Chapter Ten - UNAMdepa.fquim.unam.mx/amyd/archivero/Unidad-ChemicalEquilibrium_3… · Chapter Ten THE CONNECTION BETWEEN KINETICS AND EQUILIBRIUM 10.1 Reactions That DonÕt Go to

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Chapter TenTHE CONNECTION BETWEEN KINETICS AND EQUILIBRIUM

10.1 Reactions That Don’t Go to Completion

10.2 Gas-Phase Reactions

10.3 The Rate of a Chemical Reaction

10.4 The Collision Theory Model of Gas-Phase Reactions

10.5 Equilibrium Constant Expressions

10.6 Reaction Quotients: A Way to Decide Whether a Reaction Is atEquilibrium

10.7 Changes in Concentration That Occur as a Reaction Comes toEquilibrium

10.8 Hidden Assumptions That Make Equilibrium Calculations Easier

10.9 What Do We Do When the Assumption Fails?

10.10 The Effect of Temperature on an Equilibrium Constant

10.11 Le Châtelier’s Principle

10.12 Le Châtelier’s Principle and the Haber Process

10.13 What Happens When a Solid Dissolves in Water?

10.14 The Solubility Product Expression

10.15 The Relationship between Ksp and the Solubility of a Salt

10.16 The Role of the Ion Product (Qsp) in Solubility Calculations

10.17 The Common-Ion Effect

c10TheConnectionBetweenKineticsandEquilibrium.qxd 8/9/10 8:13 PM Page 408

10.1 Reactions That Don’t Go to CompletionSuppose that you were asked to describe the steps involved in calculating the massof the finely divided white solid produced when a 2.00-g strip of magnesium metalis burned. You might organize your work as follows.

● Start by assuming that the magnesium reacts with oxygen in the atmospherewhen it burns.

● Predict that the formula of the product is MgO.

● Use this formula to generate the following balanced equation.

● Use the mass of a mole of magnesium to convert grams of magnesium intomoles of magnesium.

● Use the balanced equation to convert moles of magnesium into moles ofmagnesium oxide.

● Finally, use the mass of a mole of magnesium oxide to convert moles ofMgO into grams of MgO.

Before you read any further, ask yourself, “How confident am I in the answer?”Before we can trust the answer, we have to consider whether there are any

hidden assumptions behind the calculation and then check the validity of theseassumptions. Three assumptions were made in the calculation.

● We assumed that the metal strip was pure magnesium.

● We assumed that the magnesium reacts only with the oxygen in the atmos-phere to form MgO, ignoring the possibility that some of the magnesiummight react with the nitrogen in the atmosphere to form Mg3N2.

● We assumed that the reaction didn’t stop until all of the magnesium metalhad been consumed.

It is relatively easy to correct for the fact that the starting material may not bepure magnesium by weight. We can also correct for the fact that as much as 5%of the product of the reaction is Mg3N2 instead of MgO. But it is the third assump-tion that is of particular importance in this chapter.

The assumption that chemical reactions proceed until the last of the limit-ing reagent is consumed is fostered by calculations such as predicting the amountof MgO produced by burning a known amount of magnesium. It is reinforced bydemonstrations such as the reaction in which a copper penny dissolves in con-centrated nitric acid, which seems to continue until the copper penny disappears.

0.0823 mol MgO *

40.30 g MgO

1 mol MgO= 3.32 g MgO

0.0823 mol Mg *

2 mol MgO

2 mol Mg= 0.0823 mol MgO

2.00 g Mg *

1 mol Mg

24.31 g Mg= 0.0823 mol Mg

2 Mg(s) + O2(g) ¡ 2 MgO(s)

10.1 REACTIONS THAT DON’T GO TO COMPLETION 409

A copper penny reacting with nitricacid to form NO2 gas.


Chemical reactions, however, don’t always go to completion. The followingequation provides an example of a chemical reaction that seems to stop prematurely.

At 25°C, when 1 mol of NO2 is added to a 1.00-L flask, the reaction seems to stopwhen 95% of the NO2 has been converted to N2O4. Once it has reached that point,the reaction doesn’t go any further. As long as the reaction is left at 25°C, about5% of the NO2 that was present initially will remain in the flask. Reactions thatseem to stop before the limiting reagent is consumed are said to reach equilibrium.

It is useful to recognize the difference between reactions that come to equi-librium and those that stop when they run out of the limiting reagent. The reac-tion between a copper penny and nitric acid is an example of a reaction that con-tinues until it has essentially run out of the limiting reagent. We indicate this bywriting the equation for the reaction with a single arrow pointing from the reac-tants to the products.

We indicate that a reaction comes to equilibrium by writing a pair of arrows point-ing in opposite directions between the two sides of the equation.

In order to work with reactions that come to equilibrium, we need a way tospecify the amount of each reactant or product that is present in the system at anymoment in time. By convention, this information is specified in terms of theconcentration of each component of the system in units of moles per liter. Thisquantity is indicated by a symbol that consists of the formula for the reactant orproduct written in parentheses. For example,

(NO2) � concentration of NO2 in moles per liter at some moment in time

We then need a way to describe the system when it is at equilibrium. This is doneby writing the symbols for each component of the system in square brackets.

[NO2] � concentration of NO2 in moles per liter if, and only if, the reaction is at equilibrium

The fact that some reactions come to equilibrium raises a number of inter-esting questions.

● Why do these reactions seem to stop before all of the starting materials areconverted into the products of the reaction?

● What is the difference between reactions that seem to go to completion andreactions that reach equilibrium?

● Is there any way to predict whether a reaction will go to completion or reachequilibrium?

● How does a change in the conditions of the reaction influence the amountof product formed at equilibrium?

Before we can understand how and why a chemical reaction comes to equilibrium,we have to build a model of the factors that influence the rate of a chemical

2 NO2(g) uv N2O4(g)

Cu(s) + 4 HNO3(aq) ¡ Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)

2 NO2(g) Δ N2O4(g)

410 CHAPTER 10 / THE CONNECTION BETWEEN KINETICS AND EQUILIBRIUM


reaction. We’ll then apply this model to the simple chemical reactions that occurin the gas phase. The next chapter will include interactions between the compo-nents of the reaction and the solvent in which the reaction is run.

10.2 Gas-Phase ReactionsThe simplest chemical reactions are those that occur in the gas phase in a singlestep, such as the following reaction in which a compound known as cis-2-buteneis converted into its isomer, trans-2-butene.

The two isomers of 2-butene have different physical properties. cis-2-Butenemelts at �139°C, whereas the trans isomer melts at �106°C. The cis isomer boilsat 4°C, whereas the trans form boils at 1°C. One of these isomers can be con-verted into the other by rotating one end of the C“C double bond relative to theother. Rotation around C“C double bonds, however, doesn’t occur at room tem-perature. We can therefore obtain a pure sample of the starting material—cis-2-butene––and wait as it is transformed into the product—trans-2-butene—when weheat the sample.

Table 10.1 shows the amount of both cis- and trans-2-butene present in thesystem at various moments in time when a 1.000-mol sample of the cis isomeris heated to 400°C in a 10.0-L flask. Initially, there is no trans-2-butene in thesystem. With time, however, the amount of cis-2-butene gradually decreases asthis compound is converted into the trans isomer.

Because one molecule of trans-2-butene is produced each time a moleculeof the cis isomer is consumed in this reaction, the total number of moles of thetwo isomers must always be the same as the number of moles of the cis isomerthat were present at the start of the experiment. The amount of the trans isomerpresent at any moment in time can therefore be calculated from the amount ofthe cis isomer that remains in the system, and vice versa.

cis-2-Butene trans-2-ButeneH H

CH3

C C

CH3

CH3H

C C

HCH3

ED

10.2 GAS-PHASE REACTIONS 411

➤ CHECKPOINTAssume that N2O4(g) is added to anevacuated container at 25°C. Whatcompounds would be present in thecontainer at equilibrium?

Molecular model of trans-2-butene

Molecular model of cis-2-butene

Table 10.1Experimental Data for the Isomerization of cis-2-Butene to trans-2-Buteneat 400°C in a 10.0-L Flask

Time Moles of cis-2-Butene Moles of trans-2-Butene

0 1.000 05.00 days 0.919 0.08110.00 days 0.848 0.15215.00 days 0.791 0.20920.00 days 0.741 0.25940.00 days 0.560 0.44060.00 days 0.528 0.472120.00 days 0.454 0.5461 year 0.441 0.5592 years 0.441 0.5593 years 0.441 0.559


In this case, we started with 1.000 mol of the cis isomer. At any moment intime, the number of moles of the cis isomer that remain in the system is 1.000 �x,

and the number of moles of the trans isomer is equal to x.

Figure 10.1 shows a plot of the number of moles of the cis and trans iso-mers of 2-butene versus time. This figure shows that there is no change in thenumber of moles of either component once the reaction reaches the point at whichthe system contains 0.441 mol of cis-2-butene and 0.559 mol of trans-2-butene.No matter how long we wait, no more cis-2-butene is converted into trans-2-butene. This indicates that the reaction has come to equilibrium.

Chemical reactions at equilibrium are described in terms of the number ofmoles per liter of each component of the system. For example,

[cis-2-butene] � concentration of cis-2-butene at equilibrium in units of moles per liter

[trans-2-butene] � concentration of trans-2-butene at equilibrium in units of moles per liter

The concentrations of cis- and trans-2-butene at equilibrium depend on the initialconditions of the experiment. But, at a given temperature, the ratio of the equilib-rium concentrations of the two components of the reaction is always the same. Itdoesn’t matter whether we start with a great deal of cis-2-butene or a relatively smallamount, or whether we start with a pure sample of cis-2-butene or one that alreadycontains some of the trans isomer. When the reaction reaches equilibrium at 400°C,the concentration of the trans isomer divided by that of the cis isomer is always 1.27.

The equation that describes the relationship between the concentrations ofthe two components of the reaction at equilibrium is known as the equilibriumconstant expression, where Kc is the equilibrium constant for the reaction.

The subscript c in the equilibrium constant indicates that the constant has beencalculated from the concentrations of the reactants and products in units of molesper liter.

Kc =

[trans-2-butene]

[cis-2-butene]= 1.27

ntrans-2-butene = x

ncis-2-butene = 1.000 - x


Fig. 10.1 A plot of the number of moles of the cisand trans isomers of 2-butene versus time at 400°C.Initially there is no trans-2-butene, but as timepasses the concentration of the cis isomer decreasesand the concentration of the trans isomer increases.

10 2 3

1.000

0.800

0.600

0.400

0.200

0

Mol

es

Time (years)

cis

trans

➤ CHECKPOINTAssume that 1.0 mol of trans-2-buteneis placed in an empty flask at 400°C.What will be the ratio of the concentra-tion of trans-2-butene to cis-2-butenewhen this system reaches equilibrium?


10.3 THE RATE OF A CHEMICAL REACTION 413

E x e r c i s e 1 0 . 1Calculate the equilibrium constant for the conversion of cis-2-butene to trans-2-butene for the following sets of experimental data.

(a) A 5.00-mol sample of the cis isomer was added to a 10.0-L flask andheated to 400°C until the reaction came to equilibrium. At equilibrium, thesystem contained 2.80 mol of the trans isomer.

(b) A 0.100-mol sample of the cis isomer was added to a 25.0-L flask andheated to 400°C until the reaction came to equilibrium. At equilibrium, thesystem contained 0.0559 mol of the trans isomer.

Solution(a) If a sample that started with 5.00 mol of cis-2-butene forms 2.80 mol of

trans-2-butene at equilibrium, then 2.20 mol of the cis isomer must remainafter the reaction reaches equilibrium. Because the experiment was donein a 10.0-L flask, the equilibrium concentrations of the two components ofthe reaction have the following values.

The equilibrium constant, Kc, for the reaction is therefore 1.27.

(b) In this case, the system comes to equilibrium at the following concentra-tions of the cis and trans isomers.

Even though the concentrations of the two isomers at equilibrium are verydifferent from the values obtained in the previous experiment, the ratioof the concentrations at equilibrium is exactly the same.

Kc =

[trans-2-butene]

[cis-2-butene]=

0.00224 M

0.00176 M= 1.27

[cis-2-butene] =

0.0441 mol25.0 L

= 0.00176 M

[trans-2-butene] =

0.0559 mol25.0 L

= 0.00224 M

Kc =

[trans-2-butene]

[cis-2-butene]=

0.280 M

0.220 M= 1.27

[cis-2-butene] =

2.20 mol10.0 L

= 0.220 M

[trans-2-butene] =

2.80 mol10.0 L

= 0.280 M

10.3 The Rate of a Chemical ReactionExperiments such as the one that gave rise to the data in Table 10.1 are classified asmeasurements of chemical kinetics (from the Greek stem meaning “to move”).The goal of these experiments is to describe the rate of reaction, that is, the rateat which the reactants are transformed into the products of the reaction.


The term rate is often used to describe the change in a quantity that occursper unit of time. The rate of inflation, for example, is the change in the averagecost of a collection of standard items per year. The rate at which an object trav-els through space is the distance traveled per unit of time in units of miles perhour or kilometers per second.

In chemical kinetics, the distance traveled is the change in the concentra-tion of one of the components of the reaction. The rate of a reaction is thereforethe change in the concentration of one of the components, ¢(X), that occurs dur-ing a given time, ¢t. The concentration of X is written in parentheses here becausethe system isn’t at equilibrium.

By convention, the symbol ¢ represents a change calculated by subtracting theinitial conditions from the final conditions. Thus, the ¢(X) represents the finalconcentration minus the initial concentration.

If X is one of the products of the reaction, then ¢(X) is positive. If X is a reac-tant, ¢(X) is negative and a minus sign is added to the rate equation to turn therate into a positive number.

Let’s use the data in Table 10.1 to calculate the rate at which cis-2-buteneis transformed into trans-2-butene during each of the following periods.

● During the first time interval, when the number of moles of cis-2-butene inthe 10.0-L flask falls from 1.000 to 0.919

● During the second interval, when the amount of cis-2-butene falls from0.919 to 0.848 mol

● During the third interval, when the amount of cis-2-butene falls from 0.848to 0.791 mol

Before we can calculate the rate of the reaction during each of these time inter-vals, we have to remember that the rate of reaction is defined in terms of changesin the number of moles per liter (M) of one of the components of the reaction,not the number of moles of that reactant. Thus we have to recognize that 1.000mol of cis-2-butene in a 10.0-L flask corresponds to a concentration of 0.1000 M.

During the first time period, the rate of the reaction is 1.6 � 10�3 M/day.

During the second time period, the rate of reaction is slightly smaller.

During the third time period, the rate of reaction is even smaller.

rate = -

¢(X)

¢t= -

(0.0791 M - 0.0848 M)

(15.00 days - 10.00 days)= 1.1 * 10- 3 M/day

rate = -

¢(X)

¢t= -

(0.0848 M - 0.0919 M)

(10.00 days - 5.00 days)= 1.4 * 10- 3 M/day

rate = -

¢(X)

¢t= -

(0.0919 M - 0.1000 M)

(5.00 days - 0 days)= 1.6 * 10- 3 M/day

rate = -

¢(X)

¢t (if X is one of the reactants)

¢(X) = (X)final - (X)initial

rate =

¢(X)

¢t (where X is one of the products)



These calculations illustrate an important point: The rate of the reaction isn’t con-stant; it changes with time. The rate of the reaction gradually decreases as thestarting materials are consumed, which means that the rate of reaction changeswhile it is being measured.

We can minimize the error this introduces into our measurements by meas-uring the rate of reaction over periods of time that are short compared with thetime it takes for the reaction to occur. We might try, for example, to measure theinfinitesimally small change in reactant concentration, d(X), that occurs over aninfinitesimally short period of time, dt. This ratio is known as the instantaneousrate of reaction.

The instantaneous rate of reaction at any moment in time can be calculatedfrom a graph of the concentration of the reactant (or product) versus time. Fig-ure 10.2 shows how the rate of reaction for the isomerization of cis-2-butene canbe calculated from such a graph. The rate of reaction at any moment is equal tothe negative of the slope of a tangent drawn to the curve at that moment.

An interesting result is obtained when the instantaneous rate of reaction iscalculated at various points along the curve in Figure 10.2. The rate of reactionat every point on the curve is directly proportional to the concentration of cis-2-butene at that moment in time.

rate � k(cis-2-butene)

This equation, which is determined from experimental data, describes the rate ofthe reaction. It is therefore called the rate law for the reaction. The proportion-ality constant k is known as the rate constant.

rate = -

d(X)

dt

10.4 THE COLLISION THEORY MODEL OF GAS-PHASE REACTIONS 415

Fig. 10.2 The rate of reaction at a given time for theisomerization of cis-2-butene is the negative of theslope of a tangent drawn to the concentration curveat that particular point in time.

100 20 30 40

0.100

0.090

0.080

0.070

0.060

0.050

Con

cent

rati

on c

is-2

-but

ene

(M)

Time (days)

➤ CHECKPOINTThe following data were obtained forthe rate constant for the decompositionof one of the metabolites that suppliesenergy in living systems.

Temperature Rate Constant

15202530

What do these data suggest happens tothe rate of the reaction as the temperatureat which the reaction is run increases?

1.6 * 10-18.1 * 10-24.5 * 10-22.5 * 10-2

(s-1)(°C)

10.4 The Collision Theory Model of Gas-Phase ReactionsOne way to understand why some reactions come to equilibrium is to consider asimple gas-phase reaction that occurs in a single step, such as the transfer of achlorine atom from ClNO2 to NO to form NO2 and ClNO.

ClNO2(g) + NO(g) uv NO2(g) + ClNO(g)


This reaction can be understood by writing the Lewis structures for the four com-ponents of the reaction. Because they contain an odd number of electrons, bothNO and NO2 can combine with a neutral chlorine atom to form a molecule inwhich all of the electrons are paired. The reaction therefore involves the transferof a chlorine atom from one molecule to another, as shown in Figure 10.3.

