PRINCIPLES OF CHEMICAL EQUILIBRIUM -...

PRINCIPLES OFCHEMICAL

EQUILIBRIUM

Until now, we have stressed reactions that go to completionand the concepts of stoichiometry that allow us to calculatethe outcomes of such reactions. We have made occasional

references to situations involving both a forward and a reversereaction—reversible reactions—but in this chapter, we will look atthem in a detailed and systematic way.

Our emphasis will be on the equilibrium condition reached whenforward and reverse reactions proceed at the same rate. Our maintool in dealing with equilibrium will be the equilibrium constant.We will begin with some key relationships involving equilibriumconstants; then we will make qualitative predictions about the con-dition of equilibrium; and finally we will do various equilibriumcalculations. As we will discover throughout the remainder of thetext, the equilibrium condition plays a role in numerous naturalphenomena and affects the methods used to produce many impor-tant industrial chemicals.

15-1 Dynamic Equilibrium15-2 The Equilibrium Constant

Expression15-3 Relationships Involving

Equilibrium Constants15-4 The Magnitude of an

Equilibrium Constant15-5 The Reaction Quotient, Q:

Predicting the Directionof Net Change

15-6 Altering EquilibriumConditions: Le Châtelier’sPrinciple

15-7 Equilibrium Calculations:Some IllustrativeExamples

➣ FOCUS ON The NitrogenCycle and the Synthesis ofNitrogen Compounds

622

C O N T E N T S

A vital natural reaction is in progress in the lightning bolt seen here:. Usually this reversible reaction does not occur

to any significant extent in the forward direction, but in the high-temperaturelightning bolt it does. At equilibrium at high temperatures, measurableconversion of and to occurs. In this chapter we study theequilibrium condition in a reversible reaction and the factors affecting it.

NO(g)O2(g)N2(g)

N2(g) + O2(g) ∆ 2 NO(g)

Point out that equilibriumis less common thannonequilibrium. For

example, your heart and otherbiochemical processes areoscillating back and forth in cycles.We certainly don’t want our heartsto come to equilibrium.

There are manydemonstrations that show

equilibrium: roll a marble in aparabolic clear dish to show theestablishment of physicalequilibrium; pour intoordinary tap water, and thechloride ion in the water willprecipitate out leaving the watercloudy; burn a match, and when itgoes out, equilibrium isestablished. Point out thatequilibrium in chemical reactions isobtained by minimizing the freeenergy (see Chapter 19).

AgNO3

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DYNAMIC EQUILIBRIUM

Let’s begin by describing three simple physical and one chemical phenomenathat will help us to establish the core attribute of a system at equilibrium—two opposing processes take place at equal rates.

1. When a liquid vaporizes within a closed container, after a time, vapormolecules condense to the liquid state at the same rate at which liquidmolecules vaporize. Even though molecules continue to pass back andforth between liquid and vapor (a dynamic process), the pressure exertedby the vapor remains constant with time. The vapor pressure of a liquid is aproperty resulting from an equilibrium condition.

2. When a solute is added to a solvent, the system may reach a point at whichthe rate of dissolution is just matched by the rate at which dissolved solutecrystallizes—that is, the solution is saturated. Even though solute particlescontinue to pass back and forth between the saturated solution and theundissolved solute, the concentration of dissolved solute remains constant.The solubility of a solute is a property resulting from an equilibrium condition.

3. When an aqueous solution of iodine, is shaken with pure carbon tetra-chloride, the molecules move into the layer. As the concen-tration of builds up in the the rate of return of to the water layerbecomes significant. When molecules pass between the two liquids atequal rates—a condition of dynamic equilibrium—the concentration of in each layer remains constant. At this point, the concentration of in the

is about 85 times greater than in the (Fig. 15-1). The ratio of con-centrations of a solute in two immiscible solvents is called the distributioncoefficient. The distribution coefficient, which represents the partitioning of asolute between two immiscible solvents, is a property resulting from an equilibri-um condition.

4. When gaseous phosphorus pentachloride is heated, it decomposes to phos-phorus trichloride and chlorine gases: Consider a sample of initially exerting a pressure of 1.0 atm in aclosed container at 250 °C. The gas pressure in the container first risesrapidly and then ever more slowly, reaching a maximum, unchanging pres-sure of 1.7 atm. Because two moles of gas are produced for each mole of

that decomposes, if the reaction went to completion the final pres-sure would have been 2.0 atm. We conclude that the decomposition of isa reversible reaction that reaches an equilibrium condition.

The properties in the first three situations just described—vapor pressure,solubility, and distribution coefficient—are examples of physical equilibria.The fourth situation is an example of chemical equilibrium. All four are de-scribed through a general quantity known as an equilibrium constant, the sub-ject of the next section.

THE EQUILIBRIUM CONSTANT EXPRESSION

Methanol (methyl alcohol) is synthesized from a carbon monoxide–hydrogenmixture called synthesis gas. This reaction is likely to become increasingly im-portant as methanol and its mixtures with gasoline find greater use as motorfuels. Methanol has a high octane rating, and its combustion produces muchless air pollution than does gasoline.

Methanol synthesis is a reversible reaction, which means that at the sametime is being formed,

(15.1)

it decomposes in the reverse reaction

(15.2)CH3OH(g) ¡ CO(g) + 2 H2(g)

CO(g) + 2 H2(g) ¡ CH3OH(g)

CH3OH(g)

15-2

PCl5

PCl5(g)

PCl5(g)PCl5(g) ¡ PCl3(g) + Cl2(g).

H2OCCl4

I2

I2

I2

I2CCl4 ,I2

CCl4I2CCl4(l),I2 ,

15-1

15-2 The Equilibrium Constant Expression 623

Dynamic Equilibriumanimation

Paiva, João C. M., Victor M. S.Gil. “The Complexity of

Teaching and Learning ChemicalEquilibrium.” J. Chem. Educ.2000: 77, 1560 (December 2000).

Silverstein, Todd P.“Equilibrium: A Teaching/

Learning Activity.” J. Chem. Educ.2000: 77, 1410 (November 2000).

(a) (b)

FIGURE 15-1Dynamic equilibrium in aphysical process(a) A yellow-brown saturatedsolution of in water(top layer) is brought intocontact with colorless (bottom layer). (b) mole-cules distribute themselvesbetween the and When equilibrium is reached,

in the (violet,bottom layer) is about 85times greater than in thewater (colorless, top layer).

CCl4[I2]

CCl4 .H2O

I2

CCl4(l)

I2

▲

The two opposingprocesses operate at themicroscopic level.

Methanol is actively beingconsidered as an alternativefuel to gasoline.

▲

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 623

Are You Wondering . . .How we know that an equilibrium is dynamic—that forward and reversereactions continue even after equilibrium is reached?

Suppose we have an equilibrium mixture of AgI(s) and its saturated aqueous solution.

Now let’s add to this mixture some saturated solution of AgI made from AgIcontaining radioactive iodine-131 as iodide ion, as illustrated in Figure 15-2. Ifboth the forward and reverse processes stopped at equilibrium, radioactivitywould be confined to the solution. What we find, though, is that radioactivityshows up in the solid in contact with the saturated solution. Over time, the ra-dioactive “hot” spots distribute themselves throughout the solution and undis-solved solid. The only way this can happen is if the dissolving of the solid soluteand its crystallization from the saturated solution continue indefinitely. The equi-librium condition is dynamic.

AgI(s) ∆ AgI(satd aq)

624 Chapter 15 Principles of Chemical Equilibrium

Initially, only the forward reaction (15.1) occurs, but as soon as someforms, the reverse reaction (15.2) begins. With passing time, the for-

ward reaction slows because of the decreasing concentrations of CO and and the reverse reaction speeds up as more accumulates. Eventually,the forward and reverse reactions proceed at equal rates, and the reaction mix-ture reaches a condition of dynamic equilibrium, which we can represent witha double arrow

(15.3)CO(g) + 2 H2(g) ∆ CH3OH(g)

∆ .

CH3OHH2

CH3OH

Saturated solution onlyadded to beaker

(a)

(b)

FIGURE 15-2Dynamic equilibrium illustrated(a) A saturated solution of radioactive AgI is added to a saturated solutionof AgI. (b) The radioactive iodide ions distribute themselves throughout thesolution and the solid AgI, showing that the equilibrium is dynamic.

▲

In anticipation of entropy,point out that a mixingprocess and a dissolution

process lead to an increase indisorder.

Remind students that therates of reactions (forwardand reverse) are affected by

the concentration of reactants.

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 624


Silverstein, Todd P. “The RealReason Why Oil and Water

Don’t Mix.” J. Chem. Educ. 1998:75, 116 (January 1998).

Jordan, A. D. “Liquid-LiquidEquilibrium: Verification of

the Lever Rule.” J. Chem. Educ.2000: 77, 395 (March 2000).

Harrison, J. A., Buckley, P. D.“Simulating Dynamic

Equilibria: A Class Experiment.”J. Chem. Educ. 2000: 77, 1013(August 2000).

TABLE 15.1 Three Approaches to Equilibrium in the Reactiona

CO(g)

Experiment 1

Initial amounts, mol 1.000 1.000 0.000Equilibrium amounts, mol 0.911 0.822 0.0892Equilibrium concentrations, mol/L 0.0911 0.0822 0.00892

Experiment 2


Experiment 3


The concentrations printed in blue are used in the calculations in Table 15.2.aReaction carried out in a 10.0-L flask at 483 K.

CH3OH(g)H2(g)

CO(g) � 2 H2(g) ∆ CH3OH(g)

2.00

1.80

1.60

1.40

1.20

1.00

0.80

0.60

0.40

0.20

TimeExperiment 1

Mol

es o

f re

acta

nts

and

prod

ucts

2.00

1.80

1.60

1.40

1.20

1.00

0.80

0.60

0.40

0.20

TimeExperiment 2

Mol

es o

f re

acta

nts

and

prod

ucts

2.00

1.80

1.60

1.40

1.20

1.00

0.80

0.60

0.40

0.20

TimeExperiment 3

Mol

es o

f re

acta

nts

and

prod

ucts

te � time for equilibrium to be reached mol CO mol H2 mol CH3OH

te

te

te

FIGURE 15-3Three approaches to equilibriumin the reaction

The initial and equilibrium amountsfor each of these three cases arelisted in Table 15.1. forequilibrium to be reached.

te = time

CO(g) � 2 H2(g) ∆ CH3OH(g)

▲

One consequence of the equilibrium condition is that the amounts of thereactants and products remain constant with time. These equilibrium amounts,however, depend on the quantities of reactants and products present initially.For example, Table 15.1 lists data for three hypothetical experiments. All threeexperiments are conducted in a 10.0-L flask at 483 K. In the first experiment,only CO and are present initially; in the second, only and in thethird, CO, and The data from Table 15.1 are plotted in Figure 15-3,and from these graphs we see that

• in no case is any reacting species completely consumed;• in all three cases the equilibrium amounts of reactants and products

appear to have nothing in common.

CH3OH.H2,CH3OH;H2

Chemical Equilibriumactivity

Point out to students thatthe equilibrium state can beobtained by starting with

any combination of reactantsand/or products.

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Although it is not obvious from a cursory inspection of the data, a particu-lar ratio involving equilibrium concentrations of product and reactants has aconstant value, independent of how the equilibrium is reached. This ratio,which is central to the study of chemical equilibrium, can be derived theoreti-cally using concepts presented later in the text, but it can also be establishedempirically, that is, by trial and error. Three reasonable attempts at formulat-ing the desired ratio for reaction (15.3) are outlined in Table 15.2, and the ratiothat works is identified.

For the methanol synthesis reaction, the ratio of equilibrium concentrationsin the following equation has a constant value of 14.5 at 483 K.

(15.4)

This ratio is called the equilibrium constant expression and its numericalvalue is the equilibrium constant.

THE EQUILIBRIUM CONSTANT AND ACTIVITIES

Let’s think about expression (15.4) for a moment. We have shown that we ob-tain a numerical value of 14.5 by substituting equilibrium concentrations fromTable 15.2 (Trial 3) into the equilibrium constant expression. But what hap-pened to the units? The numerator in expression (15.4) has the unit mol andthe denominator, The resultant unit to accompany the numericalvalue looks like it should be In later chapters we will encounter sev-eral expressions that require the logarithm of an equilibrium constant, but wecan only take the logarithms of dimensionless numbers, not of numbers thatcarry units. We took the easy way out in equation (15.4) by just “dropping” thetroublesome units. But there is a more satisfactory way to eliminate the units.

We need to replace equation (15.4) by the following expression

(15.5)

where a represents the activity of each reactant or product denoted through asubscript formula. Activity, a thermodynamic concept introduced by G. N.Lewis,* is the dimensionless ratio where [X] represents a particularconcentration and corresponds to the concentration in a chosen referencestate. Our usual choice of reference state for a substance in solution is aconcentration of one mole per liter For gases, activity can also beexpressed as the dimensionless ratio, where P is a particular partialP>P°,

(1 mol L-1).

c°[X]>c°,

K = ¢ aCH3OH

aCO(aH2)2 ≤

eq= 14.5

L2 mol-2.mol3 L-3.

L-1

K = ¢ [CH3OH]

[CO]([H2])2 ≤eq

= 14.5

TABLE 15.2

Expt Trial 1: Trial 2: Trial 3:

1

2

3

Equilibrium concentration data are from Table 15.1. In Trial 1, the equilibrium concentration of is placed in the numeratorand the product of the equilibrium concentrations, in the denominator. In Trial 2, each concentration is multiplied byits stoichiometric coefficient. In Trial 3, each concentration is raised to a power equal to its stoichiometric coefficient. Trial 3 hasessentially the same value for each experiment. This value is the equilibrium constant Kc .

[CO][H2],CH3OH

0.0620

0.138 * 10.17622 = 14.50.0620

0.138 * 12 * 0.1762 = 1.2800.0620

0.138 * 0.176= 2.55

0.0247

0.0753 * 10.15122 = 14.40.0247

0.0753 * 12 * 0.1512 = 1.090.0247

0.0753 * 0.151= 2.17

0.00892

0.0911 * 10.082222 = 14.50.00892

0.0911 * 12 * 0.08222 = 0.5960.00892

0.0911 * 0.0822= 1.19

[CH3OH]

[CO][H2]2[CH3OH]

[CO](2 : [H2])[CH3OH][CO][H2]

*G. N. Lewis and M. Randall, Thermodynamics, McGraw Hill, New York, 1923.

In the three experiments, pointout that Table 15.1 is consistent

with the equilibriumconcentrations of the two reactantsand the one product in Figure 15-3.Put the equilibrium concentrationsinto the equilibrium expression(15.4) to show that this equationis satisfied.

The representation of theequilibrium expression interms of concentrations is

only valid at low concentrations,usually less than a few molesper liter.

Stress that “equilibrium”means equal forwardand reverse reactions,

not equal concentrations ofreactants and products.

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 626


pressure and is the partial pressure in the reference state; and our usualchoice of is 1 bar (essentially equal to 1 atm). Pure solids and pure liquidsare assigned activities of 1.

Now, by returning to the methanol-synthesis example, let us establish the re-lationship between an equilibrium constant expressed in activities and the cor-responding one expressed in concentrations. We begin by writing the activityof each species, using the [ ] symbol for the equilibrium concentration and for the concentration in the reference state.

Then we choose the value substitute these relationshipsinto the equilibrium constant expression (15.5), and we have

(15.6)

Notice that we have arrived at exactly the expression in equation (15.4)! In thiscase, however, the unwanted units have been properly cancelled rather thanjust conveniently dropped; that is, inside the large parentheses iscancelled by on the outside.

Here is another reason for basing equilibrium constants on activities: Underconditions where gases do not obey the ideal gas law (Section 6-9) or solutionsdepart from ideal behavior (Section 13-3), equilibrium constant values mayvary with total concentration or pressure. This problem is eliminated whenactivities are used. As we learned in Section 13-9, activities are “effective” or“active” concentrations. In this text we will generally assume that systems areideal, and that activities can be replaced by concentrations or partial pressures.

mol2 L-2L2 mol-2

K = § [CH3OH]c°

[CO]c°

¢ [H2]c°≤2¥

eq

= 1c°22¢ [CH3OH]

[CO]([H2])2 ≤eq

= 14.5

c° = 1 mol L-1,

aH2 =[H2]

c°

aCO =[CO]

c°

aCH3OH =[CH3OH]

c°

c0

P°P°

E X A M P L E 1 5 - 1Relating Equilibrium Concentrations of Reactants and Products. These equilibrium concentrations are measured inreaction (15.3) at 483 K: and What is the equilibrium concentration of

SolutionH2?[CH3OH] = 1.56 M.[CO] = 1.03 M

Write the equilibrium constant expression in terms of activitiesK = ¢ aCH3OH

aCO(aH2)2 ≤

eq= 14.5

Assume that the reaction conditions are such that the activities canbe replaced by their concentration values, allowing concentrationunits to be canceled as in expression (15.6).

