two) equals the number of mol-ecules of gaseous reactants; therefore, changing the pressure will not influencethe position of the equilibrium.

Keep in mind that pressure-volume changes do not change the value of K aslong as the temperature remains constant. Rather, they change the partial pressuresof the gaseous substances. In Sample Exercise 15.8 we calculated for anKp

H2(g) + I2(g) ∆ 2 HI(g),NH3

N2(g) + 3 H2(g) ∆ 2 NH3(g),

2 SO2(g) + O2(g) ∆ 2 SO3(g)

BROWMC15_628-667PR2 1/26/05 15:08 Page 652


equilibrium mixture at 472°C that contained andThe value of is Consider what happens when we

suddenly reduce the volume of the system by one-half. If there were no shift inequilibrium, this volume change would cause the partial pressures of all substancesto double, giving and Thereaction quotient would then no longer equal the equilibrium constant.

Because the system is no longer at equilibrium. Equilibrium will bereestablished by increasing and decreasing and until

Therefore, the equilibrium shifts to the right asLe Châtelier’s principle predicts.

It is possible to change the total pressure of the system without changing itsvolume. For example, pressure increases if additional amounts of any of the re-acting components are added to the system. We have already seen how to dealwith a change in concentration of a reactant or product. The total pressure with-in the reaction vessel might also be increased by adding a gas that is not in-volved in the equilibrium. For example, argon might be added to the ammoniaequilibrium system. The argon would not alter the partial pressures of any ofthe reacting components and therefore would not cause a shift in equilibrium.

Effect of Temperature ChangesChanges in concentrations or partial pressures cause shifts in equilibrium with-out changing the value of the equilibrium constant. In contrast, almost everyequilibrium constant changes in value as the temperature changes. For example,consider the equilibrium established when cobalt(II) chloride is dis-solved in hydrochloric acid, HCl(aq):

Pale pink Deep blue [15.22]

The formation of from is an endothermic process. Wewill discuss the significance of this enthalpy change shortly. Because ispink and is blue, the position of this equilibrium is readily apparent fromthe color of the solution. Figure 15.14(a) » shows a room-temperature solution of

in HCl(aq). Both and are present in significantamounts in the solution; the violet color results from the presence of both the pinkand blue ions. When the solution is heated [Figure 15.14(b)], it becomes intenselyblue in color, indicating that the equilibrium has shifted to form more Cooling the solution, as in Figure 15.14(c), leads to a pink solution, indicating thatthe equilibrium has shifted to produce more How can we explain thedependence of this equilibrium on temperature?

We can deduce the rules for the temperature dependence of the equilibri-um constant by applying Le Châtelier’s principle. A simple way to do this isto treat heat as if it were a chemical reagent. In an endothermic reaction we canconsider heat as a reactant, whereas in an exothermic reaction we can considerheat as a product.

Endothermic:

Exothermic:

When the temperature of a system at equilibrium is increased, it is as if we have addeda reactant to an endothermic reaction or a product to an exothermic reaction. Theequilibrium shifts in the direction that consumes the excess reactant (or product),namely heat.

Reactants ∆ products + heat

Reactants + heat ∆ products

Co(H2O)6

2+.

CoCl4

2-.

CoCl4

2-Co(H2O)6

2+CoCl2

CoCl4

2-Co(H2O)6

2+Co(H2O)6

2+CoCl4

2-

Co(H2O)6

2+(aq) + 4 Cl-(aq) ∆ CoCl4

2-(aq) + 6 H2O(l) ¢H 7 0

(CoCl2)

Qp = Kp = 2.79 * 10-5.PH2PN2PNH3

Qp 6 Kp ,

Qp =(PNH3)

2

PN2(PH2)3 =

(0.332)2

(4.92)(14.76)3 = 6.97 * 10-6 Z Kp

PNH3 = 0.332 atm.PH2 = 14.76 atm, PN2 = 4.92 atm,

2.79 * 10-5.Kp0.166 atm NH3.2.46 atm N2,7.38 atm H2,

ACTIVITYLe Châtelier’s Principle

Lee R. Summerlin, and James L. Ealy, Jr.,“Effect of Temperature Change onEquilibrium: Cobalt Complex,” ChemicalDemonstrations, A Sourcebook for Teachers,Vol. 1 (Washington: American ChemicalSociety, 1988), pp. 79–80.

Color changes in a mixture of and as a sealed tube of gas is heated or cooledare used to demonstrate Le Châtelier’sprinciple. Lee R. Summerlin and JamesL. Ealy, Jr., “Equilibrium in the Gas Phase,”Chemical Demonstrations, A Sourcebook forTeachers (Washington: American ChemicalSociety, 1988), pp. 85–86.

N2O4NO2

BROWMC15_628-667PR2 1/26/05 15:08 Page 653


Co(H2O)62�

CoCl42�

Heating the solution shifts the equilibrium to the right, forming

more blue CoCl42�.

At room temperature both the pink Co(H2O)6

2� and blue CoCl42� ions are present in

significant amounts, giving a violet color to the solution.

Cooling the solution shifts the equilibrium to the left, toward

pink Co(H2O)62�.

EFFECT OF TEMPERATURE CHANGESAlmost every equilibrium constant changes in value as the temperature changes. In an endothermic reaction,heat is absorbed as reactants are converted to products, and the equilibrium shifts to the right and K increases.

Lowering the temperature shifts the equilibrium to the side that produces heat, the left, decreasing K.

Á Figure 15.14 Temperature and equilibrium. (The reaction shown is )CoCl4

2�(aq) � 6 H2O(l ).4 Cl-(aq) ∆Co(H2O)6

2�(aq) � 4 Cl-(aq) ∆


Use Le Châtelier’s principle to explain why the equilibrium vapor pressure of a liquidincreases with increasing temperature.

In an endothermic reaction, such as Equation 15.22, heat is absorbed asreactants are converted to products; thus, increasing the temperature causesthe equilibrium to shift to the right, in the direction of products, and Kincreases. For Equation 15.22, increasing the temperature leads to the forma-tion of more as observed in Figure 15.14(b).

In an exothermic reaction the opposite occurs. Heat is absorbed as productsare converted to reactants; therefore the equilibrium shifts to the left and K de-creases. We can summarize these results as follows:

CoCl4

2-,

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Endothermic: Increasing T results in an increase in K.

Exothermic: Increasing T results in a decrease in K.

Cooling a reaction has the opposite effect. As we lower the temperature, theequilibrium shifts to the side that produces heat. Thus, cooling an endothermicreaction shifts the equilibrium to the left, decreasing K. We observed this effectin Figure 15.14(c). Cooling an exothermic reaction shifts the equilibrium to theright, increasing K.

SAMPLE EXERCISE 15.13 | Using Le Châtelier’s Principle to Predict Shifts in Equilibrium

Consider the equilibrium

In which direction will the equilibrium shift when (a) is added, (b) is re-moved, (c) the total pressure is increased by addition of (d) the volume isincreased, (e) the temperature is decreased?

SolutionAnalyze: We are given a series of changes to be made to a system at equilibrium and areasked to predict what effect each change will have on the position of the equilibrium.Plan: Le Châtelier’s principle can be used to determine the effects of each of thesechanges.Solve: (a) The system will adjust to decrease the concentration of the added so the equilibrium shifts to the right, in the direction of products.

(b) The system will adjust to the removal of by shifting to the side that pro-duces more thus, the equilibrium shifts to the right.

(c) Adding will increase the total pressure of the system, but is not in-volved in the reaction. The partial pressures of and are therefore un-changed, and there is no shift in the position of the equilibrium.

(d) If the volume is increased, the system will shift in the direction that occupiesa larger volume (more gas molecules); thus, the equilibrium shifts to the right.(This is the opposite of the effect observed in Figure 15.13, where the volume wasdecreased.)