Figure 10.4 combines a plot of the concentration of ClNO2 versus time asthis reactant is consumed with a plot of the concentration of NO2 versus time asthis product of the reaction is formed. The data in Figure 10.4 are consistent withthe following rate law for the reaction.

According to this rate law, the rate at which ClNO2 and NO are converted intoNO2 and ClNO is proportional to the product of the concentrations of the tworeactants. Initially, the rate of reaction is relatively fast. As the reactants are con-verted to products, however, the ClNO2 and NO concentrations become smaller,and the reaction slows down.

We might expect the reaction to stop when it runs out of either ClNO2 orNO. In practice, the reaction seems to stop before this happens. This is a veryfast reaction––the concentration of ClNO2 drops by a factor of 2 in about 1 sec-ond. And yet, no matter how long we wait, some residual ClNO2 and NO remainin the reaction flask.

Figure 10.4 divides the plot of the change in the concentrations of NO2 andClNO2 into a kinetic region and an equilibrium region. The kinetic region marksthe period during which the concentrations of the components of the reaction areconstantly changing. The equilibrium region is the period after which the reac-tion seems to stop, when there is no further significant change in the concentra-tions of the components of the reaction.

The fact that this reaction seems to stop before all of the reactants areconsumed can be explained with the collision theory of chemical reactions. Thecollision theory assumes that ClNO2 and NO molecules must collide before achlorine atom can be transferred from one molecule to the other. This assumption

rate = k(ClNO2)(NO)


Fig. 10.3 The reaction between ClNO2 and NO to formNO2 and ClNO is a simple, one-step reaction that involvesthe transfer of a chlorine atom.

+ Cl + NNN O

Cl

O

N

O

O

O

O

(NO2)

(ClNO2)

0 5 10 15 20 25

1 × 10–4

2 × 10–4

Con

cent

rati

on (

mol

/L)

Time (s)

KineticRegion

EquilibriumRegion

Fig. 10.4 Plot of the change in the concentration ofClNO2 superimposed on a plot of the change inconcentration of NO2 as ClNO2 reacts with NO toproduce NO2 and ClNO.The graph can be dividedinto a kinetic and an equilibrium region.


explains why the rate of the reaction is proportional to the concentration of bothClNO2 and NO.

rate � k(ClNO2)(NO)

The number of collisions per second between ClNO2 and NO molecules dependson their concentrations. As ClNO2 and NO are consumed in the reaction, the num-ber of collisions per second between molecules of these reactants becomessmaller, and the reaction slows down.

Suppose that we start with a mixture of ClNO2 and NO, but no NO2 orClNO. The only reaction that can occur at first is the transfer of a chlorine atomfrom ClNO2 to NO.

Eventually, NO2 and ClNO build up in the reaction flask, and collisions betweenthese molecules can result in the transfer of a chlorine atom in the oppositedirection.

The collision theory model of chemical reactions assumes that the rate ofa simple, one-step reaction is proportional to the product of the concentrationsof the substances consumed in that reaction. The rate of the forward reaction istherefore proportional to the product of the concentrations of the two startingmaterials.

The rate of the reverse reaction, on the other hand, is proportional to the productof concentrations of the compounds formed in the reaction.

Initially, the rate of the forward reaction is much larger than the rate of thereverse reaction because the system contains ClNO2 and NO but virtually no NO2

or ClNO.

As ClNO2 and NO are consumed, the rate of the forward reaction slows down.At the same time, NO2 and ClNO accumulate, and the reverse reaction speeds up.

The system eventually reaches a point at which the rates of the forward andreverse reactions are the same.

At this point, the concentrations of reactants and products remain the same, nomatter how long we wait. On the molecular scale, ClNO2 and NO are consumedin the forward reaction at the same rate at which they are produced in the reversereaction. The same thing happens to NO2 and ClNO. When the rates of the for-ward and reverse reactions are the same, there is no longer any change in the con-centrations of the reactants or products of the reaction on the macroscopic scale.In other words, the reaction is at equilibrium.

Eventually: rateforward = ratereverse

Initially: rateforward W ratereverse

ratereverse = kr (NO2)(ClNO)

rateforward = kf (ClNO2)(NO)

ClNO2(g) + NO(g) — NO2(g) + ClNO(g)

ClNO2(g) + NO(g) ¡ NO2(g) + ClNO(g)

10.4 THE COLLISION THEORY MODEL OF GAS-PHASE REACTIONS 417


We can now see that there are two ways to define equilibrium.

● A system in which there is no change in the concentrations of the reactantsand products of a reaction with time.

● A system in which the rates of the forward and reverse reactions are thesame.

The first definition is based on the results of experiments that tell us that somereactions seem to stop prematurely––they reach a point at which no more reac-tants are converted to products before the limiting reagent is consumed. The otherdefinition is based on a theoretical model of chemical reactions that explains whyreactions reach equilibrium.

10.5 Equilibrium Constant ExpressionsReactions don’t stop when they come to equilibrium. But the rate of the forwardand reverse reactions at equilibrium are the same, so there is no net change in theconcentrations of the reactants or products, and the reactions appear to stop onthe macroscopic scale. Chemical equilibrium is an example of a dynamic balancebetween the forward and reverse reactions, not a static balance.

Let’s look at the logical consequences of the assumption that the reactionbetween ClNO2 and NO eventually reaches equilibrium at a given temperature.

The rates of the forward and reverse reactions are the same when the system isat equilibrium.

Substituting the rate laws for the forward and reverse reactions into the equalitygives the following result.

But this equation is valid only when the system is at equilibrium, so weshould replace the (ClNO2), (NO), (NO2), and (ClNO) terms with symbols indi-cating that the reaction is at equilibrium. By convention, we use square brack-ets for this purpose. The equation describing the balance between the forwardand reverse reactions when the system is at equilibrium should therefore be writ-ten as follows.

Rearranging the equation gives the following result.

Because kf and kr are constants, the ratio of kf divided by kr must also be a con-stant. This ratio is the equilibrium constant for the reaction, Kc. As we have

kf

kr=

[NO2][ClNO]

[ClNO2][NO]

At equilibrium: kf [ClNO2][NO] = kr [NO2][ClNO]

At equilibrium: kf (ClNO2)(NO) = kr (NO2)(ClNO)

At equilibrium: rateforward = ratereverse



➤ CHECKPOINTIf the rate of the reverse reaction isgreater than the rate of the forward re-action at a given moment in time, doesthis mean that the reverse rate constantis larger than the forward rate constant?


seen, the ratio of the concentrations of the reactants and products is known as theequilibrium constant expression.

Equilibrium constant expression

Equilibrium constant

No matter what combination of concentrations of reactants and products westart with, the reaction reaches equilibrium when the ratio of the concentrationsdefined by the equilibrium constant expression is equal to the equilibrium con-stant for the reaction at a given temperature. We can start with a lot of ClNO2

and very little NO, or a lot of NO and very little ClNO2. It doesn’t matter. Whenthe reaction reaches equilibrium, the relationship between the concentrations ofthe reactants and products described by the equilibrium constant expression willalways be the same. At 25°C, this reaction always reaches equilibrium when theratio of the concentrations is 1.3 � 104.

Kc is always reported without units. However, any calculations using Kc requirethe concentration of the products and reactants to be in units of molarity(moles/liter).

What happens if we approach equilibrium from the other direction? In thiscase, we start with a system that contains the products of the reaction––NO2 andClNO––and then let the reaction come to equilibrium. The rate laws for the for-ward and reverse reactions will still be the same.

Now, however, the rate of the forward reaction initially will be much smaller thanthe rate of the reverse reaction.

As time passes, however, the rate of the reverse reaction will slow down and therate of the forward reaction will speed up until they become equal. At that point,the reaction will have reached equilibrium.

Rearranging the equation gives us the same equilibrium constant expression.

We get the same equilibrium constant expression and the same equilibrium con-stant no matter whether we start with only reactants, only products, or a mixtureof reactants and products.

Kc =

kf

kr=

[NO2][ClNO]

[ClNO2][NO]=

[products]

[reactants]


Initially: rateforward V ratereverse

ratereverse = kr (NO2)(ClNO)

rateforward = kf (ClNO2)(NO)

Kc =

[NO2][ClNO]

[ClNO2][NO]= 1.3 * 104 (at 25°C)

Kc =

k f

k r=

[NO2][ClNO]

[ClNO2][NO]

10.5 EQUILIBRIUM CONSTANT EXPRESSIONS 419


Any reaction that reaches equilibrium, no matter how simple or complex,has an equilibrium constant expression that satisfies the following rules.

Rules for Writing Equilibrium Constant Expressions

● Even though chemical reactions that reach equilibrium occur in both direc-tions, the substances on the right side of the equation are assumed to be the“products” of the reaction and the substances on the left side of the equa-tion are assumed to be the “reactants.”

● The products of the reaction are always written above the line, in thenumerator.

● The reactants are always written below the line, in the denominator.

● For systems in which all species are either gases or aqueous solutions, theequilibrium constant expression contains a term for every reactant and everyproduct of the reaction.

● The numerator of the equilibrium constant expression is the product of theconcentrations of each species on the right side of the equation raised to a


E x e r c i s e 1 0 . 2The rate constants for the forward and reverse reactions for the following reac-tion have been measured. At 25°C, kf is 7.3 � 103 liters per mole-second andkr is 0.55 liter per mole-second. Calculate the equilibrium constant for the reac-tion at this temperature.

SolutionWe start by recognizing that the rates of the forward and reverse reactions atequilibrium are the same.

We then substitute the rate laws for the reaction into the equality.

We then rearrange the equation to get the equilibrium constant expression forthe reaction.

The equilibrium constant for the reaction is therefore equal to the rate constantfor the forward reaction divided by the rate constant for the reverse reaction.

Kc =

kf

kr=

7.3 * 103 L/mol-s0.55 L/mol-s

= 1.3 * 104

Kc =

kf

kr=

[NO2][ClNO]

[ClNO2][NO]


At equilibrium: rateforward = ratereverse


➤ CHECKPOINTIf the forward rate constant for a givenreaction is twice the reverse rate con-stant, what is the equilibrium constantfor the reaction?


power equal to the coefficient for that component in the balanced equationfor the reaction.

● The denominator of the equilibrium constant expression is the product ofthe concentrations of each species on the left side of the equation raised toa power equal to the coefficient for that component in the balanced equa-tion for the reaction.

10.5 EQUILIBRIUM CONSTANT EXPRESSIONS 421

E x e r c i s e 1 0 . 4The first step in the series of reactions by which the sugar known as glucoseis metabolized involves the formation of a compound known as glucose-6-phosphate. In theory, this could occur by the direct reaction between glucoseand inorganic phosphate , which is labeled Pi by biochemists.

Write the equilibrium constant expression for this reaction.

SolutionNo matter whether a reaction occurs in the gas phase or, as in this reaction, ina cell in the body, the same principles apply. All the species in this reactionexist in solution, and hence all can be expressed as concentrations in moles perliter. The equilibrium constant expression for this reaction would therefore bewritten as follows.

Kc =

[glucose-6-phosphate]

[glucose][Pi]

glucose (aq) + Pi (aq) uv glucose-6-phosphate (aq)

(PO43 -)

E x e r c i s e 1 0 . 3Write equilibrium constant expressions for the following reactions.

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

(b) 2 SO3(g) 2 SO2(g) � O2(g)

(c) N2(g) � 3 H2(g) 2 NH3(g)

SolutionIn each case, the equilibrium constant expression is the product of the con-centrations of the species on the right side of the equation divided by theproduct of the concentrations of those on the left side of the equation. All con-centrations are raised to the power equal to the coefficient for the species inthe balanced equation.

(a)

(b)

(c) KC =

[NH3]2

[N2][H2]3

KC =

[SO2]2[O2]

[SO3]2

KC =

[N2O4]

[NO2]2

uv

uv

uv


What happens to the magnitude of the equilibrium constant for a reactionwhen we turn the equation around? Consider the following reaction, for example.

The equilibrium constant expression for the equation is written as follows.

This reaction, however, can also be represented by an equation written in theopposite direction.

The equilibrium constant expression is now written as follows.

Each of these equilibrium constant expressions is the inverse of the other. We cantherefore calculate by dividing Kc into 1.

We can also calculate equilibrium constants by combining two or more reac-tions for which the values of Kc are known. For example, we know the equilib-rium constants for the following gas-phase reactions at 200°C.

We can combine these reactions to obtain an equation for the reaction betweenN2 and O2 to form NO2.

The equilibrium constant expression for the overall reaction is equal to the prod-uct of the equilibrium constant expressions for the two steps in the reaction.

The equilibrium constant for the overall reaction is therefore equal to the prod-uct of the equilibrium constants for the individual reactions.

Kc = Kc1 * Kc2 = (2.3 * 10- 19)(3 * 106) = 7 * 10- 13

Kc = Kc1 * Kc2 =

[NO]2

[N2][O2]*

[NO2]2

[NO]2[O2]=

[NO2]2

[N2][O2]2

N2(g) + 2 O2(g) uv 2 NO2(g) Kc = ?

+ 2 NO(g) + O2(g) uv 2 NO2(g)

N2(g) + O2(g) uv 2 NO(g)

2 NO(g) + O2(g) uv 2 NO2(g) Kc2 =

[NO2]2

[NO]2[O2]= 3 * 106

N2(g) + O2(g) uv 2 NO(g) Kc1 =

[NO]2

[N2][O2]= 2.3 * 10- 19

K¿c =

1Kc

=

1

1.3 * 104= 7.7 * 10- 5

K¿c

K¿c =

[ClNO2][NO]

[NO2][ClNO]

NO2(g) + ClNO(g) uv ClNO2(g) + NO(g)

Kc =

[NO2][ClNO]

[ClNO2][NO]= 1.3 * 104 (at 25°C)




10.6 REACTION QUOTIENTS: A WAY TO DECIDE WHETHER A REACTION IS AT EQUILIBRIUM 423

E x e r c i s e 1 0 . 5Given

Determine Kc for the reaction

SolutionWe need to manipulate the chemical equations so that they sum together to givethe equation for which we wish to determine Kc. We need 2 N2O as a product sothe first reaction must be doubled resulting in squaring its equilibrium constant.

In the second reaction we need NO as a reactant so the reaction must bereversed, resulting in taking the inverse of its equilibrium constant.

The two new equations may now be added together.

Kc = 3.1 * 10- 17

N2(g) + 2 NO(g) uv 2 N2O(g) Kc = K¿c1 * K¿c2 = (7.3 * 10- 36)(4.3 * 1018)

2 N2(g) + O2(g) + 2 NO(g) uv 2 N2O(g) + N2(g) + O2(g)

2 NO(g) uv N2(g) + O2(g) K¿c2 = 4.3 * 1018

2 N2(g) + O2(g) uv 2 N2O(g) K¿c1 = 7.3 * 10- 36

2 NO(g) uv N2(g) + O2(g) K¿c2 = 4.3 * 1018

2 NO(g) uv N2(g) + O2(g) K¿c2 = 1>(2.3 * 10- 19)

2 N2(g) + O2(g) uv 2 N2O(g) K¿c1 = 7.3 * 10- 36

2 * (N2(g) +1⁄2 O2(g) uv N2O(g)) K¿c1 = (2.7 * 10- 18)2

N2(g) + 2 NO(g) uv 2 N2O(g) Kc = ?

N2(g) + O2(g) uv 2 NO(g) Kc2 = 2.3 * 10- 19

N2(g) +1⁄2 O2(g) uv N2O(g) Kc1 = 2.7 * 10- 18

10.6 Reaction Quotients: A Way to Decide Whether a Reaction Is at EquilibriumWe now have a model that describes what happens when a reaction reaches equi-librium. At the molecular level, the rate of the forward reaction is equal to the rateof the reverse reaction. Because the reaction proceeds in both directions at the samerate, there is no apparent change in the concentrations of the reactants or the prod-ucts on the macroscopic scale (i.e., the level of objects visible to the naked eye).

This model can also be used to predict the direction in which a reaction hasto shift to reach equilibrium. If the concentrations of the reactants are too largefor the reaction to be at equilibrium, the rate of the forward reaction will be fasterthan that of the reverse reaction, and some of the reactants will be converted toproducts until equilibrium is achieved. Conversely, if the concentrations of thereactants are too small, the rate of the reverse reaction will exceed that of the


forward reaction, and the reaction will convert some of the excess products backinto reactants until the system reaches equilibrium.

We can determine the direction in which a reaction has to shift to reachequilibrium by comparing the reaction quotient (Qc) for the reaction at somemoment in time with the equilibrium constant (Kc) for the reaction. The reactionquotient expression is written in much the same way as the equilibrium constantexpression. But the concentrations used to calculate Qc describe the system at anymoment in time, whereas the concentrations used to calculate Kc describe the sys-tem only when it is at equilibrium.

To illustrate how the reaction quotient is used, consider the following gas-phase reaction.

The equilibrium constant expression for the reaction is written as follows.

By analogy, we can write the expression for the reaction quotient as follows.

There are three important differences between the equilibrium constant expres-sion and the reaction quotient expression. First, we use brackets, such as [HI], inthe equilibrium constant expression to indicate that the reaction is at equilibrium.We then use parentheses, such as (HI), in the reaction quotient expression to indi-cate that the reaction quotient can be calculated at any moment in time. The mostimportant difference between these expressions revolves around the results of thecalculation. There is only one possible value for Kc because the equilibrium con-stant expression is valid only when the reaction is at equilibrium. Qc, on the otherhand, can take on any value from zero to infinity.

If the system contains a large amount of HI and very little H2 and I2, thereaction quotient is very large. If the system contains relatively little HI and alarge amount of H2 and/or I2, the reaction quotient is very small.

At any moment in time, there are three possibilities.

● Qc is smaller than Kc. The system contains too much reactant and notenough product to be at equilibrium. The value of Qc must increase in orderfor the reaction to reach equilibrium. Thus the reaction has to convert someof the reactants into products to come to equilibrium.