K = ¢ [CH3OH]

[CO]([H2])2 ≤eq

= 14.5

Substitute the known equilibrium concentrations into the equilibri-um constant expression (15.6). K =

[CH3OH]

[CO][H2]2 =1.56

1.03[H2]2 = 14.5

Solve for the unknown concentration, (An implicit calcula-tion to restore the concentration unit is

.)0.322 * 1.00 M = 0.322 M[H2] = aH2 * c° =

[H2].

[H2] = 20.104 = 0.322 M

[H2]2 =1.56

1.03 * 14.5= 0.104

Practice Example A: In another experiment, equal concentrations of and CO are found at equilibrium inreaction (15.3). What must be the equilibrium concentration of

Practice Example B: At a certain temperature, for the reaction If the equilibrium concentrations of and are 0.015 M and 2.00 M, respectively, what is the equilibrium concen-tration of H2 ?

NH3N2

N2(g) + 3 H2(g) ∆ 2 NH3(g).K = 1.8 * 104

H2 ?CH3OH

KEEP IN MINDthat any concentrationssubstituted into anequilibrium constantexpression, or obtainedfrom it, must be equilibriumconcentrations.

▲

Point out to students thatthe equilibrium constant“14.5” calculated for the

methanol synthesis reaction isspecific to a 483 K reactiontemperature.

PETRMC15_622-662-hr 12/23/05 4:05 PM 27

Are You Wondering . . .If there is a relationship between the equilibrium constant and rateconstants?

Given the requirement that the rates of the forward and reverse reactions becomeequal at equilibrium, it seems that there should be a relationship. The rate laws forthe forward and reverse reactions include numerical constants (rate constants) andconcentration terms raised to powers. Moreover, if we set the rate equations equalto one another at equilibrium, we should be able to derive an expression involvinga ratio of concentration terms (raised to powers) and a ratio of constants. A majordifficulty, however, is that the exponents of the concentration terms in the equilib-rium constant expression must be the same as the coefficients in the balanced equa-tion, whereas the exponents in rate laws are generally not the same as thesecoefficients. The trick to discovering the relationship between rate constants and anequilibrium constant is to work with the detailed mechanism for the reaction in themanner outlined in Exercise 83. Still, it is generally easier to obtain K directly frommeasurements on equilibrium conditions than to attempt a calculation based onrate constants. Moreover, in Chapters 19 and 20 we will learn about much moredirect measurements and calculations leading to values of equilibrium constants.


A GENERAL EXPRESSION FOR KBefore proceeding to other matters, let us emphasize that the equilibriumconstant expression for the methanol synthesis reaction summarized throughexpression (15.6) is just a specific example of a more general case. For thehypothetical, generalized reaction

The equilibrium constant expression has the form

(15.7)

The numerator of an equilibrium constant expression is the product of theactivities of the species on the right side of the equation with eachactivity raised to a power given by the stoichiometric coefficient Thedenominator is the product of the activities of the species on the left side of theequation and again, with each activity raised to a power given bythe stoichiometric coefficient As previously noted, where equilibri-um systems are sufficiently close to ideal in their behavior, equilibrium con-centrations are acceptable approximations to true activities.

The numerical value of an equilibrium constant, K, depends on the particu-lar reaction and on the temperature. We will explore the significance of thesenumerical values in Section 15-4.

CONCEPT ASSESSMENT ✓Consider a hypothetical reaction in which one molecule, A, is converted to its isomer,B, that is, the reversible reaction Start with a flask containing 54 moleculesof A, represented by open circles. Convert the appropriate number of open circles tofilled circles to represent the isomer B and portray the equilibrium condition if

Repeat the process for and then for K = 1.K = 0.5K = 0.02.

A ∆ B.

(a, b, Á ).(aA, aB, Á ),

(g, h, Á ).(aG, aH, Á ),

K =(aG)

g(aH)h Á

(aA)a(aB)b Á =[G]g[H]h Á

[A]a[B]b Á

aA + bB Á ∆ gG + hH Á

Remind students not toconfuse equilibriumconstant expressions

with rate equations (Chapter 14).The former contains stoichiometriccoefficients whereas the latterdoes not. However, a similarity isthat both the rate constant and theequilibrium constant vary withtemperature.

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15-3 Relationships Involving Equilibrium Constants 629

RELATIONSHIPS INVOLVING EQUILIBRIUM

CONSTANTS

Before assessing an equilibrium situation, it may be necessary to make somepreliminary calculations or decisions to get the appropriate equilibrium con-stant expression. This section presents some useful ideas in working withequilibrium constants.

RELATIONSHIP OF K TO THE BALANCED CHEMICAL EQUATION

We must always make certain that the expression for K matches the corre-sponding balanced equation. In doing so, the following hold true.

• When we reverse an equation, we invert the value of K.• When we multiply the coefficients in a balanced equation by a common

factor we raise the equilibrium constant to the correspondingpower

• When we divide the coefficients in a balanced equation by a common fac-tor we take the corresponding root of the equilibrium constant(square root, cube root, ).

Suppose that in discussing the synthesis of from CO and wehad written the reverse of equation (15.3)—that is,

Now, according to the generalized equilibrium constant expression (15.7), weshould write

In the preceding expression, the terms printed in blue are the equilibrium con-stant expression and K value originally written as expression (15.4). We seethat

Suppose that for a certain application we want an equation based on syn-thesizing two moles of

Here, That is,

K– =[CH3OH]2

[CO]2[H2]4 = ¢ [CH3OH]

[CO][H2]2 ≤2= (K)2 = (14.5)2 = 2.10 * 102

K– = K2.

2 CO(g) + 4 H2(g) ∆ 2 CH3OH(g) K– = ?

CH3OH(g).

K¿ = 1>K.

K¿ =[CO][H2]2

[CH3OH]=

1[CH3OH]

[CO][H2]2

=1K

=1

14.5= 0.0690

CH3OH(g) ∆ CO(g) + 2 H2(g) K¿ = ?

H2,CH3OH

Á(2, 3, Á ),

(2, 3, Á ).(2, 3, Á ),

15-3

E X A M P L E 1 5 - 2Relating K to the Balanced Chemical Equation. The following K value is given at 298 K for the synthesis of from its elements.

What is the value of K at 298 K for the following reaction?

NH3(g) ∆12

N2(g) +32

H2(g) K = ?

N2(g) + 3 H2(g) ∆ 2 NH3(g) K = 3.6 * 108

NH3(g)

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Practice Example A: Use data from Example 15-2 to determine the value of K at 298 K for the reaction

Practice Example B: For the reaction at 184 ° C, What is the value

of K at 184 °C for the reaction 2 NO2(g) ∆ 2 NO(g) + O2(g)?

K = 7.5 * 102.NO(g) +12

O2(g) ∆ NO2(g)

13

N2(g) + H2(g) ∆23

NH3(g)

CONCEPT ASSESSMENT ✓Can you conclude whether the numerical value of K for the reaction

is greater or less than the numerical value of K for the reaction

Explain.

COMBINING EQUILIBRIUM CONSTANT EXPRESSIONS

In Section 7-7, through Hess’s law, we showed how to combine a series ofequations into a single overall equation. The enthalpy change of the overall re-action was obtained by adding together the enthalpy changes of the individ-ual reactions. A similar procedure can be used with equilibrium constants, butwith this important difference:

When individual equations are combined (that is, added), their equilibrium constantsare multiplied to obtain the equilibrium constant for the overall reaction.

Suppose we seek the equilibrium constant for the reaction

(15.8)

and know the K values of these two equilibria.

(15.9)

(15.10)

Equation (15.8) is obtained by reversing equation (15.9) and adding it to (15.10).This requires that we also take the reciprocal of the K value of equation (15.9).

(a)

(b)

Overall: N2O(g) +12

O2(g) ∆ 2 NO(g) K(overall) = ?

N2(g) + O2(g) ∆ 2 NO(g) K(b) = 4.7 * 10-31

= 3.7 * 1017

N2O(g) ∆ N2(g) +12

O2(g) K(a) = 1>(2.7 * 10-18)

N2(g) + O2(g) ∆ 2 NO(g) K = 4.7 * 10-31

N2(g) +12

O2(g) ∆ N2O(g) K = 2.7 * 10-18

N2O(g) +12

O2(g) ∆ 2 NO(g) K = ?

ICl(g) ∆12

I2(g) +12

Cl2(g)?

I2(g) + Cl2(g)2 ICl(g) ∆

SolutionFirst, reverse the given equation. This puts on theleft side of the equation, where we need it.

NH3(g)2 NH3(g) ∆ N2(g) + 3 H2(g)

The equilibrium constant becomesK¿ K¿ = 1>13.6 * 1082 = 2.8 * 10-9

Then, to base the equation on 1 mol divide all coef-ficients by 2.

NH3(g), NH3(g) ∆12

N2(g) +32

H2(g)

This requires the square root of K¿. K = 32.8 * 10-9 = 5.3 * 10-5

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The overall equation is expression (15.8), and according to the generalexpression (15.7),

CONCEPT ASSESSMENT ✓You wish to calculate K for the reaction

and you have available a K value for the reaction

What additional K value do you need, assuming that all K values are at the sametemperature?

EQUILIBRIA INVOLVING GASES: THE EQUILIBRIUM CONSTANTS, AND

Mixtures of gases are as much solutions as are mixtures in a liquid solvent.Thus, concentrations in a gaseous mixture can be expressed on a mole-per-liter basis, and this is what we have done in writing K expressions using con-centration units for our reference state. In order to describe gaseous reactantsand products through their partial pressures in atmospheres, which is com-monly done, we need to switch our reference state. We will do this next.

A key step in the manufacture of sulfuric acid is the following reversiblereaction.

(15.11)

Since all of the reaction species are in the gas phase it seems reasonable to usea partial-pressure reference state. We begin by writing the equilibrium con-stant expression in terms of activities

(15.12)

where the activities are

The reference-state partial pressure is bar, which we will take to beessentially the same as 1 atm. Substituting these relationships into equation(15.12), we obtain

(15.13)

As expected, the equilibrium constant is dimensionless. To designate that apartial-pressure reference state was used we add the subscript “p” to the equi-librium constant, that is, by writing

To establish the equilibrium constant based on concentrations, we first usethe ideal gas law, to relate gas concentrations and partial pressures

[SO3] =nV

=PSO3

RT[O2] =

nV

=PO2

RT;[SO2] =

nV

=PSO2

RT;

PV = nRT,

Kp.

Kp = § ¢PSO3

P°≤2

¢PSO2

P°≤2

PO2

P°

¥eq

= P°¢ (PSO3)2

(PSO2)2PO2

≤eq

P° = 1

aSO3 =PSO3

P°aO2 =

PO2

P°;aSO2 =

PSO2

P°;

K = ¢ (aSO3)2

(aSO2)2aO2

≤eq

2 SO2(g) + O2(g) ∆ 2 SO3(g)

KpKc

CO2(g) + H2(g) ∆ CO(g) + H2O(g)

CH4(g) + 2 H2O(g) ∆ CO2(g) + 4 H2(g)

= 3.7 * 1017 * 4.7 * 10-31 = 1.7 * 10-13

K(b)K(a)33

K(overall) =[NO]2

[N2O][O2]1>2 =[N2][O2]1>2

[N2O]*

[NO]2

[N2][O2]= K(a) * K(b)

Point out that the reactionof sulfur dioxide withoxygen gas to form sulfur

trioxide is a reaction that takesplace in the atmosphere. Thisresults in the formation of sulfuricacid which contributes to acid rain.

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 631


and then express activities in terms of those concentrations, where the refer-ence state is

Substituting into the equilibrium constant expression (15.12), we obtain

(15.14)

As expected the final expression is dimensionless. Because we have desig-nated that the reference state is based on we can attach thesubscript “c” to the equilibrium constant, , which is the ratio of concentra-tions in the final parenthetical term of expression (15.14). Finally, observe that(15.14) has a different value than (15.13). This result is not wrong. It simplystems from a difference in the reference state in the two cases. In working withequilibrium constant expressions, therefore, it is important to state the experi-mental conditions and the reference state clearly. Choosing and defining areference state is an important concept that we will see again in our discussionof thermodynamics in Chapter 19.

Eliminating the reference-state terms having a value of 1, the relationshipbetween and for reaction (15.11) is

(15.15)

If we carried out a similar derivation for the general reaction,

the result would be

(15.16)

where is the difference in the stoichiometric coefficients of gaseous prod-ucts and reactants; that is, In reaction(15.11), which is what we used in equation (15.15).

In summary, although we do not use units for the equilibrium constants, wedo need to use the correct dimensionless values of concentrations or pressuresfor terms within equilibrium constant expressions. These are molarity for expressions and pressures in atmospheres for expressions. These choices thenrequire that we use a value of in equation (15.16).R = 0.08206 L atm mol-1 K-1

Kp

Kc

¢ngas = 2 - 12 + 12 = -1,¢ngas = 1g + h + Á2 - 1a + b + Á2.¢ngas

Kp = Kc1RT2¢ngas

aA(g) + bB(g) + Á ∆ gG(g) + hH(g) + Á

Kp = Kc1RT2-1

KcKp

Kc

c° = 1 mol L-1,

Kp = P°¢ (PSO3)2

(PSO2)2PO2

≤eq

= P°¢ ([SO3]RT)2

([SO2]RT)2[O2]RT≤

eq=

P°RT

¢ [SO3]2

[SO2]2[O2]≤

eq

aSO3 =[SO3]

c°=

PSO3

RTc°

aO2 =[O2]

c°=

PO2

RTc°

;aSO2 =[SO2]

c°=

PSO2

RTc°

;

c° = 1 mol L-1.

KEEP IN MINDthat only if That is, because any number raised tothe zero power equals one.

Kp = Kc1RT20 = Kc

¢ngas = 0.Kp = Kc

▲

E X A M P L E 1 5 - 3Illustrating the Dependence of K on the Reference State. Complete the calculation of for reaction (15.11) know-ing that

Solution

Kc = 2.8 * 102 (at 1000 K).Kp

Write the equation relating the two equilibrium constants withdifferent reference states. Kc = RT * Kp

Rearrange the expression to obtain the quantity desired, Kp. Kp =Kc

RT

Substitute the given data and solve. Kp =2.8 * 102

0.08206 * 1000= 3.4

Practice Example A: For the reaction at 298 K, What is the valueof for this reaction?

Practice Example B: At 1065 °C, for the reaction What is the

value of for the reaction at 1065 °C?H2(g) +12

S2(g) ∆ H2S(g)Kc

2 H2S(g) ∆ 2 H2(g) + S2(g), Kp = 1.2 * 10-2.

Kp

Kc = 2.8 * 10-9.2 NH3(g) ∆ N2(g) + 3 H2(g)

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 632


Still another way to thinkabout solids and liquids isthrough their densities.Density, the mass per unitvolume of a substance, can beexpressed in moles per literby converting the unitvolume from milliliter to literand dividing the mass ingrams by the molar mass.The resultant molar density(mol/L) is a concentrationterm and, at a fixedtemperature, is a constantthat would be incorporated inthe value.Kc

▲

Of course, the relationshipbetween concentration andpressure for gases is based here

on the ideal gas law: or[gas] = (ngas>V) = P>RT.

PV = ngasRT,

EQUILIBRIA INVOLVING PURE LIQUIDS AND SOLIDS

Up to this point in the chapter, all our examples have involved gaseous reac-tions. In subsequent chapters, we will emphasize equilibria in aqueous solu-tions. Gas-phase reactions and reactions in aqueous solution are homogeneousreactions: They occur within a single phase. Let’s extend our coverage nowto include reactions involving one or more condensed phases—solids andliquids—in contact with a gas or solution phase. These are called heterogeneousreactions. One of the most important ideas about heterogeneous reactions is that

Equilibrium constant expressions do not contain concentration terms for solid orliquid phases of a single component (that is, for pure solids and liquids).

We can think about this statement in either of two ways: (1) An equilibriumconstant expression includes terms only for reactants and products whose con-centrations and/or partial pressures can change during a chemical reaction. Theconcentration of the single component within a pure solid or liquid phasecannot change. (2) Alternatively, recall that the activities of pure liquids andsolids are set equal to 1; thus the effect on the numerical value of the equilibri-um constant is the same as not including terms for pure solids and liquids at all.

The water–gas reaction, used to make combustible gases from coal, has re-acting species in both gaseous and solid phases.

Although solid carbon must be present for the reaction to occur, the equilibri-um constant expression contains terms only for the species in the homoge-neous gas phase: CO, and

The activity of solid carbon is and we have implicitly divided througheach of the remaining concentrations by the reference-state concentration, mol to obtain a dimensionless

The decomposition of calcium carbonate (limestone) is also a heterogeneousreaction. The equilibrium constant expression contains just a single term.

(15.17)

We can write for reaction (15.17) by using equation (15.16), with

(15.18)

Equation (15.18) indicates that the equilibrium pressure of in contactwith and CaO(s) is a constant equal to Its value is independent ofthe quantities of and CaO (as long as both solids are present). Figure 15-4offers a conceptualization of this decomposition reaction.

