Chemical Equilibrium - Wikispaces12chemab.wikispaces.com/file/view/Tro+-+chapter+14.pdf · Chemical Equilibrium 612 ... 614 Chapter 14 Chemical Equilibrium understand that the concentrations

Chemical Equilibrium

612

14.1 Fetal Hemoglobin and Equilibrium 613

14.2 The Concept of Dynamic Equilibrium 615

14.3 The Equilibrium Constant (K) 618

14.4 Expressing the Equilibrium Constant in Terms of Pressure 622

14.5 Heterogeneous Equilibria: Reactions Involving Solids and Liquids 625

14.6 Calculating the Equilibrium Constant from Measured Equilibrium Concentrations 626

14.7 The Reaction Quotient: Predicting the Direction of Change 629

14.8 Finding Equilibrium Concentrations 631

14.9 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances 641

IN CHAPTER 13, we examined how fast a chemical reaction occurs. In this chapter we

examine how far a chemical reaction goes. The speed of a chemical reaction is determined

by kinetics. The extent of a chemical reaction is determined by thermodynamics. In this

chapter, we focus on describing and quantifying how far a chemical reaction goes based on an

experimentally measurable quantity called the equilibrium constant. A reaction with a large

equilibrium constant proceeds nearly to completion—nearly all the reactants react to form

products. A reaction with a small equilibrium constant barely proceeds at all—nearly all the

reactants remain as reactants, hardly forming any products. For now we simply accept the

equilibrium constant as an experimentally measurable quantity and learn how to use it to predict

and quantify the extent of a reaction. In Chapter 17, we will examine the reasons underlying the

magnitude of equilibrium constants.

Every system in chemical equilibrium, under the influence of

a change of any one of the factors of equilibrium, undergoes a

transformation . . . [that produces a change] . . . in the opposite

direction of the factor in question.—Henri Le Châtelier (1850–1936)

14

613

14.1 Fetal Hemoglobin and EquilibriumHave you ever wondered how a baby in the womb gets oxygen? Unlike you and me, afetus does not breathe air. Yet, like you and me, a fetus needs oxygen. Where does thatoxygen come from? After we are born, we inhale air into our lungs and that air diffusesinto capillaries, where it comes into contact with the blood. Within red blood cells, a pro-tein called hemoglobin (Hb) reacts with oxygen according to the chemical equation:

The double arrows in this equation indicate that the reaction can occur in both the for-ward and reverse directions and can reach chemical equilibrium. We encountered thisterm in Chapters 11 and 12, and we define it more carefully in the next section. For now,

Hb + O2 Δ HbO2

A developing fetus gets oxygen from the mother’s blood because the reaction between oxygen and fetal hemoglobin has a larger equilibrium constant than the reaction between oxygen andmaternal hemoglobin.

HbO2�Hb O2

Reaction shifts left.

Muscles:low 3O24

HbO2�Hb O2

Reaction shifts right.

Lungs:high 3O24

614 Chapter 14 Chemical Equilibrium

understand that the concentrations of the reactants and products in a reaction at equilibri-um are described by the equilibrium constant, K. A large value of K means that the reac-tion lies far to the right at equilibrium—a high concentration of products and a lowconcentration of reactants. A small value of K means that the reaction lies far to the left atequilibrium—a high concentration of reactants and a low concentration of products. Inother words, the value of K is a measure of how far a reaction proceeds—the larger thevalue of K, the more the reaction proceeds toward the products.

The equilibrium constant for the reaction between hemoglobin and oxygen is suchthat hemoglobin efficiently binds oxygen at typical lung oxygen concentrations, but canalso release oxygen under the appropriate conditions. Any system at equilibrium, includ-ing the hemoglobin–oxygen system, responds to changes in ways that maintain equilibri-um. If any of the concentrations of the reactants or products change, the reaction shifts tocounteract that change. For the hemoglobin system, as blood flows through the lungswhere oxygen concentrations are high, the equilibrium shifts to the right—hemoglobinbinds oxygen:

� Hemoglobin is the oxygen-carryingprotein in red blood cells. Oxygen bindsto iron atoms, which are depicted herein white.

As blood flows out of the lungs and into muscles and organs where oxygen concentra-tions have been depleted (because muscles and organs use oxygen), the equilibrium shiftsto the left—hemoglobin releases oxygen:

In order to maintain equilibrium, hemoglobin binds oxygen when the surrounding oxy-gen concentration is high, but releases oxygen when the surrounding oxygen concentra-tion is low. In this way, hemoglobin transports oxygen from the lungs to all parts of thebody that use oxygen.

A fetus has its own circulatory system. The mother’s blood never flows into thefetus’s body and the fetus cannot get any air in the womb. How, then, does the fetus getoxygen? The answer lies in the properties of fetal hemoglobin (HbF), which is slightlydifferent from adult hemoglobin. Like adult hemoglobin, fetal hemoglobin is in equilib-rium with oxygen:

However, the equilibrium constant for fetal hemoglobin is larger than the equilibriumconstant for adult hemoglobin, meaning that the reaction tends to go farther in the direc-tion of the product. Consequently, fetal hemoglobin loads oxygen at a lower oxygen con-centration than does adult hemoglobin. In the placenta, fetal blood flows in closeproximity to maternal blood. Although the two never mix, because of the different equi-librium constants, the maternal hemoglobin releases oxygen which the fetal hemoglobinthen binds and carries into its own circulatory system (Figure 14.1 �). Nature has thusevolved a chemical system through which the mother’s hemoglobin can in effect hand offoxygen to the hemoglobin of the fetus.

HbF + O2 Δ HbFO2


Placenta

Fetus

Maternalvein

Maternalartery

Fetalvein

Fetalartery

Uterus

Umbilicalcord

Maternalblood

PlacentaFetus

Nutrients and waste materials are exchanged betweenfetal and maternal blood through the placenta.

� FIGURE 14.1 Oxygen Exchange between the Maternal and Fetal Circulation In the placenta, the bloodof the fetus comes into close proximitywith that of the mother although thetwo do not mix directly. Because thereaction of fetal hemoglobin with oxy-gen has a larger equilibrium constantthan the reaction of maternal hemoglo-bin with oxygen, the fetus receivesoxygen from the mother’s blood.

14.2 The Concept of Dynamic EquilibriumRecall from the previous chapter that reaction rates generally increase with increasingconcentration of the reactants (unless the reaction order is zero) and decrease with de-creasing concentration of the reactants. With this in mind, consider the reaction betweenhydrogen and iodine:

In this reaction, and react to form 2 HI molecules, but the 2 HI molecules can alsoreact to re-form and A reaction such as this one—that can proceed in both the for-ward and reverse directions—is said to be reversible. Suppose we begin with only and in a container (Figure 14.2a � on the next page). What happens? Initially and begin to react to form HI (Figure 14.2b). However, as and react their concentrationsdecrease, which in turn decreases the rate of the forward reaction. At the same time, HIbegins to form. As the concentration of HI increases, the reverse reaction begins to occurat a faster and faster rate. Eventually the rate of the reverse reaction (which has been in-creasing) equals the rate of the forward reaction (which has been decreasing). At thatpoint, dynamic equilibrium is reached (Figure 14.2c, d).

Dynamic equilibrium for a chemical reaction is the condition in which therate of the forward reaction equals the rate of the reverse reaction.

Dynamic equilibrium is called “dynamic” because the forward and reverse reactions arestill occurring; however, they are occurring at the same rate. When dynamic equilibriumis reached, the concentrations of and HI no longer change. They remain the samebecause the reactants and products form at the same rate that they are depleted. However,just because the concentrations of reactants and products no longer change at equilibriumdoes not mean that the concentrations of reactants and products are equal to one anotherat equilibrium. Some reactions reach equilibrium only after most of the reactants haveformed products. Others reach equilibrium when only a small fraction of the reactantshave formed products. It depends on the reaction.

I2,H2,

I2H2

I2H2I2

H2

I2.H2

I2H2

H2(g) + I2(g) Δ 2 HI(g)

Nearly all chemical reactions are atleast theoretically reversible. In manycases, however, the reversibility is sosmall that it can be ignored.


� FIGURE 14.2 Dynamic Equilibrium Equilibrium is reached in a chemical reaction when theconcentrations of the reactants and products no longer change. The molecular images on thebottom depict the progress of the reaction The graph on the topshows the concentrations of and HI as a function of time. When equilibrium is reached,both the forward and reverse reactions continue, but at equal rates, so the concentrations of thereactants and products remain constant.

I2,H2,H2(g) + I2(g) Δ 2 HI(g).

0

30

45 15

0

30

45 15

0

30

45 15

0

30

45 15

Con

cen

trat

ion

Time

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

Dynamic equilibrium

Dynamic Equilibrium

Time

A reversible reaction

H2(g) 2 HI(g)

3HI4

3H24

3I24

I2(g)

�

�

Dynamic equilibrium: Rate of forward reaction � rate of reverse reaction. Concentrations of reactant(s) and product(s) no longer change.

As concentration of product increases, and concentrations of reactants decrease, rate of forward reaction slows down and rate of reversereaction speeds up.


Dynamic Equilibrium: An Analogy

Country A Country B

Country A Country B

InitialNet movement from A to B

EquilibriumEqual movement in both directions

� FIGURE 14.3 A Population Analogyfor Chemical Equilibrium BecauseCountry A is initially overpopulated,people migrate from Country A toCountry B. As the population of Coun-try A falls and that of Country B rises,the rate of migration from Country Ato Country B decreases and the rate ofmigration from Country B to CountryA increases. Eventually the two ratesbecome equal. Equilibrium has beenreached.

We can better understand dynamic equilibrium with a simple analogy. Imagine twoneighboring countries (A and B) with a closed border between them (Figure 14.3 �).Country A is overpopulated and Country B is underpopulated. One day, the border be-tween the two countries opens, and people immediately begin to leave Country A forCountry B.

The population of Country A goes down as the population of Country B goes up. As peo-ple leave Country A, however, the rate at which they leave slows down, because as Coun-try A becomes less populated, the pool of potential emigrants gets smaller. On the otherhand, as people move into Country B, it gets more crowded and some people begin tomove from Country B to Country A.

As the population of Country B continues to grow, the rate of people moving out ofCountry B gets faster. Eventually the rate of people moving out of Country A (which hasbeen slowing down as people leave) equals the rate of people moving out of Country B(which has been increasing as Country B gets more crowded). Dynamic equilibrium hasbeen reached.

Notice that when the two countries reach dynamic equilibrium, their populations nolonger change because the number of people moving out of either country equals thenumber of people moving in. However, one country—because of its charm or the avail-ability of good jobs or lower taxes, or for whatever other reason—may have a higher pop-ulation than the other country, even when dynamic equilibrium is reached.

Similarly, when a chemical reaction reaches dynamic equilibrium, the rate of the for-ward reaction equals the rate of the reverse reaction, and the relative concentrations of re-actants and products become constant. But the concentrations of reactants and productswill not necessarily be equal at equilibrium, just as the populations of the two countriesare not necessarily equal at equilibrium.

Country A Δ Country B

Country A — Country B

Country A ¡ Country B

In this notation, [A] represents the molar concentration of A. The equilibrium con-stant quantifies the relative concentrations of reactants and products at equilibrium. Therelationship between the balanced chemical equation and the expression of the equilibri-um constant is known as the law of mass action.

Expressing Equilibrium Constants for Chemical ReactionsTo express an equilibrium constant for a chemical reaction, examine the balanced chem-ical equation and apply the law of mass action. For example, suppose we want to expressthe equilibrium constant for the following reaction:

The equilibrium constant is raised to the fourth power multiplied by raised tothe first power divided by raised to the second power:

Notice that the coefficients in the chemical equation become the exponents in the expres-sion of the equilibrium constant.

K =[NO2]4[O2]

[N2O5]2

[N2O5][O2][NO2]

2 N2O5(g) Δ 4 NO2(g) + O2(g)


14.3 The Equilibrium Constant (K)We have just learned that the concentrations of reactants and products are not equal atequilibrium—rather, the rates of the forward and reverse reactions are equal. So whatabout the concentrations? The equilibrium constant is a way to quantify the concentra-tions of the reactants and products at equilibrium.

Consider the following general chemical equation,

where A and B are reactants, C and D are products, and a, b, c, and d are the respectivestoichiometric coefficients in the chemical equation. The equilibrium constant (K) forthe reaction is defined as the ratio—at equilibrium—of the concentrations of the productsraised to their stoichiometric coefficients divided by the concentrations of the reactantsraised to their stoichiometric coefficients.

aA + bB Δ cC + dD

We distinguish between the equilibriumconstant (K) and the kelvin unit oftemperature (K) by italicizing theequilibrium constant.

K =3A4a3B4b3C4c3D4d

Products

Reactants

Law of Mass Action

The equilibrium constant is the concentrations of the products raised to theirstoichiometric coefficients divided by the concentrations of the reactantsraised to their stoichiometric coefficients.

K =[CO][H2]2

[CH3OH]

FOR PRACTICE 14.1Express the equilibrium constant for the combustion of propane as shown by thebalanced chemical equation:

C3H8(g) + 5 O2(g) Δ 3 CO2(g) + 4 H2O(g)

EXAMPLE 14.1 Expressing Equilibrium Constants for ChemicalEquationsExpress the equilibrium constant for the following chemical equation.

SOLUTION

CH3OH(g) Δ CO(g) + 2 H2(g)

Br2(g)H2(g) �

�

2 HBr(g)

K � � large number[HBr]2

[H2][Br2]


The Significance of the Equilibrium ConstantWe now know how to express the equilibrium constant, but what does it mean? What, forexample, does a large equilibrium constant imply about a reaction? A largeequilibrium constant indicates that the numerator (which specifies the amounts of prod-ucts at equilibrium) is larger than the denominator (which specifies the amounts of reac-tants at equilibrium). Therefore the forward reaction is favored. For example, considerthe reaction:

The equilibrium constant is large, indicating that the equilibrium point for the reactionlies far to the right—high concentrations of products, low concentrations of reactants(Figure 14.4 �). Remember that the equilibrium constant says nothing about how fast areaction reaches equilibrium, only how far the reaction has proceeded once equilibrium isreached. A reaction with a large equilibrium constant may be kinetically very slow andtake a long time to reach equilibrium.

K = 1.9 * 1019 (at 25 °C)H2(g) + Br2(g) Δ 2 HBr(g)

(K W 1)

� FIGURE 14.4 The Meaning of aLarge Equilibrium Constant If theequilibrium constant for a reaction islarge, the equilibrium point of the reac-tion lies far to the right—the concen-tration of products is large and theconcentration of reactants is small.

Conversely, what does a small equilibrium constant mean? It indicatesthat the reverse reaction is favored and that there will be more reactants than productswhen equilibrium is reached. For example, consider the following reaction:

The equilibrium constant is very small, indicating that the equilibrium point for the reac-tion lies far to the left—high concentrations of reactants, low concentrations of products(Figure 14.5 �). This is fortunate because and are the main components of air. Ifthis equilibrium constant were large, much of the and in air would react to formNO, a toxic gas.