● Qc is equal to Kc. If this is true, then the reaction is at equilibrium.

● Qc is larger than Kc. The system contains too much product and not enoughreactant to be at equilibrium. The value of Qc must decrease before the reac-tion can come to equilibrium. Thus the reaction must convert some of theproducts into reactants to reach equilibrium.

Qc =

(HI)2

(H2)(I2)

Kc =

[HI]2

[H2][I2]= 60 (at 350°C)

H2 + I2(g) uv 2 HI(g)


The kinetics of reactions that involvemolecular iodine can be monitoredby watching the formation ordisappearance of the intense violetcolor of elemental iodine.

E x e r c i s e 1 0 . 6Assume that the concentrations of H2, I2, and HI can be measured for the fol-lowing reaction at any moment in time.

H2 + I2(g) uv 2 HI(g) Kc = 60 (at 350°C)


10.7 CHANGES IN CONCENTRATION THAT OCCUR AS A REACTION COMES TO EQUILIBRIUM 425

For each of the following sets of concentrations, determine whether the reac-tion is at equilibrium. If it isn’t, decide in which direction it must go to reachequilibrium.

(a) (H2) � (I2) � (HI) � 0.010 M

(b) (HI) � 0.30 M, (H2) � 0.010 M, (I2) � 0.15 M

(c) (H2) � (HI) � 0.10 M, (I2) � 0.0010 M

Solution(a) The best way to decide whether the reaction is at equilibrium is to com-

pare the reaction quotient with the equilibrium constant for the reaction.

The reaction quotient in this case is smaller than the equilibrium constant.The only way to get the system to equilibrium is to increase the magni-tude of the reaction quotient. This can be done by converting some of theH2 and I2 into HI. The reaction therefore has to shift to the right to reachequilibrium.

(b) The reaction quotient for this set of concentrations is equal to the equilib-rium constant for the reaction.

The reaction is therefore at equilibrium.

(c) The reaction quotient for this set of concentrations is larger than the equi-librium constant for the reaction.

To reach equilibrium, the concentrations of the reactants and products mustchange until the reaction quotient is equal to the equilibrium constant. Thisinvolves converting some of the HI back into H2 and I2. Thus the reaction hasto shift to the left to reach equilibrium.

Qc =

(HI)2

(H2)(I2)=

(0.10)2

(0.10)(0.0010)= 1.0 * 102

7 Kc

Qc =

(HI)2

(H2)(I2)=

(0.30)2

(0.010)(0.15)= 60 = Kc

Qc =

(HI)2

(H2)(I2)=

(0.010)2

(0.010)(0.010)= 1.0 6 Kc

10.7 Changes in Concentration That Occur as a Reaction Comesto EquilibriumThe values of Qc and Kc for a reaction tell us whether a reaction is at equilibriumat any moment in time. If it isn’t, the relative sizes of Qc and Kc tell us the direc-tion in which the reaction must shift to reach equilibrium. Now we need a wayto predict how far the reaction has to go to reach equilibrium. To illustrate howthis is done, let’s look at the following reaction, which is known as the water-gasshift reaction.

CO(g) + H2O(g) Δ CO2(g) + H2(g)


“Water gas” is a mixture of CO and H2 prepared by blowing alternating blasts ofsteam and either air or oxygen through a bed of white-hot coal. The exothermicreactions between coal and oxygen to produce CO and CO2 provide enough energyto drive the reaction between steam and coal to form “water gas.” Water gas is alsoknown as “syngas” because it can be used as the starting materials to produce syn-thetic natural gas or liquid fuels such as gasoline. The water-gas shift reaction canbe used to produce high-purity hydrogen for the synthesis of ammonia.

Suppose that you are faced with the following problem.

The reaction between CO and H2O in the presence of a catalyst at atemperature of 400°C produces a mixture of CO2 and H2 via the water-gas shiftreaction.

The equilibrium constant for this reaction is 0.080 at an elevated tempera-ture. Assume that the initial concentration of both CO and H2O is 0.100 mole perliter and that there is no CO2 or H2 in the system when we start. Calculate theconcentrations of CO, H2O, CO2, and H2 when the reaction reaches equilibrium.

The first step toward solving this problem involves organizing the information sothat it provides clues as to how to proceed. The problem contains the followinginformation: (1) a balanced equation, (2) an equilibrium constant for the reaction,(3) a description of the initial conditions, and (4) an indication of the goal of thecalculation, namely, to figure out the equilibrium concentrations of the four com-ponents of the reaction.

The following format offers a useful way to summarize this information.

Initial: 0.100 M 0.100 M 0 0Equilibrium: ? ? ? ?

We start with the balanced equation and the equilibrium constant for the reactionand then add what we know about the initial and equilibrium concentrations ofthe various components of the reaction. Initially, the flask contains 0.100 mol/Lof CO and H2O in the gas phase and no CO2 or H2. Our goal is to calculate theequilibrium concentrations of the four substances.

Before we do anything else, we have to decide whether the reaction is atequilibrium. We can do this by comparing the reaction quotient for the initial con-ditions with the equilibrium constant for the reaction.

Although the equilibrium constant is small (Kc � 8.0 � 10�2), the reaction quo-tient is even smaller (Qc � 0). The only way for the reaction to get to equilib-rium is for some of the CO and H2O to react to form CO2 and H2.

Because the reaction isn’t at equilibrium, one thing is certain: The concen-trations of CO, H2O, CO2, and H2 will all change as the reaction comes to equi-librium. Because the reaction has to shift to the right to reach equilibrium, theconcentrations of CO and H2O will become smaller, while the concentrations ofCO2 and H2 will become larger.

Qc =

(CO2)(H2)

(CO)(H2O)=

(0)(0)

(0.100)(0.100)= 0 6 Kc

CO(g) + H2O(g) Δ CO2(g) + H2(g) Kc =

[CO2][H2]

[CO][H2O]= 0.080

CO(g) + H2O(g) Δ CO2(g) + H2(g)



At first glance, the problem appears to have four unknowns––the equilib-rium concentrations of CO, H2O, CO2, and H2––and we only have one equation,the equilibrium constant expression. Because it is impossible to solve one equa-tion for four unknowns, we need to look for relationships between the unknownsthat can simplify the problem. One way of achieving this goal is to look at therelationship between the changes that occur in the concentrations of CO, H2O,CO2, and H2 as the reaction approaches equilibrium.


E x e r c i s e 1 0 . 7Calculate the increase in the CO2 and H2 concentrations that occurs when theconcentrations of CO and H2O decrease by 0.022 mol/L as the following reac-tion comes to equilibrium from a set of initial conditions in which the con-centration of both CO and H2O were 0.100 M.

SolutionThis reaction has a 1:1:1:1 stoichiometry, as shown in Figure 10.5. For everymole of CO that reacts with a mole of H2O, we get 1 mole of CO2 and 1 mole of H2. Thus the magnitude of the change in the concentration of COand H2O that occurs as the reaction comes to equilibrium is equal to the mag-nitude of the change in the CO2 and H2 concentrations. The only differenceis the sign of this change. The concentrations of CO and H2O become smaller,whereas the concentrations of CO2 and H2 become larger as the reactioncomes to equilibrium. If 0.022 mol/L of CO and H2O are consumed as thereaction comes to equilibrium, 0.022 mol/L of CO2 and H2 must be formedat the same time.

CO(g) + H2O(g) Δ CO2(g) + H2(g)

Fig. 10.5 The reaction between CO and H2O in the gasphase to form CO2 and H2 is a reversible reaction with a1:1:1:1 stoichiometry.

Exercise 10.7 raises an important point. There is a relationship between thechanges in the concentrations of the four components of the water-gas shift reac-tion as it comes to equilibrium because of the stoichiometry of the reaction. Let’snow continue to examine the problem posed at the beginning of this section.

It would be useful to have a symbol that represents the change that occursin the concentration of one of the components of a reaction as it goes from theinitial conditions to equilibrium. Let’s define ¢(X) as the magnitude of the changein the concentration of X as the reaction comes to equilibrium. Thus, ¢(CO) isthe magnitude of the change in the concentration of CO that occurs when thiscompound reacts with water in the gas phase to form CO2 and H2.

The important quantities for reactions that come to equilibrium are theconcentrations of each reactant and/or product at equilibrium. Let’s look, onceagain, at the water-gas shift reaction. By definition, the concentration of CO at

➤ CHECKPOINTAssume that 1.00 mol of PCl3 and 1.00 mol of Cl2 are added to an empty1.00-L container. Furthermore, assumethat the concentration of PCl3 decreasesby 0.96 mol/L as the reaction comes toequilibrium. What are the concentra-tions of PCl5 and Cl2 at equilibrium?

PCl5 Δ PCl3 + Cl2

O O P PSS OQqC CO O O OH H� H O� HΔ


equilibrium is equal to the initial concentration of CO minus the amount of COconsumed as the reaction comes to equilibrium.

[CO] � (CO)i � ¢(CO)Concentration of Initial concentration CO consumed as

CO in moles per liter of CO in the reaction comesat equilibrium moles per liter to equilibrium

In a similar fashion, we can define the concentration of H2O at equilibrium as theinitial concentration of water minus the amount of water consumed as the reac-tion comes to equilibrium.

[H2O] � (H2O)i � ¢(H2O)Concentration of Initial concentration H2O consumed as

H2O in moles per liter of H2O in the reaction comesat equilibrium moles per liter to equilibrium

We can then define ¢(CO2) and ¢(H2) as the changes that occur in the con-centrations of CO2 and H2 as the reaction comes to equilibrium. Because CO2

and H2 are formed as this reaction comes to equilibrium, the concentrations ofboth substances at equilibrium will be larger than their initial concentrations.

Because of the 1:1:1:1 stoichiometry of the reaction, the magnitude of the changein the concentrations of CO and H2O as the reaction comes to equilibrium is equalto the changes in the concentrations of CO2 and H2, as we saw in Exercise 10.7.

We can therefore rewrite the equations that define the equilibrium concen-trations of CO, H2O, CO2, and H2 in terms of a single unknown: ¢C.

Substituting what we know about the initial concentrations of CO, H2O, CO2, andH2 into the equations that define the concentration of each component of the reac-tion at equilibrium gives the following result.

We can now summarize what we know about the water-gas shift reaction asfollows.

Initial: 0.100 M 0.100 M 0 0Change: �¢C �¢C �¢C �¢CEquilibrium: 0.100 � ¢C 0.100 � ¢C ¢C ¢C

CO(g) + H2O(g) Δ CO2(g) + H2(g)

[CO2] = [H2] = 0 + ¢C

[CO] = [H2O] = 0.100 - ¢C

[H2] = (H2)i + ¢C

[CO2] = (CO2)i + ¢C

[H2O] = (H2O)i - ¢C

[CO] = (CO)i - ¢C

¢(CO) = ¢(H2O) = ¢(CO2) = ¢(H2)

[H2] = (H2)i + ¢(H2)

[CO2] = (CO2)i + ¢(CO2)



We now have only one unknown, ¢C, and we need only one equation to solvefor one unknown. The obvious equation to turn to is the equilibrium constantexpression for the reaction.

Substituting what we know about the equilibrium concentrations of CO, H2O,CO2, and H2 into the equation gives the following result.

This equation can be expanded and then rearranged to give an equation

that can be solved with the quadratic formula.

Although two answers come out of the calculation, only the positive rootmakes any physical sense because we can’t have a negative concentration. Thusthe magnitude of the change in the concentrations of CO, H2O, CO2, and H2 asthe reaction comes to equilibrium is 0.022 mol/L.

Substituting the value of ¢C back into the equations that define the equilibriumconcentrations of CO, H2O, CO2, and H2 gives the following results for the ques-tion posed at the beginning of this section.

In other words, slightly less than one-quarter of the CO and H2O react to formCO2 and H2 when the reaction comes to equilibrium.

To check whether the results of the calculation represent legitimate valuesfor the equilibrium concentrations of the four components of this reaction, we cansubstitute these values into the equilibrium constant expression.

The results of our calculation must be legitimate because the equilibrium constantcalculated from these concentrations is equal to the value of Kc given in the prob-lem, within experimental error.

[CO2][H2]

[CO][H2O]=

[0.022][0.022]

[0.078][0.078]= 0.080

[CO2] = [H2] = 0 M + 0.022 M = 0.022 M

[CO] = [H2O] = 0.100 M - 0.022 M = 0.078 M

¢C = 0.022

¢C = 0.022 or -0.039

¢C =

-b ; 2b2- 4ac

2a=

- (0.016) ; 2(0.016)2- 4(0.92)(-0.00080)

2(0.92)

0.92[¢C]2+ 0.016[¢C] - 0.00080 = 0

[¢C][¢C]

[0.100 - ¢C][0.100 - ¢C]= 0.080

Kc =

[CO2][H2]

[CO][H2O]= 0.080



10.8 Hidden Assumptions That Make Equilibrium Calculations EasierThe water-gas shift reaction has a stoichiometry that can be described as 1:1:1:1.For each mole of carbon monoxide consumed in the reaction, one mole of wateris consumed and one mole of carbon dioxide and one mole of hydrogen are pro-duced. It is easy to imagine gas-phase reactions with a more complex stoichiom-etry. Consider the equilibrium between sulfur trioxide and a mixture of sulfurdioxide and oxygen, for example.

Sulfur trioxide decomposes to give sulfur dioxide and oxygen with an equi-librium constant of 1.6 � 10�10 at 300°C.

2 SO3(g) uv 2 SO2(g) + O2(g)


E x e r c i s e 1 0 . 8Suppose that 1.00 mol of cis-2-butene was placed in a 1.00-L flask that con-tained no trans-2-butene at 400ºC. What would be the concentrations of thecis-2-butene and trans-2-butene at equilibrium if Kc � 1.27 for the reaction inwhich the cis-2-butene isomer is converted into trans-2-butene?

SolutionWe start, as always, by representing the information in the problem in the fol-lowing format.

Initial: 1.00 M 0 MEquilibrium:

We then write the equilibrium constant expression for the reaction.

Substituting the expressions for the equilibrium concentrations of cis-2-buteneand trans-2-butene into this equation gives

We then solve for ¢C and use the results of this calculation to determine theconcentrations of butane and isobutane at equilibrium.

[cis-2-butene] = 1.00 - ¢C = 0.441 M

[trans-2-butene] = ¢C = 0.559 M

[¢C]

[1.00 - ¢C]= 1.27

K =

[trans-2-butene]

[cis-2-butene]= 1.27

¢C1.00 - ¢C

cis-2-butene uv trans-2-butene

H H

C C

CH3H

C C

HCH3

CH3 CH3

ED

trans-2-butenecis-2-butene

➤ CHECKPOINTDo the equilibrium concentrationsfound in Exercise 10.8 correctly repro-duce the Kc?


Calculate the equilibrium concentrations of the three components of the sys-tem if the initial concentration of SO3 is 0.100 M.

Once again, the first step in the problem involves building a representation of theinformation in the problem.

Initial: 0.100 M 0 0Equilibrium: ? ? ?

We then compare the reaction quotient for the initial conditions with the equilib-rium constant for the reaction.

Because the initial concentrations of SO2 and O2 are zero, the reaction has to shiftto the right to reach equilibrium. As might be expected, some of the SO3 has todecompose to SO2 and O2.

The stoichiometry of the reaction is more complex than the reaction in theprevious section, but the changes in the concentrations of the three componentsof the reaction are still related, as shown in Figure 10.6. For every 2 moles ofSO3 that decompose, we get 2 moles of SO2 and 1 mole of O2.

The signs of the ¢C terms in the problem are determined by the fact thatthe reaction has to shift from left to right to reach equilibrium. The coefficientsin the ¢C terms mirror the coefficients in the balanced equation for the reaction.Because twice as many moles of SO2 are produced as moles of O2, the changein the concentration of SO2 as the reaction comes to equilibrium must be twiceas large as the change in the concentration of O2. Because two moles of SO3 areconsumed for every mole of O2 produced, the change in the SO3 concentrationmust be twice as large as the change in the concentration of O2.

Initial: 0.100 M 0 0Change:Equilibrium:

Substituting what we know about the problem into the equilibrium constantexpression for the reaction gives the following equation.

This equation is a bit more of a challenge to expand, but it can be rearranged togive the following cubic equation.

Solving cubic equations is difficult, however. This problem is therefore an exam-ple of a family of problems that are difficult, if not impossible, to solve exactly.Such problems are solved with a general strategy that consists of making anassumption or approximation that turns them into simpler problems.

4[¢C]3- (6.4 * 10- 10)[¢C]2

+ (6.4 * 10- 11)[¢C] - (1.6 * 10- 12) = 0

Kc =

[SO2]2[O2]

[SO3]2

=

[2¢C]2[¢C]

[0.100 - 2¢C]2= 1.6 * 10- 10

¢C2¢C0.100 � 2¢C�¢C�2¢C�2¢C

2 SO3(g) uv 2 SO2(g) + O2(g) Kc = 1.6 * 10-10

QC =

(SO2)2(O2)

(SO3)2

=

(0)2(0)

(0.100)2= 0 6 KC

2 SO3(g) uv 2 SO2(g) + O2(g) Kc = 1.6 * 10-10

10.8 HIDDEN ASSUMPTIONS THAT MAKE EQUILIBRIUM CALCULATIONS EASIER 431

Fig. 10.6 The stoichiometry of thisreaction requires that the change inconcentrations of both SO3 and SO2

must be twice as large as the changein the concentration of O2 that occursas the reaction comes to equilibrium.

S +

+ +

O

O

O

S

O

S

O

O

O

O

O

S

O

O

O

➤ CHECKPOINTThe three lines of information givenunder the balanced chemical equationhelp to organize the concentrations ofreactants and products when solving anequilibrium problem. Which relation-ship between concentrations (initial,change, equilibrium) must follow thestoichiometry of the balanced chemicalequation?


What assumption can be made to simplify the problem? Let’s go back tothe first thing we did after building a representation for the problem. We startedour calculation by comparing the reaction quotient for the initial concentrationswith the equilibrium constant for the reaction.