(e) The reaction is endothermic, so we can imagine heat as a reagent on the reac-tant side of the equation. Decreasing the temperature will shift the equilibrium in thedirection that produces heat, so the equilibrium shifts to the left, toward the forma-tion of more Note that only this last change also affects the value of the equi-librium constant, K.

PRACTICE EXERCISEFor the reaction

in which direction will the equilibrium shift when (a) is removed, (b) the tem-perature is decreased, (c) the volume of the reaction system is increased, (d) isadded?Answers: (a) right, (b) left, (c) right, (d) left

SAMPLE EXERCISE 15.14 | Predicting the Effect of Temperature on K(a) Using the standard heat of formation data in Appendix C, determine the

standard enthalpy change for the reaction

(b) Determine how the equilibrium constant for this reaction should changewith temperature.

SolutionAnalyze: We are asked to determine the standard enthalpy change of a reaction andhow the equilibrium constant for the reaction varies with temperature.

N2(g) + 3 H2(g) ∆ 2 NH3(g)

PCl3(g)Cl2(g)

PCl5(g) ∆ PCl3(g) + Cl2(g) ¢H° = 87.9 kJ

N2O4.

N2O4NO2

N2N2

NO2 ;NO2

N2O4,

N2(g),NO2N2O4

N2O4(g) ∆ 2 NO2(g) ¢H° = 58.0 kJ

A collection of demonstrations dealing withequilibria associated with a sodiumbicarbonate solution. Ron DeLorenzo, “From Chicken Breath to the Killer Lakes ofCameroon: Uniting Seven InterestingPhenomena with a Single ChemicalUnderpinning,” J. Chem. Educ., Vol. 78,2001, pp. 191–194.

Reed A. Howald, “The Fizz Keeper, a CaseStudy in Chemical Education, Equilibrium,and Kinetics,” J. Chem. Educ., Vol. 76, 1999,208–209.

BROWMC15_628-667PR2 1/26/05 15:08 Page 655


TABLE 15.2 Variation in for the Equilibrium as a Function of Temperature

Temperature (°C)

300400450500550600 2.25 * 10-6

5.38 * 10-61.45 * 10-54.51 * 10-51.64 * 10-44.34 * 10-3

Kp

N2 + 3 H2 ∆ 2 NH3

Kp

Plan: (a) We can use standard enthalpies of formation to calculate for thereaction. (b) We can then use Le Châtelier’s principle to determine what effect tem-perature will have on the equilibrium constant.Solve: (a) Recall that the standard enthalpy change for a reaction is given by thesum of the standard molar enthalpies of formation of the products, each multipliedby its coefficient in the balanced chemical equation, less the same quantities for thereactants. At 25°C, for is The values for and

are zero by definition because the enthalpies of formation of the elements intheir normal states at 25°C are defined as zero (Section 5.7). Because 2 mol of isformed, the total enthalpy change is

(b) Because the reaction in the forward direction is exothermic, we can considerheat a product of the reaction. An increase in temperature causes the reaction to shiftin the direction of less and more and This effect is seen in the values for

presented in Table 15.2 «. Notice that changes markedly with changes in tem-perature and that it is larger at lower temperatures.Comment: The fact that for the formation of from and decreases withincreasing temperature is a matter of great practical importance. To form at areasonable rate requires higher temperatures. At higher temperatures, however, theequilibrium constant is smaller, and so the percentage conversion to is smaller.To compensate for this, higher pressures are needed because high pressure favors

formation.

PRACTICE EXERCISEUsing the thermodynamic data in Appendix C, determine the enthalpy change forthe reaction

Use this result to determine how the equilibrium constant for the reaction shouldchange with temperature.Answer: the equilibrium constant will increase with increasingtemperature

The Effect of CatalystsWhat happens if we add a catalyst to a chemical system that is at equilibrium?As shown in Figure 15.15 «, a catalyst lowers the activation barrier between thereactants and products. The activation energy of the forward reaction is low-ered to the same extent as that for the reverse reaction. The catalyst thereby in-creases the rates of both the forward and reverse reactions. As a result, a catalystincreases the rate at which equilibrium is achieved, but it does not change the composi-tion of the equilibrium mixture. The value of the equilibrium constant for a reac-tion is not affected by the presence of a catalyst.

The rate at which a reaction approaches equilibrium is an im-portant practical consideration. As an example, let’s again consid-er the synthesis of ammonia from and In designing aprocess for ammonia synthesis, Haber had to deal with a rapid de-crease in the equilibrium constant with increasing temperature, asshown in Table 15.2. At temperatures sufficiently high to give asatisfactory reaction rate, the amount of ammonia formed was toosmall. The solution to this dilemma was to develop a catalyst thatwould produce a reasonably rapid approach to equilibrium at asufficiently low temperature, so that the equilibrium constant wasstill reasonably large. The development of a suitable catalyst thusbecame the focus of Haber’s research efforts.

After trying different substances to see which would be mosteffective, Haber finally settled on iron mixed with metal oxides.Variants of the original catalyst formulations are still used. Thesecatalysts make it possible to obtain a reasonably rapid approach toequilibrium at temperatures around 400°C to 500°C and with gas

H2.N2

¢H° = 508 kJ;

2 POCl3(g) ∆ 2 PCl3(g) + O2(g)

NH3

NH3

NH3

H2N2NH3Kp

KpKp

H2.N2NH3

(2 mol)(-46.19 kJ>mol) - 0 = -92.38 kJ

NH3

N2(g)H2(g)¢Hf°-46.19 kJ>mol.NH3(g)¢Hf°

¢H°

Ene

rgy

Reaction pathway

B

A

Ef

rr

rf

Er

¥ Figure 15.15 Effect of a catalyst onequilibrium. At equilibrium for the hypotheticalreaction the forward reaction rate, equals the reverse reaction rate, The violetcurve represents the path over the transition statein the absence of a catalyst. A catalyst lowers theenergy of the transition state, as shown by thegreen curve. Thus, the activation energy islowered for both the forward and the reversereactions. As a result, the rates of forward andreverse reactions in the catalyzed reaction areincreased.

rr .rf ,A ∆ B,

Recall that catalysts affect the rate of areaction but are not consumed by thereaction. See Chapter 14.

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pressures of 200 to 600 atm. The high pressures are needed to obtain a satisfac-tory degree of conversion at equilibrium. You can see from Figure 15.10 that ifan improved catalyst could be found—one that would lead to sufficiently rapidreaction at temperatures lower than 400°C to 500°C—it would be possible toobtain the same degree of equilibrium conversion at much lower pressures.This would result in great savings in the cost of equipment for ammonia syn-thesis. In view of the growing need for nitrogen as fertilizer, the fixation of ni-trogen is a process of ever-increasing importance.


Does the addition of a catalyst have any effect on the position of an equilibrium?

SAMPLE INTEGRATIVE EXERCISE | Putting Concepts TogetherAt temperatures near 800°C, steam passed over hot coke (a form of carbon obtainedfrom coal) reacts to form CO and

The mixture of gases that results is an important industrial fuel called water gas. (a) At800°C the equilibrium constant for this reaction is What are the equilibri-um partial pressures of CO, and in the equilibrium mixture at this tempera-ture if we start with solid carbon and 0.100 mol of in a 1.00-L vessel? (b) What isthe minimum amount of carbon required to achieve equilibrium under these condi-tions? (c) What is the total pressure in the vessel at equilibrium? (d) At 25°C the valueof for this reaction is Is the reaction exothermic or endothermic? (e) Toproduce the maximum amount of CO and at equilibrium, should the pressure ofthe system be increased or decreased?

Solution (a) To determine the equilibrium partial pressures, we use the ideal gasequation, first determining the starting partial pressure of hydrogen.