Summarizing the Significance of the Equilibrium Constant:� Reverse reaction is favored; forward reaction does not proceed very far.� Neither direction is favored; forward reaction proceeds about halfway.� Forward reaction is favored; forward reaction proceeds essentially to

completion.K W 1K L 1K V 1

O2N2

O2N2

N2(g) + O2(g) Δ 2 NO(g) K = 4.1 * 10-31 (at 25 °C)

(K V 1)

O2(g)N2(g) �

�

2 NO(g)

K � � small number[NO]2

[N2][O2]

� FIGURE 14.5 The Meaning of aSmall Equilibrium Constant If theequilibrium constant for a reaction issmall, the equilibrium point of the re-action lies far to the left—the concen-tration of products is small and theconcentration of reactants is large.


CHEMISTRY AND MEDICINE Life and Equilibrium

of our blood does not become the same as the acidity of thesurrounding water. Living things, even the simplest ones,maintain some measure of disequilibrium with theirenvironment.

We must add one more concept, however, to completeour definition of life with respect to equilibrium. A cup of hotwater is in disequilibrium with its environment with respect totemperature, yet it is not alive. The cup of hot water has nocontrol over its disequilibrium, however, and will slowlycome to equilibrium with its environment. In contrast, livingthings—as long as they are alive—maintain and control theirdisequilibrium. Your body temperature, for example, is notonly in disequilibrium with your surroundings—it is incontrolled disequilibrium. Your body maintains your tempera-ture within a specific range that is not in equilibrium with thesurrounding temperature.

So, one criterion for life is that living things are incontrolled disequilibrium with their environment. Maintainingthat disequilibrium is a main activity of living organisms, re-quiring energy obtained from their environment. Plants derivethat energy from sunlight; animals eat plants (or other animalsthat have eaten plants), and thus they too ultimately derivetheir energy from the sun. A living thing comes into equilibri-um with its surroundings only after it dies.

� What makes thesecells alive?

Have you ever tried to define life? If you have, you prob-ably know that a definition is elusive. How are livingthings different from nonliving things? You may try to

define living things as those things that can move. But ofcourse many living things do not move (most plants, forexample), and some nonliving things, such as glaciers andEarth itself, do move. So motion is neither unique to nor de-finitive of life. You may try to define living things as thosethings that can reproduce. But again, many living things,such as mules or sterile humans, cannot reproduce; yet theyare alive. In addition, some nonliving things, such as crystals,for example, reproduce (in some sense). So what is uniqueabout living things?

One definition of life uses the concept of equilibrium—living things are not in equilibrium with their sur-

roundings. Our body temperature, forexample, is not the same as the tem-

perature of our surroundings.If we jump into a swim-

ming pool, the acidity

Conceptual Connection 14.1 Equilibrium ConstantsThe equilibrium constant for the reaction is 10. A reaction mixture ini-tially contains 11 mol of A and 0 mol of B in a fixed volume of 1 L. When equilibrium isreached, which statement is true?

(a) The reaction mixture will contain 10 mol of A and 1 mol of B.

(b) The reaction mixture will contain 1 mol of A and 10 mol of B.

(c) The reaction mixture will contain equal amounts of A and B.

ANSWER: (b) The reaction mixture will contain 1 mol of A and 10 mol of B so that [B]>[A] = 10.

A(g) Δ B(g)

Relationships between the Equilibrium Constant and the Chemical EquationIf a chemical equation is modified in some way, then the equilibrium constant for theequation must be changed to reflect the modification. The following modifications arecommon.

1. If you reverse the equation, invert the equilibrium constant. For example, con-sider this equilibrium equation:

A + 2 B Δ 3 C


The expression for the equilibrium constant of this reaction is given by

If we reverse the equation:

then, according to the law of mass action, the expression for the equilibrium constantbecomes:

2. If you multiply the coefficients in the equation by a factor, raise the equilibriumconstant to the same factor. Consider again this chemical equation and correspon-ding expression for the equilibrium constant:

If we multiply the equation by n, we get:

Applying the law of mass action, the expression for the equilibrium constantbecomes:

3. If you add two or more individual chemical equations to obtain an overall equa-tion, multiply the corresponding equilibrium constants by each other to obtainthe overall equilibrium constant. Consider these two chemical equations and theircorresponding equilibrium constant expressions:

The two equations sum as follows:

According to the law of mass action, the equilibrium constant for this overall equa-tion is then:

Notice that is the product of and

=[C]3

[A]

=[B]2

[A]*

[C]3

[B]2

Koverall = K1 * K2

K2 :K1Koverall

Koverall =[C]3

[A]

A Δ 2 B

2 B Δ 3 C

A Δ 3 C

A Δ 2 B K1 =[B]2

[A]

2 B Δ 3 C K2 =[C]3

[B]2

K¿ =[C]3n

[A]n[B]2n = a[C]3

[A][B]2 bn

= Kn

n A + 2n B Δ 3n C

A + 2 B Δ 3 C K =[C]3

[A][B]2

Kreverse =[A][B]2

[C]3 =1

Kforward

3 C Δ A + 2 B

Kforward =[C]3

[A][B]2

If n is a fractional quantity, raise K tothe same fractional quantity.


Begin by reversing the given reaction and taking the in-verse of the value of K.

2 NH3(g) Δ N2(g) + 3 H2(g) Krev =1

3.7 * 108

N2(g) + 3 H2(g) Δ 2 NH3(g) K = 3.7 * 108

Next, multiply the reaction by and raise the equilibri-um constant to the power.1

2

12

K¿ = K1>2reverse = a

1

3.7 * 108 b1>2

NH3(g) Δ 12N2(g) + 3

2H2(g)

Calculate the value of K¿. K¿ = 5.2 * 10-5

FOR PRACTICE 14.2Consider the following chemical equation and equilibrium constant at

Compute the equilibrium constant for the following reaction at

FOR MORE PRACTICE 14.2Predict the equilibrium constant for the first reaction given the equilibrium constantsfor the second and third reactions:

CO(g) + 2 H2(g) Δ CH3OH(g) K3 = 1.4 * 107

CO(g) + H2O(g) Δ CO2(g) + H2(g) K2 = 1.0 * 105

CO2(g) + 3 H2(g) Δ CH3OH(g) + H2O(g) K1 = ?

2 CO2(g) + 2 CF4(g) Δ 4 COF2(g) K¿ = ?

25 °C:

2 COF2(g) Δ CO2(g) + CF4(g) K = 2.2 * 106

25 °C:

14.4 Expressing the Equilibrium Constant in Terms of PressureSo far, we have expressed the equilibrium constant only in terms of the concentrations ofthe reactants and products. For gaseous reactions, the partial pressure of a particular gasis proportional to its concentration. Therefore, we can also express the equilibrium con-stant in terms of the partial pressures of the reactants and products. Consider the gaseousreaction:

From this point on, we designate as the equilibrium constant with respect to concen-tration in molarity. For the above reaction, is expressed using the law of mass action:

Kc =[SO2]2[O2]

[SO3]2

Kc

Kc

2 SO3(g) Δ 2 SO2(g) + O2(g)

EXAMPLE 14.2 Manipulating the Equilibrium Constant to Reflect Changes in the Chemical EquationConsider the chemical equation and equilibrium constant for the synthesis of ammoniaat

Calculate the equilibrium constant for the following reaction at

SOLUTIONYou want to manipulate the given reaction and value of K to obtain the desired reactionand value of K. You can see that the given reaction is the reverse of the desired reaction,and its coefficients are twice those of the desired reaction.

NH3(g) Δ 12N2(g) + 3

2H2(g) K¿ = ?

25 °C:

N2(g) + 3 H2(g) Δ 2 NH3(g) K = 3.7 * 108

25 °C:

14.4 Expressing the Equilibrium Constant in Terms of Pressure 623

We now designate as the equilibrium constant with respect to partial pressures in at-mospheres. The expression for takes the form of the expression for except that weuse the partial pressure of each gas in place of its concentration. For the reactionabove, we write as follows:

where is simply the partial pressure of gas A in units of atmospheres.Since the partial pressure of a gas in atmospheres is not the same as its concentration

in molarity, the value of for a reaction is not necessarily equal to the value of However, as long as the gases are behaving ideally, we can derive a relationship betweenthe two constants. The concentration of an ideal gas A is the number of moles of A divided by its volume in liters:

From the ideal gas law, we can relate the quantity to the partial pressure of A asfollows:

Since we can write

[14.1]

Now consider the following general equilibrium chemical equation:

According to the law of mass action, we write as follows:

Substituting for each concentration term, we get:

Rearranging,

Finally, if we let which is the sum of the stoichiometric coeffi-cients of the gaseous products minus the sum of the stoichiometric coefficients of thegaseous reactants, we get the following general result:

[14.2]

Notice that if the total number of moles of gas is the same after the reaction as before,then and is equal to Kc.Kp¢n = 0,

Kp = Kc(RT)¢n

¢n = c + d - (a + b),

Kp = Kc(RT)c+d- (a+b)

= Kpa1

RTbc+d- (a+b)

Kc =aPC

RTbc

aPD

RTbd

aPA

RTba

aPB

RTbb =PcCP

dDa

1

RTbc+d

PaAPbBa

1

RTba+b =

PcCPdD

PaAPbBa

1

RTbc+d- (a+b)

[X] = PX>RT

Kc =[C]c[D]d

[A]a[B]b

Kc

aA + bB Δ cC + dD

PA = [A]RT or [A] =PA

RT

[A] = nA>V,

PA =nA

VRT

PAV = nART

nA>V

[A] =nA

V

(V)(nA)

Kc.Kp

PA

Kp =(PSO2

)2PO2

(PSO3)2

Kp

SO3

Kc,Kp

Kp


CHECK The easiest way to check this answer is to substitute it back into Equation 14.2and confirm that you get the original value for

FOR PRACTICE 14.3Consider the following reaction and corresponding value of

What is the value of at this temperature?Kp

H2(g) + I2(g) Δ 2 HI(g) Kc = 6.2 * 102 at 25 °C

Kc :

= 2.2 * 1012

= 5.4 * 1013a0.08206 L # atm

mol # K * 298 Kb-1

Kp = Kc(RT)¢n

Kp.

Units of KThroughout this book, we express concentrations and partial pressures within the equilib-rium constant in units of molarity and atmospheres, respectively. When expressingthe value of the equilibrium constant, however, we have not included the units. Formally,the values of concentration or partial pressure that we substitute into the equilibrium con-stant expression are ratios of the concentration or pressure to a reference concentration(exactly 1 M) or a reference pressure (exactly 1 atm). For example, within the equilibriumconstant expression, a pressure of 1.5 atm becomes

Similarly, a concentration of 1.5 M becomes

1.5 M

1 M = 1.5

1.5 atm

1 atm= 1.5

SOLVE Solve the equation for

Calculate

Substitute the required quantities to calculate The temperaturemust be in kelvins. The units are dropped when reporting as described in the section that follows.

Kc

Kc.

¢n.

Kc. SOLUTION

= 5.4 * 1013

Kc =2.2 * 1012

a0.08206 L # atm

mol # K * 298 Kb-1

¢n = 2 - 3 = -1

Kc =Kp

(RT)¢n

SORT You are given for the reaction and asked to find Kc.Kp GIVEN:

FIND: Kc

Kp = 2.2 * 1012

STRATEGIZE Use Equation 14.2 to relate and Kc.Kp EQUATION Kp = Kc(RT)¢n

EXAMPLE 14.3 Relating and Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxidein the atmosphere according to the equation:

Find for this reaction.Kc

2 NO(g) + O2(g) Δ 2 NO2(g) Kp = 2.2 * 1012 at 25 °C

KcKp

14.5 Heterogeneous Equilibria: Reactions Involving Solids and Liquids 625

A Heterogeneous Equilibrium

2 CO(g) CO2(g) C(s)�

Same 3CO24and 3CO4

Sametemperature

C(s)C(s)

� FIGURE 14.6 Heterogeneous Equi-librium The concentration of solidcarbon (the number of atoms per unitvolume) is constant as long as somesolid carbon is present. The same istrue for pure liquids. Thus, the concen-trations of solids and pure liquids arenot included in equilibrium constantexpressions.

As long as concentration units are expressed in molarity for and pressure units are ex-pressed in atmospheres for we can skip this formality and enter the quantitites directlyinto the equilibrium expression, dropping their corresponding units.

14.5 Heterogeneous Equilibria: Reactions Involving Solids and LiquidsMany chemical reactions involve pure solids or pure liquids as reactants or products.Consider, for example, the reaction:

We might expect the expression for the equilibrium constant to be:

However, since carbon is a solid, its concentration is constant (if you double theamount of carbon its concentration remains the same). The concentration of a soliddoes not change because a solid does not expand to fill its container. Its concentration,therefore, depends only on its density, which is constant as long as some solid is present(Figure 14.6 �). Consequently, pure solids—those reactants or products labeled in thechemical equation with an (s)—are not included in the equilibrium expression (becausetheir constant value is incorporated into the value of K). The correct equilibriumexpression is:

Similarly, the concentration of a pure liquid does not change. So, pure liquids—reactantsor products labeled in the chemical equation with an —are also excluded from theequilibrium expression. For example, consider the equilibrium expression for the reac-tion between carbon dioxide and water:

Since is pure liquid, it is omitted from the equilibrium expression:

Kc =[H+][HCO3

-]

[CO2]

H2O(/)

CO2(g) + H2O(/) Δ H+(aq) + HCO3-(aq)

(/)

Kc =[CO2]

[CO]2

Kc =[CO2][C]

[CO]2

2 CO(g) Δ CO2(g) + C(s)

Kp,Kc


Conceptual Connection 14.2 Heterogeneous Equilibria, and For which reaction does

(a)

(b)

(c)

ANSWER: (b) Since for gaseous reactants and products is zero, equals Kc.Kp¢n

NH4NO3(s) Δ N2O(g) + 2 H2O(g)

NiO(s) + CO(g) Δ Ni(s) + CO2(g)

2 Na2O2(s) + 2 CO2(g) Δ 2 Na2CO3(s) + O2(g)

Kp = Kc ?

KcKp,

14.6 Calculating the Equilibrium Constant from Measured Equilibrium ConcentrationsThe most direct way to obtain an experimental value for the equilibrium constant of a re-action is to measure the concentrations of the reactants and products in a reaction mixtureat equilibrium. Consider the following reaction:

Suppose a mixture of and is allowed to come to equilibrium at The meas-ured equilibrium concentrations are What is the value of the equilibrium constant at this temperature? The expression for can be written from the balanced equation.

To calculate the value of substitute the correct equilibrium concentrations into theexpression for

The concentrations within should always be written in moles per liter (M); however,as noted in Section 14.4, the units are not normally included when expressing the value ofthe equilibrium constant, so is unitless.

For any reaction, the equilibrium concentrations of the reactants and products willdepend on the initial concentrations (and in general vary from one set of initial concentra-tions to another). However, the equilibrium constant will always be the same at a given

Kc

Kc

= 5.0 * 101

=0.782

(0.11)(0.11)

Kc =[HI]2

[H2][I2]

Kc

Kc,

Kc =[HI]2

[H2][I2]

Kc

[H2] = 0.11 M, [I2] = 0.11 M, and [HI] = 0.78 M.445 °C.I2H2

H2(g) + I2(g) Δ 2 HI(g)

Since equilibrium constants depend on temperature, many equilibriumproblems will state the temperatureeven though it has no formal part in the calculation.

FOR PRACTICE 14.4Write an equilibrium expression for the equation:

4 HCl(g) + O2(g) Δ 2 H2O(/) + 2 Cl2(g)

(Kc)

Since and are both solids, they are omitted from the equilibrium expression.