We then concluded that the reaction quotient (Qc � 0) was smaller than the equi-librium constant (Kc � 1.6 � 10�10) and decided that some of the SO3 wouldhave to decompose in order for the reaction to come to equilibrium.

But what are the relative sizes of the reaction quotient and the equilibriumconstant for this reaction? The initial values of Qc and Kc are both relatively small,which means that the initial conditions are reasonably close to equilibrium. As aresult, the reaction doesn’t have far to go to reach equilibrium. It is therefore rea-sonable to assume that ¢C is relatively small in this problem.

It is essential to understand the nature of the assumption being made. Wearen’t assuming that ¢C is zero. If we did that, some of the unknowns would dis-appear from the equation! We are only assuming that ¢C is so small compared withthe initial concentration of SO3 that it doesn’t make a significant difference when2¢C is subtracted from that number. We can write the assumption as follows.

Let’s now go back to the equation we are trying to solve.

By assuming that 2¢C is very much smaller than 0.100, we can replace this equa-tion with the following approximate equation.

We do not assume that ¢C is zero, in which case the 2¢C and ¢C terms in thenumerator would disappear. We are assuming that ¢C is much smaller than 0.100.

Expanding this equation gives an equation that is much easier to solve for ¢C.

Before we can go any further, we have to check our assumption that 2¢C is sosmall compared with 0.100 that it doesn’t make a significant difference when itis subtracted from that number. Is the assumption valid? Is 2¢C small enoughcompared with 0.100 to be ignored?

¢C is so small that 2¢C is smaller than the experimental error in the measure-ment of the initial concentration of SO3 and can therefore be legitimately ignored.As a general rule, the change in concentration is small enough to be ignored if

0.100M - 2(0.000074M) L 0.100M

¢C L 7.4 * 10- 5M

4¢C3L 1.6 * 10- 12

[2¢C]2[¢C]

[0.100]2L 1.6 * 10- 10

[2¢C]2[¢C]

[0.100 - 2¢C]2= 1.6 * 10- 10

0.100 M - 2¢C L 0.100 M

QC =

(SO2)2(O2)

(SO3)2

=

(0)2(0)

(0.100)2= 0 6 KC



the change in concentration is less than 5% of the initial concentration. In thisexample, the change in the concentration of SO3 as the reaction comes to equi-librium is only 0.15%, which is much smaller than 5%.

We can now use the approximate value of ¢C to calculate the equilibriumconcentrations of SO3, SO2, and O2.

The equilibrium between SO3 and mixtures of SO2 and O2 therefore stronglyfavors SO3, not SO2.

We can check the results of our calculation by substituting these results intothe equilibrium constant expression for the reaction.

The value of the equilibrium constant that comes out of the calculation agreesclosely with the value given in the problem. Our assumption that 2¢C is negli-gibly small compared with the initial concentration of SO3 is therefore valid, andwe can feel confident in the answers it provides.

We can also use the equilibrium expression to solve for the concentrationof products and reactants at equilibrium when a mixture of both products andreactants is present initially. Consider the same reaction, in which SO3 decom-poses to form SO2 and O2. But this time let’s assume that the initial concentra-tions of both SO3 and O2 are 0.100 M. We start, as always, by arranging therelevant information in the problem in the following format.

Initial: 0.100M 0 0.100MChange: �2�C �2�C ��CEquilibrium: 0.100–2�C �2�C 0.100 � �C

We then compare the reaction quotient for the initial conditions with the equilib-rium constant for the reaction.

The reaction must proceed to the right to reach equilibrium because thereis no SO2 present initially. Substituting what we know about the concentrationsof the three components of the reaction at equilibrium into the equilibrium con-stant expression gives the following result.

Kc =

[SO2]2[O2]

[SO3]2

=

[2¢C]2[0.100 + ¢C]

[0.100 - 2¢C]2= 1.6 * 10- 10

QC =

(SO2)2(O2)

(SO3)2

=

(0)2(0.100)

(0.100)2= 0 6 KC

2 SO3(g) uv 2 SO2(g) + O2(g) Kc = 1.6 * 10-10

Kc =

[SO2]2[O2]

[SO3]2

=

[1.5 * 10- 4]2[7.4 * 10- 5]

[0.100]2= 1.7 * 10- 10

[O2] = ¢C L 7.4 * 10-5M

[SO3] = 2¢C L 1.5 * 10-4M

[SO3] = 0.100 M - 2¢C L 0.100M

2¢C

0.100* 100% =

2(0.000074)

0.100* 100% = 0.15% 6 5%

10.8 HIDDEN ASSUMPTIONS THAT MAKE EQUILIBRIUM CALCULATIONS EASIER 433


Because both Qc and Kc are relatively small, we can assume that ¢C and 2¢Cwill be small compared with the initial concentrations of SO3 and O2, which givesus the following approximate equation.

Rearranging this equation gives the following result,

which can be solved for the approximate value of ¢C.

¢C is much smaller than the initial concentration of either O2 or SO3, confirm-ing the validity of our approximation. We can therefore use this approximate valueof ¢C to calculate the equilibrium concentrations of the three components of thereaction.

The concentration of SO2 (4.0 � 10�6 M) produced in this calculation when O2

was initially present is considerably smaller than the concentration of SO2 (1.5 �10�4 M) produced in the previous calculations when there was no O2 or SO2 ini-tially present.

10.9 What Do We Do When the Assumption Fails?What do we do when we encounter a problem for which the assumption that ¢Cis small compared with the initial concentrations cannot possibly be valid? Con-sider the following problem, for example, which plays an important role in thechemistry of the atmosphere.

The equilibrium constant for the reaction between nitrogen oxide and oxy-gen to form nitrogen dioxide is 3.0 � 106 at 200°C

Assume initial concentrations of 0.100 M for NO and 0.050 M for O2. Cal-culate the concentrations of the three components of the reaction at equilibrium

We start, as always, by representing the information in the problem as follows.

Initial: 0.100 M 0.050 M 0Equilibrium: ? ? ?

The first step is always the same: Compare the initial value of the reaction quo-tient with the equilibrium constant.

2 NO(g) + O2(g) uv 2 NO2(g) Kc = 3.0 * 106

2 NO(g) + O2(g) uv 2 NO2(g)

[O2] = 0.100 M + 2¢C L 0.100 M

[SO2] = 2¢C L 4.0 * 10- 6M

[SO3] = 0.100 M - 2¢C L 0.100 M

¢C L 2.0 * 10- 6M

4[¢C]2= 1.6 * 10- 11

[2¢C]2[0.100]

[0.100]2L 1.6 * 10- 10


➤ CHECKPOINTFor which of the following equilibriumconstants could it be safely assumedthat is small compared with the ini-tial concentration of A when A decom-poses to form B and C?

(a)(b)(c)(d) K = 1.0 * 10-10

K = 1.0 * 10-1K = 1.0 * 10-5K = 1.0 * 105

A(g) uv B(g) + C(g)

¢C


The relationship between the initial reaction quotient (Qc � 0) and the equilib-rium constant (Kc � 3.0 � 106) tells us something we may already have sus-pected: The reaction must shift to the right to reach equilibrium.

Some might ask: Why calculate the initial value of the reaction quotient forthe reaction? Isn’t it obvious that the reaction has to shift to the right to produceat least some NO2? Yes, it is. But calculating the value of Qc does more than tellus in which direction the reaction has to shift to reach equilibrium. It also givesus an indication of how far the reaction has to go to reach equilibrium.

In this case, Qc is so very much smaller than Kc for the reaction that wehave to conclude that the initial conditions are very far from equilibrium. It wouldtherefore be a mistake to expect that ¢C is small when compared with the initialconcentrations of the reactants.

We can’t assume that ¢C is negligibly small in this problem, but we canredefine the problem so that the assumption becomes valid. The key to achievingthis goal is to remember the conditions under which we can assume that ¢C issmall enough to be ignored. This assumption is valid only when Qc is of the samerelative order of magnitude as Kc (i.e., when Qc and Kc are both much larger than1 or much smaller than 1). We can solve problems for which Qc isn’t close to Kc

by redefining the initial conditions so that Qc becomes close to Kc (Figure 10.7).To show how this can be done, let’s return to the problem given in this section.

The equilibrium constant for the reaction between NO and O2 to form NO2

is much larger (Kc � 3.0 � 106) than Qc. This means that the equilibrium favorsthe products of the reaction. The best way to handle the problem is to drive thereaction as far as possible to the right, and then let it come back to equilibrium.Let’s therefore define an intermediate set of conditions that correspond to whatwould happen if we push the reaction as far as possible to the right.

Initial: 0.100 M 0.050 M 0Change: �0.100 M �0.050 M �0.100 MIntermediate: 0 0 0.100 M

We can see where this gets us by calculating the reaction quotient for the inter-mediate conditions.

The reaction quotient is now larger than the equilibrium constant, and the reac-tion has to shift back to the left to reach equilibrium. Some of the NO2 must nowdecompose to form NO and O2.

The relationship between the changes in the concentrations of the three com-ponents of this reaction is determined by the stoichiometry of the reaction, asshown in Figure 10.8. We therefore set up the problem as follows.

Intermediate: 0 0 0.100 MChange: �2¢C �¢C �2¢CEquilibrium: 2¢C ¢C 0.100�2¢C

2 NO(g) + O2(g) uv 2 NO2(g) Kc = 3.0 * 106

Qc =

(NO2)2

(NO)2(O2)=

(0.100)2

(0)2(0)= q

2 NO(g) + O2(g) uv 2 NO2(g) Kc = 3.0 * 106

Qc = (NO2)

2

(NO)2(O2) =

(0)2

(0.100)2(0.050) = 0 V Kc

10.9 WHAT DO WE DO WHEN THE ASSUMPTION FAILS? 435

Fig. 10.7 When the initial conditionsare very far from equilibrium, it isoften useful to redefine the problem.This involves driving the reaction asfar as possible in the directionfavored by the equilibrium constant.When the reaction returns toequilibrium from the intermediateconditions, changes in theconcentrations of the components ofthe reaction are often small enoughcompared with the initialconcentration to be ignored.

Reactants

EquilibriumIntermediate

Redefine problemso that Δ is small

+

O

O

OO

N

N

+ON

O

O

N+

O

Fig. 10.8 Once again, thestoichiometry of the reactiondetermines the relationship amongthe magnitudes of the changes in theconcentrations of the threecomponents of the reaction as itcomes to equilibrium.


We then substitute what we know about the reaction into the equilibrium constantexpression.

Because the reaction quotient for the intermediate conditions and the equilibriumconstant are both relatively large, we can assume that the reaction doesn’t havevery far to go to reach equilibrium. In other words, we assume that 2¢C is smallcompared with the intermediate concentration of NO2, and we derive the follow-ing approximate equation.

We then solve the equation for an approximate value of ¢C.

We now check our assumption that 2¢C is small enough compared with the inter-mediate concentration of NO2 to be ignored.

The value of 2¢C is less than 2% of the intermediate concentration of NO2, whichmeans that it can be legitimately ignored in the calculation.

Since the approximation is valid, we can use the value of ¢C to calculatethe equilibrium concentrations of NO2, NO, and O2.

The assumption that ¢C is small compared with the initial concentrations of thereactants or products works best under the following conditions.

● When both Qc and Kc are both much smaller than one.

● When both Qc and Kc are both much larger than one.

10.10 The Effect of Temperature on an Equilibrium ConstantThe temperature at which the reaction was run has been reported each time anequilibrium constant has been given in this chapter. If the equilibrium constant isreally constant, why do we have to worry about the temperature of the reaction?

Although the value of Kc for a reaction is constant at a given temperature,it can change with temperature. Consider the equilibrium between NO2 and itsdimer, N2O4, for example.

2 NO2(g) uv N2O4(g)

[O2] = ¢C L 0.00094 M

[NO] = 2¢C L 0.0019 M

[NO2] = 0.100 - 2¢C L 0.098 M

2(0.00094)

0.100* 100% = 1.9%

¢C L 9.4 * 10- 4 M

30.1004232¢C423¢C4 L 3.0 * 106

Kc =

[NO2423NO]23O24 =

30.100 - 2¢C4232¢C423¢C4 = 3.0 * 106



The equilibrium constant for this reaction decreases significantly with increasingtemperature, as shown in Table 10.2.

As noted in Exercise 10.3, the equilibrium constant for this reaction is equalto the concentration of the product of the reaction divided by the square of theconcentration of the reactant.

As a result, an increase in the equilibrium constant implies a shift toward the prod-uct of the reaction at equilibrium. In other words, at low temperatures, the equilib-rium favors the dimer, N2O4. At high temperatures, the equilibrium favors NO2. Thefact that equilibrium constants are temperature dependent explains why you mayfind different values for the equilibrium constant for the same chemical reaction.

10.11 Le Châtelier’s PrincipleIn 1884, the French chemist and engineer Henry-Louis Le Châtelier proposed oneof the central concepts of chemical equilibria. Le Châtelier’s principle suggeststhat a change in one of the variables that describe a system at equilibrium pro-duces a shift in the position of the equilibrium. The following rules summarizeLe Châtelier’s principle for closed systems at either constant pressure or con-stant temperature.

A closed system is one in which no matter can pass into or out of the system.

An increase in temperature for a closed system at constant pressure shifts theequilibrium in the direction in which the system absorbs heat from its sur-roundings.

For example, in an endothermic reaction, the equilibrium shifts to the right astemperature increases. In an exothermic reaction, the equilibrium shifts to the leftas temperature increases.

An increase in pressure for a closed system at constant temperature shifts theequilibrium in the direction in which the volume of the system decreases.1

Our attention so far has been devoted to describing what happens as a sys-tem comes to equilibrium. Le Châtelier’s principle describes what happens to asystem at equilibrium when something momentarily takes it away from equilib-rium. This section focuses on three ways in which we can change the conditionsof a chemical reaction at equilibrium: (1) changing the concentration of one ofthe components of the reaction, (2) changing the pressure or volume of the sys-tem, and (3) changing the temperature.

CHANGES IN CONCENTRATION

To illustrate what happens when we change the concentration of one of the reac-tants or products of a reaction at equilibrium, let’s consider a system that consistsof 0.500 mol of cis-2-butene in a 1.0 L flask at 400�C.

Kc =

3N2O443NO242

10.11 LE CHÂTELIER’S PRINCIPLE 437

Table 10.2Temperature Dependence ofthe Equilibrium Constant forthe Formation of N2O4

Temperature (�C) Kc

�78 4.0 � 108

0 1.4 � 103

25 1.7 � 102

100 2.1

1Ira N. Levine, Physical Chemistry 6th Edition, McGraw-Hill, New York, pp. 195–196 (2009).


Initial: 0.500 M 0Equilibrium: 0.500 � ¢C ¢C

Substituting the expressions for the equilibrium concentration of cis-2-butene andtrans-2-butene into the equilibrium constant expression gives:

This equation can then be solved for the equilibrium concentrations of cis-2-butene and trans-2-butene.

Now suppose that 0.200 mol of trans-2-butene is added to the reactionwhile it is at equilibrium. The system is no longer at equilibrium because theconcentration of trans-2-butene is now 0.480 M. The reaction quotient at theinstant the trans-2-butene is added is larger than the equilibrium constant forthe reaction.

The system must shift to reestablish the equilibrium ratio of 1.27, and so someof the trans-2-butene will be converted into cis-2-butene. When equilibrium isreestablished,

Initial: 0.220 M 0.480Equilibrium: 0.220 � ¢C 0.480 � ¢C

Substituting the new conditions for equilibrium into the equilibrium constantexpression gives the following equation:

from which ¢C can be found to be 0.0885 M.

The calculation can be checked to determine if the ratio [trans-2-butene]/ [cis-2-butene] correctly gives Kc.

Kc =

[0.392]

[0.309]= 1.27

[cis-2-butene] = 0.220 + 0.0885 = 0.309 M

[trans-2-butene] = 0.480 - 0.0885 = 0.392 M

Kc =

[trans-2-butene]

[cis-2-butene]=

[0.480 - ¢C]

[0.220 + ¢C]= 1.27

cis-2-butene (g) Δ trans-2-butene

Qc =

[trans-2-butene]

[cis-2-butene]=

0.4800.220

= 2.18 7 Kc

[cis-2-butene] = 0.500 - ¢C = 0.220 M

[trans-2-butene] = ¢C = 0.280 M

Kc =

[trans-2-butene]

[cis-2-butene]=

[¢C]

[0.500 - ¢C]= 1.27

H H

C C

CH3H

C C

HCH3

CH3 CH3

ED

trans-2-butenecis-2-butene



By comparing the new equilibrium concentrations with those obtainedbefore adding trans-2-butene, we can see the effect of increasing the concentra-tion of the trans-2-butene on the equilibrium mixture.

Before After

[trans-2-butene] � 0.480 [trans-2-butene] � 0.392[cis-2-butene] � 0.220 [cis-2-butene] � 0.309

The addition of trans-2-butene shifted the equilibrium in such a way as to pro-duce more cis-2-butene and remove some of the added trans-2-butene. Additionof more cis-2-butene would have the opposite effect, causing the equilibrium toshift toward the products.

The addition of a reactant or product will shift the equilibrium to reduce theamount of added reactant or product. The removal of a reactant or product willcause the equilibrium to shift to produce that reactant or product. Reaction quo-tients (Qc) introduced in Section 10.6 provide a way to understand the shift inequilibrium that occurs when one component of the reaction is either added orremoved from the reaction mixture.

The equilibrium between cis-2-butene and trans-2-butene can be used toshow how to predict the effect of changes in either the concentration or volumeof a system at constant temperature. Consider a syringe containing an equilibriummixture of the two gases. What is the effect on the equilibrium of pulling theplunger partway out of the syringe, thereby increasing the volume of the system?

The first step toward answering this question involves considering the defi-nition of the terms in the equilibrium constant expression for the reaction. By def-inition, the concentration of either component of the reaction is equal to the num-ber of moles of that substance divided by the volume of the sample.