CaCO3

Kp.CaCO3(s)CO2(g)

Kp = PCO2 and Kp = Kc(RT)

¢ngas = 1.Kp

CaCO3(s) ∆ CaO(s) + CO2(g) Kc = [CO2]

Kc .L-1,c0 = 1

aC(s) = 1,

Kc =aCOaH2

aC(s)aH2O=

[CO][H2][H2O]

H2.H2O,

C(s) + H2O(g) ∆ CO(g) + H2(g)

Drop two pieces of limestone, onelarge and the other small, into a

dilute solution of HCl, seal thebeakers, and leave them until nextclass. Some carbon dioxide will beseen bubbling off. By the next class,the reaction will have stopped. Pointout that the pressure of in bothbeakers is the same, but one piece oflimestone is still bigger than the other.That is, the concentration of doesnot depend upon the amount of solidpresent. An improvement of theexperiment would be to attach the twoclosed vessels to a U-tube manometerto show the pressures are equal. Relatethis to Figure 15-4.

CO2

CO2

(a) (b)

FIGURE 15-4Equilibrium in the reaction

(a) Decomposition of uponheating in a closed vessel yields a fewgranules of CaO(s), together with which soon exerts its equilibrium partialpressure. (b) Introduction of additional

and/or more CaO(s) has noeffect on the partial pressure of the

which remains the same as in (a).CO2(g),

CaCO3(s)

CO2(g),

CaCO3(s)

CaCO3(s) ∆ CaO(s) � CO2(g)

▲

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 633


THE MAGNITUDE OF AN EQUILIBRIUM CONSTANT

In principle, every chemical reaction has an equilibrium constant, but oftenthe constants are not used. Why is this so? Table 15.3 lists equilibrium con-stants for several reactions mentioned in this chapter or previously in the text.The first of these reactions is the synthesis of from its elements. We havealways assumed that this reaction goes to completion, that is, that the reversereaction is negligible and the overall reaction proceeds only in the forward di-rection. If a reaction goes to completion, one (or more) of the reactants is usedup. A term in the denominator of the equilibrium constant expression ap-proaches zero and makes the value of the equilibrium constant very large. Avery large numerical value of K signifies that the forward reaction, as written,goes to completion or very nearly so. Because the value of for the water syn-thesis reaction is we are entirely justified in saying that the reactiongoes to completion at 298 K.

1.4 * 1083,Kp

H2O

15-4

One of our examples in Section 15-1 was liquid–vapor equilibrium. This is aphysical equilibrium because no chemical reactions are involved. Consider theliquid–vapor equilibrium for water.

So, equilibrium vapor pressures such as are just values of As we haveseen before, these values do not depend on the quantities of liquid or vapor atequilibrium, as long as some of each is present.

Kp.PH2O

Kc = [H2O(g)] Kp = PH2O Kp = KcRT

H2O(l) ∆ H2O(g)

Write the equilibrium constant expression in terms ofactivities. Note that activities for the iodine and sulfurare not included, since the activity of a pure solid is 1.

K =(aHI)2

(aH2S)

Substitute partial pressures for the activities into theequilibrium constant expression. Kp =

(PHI)2

(PH2S)

Substitute the given equilibrium data into the equilib-rium constant expression. Kp =

13.65 * 10-3229.96 * 10-1 = 1.34 * 10-5

Practice Example A: Teeth are made principally from the mineral hydroxyapatite, which canbe dissolved in acidic solution such as that produced by bacteria in the mouth. The reaction that occurs is

Write the equilibrium constant expressionfor this reaction.

Practice Example B: The steam–iron process is used to generate mostly for use in hydrogenating oils. Ironmetal and steam react to produce and Write expressions for and for this reversiblereaction. How are the values of and related to each other? Explain.KpKc

KpKcH2(g).Fe3O4(s)[H2O(g)]H2(g),

Kc

5 Ca2+(aq) + 3 HPO4

2-(aq) + H2O(l).4 H+(aq) ∆Ca5(PO4)3OH(s) +

Ca5(PO4)3OH,

Point out the huge range ofvalues that equilibriumconstants can take.

E X A M P L E 1 5 - 4Writing Equilibrium Constant Expressions for Reactions Involving Pure Solids or Liquids. At equilibrium in thefollowing reaction at 60 °C, the partial pressures of the gases are found to be atm and

What is the value of for the reaction?

Solution

H2S(g) + I2(s) ∆ 2 HI(g) + S(s) Kp = ?

KpPH2S = 9.96 * 10-1 atm.PHI = 3.65 * 10-3

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 634

15-4 The Magnitude of an Equilibrium Constant 635

Equilibrium ConstantactivityTABLE 15.3 Equilibrium Constants of Some Common Reactions

Reaction Equilibrium constant,

at 298 K

at 298 K1.0 at about 1200 K

3.4 at 1000 K

at 298 K10.0 at about 1100 K1.6 * 10-21C(s) + H2O(g) ∆ CO(g) + H2(g)

2 SO2(g) + O2(g) ∆ 2 SO3(g)

1.9 * 10-23CaCO3(s) ∆ CaO(s) + CO2(g)

1.4 * 10832 H2(g) + O2(g) ∆ 2 H2O(l)

Kp

If the equilibrium constant is so large, why is a mixture of hydrogen andoxygen gases stable at room temperature? The value of the equilibrium con-stant relates to thermodynamic stability: is much more thermodynam-ically stable than a mixture of and because it lies at a lower energystate. As noted in Chapter 14, however, the rate of a chemical reaction isstrongly governed by the activation energy, Because is very high for thesynthesis of from and the rate of reaction is inconsequen-tial at 298 K. To get the reaction to occur at a measurable rate, we must eitherraise the temperature or use a catalyst. A chemist would say that the synthesisof at 298 K is a kinetically controlled reaction (as opposed tothermodynamically controlled).

From Table 15.3, we see that for the decomposition of (lime-stone) is very small at 298 K (only ). To account for a very small nu-merical value of an equilibrium constant, the numerator must be very small(approaching zero). A very small numerical value of K signifies that the for-ward reaction, as written, does not occur to any significant extent. Although lime-stone does not decompose at ordinary temperatures, the partial pressure of

in equilibrium with and CaO(s) increases with temperature.It becomes 1 atm at about 1200 K. An important application of this decompo-sition reaction is in the commercial production of quicklime (CaO).

The conversion of and to at 1000 K has an equilibriumconstant such that we expect significant amounts of both reactants and prod-ucts to be present at equilibrium (see Table 15.3). Both the forward and reversereactions are important. A similar situation exists for the reaction of C(s) and

at 1100 K, but not at 298 K where the forward reaction does not occurto any significant extent

In light of the several cases from Table 15.3, we can conclude that

A reaction is most likely to reach a state of equilibrium in which significant quanti-ties of both reactants and products are present if the numerical value of K is neithervery large nor very small, as a very rough approximation, in the range of about to

Thus, we see that equilibrium calculations are not required for all reactions.At times, we can use simple stoichiometric calculations to determine the out-come of a reaction, and in some cases there may be no reaction at all.

CONCEPT ASSESSMENT ✓Why is having a balanced equation a necessary condition for predicting the outcomeof a chemical reaction, but often not a sufficient condition?

1010.10-10

1Kp = 1.6 * 10-212.H2O(g)

SO3(g)O2(g)SO2(g)

CaCO3(s)CO2(g)

1.9 * 10-23CaCO3(s)Kp

H2O(l)

O2(g),H2(g)H2O(l)EaEa .

O2(g)H2(g)H2O(l)

Equilibrium constantsusually vary in value to alarge extent with

temperature.

The products formed ina reaction under kineticcontrol are determined by

reaction rates. The productsformed in a reaction underthermodynamic control will dependon the stability of the products.

PETRMC15_622-662-hr 12/23/05 2:31 PM Page 635


THE REACTION QUOTIENT, Q: PREDICTING THE

DIRECTION OF NET CHANGE

Let’s return briefly to the set of three experiments that we discussed in Section15-2, involving the reaction

Experiment 1 starts with just the reactants, CO and An overall, or net,change has to occur in which some forms. Only in this way can anequilibrium condition be reached in which all reacting species are present. Wesay that a net change occurs in the forward direction (to the right).

Experiment 2 starts with just the product, Here, some of themust decompose back to CO and before equilibrium can be estab-

lished. We say that a net change occurs in the reverse direction (to the left).Experiment 3 starts with all the reacting species present: CO, and

In this system, it is not obvious in what direction a net change occursto establish equilibrium.

The ability to predict the direction of net change in establishing equilibriumis important for two reasons.

• At times we do not need detailed equilibrium calculations. We may needonly a qualitative description of the changes that occur in establishingequilibrium from a given set of initial conditions.

• In some equilibrium calculations, it is helpful to determine the directionof net change as a first step.

For any set of initial activities in a reaction mixture, we can set up a ratio ofactivities having the same form as the equilibrium constant expression. Thisratio is called the reaction quotient and is designated Q. For a hypotheticalgeneralized reaction, the reaction quotient, first written in terms of activities,and then as concentrations assuming a concentration reference state, is

(15.19)

If a reaction is at equilibrium, but our primary interest in the relation-ship between Q and K is for a reaction mixture that is not at equilibrium. Tosee what this relationship is, let’s turn again to the experiments in Table 15.1.

In Experiment 1, the initial concentrations of CO and areInitially there is no The value of is

(15.20)

We know that a net reaction occurs to the right, producing some As it does, the numerator in expression (15.20) increases, the denominatordecreases, and the value of increases; eventually

If a net change occurs from left to right (the direction of the forwardreaction).

In Experiment 2, the initial concentration of is Initially, there is no CO or The value of is

(15.21)

We know that a net reaction occurs to the left, producing some CO and Asit does, the numerator in expression (15.21) decreases, the denominatorincreases, and the value of decreases; eventually

If a net change occurs from right to left (the direction of the reversereaction).

Qc 7 Kc ,

Qc = Kc .Qc

H2.

Qc =[CH3OH]init

[CO]init[H2]init2 =

0.1000 * 0

= q

QcH2.0.100 M.1.000 mol>10.0 L =CH3OH

Qc 6 Kc ,

Qc = Kc .Qc

CH3OH.

Qc =[CH3OH]init

[CO]init[H2]init2 =

0

(0.100)(0.100)2 = 0

QcCH3OH.1.000 mol>10.0 L = 0.100 M.H2

Q = K,

Q =(ainit)

g(ainit)h Á

(ainit)a(ainit)b Á Qc =[G]init

g[H]init

h Á

[A]inita [B]init

b Á

CH3OH.H2,

H2CH3OHCH3OH.

CH3OHH2.

CO(g) + 2 H2(g) ∆ CH3OH(g) Kc = 14.5

15-5

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 636

15-5 The Reaction Quotient, Q: Predicting the Direction of Net Change 637

Now let us turn to a case where direction of net change is not immediatelyobvious. In Experiment 3, the initial concentrations of all three species are

The value of is

Because (100 compared with 14.5), a net change occurs in the reversedirection. Note that you can verify this conclusion from Figure 15-3. Theamounts of CO and at equilibrium are greater than they were initially, andthe amount of is less.

The criteria for predicting the direction of a net chemical change in a re-versible reaction are summarized in Figure 15-5 and applied in Example 15-5.

CH3OHH2

Qc 7 Kc

Qc =[CH3OH]init

[CO]init[H2]init2 =

0.100

(0.100)(0.100)2 = 100

Qc1.000 mol>10.0 L = 0.100 M.

Initial condition: Pure reactants

“Left” of equilibrium

“Right” of equilibrium

Pureproducts

Equilibrium

Reaction quotient, Qc � 0 � Kc � Kc � �� KcReaction proceeds

to the right to the left

(a) (b) (c) (d) (e)

FIGURE 15-5Predicting the direction of net change in a reversible reactionFive possibilities for the relationship of initial and equilibrium conditions are shown.From Table 15.1 and Figure 15-3, Experiment 1 corresponds to initial condition (a);Experiment 2 to condition (e); and Experiment 3 to (d). The situation in Example 15-5also corresponds to condition (d).

▲

E X A M P L E 1 5 - 5Predicting the Direction of a Net Chemical Change in Establishing Equilibrium. To increase the yield of inthe water–gas reaction—the reaction of C(s) and to form CO(g) and —a follow-up reaction called the“water–gas shift reaction” is generally used. In this reaction, some of the CO(g) of the water gas is replaced by

at about 1100 K. The following amounts of substances are brought together and allowed to react at thistemperature: 1.00 mol CO, 1.00 mol 2.00 mol and 2.00 mol Compared with their initial amounts, whichof the substances will be present in a greater amount and which in a lesser amount when equilibrium is established?

Solution

H2.CO2,H2O,Kc = 1.00

CO(g) + H2O(g) ∆ CO2(g) + H2(g)

H2(g).H2(g)H2O(g)

H2(g)

Our task is to determine the direction of net change by evaluatingWrite down the expression for Qc .Qc . Qc =

[CO2][H2][CO][H2O]

Substitute concentrations into the expression for by assumingan arbitrary volume V (which cancels out in the calculation).

Qc ,Qc =

12.00>V212.00>V211.00>V211.00>V2 = 4.00

Compare to .KcQc 4.00 7 1.00

Because (that is, ), a net change occurs to theleft. When equilibrium is established, the amounts of CO andwill be greater than the initial quantities and the amounts ofand will be less.H2

CO2

H2O4.00 7 1.00Qc 7 Kc

Practice Example A: In Example 15-5, equal masses of CO, and are mixed at a temperature ofabout 1100 K. When equilibrium is established, which substance(s) will show an increase in quantity and which willshow a decrease compared with the initial quantities?

Practice Example B: For the reaction at 261 °C. If a vessel is filledwith these gases such that the initial partial pressures are in whichdirection will a net change occur?

PPCl5 = 19.7 atm,PCl2 = 0.88 atm,PPCl3 = 2.19 atm,Kc = 0.0454PCl5(g) ∆ PCl3(g) + Cl2(g),

H2H2O, CO2,

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 637


A gas at equilibrium is agood example of LeChâtelier’s principle. Use

and imaginechanging any one of P, [gas], or T.The ideal gas law must be satisfied,so each time one parameter ischanged, the others must readjust.

P = [gas]RT

Huddle, Benjamin P.“Conceptual Questions” on Le

Châtelier’s Principle. J. Chem. Educ.2000: 75, 1175 (September 1998).

Thomsen, Volker B. E. “LeChâtelier’s Principle in the

Sciences.” J. Chem. Educ. 2000: 77,173 (February 2000).

CONCEPT ASSESSMENT ✓A mixture of 1.00 mol each of CO(g), and is placed in a 10.0-L flask ata temperature at which in the reaction

When equilibrium is established, (a) the amount of will be 1.00 mol; (b) theamounts of all reactants and products will be greater than 1.00 mol; (c) the amountsof all reactants and products will be less than 1.00 mol; (d) the amount of willbe greater than 1.00 mol and the amounts of CO(g), and will be lessthan 1.00 mol; (e) the amounts of reactants and products cannot be predicted and canonly be determined by analyzing the equilibrium mixture.

ALTERING EQUILIBRIUM CONDITIONS:LE CHÂTELIER’S PRINCIPLE

At times, we want only to make qualitative statements about a reversible reac-tion: the direction of a net change, whether the amount of a substance willhave increased or decreased when equilibrium is reached, and so on. Also, wemay not have the data needed for a quantitative calculation. In these cases, wecan use a statement attributed to the French chemist Henri Le Châtelier (1884).Le Châtelier’s principle is hard to state unambiguously, but its essentialmeaning is stated here.

When an equilibrium system is subjected to a change in temperature, pressure,or concentration of a reacting species, the system responds by attaining a new equi-librium that partially offsets the impact of the change.

As we will see in the examples that follow, it is generally not difficult to pre-dict the outcome of changing one or more variables in a system at equilibrium.

EFFECT OF CHANGING THE AMOUNTS OF REACTING SPECIES

ON EQUILIBRIUM

Let’s return to reaction (15.11)

2 SO2(g) + O2(g) ∆ 2 SO3(g) Kc = 2.8 * 102 at 1000 K

15-6

H2(g)H2O(g),CO2(g)

H2(g)

CO(g) + H2O(g) ∆ CO2(g) + H2(g)

Kp = 10.0CO2(g)H2O(g),

�2 SO2(g) � O2(g) ∆ 2 SO3(g)

▲

Note that there is a timedelay between theapplication of a stress and

the establishment of equilibrium.This is similar to Figure 15-3.Imagine if the applied stress is verysmall. In this case equilibrium isalways maintained because thechange is too small to notice. Usethis to introduce the concept ofreversible processes.

Suppose we start with certain equilibrium amounts of and assuggested by Figure 15-6(a). Now let’s create a disturbance in the equilibriummixture by forcing an additional 1.00 mol into the 10.0-L flask (Fig. 15-6b).How will the amounts of the reacting species change to reestablish equilibrium?