We then construct a table of starting partial pressures and their changes as equilibri-um is achieved:

Initial 8.81 atm 0 atm 0 atmChangeEquilibrium x atm x atm

There are no entries in the table under C(s) because the reactant, being a solid, doesnot appear in the equilibrium-constant expression. Substituting the equilibrium par-tial pressures of the other species into the equilibrium-constant expression for the re-action gives

Multiplying through by the denominator gives a quadratic equation in x:

Solving this equation for x using the quadratic formula yields Hence,the equilibrium partial pressures are and

(b) Part (a) shows that of must react in order for the system toachieve equilibrium. We can use the ideal-gas equation to convert this partial pres-sure into a mole amount.

n =PVRT

=(6.14 atm)(1.00 L)

(0.0821 L-atm>mol-K)(1073 K)= 0.0697 mol

H2Ox = 6.14 atmPH2O = 18.81 - x2 = 2.67 atm.

PH2 = x = 6.14 atm,PCO = x = 6.14 atm,x = 6.14 atm.

x2 + 14.1x - 124.22 = 0 x2 = 114.1218.81 - x2

Kp =PCOPH2

PH2O=

1x21x218.81 - x2 = 14.1

8.81 - x atm+x+x-x

H2(g)+CO(g)∆H2O(g)+C(s)

PH2O =nH2O RT

V=

(0.100 mol)(0.0821 L-atm>mol-K)(1073 K)

1.00 L= 8.81 atm

H2

1.7 * 10-21.Kp

H2OH2H2O,

Kp = 14.1.

C(s) + H2O(g) ∆ CO(g) + H2(g)

H2 :

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CHEMISTRY AT WORK | Controlling Nitric Oxide Emissions

0 1000 2000

1 � 10�15

1 � 10�10

1 � 10�5

1

Temperature (K)

Kp

Cylinder temperatureduring combustionExhaust gas

temperature

Á Figure 15.16 Equilibrium and temperature. The graph shows howthe equilibrium constant for the reaction varies as a function of temperature. The equilibrium constant increases withincreasing temperature because the reaction is endothermic. It is necessary touse a log scale for because the values vary over such a large range.Kp

12 N2(g) + 1

2 O2(g) ∆ NO(g)

slow to permit much loss of NO before the gases are cooledstill further.

As discussed in the “Chemistry at Work” box in Section14.7, one of the goals of automotive catalytic converters is toachieve the rapid conversion of NO to and at the tem-perature of the exhaust gas. Some catalysts for this reactionhave been developed that are reasonably effective under thegrueling conditions found in automotive exhaust systems.Nevertheless, scientists and engineers are continually search-ing for new materials that provide even more effective cataly-sis of the decomposition of nitrogen oxides.

O2N2

T he formation of NO from and provides another in-teresting example of the practical importance of changes

in the equilibrium constant and reaction rate with tempera-ture. The equilibrium equation and the standard enthalpychange for the reaction are

[15.23]

The reaction is endothermic; that is, heat is absorbedwhen NO is formed from the elements. By applying Le Châte-lier’s principle, we deduce that an increase in temperature willshift the equilibrium in the direction of more NO. The equilib-rium constant for formation of 1 mol of NO from the ele-ments at 300 K is only about In contrast, at a muchhigher temperature of about 2400 K the equilibrium constantis times as large, about 0.05. The manner in which for Equation 15.23 varies with temperature is shown inFigure 15.16 ».

This graph helps to explain why NO is a pollution prob-lem. In the cylinder of a modern high-compression auto en-gine, the temperatures during the fuel-burning part of thecycle may be on the order of 2400 K. Also, there is a fairly largeexcess of air in the cylinder. These conditions favor the forma-tion of some NO. After the combustion, however, the gases arequickly cooled. As the temperature drops, the equilibrium inEquation 15.23 shifts strongly to the left (that is, in the direc-tion of and ). The lower temperatures also mean that therate of the reaction is decreased, however, so the NO formed athigh temperatures is essentially “frozen” in that form as thegas cools.

The gases exhausting from the cylinder are still quite hot,perhaps 1200 K. At this temperature, as shown in Figure 15.16,the equilibrium constant for formation of NO is much smaller.However, the rate of conversion of NO to and is tooO2N2

O2N2

Kp1013

10-15.Kp

12 N2(g) + 1

2 O2(g) ∆ NO(g) ¢H° = 90.4 kJ

O2N2

Thus, 0.0697 mol of and the same amount of C must react to achieve equi-librium. As a result, there must be at least 0.0697 mol of C (0.836 g C) present amongthe reactants at the start of the reaction.

(c) The total pressure in the vessel at equilibrium is simply the sum of the equi-librium partial pressures:

(d) In discussing Le Châtelier’s principle, we saw that endothermic reactions ex-hibit an increase in with increasing temperature. Because the equilibrium constantfor this reaction increases as temperature increases, the reaction must be endothermic.From the enthalpies of formation given in Appendix C, we can verify our predictionby calculating the enthalpy change for the reaction,

The positive sign for indicates that the reac-tion is endothermic.

(e) According to Le Châtelier’s principle, a decrease in the pressure causes agaseous equilibrium to shift toward the side of the equation with the greater numberof moles of gas. In this case there are two moles of gas on the product side and onlyone on the reactant side. Therefore, the pressure should be reduced to maximize theyield of the CO and H2.

¢H°¢H°(C) - ¢Hf°(H2O) = +131.3 kJ.¢H° = ¢Hf°(CO) + ¢Hf°(H2) -

Kp

Ptotal = PH2O + PCO + PH2 = 2.67 atm + 6.14 atm + 6.14 atm = 14.95 atm

H2O

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Summary and Key Terms 659

S U M M A RY A N D K E Y T E R M S

Introduction and Section 15.1 A chemical reactioncan achieve a state in which the forward and reverseprocesses are occurring at the same rate. This condition iscalled chemical equilibrium, and it results in the forma-tion of an equilibrium mixture of the reactants and prod-ucts of the reaction. The composition of an equilibriummixture does not change with time.

Section 15.2 An equilibrium that is used throughoutthis chapter is the reaction of with to form

This reaction isthe basis of the Haber process for the production of am-monia. The relationship between the concentrations ofthe reactants and products of a system at equilibrium isgiven by the law of mass action. For a general equilibri-um equation of the form the equilibrium-constant expression is written as

The equilibrium-constant expression depends only onthe stoichiometry of the reaction. For a system at equilib-rium at a given temperature, will be a constant calledthe equilibrium constant. When the equilibrium systemof interest consists of gases, it is often convenient to ex-press the concentrations of reactants and products interms of gas pressures:

and are related by the expression

Section 15.3 The value of the equilibrium constantchanges with temperature. A large value of indicatesthat the equilibrium mixture contains more productsthan reactants. A small value for the equilibrium constantmeans that the equilibrium lies toward the reactant side.The equilibrium-constant expression and the equilibriumconstant of the reverse of a reaction are the reciprocals ofthose of the forward reaction. If a reaction is the sum oftwo or more reactions, its equilibrium constant will be theproduct of the equilibrium constants for the individualreactions.

Kc

Kp = Kc(RT)¢n.KpKc

Kp =(PD)d(PE)e

(PA)a(PB)b

Kc

Kc =[D]d[E]e

[A]a[B]b

aA + bB ∆ dD + eE,

NH3(g): N2(g) + 3 H2(g) ∆ 2 NH3(g).H2(g)N2(g)

Section 15.4 Equilibria for which all substances are inthe same phase are called homogeneous equilibria; inheterogeneous equilibria two or more phases are pre-sent. Because the concentrations of pure solids and liq-uids are constant, these substances are left out of theequilibrium-constant expression for a heterogeneousequilibrium.

Section 15.5 If the concentrations of all species in anequilibrium are known, the equilibrium-constant expres-sion can be used to calculate the value of the equilibriumconstant. The changes in the concentrations of reactantsand products on the way to achieving equilibrium aregoverned by the stoichiometry of the reaction.

Section 15.6 The reaction quotient, Q, is found by sub-stituting reactant and product partial pressures or con-centrations into the equilibrium-constant expression. Ifthe system is at equilibrium, If however,the system is not at equilibrium. When the reac-tion will move toward equilibrium by forming moreproducts (the reaction moves from left to right); when

the reaction will proceed from right to left.Knowing the value of K makes it possible to calculate theequilibrium amounts of reactants and products, often bythe solution of an equation in which the unknown is thechange in a partial pressure or concentration.