Because the volumes cancel, this equilibrium is not affected by any change in volume.What would happen to this system if more trans-2-butene was added to the

mixture? The reaction quotient would now be larger than the equilibrium constantfor the reaction because the value of Qc for the reaction would be equal to theratio of moles of trans-2-butene to moles of cis-2-butene, and the equilibrium hasbeen perturbed by adding more trans-2-butene.

So in order to restore equilibrium, ntrans must decrease and ncis must increase.For reactions in which the sum of the coefficients of the gaseous products

is equal to the sum of the coefficients for the gaseous reactants, no effect will beobserved due to volume changes alone. If the sums of the coefficients are not thesame, any change in volume will produce a shift in the reaction direction.

Qc =

ntrans

ncis7 Kc

Kc =

[trans-2-butene]

[cis-2-butene]=

[ntrans>V]

[ncis>V]=

ntrans

ncis


E x e r c i s e 1 0 . 9The following reaction is at equilibrium in a syringe. The plunger is pushed intothe syringe, thereby decreasing the volume of the system at constant temperature.

3 O2(g) uv 2 O3(g)


CHANGES IN PRESSURE

Sometimes it is convenient to discuss gas-phase chemical reactions in terms ofthe partial pressures of individual species rather than their concentrations. How-ever, this makes no difference to the general conclusions about equilibrium thathave been discussed. The reason is that partial pressures are related directly toconcentrations through the ideal gas law equation.

The effect of changing the pressure on a gas-phase reaction depends on thestoichiometry of the reaction. We can demonstrate this by looking at the result ofincreasing the total pressure on the following reaction at equilibrium.

Let’s start with a system that initially contains 2.5 atm of N2 and 7.5 atm of H2

at 500�C, and allow the reaction to come to equilibrium. Let’s then compress thesystem by increasing the pressure by a factor of 10 and allow the system to return

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


In which direction would the reaction have to shift to get back to equilibrium?

SolutionThe equilibrium constant expression for this reaction can be analyzed as follows.

When the volume is changed, the system will no longer be at equilibrium. Todetermine the direction that the reaction must shift in order to reestablish equi-librium, we need the Qc expression for this reaction.

We can define Qn as the ratio of the moles of product raised to their appro-priate powers divided by the moles of reactants raised to their appropriate pow-ers. For this reaction,

Qc � Qn � V

Pushing the plunger into the syringe decreases the volume but does notimmediately change the moles of O2 or O3. When the volume, V, is decreasedat constant temperature, Qc � Kc because the volume term is smaller but themoles of O2 and O3 are initially the same. The reaction must therefore shift tothe right in order to establish a new equilibrium. This means that some of theO2 will be consumed and some O3 will be produced in order to increase thevalue of Qc until it is equal to Kc and the system is at a new equilibrium.

A decrease in volume at constant temperature for a closed system alsocorresponds to an increase in pressure. This is consistent with Le Chatelier’sprinciple that an increase in pressure at constant temperature shifts the equi-librium in the direction in which the volume of the system decreases. The vol-ume of the system decreases because the volume of two moles of O3 is lessthan the volume of three moles of O2.

Qc =

(O3)2

(O2)3

=

[nO3>V]2

[nO2>V]3

=

[nO3]2

[nO2]3

* V

Kc =

[O3]2

[O2]3

=

[nO3>V]2

[nO2>V]3

=

[nO3]2

[nO2]3

* V


to equilibrium. The partial pressures at equilibrium of all three components of thereaction change when the system is compressed.

Before Compression After Compression

Before the system was compressed, the partial pressure of NH3 was only about1% of the total pressure. After the system is compressed, the partial pressure ofNH3 is almost 10% of the total.

These data provide another example of Le Châtelier’s principle. A reactionat equilibrium was subjected to an increase in the total pressure on the system.The reaction then shifted toward the products because this reduced the total num-ber of molecules in the gaseous mixture, as shown in Figure 10.9. This in turndecreased the total pressure exerted by the gases.

Whenever the pressure exerted on a system, at constant temperature, con-taining gaseous reactants or products is changed, the equilibrium will shift. If thepressure is increased, the equilibrium will shift in the direction of fewer moles ofgas. If the pressure is decreased, it will shift to produce more moles of gas.

CHANGES IN TEMPERATURE

Changes in the concentrations of the reactants or products of a reaction shift theposition of the equilibrium, but they don’t change the equilibrium constant for thereaction. Similarly, a change in the pressure on a reaction shifts the position ofthe equilibrium without changing the magnitude of the equilibrium constant.Changes in the temperature of the system, however, affect the position of the equi-librium by changing the magnitude of the equilibrium constant for the reaction,as shown in Section 10.10.

The reaction in which NO2 dimerizes to form N2O4 provides an example ofthe effect of changes in temperature on the equilibrium constant for a reactionand the resulting shift in equilibrium. The reaction is exothermic.

The equilibrium constant therefore decreases with increasing temperature,as shown in Table 10.2. This results in a shift in the equilibrium toward the left

2 NO2(g) uv N2O4(g) ¢H° = -57.2 kJ/molrxn

PH2� 62 atmPH2

� 7.3 atmPN2

� 21 atmPN2� 2.4 atm

PNH3� 8.4 atmPNH3

� 0.12 atm


H — H

H — H

H — H

H — HH — H

H — HH — H

H — HH — HN

HH

H

N N

N NN N

H — H

H — H

H — H

N

HH

HN

HH

HN

HH

HN

HH

HN

HH

H

N N

Fig. 10.9 The total number of molecules in thesystem decreases when N2 reacts with H2 to formNH3. Shifting the equilibrium toward NH3

decreases the total pressure of the gaseous mixture.

➤ CHECKPOINTWhich way will the equilibrium for thefollowing reaction shift if more P2 isadded to the system at equilibrium?Which way will the equilibrium shift ifthe pressure is increased? The tempera-ture remains constant.

2 P2(g) uv P4(g)


to increase the concentration of NO2 at equilibrium. For endothermic reactions,an increase in temperature will cause an increase in the equilibrium constant andtherefore a shift in equilibrium toward the products.


E x e r c i s e 1 0 . 1 0Predict the effect of the following changes on the equilibrium for the decom-position of SO3 to form SO2 plus O2 for a sample contained in a syringe.

(a) Increasing the pressure by decreasing the volume of the syringe at con-stant temperature.

(b) Decreasing the pressure by increasing the volume of the syringe at con-stant temperature.

(c) Adding an inert gas at constant volume and constant temperature.

(d) Adding an inert gas at constant temperature and constant pressure.

(e) Adding O2 at constant volume and constant temperature.

(f) Adding SO3 at constant pressure and constant temperature.

(g) Increasing the temperature at constant pressure.

SolutionWe start by analyzing the equilibrium constant and reaction quotient expres-sions for the reaction.

(a) Because the volume is decreased, Qc Kc. The reaction therefore shiftsto the left.

(b) Because the volume is increased, Qc � Kc. The reaction therefore shifts tothe right.

(c) Because the volume is constant, Qc � Kc and there is no change in theequilibrium. Although the addition of an inert gas at constant volume andtemperature does increase the pressure, the total moles also increase. Asshown in Chapter 6

from which we see:

Thus the ratio Ptot/ntot is unchanged because T and V are constant.

ptot

ntot=

RT

V

Ptot * V = ntot * RT

Qc = Qn *

1V

Qc =

(SO2)2(O2)

(SO3)2

=

(nSO2>V)2(nO2

>V)

(nSO3>V)2

=

(nSO2)2(nO2

)

(nSO3)2

*

1V

Kc =

[SO2]2[O2]

[SO3]2

=

[nSO2>V]2[nO2

>V ]

[nSO3>V]2

=

[nSO2]2[nO2

]

[nSO3]2

*

1V

2 SO3(g) uv 2 SO2(g) + O2(g) ¢H° = 197.84 kJ/molrxn


For a general gas phase reaction

the relationship between Qc and Qn can be generalized as

where ¢n � (c � d) � (a � b).

10.12 Le Châtelier’s Principle and the Haber ProcessAmmonia has been produced commercially from N2 and H2 since 1913, whenBadische Anilin und Soda Fabrik (BASF) built a plant that used the Haber processto make 30 metric tons of synthetic ammonia per day.

Until that time, the principal source of nitrogen for use in farming had been ani-mal and vegetable waste. Today, almost 20 million tons of ammonia worth $2.5billion is produced in the United States each year, about 80% of which is usedfor fertilizers. Ammonia is usually applied directly to the fields as a liquid at ornear its boiling point of �33.35�C. By using this so-called anhydrous ammonia,farmers can apply a fertilizer that contains 82% nitrogen by weight.

The Haber process was the first example of the use of Le Châtelier’s prin-ciple to optimize the yield of an industrial chemical. An increase in the pressureat which the reaction is run favors the products of the reaction because there is anet reduction in the number of molecules in the system as N2 and H2 combine toform NH3. Because the reaction is exothermic, the equilibrium constant increasesas the temperature of the reaction decreases.

Table 10.3 shows the mole percent of NH3 at equilibrium when the reactionis run at different combinations of temperature and pressure. The mole percent ofNH3 under a particular set of conditions is equal to the number of moles of NH3

at equilibrium divided by the total number of moles of all three components ofthe reaction times 100. As the data in Table 10.3 demonstrate, the best yields ofammonia are obtained at low temperatures and high pressures.

N2(g) + 3 H2(g) uv 2 NH3(g) ¢H° = -92.2 kJ/molrxn

QC =

ncC

naA

nd

D

nbB

*

1

V¢n= Qn *

1

V¢n

aA + bB uv cC + dD

10.12 LE CHÂTELIER’S PRINCIPLE AND THE HABER PROCESS 443

(d) Because the pressure remains constant, the volume must increase. If thevolume increases, Qc � Kc and the reaction shifts to the right.

(e) Adding O2 increases the numerator in Qc so that Qc Kc, which meansthe reaction shifts to the left.

(f) If SO3 is added at constant pressure, the volume must increase. This changecombined with the effect of the increase in the number of moles of SO3

makes Qc smaller, so that Qc � Kc. The reaction therefore shifts to theright.

(g) The reaction is endothermic. Therefore, according to the first rule intro-duced in Section 10.11, the equilibrium will shift toward the products ofthe reaction.


Unfortunately, low temperatures slow down the rate of the reaction, and thecost of building plants rapidly escalates as the pressure at which the reaction isrun is increased. When commercial plants are designed, a temperature is chosenthat allows the reaction to proceed at a reasonable rate without decreasing theequilibrium concentration of the product by too much. The pressure is alsoadjusted so that it favors the production of ammonia without excessively increas-ing the cost of building and operating the plant. The optimum conditions for run-ning the reaction at present are a pressure between 140 atm and 340 atm and atemperature between 400�C and 600�C.

Despite all efforts to optimize reaction conditions, the percentages of hydro-gen and nitrogen converted to ammonia are still relatively small. Another form ofLe Châtelier’s principle is therefore used to drive the reaction to completion. Peri-odically, the reaction mixture is cycled through a cooling chamber. The boilingpoint of ammonia (BP � �33�C) is much higher than that of either hydrogen(BP � �252.8�C) or nitrogen (BP � �195.8�C). Ammonia can be removed fromthe reaction mixture, forcing the equilibrium to the right. The remaining hydro-gen and nitrogen gases are then recycled through the reaction chamber, wherethey react to produce more ammonia.


A photograph of the first high-pressure reactor for thesynthesis of ammonia.

Table 10.3Mole Percentage of NH3 at Equilibrium

Pressure (atm)

Temperature (�C) 200 300 400 500

400 38.74 47.85 58.87 60.61450 27.44 35.93 42.91 48.84500 18.86 26.00 32.25 37.79550 12.82 18.40 23.55 28.31600 8.77 12.97 16.94 20.76


10.13 What Happens When a Solid Dissolves in Water?Silver chloride is categorized as an insoluble ionic compound because the maxi-mum amount of silver chloride that will dissolve in water is less than 0.1M. How-ever, if we add small amounts of silver chloride solid to water, the salt dissolvesto form Ag� and Cl� ions. The addition of solid silver chloride to a liter of wateris shown in Table 10.4.

If we added very small amounts of silver chloride (1.0 � 10�6 moles), we findthat no solid will remain. All of the silver chloride has dissolved to form Ag�

and Cl�. In the same manner when we add 5.0 � 10�6 or 1.0 � 10�5 moles ofsolid AgCl, we observe that all of the solid dissolves. However, as the concen-trations of the ions become larger, the reverse reaction starts to compete with theforward reaction, which leads to a decrease in the rate at which Ag� and Cl� ionsenter the solution.

As we continue to add solid AgCl, the Ag� and Cl� ion concentrationsbecome large enough that the rate at which precipitation (formation of solidAgCl) occurs exactly balances the rate at which AgCl dissolves. At this point thesolid is in equilibrium with its ions in solution, and no additional solid will dis-solve. As we add more solid (5.0 � 10�5 to 5.0 � 10�4 moles in Table 10.4),we observe that the concentrations of Ag� and Cl� do not change and that solidAgCl that did not dissolve is in the bottom of the beaker.

When the system reaches equilibrium, it is called a saturated solutionbecause it contains the maximum concentration of ions that can exist in equi-librium with the solid salt. The amount of salt that must be added to a givenvolume of solvent to form a saturated solution is called the solubility of thesalt. A set of solubility rules for ionic compounds in water can be found inTable 8.9.

When an ionic compound dissolves in water it breaks up into its ions(cations and anions).

ionic compound uvH2O

cations(aq) + anions(aq)

AgCl(s) uvH2O

Ag+(aq) + Cl-(aq)

10.13 WHAT HAPPENS WHEN A SOLID DISSOLVES IN WATER? 445

Table 10.4 Solubility of AgCl(s) in 1.0 L of Water

Moles of AgCl(s) Moles of AgCl(s) Remaining Ag� Concentration Cl� Concentration

Added to 1.0 L H2O in the Beaker in the Solution in the Solution

1.0 � 10�6 0 1.0 � 10�6 1.0 � 10�6

5.0 � 10�6 0 5.0 � 10�6 5.0 � 10�6

1.0 � 10�5 0 1.0 � 10�5 1.0 � 10�5

5.0 � 10�5 3.7 � 10�5 1.3 � 10�5 1.3 � 10�5

1.0 � 10�4 8.7 � 10�5 1.3 � 10�5 1.3 � 10�5

5.0 � 10�4 4.9 � 10�4 1.3 � 10�5 1.3 � 10�5

Estimates based on equilibrium calculations.


10.14 The Solubility Product ExpressionSilver chloride is so insoluble in water ( 0.002 g/L) that a saturated solution con-tains only about 1.3 � 10�5 moles of AgCl per liter of water.

The rules for writing equilibrium constant expressions given in Section 10.5 donot address the inclusion of either pure liquids or pure solids in an equilibriumconstant expression.

● The concentrations of solids are never included in an equilibrium constantexpression because the concentration of a solid is constant (it does notchange).

● The concentrations of liquids are included in an equilibrium constant expres-sion only when the concentration of the liquid changes during the chemicalreaction.

The dissolution of AgCl in water will serve as an example of why the concen-tration of a solid is not included in an equilibrium expression. Strict adherence tothe rules discussed in Section 10.5 gives the following expression.

Kc =

[Ag+][Cl-]

[AgCl]

AgCl(s) uvH2O

Ag+(aq) + Cl-(aq)

L


E x e r c i s e 1 0 . 1 1Write chemical equations that describe the process in which the following ioniccompounds dissolve in water.

(a) Cu2S(s)

(b) SrF2(s)

(c) PbCO3(s)

(d) Ag2SO4(s)

(e) Cr(OH)3(s)

Solution

(a)

(b)

(c)

(d)

(e) Cr(OH)3(s) uvH2O

Cr3 +(aq) + 3 OH-(aq)

Ag2SO4(s) uvH2O

2 Ag+(aq) + SO42 -(aq)

PbCO3(s) uvH2O

Pb2 +(aq) + CO32 -(aq)

SrF2(s) uvH2O

Sr2 +(aq) + 2 F-(aq)

Cu2S(s) uvH2O

2 Cu+(aq) + S2 -(aq)


(Water isn’t included in the equilibrium constant expression because it is neitherconsumed nor produced in this reaction, even though it is a vital component ofthe system.)

Two of the terms in this expression are easy to interpret. The [Ag�] and [Cl�]terms represent the concentrations of the Ag� and Cl� ions in units of moles perliter when the solution is at equilibrium. The third term––[AgCl]––is more ambigu-ous. It doesn’t represent the concentration of AgCl dissolved in water because weassume that AgCl dissociates into Ag� ions and Cl� ions when it dissolves.

The [AgCl] term has to be translated quite literally as the number of molesof AgCl in a liter of the solid AgCl that lies at the bottom of the beaker. Thisquantity is a constant, however. The number of moles per liter in solid AgCl isthe same at the start of the reaction as it is when the reaction reaches equilibrium.Since the [AgCl] term is a constant, which has no effect on the equilibrium, it isbuilt into the equilibrium constant for the reaction.

The concentration of solid silver chloride in units of moles per liter can becalculated from the density of the solid and its molar mass. The density of solidsilver chloride is 5.56 g/cm3, and the molar mass of silver chloride is 143.32 g/mol.

The concentration of the solid does not change, even though the amount of thesolid may become smaller as some of the solid dissolves. Although both the num-ber of moles of solid and the volume of the solid decrease when it dissolves, theconcentration doesn’t change because there is no change in the ratio of moles ofsolid to liters of solid.

Because the [AgCl] term is a constant, it has no effect on the equilibriumand is built into the equilibrium constant for the reaction.

This equation suggests that the product of the equilibrium concentrations of theAg� and Cl� ions in this solution is equal to a constant. Since this constant isproportional to the solubility of the salt, it is called the solubility product equi-librium constant for the reaction, or Ksp.