According to Le Châtelier’s principle, if the system is to partially offset anaction that increases the equilibrium concentration of one of the reactingspecies, it must do so by favoring the reaction in which that species is con-sumed. In this case, this is the reverse reaction—conversion of some of theadded to and In the new equilibrium, there are greater amountsof all the substances than in the original equilibrium, but the additionalamount of is less than the 1.00 mol that was added.

Another way to look at the matter is to evaluate the reaction quotient im-mediately after adding the

Original equilibrium Immediately following disturbance

Qc =[SO3]

[SO2]2[O2]7 KcQc =

[SO3]

[SO2]2[O2]= Kc

SO3.

SO3

O2.SO2SO3

SO3

SO3,SO2, O2,

KEEP IN MINDthat volume terms cancel ina reaction quotient orequilibrium constantexpression whenever the sumof the exponents in thenumerator equals that in thedenominator. This cansimplify problem solvingin some instances.

▲

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 638

15-6 Altering Equilibrium Conditions: Le Châtelier’s Principle 639

0.68 mol SO3

1.00 mol SO3

Add:

10.0 L 10.0 L0.32 mol SO2

0.16 mol O2

1.46 mol SO3

0.54 mol SO2

0.27 mol O2

(a) (c)

(b)

FIGURE 15-6Changing equilibrium conditions by changing the amount of a reactant

(a) The original equilibrium condition. (b) Disturbance caused by adding 1.00 mol(c) The new equilibrium condition. The amount of in the new equilibrium

mixture, 1.46 mol, is greater than the original 0.68 mol but it is not as great asimmediately after the addition of 1.00 mol The effect of adding to anequilibrium mixture is partially offset when equilibrium is restored.

SO3SO3 .

SO3SO3 .

2 SO2(g) � O2(g) ∆ 2 SO3(g), Kc � 2.8 : 102 at 1000 K

▲

Le Châtelier’s Principlemovie

Adding any quantity of to a constant-volume equilibrium mixturemakes larger than A net change occurs in the direction that reduces

that is, to the left, or in the reverse direction. Notice that reactionin the reverse direction increases and further decreasing thevalue of

E X A M P L E 1 5 - 6Applying Le Châtelier’s Principle: Effect of Adding More of a Reactant to anEquilibrium Mixture. Predict the effect of adding more to a constant-volumeequilibrium mixture of and

SolutionIncreasing stimulates the forward reaction and a shift in the equilibrium con-dition to the right. However, only a portion of the added is consumed in this re-action. When equilibrium is reestablished, there will be more than was presentoriginally, and also more but the amount of will be smaller. Some of theoriginal must be consumed in converting some of the added to

Practice Example A: Given the reaction whatis the effect of adding to a constant-volume equilibrium mixture?

Practice Example B: Calcination of limestone (decomposition by heating),is the commercial source of quicklime, CaO(s).

After this equilibrium has been established in a constant-temperature, constant-volume container, what is the effect on the equilibrium amounts of materials causedby adding some (a) CaO(s); (b) (c) CaCO3(s)?CO2(g);

CaCO3(s) ∆ CaO(s) + CO2(g),

O2(g)2 CO(g) + O2(g) ∆ 2 CO2(g),

NH3.H2N2

N2NH3,H2

H2

[H2]

N2(g) + 3 H2(g) ∆ 2 NH3(g)

NH3.N2, H2,H2(g)

Qc .[O2],[SO2]

[SO3],Kc .Qc

SO3

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 639


EFFECT OF CHANGES IN PRESSURE OR VOLUME ON EQUILIBRIUM

There are three ways to change the pressure of a constant-temperature equi-librium mixture.

1. Add or remove a gaseous reactant or product. The effect of these actionson the equilibrium condition is simply that due to adding or removing areaction component, as described previously.

2. Add an inert gas to the constant-volume reaction mixture. This has theeffect of increasing the total pressure, but the partial pressures of the react-ing species are all unchanged. An inert gas added to a constant-volumeequilibrium mixture has no effect on the equilibrium condition.

3. Change the pressure by changing the volume of the system. Decreasingthe volume of the system increases the pressure, and increasing the systemvolume decreases the pressure. Thus, the effect of this type of pressurechange is simply that of a volume change.

Let’s explore the third situation first. Consider, again, the formation offrom and

The equilibrium mixture in Figure 15-7(a) has its volume reduced to one-tenthof its original value by increasing the external pressure. To see how the equi-librium amounts of the gases change, let’s first rearrange the equilibrium con-stant expression to the form

(15.22)

From equation (15.22), we see that if V is reduced by a factor of 10, the ratio(nSO3)

2

(nSO2)2(nO2)

Kc =[SO3]2

[SO2]2[O2]=

(nSO3>V)2

(nSO2>V)2(nO2>V)=

(nSO3)2

(nSO2)2(nO2)

* V = 2.8 * 102

2 SO2(g) + O2(g) ∆ 2 SO3(g) Kc = 2.8 * 102 at 1000 K

O2(g).SO2(g)SO3(g)

1.00 L10.0 L

(a) (b)

0.68 mol SO3

0.32 mol SO2

0.16 mol O2

0.83 mol SO3

0.17 mol SO2

0.085 mol O2

FIGURE 15-7Effect of pressure change on equilibrium in the reaction

An increase in external pressure causes a decrease in the reaction volume and ashift in equilibrium “to the right.” (See Exercise 65 for a calculation of the newequilibrium amounts.)

2 SO2(g) � O2(g) ∆ 2 SO3(g)

▲

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 640


must increase by a factor of 10. In this way, the value of is restored, as it mustbe to restore equilibrium. There is only one way in which the ratio of moles ofgases will increase in value: The number of moles of must increase, and thenumbers of moles of and must decrease. The equilibrium shifts inthe direction producing more —to the right.

Notice that three moles of gas on the left produce two moles of gas on theright in reaction (15.11). When compared at the same temperature and pres-sure, two moles of occupy a smaller volume than does a mixture of twomoles of and one mole of Given this fact and the observationfrom equation (15.22) that a decrease in volume favors the production of addi-tional we can formulate a statement that is especially easy to apply.

When the volume of an equilibrium mixture of gases is reduced, a net change occursin the direction that produces fewer moles of gas. When the volume is increased, a netchange occurs in the direction that produces more moles of gas.

Figure 15-7 suggests a way of decreasing the volume of gaseous mixture atequilibrium—by increasing the external pressure. One way to increase the vol-ume is to lower the external pressure. Another way is to transfer the equilibri-um mixture from its original container to one of larger volume. A thirdmethod is to add an inert gas at constant pressure; the volume of the mixturemust increase to make room for the added gas. The effect on the equilibrium,however, is the same for all three methods: Equilibrium shifts in the directionof the reaction producing the greater number of moles of gas.

Equilibria between condensed phases are not affected much by changes inexternal pressure because solids and liquids are not easily compressible. Also,we cannot assess whether the forward or reverse reaction is favored by thesechanges by examining only the chemical equation.

E X A M P L E 1 5 - 7Applying Le Châtelier’s Principle: The Effect of Changing Volume. An equilibriummixture of and is transferred from a 1.50-L flask to a 5.00-L flask.In which direction does a net change occur to restore equilibrium?

SolutionWhen the gaseous mixture is transferred to the larger flask, the partial pressure ofeach gas and the total pressure drop. Whether we think in terms of a decrease inpressure or an increase in volume, we reach the same conclusion. Equilibrium shiftsin such a way as to produce a larger number of moles of gas. Some of the orig-inally present decomposes back to and A net change occurs in the directionof the reverse reaction—to the left—in restoring equilibrium.

Practice Example A: The reaction is at equilibrium ina 3.00-L cylinder. What would be the effect on the concentrations of and

if the pressure were doubled (that is, cylinder volume decreased to 1.50 L)?

Practice Example B: How is the equilibrium amount of produced in thewater–gas shift reaction affected by changing the total gas pressure or the systemvolume? Explain.

CO(g) + H2O(g) ∆ CO2(g) + H2(g)

H2(g)

NO2(g)N2O4(g)

N2O4(g) ∆ 2 NO2(g)

H2.N2

NH3

N2(g) + 3 H2(g) ∆ 2 NH3(g)

NH3(g)N2(g), H2(g),

SO3,

O2(g).SO2(g)SO3(g)

SO3

O2SO2

SO3

Kc

Equilibrium animationNO2 ¬ N2O4

KEEP IN MINDthat an inert gas has no effecton an equilibrium conditionif the gas is added to a systemmaintained at constant vol-ume, but it can have an effectif added at constant pressure.

▲

Remind students that thepressure of a particular gasinside a container remains

unchanged even if an inert gas isadded to the container as long asthe volume of the container is notchanged. Review Dalton's Law ofPartial Pressures.

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 641


Note that an exothermicreaction is aided bylowering the temperature

and an endothermic reaction isaided by raising the temperature.You can then point out thedifference between the system andthe surroundings.

CONCEPT ASSESSMENT ✓The following reaction is brought to equilibrium at 700 °C.

Indicate whether each of the following statements is true, false, or not possible toevaluate from the information given.

(a) If the equilibrium mixture is allowed to expand into an evacuated larger con-tainer, the mole fraction of will increase.

(b) If several moles of Ar(g) are forced into the reaction container, the amounts ofand will increase.

(c) If the equilibrium mixture is cooled to 100 °C, the mole fractions of the fourgases will likely change.

(d) If the equilibrium mixture is forced into a slightly smaller container, the partialpressures of the four gases will all increase.

EFFECT OF TEMPERATURE ON EQUILIBRIUM

We can think of changing the temperature of an equilibrium mixture in termsof adding heat (raising the temperature) or removing heat (lowering the tem-perature). According to Le Châtelier’s principle, adding heat favors the reac-tion in which heat is absorbed (endothermic reaction). Removing heat favorsthe reaction in which heat is evolved (exothermic reaction). Stated in terms ofchanging temperature,

Raising the temperature of an equilibrium mixture shifts the equilibrium condition inthe direction of the endothermic reaction. Lowering the temperature causes a shift in thedirection of the exothermic reaction.

The principal effect of temperature on equilibrium is in changing the valueof the equilibrium constant. In Chapter 19, we will learn how to calculate equi-librium constants as a function of temperature. For now, we will limit our-selves to making qualitative predictions.

E X A M P L E 1 5 - 8Applying Le Châtelier’s Principle: Effect of Temperature on Equilibrium.Consider the reaction

Will the amount of formed from given amounts of and begreater at high or low temperatures?

SolutionRaising the temperature favors the endothermic reaction, the reverse reaction. Low-ering the temperature favors the forward (exothermic) reaction. Therefore, an equi-librium mixture would have a higher concentration of at lower temperatures.The conversion of to is favored at low temperatures.

Practice Example A: The reaction has Will the amount of formed from be greater at high or low temperatures?

Practice Example B: The enthalpy of formation of is Will the concentration of in an equilibrium mixture with

its elements be greater at or at 300 °C? Explain.100NH3-46.11 kJ>mol NH3.

¢Hf°[NH3(g)] =NH3

N2O4(g)NO2(g)¢H° = +57.2 kJ.N2O4(g) ∆ 2 NO2(g)

SO3SO2

SO3

O2(g)SO2(g)SO3(g)

2 SO2(g) + O2(g) ∆ 2 SO3(g) ¢H° = -197.8 kJ

CH4H2S

H2

2 H2S(g) + CH4(g) ∆ CS2(g) + 4 H2(g)

Remind students not toconfuse shifts inequilibrium with changes

in reaction rates that result fromtemperature changes. Equilibriaof exothermic and endothermicreactions will shift differentlywhen temperatures are increased,but the rates of exothermic andendothermic reactions both increasewith increasing temperature.

Temperature Dependenceof Equilibrium animation

Stress to students thatchanging the temperatureof a reaction will result in

a shift of the equilibrium and anew equilibrium constant.

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 642


Sulfuric acid is produced from

The catalyst used to speed up the conversion of to in the commercialproduction of sulfuric acid is V2O5(s).

SO3SO2

SO3(g) � H2O(l) ∆ H2SO4(aq)

SO3▲

EFFECT OF A CATALYST ON EQUILIBRIUM

Adding a catalyst to a reaction mixture speeds up both the forward and re-verse reactions. Equilibrium is achieved more rapidly, but the equilibriumamounts are unchanged by the catalyst. Consider again reaction (15.11)

For a given set of reaction conditions, the equilibrium amounts of andhave fixed values. This is true whether the reaction is carried out by a slow

homogeneous reaction, catalyzed in the gas phase, or conducted as a heteroge-neous reaction on the surface of a catalyst. Stated another way, the presence of acatalyst does not change the numerical value of the equilibrium constant.

We now have two thoughts about a catalyst to reconcile: one from the pre-ceding chapter and one from this discussion.

• A catalyst changes the mechanism of a reaction to one with a lower acti-vation energy.

• A catalyst has no effect on the condition of equilibrium in a reversiblereaction.

Taken together, these two statements must mean that an equilibrium condi-tion is independent of the reaction mechanism. Thus, even though equilibriumhas been described in terms of opposing reactions occurring at equal rates, wedo not have to be concerned with the kinetics of chemical reactions to workwith the equilibrium concept. This observation is still another indication thatthe equilibrium constant is a thermodynamic quantity, as we shall describemore fully in Chapter 19.

CONCEPT ASSESSMENT ✓Two students are performing the same experiment in which an endothermic reactionrapidly attains a condition of equilibrium. Student A does the reaction in a beakerresting on the surface of the lab bench while student B holds the beaker in which thereaction occurs. Assuming that all other environmental variables are the same, whichstudent should end up with more product? Explain.

SO3

SO2, O2,

2 SO2(g) + O2(g) ∆ 2 SO3(g) Kc = 2.8 * 102 at 1000 K

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 643


Good Web sites forequilibrium calculations:http://www.chem.vt.edu/

RVGS/ACT/notes/chem-eqm.html(noncommercial);http://stokes.lance.colcostate.edu/tools/equil.html (noncommercial).

Nitrogen Dioxideand DinitrogenTetroxide movie

Leenson, I. A. “ApproachingEquilibrium in the

System: A CommonMistake in Textbooks.” J. Chem.Educ. 2000: 77, 1652(December 2000).

N2O4 ¬ NO2

EQUILIBRIUM CALCULATIONS: SOME

ILLUSTRATIVE EXAMPLES

We are now ready to tackle the problem of describing, in quantitative terms,the condition of equilibrium in a reversible reaction. Part of the approach weuse may seem unfamiliar at first—it has an algebraic look to it. But as you ad-just to this “new look,” do not lose sight of the fact that we continue to usesome familiar and important ideas—molar masses, molarities, and stoichio-metric factors from the balanced equation, for example.

The five numerical examples that follow apply the general equilibriumprinciples described earlier in the chapter. The first four involve gases, whilethe fifth deals with equilibrium in an aqueous solution. (The study of equilib-ria in aqueous solutions is the principal topic of the next three chapters.) Eachexample includes a brief section labeled “comments,” which is printed on ayellow background. Think of the comments as the basic methodology of equi-librium calculations. You may want to refer back to these comments from timeto time while you are studying later chapters.

Example 15-9 is relatively straightforward. It demonstrates how to deter-mine the equilibrium constant of a reaction when the equilibrium concentra-tions of the reactants and products are known.

15-7

O

O

ON

O O

O

N

O

O

NN

The Lewis structures of and Nitrogen dioxide is a free radical that combinesexothermically to form dinitrogen tetroxide.

NO2(g)N2O4▲

E X A M P L E 1 5 - 9Determining a Value of from the Equilibrium Quantities of Substances. Dinitrogen tetroxide, is animportant component of rocket fuels—for example, as an oxidizer of liquid hydrazine in the Titan rocket. At 25 °C,

is a colorless gas that partially dissociates into a red-brown gas. The color of an equilibrium mixture ofthese two gases depends on their relative proportions, which in turn depends on the temperature (Fig. 15-8).

Equilibrium is established in the reaction at 25 °C. The quantities of the two gases presentin a 3.00-L vessel are 7.64 g and 1.56 g What is the value of for this reaction?KcNO2.N2O4

N2O4(g) ∆ 2 NO2(g)

NO2,N2O4

N2O4(l),Kc

(a) (b)

Use these examples to pointout the importance ofsignificant figures.

FIGURE 15-8The equilibrium (a) At dry ice temperatures, exists as a solid.The gas in equilibrium with the solid is mostlycolorless with only a trace of brown (b) When warmed to room temperature and above,the melts and vaporizes. The proportion of

at equilibrium increases over that at lowtemperatures, and the equilibrium mixture of

and has a red-brown color.NO2(g)N2O4(g)

NO2(g)N2O4

NO2.N2O4,

N2O4

N2O4(g) ∆ 2 NO2(g)▲

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 644

15-7 Equilibrium Calculations: Some Illustrative Examples 645

Example 15-10 is somewhat more involved than Example 15-9. We are stillinterested in determining the equilibrium constant for a reaction, but we do nothave the same sort of information as in Example 15-9. We are given the initialconcentrations of all the reactants and products, but the equilibrium concentra-tion of only one substance. This case requires a little algebra and some carefulbookkeeping. We will introduce a tabular system, sometimes called an ICE table,for keeping track of changing concentrations of reactants and products. The tablecontains the initial, change in, and equilibrium concentration of each species. It is ahelpful device that we will use throughout the next three chapters.