CaO(s)CaCO3(s) Kc = [CO2]

EXAMPLE 14.4 Writing Equilibrium Expressions for ReactionsInvolving a Solid or a LiquidWrite an expression for the equilibrium constant for the chemical equation:

SOLUTION

CaCO3(s) Δ CaO(s) + CO2(g)

(Kc)

14.6 Calculating the Equilibrium Constant from Measured Equilibrium Concentrations 627

TABLE 14.1 Initial and Equilibrium Concentrations for the Reactionat 445 °C

Initial Concentrations Equilibrium Concentrations Equilibrium Constant

[HI] [HI]

0.50 0.50 0.0 0.11 0.11 0.78

0.0 0.0 0.50 0.055 0.055 0.39

0.50 0.50 0.50 0.165 0.165 1.17

1.0 0.50 0.0 0.53 0.033 0.934

0.50 1.0 0.0 0.033 0.53 0.9340.9342

(0.033)(0.53)= 50

0.9342

(0.53)(0.033)= 50

1.172

(0.165)(0.165)= 50

0.392

(0.055)(0.055)= 50

0.782

(0.11)(0.11)= 50

Kc =[HI]2

[H2][I2][I2][H2][I2][H2]

H2(g) � I2(g ) Δ 2 HI(g )

temperature, regardless of the initial concentrations. For example, Table 14.1 showsseveral different equilibrium concentrations of and HI, each from a different set ofinitial concentrations. Notice that the equilibrium constant is always the same, regardlessof the initial concentrations. Whether you start with only reactants or only products, thereaction reaches equilibrium concentrations in which the equilibrium constant is thesame. No matter what the initial concentrations are, the reaction will always go in a direc-tion so that the equilibrium concentrations—when substituted into the equilibriumexpression—give the same constant, K.

So far, we have calculated equilibrium constants from values of the equilibriumconcentrations of all the reactants and products. In most cases, however, we need onlyknow the initial concentrations of the reactant(s) and the equilibrium concentration ofany one reactant or product. The other equilibrium concentrations can be deducedfrom the stoichiometry of the reaction. For example, consider the following simplereaction:

Suppose that we have a reaction mixture in which the initial concentration of A is 1.00 M andthe initial concentration of B is When equilibrium is reached, the concentration of Ais 0.75 M. Since [A] has changed by we can deduce (based on the stoichiometry)that [B] must have changed by We can summarize the initialconditions, the changes, and the equilibrium conditions in the following table:

2 * (+0.25 M) or +0.50 M.-0.25 M,

0.00 M.

A(g) Δ 2 B(g)

H2, I2,

[A] [B]

Initial 1.00 0.00

Change

Equilibrium 0.75 0.50

+0.50-0.25

This type of table is often referred to as an ICE table To calculate the equilibrium constant, we use the balanced equation to

write an expression for the equilibrium constant and then substitute the equilibrium con-centrations from the ICE table:

In the examples that follow, we show the general procedure for solving these kinds ofequilibrium problems in the left column and work two examples exemplifying the proce-dure in the center and right columns.

K =[B]2

[A]=

(0.50)2

(0.75)= 0.33

E = equilibrium).(I = initial, C = change,

2. For the reactant or product whoseconcentration is known both ini-tially and at equilibrium, calculatethe change in concentration thatoccurred.

CO(g) + 2 H2(g) Δ CH3OH(g) 2 CH4(g) Δ C2H2(g) + 3 H2(g)

3. Use the change calculated in step 2and the stoichiometric relation-ships from the balanced chemicalequation to determine the changesin concentration of all other reac-tants and products. Since reactantsare consumed during the reaction, thechanges in their concentrations arenegative. Since products are formed,the changes in their concentrationsare positive.


4. Sum each column for each reac-tant and product to determine theequilibrium concentrations.

5. Use the balanced equation to writean expression for the equilibriumconstant and substitute the equi-librium concentrations to calcu-late K.

= 26

=0.35

(0.15)(0.30)2

Kc =[CH3OH]

[CO][H2]2

= 0.020

=(0.035)(0.105)3

0.0452

Kc =[C2H2][H2]3

[CH4]2

[CO]

Initial 0.500 1.00 0.00

Change

Equil 0.15

-0.35

[CH3OH][H2]

Initial 0.115 0.00 0.00

Change

Equil 0.035

+0.035

[H2][C2H2][CH4]

Initial 0.115 0.00 0.00

Change

Equil 0.035

+0.105+0.035-0.070

[H2][C2H2][CH4]

[CO]

Initial 0.500 1.00 0.00

Change

Equil 0.15 0.30 0.35

+0.35-0.70-0.35

[CH3OH][H2]

Initial 0.115 0.00 0.00

Change

Equil 0.045 0.035 0.105

+0.105+0.035-0.070

[H2][C2H2][CH4]

[CO]

Initial 0.500 1.00 0.00

Change

Equil 0.15

+0.35-0.70-0.35

[CH3OH][H2]


PROCEDURE FOR...Finding Equilibrium Constantsfrom ExperimentalConcentration MeasurementsTo solve these types of problems, followthe procedure outlined below.

EXAMPLE 14.5Finding Equilibrium Constants from ExperimentalConcentration MeasurementsConsider the following reaction:

A reaction mixture at initiallycontains and

At equilibrium, the CO con-centration is found to be Whatis the value of the equilibrium constant?

0.15 M.1.00 M.

=[H2][CO] = 0.500 M780 °C

CO(g) + 2 H2(g) Δ CH3OH(g)

EXAMPLE 14.6Finding EquilibriumConstants from ExperimentalConcentration MeasurementsConsider the following reaction:

A reaction mixture at initiallycontains At equilib-rium, the mixture contains

What is the value of the equi-librium constant?0.035 M.

=[C2H2][CH4] = 0.115 M.

1700 °C

2 CH4(g) Δ C2H2(g) + 3 H2(g)

1. Using the balanced equation as aguide, prepare an ICE table showingthe known initial concentrationsand equilibrium concentrations ofthe reactants and products. Leavespace in the middle of the table for de-termining the changes in concentra-tion that occur during the reaction.


[CO] [H2] [CH3OH]

Initial 0.500 1.00 0.00

Change

Equil 0.15

Initial 0.115 0.00 0.00

Change

Equil 0.035

[H2][C2H2][CH4]

14.7 The Reaction Quotient: Predicting the Direction of Change 629

FOR PRACTICE 14.5The above reaction between CO and

is carried out at a different temper-ature with initial concentrations of

and At equilibrium, the concentration of

is 0.11 M. Find the equilibri-um constant at this temperature.CH3OH

[H2] = 0.49 M.[CO] = 0.27 M

H2

FOR PRACTICE 14.6The above reaction of is carried outat a different temperature with an initialconcentration of At equilibrium, the concentration of

is 0.012 M. Find the equilibriumconstant at this temperature.H2

[CH4] = 0.087 M.

CH4

14.7 The Reaction Quotient: Predicting the Direction of ChangeWhen the reactants of a chemical reaction mix, they generally react to form products—wesay that the reaction proceeds to the right (toward the products). The amount of productsformed when equilibrium is reached depends on the magnitude of the equilibrium con-stant, as we have seen. However, what if a reaction mixture not at equilibrium containsboth reactants and products? Can we predict the direction of change for such a mixture?

In order to compare the progress of a reaction to the equilibrium state of the reaction,we use a quantity called the reaction quotient. The definition of the reaction quotienttakes the same form as the definition of the equilibrium constant, except that the reactionneed not be at equilibrium. So, for the general reaction

we define the reaction quotient as the ratio—at any point in the reaction—of theconcentrations of the products raised to their stoichiometric coefficients divided by theconcentrations of the reactants raised to their stoichiometric coefficients. For gases withamounts measured in atmospheres, the reaction quotient uses the partial pressures inplace of concentrations and is called

The difference between the reaction quotient and the equilibrium constant is that, at agiven temperature, the equilibrium constant has only one value and it specifies the rela-tive amounts of reactants and products at equilibrium. The reaction quotient, however,depends on the current state of the reaction and has many different values as the reactionproceeds. For example, in a reaction mixture containing only reactants, the reaction quo-tient is zero

In a reaction mixture containing only products, the reaction quotient is infinite

In a reaction mixture containing both reactants and products, each at a concentration of1 M, the reaction quotient is one

The reaction quotient is useful because the value of Q relative to K is a measure of theprogress of the reaction toward equilibrium. At equilibrium, the reaction quotient is

Qc =(1)c(1)d

(1)a(1)b= 1

(Qc = 1):

Qc =[C]c[D]d

[0]a[0]b= q

(Qc = q):

Qc =[0]c[0]d

[A]a[B]b= 0

(Qc = 0):

Qc =[C]c[D]d

[A]a[B]b Qp =

PcCPdD

PaAPbB

Qp.

(Qc)

aA + bB Δ cC + dD

Q, K, and the Direction of a Reaction

3.5

3.0

2.5

2.0

1.5

1.0

0.5

0.0

Q o

r K

Concentration (M)ProductsReactants

1.000.000.00

0.750.25

0.500.50

0.250.751.00

3B43A4

Q � KReaction runs to left

(A B)

K � 1.45

Q �3B4

3A4Q � at 3A4 � 0, 3B4 � 1A(g) B(g)

Q � KReaction runs to right

(A B)

Q � KReaction is at equilibrium

(A B)

� FIGURE 14.7 Q, K, and the Directionof a Reaction The graph shows a plotof Q as a function of the concentra-tions of the reactants and products in asimple reaction in which

and the sum of the reactantand product concentrations is 1 M. Thefar left of the graph represents pure re-actant and the far right represents pureproduct. The midpoint of the graphrepresents an equal mixture of A andB. When Q is less than K, the reactionmoves in the forward direction

When Q is greater thanK, the reaction moves in the reverse di-rection When Q is equalto K, the reaction is at equilibrium.

(A — B).

(A ¡ B).

K = 1.45A Δ B,


equal to the equilibrium constant. Figure 14.7 � shows a plot of Q as a function of theconcentrations of A and B for the simple reaction which has an equilib-rium constant of The following points are representative of three possibleconditions:

K = 1.45.A(g) Δ B(g),

Q K Predicted Direction of Reaction

0.55 1.45 To the right (toward products)

2.55 1.45 To the left (toward reactants)

1.45 1.45 No change (at equilibrium)

For the first set of values in the table, Q is less than K and must therefore get largeras the reaction proceeds toward equilibrium. Q becomes larger as the reactant concentra-tion decreases and the product concentration increases—the reaction proceeds to theright. For the second set of values, Q is greater than K and must therefore get smaller asthe reaction proceeds toward equilibrium. Q gets smaller as the reactant concentration in-creases and the product concentration decreases—the reaction proceeds to the left. In thethird set of values, implying that the reaction is at equilibrium—the reaction willnot proceed in either direction.

Summarizing Direction of Change Predictions:The reaction quotient (Q) is a measure of the progress of a reaction toward equilibrium.

� Reaction goes to the right (toward products).

� Reaction goes to the left (toward reactants).

� Reaction is at equilibrium.Q = KQ 7 KQ 6 K

Q = K,


FOR PRACTICE 14.7Consider the following reaction and its equilibrium constant:

A reaction mixture contains and Calculateand determine the direction in which the reaction will proceed.Qc

[N2O4] = 0.0331 M.[NO2] = 0.0255 M

N2O4(g) Δ 2 NO2(g) Kc = 5.85 * 10-3

To determine the progress of the reaction relative to the equilibrium state,calculate Q.

= 10.8

=0.3552

(0.114)(0.102)

Qp =P2

ICl

PI2PCl2

Compare Q to K. Qp = 10.8; Kp = 81.9

Since the reaction is not at equilibriumand will proceed to the right.

Qp 6 Kp,

EXAMPLE 14.7 Predicting the Direction of a Reaction by Comparing Q and KConsider the following reaction and its equilibrium constant:

A reaction mixture contains and Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed?

SOLUTION

PICl = 0.355 atm.PCl2 = 0.102 atm,PI2= 0.114 atm,

I2(g) + Cl2(g) Δ 2 ICl(g) Kp = 81.9

Conceptual Connection 14.3 Q and KFor the reaction a reaction mixture at a certain temperatureinitially contains both and in their standard states (see the definition of stan-dard state in Section 6.9). If which statement is true of the reaction mixturebefore any reaction occurs?

(a) the reaction is at equilibrium.

(b) the reaction will proceed to the right.

(c) the reaction will proceed to the left.

ANSWER: (c) Since and are both in their standard states, they each have a partial pres-sure of 1.0 atm. Consequently, Since and the reaction proceeds tothe left.

Qp 7 Kp,Kp = 0.15,Qp = 1.NO2N2O4

Q 7 K;

Q 6 K;

Q = K;

Kp = 0.15,NO2N2O4

2 NO2(g),ΔN2O4(g)

14.8 Finding Equilibrium ConcentrationsIn Section 14.6, we learned how to calculate an equilibrium constant from the equilibri-um concentrations of the reactants and products. Just as commonly we will want to cal-culate equilibrium concentrations of reactants or products from the equilibrium constant.These kinds of calculations are important because they allow us to calculate the amountof a reactant or product at equilibrium. For example, in a synthesis reaction, we mightwant to know how much of the product forms when the reaction reaches equilibrium. Orfor the hemoglobin–oxygen equilibrium discussed in Section 14.1, we might want toknow the concentration of oxygenated hemoglobin present under certain oxygen concen-trations within the lungs or muscles.

We can divide these types of problems into the following two categories: (1) findingequilibrium concentrations when we know the equilibrium constant and all but one of


SORT You are given the equilibrium constant of a chemical reaction, to-gether with the equilibrium concentrations of the reactant and one product.You are asked to find the equilibrium concentration of the other product.

GIVEN:

FIND: [CO2]

Kc = 2.00 [CF4] = 0.118 M

[COF2] = 0.255 M

EXAMPLE 14.8 Finding Equilibrium Concentrations When YouKnow the Equilibrium Constant and All but One of the EquilibriumConcentrations of the Reactants and ProductsConsider the following reaction:

In an equilibrium mixture, the concentration of is 0.255 M and the concentrationof is 0.118 M. What is the equilibrium concentration of CO2 ?CF4

COF2

2 COF2(g) Δ CO2(g) + CF4(g) Kc = 2.00 at 1000 °C

the equilibrium concentrations of the reactants and products; and (2) finding equilibri-um concentrations when we know the equilibrium constant and only initial concentra-tions. The second category of problem is more difficult than the first. Let’s examineeach separately.

Finding Equilibrium Concentrations When We Know the Equilibrium Constant and All but One of the EquilibriumConcentrations of the Reactants and ProductsWe can use the equilibrium constant to calculate the equilibrium concentration of one ofthe reactants or products, given the equilibrium concentrations of the others. To solve thistype of problem, we can follow our general problem-solving procedure.

CHECK Check your answer by mentally substituting the given values of andas well as your calculated value for back into the equilibrium expression.

was found to be roughly equal to one. and Therefore is approximately 2, as given in the problem.

FOR PRACTICE 14.8Diatomic iodine decomposes at high temperature to form I atoms according to thefollowing reaction:

In an equilibrium mixture, the concentration of is 0.10 M. What is the equilibriumconcentration of I?

I2

I2(g) Δ 2 I(g) Kc = 0.011 at 1200 °C

(I2)

Kc

[CF4] L 0.12.[COF2]2 L 0.06[CO2]

Kc =[CO2][CF4]

[COF2]2

[CO2][CF4][COF2]

STRATEGIZE You can calculate the concentration of the product using thegiven quantities and the expression for Kc.