The Ksp expression for a salt is the product of the concentrations of the ions,with each concentration raised to a power equal to the coefficient of that ion inthe balanced equation for the solubility equilibrium. Solubility product constantsfor a number of sparingly soluble salts are given in Table B.10 in Appendix B.

Ksp = [Ag+][Cl-] = [1.34 * 10- 5][1.34 * 10- 5] = 1.8 * 10- 10

[Ag+][Cl-] = Kc * 38.8 M = Ksp

[Ag+][Cl-] = Kc * [AgCl]

M = a5.56 g

cm3b a 1 cm3

1 mLb a1000 mL

Lb a 1 mol

143.32 gb = 38.8 M

10.14 THE SOLUBILITY PRODUCT EXPRESSION 447

[AgCl] = concentration in moles per literof AgCl in the solid at the bottom of thecontainer

[Ag+] = concentrationof Ag+ ions at equilibrium

Ag+

Cl–Ag+

Cl–

[Cl–] = concentrationof Cl– ions at equilibrium

E x e r c i s e 1 0 . 1 2Calcium fluoride (CaF2) was considered as a possible source of the fluorideion when toothpaste was first fluoridated. Write the Ksp expression for a satu-rated solution of CaF2 in water.

SolutionThe Ksp expression for a salt is the product of the concentrations of the ionsformed when this salt dissolves in water, with each concentration raised to a


10.15 The Relationship between Ksp and the Solubility of a SaltKsp is called the solubility product because it is literally the product of the con-centrations of the ions in moles per liter raised to their appropriate powers. Thesolubility product of a salt can therefore be estimated from its solubility, or viceversa. It is important to remember that equilibrium calculations are models ofwhat is happening during a chemical process. Particularly in the case of solubil-ity calculations, these models are not exact and give only estimates. However, sol-ubility calculations can still help us better understand the equilibrium associatedwith an ionic solid and its ions in solution, even though we realize the results ofthe calculations are not exact.

Photographic films are based on the sensitivity of AgBr to light. When lighthits a crystal of AgBr, a small fraction of the Ag� ions are reduced to silver metal.The rest of the Ag� ions in these crystals are reduced to silver metal when thefilm is developed. AgBr crystals that don’t absorb light are then removed fromthe film to “fix” the image. Let’s calculate the solubility of AgBr in water in gramsper liter, to see whether AgBr can be removed by simply washing the film.

We start with the balanced equation for the equilibrium.

AgBr(s) uvH2O

Ag+(aq) + Br-(aq)


power equal to the coefficient of that ion in the balanced equation for the sol-ubility equilibrium.

We start with a balanced equation for the equilibrium we want to describe.

Because two F� ions are produced for each Ca2� ion when this salt dissolvesin water, the Ksp expression for CaF2 is

Ksp = [Ca2 +][F-]2

CaF2 (s) uvH2O

Ca2 +(aq) + 2 F-(aq)

Calcium fluoride is a naturally occurring mineral known as fluorite.


We then write the solubility product expression for this reaction and find the valueof Ksp for this salt in Table B.10 in Appendix B.

We can’t solve one equation for two unknowns—the Ag� and Br� ion concen-trations. We are given the equilibrium constant and the initial concentrations ofthe products. It is not necessary to know the concentration of the solid becauseits concentration does not change and is incorporated as part of the equilibriumconstant, Ksp. We can therefore set up a table like those used in previous equi-librium calculations.

Initial: ____ 0 0Change: ¢C ¢C ¢CEquilibrium: ____ ¢C ¢C

Substituting this equality into the Ksp expression gives the following result.

Ksp � [Ag�][Br�] � 5.0 � 10�13 � ¢C2

Taking the square root of both sides of this equation gives the equilibrium con-centrations of the Ag� and Br� ions.

Once we know how many moles of AgBr dissolve in a liter of water, we can cal-culate the solubility in grams per liter.

The solubility of AgBr in water is only 0.00013 gram per liter. It therefore isn’tpractical to try to wash the unexposed AgBr off photographic film with water.

Solubility product calculations with 1:1 salts such as AgBr are relativelyeasy to perform. In order to extend such calculations to compounds with morecomplex formulas, we have to understand the relationship between the solubilityof a salt and the concentrations of its ions at equilibrium. We will use the sym-bol ¢C to represent the solubility of a salt in a saturated solution at equilibriumin units of moles per liter.

7.1 * 10- 7 mol AgBr

1 L*

187.8 g AgBr

1 mol= 1.3 * 10- 4 g AgBr

L

[Ag+] = [Br-] = 7.1 * 10- 7M

[Ag+]2= 5.0 * 10- 13

Because [Ag+] = [Br-]

AgBr(s) uv Ag+(aq) + Br-(aq)

Ksp = [Ag+][Br-] = 5.0 * 10- 13

10.15 THE RELATIONSHIP BETWEEN KSP AND THE SOLUBILITY OF A SALT 449

E x e r c i s e 1 0 . 1 3Write equations that describe the relationship between the solubility of CaF2

and the equilibrium concentrations of the Ca2� and F� ions in a saturated solu-tion as a first step toward evaluating its use as a fluoridating agent.



SolutionAs always, we start with the balanced equation for the reaction.

Initial: 0 0Change:Equilibrium:

No values are given for the initial and/or equilibrium concentrations of CaF2

because it is a solid and the concentration of a solid does not change.Salts dissociate into their ions when they dissolve in water. For every

mole of CaF2 that dissolves, we get a mole of Ca2� ions. The equilibrium con-centration of the Ca2� ion is therefore equal to the solubility of this compoundin moles per liter.

For every mole of CaF2 that dissolves, we get twice as many moles of F� ions.The F� ion concentration at equilibrium is therefore equal to twice the solu-bility of the compound in moles per liter.

[F-] = 2¢C

[Ca2 +] = ¢C

2¢C¢C2¢C¢C

CaF2 (s) uvH2O

Ca2 +(aq) + 2 F-(aq)

CaF2(s)

[Ca2+] = �CCa2+ F–

F–

Ca2+F–

[F–] = 2�CF–

F–

F–

Ca2+

E x e r c i s e 1 0 . 1 4Use the Ksp for calcium fluoride to calculate its solubility in grams per liter.Use the results of this calculation to explain why calcium fluoride wasn’t usedas a source of F� ions in toothpaste, in spite of the fact that Ca2� ions aregood for bones (CaF2: Ksp � 4.0 � 10�11).

SolutionAccording to Exercise 10.12, the solubility product expression for CaF2 is writ-ten as follows.

Exercise 10.13 gave us the following equations for the relationship betweenthe solubility of this salt and the concentrations of the Ca2� and F� ions.

Substituting this information into the Ksp expression gives the following result.

This equation can be solved for the solubility of CaF2 in units of moles per liter.

¢C = 2.2 * 10- 4M

4¢C3= 4.0 * 10- 11

[¢C][2¢C]2= 4.0 * 10- 11

[Ca2 +][F-]2= 4.0 * 10- 11

[F-] = 2¢C

[Ca2 +] = ¢C

Ksp = [Ca2 +][F-]2


10.16 The Role of the Ion Product [Qsp] in Solubility CalculationsConsider a saturated solution of AgCl in water.

Because AgCl is a 1:1 salt, the concentrations of the Ag� and Cl� ions in thissolution are equal.

Imagine what happens when a few crystals of solid AgNO3 are added to asaturated solution of AgCl in water. According to the solubility rules in Table 8.9,silver nitrate is a soluble salt. It therefore dissolves and dissociates into Ag� andNO3

� ions. As a result, there are two sources of the Ag� ion in this solution.

Adding AgNO3 to a saturated AgCl solution therefore increases the Ag� ion con-centration. When this happens, the solution is no longer at equilibrium becausethe product of the concentrations of the Ag� and Cl� ions is too large. In otherwords, the ion product (Qsp) for the solution is larger than the solubility prod-uct (Ksp) for AgCl.

When Qsp is larger than Ksp, the reaction has to shift toward the solid AgCl tocome to equilibrium. Thus, AgCl will precipitate from the solution as shown bythe following equation.

Ag�(aq) � Cl� (aq) AgCl(s)

The ion product is literally the product of the concentrations of the ions raisedto their appropriate powers. When the ion product is equal to the solubilityproduct for the salt, the system is at equilibrium. Silver chloride will precipi-tate from solution until the concentrations of the Ag� and Cl� ions decreaseto the point at which the ion product is equal to Ksp and a new equilibrium hasbeen established.

¡

Qsp = (Ag+)(Cl-) 7 Ksp

AgCl(s) uvH2O

Ag+(aq) + Cl-(aq)

AgNO3(s) ¡ Ag+(aq) + NO3-(aq)

Saturated solution of AgCl in water: [Ag+] = [Cl-]

AgCl(s) uvH2O

Ag+(aq) + Cl-(aq)

10.16 THE ROLE OF THE ION PRODUCT [QSP] IN SOLUBILITY CALCULATIONS 451

Once we know how many moles of CaF2 dissolve in a liter, we can cal-culate the solubility in units of grams per liter.

The solubility of calcium fluoride is fairly small: 0.017 gram per liter. Stan-nous fluoride, or tin (II) fluoride, is over 10,000 times as soluble, so SnF2 waschosen as the first fluoridating agent used in fluoride toothpastes.

2.2 * 10- 4 mol CaF2

1 L*

78.1 g CaF2

1 mol= 0.017

g CaF2

L


After the excess ions precipitate from solution as solid AgCl, the reactioncomes back to equilibrium. When equilibrium is reestablished, however, the con-centrations of the Ag� and Cl� ions are no longer the same. Because there aretwo sources of the Ag� ion in this solution, there will be more Ag� ion at equi-librium than Cl� ion.

Now imagine what happens when a few crystals of NaCl are added to asaturated solution of AgCl in water. There are two sources of the chloride ion inthis solution.

Once again, the ion product is larger than the solubility product.

Therefore, AgCl will precipitate from solution until the product of the concen-trations of the Ag� and Cl� ions is equal to the Ksp and a new equilibrium hasbeen established. This time, when the reaction comes back to equilibrium, therewill be more Cl� ion in the solution than Ag� ion.

Figure 10.10 shows a small portion of the possible combinations of the Ag�

and Cl� ion concentrations in an aqueous solution. The solid line in this graph iscalled the saturation curve for AgCl. Any point along this curve corresponds to asystem at equilibrium because the product of the Ag� and Cl� ion concentrationsfor these solutions is equal to Ksp for AgCl.

Point A in Figure 10.10 represents a saturated solution at equilibrium thatcould be produced by dissolving two sources of the Ag� ion––such as AgNO3

and AgCl––in water. Point B represents a saturated solution of AgCl in pure water,in which the [Ag�] and [Cl�] terms are equal. Point C describes a solution at

Saturated solution of AgCl to which NaCl has been added: [Ag+] 6 [Cl-]

Qsp = (Ag+)(Cl-) 7 Ksp

AgCl(s) uvH2O

Ag+(aq) + Cl-(aq)

NaCl(s) ¡

H2ONa+(aq) + Cl-(aq)

Saturated solution of AgCl to which AgNO3 has been added: [Ag+] 7 [Cl-]


Fig. 10.10 Possible combinations of Ag� and Cl�

ion concentrations in an aqueous solution.

1 × 10–4

8 × 10–5

6 × 10–5

4 × 10–5

2 × 10–5

2 × 10–5

Concentration of Cl–(mol/L)

Ag+ + Cl– AgCl

Qsp>Ksp

Qsp<Ksp

Con

cent

rati

on o

f A

g+(m

ol/L

)

4 × 10–5 6 × 10–5

D

C

8 × 10–5 1 × 10–4

A

E

B


equilibrium that was prepared by dissolving two sources of the Cl� ion in water,such as NaCl and AgCl.

Points that don’t lie on the solid line in Figure 10.10 represent solutions thataren’t at equilibrium. Any point below the solid line (such as point D) representsa solution for which the ion product is smaller than the solubility product.

If more AgCl were added to the solution at point D, it would dissolve.

Points above the solid line (such as point E) represent solutions for which the ionproduct is larger than the solubility product.

The solution described by point E will eventually come to equilibrium afterenough solid AgCl has precipitated.

Table 10.5 shows the addition of small amounts of soluble AgNO3 to a solutionthat is not saturated in Ag� and Cl�, but instead contains only dissolved NaCl,Na�(aq) and Cl�(aq).

Initially, when small amounts (1.0 � 10�5 to 1.0 � 10�4 moles) of AgNO3

are added to the NaCl solution, the resulting concentrations of Ag� and Cl� yieldQsp values that are less than the Ksp of AgCl. Therefore, no solid precipitant forms.However, as additional AgNO3 is added (5.0 � 10�4 to 5.0 � 10�5), the solu-tion becomes saturated with Ag� and Cl� and solid AgCl begins to form. TheQsp values have become greater than the Ksp of AgCl.

10.17 The Common-Ion EffectWhen AgNO3 is added to a saturated solution of AgCl, it is often described as asource of a common ion, the Ag� ion. By definition, a common ion is an ion thatenters the solution from two different sources. Solutions to which both NaCl and

If Qsp 7 Ksp: Ag+(aq) + Cl-(aq) ¡ AgCl(s)

point E: Qsp 7 Ksp

if Qsp 6 Ksp: AgCl(s) ¡ Ag+(aq) + Cl-(aq)

point D: Qsp 6 Ksp

10.17 THE COMMON-ION EFFECT 453

Table 10.5The Addition of solid AgNO3 to a 1.0 L Solution of 1.0 � 10�6 M Cl�

Moles of AgNO3 Moles of AgCl Concentration of Ag� Concentration Added Formed in Solution of Cl� in Solution

1.0 � 10�5 0 1.0 � 10�5 1.0 � 10�6

5.0 � 10�5 0 5.0 � 10�5 1.0 � 10�6

1.0 � 10�4 0 1.0 � 10�4 1.0 � 10�6

5.0 � 10�4 6.4 � 10�7 5.0 � 10�4 3.6 � 10�7

1.0 � 10�3 8.2 � 10�7 1.0 � 10�3 1.8 � 10�7

5.0 � 10�3 1.0 � 10�6 5.0 � 10�3 3.6 � 10�8

Estimates based on equilibrium calculations.


AgCl have been added also contain a common ion––in this case, the Cl� ion.This section focuses on a phenomenon known as the common-ion effect––theeffect of common ions on solubility product equilibria.


E x e r c i s e 1 0 . 1 5Calculate the solubility of AgCl in pure water (AgCl: Ksp � 1.8 � 10�10).

SolutionThe solubility product expression for AgCl is written as follows.

Because there is only one source of the Ag� and Cl� ions in this solution, theconcentrations of these ions at equilibrium must be the same. Furthermore, theconcentrations of both ions are equal to the solubility of AgCl in units of molesper liter: ¢C.

Substituting this information into the solubility product expressions leads tothe conclusion that the solubility of AgCl is equal to the square root of Ksp

for this salt.

¢C = 1.3 * 10- 5 M

¢C2= 1.8 * 10- 10

[¢C][¢C] = 1.8 * 10- 10

[Ag+] = [Cl-] = ¢C

Ksp = [Ag+][Cl-] = 1.8 * 10- 10

The common-ion effect can be understood by considering the following ques-tion: What happens to the solubility of AgCl when we dissolve this salt in a solu-tion that is already 0.10 M NaCl? As a rule, we can assume that salts dissociateinto their ions when they dissolve. A 0.10 M NaCl solution therefore contains 0.10mole of the Cl� ion per liter of solution. Because the Cl� ion is one of the prod-ucts of the solubility equilibrium, Le Châtelier’s principle leads us to expect thatAgCl will be even less soluble in a 0.10 M Cl� solution than it is in pure water.

E x e r c i s e 1 0 . 1 6Calculate the solubility of AgCl in 0.10 M NaCl (AgCl: Ksp � 1.8 � 10�10).

SolutionThe Ag� and Cl� ion concentrations at equilibrium will no longer be thesame because there are now two sources of the Cl� ion in this solution: AgCland NaCl.

[Ag+] Z [Cl-]


10.17 THE COMMON-ION EFFECT 455

Initially, there is no Ag� ion in the solution, but the Cl� ion concentration is0.10 M. As the reaction comes to equilibrium, some of the AgCl will dissolveand the concentrations of both the Ag� and Cl� ions will increase. Both con-centrations will increase by an amount equal to the solubility of AgCl in thissolution: ¢C.

Initial: 0 0.10 MEquilibrium: ¢C 0.10 � ¢C

We now write the solubility product expression for this reaction.

We then substitute what we know about the equilibrium concentrations of theAg� and Cl� ions into this equation.

We could expand the equation and solve it with the quadratic formula, but thatwould involve a lot of work. Let’s see if we can find an assumption that makesthe calculation easier.

What do we know about ¢C? In pure water, the solubility of AgCl is0.000013 M. In this solution, we expect it to be even smaller. It therefore seemsreasonable to expect that ¢C should be small compared with the initial con-centration of the Cl� ion.

Substituting this approximation into the solubility product expression gives thefollowing approximate equation.

Solving this approximate equation gives the following result.

The assumption used to generate the approximate equation is valid. (The Cl�

ion concentration from the dissociation of AgCl is about 50 million timessmaller than the initial Cl� ion concentration.) This assumption works verywell with common-ion problems involving insoluble salts because the Ksp val-ues for these salts are so small.

Let’s compare the results of Exercises 10.15 and 10.16.

These calculations show how the common-ion effect can be used to make an“insoluble” salt even less soluble in water.

AgCl in 0.10 M NaCl: ¢C = 1.8 * 10- 9M

AgCl in pure water: ¢C = 1.3 * 10- 5M

¢C L 1.8 * 10- 9M

[¢C][0.10] L 1.8 * 10- 10

[0.10 + ¢C] L [0.10]

[¢C][0.10 + ¢C] = 1.8 * 10- 10

Ksp = [Ag+][Cl-] = 1.8 * 10- 10

AgCl(s) uv Ag+(aq) + Cl-(aq) Ksp = 1.8 * 10- 10



E x e r c i s e 1 0 . 1 7Which salt––CaCO3 or Ag2CO3––is more soluble in water in units of molesper liter? (CaCO3: Ksp � 2.8 � 10�9, Ag2CO3: Ksp � 8.1 � 10�12)

SOLUTIONWe might expect CaCO3 to be more soluble than Ag2CO3 because it has alarger Ksp. The only way to test this prediction is to calculate the solubilitiesof both compounds.