Section 15.7 Le Châtelier’s principle states that if asystem at equilibrium is disturbed, the equilibrium willshift to minimize the disturbing influence. By this princi-ple, if a reactant or product is added to a system at equi-librium, the equilibrium will shift to consume the addedsubstance. The effects of removing reactants or productsand of changing the pressure or volume of a reaction canbe similarly deduced. For example, if the volume of thesystem is reduced, the equilibrium will shift in the direc-tion that decreased the number of gas molecules. The en-thalpy change for a reaction indicates how an increase intemperature affects the equilibrium: For an endothermicreaction, an increase in temperature shifts the equilibri-um to the right; for an exothermic reaction, a temperatureincrease shifts the equilibrium to the left. Catalysts affectthe speed at which equilibrium is reached but do not af-fect the magnitude of K.

Q 7 K,

Q 6 K,Q Z K,Q = K.

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15.5 The reaction has an equilibriumconstant The following diagrams represent re-action mixtures containing molecules (red), mole-cules (blue) and AB molecules. (a) Which reactionmixture is at equilibrium? (b) For those mixtures thatare not at equilibrium, how will the reaction proceed toreach equilibrium? [Sections 15.5 and 15.6]

15.6 The reaction has anequilibrium constant of The diagram belowshows a mixture containing A atoms (red), mole-cules, and AB molecules (red and blue). How many Batoms should be added to the diagram if the system is atequilibrium? [Section 15.6]

15.7 The following diagram represents the equilibriumstate for the reaction (a) Assuming the volume is 1 L, calculate the equilibri-um constant, for the reaction. (b) If the volume of theequilibrium mixture is decreased, will the number of ABmolecules increase or decrease? [Sections 15.5 and 15.7]

Kc ,

A2(g) + 2 B(g) ∆ 2 AB(g).

A2

Kp = 2.A2(g) + B(g) ∆ A(g) + AB(g)

(i) (ii) (iii)

B2A2

Kc = 1.5.A2 + B2 ∆ 2 AB15.1 (a) Based on the following energy profile, predict

whether or (b) Using Equation 15.5,predict whether the equilibrium constant for the processis greater than 1 or less than 1. [Section 15.1]

15.2 The following diagrams represent a hypothetical reac-tion with A represented by red spheres and Brepresented by blue spheres. The sequence from left toright represents the system as time passes. Do the dia-grams indicate that the system reaches an equilibriumstate? Explain. [Sections 15.1 and 15.2]

15.3 The following diagram represents an equilibrium mixtureproduced for a reaction of the type If thevolume is 1 L, is K greater or smaller than 1? [Section 15.2]

15.4 The following diagram represents a reaction showngoing to completion. (a) Letting spheres and

spheres, write a balanced equation for the re-action. (b) Write the equilibrium-constant expression forthe reaction. (c) Assuming that all of the molecules arein the gas phase, calculate the change in the numberof gas molecules that accompanies the reaction. (d) Howcan you calculate if you know at a particular tem-perature? [Section 15.2]

KcKp

¢n,

B = blueA = red

A + X ∆ AX.

A ¡ B,

Reaction pathway

Pote

ntia

l ene

rgy

Reactants

Products

kf 6 kr .kf 7 kr

CQ

CQ

CQ

CQ

CQ

CQ

CQ


V I S U A L I Z I N G C O N C E P T S

BROWMC15_628-667PR2 1/26/05 15:08 Page 660

Exercises 661

15.8 The following diagrams represent equilibrium mixturesfor the reaction at (1) 300 K and (2)500 K. The A atoms are red, and the B atoms are blue. Isthe reaction exothermic or endothermic? [Section 15.7]

A2 + B ∆ A + AB

(a) (b)

E X E R C I S E S

Equilibrium; The Equilibrium Constant

15.9 Suppose that the gas-phase reactions andare both elementary processes with rate con-

stants of and respectively.(a) What is the value of the equilibrium constant for theequilibrium (b) Which is greater atequilibrium, the partial pressure of A or the partial pres-sure of B? Explain.

15.10 Consider the reaction Assume thatboth the forward reaction and the reverse reaction areelementary processes and that the value of the equilibri-um constant is very large. (a) Which species predomi-nate at equilibrium, reactants or products? (b) Whichreaction has the larger rate constant, the forward or thereverse? Explain.

15.11 (a) What is the law of mass action? Illustrate the law byusing the reaction (b) What is the difference between the equilibrium-constant expression and the equilibrium constant for agiven equilibrium? (c) Describe an experiment thatcould be used to determine the value of the equilibriumconstant for the reaction in part (a).

15.12 (a) The mechanism for a certain reaction is unknown. Is it still possible to apply the law of

mass action to the reaction? Explain. (b) Write the chemicalreaction involved in the Haber process. Why is this reac-tion important to humanity? (c) Write the equilibrium-constant expression for the reaction in part (b).

15.13 Write the expression for for the following reactions.In each case indicate whether the reaction is homoge-neous or heterogeneous.(a)(b)(c)(d)(e)

15.14 Write the expressions for for the following reactions.In each case indicate whether the reaction is homoge-neous or heterogeneous.(a)(b)(c) 2 C2H4(g) + 2 H2O(g) ∆ 2 C2H6(g) + O2(g)

Ti(s) + 2 Cl2(g) ∆ TiCl4(l)N2(g) + O2(g) ∆ 2 NO(g)

Kc

2 Ag(s) + Zn2+(aq) ∆ 2 Ag+(aq) + Zn(s)HF(aq) ∆ H+(aq) + F-(aq)Ni(CO)4(g) ∆ Ni(s) + 4 CO(g)CH4(g) + 2 H2S(g) ∆ CS2(g) + 4 H2(g)3 NO(g) ∆ N2O(g) + NO2(g)

Kc

C + DA + B ∆

NO(g) + Br2(g) ∆ NOBr2(g).

A + B ∆ C + D.

A(g) ∆ B(g)?

3.1 * 10-1 s-1,3.8 * 10-2 s-1B ¡ A

A ¡ B

(d)(e)

15.15 When the following reactions come to equilibrium, doesthe equilibrium mixture contain mostly reactants or most-ly products?(a)(b)

15.16 Which of the following reactions lies to the right, favor-ing the formation of products, and which lies to the left,favoring formation of reactants?(a)(b)

15.17 If for at 500 K,what is the value of for this reaction at thistemperature?

15.18 Calculate at 303 K for if at this temperature.

15.19 The equilibrium constant for the reaction

is at 1000 K. (a) Calculate for. (b) At this tempera-

ture does the equilibrium favor NO and or does itfavor NOBr?

15.20 The equilibrium constant for the reaction

is at 184°C. (a) Calculate for (b) Does the equilibrium favor

NO and or does it favor at this temperature?15.21 At 1000 K, for the reaction

(a) What is the value of for the reaction(b) What is the value of

for the reaction (c) What is the value of for the reaction in part (b)?

15.22 Consider the following equilibrium, for whichat 480°C:

2 Cl2(g) + 2 H2O(g) ∆ 4 HCl(g) + O2(g)

Kp = 0.0752

Kc

2 SO2(g) + O2(g) ∆ 2 SO3(g)?Kp

SO3(g) ∆ SO2(g) + 12 O2(g)?