E X A M P L E 1 5 - 1 0Determining a Value of from Initial and Equilibrium Amounts of Substances:Relating and The equilibrium condition for and is im-portant in sulfuric acid production. When a 0.0200-mol sample of is introducedinto an evacuated 1.52-L vessel at 900 K, 0.0142 mol is present at equilibrium.What is the value of for the dissociation of at 900 K?

SolutionLet’s first determine and then convert to by using equation (15.16). In the ICEtable on page 646, the key term leading to the other data is the change in amount of

In progressing from 0.0200 mol to 0.0142 mol 0.0058 mol is disso-ciated. The negative sign indicates that this amount of is consumedin establishing equilibrium. In the row labeled “changes,” the changes in amounts of

and must be related to the change in amount of For this, we use the sto-ichiometric coefficients from the balanced equation: 2, 2, and 1. That is, two moles of

and one mole of are produced for every two moles of that dissociate.SO3O2SO2

SO3.O2SO2

SO3(-0.0058 mol)SO3SO3,SO3SO3 :

KpKc

2 SO3(g) ∆ 2 SO2(g) + O2(g) Kp = ?

SO3(g)Kp

SO3

SO3

SO3(g)SO2(g), O2(g),KpKc

Kp

Convert the mass of to moles.N2O4 mol N2O4 = 7.64 g N2O4 *1 mol N2O4

92.01 g N2O4= 8.303 * 10-2 mol

Convert moles of to mol>L.N2O4 [N2O4] =8.303 * 102 mol

3.00 L= 0.0277 M

Convert the mass of to moles.NO2 mol NO2 = 1.56 g NO2 *1 mol NO2

46.01 g NO2= 3.391 * 10-2 mol

Convert moles of to mol>L.NO2 [NO2] =3.391 * 10-2

3.00 L= 0.0113 M

Write the equilibrium constant expression, sub-stitute the equilibrium concentrations, and solvefor Kc .

Kc =[NO2]2

[N2O4]=

(0.0113)2

(0.0277)= 4.61 * 10-3

Practice Example A: Equilibrium is established in a 3.00-L flask at 1405 K for the reactionAt equilibrium, there is 0.11 mol 0.22 mol and 2.78 mol What is

the value of for this reaction?

Practice Example B: Equilibrium is established at 25 °C in the reaction If in a 2.26-L flask, how many grams of are also present?

Comments

1. The quantities required in an equilibrium constant expression, are equilibrium concentrations in moles perliter, not simply equilibrium amounts in moles or masses in grams. You will find it helpful to organize all theequilibrium data and carefully label each item.

Kc ,

N2O4[NO2] = 0.0236 MN2O4(g) ∆ 2 NO2(g), Kc = 4.61 * 10-3.

Kc

H2S(g).H2(g),S2(g),2 H2S(g) ∆ 2 H2(g) + S2(g).

Solution

Silverstein, Todd P.“Graphing Calculator

Strategies for Solving ChemicalEquilibrium Problems.” J. Chem.Educ. 2000: 77, 1120.(September 2000).

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 645

E X A M P L E 1 5 - 1 1Determining Equilibrium Partial and Total Pressures from a Value of Ammonium hydrogen sulfide,

used as a photographic developer, is unstable and dissociates at room temperature.

A sample of is introduced into an evacuated flask at 25 °C. What is the total gas pressure at equilibrium?

Solution

NH4HS(s)

NH4HS(s) ∆ NH3(g) + H2S(g) Kp = 0.108 at 25 °C

NH4HS(s),Kp .


The reaction:

initial amounts: 0.0200 mol 0.00 mol 0.00 molchanges:equil amounts: 0.0142 mol 0.0058 mol 0.0029 mol

equil concns:

[O2] = 1.9 * 10-3 M [SO2] = 3.8 * 10-3 M; [SO3] = 9.34 * 10-3 M;

[O2] =0.0029 mol

1.52 L [SO2] =

0.0058 mol1.52 L

; [SO3] =0.0142 mol

1.52 L;

+0.0029 mol+0.0058 mol-0.0058 mol

O2(g)�2 SO2(g)∆2 SO3(g)

Write out the equilibrium constant expression interms of pressures. Kp = (PNH3)(PH2S) = 0.108

Practice Example A: A 5.00-L evacuated flask is filled with 1.86 mol NOBr. Atequilibrium at 25 °C, there is 0.082 mol of present. Determine and for thereaction

Practice Example B: 0.100 mol and 0.100 mol are introduced into anevacuated 1.52-L flask at 900 K. When equilibrium is reached, the amount of found is 0.0916 mol. Use these data to determine for the reaction

Comments

2. The chemical equation for a reversible reaction serves both to establish theform of the equilibrium constant expression and to provide the conversionfactors (stoichiometric factors) to relate the equilibrium quantity of onespecies to equilibrium quantities of the others.

3. For equilibria involving gases, you can use either or In general, if thedata given involve amounts of substances and volumes, it is easier to workwith If data are given as partial pressures, then work with Whetherworking with or or the relationship between them, you must alwaysbase these expressions on the given chemical equation, not on equationsyou may have used in other situations.

The methods used in Examples 15-9 and 15-10 are summarized in Figure15-9. Example 15-11 demonstrates that we can often determine several piecesof useful information about an equilibrium system from just the equilibriumconstant and the reaction equation.

KpKc

Kp.Kc .

Kp.Kc

2 SO3(g) ∆ 2 SO2(g) + O2(g).Kp

SO3

O2SO2

2 NOBr(g) ∆ 2 NO(g) + Br2(g).KpKcBr2

= 3.1 * 10-4 10.0821 * 90021 = 2.3 * 10-2

Kp = Kc1RT2¢ngas = 3.1 * 10-4 10.0821 * 900212+12-2

Kc =[SO2]2[O2]

[SO3]2 =13.8 * 10-32211.9 * 10-32

19.34 * 10-322 = 3.1 * 10-4

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 646


Yes No

Establish all the equilibriumconcentrations or partial

pressures

Relate changes in initial amounts,concentrations, or partial pressures to

a known equilibrium amount,concentration, or partial pressure

Substitute concentrationsinto Kc or partial pressures

into Kp expressions

Are the experimental dataexclusively equilibrium data?

Convert the datato concentrations or

partial pressures

FIGURE 15-9Determining or from experimental dataKpKc

▲

for this reaction is just the product of the equi-librium partial pressures of and each stated in atmospheres. (There is no term for

because it is a solid.) Because these gasesare produced in equimolar amounts, PNH3 = PH2S.NH4HS

H2S(g),NH3(g)Kp

Kp = (PNH3)(PH2S) = (PNH3)(PNH3) = (PNH3)2 = 0.108

Find (Note that the unit atm appearsbecause in the equilibrium expression the refer-ence pressure was implicitly included.)P°

PNH3.

PNH3 = 20.108 = 0.329 atm PH2S = PNH3 = 0.329 atm

The total pressure is Ptot = PNH3 + PH2S = 0.329 atm + 0.329 atm = 0.658 atm

Practice Example A: Sodium hydrogen carbonate (baking soda) decomposes at elevated temperatures and is oneof the sources of when this compound is used in baking.

What is the partial pressure of when this equilibrium is established starting with

Practice Example B: If enough additional is added to the flask in Example 15-11 to raise its partial pres-sure to 0.500 atm at equilibrium, what will be the total gas pressure when equilibrium is reestablished?

Comments

4. When using expressions, look for relationships among partial pressures of the reactants. If you need torelate the total pressure to the partial pressures of the reactants, you should be able to do this with some equa-tions presented in Chapter 6 (for example, equations 6.15, 6.16, and 6.17).

Kp

NH3(g)

NaHCO3(s)?CO2(g)

2 NaHCO3(s) ∆ Na2CO3(s) + H2O(g) + CO2(g) Kp = 0.231 at 100 °C

CO2(g)

Example 15-12 brings back the ICE format, but with a twist. This time theknown values include the equilibrium constant and an initial amount of the re-actant, but no information is given about the equilibrium amount of the reactantor the product. That means that we do not know how much the initial value willchange. We show this by using an “x” in that part of the table. The setup will bequite algebraic; in fact, we must use the quadratic formula to obtain a solution.

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 647


Most students can solve aquadratic equation, but fewof us can solve polynomials

greater than a quadratic. Suggest tothe students that if they get anequation that is a cubic or higherdegree equation, it is likely that itwill simplify by using anapproximation.

For a discussion ofquadratic equations, see Appendix A.

▲

E X A M P L E 1 5 - 1 2Calculating Equilibrium Concentrations from Initial Conditions. A 0.0240-molsample of is allowed to come to equilibrium with in a 0.372-L flaskat 25 °C. Calculate the amount of present at equilibrium (Fig. 15-10).

N2O4(g) ∆ 2 NO2(g) Kc = 4.61 * 103 at 25 °C

N2O4

NO2(g)N2O4(g)

The symbol signifiesthat there are two possibleroots. In this problem, x mustbe a positive quantity smallerthan 0.0240.

;

▲

� N2O4

� NO2

(a) (b)

FIGURE 15-10Equilibrium in the reaction

at 25 °C—Example 15-12 illustratedEach “molecule” illustrated represents 0.001 mol. (a) Initially, the bulb contains0.024 mol represented by 24 “molecules.” (b) At equilibrium, some“molecules” of have dissociated to The 21 “molecules” of and 6of correspond to 0.021 mol and 0.006 mol at equilibrium.NO2N2O4NO2

N2O4NO2 .N2O4

N2O4 ,

N2O4(g) ∆ 2 NO2(g)

▲

SolutionWe need to determine the amount of that dissociates to establish equilibrium.For the first time, we introduce an algebraic unknown, x. Suppose we let number of moles of that dissociate. In the following ICE table, we enter thevalue into the row labeled “changes.” The amount of produced is be-cause the stoichiometric coefficient of is 2 and that of is 1.The reaction:initial amounts: 0.0240 mol 0.00 molchanges:equil amounts: 2x molequil concns:

=-4.28 * 10-4 ; 411.83 * 10-72 + 4.12 * 10-5

2

x =-4.28 * 10-4 ; 414.28 * 10-422 + 4 * 1.03 * 10-5

2

x2 + 14.28 * 10-42x - 1.03 * 10-5 = 0

4x2 = 4.12 * 10-5 - 11.71 * 10-32x

Kc =[NO2]2

[N2O4]=

a 2x0.372

b2

a0.0240 - x0.372

b=

4x2

0.37210.0240 - x2 = 4.61 * 10-3

[NO2] = 2x mol>0.372 L[N2O4] = 10.0240 - x mol2>0.372 L10.0240 - x2 mol

+2x mol-x mol

2 NO2(g)∆N2O4(g)N2O4NO2

+2xNO2-xN2O4

x = theN2O4

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The amount of at equilibrium is .

Practice Example A: If 0.150 mol and 0.200 mol are introduced intoa 15.0-L flask at 445 °C and allowed to come to equilibrium, how many moles ofHI(g) will be present?

Practice Example B: Suppose the equilibrium mixture of Example 15-12 istransferred to a 10.0-L flask. (a) Will the equilibrium amount of increase ordecrease? Explain. (b) Calculate the number of moles of in the new equilibri-um condition.

Comments

5. When you need to introduce an algebraic unknown, x, into an equilibriumcalculation, follow these steps.• Introduce x into the ICE setup in the row labeled “changes.”• Decide which change to label as x, that is, the amount of a reactant con-

sumed or of a product formed. Usually, we base this on the speciesthat has the smallest stoichiometric coefficient in the balanced chemicalequation.

• Use stoichiometric factors to relate the other changes to x (that is, 2x,3x, ).

• Consider that equilibrium (Ifyou have assigned the correct signs to the changes, equilibrium amountswill also be correct.)

• After substitutions have been made into the equilibrium constant expres-sion, the equation will often be a quadratic equation in x, which youcan solve by the quadratic formula. Occasionally you may encounter ahigher-degree equation. Appendix A-3 outlines a straightforward methodof dealing with these.

Our final example is similar to the previous one, but with this slight com-plication: Initially, we don’t know whether a net change occurs to the right orto the left to establish equilibrium. We can find out, though, by using the reac-tion quotient, and proceeding in the manner suggested in Figure 15-11.Also, because the reactants and products are in solution, we can work exclu-sively with concentrations in formulating the expression.

E X A M P L E 1 5 - 1 3Using the Reaction Quotient, in an Equilibrium Calculation. Solid silver isadded to a solution with these initial concentrations:

and The following reversible reaction occurs.

What are the ion concentrations when equilibrium is established?

Ag+(aq) + Fe2+(aq) ∆ Ag(s) + Fe3+(aq) Kc = 2.98

[Fe3+] = 0.300 M.[Fe2+] = 0.100 M,[Ag+] = 0.200 M,

Qc ,

Kc

Qc ,

amounts = intial amounts + “changes.–Á

N2O4

N2O4

H2(g) + I2(g) ∆ 2 HI(g) Kc = 50.2 at 445 °C

I2(g)H2(g)

0.0210 mol N2O4

10.0240 - x2 = 10.0240 - 0.00302 =N2O4

= 3.00 * 10-3 mol N2O4

=-4.28 * 10-4 + 6.43 * 10-3

2=

6.00 * 10-3

2

=-4.28 * 10-4 ; 6.43 * 10-3

2

x =-4.28 * 10-4 ; 44.14 * 10-5

2

Determine the direction ofnet change by comparing

Q and K

Let x � change in amount, concentration, or partial

pressure of one reactant orproduct to reach equilibrium

Relate changes in amounts,concentrations, or partial

pressures of other reactantsor products to the chosen x

Express equilibriumamounts, concentrations,

partial pressuresin terms of x

Substitute equilibriumconcentrations or

partial pressures into Kc or Kpexpression; solve for x;substitute the value of x

into any equation in whichx appeared to find the

desired quantities

Tabulate these data in an IC

E table

FIGURE 15-11Determining equilibriumconcentrations and partialpressures

▲

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 649


After making anapproximation, it is alwaysnecessary to check its

validity by substituting theapproximate value back into theequation and verifying the 5% rule.

SolutionBecause all reactants and products are present initially, we need to use the reactionquotient to determine the direction in which a net change occurs.

Because (15.0) is larger than (2.98), a net change must occur in the direction ofthe reverse reaction, to the left. Let’s define x as the change in molarity of Be-cause the net change occurs to the left, we designate the changes for the species onthe left side of the equation as positive and those on the right side as negative. Therelevant data are tabulated as follows.The reaction:Initial concns: 0.200 M 0.100 M 0.300 Mchanges:equil concns:

This equation, which is solved in Appendix A-3, is a quadratic equation for whichthe acceptable root is To obtain the equilibrium concentrations, we substi-tute this value of x into the terms shown in the table of data.

CHECK: If we have done the calculation correctly, we should obtain a value veryclose to that given for when we substitute the calculated equilibrium concentra-tions into the reaction quotient, We do.

Practice Example A: Excess Ag(s) is added to 1.20 M Given that

what are the equilibrium concentrations of the species in solution?

Practice Example B: A solution is prepared with and The following reaction occurs.

What are the ion concentrations when equilibrium is established?

[Hint: The algebra can be greatly simplified by extracting the square root of bothsides of an equation at the appropriate point.]

Comments

6. It is sometimes helpful to compare the reaction quotient Q to the equilibri-um constant K to determine the direction of the net change.

7. In many equilibrium calculations—often those in aqueous solutions—youcan work with molarities directly, without having to work with moles of re-actants and solution volumes.

8. Where possible, check your calculation, for instance, by substitutingcalculated equilibrium concentrations into the reaction quotient Q to see if itsnumerical value is close to that of K.

V3+(aq) + Cr2+(aq) ∆ V2+(aq) + Cr3+(aq) Kc = 7.2 * 102

[V2+] = [Cr3+] = 0.150 M.[V3+] = [Cr2+] = 0.0100 M

Ag+(aq) + Fe2+(aq) ∆ Ag(s) + Fe3+(aq) Kc = 2.98

Fe3+(aq).

Qc =[Fe3+]

[Ag+][Fe2+]=

(0.19)(0.31)(0.21)

= 2.9 (Kc = 2.98)

Qc .Kc

[Fe3+]equil = 0.300 - 0.11 = 0.19 M

[Fe2+]equil = 0.100 + 0.11 = 0.21 M

[Ag+]equil = 0.200 + 0.11 = 0.31 M

x = 0.11.