CONCEPTUAL PLAN

SOLVE Solve the equilibrium expression for and then substitute inthe appropriate values to calculate it.

[CO2] SOLUTION

= 1.10 M[CO2] = 2.00a0.2552

0.118b

[CO2] = Kc[COF2]2

[CF4]

7CO287COF28, 7CF48, Kc

7CO287CF48

7COF282Kc �


Finding Equilibrium Concentrations When We Know theEquilibrium Constant and Initial Concentrations or PressuresMore commonly, we know the equilibrium constant and only initial concentrations ofreactants and need to find the equilibrium concentrations of the reactants or products.These kinds of problems are generally more involved than those we just examined andrequire a specific procedure to solve them. The procedure has some similarities to theone used in Examples 14.5 and 14.6 in that you must set up an ICE table showing theinitial conditions, the changes, and the equilibrium conditions. However, unlike Ex-amples 14.5 and 14.6, here the changes in concentration are not known and must berepresented with the variable x. For example, consider again the following simplereaction:

Suppose that, as before (see Section 14.6), we have a reaction mixture in which the initialconcentration of A is 1.0 M and the initial concentration of B is 0.00 M. However, nowwe know the equilibrium constant, and want to find the equilibrium concen-trations. Set up the ICE table with the given initial concentrations and then represent theunknown change in [A] with the variable x as follows:

K = 0.33,

A(g) Δ 2 B(g)

[A]

Initial

Change

Equil

1.0

�x

1.0 � x

0.00

�2x

2x

[B]

Represent changesfrom initial conditions

with the variable x.

Notice that, due to the stoichiometry of the reaction, the change in [B] must be Asbefore, each equilibrium concentration is the sum of the two entries above it in the ICEtable. In order to find the equilibrium concentrations of A and B, we must find the valueof the variable x. Since we know the value of the equilibrium constant, we can use theequilibrium expression to set up an equation in which x is the only variable:

or more simply:

The above equation is a quadratic equation—it contains the variable x raised to the sec-ond power. In general, we can solve quadratic equations with the quadratic formula,which we introduce in Example 14.10. If the quadratic equation is a perfect square, how-ever, it can be solved by simpler means, as shown in Example 14.9. For both of these ex-amples, we give the general procedure in the left column and apply the procedure to thetwo different example problems in the center and right columns. Later in this section, wesee that quadratic equations can often be simplified by making some approximationsbased on our chemical knowledge.

4x2

1.0 - x= 0.33

K =[B]2

[A]=

(2x)2

1.0 - x= 0.33

+2x.

N2(g) + O2(g) Δ 2 NO(g) N2O4(g) Δ 2 NO2(g)

Initial 0.00 0.100

Change

Equil

-2x+x

[NO2][N2O4][NO]

Initial 0.200 0.200 0.00

Change

Equil

+2x-x-x

[O2][N2]


PROCEDURE FOR... Finding EquilibriumConcentrations from InitialConcentrations and theEquilibrium Constant

To solve these types of problems, followthe procedure outlined below.

EXAMPLE 14.9Finding EquilibriumConcentrations from InitialConcentrations and theEquilibrium Constant

Consider the following reaction:

A reaction mixture at 2000 °C initiallycontains and

Find the equilibrium con-centrations of the reactants and prod-ucts at this temperature.

0.200 M.[O2] =[N2] = 0.200 M

Kc = 0.10 (at 2000 °C)

N2(g) + O2(g) Δ 2 NO(g)

EXAMPLE 14.10Finding EquilibriumConcentrations from InitialConcentrations and theEquilibrium Constant

Consider the following reaction:

A reaction mixture at 100 °C initiallycontains Find theequilibrium concentrations of and

at this temperature.N2O4

NO2

[NO2] = 0.100 M.

Kc = 0.36 (at 100 °C)

N2O4(g) Δ 2 NO2(g)

1. Using the balanced equation as aguide, prepare a table showing theknown initial concentrations of thereactants and products. Leave roomin the table for the changes in concen-trations and for the equilibrium con-centrations.


2. Use the initial concentrations to calculate the reaction quotient (Q)for the initial concentrations. Com-pare Q to K to predict the directionin which the reaction will proceed. therefore the reaction will

proceed to the right.Q 6 K,

= 0

Qc =[NO]2

[N2][O2]=

0.002

(0.200)(0.200)

therefore the reaction willproceed to the left.Q 7 K,

= q

Qc =[NO2]2

[N2O4]=

(0.100)2

0.00

[NO]

Initial 0.200 0.200 0.00

Change

Equil

[O2][N2]

Initial 0.00 0.100

Change

Equil

[NO2][N2O4]

4. Sum each column for each reac-tant and product to determine theequilibrium concentrations in termsof the initial concentrations andthe variable x.


Initial 0.00 0.100

Change

Equil x 0.100 – 2x

-2x+x

[NO2][N2O4][NO]

Initial 0.200 0.200 0.00

Change

Equil 2x0.200 - x0.200 - x+2x-x-x

[O2][N2]

3. Represent the change in the con-centration of one of the reactantsor products with the variable x.Define the changes in the concen-trations of the other reactants orproducts in terms of x. It is usuallymost convenient to let x representthe change in concentration of thereactant or product with the smalleststoichiometric coefficient.

FOR PRACTICE 14.9

The above reaction is carried out at a different temperature at which

This time, however, thereaction mixture starts with only theproduct, and noreactants. Find the equilibrium con-centrations of and NO atequilibrium.

O2,N2,

[NO] = 0.0100 M,

Kc = 0.055.

FOR PRACTICE 14.10

The above reaction is carried out at thesame temperature, but this time the reac-tion mixture initially contains only thereactant, and no

Find the equilibrium concentrationsof and NO2.N2O4

NO2.[N2O4] = 0.0250 M,


5. Substitute the expressions for theequilibrium concentrations (fromstep 4) into the expression for theequilibrium constant. Using thegiven value of the equilibrium con-stant, solve the expression for thevariable x. In some cases, such asExample 14.9, you can take thesquare root of both sides of the ex-pression to solve for x. In other cases,such as Example 14.10, you mustsolve a quadratic equation to find x.

Remember the quadratic formula:

x =-b ; 2 b2 - 4ac

2a

ax2 + bx + c = 0

x = 0.027

0.063 = 2.3x

0.063 = 2x + 1 0.10x

= 2x

1 0.10(0.200) - 1 0.10x

1 0.10(0.200 - x) = 2x

1 0.10 =2x

0.200 - x

0.10 =(2x)2

(0.200 - x)2

=(2x)2

(0.200 - x)(0.200 - x)

Kc =[NO]2

[N2][O2]

x = 0.176 or x = 0.014

=0.76 ; 0.65

8

=-(-0.76) ; 2 (-0.76)2 - 4(4)(0.0100)

2(4)

x =-b ; 2 b2 - 4ac

2a

4x2 - 0.76x + 0.0100 = 0 (quadratic)

0.36x = 0.0100 - 0.400x + 4x2

0.36 =0.0100 - 0.400x + 4x2

x

=(0.100 - 2x)2

x

Kc =[NO2]2

[N2O4]

6. Substitute x into the expressionsfor the equilibrium concentrationsof the reactants and products (fromstep 4) and calculate the concentra-tions. In cases where you solved aquadratic and have two values for x,choose the value for x that gives aphysically realistic answer. For ex-ample, reject the value of x that re-sults in any negative concentrations.

= 0.054 M[NO] = 2(0.027)

= 0.173 M[O2] = 0.200 - 0.027

= 0.173 M[N2] = 0.200 - 0.027 We reject the root because it

gives a negative concentration for Using we get the followingconcentrations:

= 0.014 M[N2O4] = x

= 0.100 - 2(0.014) = 0.072 M[NO2] = 0.100 - 2x

x = 0.014,NO2.

x = 0.176

7. Check your answer by substitut-ing the computed equilibriumvalues into the equilibrium ex-pression. The calculated value ofK should match the given value ofK. Note that rounding errors couldcause a difference in the least signif-icant digit when comparing valuesof the equilibrium constant.

Since the calculated value of matches the given value (to withinone digit in the least significant fig-ure), the answer is valid.

Kc

=0.0542

(0.173)(0.173)= 0.097

Kc =[NO]2

[N2][O2]

Since the calculated value of matchesthe given value (to within one digit in theleast significant figure), the answer isvalid.

Kc

= 0.37

Kc =[NO2]2

[N2O4]=

0.0722

0.014

When the initial conditions are given in terms of partial pressures (instead of concentra-tions) and the equilibrium constant is given as instead of use the same procedure,but substitute partial pressures for concentrations, as shown in the following example.

Kc,Kp


1. Using the balanced equation as a guide, prepare a tableshowing the known initial partial pressures of thereactants and products.

I2(g) + Cl2(g) Δ 2 ICl(g)

2. Use the initial partial pressures to calculate the reactionquotient (Q). Compare Q to K to predict the direction inwhich the reaction will proceed.

therefore the reaction will proceed to the right.Q 6 K,

Kp = 81.9 (given)

Qp =P2

ICl

PI2PCl2

=0.1002

(0.100)(0.100)= 1

3. Represent the change in the partial pressure of one ofthe reactants or products with the variable x. Define thechanges in the partial pressures of the other reactants orproducts in terms of x.


(atm) (atm) (atm)

Initial 0.100 0.100 0.100

Change

Equil

PIClPCl2PI2

(atm) (atm) (atm)

Initial 0.100 0.100 0.100

Change

Equil

+2x-x-x

PIClPCl2PI2

4. Sum each column for each reactant and product todetermine the equilibrium partial pressures in terms ofthe initial partial pressures and the variable x.


5. Substitute the expressions for the equilibrium partialpressures (from step 4) into the expression for theequilibrium constant. Use the given value of theequilibrium constant to solve the expression for thevariable x.

0.805 = 11.05x

x = 0.0729

1 81.9 (0.100) - 0.100 = 2x + 1 81.9 x

1 81.9 (0.100) - 1 81.9 x = 0.100 + 2x

1 81.9 (0.100 - x) = 0.100 + 2x

1 81.9 =(0.100 + 2x)

(0.100 - x)

81.9 =(0.100 + 2x)2

(0.100 - x)2 (perfect square)

=(0.100 + 2x)2

(0.100 - x)(0.100 - x)Kp =

P2ICl

PI2PCl2

(atm) (atm) (atm)

Initial 0.100 0.100 0.100

Change

Equil 0.100 + 2x0.100 - x0.100 - x+2x-x-x

PIClPCl2PI2

6. Substitute x into the expressions for the equilibriumpartial pressures of the reactants and products (fromstep 4) and calculate the partial pressures. PICl = 0.100 + 2(0.0729) = 0.246 atm

PCl2 = 0.100 - 0.0729 = 0.027 atmPI2

= 0.100 - 0.0729 = 0.027 atm

EXAMPLE 14.11 Finding Equilibrium Partial Pressures When YouAre Given the Equilibrium Constant and Initial Partial PressuresConsider the following reaction:

A reaction mixture at initially contains andFind the equilibrium partial pressures of and ICl at this

temperature.

SOLUTIONFollow the procedure outlined in Examples 14.5 and 14.6 (using the pressures in placeof the concentrations) to solve the problem.

Cl2,I2,PICl = 0.100 atm.PCl2 = 0.100 atm,PI2

= 0.100 atm,25 °C

I2(g) + Cl2(g) Δ 2 ICl(g) Kp = 81.9 (at 25 °C)


7. Check your answer by substituting the calculatedequilibrium partial pressures into the equilibriumexpression. The calculated value of K should match thegiven value of K.

Since the calculated value of matches the given value(within the uncertainty indicated by the significant figures),the answer is valid.

Kp

= 83Kp =P2

ICl

PI2PCl2

=0.2462

(0.027)(0.027)

FOR PRACTICE 14.11The reaction between and (from Example 14.11) is carried out at the same tem-perature, but with these initial partial pressures:

Find the equilibrium partial pressures of all three substances.PICl = 0.00 atm.PCl2 = 0.150 atm,PI2

= 0.150 atm,Cl2I2

Simplifying Approximations in Working Equilibrium ProblemsFor some equilibrium problems of the type shown in the previous three examples, we canmake an approximation that simplifies the calculations without any significant loss of accu-racy. For example, if the equilibrium constant is relatively small, the reaction will not pro-ceed very far to the right. Therefore if the initial reactant concentration is relatively large,we can make the assumption that x is small relative to the initial concentration of reactant.To see how this approximation works, consider again the simple reaction Suppose that, as before, we have a reaction mixture in which the initial concentration of Ais and the initial concentration of B is 0.0 M, and that we want to find the equilibri-um concentrations. However, suppose that in this case the equilibrium constant is muchsmaller, say The ICE table is identical to the one we set up previously:K = 3.3 * 10-5.

1.0 M

A Δ 2 B.

[A] [B]

Initial 1.0 0.0

Change

Equil 2x1.0 - x+2x-x

With the exception of the value of K, we end up with the exact quadratic equation that wehad before:

or simply,

We can multiply out this quadratic equation and solve it using the quadratic formula. Butsince K is small, the reaction will not proceed very far toward products and therefore, x will also be small. If x is much smaller than 1.0, then (the quantity in thedenominator) can be approximated by (1.0).

This approximation greatly simplifies the equation, which we can then solve easily for xas follows:

We can check the validity of this approximation by comparing the computed value ofx to the number it was subtracted from. The ratio of x to the number it is subtracted fromshould be less than 0.05 (or 5%) for the approximation to be valid. In this case, x was sub-tracted from 1.0, and therefore the ratio of the value of x to 1.0 is calculated as follows:

0.0029

1.0* 100% = 0.29%

x =A

3.3 * 10-5

4= 0.0029

4x2 = 3.3 * 10-5

4x2

1.0= 3.3 * 10-5

4x2

(1.0 - x)= 3.3 * 10-5

(1.0 - x)

4x2

1.0 - x= 3.3 * 10-5

K =[B]2

[A]=

(2x)2

1.0 - x= 3.3 * 10-5


PROCEDURE FOR...Finding EquilibriumConcentrations from InitialConcentrations in Cases with a Small Equilibrium ConstantTo solve these types of problems, followthe procedure outlined below.

EXAMPLE 14.12Finding EquilibriumConcentrations from InitialConcentrations in Cases witha Small Equilibrium ConstantConsider the reaction for the decom-position of hydrogen disulfide:

A reaction vessel initiallycontains of at

Find the equilibrium concen-trations of and S2.H2

800 °C.H2S0.0125 mol

0.500-L

Kc = 1.67 * 10-7 at 800 °C

2 H2S(g) Δ 2 H2(g) + S2(g)

EXAMPLE 14.13Finding EquilibriumConcentrations from InitialConcentrations in Cases witha Small Equilibrium ConstantConsider the reaction for the decompo-sition of hydrogen disulfide:

A reaction vessel initially con-tains of at

Find the equilibrium concentra-tions of and S2.H2

800 °C.H2S1.25 * 10-4 mol

0.500-L

Kc = 1.67 * 10-7 at 800 °C

2 H2S(g) Δ 2 H2(g) + S2(g)

1. Using the balanced equation as aguide, prepare a table showing theknown initial concentrations of the re-actants and products. (In these exam-ples, you must first calculate theconcentration of from the givennumber of moles and volume.)