Kp

SO2(g) + 12 O2(g) ∆ SO3(g)

Kp = 1.85NO2O2,

∆ 2 NO(g) + O2(g).2 NO2(g)KpKp = 1.48 * 104

2 NO(g) + O2(g) ∆ 2 NO2(g)

Br2 ,2 NOBr(g) ∆ 2 NO(g) + Br2(g)

KcKc = 1.3 * 10-2

2 NO(g) + Br2(g) ∆ 2 NOBr(g)

Kp = 34.5SO2Cl2(g)SO2(g) + Cl2(g) ∆Kc

Kp

PCl3(g) + Cl2(g) ∆ PCl5(g)Kc = 0.042H2(g) + Br2(g); Kc = 5.8 * 10-182 HBr(g) ∆

2 NO(g) + O2(g) ∆ 2 NO2(g); Kp = 5.0 * 1012

2 SO2(g) + O2(g) ∆ 2 SO3(g); Kp = 2.5 * 109N2(g) + O2(g) ∆ 2 NO(g); Kc = 1.5 * 10-10

4 HCl(aq) + O2(g) ∆ 2 H2O(l) + 2 Cl2(g)FeO(s) + H2(g) ∆ Fe(s) + H2O(g)

CQ

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(a) What is the value of for the reaction(b) What

is the value of for the reaction (c) What is the value of for the

reaction in part (b)?15.23 Consider the reactions and

for which the equi-librium constants at 100°C are and

respectively. What is the value of forthe reaction

[15.24] Consider the equilibrium

Calculate the equilibrium constant for this reaction,given the following information (at 298 K):

2 NO(g) ∆ N2(g) + O2(g) Kc = 2.1 * 1030 2 NO(g) + Br2(g) ∆ 2 NOBr(g) Kc = 2.0

Kp

N2(g) + O2(g) + Br2(g) ∆ 2 NOBr(g)

B(aq) + D(aq) ∆ E(aq)?KcKc = 8.5 * 102,

Kc = 1.9 * 10-4C(aq) + D(aq) ∆ E(aq) + A(aq)

A(aq) + B(aq) ∆ C(aq)

Kc2 HCl(g) + 12 O2(g)?

Cl2(g) + H2O(g) ∆Kp

4 HCl(g) + O2(g) ∆ 2 Cl2(g) + 2 H2O(g)?Kp 15.25 Mercury(I) oxide decomposes into elemental mercury

and elemental oxygen: (a) Write the equilibrium-constant expression for this re-action in terms of partial pressures. (b) Explain why wenormally exclude pure solids and liquids fromequilibrium-constant expressions.

15.26 Consider the equilibrium (a) Write the equilibrium-constant expres-

sion for this reaction in terms of partial pressures.(b) Why doesn’t the concentration of appear inthe equilibrium-constant expression?

Na2O

Na2SO3(s).Na2O(s) + SO2(g) ∆

2 Hg2O(s) ∆ 4 Hg(l) + O2(g).

At equilibrium (a) Calculate the equi-librium concentrations of and (b) Calcu-late

15.32 A mixture of 1.374 g of and 70.31 g of is heated in a2.00-L vessel at 700 K. These substances react as follows:

At equilibrium the vessel is found to contain 0.566 g of(a) Calculate the equilibrium concentrations of

and HBr. (b) Calculate 15.33 A mixture of 0.2000 mol of 0.1000 mol of and

0.1600 mol of is placed in a 2.000-L vessel. The fol-lowing equilibrium is established at 500 K:

(a) Calculate the initial partial pressures of and (b) At equilibrium Calculatethe equilibrium partial pressures of and CO.(c) Calculate for the reaction.

15.34 A flask is charged with 1.500 atm of and1.00 atm at 25°C, and the following equilibriumis achieved:

After equilibrium is reached, the partial pressure ofis 0.512 atm. (a) What is the equilibrium partial

pressure of (b) Calculate the value of for thereaction.

KpN2O4 ?NO2

N2O4(g) ∆ 2 NO2(g)

NO2(g)N2O4(g)

Kp

CO2, H2,PH2O = 3.51 atm.H2O.

CO2, H2,

CO2(g) + H2(g) ∆ CO(g) + H2O(g)

H2OH2,CO2,

Kc .H2, Br2 ,H2.

H2(g) + Br2(g) ∆ 2 HBr(g)

Br2H2

Kc .H2O.H2, N2,

[NO] = 0.062 M.Calculating Equilibrium Constants

15.27 Gaseous hydrogen iodide is placed in a closed con-tainer at 425°C, where it partially decomposes to hy-drogen and iodine: Atequilibrium it is found that

and Whatis the value of at this temperature?

15.28 Methanol is produced commercially by thecatalyzed reaction of carbon monoxide and hydrogen:

An equilibriummixture in a 2.00-L vessel is found to contain

0.170 mol CO, and at500 K. Calculate at this temperature.

15.29 The equilibrium isestablished at 500 K. An equilibrium mixture of thethree gases has partial pressures of 0.095 atm, 0.171 atm,and 0.28 atm for NO, and NOCl, respectively. Cal-culate for this reaction at 500 K.

15.30 Phosphorus trichloride gas and chlorine gas react to formphosphorus pentachloride gas:

A gas vessel is charged with a mixture ofand which is allowed to equilibrate at

450 K. At equilibrium the partial pressures of the threegases are and

(a) What is the value of at this tem-perature? (b) Does the equilibrium favor reactants orproducts?

15.31 A mixture of 0.10 mol of NO, 0.050 mol of and 0.10mol of is placed in a 1.0-L vessel at 300 K. The fol-lowing equilibrium is established:

2 NO(g) + 2 H2(g) ∆ N2(g) + 2 H2O(g)

H2OH2,

KpPPCl5 = 1.30 atm.PCl2 = 0.157 atm,PPCl3 = 0.124 atm,

Cl2(g),PCl3(g)PCl5(g).

PCl3(g) + Cl2(g) ∆

Kp

Cl2,

2 NO(g) + Cl2(g) ∆ 2 NOCl(g)Kc

0.302 mol H20.0406 mol CH3OH,

CO(g) + 2 H2(g) ∆ CH3OH(g).

(CH3OH)Kc

[I2] = 4.79 * 10-4 M.[H2] = 4.79 * 10-4 M,[HI] = 3.53 * 10-3 M,

2 HI(g) ∆ H2(g) + I2(g).

must the reaction proceed to reach equilibrium? (c) Atthe start of a certain reaction, only reactants are present;no products have been formed. What is the value of at this point in the reaction?

15.37 At 100°C the equilibrium constant for the reactionhas the value

Are the following mixtures of CO,COCl2,2.19 * 10-10.Kc =COCl2(g) ∆ CO(g) + Cl2(g)

Qc

Applications of Equilibrium Constants

15.35 (a) How does a reaction quotient differ from an equilib-rium constant? (b) If in which direction will areaction proceed in order to reach equilibrium? (c) Whatcondition must be satisfied so that

15.36 (a) How is a reaction quotient used to determinewhether a system is at equilibrium? (b) If howQc 7 Kc ,

Qc = Kc ?

Qc 6 Kc ,

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Exercises 663

and at 100°C at equilibrium? If not, indicate the direc-tion that the reaction must proceed to achieve equilibrium.(a)

(b)

(c)15.38 As shown in Table 15.2, for the equilibrium

is at 450°C. For each of the mixtures listedhere, indicate whether the mixture is at equilibrium at450°C. If it is not at equilibrium, indicate the direction(toward product or toward reactants) in which the mix-ture must shift to achieve equilibrium.(a)(b) no (c)

15.39 At 100°C, for the reaction

In an equilibrium mixture of the three gases, the concen-trations of and are 0.108 M and 0.052 M, re-spectively. What is the partial pressure of in theequilibrium mixture?

15.40 At 900 K the following reaction has

In an equilibrium mixture the partial pressures of and are 0.165 atm and 0.755 atm, respectively.What is the equilibrium partial pressure of in themixture?

15.41 (a) At 1285°C the equilibrium constant for the reactionis A 0.200-L vessel

containing an equilibrium mixture of the gases has0.245 g in it. What is the mass of Br(g) in the ves-sel? (b) For the reaction

at 700 K. In a 2.00-L flask containing an equi-librium mixture of the three gases, there are and What is the mass of HI in the flask?