Kc =[Fe3+]

[Ag+][Fe2+]=

(0.300 - x)(0.200 + x)(0.100 + x)

= 2.98

10.300 - x2 M10.100 + x2 M10.200 + x2 M-x M+x M+x M

Ag(s) � Fe3�(aq)∆Fe2+(aq)�Ag+(aq)

Fe3+.KcQc

Qc =[Fe3+]

[Ag+][Fe2+]=

0.300(0.200)(0.100)

= 15.0

Qc

KEEP IN MINDthat if one or more of thesubstances appearing in (or ) is not present initially,a net change must occur toproduce some of thesubstance(s). You need tocompare (or ) with (or ) only if all thesubstances appearing inthese expressions arepresent initially.

Kp

KcQpQc

Qp

Qc

▲

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 650

Integrative Example 651

Summary15-1 Dynamic Equilibrium—Equilibrium is the conditionin which the forward and reverse reaction rates of re-versible processes are equal. Chemical and physicalprocesses in equilibrium are dynamic by nature.

15-2 The Equilibrium Constant Expression—This condi-tion of dynamic equilibrium is described through anequilibrium constant expression. The form of the equilib-rium constant expression is established from the balancedchemical equation using activities to express the “effec-tive” concentrations (equation 15.7). The numerical valueobtained from the equilibrium constant expression is re-ferred to as the equilibrium constant, K. Equilibrium con-stants are unitless.

15-3 Relationships Involving Equilibrium Constants—When the equation for a reversible reaction is written inthe reverse order, the equilibrium constant expression andthe value of K are both inverted from their original form.When two or more reactions are coupled together, theequilibrium constant for the overall reaction is the productof the K values of the individual reactions. The equilibri-um constant of a reaction can have different values de-pending on the reference state used. For aconcentration reference state is used, while for a pres-sure reference state is used. The relationship between and is given by equation (15.16).

15-4 The Magnitude of an Equilibrium Constant—Themagnitude of the equilibrium constant can be used todetermine the outcome of a reaction. For large values of Kthe reaction goes to completion, with all reactants con-verted to products. A very small equilibrium constant,for example, a large negative power of ten, indicates thatpractically none of the reactants have been converted toproducts. Finally, equilibrium constants of an intermedi-

Kp

Kc

Kp ,Kc

ate value, for example, between and indicatethat some of the reactants have been converted toproducts.

15-5 The Reaction Quotient, Q: Predicting the Directionof Net Change—The reaction quotient, Q (equation15.19), has the same form as the equilibrium constantexpression; however, its numerical value is determinedusing the initial reaction activities. A comparison of thereaction quotient with the equilibrium constant makes itpossible to predict the direction of net change leading toequilibrium (Fig. 15-5). If the forward reaction isfavored, meaning that when equilibrium is established theamounts of products will have increased and the amountsof reactants will have decreased. If the reverse re-action is favored until equilibrium is established. If neither the forward nor reverse reaction is favored. Theinitial conditions are in fact equilibrium conditions.

15-6 Altering Equilibrium Conditions: Le Châtelier’sPrinciple—Le Châtelier’s principle is used to make quali-tative predictions of the effects of different variables on anequilibrium condition. This principle describes how anequilibrium condition is modified, or “shifts,” in responseto the addition or removal of reactants or changes in reac-tion volume, external pressure, or temperature. Catalysts,by speeding up the forward and reverse reactions equally,have no effect on an equilibrium condition.

15-7 Equilibrium Calculations: Some IllustrativeExamples—For quantitative equilibrium calculations, afew basic principles and algebraic techniques arerequired. A useful method employs a tabular system,called an ICE table, for keeping track of the initial concen-trations of the reactants and products, changes in theseconcentrations, and the equilibrium concentrations.

Q = K,Q 7 K,

Q 6 K,

1010,10-10

Integrative ExampleIn the manufacture of ammonia, the chief source of hydrogen gas is the following reaction for the reforming of methaneat high temperatures.

(15.23)

The following data are also given.

(a)(b)

At 1000 K, 1.00 mol each of and are allowed to come to equilibrium in a 10.0-L vessel. Calculate the number ofmoles of present at equilibrium. Would the yield of increase if the temperature were raised above 1000 K?

StrategyFirst, we should assemble the data needed to solve this problem. The amounts of substances and a reaction volume aregiven, so we should be able to work with a expression. However, because the value for the reaction of interest is notgiven, we will have to derive this value by combining the two equations for which data are given. This will yield valuesof both and for the reaction of interest.

To calculate the number of moles of at equilibrium we can use the ICE method, and to assess the effect of tempera-ture on the equilibrium yield of we can apply Le Châtelier’s principle.H2

H2

¢HKc

KcKc

H2H2

H2OCH4

¢H° = -230 kJ; Kc = 190 at 1000 KCO(g) + 3 H2(g) ∆ H2O(g) + CH4(g)¢H° = -40 kJ; Kc = 1.4 at 1000 KCO(g) + H2O(g) ∆ CO2(g) + H2(g)

CH4(g) + 2 H2O(g) ∆ CO2(g) + 4 H2(g)

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 651

FOCUS ON

Nitrogen is an element essential both in living things andin industry. The atmosphere contains a vast pool of ele-mental nitrogen, but most organisms cannot use nitrogenin that form. Instead, they require certain nitrogen-containing compounds. The conversion of elemental atmo-spheric nitrogen to nitrogen compounds is called nitrogenfixation, and in nature, this process is carried out only bycertain types of bacteria. Nitrogen fixation is part of the ni-trogen cycle, which is the path of nitrogen through the

THE NITROGEN CYCLE AND THE SYNTHESIS OFNITROGEN COMPOUNDS

environment and a variety of living organisms and backinto the environment. A simplified version of the nitrogencycle is depicted in Figure 15-12, which shows that elemen-tal nitrogen from the atmosphere is fixed, becomes part ofplants, animals, and other organisms, and is then returnedto the atmosphere. The numbers in parentheses refer to thevarious parts of the cycle described below.

A few leguminous plants, such as beans, peas, and al-falfa, have bacteria residing on their roots that convert ele-mental nitrogen from the air into compounds used tomake plant protein (1). Plants also convert nitrates in thesoil to proteins (5). Animals obtain their protein-buildingnitrogen by feeding on plants and/or other animals (2).The decay of plant and animal proteins produces ammo-nia (3). Through a series of bacterial actions, ammonia isconverted to nitrites and nitrates (4). Denitrifying bacteriadecompose nitrites and nitrates, returning and tothe atmosphere, thus completing the cycle (6). Some at-mospheric nitrogen is converted to nitrates during electri-cal storms (1a).

Reversible chemical reactions play an important role inthe nitrogen cycle. One example is the reaction of and to form NO(g).

The reaction does not occur to any measurable extent at298 K, but at 1800 K the situation is a little different. Anequilibrium mixture of and at 1800 KO2(g)N2(g)

= 1.3 * 10-4 at 1800 K

Kp = 4.7 * 10-31 at 298 K

N2(g) + O2(g) ∆ 2 NO(g)

O2(g)N2(g)

N2N2O

Nitrites

Nitrates

(6) Bacteria

(4) Bacteria

(6) Bacteria

(4) Bacteria

(3) Decay

(3) Decay

(1) Bacteria

(1a) Lightning

(5)

(2)

Atmosphericnitrogen

Ammonia

Animalprotein

Plantprotein

FIGURE 15-12The nitrogen cycle▲

Use of liquid ammonia as a fertilizer by direct injection intothe soil.▲

652

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 652

contains about 1 or 2% NO(g). NO(g) is introduced intothe nitrogen cycle through this high-temperature reactionoccurring naturally in lightning flashes during thunder-storms and artificially in high-temperature combustionprocesses, as in internal combustion engines. Further reac-tions account for the formation of in rainstormsand the introduction of nitrates into the soil.

Ammonia is an important industrial chemical. Its syn-thesis is achieved by the following reaction.

Since ammonia can be converted by soil bacteria to ni-trites and nitrates that can be used by plants, one of theprincipal uses of artificial is as a fertilizer that is in-jected directly into the soil. is also used in the produc-tion of other nitrogen compounds, such as urea, hydrazine,ammonium sulfate, ammonium nitrate, ammonium dihy-drogen phosphate, and ammonium hydrogen phosphate.Several of these compounds are used as fertilizers, andothers are used in the manufacture of explosives, pharma-ceuticals, and plastics. So much nitrogen is now beingfixed artificially that fixed nitrogen is accumulating in theenvironment somewhat faster than it is being returned tothe atmosphere. This causes environmental problems suchas the buildup of nitrates in groundwater.

Let us look at the commercial synthesis of ammonia,called the Haber–Bosch process, from the standpoint ofchemical equilibrium and kinetics. In the synthesis reac-tion, two moles of gaseous product are formed for everyfour moles of gaseous reactants. Carrying out the reactionat high pressure favors the production of Becausethe forward reaction is exothermic, the equilibrium yield

NH3.

NH3

NH3

¢H° = -92.22 kJ Kp = 6.2 * 105 at 298 K

N2(g) + 3 H2(g) ∆ 2 NH3(g)

3 NO2(g) + H2O(l) ¡ 2 HNO3(aq) + NO(g) 2 NO(g) + O2(g) ¡ 2 NO2(g)

HNO3(aq)

Am

mon

ia a

t equ

ilibr

ium

, %(R

atio

H2/

N2

� 3

:1)

Tem

pera

ture

, �C

100

100

200

300

100

400

500

600

700

300 500 700 900

“Theoretical”conditions

Pressure, atm

Commercialoperating conditions

20

30

40

50

60

70

80

90

10

FIGURE 15-13Equilibrium conversion of and to as a function of temperature and pressure

NH3(g)H2(g)N2(g)▲

of is greatest at low temperatures. Thus, the opti-mum conditions for the equilibrium production of isgreatest at low temperatures. Thus, the optimum condi-tions for the equilibrium production of are high pres-sures and low temperatures. However, these “optimum”conditions do not take into account the rate of reaction.

Although the equilibrium production of is fa-vored at low temperatures, equilibrium is achieved soslowly that the synthesis is not feasible at those tempera-tures. One way to speed up the reaction is to raise the tem-perature, even though doing so decreases the equilibriumconcentration of Another way is to use a catalyst.The usual industrial operating conditions are a tempera-ture of about 550 °C, pressures ranging from 150 to 350atm, and a catalyst—usually iron in the presence of MgO, CaO, and Figure 15-13 suggests the dramaticdifference between the theoretical optimum conditionsand the actual operating conditions.

Another essential feature of the Haber–Bosch methodis to remove as it forms. This is done by liquefyingthe In fact, the mixture need not be brought toequilibrium at all, and nearly 100% conversion of and

to is achieved.NH3H2

N2

NH3(g).NH3

K2O.Al2O3,

NH3.

NH3

NH3

NH3

NH3

During electrical storms, and combine to producesmall quantities of NO(g) in a reversible chemical reaction.

O2(g)N2(g)▲

653

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SolutionWe combine equations (a) and (b) to obtain the data needed in this problem.

(a)(b)

Overall:

Next we set up an ICE table in which x represents the number of moles of consumed in reaching equilibrium.

The reaction:initial amounts: 1.00 mol 1.00 mol 0.00 mol 0.00 molchanges: mol mol x mol molequil amounts: mol mol x mol molequil concns, M: 4x>10.0x>10.011.00 - 2x2>10.011.00 - x2>10.0

4x11.00 - 2x211.00 - x24x-2x-x

4 H2(g)�CO2(g)∆2 H2O(g)�CH4(g)

CH4

CH4(g) + 2 H2O(g) ∆ CO2(g) + 4 H2(g) ¢H = 190 kJ Kc = 1.4>190 = 7.4 * 10-3

CH4(g) + H2O(g) ∆ CO(g) + 3 H2(g) ¢H = 230 kJ Kc = 1>190 CO(g) + H2O(g) ∆ CO2(g) + H2(g) ¢H = -40 kJ Kc = 1.4

Exercises

Writing Equilibrium Constant Expressions

1. Based on these descriptions, write a balanced equa-tion and the corresponding expression for each re-versible reaction.(a) Carbonyl fluoride, decomposes intogaseous carbon dioxide and gaseous carbon tetra-fluoride.(b) Copper metal displaces silver(I) ion from aqueoussolution, producing silver metal and an aqueous solu-tion of copper(II) ion.(c) Peroxodisulfate ion, oxidizes iron(II) ionto iron(III) ion in aqueous solution and is itself re-duced to sulfate ion.

S2O8

2-,

COF2(g),

Kc

2. Based on these descriptions, write a balanced equa-tion and the corresponding expression for each re-versible reaction.(a) Oxygen gas oxidizes gaseous ammonia togaseous nitrogen and water vapor.(b) Hydrogen gas reduces gaseous nitrogen dioxideto gaseous ammonia and water vapor.(c) Nitrogen gas reacts with the solid sodium carbon-ate and carbon to produce solid sodium cyanide andcarbon monoxide gas.

3. Write equilibrium constant expressions, for thereactions

Kc ,

Kp

Now we set up and make substitutions into theexpression.

Kc

=x14x24

10011.00 - x211.00 - 2x22 = 7.4 * 10-3

=1x>10.0214x>10.024

311.00 - x2/10.04311.00 - 2x2>10.042

Kc =[CO2][H2]4

[CH4][H2O]2

The above equation reduces to 256x5 = 0.74311.00 - x211.00 - 2x224and then to (15.24)256x5 - 0.74311.00 - x211.00 - 2x224 = 0

The solution to this equation is mol. The numberof moles of at equilibrium is mol.

Because the reaction is endothermic the forward reaction is favored at higher temperatures.The equilibrium yield of will increase if the tempera-ture is raised above 1000 K.

H2

1¢H = 190 kJ2,4x = 0.92H2

x = 0.23

AssessmentEquation (15.24) looks impossibly difficult to solve, but it is not. It can be solved for x rather simply by the method of suc-cessive approximations. This is done in Appendix A-3, equation (A.2). An important clue as to the possible range of val-ues for x can be found in the ICE table. Note that the equilibrium amount of is meaning that or else all of the would be consumed. This marks a good place to start the approximations.H2O(g)

x 6 0.50,1.00 - 2x,H2O(g)

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 654

Exercises 655

(a)(b)(c)

4. Write equilibrium constant expressions, for thereactions(a)

(b)

(c)

5. Write an equilibrium constant, for the formationfrom its gaseous elements of (a) 1 mol HF(g);(b) (c) (d)

6. Write an equilibrium constant, for the formationfrom its gaseous elements of (a) 1 mol NOCl(g);(b) (c) (d)

7. Determine values of from the values given.(a)

(b)

(c)

8. Determine the values of from the values given.(a)

(b)

(c)

9. The vapor pressure of water at 25 °C is 23.8 mmHg.Write for the vaporization of water, with pressuresin atmospheres. What is the value of for the vapor-ization process?

Kc

Kp

Kc = 5.27 * 10-8 at 973 K2 H2S(g) + CH4(g) ∆ 4 H2(g) + CS2(g)

Kc = 0.154 at 2000 K2 CH4(g) ∆ C2H2(g) + 3 H2(g)

Kc = 4.61 * 10-3 at 25 °CN2O4(g) ∆ 2 NO2(g)

KcKp

Kp = 0.429 at 713 KSb2S3(s) + 3 H2(g) ∆ 2 Sb(s) + 3 H2S(g)

Kp = 1.48 * 104 at 184 °C2 NO(g) + O2(g) ∆ 2 NO2(g)

Kp = 2.9 * 10-2 at 303 KSO2Cl2(g) ∆ SO2(g) + Cl2(g)

KpKc

NH4Cl(s).1 mol1 mol N2H4(g);2 mol ClNO2(g);

Kp,1 mol ClF3(l).2 mol N2O(g)2 mol NH3(g);

Kc ,Na2CO3(s) + CO2(g) + H2O(g)

2 NaHCO3(s) ∆

Ag2O(s) ∆ 2 Ag(s) +12

O2(g)

CS2(g) + 4 H2(g) ∆ CH4(g) + 2 H2S(g)

Kp,MgCO3(s) + 2 OH-(aq)

Mg(OH)2(s) + CO3

2-(aq) ∆Zn(s) + 2 Ag+(aq) ∆ Zn2+(aq) + 2 Ag(s)2 NO(g) + O2(g) ∆ 2 NO2(g) 10. If for the equilibrium established

between liquid benzene and its vapor at 25 °C, what isthe vapor pressure of at 25 °C, expressed in mil-limeters of mercury?

11. Determine for the reaction

from the following information (at 298 K).