The solubility product expression for CaCO3 has the following form.

Because CaCO3 is a 1:1 salt, the concentrations of the Ca2� and CO32� would

be the same at equilibrium.

Substituting this information into the solubility product expressions leads to theconclusion that the solubility of CaCO3 is equal to the square root of Ksp forthis salt.

Ag2CO3 is a 2:1 salt, for which the following solubility product expres-sion is written.

The CO32� ion concentration is equal to the solubility of the salt, but the Ag� ion

concentration is twice as large. If we define ¢C as the solubility of the salt, then:

Substituting this information into the Ksp expression gives the following results.

This equation can be solved for the solubility of Ag2CO3.

In spite of the fact that Ksp for CaCO3 is larger than Ksp for Ag2CO3, CaCO3

is less soluble than Ag2CO3.

CaCO3: Solubility = 5.3 * 10- 5 M

Ag2CO3: Solubility = 1.3 * 10- 4 M

¢C = 1.3 * 10- 4 M

4[¢C]3= 8.1 * 10- 12

[2¢C]2[¢C] = 8.1 * 10- 12

[CO32 -] = ¢C

[Ag+] = 2¢C

Ksp = [Ag+]2[CO32 -]

¢C = 5.3 * 10- 5M

¢C2= 2.8 * 10- 9

[¢C][¢C] = 2.8 * 10- 9

[Ca2 +] = [CO32 -] = ¢C

Ksp = [Ca2 +][CO32 -]


10.18 Selective PrecipitationSolutions that contain a mixture of many ions can be qualitatively analyzed bytaking advantage of the different solubilities of the ions. For example, suppose asolution contains Ag�, Cd2�, and Ba2� ions. Table B.10 provides solubility prod-uct constants for these and many other insoluble compounds. From this table itcan be determined that of the three ions in this solution, only Ag� can be pre-cipitated as the chloride, AgCl. Thus if a solution of NaCl is added, AgCl willprecipitate and can be filtered away and separated from the remaining solution.Similarly, Ba2� forms a precipitate with the sulfate ion SO4

2�, and if a Na2SO4

solution is carefully added, the Ba2� ion will precipitate as BaSO4 and can beseparated by filtration. Only the Cd2� ion would then remain in solution, and ifdesired, this ion could be precipitated as the carbonate CdCO3 by adding aNa2CO3 solution.

10.18 SELECTIVE PRECIPITATION 457

E x e r c i s e 1 0 . 1 8A solution contains Cd2� and Cr3� ions both at a concentration of 1.0 � 10�2

mol/L. If a solution of NaOH is slowly added to the mixture, the insolublehydroxide of each ion will be formed. Which will precipitate first?

SolutionKsp values are given in Table B.10 as 2.5 � 10�14 for cadmium hydroxide and6.3 � 10�31 for the hydroxide of chromium.

We begin by writing the Ksp relations for both hydroxides:

Substituting the Cd2� and Cr3� concentrations into the Ksp expressions gives

[1.0 � 10�2] [ OH�]2 � 2.5 � 10�14

[1.0 � 10�2] [OH�]3 � 6.3 � 10�31

The concentrations of hydroxide needed to cause precipitation can then be cal-culated to be

Cd (OH)2 [OH�] � 1.6 � 10�6 M

Cr (OH)3 [OH�] � 4.0 � 10�10 M

A lower concentration of hydroxide ion will cause a precipitate to first formwith Cr�3.

Ksp = [Cr3 +] [OH-]3= 6.3 * 10- 31

Ksp = [Cd2 +] [OH-]2= 2.5 * 10- 14

E x e r c i s e 1 0 . 1 9A solution contains Pb2�, Ca2�, and Sn2� ions. Devise a scheme to qualita-tively separate each species from the solution.



SolutionFrom the Ksp values listed in Table B.10, it can be determined that only thePb2� ion forms a precipitate with Cl� ion. Thus if a solution of NaCl is addedto the mixture, PbCl2 will precipitate and can be removed by filtration. TheSn2� ion does not form a precipitate when mixed with the CO3

2� ion, but theCa2� ion does. If a solution of Na2CO3 is added to the solution after removalof Pb2�, only CaCO3 would precipitate, and this precipitate could be removedby filtration. If it is desired, Sn2� can be precipitated as the hydroxide,Sn(OH)2, by addition of a NaOH solution.

Key TermsChemical kineticsCollision theoryCommon-ion effectEquilibriumEquilibrium constant (Kc)Equilibrium constant expressionEquilibrium region

Haber processIon product (Qsp)Kinetic regionLe Châtelier’s principlePrecipitationRate constantRate law

Rate of reactionReaction quotient (Qc)Saturated solutionSolubilitySolubility product equilibrium

constant (Ksp)

ProblemsReactions That Don’t Go to Completion

1. Describe the difference between reactions that go tocompletion and reactions that come to equilibrium.

2. Describe the meaning of the symbols [NO] and (NO).

Gas-Phase Reactions3. Define the terms equilibrium constant and equilibrium

constant expression.4. If 10.0 mol of trans-2-butene is placed into an empty

flask at 400�C, what will be the equilibrium ratio oftrans-2-butene to cis-2-butene? What if 15.0 mol isplaced into an empty flask at the same temperature?

5. If Kc is greater than 1 for the reaction A B, willthe equilibrium concentrations of the products besmaller or larger than the equilibrium concentrations ofthe reactants? What if Kc is less than 1?

6. Is the following statement true or false? The equilib-rium concentrations depend on the initial concentra-tions, but the ratio of the equilibrium concentrationsspecified by the equilibrium constant expression is in-dependent of the initial concentrations.

The Rate of a Chemical Reaction7. Translate the following equation into an English sen-

tence that carries the same meaning.

rate of reaction = -

¢(X)

¢t

uv

8. What does the rate law tell us about a chemical reac-tion? Does the rate law tell us anything about the ratioof reactants to products at equilibrium?

9. What is a rate constant? How does it differ from therate law?

The Collision Theory Model of Gas-PhaseReactions10. Use the collision theory to explain why the rate of the

reaction of ClNO2 with NO to form ClNO and NO2

depends on the concentrations of the reactants ClNO2

and NO.11. Explain how the rates of the forward and reverse reac-

tions change as the reaction between ClNO2 and NOproceeds to equilibrium. Assume that no ClNO or NO2

are present initially.12. Sketch a graph of what happens to the concentrations

of N2, H2, and NH3 versus time as the following reac-tion comes to equilibrium.

Assume that the initial concentrations of N2 and H2 areboth 1.00 mol/L and that no NH3 is present initially. Labelthe kinetic and the equilibrium regions of the graph.

13. Give two ways to define equilibrium.14. On the molecular level, do chemical reactions stop at

equilibrium? Explain why or why not.

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


Equilibrium Constant Expressions15. Which of the following is the correct equilibrium con-

stant expression for the following reaction?

(a) (b)

(c) (d)

(e)

16. Which of the following is the correct equilibrium con-stant expression for the following reaction?

(a) (b)

(c) (d)

(e)

17. Write equilibrium constant expressions for the follow-ing reactions.(a)(b)(c)

18. Write equilibrium constant expressions for the follow-ing reactions.(a)(b)(c)

19. Write equilibrium constant expressions for the follow-ing reactions.(a)(b)Calculate the value of Kc at 500 K for reaction (a) if thevalue of Kc for reaction (b) is 6.2 � 105 at 500 K.

20. Use the equilibrium constants for reaction (a) and (b) at200�C to calculate the equilibrium constant for reac-tion (c) at that temperature.(a)

(b)

(c)21. Use the equilibrium constants for reactions (a) and

(b) at 1000 K to calculate the equilibrium constant

2 NO2(g) uv N2(g) + 2 O2(g) Kc = ?

2 NO2(g) uv 2 NO(g) + O2(g)Kc = 3.4 * 10- 7

2 NO(g) uv N2(g) + O2(g)Kc = 4.3 * 1018

2 NO(g) + O2(g) uv 2 NO2(g)2 NO2(g) uv 2 NO(g) + O2(g)

2 NO(g) + O2(g) uv 2 NO2(g)2 NOCl(g) uv 2 NO(g) + Cl2(g)2 NO(g) + 2 H2(g) uv N2(g) + 2 H2O(g)

2 SO3(g) + 2 Cl2(g) uv 2 SO2Cl2(g) + O2(g)2 SO2(g) + O2(g) uv 2 SO3(g)O2(g) + 2 F2(g) uv 2 OF2(g)

Kc =

[2NO]2[O2]

[2NO2]2

Kc =

[NO]2[O2]

[NO2]2

Kc =

[NO2]2

[NO]2[O2]

Kc =

[NO][O2]

[NO2]Kc =

[NO2]

[NO][O2]

2 NO2(g) uv 2 NO(g) + O2(g)

Kc =

[Cl2][F2]3

[ClF3]2

Kc =

[ClF3]2

[Cl2][F2]3

Kc =

[ClF3]

[Cl2][F2]

Kc =

[Cl2] + 3[F2]

2[ClF3]Kc =

2[ClF3]

[Cl2] + 3[F2]

Cl2(g) + 3 F2(g) uv 2 ClF3(g)

for reaction (c), the water-gas shift reaction, at thattemperature.(a)

(b)

(c)

22. Calculate Kc for the following reaction at 400 K if1.000 mol/L of NOCl decomposes at that temperatureto give equilibrium concentrations of 0.0222 M NO,0.0111 M Cl2, and 0.978 M NOCl.

23. Taylor and Crist [Journal of the American ChemicalSociety, 63, 1381 (1941)] studied the reaction betweenhydrogen and iodine to form hydrogen iodide.

They obtained the following data for the concentrationsof H2, I2, and HI at equilibrium in units of moles per liter.

Trial [H2] [I2] [HI]

I 0.0032583 0.0012949 0.015869II 0.0046981 0.0007014 0.013997III 0.0007106 0.0007106 0.005468

Calculate the value of Kc for each of the trials. Realiz-ing that there will be deviation due to experimental er-ror, is Kc constant for the reaction?

Reaction Quotients: A Way to Decide Whether a Reaction Is at Equilibrium24. Suppose that the reaction quotient (Qc) for the follow-

ing reaction at some moment in time is 1.0 � 10�8 andthe equilibrium constant for the reaction (Kc) at thesame temperature is 3 � 10�7.

Which of the following is a valid conclusion?(a) The reaction is at equilibrium.(b) The reaction must shift toward the products to

reach equilibrium.(c) The reaction must shift toward the reactants to

reach equilibrium.25. Which of the following statements correctly describes

a system for which Qc is larger than Kc?(a) The reaction is at equilibrium.(b) The reaction must shift to the right to reach equi-

librium

2 NO2(g) uv 2 NO(g) + O2(g)

H2(g) + I2(g) uv 2 HI(g)

2 NOCl(g) uv 2 NO(g) + Cl2(g)

CO(g) + H2O(g) uv CO2(g) + H2(g)Kc = ?

H2O(g) uv H2(g) +1⁄2 O2(g)

Kc = 7.1 * 10- 12

CO(g) +1⁄ 2 O2(g) uv CO2(g)

Kc = 1.1 * 1018

PROBLEMS 459


(c) The reaction must shift to the left to reach equilib-rium.

(d) The reaction can never reach equilibrium.26. Under which set of conditions will the following reac-

tion shift to the right to reach equilibrium?

(a) Kc 6 1 (b) Kc 7 1 (c) Qc 6 Kc

(d) Qc � Kc (e) Qc 7 Kc

27. Carbon monoxide reacts with chlorine to form phosgene.

The equilibrium constant, Kc, for the reaction is 1.5 � 104

at 300�C. Is the system at equilibrium at the followingconcentrations: 0.0040 M COCl2, 0.00021 M CO, and0.00040 M Cl2? If not, in which direction does thereaction have to shift to reach equilibrium?

Changes in Concentration That Occur as aReaction Comes to Equilibrium28. Explain why the change in the N2 concentration that

occurs when the following reaction comes to equilib-rium is related to the change in the H2 concentration.

Derive an equation that describes the relationship be-tween the changes in the concentrations of the tworeagents.

29. When confronted with the task in the previous prob-lem, the following incorrect answer is often given.

Explain why this equation is wrong. Write the correctform of the relationship.

30. Calculate the changes in the CO and Cl2 concentra-tions that occur if the concentration of COCl2 de-creases by 0.250 mol/L as the following reactioncomes to equilibrium.

31. Calculate the changes in the N2 and H2 concentra-tions that occur if the concentration of NH3 de-creases by 0.234 mol/L as the following reactioncomes to equilibrium.

32. Which of the following equations describes the rela-tionship between the magnitude of the changes in the

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

COCl2(g) uv CO(g) + Cl2(g)

¢(N2) = 3¢(H2)

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

CO(g) + Cl2(g) uv COCl2(g)

2 SO2(g) + O2(g) uv 2 SO3(g)

NO2 and O2 concentrations as the following reactioncomes to equilibrium?

(a) ¢(NO2) � ¢(O2) (b) ¢(NO2) � 2¢(O2)(c) ¢(O2) � 2¢(NO2)

33. Which of the following equations correctly describesthe relationship between the changes in the Cl2 andF2 concentrations as the following reaction comes toequilibrium?

(a) ¢(Cl2) � ¢(F2) (b) ¢(Cl2) � 2¢(F2)(c) ¢(Cl2) � 3¢(F2) (d) ¢(F2) � 2¢(Cl2)(e) ¢(F2) � 3¢(Cl2)

34. Which of the following describes the change that oc-curs in the concentration of H2O when ammonia reactswith oxygen to form nitrogen oxide and water accord-ing to the following equation if the change in the NH3

concentration is ¢C?

(a) ¢C (b) 1.5¢C (c) 2¢C(d) 4¢C (e) 6¢C

35. Calculate the concentrations of H2 and NH3 at equilib-rium if a reaction that initially contained 1.000 M con-centrations of both N2 and H2 is found to have an N2

concentration of 0.922 M at equilibrium.

Initial: 1.000 M 1.000 M 0 MEquilibrium: 0.922 ? ?

36. Calculate the equilibrium constant for the reaction inthe previous problem.

Hidden Assumptions That Make EquilibriumCalculations Easier37. Calculate the equilibrium concentrations of N2O4

and NO2 when 0.100 M N2O4 decomposes to formNO2 at 25�C.

38. Without doing detailed equilibrium calculations, esti-mate the equilibrium concentration of N2O4 presentwhen 1.00 M NO2 reacts to form N2O4 at 25�C.

39. Calculate the equilibrium concentrations of N2, H2,and NH3 present when a mixture that was initially

N2O4(g) uv 2 NO2(g) Kc = 5.8 * 10- 5

N2O4(g) uv 2 NO2(g) Kc = 5.8 * 10- 5

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

4 NH3(g) + 5 O2(g) uv 4 NO(g) + 6 H2O(g)

Cl2(g) + 3 F2(g) uv 2 ClF3(g)

2 NO(g) + O2(g) uv 2 NO2(g)



0.10 M N2, 0.10 M H2, and 0.10 M NH3 comes to equi-librium at 500�C.

40. Calculate the equilibrium concentrations of CO, H2O,CO2, and H2 present in the water-gas shift reaction at800�C if the initial concentrations of CO and H2O are1.00 M.

41. Calculate the equilibrium concentrations of N2, O2,and NO present when a mixture that was initially 0.100M in N2 and 0.090 M in O2 comes to equilibrium at600�C.

42. Sulfuryl chloride decomposes to sulfur dioxide andchlorine. Calculate the concentrations of the threecomponents of the system at equilibrium if 6.75 g ofSO2Cl2 in a 1.00-L flask decomposes at 25�C.

43. Without detailed equilibrium calculations, estimate theconcentrations of NO and NOCl at equilibrium if amixture that was initially 0.50 M in NO and 0.10 M inCl2 combined to form nitrosyl chloride, NOCl.

44. Calculate the concentrations of PCl5, PCl3, and Cl2 thatare present when the following gas-phase reactioncomes to equilibrium. Calculate the percent of the PCl5that decomposes when the reaction comes to equilib-rium. Kc � 0.0013 at 450 K.

Initial: 1.0 M 0 0

45. Calculate the concentrations of PCl5, PCl3, and Cl2present when the following gas-phase reaction comesto equilibrium. Calculate the percent decomposition inthe reaction and explain any difference between the re-sults of this calculation and the results obtained in Sec-tion 10.7. Assume that Kc � 0.0013 for this reaction at450 K.

Initial: 1.00 M 0 0.20 M

PCl5(g) uv PCl3(g) + Cl2(g)

PCl5(g) uv PCl3(g) + Cl2(g)

Kc = 2.1 * 103 (at 500 K)

2 NO(g) + Cl2(g) uv 2 NOCl(g)

Kc = 1.4 * 10- 5

SO2Cl2(g) uv SO2(g) + Cl2(g)

N2(g) + O2(g) uv 2 NO(g) Kc = 3.3 * 10- 10

Kc = 0.72 (at 800 �C)

CO(g) + H2O(g) uv CO2(g) + H2(g)

Kc = 0.040 (at 500 �C)

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

46. Calculate the concentrations of NO, NO2, and O2 pres-ent when the following gas-phase reaction reachesequilibrium. Assume that Kc � 3.4 � 10�7 for this re-action at 200�C.

Initial: 0.100 M 0 0

47. Calculate the concentrations of NO, NO2, and O2 pres-ent when the following gas-phase reaction reachesequilibrium. Assume that Kc � 3.4 � 10�7 for this re-action at 200�C.