15.42 (a) At 800 K the equilibrium constant for is If an equilibrium mixture in a 10.0-Lvessel contains of I(g), how many grams of

are in the mixture? (b) For at 700 K. In a 2.00-L vessel the

equilibrium mixture contains 1.57 g of and 0.125 g ofHow many grams of are in the vessel?

15.43 At 2000°C the equilibrium constant for the reaction

2 NO(g) ∆ N2(g) + O2(g)

SO2O2.SO3

Kp = 3.0 * 1042 SO3(g),2 SO2(g) + O2(g) ∆I2

2.67 * 10-2 gKc = 3.1 * 10-5.

I2(g) ∆ 2 I(g)4.36 g I2 .

0.056 g H2

Kc = 55.3H2(g) + I2(g) ∆ 2 HI(g),

Br2(g)

Kc = 1.04 * 10-3.Br2(g) ∆ 2 Br(g)

SO3

O2

SO2

2 SO2(g) + O2(g) ∆ 2 SO3(g)

Kp = 0.345:

Cl2

SO2SO2Cl2

SO2Cl2(g) ∆ SO2(g) + Cl2(g)

Kc = 0.07826 atm NH3, 42 atm H2, 202 atm N2.

N2 ;35 atm NH3, 595 atm H2,105 atm NH3, 35 atm N2, 495 atm H2 ;

4.51 * 10-5

N2(g) + 3 H2(g) ∆ 2 NH3(g)

Kp

[Cl2] = 1.48 * 10-6 M.[CO] =[COCl2] = 0.0100 M,[Cl2] = 2.25 * 10-6 M;[CO] = 1.1 * 10-7 M,

[COCl2] = 4.50 * 10-2 M,6.62 * 10-6 M;[Cl2] =3.3 * 10-6 M,[CO] =[COCl2] = 2.00 * 10-3 M,

Cl2 is If the initial concentration of NO is0.200 M, what are the equilibrium concentrations of NO,

and 15.44 For the equilibrium

at 400 K, If 0.30 mol of and areintroduced into a 1.0-L container at 400 K, what will bethe equilibrium concentrations of and BrCl?

15.45 At 373 K, for the equilibrium

If the pressures of NOBr(g) and NO(g) are equal, what isthe equilibrium pressure of

15.46 At 218°C, for the equilibrium

Calculate the equilibrium concentrations of andif a sample of solid is placed in a closed

vessel and decomposes until equilibrium is reached.15.47 Consider the reaction

At 25°C the equilibrium constant is forthis reaction. (a) If excess is mixed with waterat 25°C to produce a saturated solution of whatare the equilibrium concentrations of and (b) If the resulting solution has a volume of 3.0 L, whatis the minimum mass of needed to achieveequilibrium?

15.48 At 80°C, for the reaction

(a) Calculate the equilibrium concentrations of andif a solid sample of is placed in a closed

vessel and decomposes until equilibrium is reached.(b) If the flask has a volume of 0.500 L, what is the mini-mum mass of that must be added to theflask to achieve equilibrium?

15.49 For the reaction at 150°C. Suppose that 0.500 mol IBr in a 1.00-L flask isallowed to reach equilibrium at 150°C. What are theequilibrium concentrations of IBr, and

15.50 At 25°C the reaction

has an equilibrium constant What arethe equilibrium concentrations of and in asaturated solution of CaCrO4 ?

CrO4

2-Ca2+Kc = 7.1 * 10-4.

CaCrO4(s) ∆ Ca2+(aq) + CrO4

2-(aq)

Br2 ?I2 ,

I2(g) + Br2(g) ∆ 2 IBr(g), Kc = 280

PH3BCl3(s)

PH3BCl3BCl3

PH3

PH3BCl3(s) ∆ PH3(g) + BCl3(g)

Kc = 1.87 * 10-3

CaSO4(s)

SO4

2-?Ca2+CaSO4,

CaSO4(s)Kc = 2.4 * 10-5

CaSO4(s) ∆ Ca2+(aq) + SO4

2-(aq)

NH4HSH2SNH3

NH4HS(s) ∆ NH3(g) + H2S(g)

Kc = 1.2 * 10-4Br2(g)?

2 NOBr(g) ∆ 2 NO(g) + Br2(g)

Kp = 0.416Br2, Cl2 ,

0.30 mol Cl2Br2Kc = 7.0.

Br2(g) + Cl2(g) ∆ 2 BrCl(g)

O2 ?N2,

Kc = 2.4 * 103.

is increased by adding a noble gas; (f) is removedfrom the system.

15.52 For the following reaction,

How is the equilibrium yield of affected by(a) increasing (b) increasing temperature, (c) remov-ing (d) decreasing the total pressure, (e) removingpart of the (f) adding a catalyst?C6H12O6,

CO2,PCO2,

C6H12O6

6 CO2(g) + 6 H2O(l) ∆ C6H12O6(s) + 6 O2(g)

¢H° = 2816 kJ:

SO3(g)Le Châtelier’s Principle

15.51 Consider the following equilibrium, for which

How will each of the following changes affect an equi-librium mixture of the three gases? (a) is added tothe system; (b) the reaction mixture is heated; (c) thevolume of the reaction vessel is doubled; (d) a catalyst isadded to the mixture; (e) the total pressure of the system

O2(g)

2 SO2(g) + O2(g) ∆ 2 SO3(g)

¢H 6 0:

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15.53 How do the following changes affect the value of theequilibrium constant for a gas-phase exothermic reac-tion: (a) removal of a reactant or product, (b) decreasein the volume, (c) decrease in the temperature, (d) addi-tion of a catalyst?

15.54 For a certain gas-phase reaction, the fraction of productsin an equilibrium mixture is increased by increasingthe temperature and increasing the volume of the reac-tion vessel. (a) What can you conclude about thereaction from the influence of temperature on the equi-librium? (b) What can you conclude from the influenceof increasing the volume?

15.55 Consider the following equilibrium between oxides ofnitrogen

3 NO(g) ∆ NO2(g) + N2O(g)

(a) Use data in Appendix C to calculate for this re-action. (b) Will the equilibrium constant for the reactionincrease or decrease with increasing temperature? Ex-plain. (c) At constant temperature would a change in thevolume of the container affect the fraction of products inthe equilibrium mixture?

15.56 Methanol can be made by the reaction of COwith

(a) Use thermochemical data in Appendix C to calculatefor this reaction. (b) In order to maximize the equi-

librium yield of methanol, would you use a high or lowtemperature? (c) In order to maximize the equilibriumyield of methanol, would you use a high or low pressure?

¢H°

CO(g) + 2 H2(g) ∆ CH3OH(g)

H2 :(CH3OH)

¢H°

pressure in the flask at equilibrium? (b) What is thevalue of

15.63 As shown in Table 15.2, the equilibrium constant forthe reaction is

at 300°C. Pure is placed in a 1.00-L flaskand allowed to reach equilibrium at this temperature.There are in the equilibrium mixture. (a) Whatare the masses of and in the equilibrium mixture?(b) What was the initial mass of ammonia placed in thevessel? (c) What is the total pressure in the vessel?

15.64 For the equilibrium

at 150°C. If 0.025 atm of IBr is placed ina 2.0-L container, what is the partial pressure of this sub-stance after equilibrium is reached?

15.65 For the equilibrium

at 60°C. (a) Calculate (b) Some solidis added to a closed 0.500-L vessel at 60°C; the

vessel is then charged with 0.0128 mol of What is the equilibrium concentration of

[15.66] Solid is introduced into an evacuated flask at24°C. The following reaction takes place:

At equilibrium the total pressure (for and taken together) is 0.614 atm. What is for this equilib-rium at 24°C?

[15.67] A 0.831-g sample of is placed in a 1.00-L container andheated to 1100 K. The decomposes to and

At equilibrium the total pressure in the container is1.300 atm. Find the values of and for this reactionat 1100 K.