12. Given the equilibrium constant values

Determine a value of for the reaction

13. Use the following data to estimate a value of at 1200 Kfor the reaction

14. Determine for the reaction given the following data at

298 K.

NO2(g) +12

Cl2(g) ∆ NO2Cl(g) Kp = 0.3

NOCl(g) +12

O2(g) ∆ NO2Cl(g) Kp = 1.1 * 102

12

N2(g) + O2(g) ∆ NO2(g) Kp = 1.0 * 10-9

Cl2(g) ∆ 2 NOCl(g),N2(g) + O2(g) +Kc

Kc = 1 * 108 C(graphite) +12

O2(g) ∆ CO(g)

CO2(g) + H2(g) ∆ CO(g) + H2O(g) Kc = 1.4 C(graphite) + CO2(g) ∆ 2 CO(g) Kc = 0.64

2 H2(g) + O2(g) ∆ 2 H2O(g)Kp

2 N2O(g) + 3 O2(g) ∆ 2 N2O4(g)Kc

12

N2(g) + O2(g) ∆ NO2(g) Kc = 4.1 * 10-9

N2O4(g) ∆ 2 NO2(g) Kc = 4.6 * 10-3

N2(g) +12

O2(g) ∆ N2O(g) Kc = 2.7 * 10-18

NO(g) +12

Br2(g) ∆ NOBr(g) Kc = 1.4

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

12

N2(g) +12

O2(g) +12

Br2(g) ∆ NOBr(g)

Kc

C6H6

Kc = 5.12 * 10-3

Experimental Determination of Equilibrium Constants

15. is introduced into a 250.0-mLflask, and equilibrium is established at 284 °C:

The quantity of present at equilibrium is found to be What is the value of for the dissociation reactionat 284 °C?

16. A mixture of and in a 0.500-L flaskcomes to equilibrium at

The equilibrium amount of found isDetermine the value of at 1670 K.

17. The two common chlorides of phosphorus, andboth important in the production of other phos-PCl5 ,

PCl3

Kp8.00 * 10-6 mol.S2(g)2 H2S(g).

1670 K: 2 H2(g) + S2(g) ∆1.06 g H2S1.00 g H2

Kc

9.65 * 10-4 mol.Cl2(g)PCl5(g) ∆ PCl3(g) + Cl2(g).

1.00 * 10-3 mol PCl5 phorus compounds, coexist in equilibrium throughthe reaction

At 250 °C, an equilibrium mixture in a 2.50-L flaskcontains 0.105 g 0.220 g and What are the values of (a) and (b) for this reac-tion at 250 °C?

18. A 0.682-g sample of ICl(g) is placed in a 625-mL reactionvessel at 682 K. When equilibrium is reached betweenthe ICl(g) and and formed by its dissocia-tion, is present. What is for this reaction?Kc0.0383 g I2

Cl2(g)I2(g)

KpKc

2.12 g Cl2 .PCl3 ,PCl5 ,

PCl3(g) + Cl2(g) ∆ PCl5(g)

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 655


Equilibrium Relationships

19. Equilibrium is established at 1000 K, where for the reaction Theequilibrium amount of in a 0.185-L flask is0.00247 mol. What is the ratio of to in thisequilibrium mixture?

20. For the dissociation of at about 1200 °C,What volume flask

should we use if we want 0.37 mol I to be present forevery at equilibrium?

21. In the Ostwald process for oxidizing ammonia, a vari-ety of products is possible— NO, and —depending on the conditions. One possibility is

For the decomposition of at 700 K,

(a) Write a chemical equation for the oxidation ofto

(b) Determine for the chemical equation you havewritten.

Kp

NO2(g).NH3(g)

Kp = 0.524NO2(g) ∆ NO(g) +12

O2(g)

NO2

Kp = 2.11 * 1019 at 700 K

NH3(g) +54

O2(g) ∆ NO(g) +32

H2O(g)

NO2N2, N2O,

1.00 mol I2

Kc = 1.1 * 10-2.I2(g) ∆ 2 I(g),I2(g)

[SO3][SO2]O2(g)

2 SO2(g) + O2(g) ∆ 2 SO3(g).Kc = 281 22. At 2000 K, for the reaction

If a 1.00-L equilibrium mixture at2000 K contains 0.10 mol each of and (a) what is the mole fraction of present?(b) Is the conversion of to favored athigh or low pressures?(c) If the equilibrium mixture at 2000 K is transferredfrom a 1.00-L flask to a 2.00-L flask, will the numberof moles of increase, decrease, or remainunchanged?

23. An equilibrium mixture at 1000 K contains0.224 mol CO, and

(a) Show that for this reaction, is independent ofthe reaction volume, V.(b) Determine the value of and

24. For the reaction at 600 K. Explain which of the fol-

lowing situations might be found at equilibrium:(a) (b)

(c)(d) PH2>PCO.PCO2>PH2O =

(PCO2)(PH2) = (PCO)(PH2O);PCO2>PCO ;PH2>PH2O =PCO = PH2O = PCO2 = PH2 ;

H2(g), Kp = 23.2+CO(g) + H2O(g) ∆ CO2(g)

Kp.Kc

Kc

CO2(g) + H2(g) ∆ CO(g) + H2O(g)

0.224 mol H2O.0.276 mol CO2,0.276 mol H2,

C2H2(g)

C2H2(g)CH4(g)C2H2(g)

H2(g),CH4(g)C2H2(g) + 3 H2(g).

2 CH4(g) ∆Kc = 0.154

Direction and Extent of Chemical Change

25. Can a mixture of andbe maintained indefinitely in a 7.2-L flask

at a temperature at which in this reaction?Explain.

26. Is a mixture of 0.0205 mol and 0.750 molin a 5.25-L flask at 25 °C at equilibrium? If

not, in which direction will the reaction proceed—toward products or reactants?

27. In the reaction 0.455mol 0.183 mol and 0.568 mol are intro-duced simultaneously into a 1.90-L vessel at 1000 K.(a) Is this mixture at equilibrium?(b) If not, in which direction will a net change occur?

28. In the reaction Equal masses of each reac-

tant and product are brought together in a reactionvessel at 588 K.(a) Can this mixture be at equilibrium?(b) If not, in which direction will a net change occur?

29. A mixture consisting of 0.150 mol and 0.150 mol is brought to equilibrium at 445 °C in a 3.25-L flask.What are the equilibrium amounts of and HI?

30. Starting with 0.280 mol and 0.160 mol howmany moles of and are present whenequilibrium is established at 248 °C in a 2.50-L flask?

Kc = 2.5 * 10-2 at 248 °C

SbCl5(g) ∆ SbCl3(g) + Cl2(g)

Cl2SbCl5 , SbCl3 ,Cl2 ,SbCl3

H2(g) + I2(g) ∆ 2 HI (g) Kc = 50.2 at 445 °C

H2, I2 ,

I2H2

K = 31.4 at 588 K.H2(g),CO(g) + H2O(g) ∆ CO2(g) +

SO3O2,SO2,2 SO2(g) + O2(g) ∆ 2 SO3(g),

Kc = 4.61 * 10-3 at 25 °CN2O4(g) ∆ 2 NO2(g)

N2O4(g)NO2(g)

2 SO2(g) + O2(g) ∆ 2 SO3(g)

Kc = 1001.8 mol SO3

2.2 mol O2, 3.6 mol SO2, 31. Starting with 0.3500 mol CO(g) and 0.05500 molin a 3.050-L flask at 668 K, how many moles

of will be present at equilibrium?

32. 1.00 g each of CO, and are sealed in a 1.41-Lvessel and brought to equilibrium at 600 K. Howmany grams of will be present in the equilibriummixture?

33. Equilibrium is established in a 2.50-L flask at 250 °Cfor the reaction

How many moles of and are present atequilibrium, if(a) 0.550 mol each of and are initially intro-duced into the flask?(b) 0.610 mol alone is introduced into the flask?

34. For the following reaction, at 1000 °C.

If a 5.00-L mixture contains 0.145 mol 0.262 moland 0.074 mol at a temperature of 1000 °C,

(a) Will the mixture be at equilibrium?(b) If the gases are not at equilibrium, in what direc-tion will a net change occur?(c) How many moles of each gas will be present atequilibrium?

CF4CO2,COF2,

2 COF2(g) ∆ CO2(g) + CF4(g)

Kc = 2.00PCl5

PCl3PCl5

Cl2PCl5 , PCl3 ,

PCl5(g) ∆ PCl3(g) + Cl2(g) Kc = 3.8 * 10-2

Kc = 23.2CO(g) + H2O(g) ∆ CO2(g) + H2(g)

CO2

H2H2O,

Kc = 1.2 * 103 at 668 K

CO(g) + Cl2(g) ∆ COCl2(g)

Cl2(g)COCl2(g)

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Exercises 657

35. In the following reaction,

A reaction is allowed to occur in a mixture of 17.2 g23.8 g 48.6 g

and 71.2 g (a) In what direction will a net change occur?(b) How many grams of each substance will be pre-sent at equilibrium?

36. The equilibrium mixture in the flask onthe left in the figure is allowed to expand into theevacuated flask on the right. What is the compositionof the gaseous mixture when equilibrium is reestab-lished in the system consisting of the two flasks?

37. Formamide, used in the manufacture of pharmaceuti-cals, dyes, and agricultural chemicals, decomposes athigh temperatures.

If 0.186 mol dissociates in a 2.16-L flask at400 K, what will be the total pressure at equilibrium?

38. A mixture of 1.00 mol and 1.00 molis introduced into a 2.50-L flask in which

the partial pressure of is 2.10 atm and that ofis 715 mmHg. When equilibrium is estab-

lished at 100 °C, will the partial pressures of and be greater or less than their initial partialpressures? Explain.

39. Cadmium metal is added to 0.350 L of an aqueous so-lution in which What are the concen-trations of the different ionic species at equilibrium?What is the minimum mass of cadmium metal re-quired to establish this equilibrium?

40. Lead metal is added to What areand when equilibrium is estab-

lished in the reaction?

Kc = 3.2 * 10-10

Pb(s) + 2 Cr3+(aq) ∆ Pb2+(aq) + 2 Cr2+(aq)

[Cr3+][Pb2+], [Cr2+],0.100 M Cr3+(aq).

Kc = 0.288

2 Cr3+(aq) + Cd(s) ∆ 2 Cr2+(aq) + Cd2+(aq)

[Cr3+] = 1.00 M.

Kp = 0.23 at 100 °C

2 NaHCO3(s) ∆ Na2CO3(s) + CO2(g) + H2O(g)

H2O(g)CO2(g)

H2O(g)CO2

Na2CO3(s)NaHCO3(s)

HCONH2(g)

Kc = 4.84 at 400 K

HCONH2(g) ∆ NH3(g) + CO(g)

0.971 mol N2O40.0580 mol NO2

0.750 L25 �C

2.25 L25 �C

Kc = 4.61 * 10-3 at 25 °CN2O4(g) ∆ 2 NO2(g)

N2O4–NO2

H2O.CH3COOC2H5,CH3COOH,C2H5OH,

C2H5OH + CH3COOH ∆ CH3COOC2H5 + H2O

Kc = 4.0. 41. One sketch below represents an initial nonequilibri-um mixture in the reversible reaction

Which of the other three sketches best represents anequilibrium mixture? Explain.

42. One sketch below represents an initial nonequilibri-um mixture in the reversible reaction

Which of the other three sketches best represents anequilibrium mixture? Explain.

Initial mixture (a)

(c)(b)

Kc = 3.02 NO(g) + Br2(g) ∆ 2 NOBr(g)

Initial mixture (a)

(b) (c)

Kc = 4.0SO2(g) + Cl2(g) ∆ SO2Cl2(g)

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Partial Pressure Equilibrium Constant, Kp

43. Refer to Example 15-4. at 747.6 mmHg pres-sure and a 1.85-g sample of are introduced into a725-mL flask at 60 °C. What will be the total pressurein the flask at equilibrium?

44. A sample of is placed in a 2.58-L flask con-taining 0.100 mol What will be the total gaspressure when equilibrium is established at 25 °C?

45. The following reaction is used in some self-containedbreathing devices as a source of

Suppose that a sample of is added to an evacu-ated flask containing and equilibrium is estab-lished. If the equilibrium partial pressure of isCO2(g)

KO2(s)CO2(g)

Kp = 28.5 at 25 °C

4 KO2(s) + 2 CO2(g) ∆ 2 K2CO3(s) + 3 O2(g)

O2(g).

Kp = 0.108 at 25 °C

NH4HS(s) ∆ NH3(g) + H2S(g)

NH3(g).NH4HS(s)

Kp = 1.34 * 105 at 60 °C

H2S(g) + I2(s) ∆ 2 HI(g) + S(s)

I2(s)H2S(g) found to be 0.0721 atm, what are the equilibrium par-

tial pressure of and the total gas pressure?46. Concerning the reaction in Exercise 45, if

and are maintained in contact with air at1.00 atm and 25 °C, in which direction will a netchange occur to establish equilibrium? Explain.[Hint: Recall equation (6.17) and the composition ofair (Table 6.3.)]

47. 1.00 mol each of CO and are introduced into anevacuated 1.75-L flask, and the following equilibriumis established at 668 K.

For this equilibrium, calculate (a) the partial pressureof (b) the total gas pressure.

48. For the reaction at 184 °C. What is the value of for

this reaction at 184 °C?

NO(g) +12

O2(g) ∆ NO2(g)

KpKc = 1.8 * 10-62 NO2(g) ∆ 2 NO(g) + O2(g),

COCl2(g);

CO(g) + Cl2(g) ¡ COCl2(g) Kp = 22.5

Cl2

K2CO3(s)KO2(s)

O2(g)

Le Châtelier’s Principle

49. Continuous removal of one of the products of a chem-ical reaction has the effect of causing the reaction to goto completion. Explain this fact in terms of Le Châte-lier’s principle.

50. We can represent the freezing of at 0 °C as Ex-

plain why increasing the pressure on ice causes it tomelt. Is this the behavior you expect for solids ingeneral? Explain.

51. Explain how each of the following affects the amountof present in an equilibrium mixture in the reaction

(a) Raising the temperature of the mixture; (b) intro-ducing more (c) doubling the volume of thecontainer holding the mixture; (d) adding an appro-priate catalyst.

52. In the gas phase, iodine reacts with cyclopenteneby a free radical mechanism to produce cy-

clopentadiene and hydrogen iodide. Explainhow each of the following affects the amount of HI(g)present in the equilibrium mixture in the reaction

(a) Raising the temperature of the mixture; (b) intro-ducing more (c) doubling the volume of thecontainer holding the mixture; (d) adding an appro-priate catalyst; (e) adding an inert gas such as He to aconstant-volume reaction mixture.

C5H6(g);

¢H° = 92.5 kJ

I2(g) + C5H8(g) ∆ C5H6(g) + 2 HI(g)

(C5H6)(C5H8)

H2O(g);

¢H° = -150 kJ

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

H2

(l, d = 1.00 g>cm3) ∆ H2O(s, d = 0.92 g>cm3).H2OH2O(l)

53. The reaction occurs in high-temperature combustion

processes carried out in air. Oxides of nitrogen produ-ced from the nitrogen and oxygen in air are intimatelyinvolved in the production of photochemical smog.What effect does increasing the temperature have on(a) the equilibrium production of NO(g); (b) the rateof this reaction?

54. Use data from Appendix D to determine whether theforward reaction is favored by high temperatures orlow temperatures.(a)(b)(c)

55. If the volume of an equilibrium mixture of and is reduced by doubling the pres-

sure, will have increased, decreased, or remainedthe same when equilibrium is reestablished? Explain.

56. For the reaction

(a) Will increase, decrease, or remain constantwith temperature? Explain.(b) If a constant-volume mixture at equilibrium at 298K is heated to 400 K and equilibrium reestablished,will the number of moles of D(g) increase, decrease, orremain constant? Explain.

Kp

A(s) ∆ B(s) + 2 C(g) +12

D(g) ¢H° = 0

N2(g) + 3 H2(g) ∆ 2 NH3(g)

PN2

NH3(g)H2(g),N2(g),

4 NOCl(g) + 2 H2O(g)2 N2(g) + 3 O2(g) + 4 HCl(g) ∆SO2(g) + 2 H2S(g) ∆ 2 H2O(g) + 3 S(s)PCl3(g) + Cl2(g) ∆ PCl5(g)

+181 kJ,¢H° =N2(g) + O2(g) ∆ 2 NO(g),

PETRMC15_622-662-hr 12/23/05 2:32 PM Page 658

(a)

(b)

(c)

(d) COCl2(g) ∆ CO(g) + Cl2(g) ¢H° = +108.3 kJ

N2H4(g) ∆ N2(g) + 2 H2(g) ¢H° = -95.4 kJ

SO3(g) ∆ SO2(g) +12

O2(g) ¢H° = +98.9 kJ

NO(g) ∆12

N2(g) +12

O2(g) ¢H° = -90.2 kJ

Integrative and Advanced Exercises 659

57. What effect does increasing the volume of the systemhave on the equilibrium condition in each of the fol-lowing reactions?(a)(b)(c)

58. For which of the following reactions would you ex-pect the extent of the forward reaction to increasewith increasing temperatures? Explain.

4 NH3(g) + 5 O2(g) ∆ 4 NO(g) + 6 H2O(g)Ca(OH)2(s) + CO2(g) ∆ CaCO3(s) + H2O(g)C(s) + H2O(g) ∆ CO(g) + H2(g)

Integrative and Advanced Exercises59. Explain why the percent of molecules that dissociate

into atoms in reactions of the type always increases with an increase in temperature.