H2S

2 H2S(g) Δ 2 H2(g) + S2(g)

[H2S] =0.0125 mol

0.500 L= 0.0250 M

2 H2S(g) Δ 2 H2(g) + S2(g)

= 2.50 * 10-4 M

[H2S] =1.25 * 10-4 mol

0.500 L

Initial 0.00 0.00

Change

Equil

2.50 * 10-4

[S2][H2][H2S]

Initial 0.0250 0.00 0.00

Change

Equil

[S2][H2][H2S]

The approximation is therefore valid. In the two side-by-side examples that follow, wetreat two nearly identical problems—the only difference is the initial concentration of thereactant. In Example 14.12, the initial concentration of the reactant is relatively large,the equilibrium constant is small, and the x is small approximation works well. In Exam-ple 14.13, however, the initial concentration of the reactant is much smaller, and eventhough the equilibrium constant is the same, the x is small approximation does not work(because the initial concentration is also small). In cases such as this, we have a couple ofoptions to solve the problem. We can either solve the equation exactly (using the quadraticformula, for example), or we can use the method of successive approximations, which isintroduced in Example 14.13. In this method, we essentially solve for x as if it weresmall, and then substitute the value obtained back into the equation to solve for x again.This can be repeated until the calculated value of x stops changing with each iteration, anindication that we have arrived at an acceptable value for x.

Note that the x is small approximation does not imply that x is zero. If that were thecase, the reactant and product concentrations would not change from their initial values.The x is small approximation just means that when x is added or subtracted to anothernumber, it does not change that number by very much. For example, we can calculate thevalue of the difference when

Since the value of 1.0 is known only to two significant figures, subtracting the small xdoes not change the value at all. This situation is similar to weighing yourself on a bath-room scale with and without a penny in your pocket. Unless your scale is unusually pre-cise, removing the penny from your pocket will not change the reading on the scale. Thisdoes not imply that the penny is weightless, only that its weight is small when comparedto your body weight. The weight of the penny can thus be neglected in reading yourweight with no detectable loss in accuracy.

= 1.0= 0.99971.0 - x = 1.0 - 3.0 * 10-4

x = 3.0 * 10-4:1.0 - x


2. Use the initial concentrations to cal-culate the reaction quotient (Q). Com-pare Q to K to predict the directionthat the reaction will proceed.

By inspection, the reactionwill proceed to the right.

Q = 0; By inspection, the reaction willproceed to the right.

Q = 0;

3. Represent the change in the concen-tration of one of the reactants orproducts with the variable x. Definethe changes in the concentrations ofthe other reactants or products withrespect to x.

2 H2S(g) Δ 2 H2(g) + S2(g) 2 H2S(g) Δ 2 H2(g) + S2(g)

4. Sum each column for each reactantand product to determine the equi-librium concentrations in terms ofthe initial concentrations and thevariable x.

2 H2S(g) Δ 2 H2(g) + S2(g) 2 H2S(g) Δ 2 H2(g) + S2(g)

[H2S] [H2] [S2]

Initial 0.0250 0.00 0.00

Change

Equil x2x0.0250 - 2x

+x+2x-2x

[H2S] [H2] [S2]

Initial 0.00 0.00

Change

Equil

+x+2x-2x

2.50 * 10-4

[H2S] [H2] [S2]

Initial 0.00 0.00

Change

Equil x2x2.50 * 10-4 - 2x

+x+2x-2x

2.50 * 10-4

5. Substitute the expressions for theequilibrium concentrations (from step4) into the expression for the equilib-rium constant. Use the given value ofthe equilibrium constant to solve theexpression for the variable x.

x = 2.97 * 10-4

x3 =6.25 * 10-4(1.67 * 10-7)

4

6.25 * 10-4(1.67 * 10-7) = 4x3

1.67 * 10-7 =4x3

6.25 * 10-4

1.67 * 10-7 =4x3

(0.0250 - 2x)2

=(2x)2(x)

(0.0250 - 2x)2

Kc =[H2]2[S2]

[H2S]2

x = 1.38 * 10-5

x3 =6.25 * 10-8(1.67 * 10-7)

4

6.25 * 10-8(1.67 * 10-7) = 4x3

1.67 * 10-7 =4x3

6.25 * 10-8

1.67 * 10-7 =4x3

(2.50 * 10-4 - 2x)2

=(2x)2x

(2.50 * 10-4 - 2x)2

Kc =[H2]2[S2]

[H2S]2

In this case, the resulting equation iscubic in x. Although cubic equationscan be solved, the solutions are notusually simple. However, since theequilibrium constant is small, weknow that the reaction does not pro-ceed very far to the right. Therefore, xwill be a small number and can bedropped from any quantities inwhich it is added to or subtractedfrom another number (as long as thenumber itself is not too small).

Check whether your approximationwas valid by comparing the calcu-lated value of x to the number it wasadded to or subtracted from. The ratioof the two numbers should be lessthan 0.05 (or 5%) for the approxima-tion to be valid. If approximation isnot valid, proceed to step 5a.

Checking the x is small approximation: Checking the x is small approximation:

1.38 * 10-5

2.50 * 10-4 * 100% = 5.52%

1.67 � 10�7 �(0.0250 � 2x)2

4x3x is small.

1.67 � 10�7 �

x is small.

(2.50 � 10�4 � 2x)24x3

2.97 * 10-4

0.0250* 100% = 1.19%

The x is small approximation is valid,proceed to step 6.

The approximation does not satisfy therule (although it is close).65%

[H2S] [H2] [S2]

Initial 0.0250 0.00 0.00

Change

Equil

+x+2x-2x


If we substitute this value of x backinto the cubic equation and solve it, weget which is nearlyidentical to Therefore,we have arrived at the best approxima-tion for x.

1.27 * 10-5.x = 1.28 * 10-5,

Substitute the value obtained for x instep 5 back into the original cubicequation, but only at the exact spotwhere x was assumed to be negligi-ble and then solve the equation for xagain. Continue this procedure untilthe value of x obtained from solvingthe equation is the same as the onethat is substituted into the equation.

[S2] = 1.28 * 10-5 M = 2.56 * 10-5 M

[H2] = 2(1.28 * 10-5)= 2.24 * 10-4 M

2.50 * 10-4 - 2(1.28 * 10-5)[H2S] =

[S2] = 2.97 * 10-4 M= 5.94 * 10-4 M

[H2] = 2(2.97 * 10-4)= 0.0244 M

[H2S] = 0.0250 - 2(2.97 * 10-4)

7. Check your answer by substitutingthe calculated equilibrium valuesinto the equilibrium expression. Thecalculated value of K should matchthe given value of K. Note that theapproximation method and roundingerrors could cause a difference of upto about 5% when comparing valuesof the equilibrium constant.

6. Substitute x into the expressions forthe equilibrium concentrations of thereactants and products (from step 4)and calculate the concentrations.

The calculated value of K is equal tothe given value. Therefore the answeris valid.

= 1.67 * 10-7

Kc =(2.56 * 10-5)2(1.28 * 10-5)

(2.24 * 10-4)2

The calculated value of K is closeenough to the given value when weconsider the uncertainty introducedby the approximation. Therefore theanswer is valid.

= 1.76 * 10-7

Kc =(5.94 * 10-4)2(2.97 * 10-4)

(0.0244)2

FOR PRACTICE 14.12The reaction in Example 14.12 is car-ried out at the same temperature withthe following initial concentrations:

and Find the equi-librium concentration of [S2].

[S2] = 0.00 M.[H2] = 0.100 M,[H2S] = 0.100 M,

FOR PRACTICE 14.13The reaction in Example 14.13 is car-ried out at the same temperature withthe following initial concentrations:

and Find the equilibrium concentration of[S2].

[S2] = 0.00 M.[H2] = 0.00 M,[H2S] = 1.00 * 10-4

M,

Conceptual Connection 14.4 The x is small ApproximationFor the generic reaction, A(g) , consider each value of K and initial concen-tration of A. For which set will the x is small approximation most likely apply?

(a) (b)

(c) (d)

ANSWER: (a) The x is small approximation is most likely to apply to a reaction with a small equi-librium constant and an initial concentration of reactant that is not too small. The bigger the equi-librium constant and the smaller the initial concentration of reactant, the less likely that the x issmall approximation will apply.

K = 1.0 * 10-2; [A] = 0.00250 MK = 1.0 * 10-5; [A] = 0.00250 M

K = 1.0 * 10-2; [A] = 0.250 MK = 1.0 * 10-5; [A] = 0.250 M

Δ B(g)

5a. If the approximation is not valid, youcan either solve the equation exactly(by hand or with your calculator), oruse the method of successive approxi-mations. In this case, we use themethod of successive approximations.

1.67 � 10�7 �(2.50 � 10�4 � 2x)2

4x3

x � 1.38 � 10�5

1.67 * 10-7 =4x3

(2.50 * 10-4 - 2.76 * 10-5)2

x = 1.27 * 10-5


System responds tominimize disturbance.

Equilibrium is disturbed whenthe population in Country B

grows.

Le Châtelier’s Principle: An Analogy

Country ACountry B

Country ACountry B

� FIGURE 14.8 A Population Analogyfor Le Châtelier’s Principle A babyboom in Country B shifts the equilibri-um to the left. People leave Country B(because it has become too crowded)and migrate to Country A until equi-librium is reestablished.

14.9 Le Châtelier’s Principle: How a System at Equilibrium Responds to DisturbancesWe have seen that a chemical system not in equilibrium tends to progress toward equilib-rium and that the relative concentrations of the reactants and products at equilibrium arecharacterized by the equilibrium constant, K. What happens, however, when a chemicalsystem already at equilibrium is disturbed? Le Châtelier’s principle states that thechemical system will respond to minimize the disturbance.

Le Châtelier’s principle: When a chemical system at equilibrium is dis-turbed, the system shifts in a direction that minimizes the disturbance.

In other words, a system at equilibrium tends to maintain that equilibrium—it bouncesback when disturbed.

We can understand Le Châtelier’s principle by returning to our two neighboringcountries analogy. Suppose the populations of Country A and Country B are at equilibri-um. This means that the rate of people moving out of Country A and into Country B isequal to the rate of people moving into Country A and out of Country B, and the popula-tions of the two countries are stable.

Now imagine disturbing the balance (Figure 14.8 �). Suppose there is a notable increasein the birthrate in Country B. What happens? After Country B becomes more crowded,the rate of people leaving Country B increases. The net flow of people is out of CountryB and into Country A. Equilibrium was disturbed by the addition of more people toCountry B, and people left Country B in response. In effect, the system responded byshifting in the direction that minimized the disturbance.

On the other hand, what happens if there is a baby boom in Country A instead? AsCountry A gets more crowded, the rate of people leaving Country A increases. The netflow of people is out of Country A and into Country B. The number of people in CountryA increased and the system responded; people moved out of Country A. Chemical

Country A Δ Country B

Pronounced “Le-sha-te-lyay.”

The two-country analogy should helpyou see the effects of disturbing asystem in equilibrium—it should not betaken as an exact parallel.

2 NO2(g)N2O4 (g)

Reaction shifts left.

Add NO2.


Le Châtelier’s Principle: Changing Concentration

Add moreNO2.

Equilibriumis disturbed.

MoreN2O4forms.

Add NO2 System shifts left.

N2O4(g) 2 NO2(g)

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

� FIGURE 14.9 Le Châtelier’s Principle: The Effect of a Concentration Change Adding causesthe reaction to shift left, consuming some of the added and forming more N2O4.NO2

NO2

systems behave similarly: When their equilibrium is disturbed, they react to counter thedisturbance. We can disturb a system in chemical equilibrium in several different ways,including changing the concentration of a reactant or product, changing the volume orpressure, and changing the temperature. We consider each of these separately.

The Effect of a Concentration Change on EquilibriumConsider the following reaction in chemical equilibrium:

Suppose we disturb the equilibrium by adding to the equilibrium mixture(Figure 14.9 �). In other words, we increase the concentration of Whathappens? According to Le Châtelier’s principle, the system shifts in a direction to mini-mize the disturbance. The reaction goes to the left (it proceeds in the reverse direction),consuming some of the added thus bringing its concentration back down, as depict-ed graphically in Figure 14.10(a) �.

NO2

NO2, the product.NO2

N2O4(g) Δ 2 NO2(g)

The reaction shifts to the left because the value of Q changes as follows:

• Before addition of

• Immediately after addition of

• Reaction shifts to left to reestablish equilibrium.

NO2 : Q 7 K.

NO2 : Q = K.

14.9 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances 643C

once

ntr

atio

n

Time

(a)

Con

cen

trat

ion

Time

(b)

[NO2][NO2]

[N2O4][N2O4]

Originalequilibrium

Q � KEquilibrium isreestablished.

Q � K

Disturbequilibriumby adding

NO2. Q K

Originalequilibrium

Q � K

Equilibrium isreestablished.

Q � K

Disturbequilibriumby adding

N2O4. Q K

Le Châtelier’s Principle: Graphical Representation

N2O4(g) 2 NO2(g)

� FIGURE 14.10 Le Châtelier’s Principle: Changing Concentration The graph shows the concentra-tions of and for the reaction in three distinct stages of the reac-tion: initially at equilibrium (left); upon disturbance of the equilibrium by addition of more (a) or (b) to the reaction mixture (center); and upon reestablishment of equilibrium (right).N2O4

NO2

N2O4(g) ¡ 2 NO2(g)N2O4NO2

2 NO2(g)N2O4(g)

Reaction shifts right.

Add N2O4.

On the other hand, what happens if we add extra increasing its con-centration? In this case, the reaction shifts to the right, consuming some of the added

and bringing its concentration back down, as shown in Figure 14.10(b) �.N2O4

N2O4 (the reactant)

The reaction shifts to the right because the value of Q changes as follows:

• Before addition of • Immediately after addition of • Reaction shifts to right to reestablish equilibrium.

In both of the cases, the system shifts in a direction that minimizes the disturbance.Lowering the concentration of a reactant (which makes ) causes the system to shiftin the direction of the reactants to minimize the disturbance. Lowering the concentration ofa product (which makes ) causes the system to shift in the direction of products.

Summarizing the Effect of a Concentration Change on Equilibrium:If a chemical system is at equilibrium:

� Increasing the concentration of one or more of the reactants (which makes )causes the reaction to shift to the right (in the direction of the products).

� Increasing the concentration of one or more of the products (which makes )causes the reaction to shift to the left (in the direction of the reactants).

� Decreasing the concentration of one or more of the reactants (which makes )causes the reaction to shift to the left (in the direction of the reactants).

� Decreasing the concentration of one or more of the products (which makes )causes the reaction to shift to the right (in the direction of the products).

Q 6 K

Q 7 K

Q 7 K

Q 6 K

Q 6 K

Q 7 K

N2O4 : Q 6 K.N2O4 : Q = K.


In considering the effect of a change involume, we are assuming that thechange in volume is carried out atconstant temperature.

EXAMPLE 14.14 The Effect of a Concentration Change on EquilibriumConsider the following reaction at equilibrium.

What is the effect of adding additional to the reaction mixture? What is the effectof adding additional

SOLUTIONAdding additional increases the concentration of and causes the reaction toshift to the left. Adding additional however, does not increase the concentrationof because is a solid and therefore has a constant concentration. Thus,adding additional has no effect on the position of the equilibrium. (Note that, aswe saw in Section 14.5, solids are not included in the equilibrium expression.)

FOR PRACTICE 14.14Consider the following reaction in chemical equilibrium:

What is the effect of adding additional to the reaction mixture? What is the effectof adding additional BrNO?