15.68 Nitric oxide (NO) reacts readily with chlorine gas asfollows:

2 NO(g) + Cl2(g) ∆ 2 NOCl(g)

KcKp

2 SO3(g) ∆ 2 SO2(g) + O2(g)

O2 :SO2SO3

SO3

Kp

H2SNH3


NH4HSPH3 ?BCl3(g).

PH3BCl3

Kc .Kp = 0.052

PH3BCl3(s) ∆ PH3(g) + BCl3(g)

Kp = 8.5 * 10-3

2 IBr(g) ∆ I2(g) + Br2(g)

H2N2

1.05 g NH3

NH34.34 * 10-3=KpN2(g) + 3 H2(g) ∆ 2 NH3(g)

Kp ?Additional Exercises

15.57 Both the forward reaction and the reverse reaction in thefollowing equilibrium are believed to be elementarysteps:

At 25°C the rate constants for the forward and reversereactions are and respectively. (a) What is the value for the equilibriumconstant at 25°C? (b) Are reactants or products moreplentiful at equilibrium?

15.58 A mixture of and is passed over a nickel cata-lyst at 1000 K. The emerging gas is collected in a 5.00-Lflask and is found to contain 8.62 g of CO, 2.60 g of 43.0 g of and 48.4 g of Assuming that equi-librium has been reached, calculate and for the re-action.

15.59 When 2.00 mol of is placed in a 2.00-L flask at303 K, 56% of the decomposes to and

Calculate for this reaction at this temperature.15.60 A mixture of S, and is held in a 1.0-L vessel at

90°C until the following equilibrium is achieved:

At equilibrium the mixture contains 0.46 g of and(a) Write the equilibrium-constant expression

for this reaction. (b) What is the value of for the reac-tion at this temperature? (c) Why can we ignore theamount of S when doing the calculation in part (b)?

15.61 A sample of nitrosyl bromide (NOBr) decomposes ac-cording to the equation

An equilibrium mixture in a 5.00-L vessel at 100°C con-tains 3.22 g of NOBr, 3.08 g of NO, and 4.19 g of (a) Calculate (b) What is the total pressure exertedby the mixture of gases?

15.62 Consider the hypothetical reaction Aflask is charged with 0.55 atm of pure A, after which it isallowed to reach equilibrium at 0°C. At equilibrium thepartial pressure of A is 0.36 atm. (a) What is the total

A(g) ∆ 2 B(g).

Kc .Br2 .

2 NOBr(g) ∆ 2 NO(g) + Br2(g)

Kc

0.40 g H2.H2S

H2(g) + S(s) ∆ H2S(g)

H2SH2,Kc

SO2Cl2(g) ∆ SO2(g) + Cl2(g)

Cl2 :SO2SO2Cl2

SO2Cl2

KpKc

H2O.CH4,H2,

H2OCH4

9.3 * 1010 M-1s-1,1.4 * 10-28 M-1s-1

CO(g) + Cl2(g) ∆ COCl(g) + Cl(g)

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Exercises 665

At 700 K the equilibrium constant for this reactionis 0.26. Predict the behavior of each of the followingmixtures at this temperature: (a)

and (b)and

(c) and

15.69 At for the reaction

A mixture of CaO, and is placed in a 10.0-Lvessel at 900°C. For the following mixtures, will theamount of increase, decrease, or remain thesame as the system approaches equilibrium?(a) 15.0 g CaO, and (b) 25.0 g CaO, and (c) 25.5 g CaO, and

15.70 Nickel carbonyl, is an extremely toxic liquidwith a low boiling point. Nickel carbonyl results fromthe reaction of nickel metal with carbon monoxide. Fortemperatures above the boiling point (42.2°C) of

the reaction is

Nickel that is more than 99.9% pure can be produced bythe carbonyl process: Impure nickel combines with CO at50°C to produce The is then heatedto 200°C, causing it to decompose back into Ni(s) andCO(g). (a) Write the equilibrium-constant expression forthe formation of (b) Given the temperaturesused for the steps in the carbonyl process, do you thinkthe formation reaction is endothermic or exothermic?(c) In the early days of automobiles, nickel-plated ex-haust pipes were used. Even though the equilibriumconstant for the formation of is very small at thetemperature of automotive exhaust gases, the exhaustpipes quickly corroded. Explain why this occurred.

15.71 NiO is to be reduced to nickel metal in an industrialprocess by use of the reaction

At 1600 K the equilibrium constant for the reaction isIf a CO pressure of 150 torr is to be

employed in the furnace and total pressure never ex-ceeds 760 torr, will reduction occur?

[15.72] At 700 K the equilibrium constant for the reaction

is A flask is charged with 2.00 atm of which then reaches equilibrium at 700 K. (a) What frac-tion of the is converted into C and (b) Whatare the partial pressures of and at equilibrium?

[15.73] The reaction has at 300°C. A flask is charged with 0.50 atm PCl3 ,0.0870

=KpPCl3(g) + Cl2(g) ∆ PCl5(g)Cl2CCl4

Cl2 ?CCl4

CCl4,Kp = 0.76.

CCl4(g) ∆ C(s) + 2 Cl2(g)

Kp = 6.0 * 102.

NiO(s) + CO(g) ∆ Ni(s) + CO2(g)

Ni(CO)4

Ni(CO)4.

Ni(CO)4Ni(CO)4(g).

Ni(s) + 4 CO(g) ∆ Ni(CO)4(g)

Ni(CO)4,

Ni(CO)4,6.48 g CO230.5 g CaCO3,5.66 g CO22.50 g CaCO3,4.25 g CO215.0 g CaCO3,

CaCO3

CO2CaCO3,

CaCO3(s) ∆ CaO(s) + CO2(g)

900°C, Kc = 0.01085.10 * 10-3 atm.

=PNOClPNO = 0.15 atm, PCl2 = 0.20 atm,PNOCl = 0.050 atm;PCl2 = 0.10 atm,0.12 atm,

=PNOPNOCl = 0.11 atm;PCl2 = 0.31 atm,PNO = 0.15 atm,

Kp and at this temperature.(a) Use the reaction quotient to determine the directionthe reaction must proceed in order to reach equilibrium.(b) Calculate the equilibrium partial pressures of thegases. (c) What effect will increasing the volume ofthe system have on the mole fraction of in the equi-librium mixture? (d) The reaction is exothermic. Whateffect will increasing the temperature of the system haveon the mole fraction of in the equilibrium mixture?

[15.74] An equilibrium mixture of and HI at 458°C con-tains and 0.775 mol HI in a5.00-L vessel. What are the equilibrium partial pressureswhen equilibrium is reestablished following the addi-tion of 0.100 mol of HI?

[15.75] Consider the hypothetical reaction for which at some temperature. A 1.00-

L reaction vessel is loaded with 1.00 mol of compoundC, which is allowed to reach equilibrium. Let the vari-able x represent the number of of compound Apresent at equilibrium. (a) In terms of x, what are theequilibrium concentrations of compounds B and C?(b) What limits must be placed on the value of x so thatall concentrations are positive? (c) By putting the equi-librium concentrations (in terms of x) into theequilibrium-constant expression, derive an equationthat can be solved for x. (d) The equation from part (c)is a cubic equation (one that has the form

). In general, cubic equationscannot be solved in closed form. However, you can esti-mate the solution by plotting the cubic equation in theallowed range of x that you specified in part (b). Thepoint at which the cubic equation crosses the x-axis isthe solution. (e) From the plot in part (d), estimate theequilibrium concentrations of A, B, and C. [Hint: Youcan check the accuracy of your answer by substitutingthese concentrations into the equilibrium expression.]

15.76 At 1200 K, the approximate temperature of automobileexhaust gases (Figure 15.16), for the reaction

is about Assuming that the exhaust gas (totalpressure 1 atm) contains 0.2% CO, and by volume, is the system at equilibrium with respect tothe above reaction? Based on your conclusion, wouldthe CO concentration in the exhaust be decreased or in-creased by a catalyst that speeds up the reaction above?