60. A 1.100-L flask at 25 °C and 1.00 atm pressure containsin contact with 100.0 mL of a saturated aque-

ous solution in which (a) What is the value of at 25 °C for the equilibri-um (b) If 0.01000 mol of radioactive is added to theflask, how many moles of the will be found inthe gas phase and in the aqueous solution when equi-librium is reestablished?[Hint: The radioactive distributes itself be-tween the two phases in exactly the same manner asthe nonradioactive ]

61. Refer to Example 15-13. Suppose that 0.100 L of theequilibrium mixture is diluted to 0.250 L with water.What will be the new concentrations when equilibri-um is reestablished?

62. In the equilibrium described in Example 15-12, thepercent dissociation of can be expressed as

What must be the total pressure of the gaseous mix-ture if is to be 10.0% dissociated at 298 K?

63. Starting with at 1.00 atm, what will be the totalpressure when equilibrium is reached in the followingreaction at 700 K?

64. A sample of air with a mole ratio of to of is heated to 2500 K. When equilibrium is establishedin a closed container with air initially at 1.00 atm, themole percent of NO is found to be 1.8%. Calculate for the reaction.

65. Derive, by calculation, the equilibrium amounts ofand listed in (a) Figure 15-6(c); (b)

Figure 15-7(b).66. The decomposition of salicylic acid to phenol and car-

bon dioxide was carried out at 200.0 °C, a temperatureat which the reactant and products are all gaseous. A0.300-g sample of salicylic acid was introduced into a50.0-mL reaction vessel, and equilibrium was estab-lished. The equilibrium mixture was rapidly cooled to

SO3SO2, O2,

N2(g) + O2(g) ∆ 2 NO(g)

Kp

79 : 21O2N2

Kp = 1.6 * 10-52 SO3(g) ∆ 2 SO2(g) + O2(g)

SO3(g)

N2O4 ∆ 2 NO2(g) Kp = 0.113 at 298 K

N2O4(g)

3.00 * 10-3 mol N2O4

0.0240 mol N2O4 initially* 100% = 12.5%

N2O4

12CO2.

14CO2

14CO2

14CO2

CO2(g) ∆ CO2(aq)?Kc

[CO2(aq)] = 3.29 * 10-2 M.CO2(g)

I2(g) ∆ 2 I(g)condense salicylic acid and phenol as solids; the

was collected over mercury and its volumewas measured at 20.0 °C and 730 mmHg. In two iden-tical experiments, the volumes of obtainedwere 48.2 and 48.5 mL, respectively. Calculate forthis reaction.

67. One of the key reactions in the gasification of coal isthe methanation reaction, in which methane is pro-duced from synthesis gas—a mixture of CO and

(a) Is the equilibrium conversion of synthesis gas tomethane favored at higher or lower temperatures?Higher or lower pressures?(b) Assume you have 4.00 mol of synthesis gas with a3:1 mol ratio of to CO(g) in a 15.0-L flask. Whatwill be the mole fraction of at equilibrium at1000 K?

68. A sample of pure is introduced into an evacu-ated flask and allowed to dissociate.

If the fraction of molecules that dissociate is de-noted by and if the total gas pressure is P, show that

69. Nitrogen dioxide obtained as a cylinder gas is always amixture of and A 5.00-g sample ob-tained from such a cylinder is sealed in a 0.500-L flask at298 K. What is the mole fraction of in this mixture?

70. What is the apparent molar mass of the gaseous mix-ture that results when is allowed to dissoci-ate at 395 °C and a total pressure of 3.00 atm?

Think of the apparent molar mass as the molar massof a hypothetical single gas that is equivalent to thegaseous mixture.

Kp = 4.44 * 10-2 at 395 °C

COCl2(g) ∆ CO(g) + Cl2(g)

COCl2(g)

N2O4(g) ∆ 2 NO2(g) Kc = 4.61 * 10-3

NO2

N2O4(g).NO2(g)

Kp =a2

P

1 - a2

a,PCl5

PCl5(g) ∆ PCl3(g) + Cl2(g)

PCl5(g)

CH4(g)H2(g)

¢H = -230 kJ; Kc = 190 at 1000 K

CO(g) + 3 H2(g) ∆ CH4(g) + H2O(g)

H2.

COOH

OH OH CO2�

Kp

CO2(g)

CO2(g)

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71. Show that in terms of mole fractions of gases and totalgas pressure the equilibrium constant expression for

is

72. For the synthesis of ammonia at 500 K,

Assume that and are mixed in the mole ratio1:3 and that the total pressure is maintained at 1.00atm. What is the mole percent at equilibrium?[Hint: Use the equation from Exercise 71.]

73. A mixture of and in the mole ratio was brought to equilibrium at 700 °C and a total pres-sure of 1 atm. On analysis, the equilibrium mixturewas found to contain The present at equilibrium was converted successively to

and then to ; was obtained. Use these data to determine at700 °C for the reaction

74. A solution is prepared having these initial concentra-tions:

The following reaction occurs among theions at 25 °C.

What will be the ion concentrations at equilibrium?75. Refer to the Integrative Example. A gaseous mixture is

prepared containing 0.100 mol each of and in a 5.00-L flask. Then theH2(g)CO2(g),H2O(g),

CH4(g),

Kc = 9.14 * 10-6

2 Fe3+(aq) + Hg2

2+(aq) ∆ 2 Fe2+(aq) + 2 Hg2+(aq)

0.03000 M.=[Hg2+]=0.5000 M; [Fe2+]=[Hg2

2+]=[Fe3+]

Kp at 700 °C = ?

2 H2S(g) + CH4(g) ∆ CS2(g) + 4 H2(g)

Kp

1.42 * 10-3 mol BaSO4BaSO4 H2SO4

CS29.54 * 10-3 mol H2S.

2 : 1CH4(g)H2S(g)

NH3

H2N2

N2(g) + 3 H2(g) ∆ 2 NH3(g), Kp = 9.06 * 10-2.

Kp =(xNH3)

2

(xN2)(xH2)2 *

1

(Ptot)2

N2(g) + 3 H2(g) ∆ 2 NH3(g)

mixture is allowed to come to equilibrium at 1000 K inreaction (15.23). What will be the equilibrium amount,in moles, of each gas?

76. Concerning the reaction in Exercise 22 and the situa-tion described in part (c) of that exercise, will the molefraction of increase, decrease, or remain un-changed when equilibrium is reestablished? Explain.

77. The formation of nitrosyl chloride is given by the fol-lowing equation:

at 298 K. In a 1.50-L flask, there are4.125 mol of NOCl and 0.1125 mol of present atequilibrium (298 K).(a) Determine the partial pressure of NO atequilibrium.(b) What is the total pressure of the system atequilibrium?

78. At 500 K, a 10.0-L equilibrium mixture contains0.424 mol 1.272 mol and 1.152 mol Themixture is quickly chilled to a temperature at which the

liquefies, and the is completely removed.The 10.0-L gaseous mixture is then returned to 500 K,and equilibrium is reestablished. How many moles of

will be present in the new equilibrium mixture?

79. Recall the formation of methanol from synthesis gas,the reversible reaction with which we began our discus-sion of the equilibrium constant expression (page 623).

A particular synthesis gas consisting of 35.0 mole per-cent CO(g) and 65.0 mole percent at a total pres-sure of 100.0 atm at 483 K is allowed to come toequilibrium. Determine the partial pressure of

in the equilibrium mixture.CH3OH(g)

H2(g)

Kc = 14.5 at 483 KCO(g) + 2 H2(g) ∆ CH3OH(g)

N2(g) + 3 H2(g) ∆ 2 NH3 Kc = 152 at 500 K

NH3(g)

NH3(l)NH3

NH3.H2,N2,

Cl2

Kc = 4.6 * 1042 NO(g) + Cl2(g) ∆ 2 NOCl(g);

C2H2(g)

Feature Problems80. A classic experiment in equilibrium studies dating

from 1862 involved the reaction in solution of ethanoland acetic acid to produce

ethyl acetate and water.

The reaction can be followed by analyzing the equilib-rium mixture for its acetic acid content.

In one experiment, a mixture of 1.000 mol acetic acidand 0.5000 mol ethanol is brought to equilibrium. Asample containing exactly one-hundredth of the equi-librium mixture requires 28.85 mL 0.1000 M for its titration. Calculate the equilibrium constant,

for the ethanol–acetic acid reaction based on thisexperiment.

81. The decomposition of HI(g) is represented by theequation

2 HI(g) ∆ H2(g) + I2(g)

Kc ,

Ba(OH)2

Ba(CH3COO)2(aq) + 2 H2O(l)

2 CH3COOH(aq) + Ba(OH)2(aq) ∆

CH3COOC2H5 + H2OC2H5OH + CH3COOH ∆

(CH3COOH)(C2H5OH)

HI(g) is introduced into five identical glassbulbs, and the five bulbs are maintained at 623 K.Each bulb is opened after a period of time and ana-lyzed for by titration with 0.0150 M

Data for this experiment are provided in the tablebelow. What is the value of at 623 K?

VolumeTime

Initial Bulb RequiredBulb Mass of Opened for TitrationNumber HI(g), g (h) (in mL)

1 0.300 2 20.962 0.320 4 27.903 0.315 12 32.314 0.406 20 41.505 0.280 40 28.68

0.0150 M Na2S2O3

Kc

Na2S4O6(aq) + 2 NaI(aq)

I2(aq) + 2 Na2S2O3(aq) ¡

Na2S2O3(aq).I2

400-cm3

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Self-Assessment Exercises 661

82. In one of Fritz Haber’s experiments to establish the con-ditions required for the ammonia synthesis reaction,pure was passed over an iron catalyst at 901 °Cand 30.0 atm. The gas leaving the reactor was bubbledthrough 20.00 mL of a HCl(aq) solution. In this way, the

present was removed by reaction with HCl.The remaining gas occupied a volume of 1.82 L at STP.The 20.00 mL of HCl(aq) through which the gas hadbeen bubbled required 15.42 mL of 0.0523 M KOH for itstitration. Another 20.00-mL sample of the same HCl(aq)through which no gas had been bubbled required18.72 mL of 0.0523 M KOH for its titration. Use thesedata to obtain a value of at 901 °C for the reaction

83. The following is an approach to establishing a rela-tionship between the equilibrium constant and rateconstants mentioned in the Are You Wondering fea-ture on page 628.• Work with the detailed mechanism for the reaction.• Use the principle of microscopic reversibility, the

idea that every step in a reaction mechanism is re-versible. (In the presentation of elementary reac-tions in Chapter 14, we treated some reaction

N2(g) + 3 H2(g) ∆ 2 NH3(g).Kp

NH3(g)

NH3(g)

steps as reversible and others as going to com-pletion. However, as noted in Table 15.3, everyreaction has an equilibrium constant even thougha reaction is generally considered to go to comple-tion if its equilibrium constant is very large.)

• Use the idea that when equilibrium is attained inan overall reaction, it is also attained in each stepof its mechanism. Moreover, we can write an equi-librium constant expression for each step in themechanism, similar to what we did with thesteady-state assumption in describing reactionmechanisms.

• Combine the expressions for the elementarysteps into a expression for the overall reaction.The numerical value of the overall can therebybe expressed as a ratio of rate constants, k.

Use this approach to establish the equilibrium constantexpression for the overall reaction,

The mechanism of the reaction appears to be thefollowing:Fast:

Slow: 2 I(g) + H2(g) ∆ 2 HI(g)

I2(g) ∆ 2 I(g)

2 HI(g)H2(g) + I2(g) ∆

Kc

Kc

Kc

Self-Assessment Exercises84. In your own words, define or explain the following

terms or symbols: (a) (b) (c)85. Briefly describe each of the following ideas or phe-

nomena: (a) dynamic equilibrium; (b) direction of anet chemical change; (c) Le Châtelier’s principle;(d) effect of a catalyst on equilibrium.

86. Explain the important distinctions between each pairof terms: (a) reaction that goes to completion and re-versible reaction; (b) and (c) reaction quotient(Q) and equilibrium constant expression (K); (d) ho-mogeneous and heterogeneous reaction.

87. In the reversible reaction an initial mixture contains 2 mol and 1 mol Theamount of HI expected at equilibrium is (a) 1 mol;(b) 2 mol; (c) less than 2 mol; (d) more than 2 mol butless than 4 mol.

88. Equilibrium is established in the reaction at a temperature where

If the number of moles of in the equilibriummixture is the same as the number of moles of (a) the number of moles of is also equal to thenumber of moles of (b) the number of moles of

is half the number of moles of (c) mayhave any of several values; (d)

89. The volume of the reaction vessel containing an equi-librium mixture in the reaction

is increased. When equilibrium isreestablished, (a) the amount of will haveincreased; (b) the amount of will have decreased;(c) the amounts of and will have remained thesame; (d) the amount of will have increased.

90. For the reaction at 184 °C. At 184 °C, the value of

for the reaction is(a) (b) (c)(d) 2.8 * 105.

5.6 * 105;7.5 * 102;0.9 * 106;NO(g) + 1

2 O2(g) ∆ NO2(g)KcKc = 1.8 * 10-6

2 NO2(g) ∆ 2 NO(g) + O2(g),SO2Cl2

Cl2SO2

SO2

Cl2

SO2(g) + Cl2(g)SO2Cl2(g) ∆

[O2] = 0.010 M.[O2]SO2 ;O2(g)

SO2(g);O2(g)

SO2(g),SO3(g)

Kc = 100.O2(g) ∆ 2 SO3(g)2 SO2(g) +

I2 .H2

H2(g) + I2(g) ∆ 2 HI(g),

Kp ;Kc

¢ngas .Qc ;Kp ;91. For the dissociation reaction

at 1065 °C. For this samereaction at 1000 K, (a) is less than (b) isgreater than (c) (d) whether is lessthan, equal to, or greater than depends on the totalgas pressure.

92. The following data are given at 1000 K:

After an initial equilibrium is establishedin a 1.00-L container, the equilibrium amount of can be increased by (a) adding a catalyst; (b) increas-ing the temperature; (c) transferring the mixture to a10.0-L container; (d) in some way other than (a), (b),or (c).

93. Equilibrium is established in the reversible reactionThe equilibrium concentrations

are Whatis the value of for this reaction?

94. The Deacon process for producing chlorine gas fromhydrogen chloride is used in situations where HCl isavailable as a by-product from other chemicalprocesses.

A mixture of HCl, and is brought toequilibrium at 400 °C. What is the effect on the equi-librium amount of if(a) additional is added to the mixture at con-stant volume;(b) HCl(g) is removed from the mixture at constantvolume;(c) the mixture is transferred to a vessel of twice thevolume;(d) a catalyst is added to the reaction mixture;(e) the temperature is raised to 500 °C?

O2(g)Cl2(g)

Cl2O2, H2O,

¢H° = -114 kJ

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

Kc

[A] = 0.55 M, [B] = 0.33 M, [C] = 0.43 M.2 A + B ∆ 2 C.

H2

Kc = 0.66.¢H° = -42 kJ;H2O(g) ∆ CO2(g) + H2(g);

CO(g) +

Kp

KcKc = Kp ;Kp ;KcKp ;Kc

Kp = 1.2 * 10-2S2(g),2 H2S(g) ∆ 2 H2(g) +

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95. An equilibrium mixture of and gases ismaintained in a 2.05-L flask at a temperature at which

for the reaction

(a) If the numbers of moles of and in theflask are equal, how many moles of are present?O2

SO3SO2

2 SO2(g) + O2(g) ∆ 2 SO3(g)

Kc = 35.5

O2SO2, SO3, (b) If the number of moles of in the flask is twicethe number of moles of how many moles of are present?

96. Using the method in Appendix E, construct a conceptmap of Section 15-6, illustrating the shift in equilibri-um due to the various stresses.

O2SO2,SO3

eMedia Exercises

97. After viewing the Dynamic Equilibrium animationof equilibrium through a physical change (Activebook15-1), describe the influence of temperature on ratesof change.

98. After viewing the Chemical Equilibrium activity(eBook 15-2), estimate the value of the equilibriumconstant of the reaction depicted. What informationdid you use to make this estimate?

99. (a) Describe how the stoichiometry of the reactionin the Chemical Equilibrium activity (Activebook 15-2 ) influences the values of concentrations at equi-librium. (b) If the initial concentrations of both reac-tants are doubled in this exercise, does this change thevalue of the equilibrium constant? (c) Does thischange, if any, affect the concentration of the productat equilibrium?

100. Using equal initial concentrations of reactant andproduct in the Equilibrium Constant activity(Activebook 15-4), describe how the value of the equi-librium constant is related to the concentration of theproduct once equilibrium has been reached.

101. (a) What would be the change in the results seenin the Equilibrium animation (Active-book 15-6 ) if the pressure were decreased by a factor oftwo instead of increased? (b) Given that this forwardreaction is exothermic, what would be the effect onthe ratio of the and concentrations byincreasing the temperature of the system?

N2O4NO2

NO2–N2O4

PETRMC15_622-662-hr 12/23/05 4:05 PM Page 662

PRINCIPLES OF CHEMICAL EQUILIBRIUM -...

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