Br2

2 BrNO(g) Δ 2 NO(g) + Br2(g)

CaCO3

CaCO3CaCO3

CaCO3,CO2CO2

CaCO3 ?CO2


The Effect of a Volume (or Pressure) Change on EquilibriumHow does a system in chemical equilibrium respond to a volume change? Remember fromChapter 5 that changing the volume of a gas (or a gas mixture) results in a change in pres-sure. Remember also that pressure and volume are inversely related: a decrease in volumecauses an increase in pressure, and an increase in volume causes a decrease in pressure. So,if the volume of a reaction mixture at chemical equilibrium is changed, the pressurechanges and the system will shift in a direction to minimize that change. For example, con-sider the following reaction at equilibrium in a cylinder equipped with a moveable piston:

What happens if we push down on the piston, lowering the volume and raising the pres-sure (Figure 14.11a �)? How can the chemical system respond to bring the pressure backdown? Look carefully at the reaction coefficients. If the reaction shifts to the right, of gas particles are converted to of gas particles. From the ideal gas law

we know that lowering the number of moles of a gas (n) results in a lowerpressure (P). Therefore, the system shifts to the right, lowering the number of gas mole-cules and bringing the pressure back down, minimizing the disturbance.

Consider the same reaction mixture at equilibrium again. What happens if, this time,we pull up on the piston, increasing the volume (Figure 14.11b)? The higher volume re-sults in a lower pressure and the system responds to bring the pressure back up. It doesthis by shifting to the left, converting of gas particles into of gas particles, in-creasing the pressure and minimizing the disturbance.

Consider again the same reaction mixture at equilibrium. What happens if, this time,we keep the volume the same, but increase the pressure by adding an inert gas to the mix-ture? Although the overall pressure of the mixture increases, the partial pressures of thereactants and products do not change. Consequently, there is no effect and the reactiondoes not shift in either direction.

Summarizing the Effect of Volume Change on Equilibrium:If a chemical system is at equilibrium:

� Decreasing the volume causes the reaction to shift in the direction that has the fewermoles of gas particles.

� Increasing the volume causes the reaction to shift in the direction that has thegreater number of moles of gas particles.

4 mol2 mol

(PV = nRT),2 mol

4 mol

N2(g) + 3 H2(g) Δ 2 NH3(g)


Le Châtelier’s Principle: Changing Pressure

4 mol of gas

N2(g) 2 NH3(g)3 H2(g)�

2 mol of gas 4 mol of gas

N2(g) 2 NH3(g)3 H2(g)�

2 mol of gas

Decreasepressure

Increasepressure

Reaction shifts right(toward side with fewermoles of gas particles).

Reaction shifts left(toward side with moremoles of gas particles).

(a) (b)

� FIGURE 14.11 Le Châtelier’s Princi-ple: The Effect of a Pressure Change(a) Decreasing the volume increasesthe pressure, causing the reaction toshift to the right (fewer moles of gas,lower pressure). (b) Increasing the vol-ume reduces the pressure, causing thereaction to shift to the left (more molesof gas, higher pressure).

� If a reaction has an equal number of moles of gas on both sides of the chemicalequation, then a change in volume produces no effect on the equilibrium.

� Adding an inert gas to the mixture at a fixed volume has no effect on the equilibrium.

EXAMPLE 14.15 The Effect of a Volume Change on EquilibriumConsider the following reaction at chemical equilibrium:

What is the effect of decreasing the volume of the reaction mixture? Increasing the vol-ume of the reaction mixture? Adding an inert gas at constant volume?

SOLUTIONThe chemical equation has 3 mol of gas on the right and zero moles of gas on the left.Decreasing the volume of the reaction mixture increases the pressure and causes the re-action to shift to the left (toward the side with fewer moles of gas particles). Increasingthe volume of the reaction mixture decreases the pressure and causes the reaction toshift to the right (toward the side with more moles of gas particles.) Adding an inert gashas no effect.

FOR PRACTICE 14.15Consider the following reaction at chemical equilibrium:

What is the effect of decreasing the volume of the reaction mixture? Increasing thevolume of the reaction mixture?

2 SO2(g) + O2(g) Δ 2 SO3(g)

2 KClO3(s) Δ 2 KCl(s) + 3 O2(g)


The Effect of a Temperature Change on EquilibriumAccording to Le Châtelier’s principle, if the temperature of a system at equilibrium ischanged, the system will shift in a direction to counter that change. So, if the temperatureis increased, the reaction will shift in the direction that tends to decrease the temperatureand vice versa. Recall from Chapter 6 that an exothermic reaction (negative ) emitsheat:

We can think of heat as a product in an exothermic reaction. In an endothermic reaction(positive ), the reaction absorbs heat.

We can think of heat as a reactant in an endothermic reaction.At constant pressure, raising the temperature of an exothermic reaction—think of

this as adding heat—is similar to adding more product, causing the reaction to shift left.For example, the reaction of nitrogen with hydrogen to form ammonia is exothermic.

Endothermic reaction: A + B + heat Δ C + D

¢H

Exothermic reaction: A + B Δ C + D + heat

¢H

In considering the effect of a change intemperature, we are assuming that theheat is added (or removed) at constantpressure.

2 NH3(g)3 H2(g)N2(g)

Reaction shifts left.Smaller K

heat��

Add heat

Raising the temperature of an equilibrium mixture of these three gases causes the reac-tion to shift left, absorbing some of the added heat and forming less products and morereactants. Note that, unlike adding additional to the reaction mixture (which doesnot change the value of the equilibrium constant), changing the temperature does changethe value of the equilibrium constant. The new equilibrium mixture will have more reac-tants and fewer products and therefore a smaller value of K.

Conversely, lowering the temperature causes the reaction to shift right, releasing heatand producing more products because the value of K has increased.

NH3

2 NO2(g)heat

Reaction shifts right. Larger K

colorless brown

N2O4(g) �

Add heat

2NH3(g)N2(g) � 3 H2(g)

Reaction shifts right.Larger K

heat�

Remove heat

In contrast, for an endothermic reaction, raising the temperature (adding heat) caus-es the reaction to shift right to absorb the added heat. For example, the following reactionis endothermic.


2 NO2(g)N2O4(g) � heat

Reaction shifts left. Smaller K

colorless brown

Remove heat

Raising the temperature of an equilibrium mixture of these two gases causes the reactionto shift right, absorbing some of the added heat and producing more products because thevalue of K has increased. Since is colorless and is brown, the effects of chang-ing the temperature of this reaction are easily seen (Figure 14.12 �). On the other hand,lowering the temperature (removing heat) of a reaction mixture of these two gases caus-es the reaction to shift left, releasing heat, forming less products, and lowering the valueof K.

NO2N2O4

Le Châtelier’s Principle: Changing Temperature

N2O4(g) heat 2 NO2(g)colorless brown

�

Higher temperature:NO2 favored

Lower temperature:N2O4 favored

� FIGURE 14.12 Le Châtelier’s Principle: The Effect of a Temperature Change Because the reactionis endothermic, raising the temperature causes a shift to the right, toward the formation of brown NO2.

Summarizing the Effect of a Temperature Change on Equilibrium:In an exothermic chemical reaction, heat is a product.

� Increasing the temperature causes an exothermic reaction to shift left (in the directionof the reactants); the value of the equilibrium constant decreases.

� Decreasing the temperature causes an exothermic reaction to shift right (in thedirection of the products); the value of the equilibrium constant increases.

In an endothermic chemical reaction, heat is a reactant.

� Increasing the temperature causes an endothermic reaction to shift right (in thedirection of the products); the equilibrium constant increases.

� Decreasing the temperature causes an endothermic reaction to shift left (in the direc-tion of the reactants); the equilibrium constant decreases.

Adding heat favors the endothermicdirection. Removing heat favors theexothermic direction.


EXAMPLE 14.16 The Effect of a Temperature Change on EquilibriumThe following reaction is endothermic:

What is the effect of increasing the temperature of the reaction mixture? Decreasing thetemperature?

SOLUTIONSince the reaction is endothermic, we can think of heat as a reactant:

Raising the temperature is equivalent to adding a reactant, causing the reaction to shiftto the right. Lowering the temperature is equivalent to removing a reactant, causing thereaction to shift to the left.

FOR PRACTICE 14.16The following reaction is exothermic.

What is the effect of increasing the temperature of the reaction mixture? Decreasingthe temperature?

2 SO2(g) + O2(g) Δ 2 SO3(g)

Heat + CaCO3(s) Δ CaO(s) + CO2(g)


CHAPTER IN REVIEW

Key TermsSection 14.2reversible (615)dynamic equilibrium (615)

Section 14.3equilibrium constant (K) (618)law of mass action (618)

Section 14.7reaction quotient (Q) (629)

Section 14.9Le Châtelier’s principle (641)

Key ConceptsThe Equilibrium Constant (14.1)The relative concentrations of the reactants and the products at equi-librium are expressed by the equilibrium constant, K. The equilibri-um constant measures how far a reaction proceeds toward products:a large K (much greater than 1) indicates a high concentration ofproducts at equilibrium and a small K (less than 1) indicates a lowconcentration of products at equilibrium.

Dynamic Equilibrium (14.2)Most chemical reactions are reversible; they can proceed in either theforward or the reverse direction. When a chemical reaction is in dy-namic equilibrium, the rate of the forward reaction equals the rate ofthe reverse reaction, so the net concentrations of reactants and prod-ucts do not change. However, this does not imply that the concentra-tions of the reactants and the products are equal at equilibrium.

The Equilibrium Constant Expression (14.3)The equilibrium constant expression is given by the law of mass ac-tion and is equal to the concentrations of the products, raised to theirstoichiometric coefficients, divided by the concentrations of the re-actants, raised to their stoichiometric coefficients. When the equa-tion for a chemical reaction is reversed, multiplied, or added toanother equation, K must be modified accordingly.

The Equilibrium Constant, K (14.4)The equilibrium constant can be expressed in terms of concentra-tions or in terms of partial pressures These two constantsare related by Equation 14.2. Concentration must always be ex-pressed in units of molarity for Partial pressures must always beexpressed in units of atmospheres for

States of Matter and the Equilibrium Constant (14.5)The equilibrium constant expression contains only partial pressuresor concentrations of reactants and products that exist as gases orsolutes dissolved in solution. Pure liquids and solids are not includedin the expression for the equilibrium constant.

Calculating K (14.6)The equilibrium constant can be calculated from equilibrium con-centrations or partial pressures by substituting measured values intothe expression for the equilibrium constant (as obtained from the lawof mass action). In most cases, the equilibrium concentrations of thereactants and products—and therefore the value of the equilibriumconstant—can be calculated from the initial concentrations of the re-actants and products and the equilibrium concentration of just onereactant or product.

Kp.Kc.

(Kp).(Kc)

Chapter in Review 649

Key Equations and RelationshipsExpression for the Equilibrium Constant, Kc (14.3)

Relationship between the Equilibrium Constant and the Chemical Equation (14.3)

1. If you reverse the equation, invert the equilibrium constant.2. If you multiply the coefficients in the equation by a factor, raise

the equilibrium constant to the same factor.3. If you add two or more individual chemical equations to obtain an

overall equation, multiply the corresponding equilibrium con-stants by each other to obtain the overall equilibrium constant.

Expression for the Equilibrium Constant, Kp (14.4)

Kp =PcCP

dD

PaAPbB

(equilibrium partial pressures only)

aA + bB Δ cC + dD

K =[C]c[D]d

[A]a[B]b (equilibrium concentrations only)

aA + bB Δ cC + dD

Relationship between the Equilibrium Constants, Kc and Kp (14.4)

The Reaction Quotient, Qc (14.7)

The Reaction Quotient, Qp (14.7)

Relationship of Q to the Direction of the Reaction (14.7)

Q = K Reaction is at equilibrium.

Q 7 K Reaction goes to the left.

Q 6 K Reaction goes to the right.

QP =PcCP

dD

PaAPbB

(partial pressures at any point in the reaction)

aA + bB Δ cC + dD

Qc =[C]c[D]d

[A]a[B]b (concentrations at any point in the reaction)

aA + bB Δ cC + dD

Kp = Kc(RT)¢n

The Reaction Quotient, Q (14.7)The ratio of the concentrations (or partial pressures) of productsraised to their stoichiometric coefficients to the concentrations of re-actants raised to their stoichiometric coefficients at any point in thereaction is called the reaction quotient, Q. Like K, Q can be ex-pressed in terms of concentrations or partial pressures Atequilibrium, Q is equal to K; therefore, the direction in which a reac-tion will proceed can be determined by comparing Q to K. If the reaction moves in the direction of the products; if the re-action moves in the reverse direction.

Finding Equilibrium Concentrations (14.8)There are two general types of problems in which K is given and one(or more) equilibrium concentrations can be found: (1) K, initial con-

Q 7 K,Q 6 K,

(Qp).(Qc)

centrations, and (at least) one equilibrium concentration are given;and (2) K and only initial concentrations are given. The first type issolved by rearranging the law of mass action and substituting givenvalues. The second type is solved by using a variable x to representthe change in concentration.

Le Châtelier’s Principle (14.9)When a system at equilibrium is disturbed—by a change in theamount of a reactant or product, a change in volume, or a change intemperature—the system shifts in the direction that minimizes thedisturbance.

Key SkillsExpressing Equilibrium Constants for Chemical Equations (14.3)

• Example 14.1 • For Practice 14.1 • Exercises 21, 22

Manipulating the Equilibrium Constant to Reflect Changes in the Chemical Equation (14.3)• Example 14.2 • For Practice 14.2 • For More Practice 14.2 • Exercises 27–30

Relating Kp and Kc (14.4)• Example 14.3 • For Practice 14.3 • Exercises 31, 32

Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid (14.5)• Example 14.4 • For Practice 14.4 • Exercises 33, 34

Finding Equilibrium Constants from Experimental Concentration Measurements (14.6)• Examples 14.5, 14.6 • For Practice 14.5, 14.6 • Exercises 35, 36, 41, 42

Predicting the Direction of a Reaction by Comparing Q and K (14.7)• Example 14.7 • For Practice 14.7 • Exercises 45–48


Calculating Equilibrium Concentrations from the Equilibrium Constant and One or More Equilibrium Concentrations (14.8)

• Example 14.8 • For Practice 14.8 • Exercises 37–44

Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant (14.8)• Examples 14.9, 14.10 • For Practice 14.9, 14.10 • Exercises 51–56

Calculating Equilibrium Partial Pressures from the Equilibrium Constant and Initial Partial Pressures (14.8)• Example 14.11 • For Practice 14.11 • Exercises 57, 58

Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant (14.8)• Examples 14.12, 14.13 • For Practice 14.12, 14.13 • Exercises 59, 60

Determining the Effect of a Concentration Change on Equilibrium (14.9)• Example 14.14 • For Practice 14.14 • Exercises 61–64

Determining the Effect of a Volume Change on Equilibrium (14.9)• Example 14.15 • For Practice 14.15 • Exercises 65, 66

Determining the Effect of a Temperature Change on Equilibrium (14.9)• Example 14.16 • For Practice 14.16 • Exercises 67, 68

EXERCISES

Review Questions1. How does a developing fetus in the womb get oxygen?

2. What is dynamic equilibrium? Why is it called dynamic?

3. Give the general expression for the equilibrium constant of thefollowing generic reaction:

4. What is the significance of the equilibrium constant? What doesa large equilibrium constant tell us about a reaction? A smallone?