15.77 Suppose that you worked at the U.S. Patent Office and apatent application came across your desk claiming thata newly developed catalyst was much superior to theHaber catalyst for ammonia synthesis because the cata-lyst led to much greater equilibrium conversion of and into than the Haber catalyst under thesame conditions. What would be your response?

NH3H2

N2

3% O212% CO2,1 * 10-13.

2 CO2(g) ∆ 2 CO(g) + O2(g)

Kp

ax3 + bx2 + cx + d = 0

mol>L

Kc = 0.252 C(g),∆A(g) + 2 B(g)

0.112 mol H2, 0.112 mol I2 ,H2, I2 ,

Cl2

Cl2

0.20 atm PCl50.50 atm Cl2 ,

CQ

CQ

CQ

CQ

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Integrative Exercises

15.78 Consider the following equilibria in aqueous solution:(i)

(ii)

(iii)(a) For each reaction, write the equilibrium-constant ex-pression for (b) Using information provided in Table4.5, predict whether is large or small

Explain your reasoning. (c) At 25°C thereaction has

If Cd were added to Table 4.5, wouldyou expect it to be above or below iron? Explain.

15.79 Silver chloride, AgCl(s), is an insoluble strong elec-trolyte. (a) Write the equation for the dissolution ofAgCl(s) in (b) Write the expression for for thereaction in part (a). (c) Based on the thermochemicaldata in Appendix C and Le Châtelier’s principle, predictwhether the solubility of AgCl in increases or de-creases with increasing temperature.

15.80 The hypothetical reaction occurs in theforward direction in a single step. The energy profile ofthe reaction is shown in the drawing. (a) Is the forwardor reverse reaction faster at equilibrium? (b) Would youexpect the equilibrium to favor reactants or products?(c) In general, how would a catalyst affect the energyprofile shown? (d) How would a catalyst affect the ratioof the rate constants for the forward and reverse reac-tions? (e) How would you expect the equilibrium con-stant of the reaction to change with increasingtemperature?

[15.81] Consider the equilibrium in which both theforward and reverse reactions are elementary (single-step) reactions. Assume that the only effect of a catalyston the reaction is to lower the activation energies of theforward and reverse reactions, as shown in Figure 15.15.Using the Arrhenius equation (Section 14.5), prove thatthe equilibrium constant is the same for the catalyzedreaction as for the uncatalyzed one.

[15.82] At 25°C the reaction

has A 5.00-L flask is charged with 0.300 g ofpure at 25°C. Solid is then added untilthere is excess unreacted solid remaining. (a) What isthe initial pressure of in the flask? (b) Why doesno reaction occur until is added? (c) What arethe partial pressures of and at equilibrium?H2SNH3

NH4HSH2S(g)

NH4HSH2S(g)Kp = 0.120.


A ∆ B,

Reaction pathway

Ene

rgy

CA � B

A + B ∆ C

H2O

KcH2O(l).

Kc = 6 * 10-2.Cd(s) + Fe2+(aq) ∆ Cd2+(aq) + Fe(s)

1Kc V 12.1Kc W 12Kc

Kc .

Zn(s) + 2 H+(aq) ∆ Zn2+(aq) + H2(g)

3 Hg(l) + 2 Al3+(aq) ∆ 3 Hg2+(aq) + 2 Al(s)

Na(s) + Ag+(aq) ∆ Na+(aq) + Ag(s)

(d) What is the mole fraction of in the gas mixtureat equilibrium? (e) What is the minimum mass, ingrams, of that must be added to the flask toachieve equilibrium?

[15.83] Write the equilibrium-constant expression for theequilibrium

The table included below shows the relative mole per-centages of and CO(g) at a total pressure of 1 atmfor several temperatures. Calculate the value of ateach temperature. Is the reaction exothermic or en-dothermic? Explain.

COTemperature (°C) (mol %) (mol %)

850 6.23 93.77950 1.32 98.68

1050 0.37 99.631200 0.06 99.94

15.84 In Section 11.5 we defined the vapor pressure of a liquidin terms of an equilibrium. (a) Write the equation repre-senting the equilibrium between liquid water and watervapor, and the corresponding expression for (b) Byusing data in Appendix B, give the value of for thisreaction at 30°C. (c) What is the value of for any liq-uid in equilibrium with its vapor at the normal boilingpoint of the liquid?

[15.85] Polyvinyl chloride (PVC) is one of the most commercial-ly important polymers (Table 12.4). PVC is made by ad-dition polymerization of vinyl chloride Vinylchloride is synthesized from ethylene in a two-step process involving the following equilibria:

Equilibrium 1:

Equilibrium 2:

The product of Equilibrium 1 is 1,2-dichloroethane, acompound in which one Cl atom is bonded to each Catom. (a) Draw Lewis structures for and

What are the bond orders in these twocompounds? (b) Use average bond enthalpies (Table8.4) to estimate the enthalpy changes in the two equilib-ria. (c) How would the yield of in Equilibrium1 vary with temperature and volume? (d) How wouldthe yield of in Equilibrium 2 vary with tempera-ture and volume? (e) Look up the normal boiling pointsof 1,2-dichloroethane and vinyl chloride in a source-book, such as the CRC Handbook of Chemistry and Physics.Based on these data, propose a reactor design (analo-gous to Figure 15.12) that could be used to maximizethe amount of produced by using the twoequilibria.

C2H3Cl

C2H3Cl

C2H4Cl2

C ¬ CC2H3Cl.C2H4Cl2

C2H4Cl2(g) ∆ C2H3Cl(g) + HCl(g)

C2H4(g) + Cl2(g) ∆ C2H4Cl2(g)

(C2H4)(C2H3Cl).

Kp

Kp

Kp .

CO2

Kp

CO2(g)

C(s) + CO2(g) ∆ 2 CO(g)

NH4HS

H2S

CQ

CQ

CQ

CQ

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eMedia Exercises 667

e M E D I A E X E R C I S E S

These exercises make use of the interactive objects available on-line in OneKey or the Companion Website, and on your Accel-erator CD. Access to these resources comes in your MediaPak.

15.86 You can choose starting concentrations for reactants inthe Chemical Equilibrium activity (15.1). (a) Write theequilibrium-constant expression for the reaction in thesimulation. (b) Calculate the value of the equilibriumconstant. (c) Does this reaction lie to the right or to theleft?

15.87 Using the equilibrium constant that you calculated in theprevious question for the reaction in the Chemical Equi-librium activity (15.1), predict what the equilibrium con-centrations of reactants and products will be if you startwith and Use thesimulation to check your answer.

15.88 The significance of an equilibrium constant’s magnitudeis illustrated in the Equilibrium Constant activity(15.2). Select the reaction, and carry out sever-al experiments with varying starting concentrations andvarying equilibrium constants. Because reactants are fa-

A ∆ B

0.0004 M SCN-(aq).0.0007 M Fe3+(aq)

vored when is very small, is it possible to enter a verysmall value for start with only A (no B), and stillhave only A at equilibrium? Explain.

15.89 In Exercise 15.34 you calculated for the reaction(a) Which of the reactions in

the Equilibrium Constant activity (15.2) best representsthe equilibrium between and (b) Using the simulation, enter the value from Exer-cise 15.34 and experiment with various starting concen-trations of reactant and products. Can you selectstarting concentrations of reactants and products suchthat no change in concentrations is observed? Explain.

15.90 Consider the and reactions in theEquilibrium Constant activity (15.2). Assume that Aand B are both gases, that is an exothermic re-action, and that is an endothermic reaction.For both reactions, state the effect that each of the fol-lowing changes would have on the value of the con-centration of B at equilibrium, and the rate of thereaction: (a) increased temperature, (b) increased pres-sure, and (c) addition of a catalyst.

Kc ,

A ∆ BA ∆ B

A ∆ 2 BA ∆ B

Kp

2 NO2(g)?N2O4(g)

N2O4(g) ∆ 2 NO2(g).Kp

Kc ,Kc

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