5. What happens to the value of the equilibrium constant for a re-action if the reaction equation is reversed? Multiplied by aconstant?

6. If two reactions sum to an overall reaction, and the equilibriumconstants for the two reactions are and what is the equi-librium constant for the overall reaction?

7. Explain the difference between and For a given reaction,how are the two constants related?

8. What units should be used when expressing concentrations orpartial pressures in the equilibrium constant? What are the unitsof and Explain.

9. Why are the concentrations of solids and liquids omitted fromequilibrium expressions?

10. Does the value of the equilibrium constant depend on the initialconcentrations of the reactants and products? Do the equilibri-um concentrations of the reactants and products depend on theirinitial concentrations? Explain.

11. Explain how you might deduce the equilibrium constant for areaction in which you know the initial concentrations of the re-actants and products and the equilibrium concentration of onlyone reactant or product.

12. What is the definition of the reaction quotient (Q) for a reac-tion? What does Q measure?

Kc ?Kp

Kp.Kc

K2,K1

aA + bB Δ cC + dD.

13. What is the value of Q when each reactant and product is in itsstandard state? (See Section 6.9 for the definition of standardstates.)

14. In what direction will a reaction proceed for each condition: (a) (b) and (c)

15. Many equilibrium calculations involve finding the equilibriumconcentrations of reactants and products given their initial con-centrations and the equilibrium constant. Outline the generalprocedure used in solving these kinds of problems.

16. In equilibrium problems involving equilibrium constants thatare small relative to the initial concentrations of reactants, wecan often assume that the quantity x (which represents how farthe reaction proceeds toward products) is small. When this as-sumption is made, the quantity x can be ignored when it is sub-tracted from a large number, but not when it is multiplied by alarge number. In other words, but Explain why a small x can be ignored in the first case, but not inthe second.

17. What happens to a chemical system at equilibrium when thatequilibrium is disturbed?

18. What is the effect of a change in concentration of a reactant orproduct on a chemical reaction initially at equilibrium?

19. What is the effect of a change in volume on a chemical reaction(that includes gaseous reactants or products) initially atequilibrium?

20. What is the effect of a temperature change on a chemical reac-tion initially at equilibrium? How does the effect differ for anexothermic reaction compared to an endothermic one?

2.5x Z 2.5.2.5 - x L 2.5,

Q = K?Q 7 K;Q 6 K;

Exercises 651

Problems by TopicEquilibrium and the Equilibrium Constant Expression21. Write an expression for the equilibrium constant of each chem-

ical equation.a.

b.

c.

d.

22. Find and fix each mistake in the equilibrium constantexpressions.

a.

b.

23. When this reaction comes to equilibrium, will the concentra-tions of the reactants or products be greater? Does the answer tothis question depend on the initial concentrations of the reac-tants and products?

24. Ethene can be halogenated by this reaction:

where can be (green), (brown), or (purple). Exam-ine the three figures representing equilibrium concentrations inthis reaction at the same temperature for the three differenthalogens. Rank the equilibrium constants for the three reactionsfrom largest to smallest.

I2Br2Cl2X2

C2H4(g) + X2(g) Δ C2H4X2(g)

(C2H4)

A(g) + B(g) Δ 2 C(g) Kc = 1.4 * 10-5

Keq =[CO][Cl2]

[COCl2]CO(g) + Cl2(g) Δ COCl2(g)

Keq =[H2][S2]

[H2S]2 H2S(g) Δ 2 H2(g) + S2(g)

2 CO(g) + O2(g) Δ 2 CO2(g)

CH4(g) + 2 H2S(g) Δ CS2(g) + 4 H2(g)

2 BrNO(g Δ 2 NO(g) + Br2(g)

SbCl5(g) Δ SbCl3(g) + Cl2(g)

25. and are combined in a flask and allowed to react accord-ing to the reaction:

Examine the figures below (sequential in time) and answer thequestions:a. Which figure represents the point at which equilibrium is

reached?

H2(g) + I2(g) Δ 2 HI(g)

I2H2

(a) (b)

(c)

(i) (ii)

(iii) (iv)

(v) (vi)

b. How would the series of figures change in the presence of acatalyst?

c. Would the final figure (vi) have different amounts of reac-tants and products in the presence of a catalyst?

26. A chemist trying to synthesize a particular compound attemptstwo different synthesis reactions. The equilibrium constants forthe two reactions are 23.3 and at room temperature.However, upon carrying out both reactions for 15 minutes, thechemist finds that the reaction with the smaller equilibrium con-stant produces more of the desired product. Explain how thismight be possible.

27. The reaction below has an equilibrium constant of

Calculate for each of the reactions and predict whether reac-tants or products will be favored at equilibrium:

a.

b.

c.

28. The reaction below has an equilibrium constant

2 COF2(g) Δ CO2(g) + CF4(g)

106 at 298 K.

Kp = 2.2 *2 CH3OH(g) Δ 2 CO(g) + 4 H2(g)

12 CO(g) + H2(g) Δ 1

2 CH3OH(g)

CH3OH(g) Δ CO(g) + 2 H2(g)

Kp


104 at 298 K.Kp = 2.26 *

2.2 * 104


Calculate for each of the reactions and predict whether reac-tants or products will be favored at equilibrium:

a.

b.

c.

29. Consider the reactions and their respective equilibrium con-stants:

Use these reactions and their equilibrium constants to predictthe equilibrium constant for the following reaction:

30. Use the reactions below and their equilibrium constants to predictthe equilibrium constant for the reaction,

Kp, Kc, and Heterogeneous Equilibria

31. Calculate for each reaction:

a.

b.

c.

32. Calculate for each reaction:

a.

b.

c.

33. Write an equilibrium expression for each chemical equation in-volving one or more solid or liquid reactants or products.

a.

b.

c.

d.

34. Find the mistake in the equilibrium expression and fix it.

Relating the Equilibrium Constant to EquilibriumConcentrations and Equilibrium Partial Pressures35. Consider the reaction:

An equilibrium mixture of this reaction at a certain temperaturewas found to have and

What is the value of the equilibriumconstant at this temperature?

36. Consider the reaction:

An equilibrium mixture of this reaction at a certain temperaturewas found to have and What is the value of the equilibrium constant at thistemperature?

(Kc)[H2S] = 0.355 M.[NH3] = 0.278 M

NH4HS(s) Δ NH3(g) + H2S(g)

(Kc)[CH3OH] = 0.185 M.

[H2] = 0.114 M,[CO] = 0.105 M,


Kc =[PCl3][Cl2]

[PCl5]PCl5(g) Δ PCl3(l) + Cl2(g)

NH3(aq) + H2O(l) Δ NH4+(aq) + OH-(aq)

HF(aq) + H2O(l) Δ H3O+(aq) + F-(aq)

2 KClO3(s) Δ 2 KCl(s) + 3 O2(g)

CO32-(aq) + H2O(l) Δ HCO3

-(aq) + OH-(aq)

Kc = 4.10 * 10-31 (at 298 K)

N2(g) + O2(g) Δ 2 NO(g)

Kc = 3.7 * 108 (at 298 K)

N2(g) + 3 H2(g) Δ 2 NH3(g)

N2O4(g) Δ 2 NO2(g) Kc = 5.9 * 10-3 (at 298 K)

Kp

Kp = 81.9 (at 298 K)I2(g) + Cl2(g) Δ 2 ICl(g)

Kp = 7.7 * 1024 (at 298 K)

CH4(g) + H2O(g) Δ CO(g) + 3 H2(g)

I2(g) Δ 2 I(g) Kp = 6.26 * 10-22 (at 298 K)

Kc

3 D(g) Δ B(g) + 2 C(g) K2 = 2.35

A(s) Δ 12B(g) + C(g) K1 = 0.0334

2 A(s) Δ 3 D(g).

N2(g) + O2(g) + Br2(g) Δ 2 NOBr(g)

2 NO(g) Δ N2(g) + O2(g) Kp = 2.1 * 1030

NO(g) + 12Br2(g) Δ NOBr(g) Kp = 5.3

2 CO2(g) + 2 CF4(g) Δ 4 COF2(g)

6 COF2(g) Δ 3 CO2(g) + 3 CF4(g)

COF2(g) Δ 12 CO2(g) + 1

2 CF4(g)

Kp 37. Consider the reaction:

Complete the table. Assume that all concentrations are equilib-rium concentrations in M.

N2(g) + 3 H2(g) Δ 2 NH3(g)

T (K) [N2] [H2] [NH3] [Kc]

500 0.115 0.105 0.439 _____

575 0.110 _____ 0.128 9.6

775 0.120 0.140 _____ 0.0584

T [H2] [I2] [HI] [Kc]

25 0.0355 0.0388 0.922 _____

340 ______ 0.0455 0.387 9.6

445 0.0485 0.0468 _____ 50.2

(°C)

38. Consider the following reaction:

Complete the table. Assume that all concentrations are equilib-rium concentrations in M.

H2(g) + I2(g) Δ 2 HI(g)


In a reaction mixture at equilibrium, the partial pressure of NOis 108 torr and that of is 126 torr. What is the partial pres-sure of NOBr in this mixture?


In a reaction at equilibrium, the partial pressure of is137 torr and that of is 285 torr. What is the partial pressureof in this mixture?


A solution is made containing an initial Mand an initial At equilibrium,

Calculate the value of the equi-librium constant


A reaction mixture is made containing an initial ofAt equilibrium, Calculate the

value of the equilibrium constant


A reaction mixture in a 3.67-L flask at a certain temperature ini-tially contains 0.763 g and 96.9 g At equilibrium, theflask contains 90.4 g HI. Calculate the equilibrium constant

for the reaction at this temperature.(Kc)

I2.H2

H2(g) + I2(g) Δ 2 HI(g)

(Kc).[Cl2] = 1.2 * 10-2 M.0.020 M.

[SO2Cl2]

SO2Cl2(g) Δ SO2(g) + Cl2(g)

(Kc).1.7 * 10-4

M.[FeSCN2+] =[SCN-] of 8.0 * 10-4 M.

1.0 * 10-3[Fe3+] of

Fe3+(aq) + SCN-(aq) Δ FeSCN2+(aq)

SO2Cl2

Cl2

SO2

Kp = 2.91 * 103 at 298 K


Br2

Kp = 28.4 at 298 K

2 NO(g) + Br2(g) Δ 2 NOBr(g)

Exercises 653


A reaction mixture in a 5.19-L flask at a certain temperaturecontains 26.9 g CO and 2.34 g At equilibrium, the flaskcontains 8.65 g Calculate the equilibrium constant

for the reaction at this temperature.

The Reaction Quotient and Reaction Direction45. Consider the reaction:

At a certain temperature, A reaction mixtureat this temperature containing solid has

Will more of the solid form orwill some of the existing solid decompose as equilibrium isreached?


A reaction mixture contains 0.112 atm of 0.055 atm of and 0.445 atm of Is the reaction mixture at equilibrium? Ifnot, in what direction will the reaction proceed?

47. Silver sulfate dissolves in water according to the reaction:

A 1.5-L solution contains 6.55 g of dissolved silver sulfate. Ifadditional solid silver sulfate is added to the solution, will it dis-solve?

48. Nitrogen dioxide dimerizes according to the reaction:

A 2.25-L container contains 0.055 mol of and 0.082 molof Is the reaction at equilibrium? If not, in whatdirection will the reaction proceed?

Finding Equilibrium Concentrations from InitialConcentrations and the Equilibrium Constant49. Consider the reaction and associated equilibrium constant:

Find the equilibrium concentrations of A and B for each valueof a and b. Assume that the initial concentration of A in eachcase is 1.0 M and that no B is present at the beginning of thereaction.a.b.c.

50. Consider the reaction and associated equilibrium constant:

Find the equilibrium concentrations of A, B, and C for eachvalue of a, b, and c. Assume that the initial concentrations of Aand B are each 1.0 M and that no product is present at the begin-ning of the reaction.a.b.c. a = 2; b = 1; c = 1 (set up equation for x; don’t solve)a = 1; b = 1; c = 1a = 1; b = 1; c = 2

aA(g) + bB(g) Δ cC(g) Kc = 5.0

a = 1; b = 2a = 2; b = 2a = 1; b = 1

aA(g) Δ bB(g) Kc = 4.0

N2O4 at 298 K.NO2

Kp = 6.7 at 298 K2 NO2(g) Δ N2O4(g)

Kc = 1.1 * 10-5 at 298 K

Ag2SO4(s) Δ 2 Ag+(aq) + SO42-(aq)

H2S.S2,H2,

Kp = 2.4 * 10-4 at 1073 K

2 H2S(g) Δ 2 H2(g) + S2(g)

and [H2S] = 0.166 M.0.166 M[NH3] =NH4HS

Kc = 8.5 * 10-3.

NH4HS(s) Δ NH3(g) + H2S(g)

(Kc)CH3OH.

H2.


51. For the reaction,

If a reaction vessel initially contains an concentration of0.0500 M at 500 K, what are the equilibrium concentrations of

and at 500 K?

52. For the reaction,

If a reaction mixture initially contains a CO concentration of0.1500 M and a concentration of 0.175 M at 1000 K, whatare the equilibrium concentrations of CO, and at1000 K?


If a mixture of solid nickel(II) oxide and 0.20 M carbon monox-ide is allowed to come to equilibrium at 1500 K, what will bethe equilibrium concentration of


If a reaction mixture initially contains 0.110 M CO and 0.110 Mwhat will be the equilibrium concentration of each of the

reactants and products?


If a solution initially contains what is theequilibrium concentration of


If a reaction mixture initially contains what isthe equilibrium concentration of at 227 °C?


A reaction mixture initially contains a partial pressure of755 torr and a partial pressure of 735 torr at 150 K. Calcu-late the equilibrium partial pressure of BrCl.


A reaction mixture initially contains a CO partial pressure of1344 torr and a partial pressure of 1766 torr at 2000 K.Calculate the equilibrium partial pressures of each of theproducts.


Find the equilibrium concentrations of A, B, and C for eachvalue of Assume that the initial concentration of A in eachKc.

A(g) Δ B(g) + C(g)

H2O

Kp = 0.0611 at 2000 K

CO(g) + H2O(g) Δ CO2(g) + H2(g)

Cl2

Br2

Kp = 1.11 * 10-4 at 150 K

Br2(g) + Cl2(g) Δ 2 BrCl(g)

Cl2

0.175 M SO2Cl2,

Kc = 2.99 * 10-7 at 227 °C


H3O+ at 25 °C?0.210 M HC2H3O2,

Kc = 1.8 * 10-5 at 25 °C

HC2H3O2(aq) + H2O(l) Δ H3O+(aq) + C2H3O2-(aq)

H2O,

Kc = 102 at 500 K

CO(g) + H2O(g) Δ CO2(g) + H2(g)

CO2?

Kc = 4.0 * 103 at 1500 K

NiO(s) + CO(g) Δ Ni(s) + CO2(g)

COCl2Cl2,Cl2

CO(g) + Cl2(g) Δ COCl2(g)

Kc = 255 at 1000 K.

NO2N2O4

N2O4

N2O4(g) Δ 2 NO2(g)

Kc = 0.513 at 500 K.


case is 1.0 M and that the reaction mixture initially contains noproducts. Make any appropriate simplifying assumptions.

a.

b.

c.


Find the equilibrium partial pressures of A and B for each valueof Assume that the initial partial pressure of B in each case is1.0 atm and that the initial partial pressure of A is 0.0 atm.Make any appropriate simplifying assumptions.

a.

b.

c.

Le Châtelier’s Principle61. Consider the reaction at equilibrium:

Predict whether the reaction will shift left, shift right, or remainunchanged after each disturbance:a. is added to the reaction mixture.b. is added to the reaction mixture.c. is removed from the reaction mixture.

62. Consider this reaction at equilibrium:

Predict whether the reaction will shift left, shift right, or remainunchanged after each disturbance.a. NO is added to the reaction mixture.b. BrNO is added to the reaction mixture.c. is removed from the reaction mixture.


Predict whether the reaction will shift left, shift right, or remainunchanged after each disturbance.a. is removed from the reaction mixture.b. is added to the reaction mixture.c. is added to the reaction mixture.d. is added to the reaction mixture.


Predict whether the reaction will shift left, shift right, or remainunchanged after each disturbance.a. C is added to the reaction mixture.b. is condensed and removed from the reaction mixture.c. CO is added to the reaction mixture.d. is removed from the reaction mixture.H2

H2O

C(s) + H2O(g) Δ CO(g) + H2(g)

O2

KClO3

KClO2

2 KClO3(s) Δ 2 KCl(s) + 3 O2(g)

Br2

2 BrNO(g) Δ 2 NO(g) + Br2(g)

COCl2

Cl2

COCl2


Kc = 1.0 * 105Kc = 1.0 * 10-4Kc = 1.0

K.

A(g) Δ 2 B(g)

Kc = 1.0 * 10-5Kc = 0.010

Kc = 1.0

65. Each of the following reactions is allowed to come to equilibri-um and then the volume is changed as indicated. Predict the ef-fect (shift right, shift left, or no effect) of the indicated volumechange.

a. (volume is increased)

b. (volume is decreased)

c. (volume is decreased)

66. Each of the following reactions is allowed to come to equilibri-um and then the volume is changed as indicated. Predict the ef-fect (shift right, shift left, or no effect) of the indicated volumechange.

a.(volume is decreased)

b. (volume is increased)

c. (volume is increased)

67. This reaction is endothermic.

Predict the effect (shift right, shift left, or no effect) of increas-ing and decreasing the reaction temperature. How does thevalue of the equilibrium constant depend on temperature?

68. This reaction is exothermic.

Predict the effect (shift right, shift left, or no effect) of increas-ing and decreasing the reaction temperature. How does thevalue of the equilibrium constant depend on temperature?

69. Coal, which is primarily carbon, can be converted to naturalgas, primarily by the exothermic reaction:

Which disturbance will favor at equilibrium?a. adding more C to the reaction mixtureb. adding more to the reaction mixturec. raising the temperature of the reaction mixtured. lowering the volume of the reaction mixturee. adding a catalyst to the reaction mixturef. adding neon gas to the reaction mixture

70. Coal can be used to generate hydrogen gas (a potential fuel) bythe endothermic reaction:

If this reaction mixture is at equilibrium, predict whether eachdisturbance will result in the formation of additional hydrogengas, the formation of less hydrogen gas, or have no effect on thequantity of hydrogen gas.a. adding more C to the reaction mixtureb. adding more to the reaction mixturec. raising the temperature of the reaction mixtured. increasing the volume of the reaction mixturee. adding a catalyst to the reaction mixturef. adding an inert gas to the reaction mixture

H2O

C(s) + H2O(g) Δ CO(g) + H2(g)

H2

CH4

C(s) + 2 H2(g) Δ CH4(g)

CH4,

C6H12O6(s) + 6 O2(g) Δ 6 CO2(g) + 6 H2O(g)

C(s) + CO2(g) Δ 2 CO(g)


PCl3(g) + Cl2(g) Δ PCl5(g)

CO(g) + H2O(g) Δ CO2(g) + H2(g)


2 H2S(g) Δ 2 H2(g) + S2(g)

I2(g) Δ 2 I(g)

Cumulative Problems71. Carbon monoxide replaces oxygen in oxygenated hemoglobin

according to the reaction:

HbO2(aq) + CO(aq) Δ HbCO(aq) + O2(aq)

a. Use the reactions and associated equilibrium constants atbody temperature to find the equilibrium constant for theabove reaction.

Hb(aq) + CO(aq) Δ HbCO(aq) Kc = 306

Hb(aq) + O2(aq) Δ HbO2(aq) Kc = 1.8

Exercises 655

b. Suppose that an air mixture becomes polluted with carbonmonoxide at a level of 0.10%. Assuming the air contains20.0% oxygen, and that the oxygen and carbon monoxide ra-tios that dissolve in the blood are identical to the ratios in theair, what would be the ratio of HbCO to in the blood-stream? Comment on the toxicity of carbon monoxide.

72. Nitrogen oxide is a pollutant in the lower atmosphere that irri-tates the eyes and lungs and leads to the formation of acid rain.Nitrogen oxide forms naturally in atmosphere according to theendothermic reaction:

Use the ideal gas law to calculate the concentrations of nitrogenand oxygen present in air at a pressure of 1.0 atm and a temper-ature of 298 K. Assume that nitrogen composes 78% of air byvolume and that oxygen composes 21% of air. Find the “natu-ral” equilibrium concentration of NO in air in units of

How would you expect this concentration tochange in an automobile engine in which combustion isoccurring?

73. The reaction has Kp = 5.78 at1200 K.a. Calculate the total pressure at equilibrium when 4.45 g of

CO2 is introduced into a 10.0-L container and heated to1200 K in the presence of 2.00 g of graphite.

b. Repeat the calculation of part a) in the presence of 0.50 g ofgraphite.

74. A mixture of water and graphite is heated to 600 K. When thesystem comes to equilibrium it contains 0.13 mol of ,0.13 mol of CO, 0.43 mol of , and some graphite. Some is added to the system and a spark is applied so that the re-acts completely with the . Find the amount of CO in the flaskwhen the system returns to equilibrium.

75. At 650 K, the reaction has . A 10.0-L container at 650 K has 1.0 g ofMgO(s) and at atm. The container is thencompressed to a volume of 0.100 L. Find the mass of that is formed.

76. A system at equilibrium contains I2(g) at a pressure of 0.21 atmand I(g) at a pressure of 0.23 atm. The system is then com-pressed to half its volume. Find the pressure of each gas whenthe system returns to equilibrium.

77. Consider the exothermic reaction:

If you were a chemist trying to maximize the amount ofproduced, which tactic might you try? Assume that the

reaction mixture reaches equilibrium.a. increasing the reaction volumeb. removing from the reaction mixture as it formsc. lowering the reaction temperatured. adding

78. Consider the endothermic reaction:

If you were a chemist trying to maximize the amount of produced, which tactic might you try? Assume that the reactionmixture reaches equilibrium.a. decreasing the reaction volumeb. removing from the reaction mixtureI2

C2H4I2

C2H4(g) + I2(g) Δ C2H4I2(g)

Cl2

C2H4Cl2

C2H4Cl2

C2H4(g) + Cl2(g) Δ C2H4Cl2(g)

MgCO3

P = 0.0260CO2

Kp = 0.026MgCO3(s) Δ MgO(s) + CO2(g)

O2

H2

O2H2OH2

CO2(g) + C(s) Δ 2 CO(g)

molecules>cm3.

N2(g) + O2(g) Δ 2 NO(g) Kp = 4.1 * 10-31 at 298 K.

HbO2

c. raising the reaction temperatured. adding to the reaction mixture


A reaction mixture at equilibrium at 175 K containsand

A second reaction mixture, also at 175 K, containsand Is the second

reaction at equilibrium? If not, what will be the partial pressureof HI when the reaction reaches equilibrium at 175 K?


A reaction mixture initially containing 0.500 M and0.500 M was found to contain 0.0011 M at a certaintemperature. A second reaction mixture at the same temperatureinitially contains and Calculate the equilibrium concentration of in the secondmixture at this temperature.

81. Ammonia can be synthesized according to the reaction:

A 200.0-L reaction container initially contains 1.27 kg of and 0.310 kg of at 725 K. Assuming ideal gas behavior, cal-culate the mass of (in g) present in the reaction mixture atequilibrium. What is the percent yield of the reaction underthese conditions?

82. Hydrogen can be extracted from natural gas according to thereaction:

An 85.0-L reaction container initially contains 22.3 kg of and 55.4 kg of at 825 K. Assuming ideal gas behavior, cal-culate the mass of (in g) present in the reaction mixture atequilibrium. What is the percent yield of the reaction underthese conditions?

83. The system described by the reaction

is at equilibrium at a given temperature when and An additional pres-

sure of is added. Find the pressure of COwhen the system returns to equilibrium.

84. A reaction vessel at 27 °C contains a mixture of SO2and When a catalyst is

added this reaction takes place:

At equilibrium the total pressure is 3.75 atm. Find the value of

85. At 70 K, decomposes to carbon and chlorine. The forthe decomposition is 0.76. Find the starting pressure of atthis temperature that will produce a total pressure of 1.0 atm atequilibrium.

86. The equilibrium constant for the reaction is 3.0. Find the amount of

that must be added to 2.4 mol of in order to form1.2 mol of at equilibrium.SO3

SO2NO2

NO2(g) Δ SO3(g) + NO(g)SO2(g) +

CCl4

KpCCl4

Kc.

2 SO2(g) + O2(g) Δ 2 SO3(g)

O2 (P = 1.00 atm).(P = 3.00 atm)

Cl2(g) = 0.40 atmPCOCl2 = 0.60 atm.PCl2 = 0.10 atm,

PCO = 0.30 atm,


H2

CO2

CH4

Kp = 4.5 * 102 at 825 K

CH4(g) + CO2(g) Δ 2 CO(g) + 2 H2(g)

NH3

H2

N2

Kp = 5.3 * 10-5 at 725 K

N2(g) + 3 H2(g) Δ 2 NH3(g)

H2O[SO2] = 0.325 M.[H2S] = 0.250 M

H2OSO2

H2S

2 H2S(g) + SO2(g) Δ 3 S(s) + 2 H2O(g)

PHI = 0.101 atm.PH2= PI2

= 0.621 atm,

PHI = 0.020 atm.PI2= 0.877 atm,PH2

= 0.958 atm,

H2(g) + I2(g) Δ 2 HI(g)

C2H4

87. A sample of is introduced into a sealed container ofvolume 0.654 L and heated to 1000 K until equilibrium isreached. The for the reaction

is at this temperature. Calculate the mass of CaO(s)that is present at equilibrium.

88. An equilibrium mixture contains andat 350 K. The volume of the container is

doubled at constant temperature. Calculate the equilibriumpressures of the two gases when the system reaches a newequilibrium.

(P = 1.1 atm)NO2

(P = 0.28 atm)N2O4,

3.9 * 10-2


Kp

CaCO3(s)


Challenge Problems91. Consider the reaction:

a. A reaction mixture at 175 K initially contains 522 torr of NOand 421 torr of At equilibrium, the total pressure in thereaction mixture is 748 torr. Calculate at this temperature.

b. A second reaction mixture at 175 K initially contains255 torr of NO and 185 torr of What is the equilibriumpartial pressure of in this mixture?


A 2.75-L reaction vessel at initially contains 0.100 molof and 0.100 mol of Calculate the total pressure (in at-mospheres) in the reaction vessel when equilibrium is reached.

93. Nitric oxide reacts with chlorine gas according to the reaction:

A reaction mixture initially contains equal partial pressures ofNO and At equilibrium, the partial pressure of NOCl wasmeasured to be 115 torr. What were the initial partial pressuresof NO and

94. At a given temperature a system containing and some ox-ides of nitrogen can be described by these reactions:

2 NO2(g) Δ N2O4(g) Kp = 0.10

2 NO(g) + O2(g) Δ 2 NO2(g) Kp = 104

O2(g)

Cl2 ?

Cl2.

Kp = 0.27 at 700 K

2 NO(g) + Cl2(g) Δ 2 NOCl(g)

O2.SO2

950 K

Kp = 0.355 at 950 K

2 SO2(g) + O2(g) Δ 2 SO3(g)

NO2

O2.

Kp

O2.

2 NO(g) + O2(g) Δ 2 NO2(g)

A pressure of 1 atm of is placed in a container at thistemperature. Predict which, if any component (other than )will be present at a pressure greater than 0.2 atm at equilibrium.

95. A sample of pure is heated to 337 °C at which temperatureit partially dissociates according to the equation

At equilibrium the density of the gas mixture is at0.750 atm. Calculate for the reaction.

96. When is heated it dissociates into and according to the reaction:

The dissociates to give and accordingthe reaction:

When 4.00 mol of is heated in a 1.00-L reaction vesselto this temperature, the concentration of at equilibrium is

Find the concentrations of all the other species inthe equilibrium system.

97. A sample of SO3 is introduced into an evacuated sealed contain-er and heated to 600 K. The following equilibrium is estab-lished:

The total pressure in the system is found to be 3.0 atm and themole fraction of O2 is 0.12. Find .Kp

2 SO3(g) Δ 2 SO2(g) + O2(g).

4.50 mol>L.O2(g)

N2O5(g)

Kc = 4.00 at the same temperature

N2O3(g) Δ N2O(g) + O2(g)

O2(g)N2O(g)N2O3(g)

Kc = 7.75 at a given temperature

N2O5(g) Δ N2O3(g) + O2(g)

O2(g)N2O3(g)N2O5(g)

Kc

0.520 g>L2 NO2(g) Δ 2 NO(g) + O2(g)

NO2

N2O4

N2O4(g)

89. Carbon monoxide and chlorine gas react to form phosgene:

If a reaction mixture initially contains 215 torr of CO and245 torr of , what is the mole fraction of when equi-librium is reached?

90. Solid carbon can react with gaseous water to form carbonmonoxide gas and hydrogen gas. The equilibrium constant forthe reaction at 700.0 K is . If a 1.55-L reac-tion vessel initially contains 145 torr of water at 700.0 K in con-tact with excess solid carbon, find the percent by mass ofhydrogen gas of the gaseous reaction mixture at equilibrium.

Kp = 1.60 * 10-3

COCl2Cl2

CO(g) + Cl2(g) Δ COCl2(g) Kp = 3.10 at 700 K

Exercises 657

98. A reaction has an equilibrium constant of. For which of the initial reaction mixtures is the x is

small approximation most likely to apply?a.b.c.d.

99. The reaction has an equilibrium constant ofat a given temperature. If a reaction vessel contains

equal initial amounts (in moles) of A and B, will the direction inwhich the reaction proceeds depend on the volume of the reac-tion vessel? Explain.

100. A particular reaction has an equilibrium constant of A reaction mixture is prepared in which all the reactants andproducts are in their standard states. In which direction will thereaction proceed?

Kp = 0.50.

Kc = 1.0A(g) Δ 2 B(g)

[A] = 0.10 M; [B] = 0.00 M[A] = 0.10 M; [B] = 0.10 M[A] = 0.00 M; [B] = 0.10 M[A] = 0.0010 M; [B] = 0.00 M

1.0 * 10-4A(g) Δ B(g) 101. Consider the reaction:

Each of the entries in the table below represents equilibriumpartial pressures of A and B under different initial conditions.What are the values of a and b in the reaction?

aA(g) Δ bB(g)

PA (atm) PB (atm)

4.0 2.0

2.0 1.4

1.0 1.0

0.50 0.71

0.25 0.50

Conceptual Problems

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