Initial: 0.100 M 0 0.050 M

48. Calculate the equilibrium concentrations of SO3, SO2,and O2 present when 0.100 mol of SO3 in a 250-mLflask at 300�C decomposes to form SO2 and O2. As-sume that Kc � 1.6 � 10�10 for this reaction at 300�C.

49. Without detailed equilibrium calculations, estimate theequilibrium concentration of SO3 when a mixture of0.100 mol of SO2 and 0.050 mol of O2 in a 250-mLflask at 300�C combine to form SO3. Assume that Kc �6.3 � 109 for this reaction at 300�C.

50. Sometimes the technique used in this chapter to sim-plify equilibrium problems is incorrectly stated asfollows: “Assume that ¢C is zero.” Explain why thisis wrong. What is the correct way of describing theassumption?

51. What is the advantage of setting up equilibrium prob-lems so that ¢C is small compared with the initialconcentrations?

52. Describe how to test whether ¢C is small enoughcompared with the initial concentrations to be legiti-mately ignored.

53. At 600�C the equilibrium constant for the following re-action is 3.3 � 10�10.

(a) Is ¢C likely to be small or large for this reaction?Explain your answer.

(b) Find Kc for the following reaction and decidewhether ¢C is likely to be large or small for the de-composition of NO.

2 NO(g) uv N2(g) + O2(g)

N2(g) + O2(g) uv 2 NO(g)

2 SO2(g) + O2(g) uv 2 SO3(g)

2 SO3(g) uv 2 SO2(g) + O2(g)

2 NO2(g) uv 2 NO(g) + O2(g)

2 NO2(g) uv 2 NO(g) + O2(g)

PROBLEMS 461


What Do We Do When the Assumption Fails?54. Describe what happens if you make the assumption

that ¢C is zero in the following equation.

Explain how to get around this problem.55. Explain why ¢C is relatively small when the reaction

quotient (Qc) is reasonably close to the equilibriumconstant for the reaction (Kc).

56. Explain why the assumption that ¢C is small com-pared with the initial concentrations of the reactantsand products is doomed to fail when the reaction quo-tient (Qc) is very different from the equilibrium con-stant for the reaction (Kc).

57. Describe the technique used to solve problems forwhich the reaction quotient is very different from theequilibrium constant.

58. Before we can solve the following problem, we have todefine a set of intermediate conditions under which theconcentration of one of the reactants or products is zero.

Initial: 0.10 M 0.10 M 0.005 M

Which of the following goals determines whether wepush the reaction as far as possible to the right or as faras possible to the left?(a) To make both QC and Kc large(b) To make both QC and Kc small(c) To bring QC as close as possible to Kc

(d) To make the difference between QC and Kc as largeas possible

The Effect of Temperature on an EquilibriumConstant59. Why is it important to specify the temperature at which

an equilibrium constant is reported?60. If an equilibrium constant gets smaller as temperature

increases, will increasing the temperature favor theproducts or the reactants?

61. If Kc decreases with decreasing temperature, will in-creasing the temperature favor the reactants or products?

Le Châtelier’s Principle62. Le Châtelier’s principle has been applied to many

fields, ranging from economics to psychology to polit-ical science. Give an example of Le Châtelier’s princi-ple in a field outside the physical sciences.

63. Predict the effect of increasing the pressure at constanttemperature on the following reactions at equilibrium.

Kc = 5.3 * 10- 6(at 250°C)

2 NO2(g) uv 2 NO(g) + O2(g)

[0.125 - ¢C][2.40 - 2¢C]2

[0.200 + 3¢C]3= 1.3 * 10- 8

(a)(b)(c)

64. Predict the effect of decreasing the pressure at constanttemperature on the following reactions at equilibrium.(a)(b)(c)

65. Predict the effect of increasing the concentration of thereagent indicated in boldface on each of the followingreactions at equilibrium. Assume that temperature andpressure are both constant.(a)(b)(c)

66. Use Le Châtelier’s principle to predict the effect of anincrease in pressure on the solubility of a gas in waterat a constant temperature.

Le Châtelier’s Principle and the Haber Process67. List as many ways as possible of increasing the yield of

ammonia in the Haber process.

68. Explain why an increase in pressure favors the forma-tion of ammonia in the Haber process.

69. Predict how an increase in the volume of the container bya factor of 2 would affect the concentrations of ammoniaand oxygen in the following reaction. T is constant.

70. How are the data in Table 10.3 consistent with LeChâtelier’s principle? Consider a fixed temperaturewith a changing pressure and a fixed pressure with achanging temperature.

What Happens When a Solid Dissolves in Water?71. Write a chemical equation that describes the relation-

ship between the concentrations of the Ag� and CrO42�

ions in a saturated solution of Ag2CrO4.72. Write an equation that describes the relationship be-

tween the concentrations of the Bi3� and S2� ions in asaturated solution of Bi2S3.

The Solubility Product Expression73. Explain why the [Ag�] and [Cl�] terms are variables

but the [AgCl] term is a constant no matter how muchAgCl is added to a saturated solution of silver chloridein water.

74. What is the correct solubility product expression forthe following reaction?

Ca3(PO4)2(s) uv 3 Ca2 +(aq) + 2 PO3 -

4 (aq)

4 NH3(g) + 5 O2(g) uv 4 NO(g) + 6 H2O(g)

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

PF5(g) uv PF3(g) + F2(g)2 SO3(g) uv 2 SO2(g) + O2(g)2 NO2(g) uv N2O4(g)

NO(g) + NO2(g) uv N2O3(g)N2(g) + O2(g) uv 2 NO(g)N2O4(g) uv 2 NO2(g)

2 NO(g) + O2(g) uv 2 NO2(g)O2(g) + 2 F2(g) uv 2 OF2(g)2 SO3(g) + 2 Cl2(g) uv 2 SO2Cl2(g) + O2(g)



(a) (b)

(c) (d)

(e)

75. Which of the following is the correct solubility productexpression for Al2 (SO4) 3?(a)(b)(c)(d)

76. Write the solubility product expression for each of thefollowing salts.(a) BaCrO4 (b) CaCO3

(c) PbF2 (d) Ag2S

The Relationship between Ksp and the Solubilityof a Salt77. Write a chemical equation that describes the dissolution

of Ag2CO3. Write a mathematical equation that describesthe relationship between the concentrations of the Ag�

and CO32� ions in a saturated solution of Ag2CO3.

78. Write a chemical equation that describes the dissolu-tion of Cu2S. Write a mathematical equation that de-scribes the relationship between the concentrations ofthe Cu� and S2� ions in a saturated solution of Cu2S.

79. Write a chemical equation that describes the dissolu-tion of SrF2. Calculate the Ksp constant for the dissolu-tion of strontium fluoride if the solubility of SrF2 inwater is 8.5 � 10�4 mol/L.

80. Silver acetate, Ag (CH3CO2), is marginally soluble inwater. What is the Ksp for silver acetate if 6.6 � 10�2

moles of Ag (CH3CO2) dissolve in 1000 mL of water?81. For the dissolution of magnesium hydroxide in water:

(a) Write a chemical equation for the dissolution process.(b) Write the Ksp equilibrium expression for this reac-

tion.(c) If Ksp is 1.8 � 10�11, calculate the solubility of

magnesium hydroxide in moles per liter.(d) Calculate the solubility of magnesium hydroxide

in grams per 100 mL.82. What is the solubility of BaF2 in water in grams per

100 mL if the Ksp is 1.0 � 10�6?83. What is the solubility in water for each of the following

salts in grams per 100 mL?

(a) Cu2S (Ksp � 2.5 � 10�48)

(b) CuS (Ksp � 6.3 � 10�36)

84. What is the solubility of Hg2S in mol/L in a solutionthat contains an S2� concentration of 0.10 M? What isthe solubility of HgS in a solution that contains an S2�

concentration of 0.10 M? Hg2S (Ksp � 1.0 � 10�47)HgS (Ksp � 4 � 10�53).

Ksp = [2 Al3 +]2[3 SO2 -

4 ]3

Ksp = [Al3 +]2[SO2 -

4 ]3

Ksp = [2 Al3 +][3 SO2 -

4 ]Ksp = [Al3 +][SO2 -

4 ]

Ksp = [Ca2 +]2[PO3 -

4 ]3

Ksp = [Ca2 +]3[PO3 -

4 ]2Ksp = [Ca2 +][PO3 -

4 ]

Ksp =

[Ca2 +]3[PO3 -

4 ]2

[Ca3(PO4)2]Ksp =

[Ca2 +][PO3 -

4 ]

[Ca3(PO4)2]

85. How many grams of AgBr will dissolve in 1.0 L of wa-ter containing a Br� concentration of 0.050 M? TableB.10 in Appendix B contains solubility product con-stant values.

86. What is the solubility of Ag2CO3 in water in mol/L?What is the solubility in mol/L of Ag2CO3 in a solutioncontaining an Ag� concentration of 0.15 M? See TableB.10 in Appendix B.

87. Which of the following equations describes the rela-tionship between the solubility product for MgF2 andthe solubility of this compound?

(a) Ksp � 2¢C (b) Ksp � ¢C2 (c) Ksp � 2¢C2

(d) Ksp � ¢C3 (e) Ksp � 4¢C3

88. Hg2Cl2 contains the Hg22� and Cl� ions. Which of the

following equations describes the relationship be-tween the solubility product and the solubility of thiscompound?

(a) Ksp � ¢C (b) Ksp � ¢C3 (c) Ksp � 4¢C3

(d) Ksp � ¢C4 (e) Ksp � 16¢C4

89. Which is more soluble, Ag2S or HgS? (Ag2S: Ksp �6.3 � 10�50; HgS: Ksp � 4 � 10�53)

90. Which is more soluble, PbSO4 or PbI2? (PbSO4: Ksp �1.6 � 10�8; PbI2: Ksp � 7.1 � 10�9)

91. Mercury forms salts that contain either the Hg2� ion orthe Hg2

2� ion. Which is more soluble, HgS or Hg2S?(HgS: Ksp � 4 � 10�53; Hg2S: Ksp � 1.0 � 10�47)

92. What is the concentration of the CN� ion in a saturatedsolution of zinc cyanide dissolved in water if the Zn2�

ion concentration is 4.0 � 10�5 M?

93. What is the concentration of the CrO42� ion in a satu-

rated solution of silver chromate dissolved in water ifthe Ag� ion concentration is 1.3 � 10�4 M?

94. What is the solubility product for strontium fluoride ifthe solubility of SrF2 in water is 0.107 gram per liter?

95. Silver acetate, Ag(CH3CO2), is marginally soluble inwater. What is the Ksp for silver acetate if 1.190 gramsof Ag(CH3CO2) dissolve in 99.40 mL of water?

96. Lithium salts, such as lithium carbonate, are used totreat manic-depressives. What is the solubility productfor lithium carbonate if 1.36 grams of Li2CO3 dissolvein 100 mL of water?

97. People who have the misfortune of going through a seriesof X rays of the gastrointestinal tract are often given a sus-pension of solid barium sulfate in water to drink. BaSO4 isused instead of other Ba2� salts, which also reflect X rays,because it is relatively insoluble in water. (Thus the pa-tient is exposed to the minimum amount of toxic Ba2�

ion.) What is the solubility product for barium sulfate if1 gram of BaSO4 dissolves in 400,000 grams of water?

98. What is the solubility of silver sulfide in water ingrams per 100 mL if the solubility product for Ag2S is6.3 � 10�50?

PROBLEMS 463


99. What is the solubility in water for each of the followingsalts in grams per 100 mL?(a) Hg2S (Ksp � 1.0 � 10�47)(b) HgS (Ksp � 4 � 10�53)

100. What is the solubility in water for each of the followingsalts in grams per 100 mL?(a) Ca3 (PO4)2 (Ksp � 2.0 � 10�29)(b) Pb3 (PO4)2 (Ksp � 8.0 � 10�43)(c) Ag3PO4 (Ksp � 1.4 � 10�16)

101. List the following salts in order of increasing solubilityin water.(a) Ag2S (Ksp � 6.3 � 10�50)(b) Bi2S3 (Ksp � 1 � 10�97)(c) CuS (Ksp � 6.3 � 10�36)(d) HgS (Ksp � 4 � 10�53)

The Role of the Ion Product (Qsp) in SolubilityCalculations102. If a solution contains an Ag� concentration of 1.0 �

10�8 M and an I� concentration of 1.0 � 10�8 M, willa precipitate form? Explain. (AgI: Ksp � 8.3 � 10�17)

103. If a solution contains a Pb2� concentration of 1.9 �10�4 M and an F� concentration of 1.9 � 10�4 M, willa precipitate form? Explain. (PbF2: Ksp � 2.7 � 10�8)

104. A 50.0-mL solution of 0.10 M Ca(NO3)2 is added to50.0 mL of a 0.25 M solution of NaOH. Will a precipi-tate form? [Ksp Ca(OH)2 � 5.5 � 10�6]

105. A 100-mL solution of 0.0015 M AgNO3 is added to50.0 mL of a 0.0030 M solution of Na2CO3. Will a pre-cipitate form? (Ag2CO3: Ksp � 8.1 � 10�12)

The Common-Ion Effect106. Define the term common-ion effect. Describe how Le

Châtelier’s principle can be used to explain thecommon-ion effect.

107. Describe what happens to the equilibrium concentra-tions of the Ag� and Cl� ions when 10 grams of NaClare added to a liter of a saturated solution of silverchloride in water.

108. Which of the following statements is true? (a) MgF2

is more soluble in 0.100 M NaF than in pure water.(b) MgF2 is less soluble in 0.100 M NaF than in purewater. (c) MgF2 is just as soluble in 0.100 M NaF as inpure water.

109. Calculate the equilibrium concentration of the Ag� ionin a solution prepared by dissolving 3.21 grams ofpotassium iodide in 350 mL of water and then addingsilver iodide until the solution is saturated with AgI.(AgI: Ksp � 8.3 � 10�17)

110. How many grams of silver sulfide will dissolve in 500mL of a 0.050 M S2� solution? (Ag2S: Ksp � 6.3 �10�50)

111. In which of the following solutions is Pb (OH)2 mostsoluble? [Pb (OH)2: Ksp � 1.2 � 10�15](a) pure water(b) 0.010 M NaOH

Integrated Problems112. Which of the following diagrams best represents the

concentrations of the reactants and products for the fol-lowing reaction at equilibrium? Explain what is wrongwith each incorrect diagram. ( represents isobutane,and represents n-butane.)

ƒ

Isobutane n-Butane

CH3CHCH3(g) uv CH3CH2CH2CH3(g) Kc = 0.4

CH3


(d) (e)

(a) (b) (c)

X

X

M XX M

M X

X

X

X

M XX M

M X

X

X

X

M XX M

M X

X

X

X

M XX M

M X

X

(a) (b)

(c) (d)

XMX

XMXXMX

M2+

M2+X–

X–X–

X–

M2+

M2+M2+X–

X–

X–

113. A sparingly soluble hypothetical ionic compound, MX2,is placed into a beaker of distilled water. Which of thefollowing diagrams best describes what happens in solu-tion? Explain what is wrong with each incorrect diagram.


114. Describe the relationship between kf and kr for the fol-lowing one-step reaction at equilibrium.

Which is true: kf � kr, kf 6 kr, or kf 7 kr? Explain your reasoning.

115. For the reaction A B, match the graphs of con-centration versus time to the appropriate set of rateconstants.

(a) kA � kB

(b) kA � 1.0/s, kB � 0.5/s(c) kA � 0.5/s, kB � 1.0/s

rateforward = kA(A) ratereverse = kB(B)

uv

Z(g) + X(g) uvkf

kr

Y(g) Kc = 1 * 10- 3

(a) When SrF2(s) is placed in water, the compounddissolves to produce equilibrium Sr2� concen-tration of 5.8 � 10�4 mol/L. What is Ksp for thereaction?

(b) If 50.00 mL of 0.100 M Sr (NO3)2 is mixed with50.00 mL of 0.100 M NaF, will a precipitate form?Explain your answer.

117. Several plots of concentration versus time for the reac-tion A B are given below. Kc � 2. Only one ofthe plots can be correct. Which one is it? Explain whatis wrong with each of the incorrect plots.

uv

PROBLEMS 465

time

A(1)

(4)

(5)

(2)

(3)

B

A

B

A

B

B

A

B

A

conc

.

time

conc

.

time

conc

.

time

conc

.

time

conc

.

116. Write the equilibrium constant expression for the dis-solving of strontium fluoride in water.

SrF2(s) uv Sr2 +(aq) + 2 F-(aq)

A

B

B

A

B

A

1

1

1.0

0.5

0.5

conc(A)o = 1.0(B)o = 0.5

conc(A)o = 0.5(B)o = 1.0

conc(A)o = 0(B)o = 1.0

time

time

time

Initialconcentrations

(a)

(b)

(c)

118. Molecular iodine dissociates into iodine atoms at 1000 K.

(a) If 0.456 mol of I2 is placed into a 2.30-L flask at1000 K, what will be the equilibrium concentra-tions of I2 and I?

I2(g) uv 2 I(g) Kc = 3.8 * 10- 5


(b) If 0.912 mol of I is placed into a 2.30-L flask con-taining no I2 at 1000 K, estimate the concentrationof I2 at equilibrium. Do no detailed equilibriumcalculations, but clearly explain your answer.

119. When equilibrium is reached for the dissolving ofsolid calcium sulfate in water at 25�C, it is found that[Ca2�] � [SO4

2�] � 4.9 � 10�3M.

(a) Calculate the equilibrium constant (Ksp) for theabove reaction.

(b) A solution contains Ca2�(aq) at a concentration of3.6 � 10�3M and SO4

2� (aq) at a concentration of8.0 � 10�3 M. Will solid CaSO4 be formed? Showyour calculations.

(c) CaSO4 is allowed to dissolve in 1.0 L of water at25�C until equilibrium is reached. Then the wateris allowed to evaporate to half its original volume.What are the equilibrium concentrations of Ca2�

and SO42� in this solution? Explain.

CaSO4(s) uv Ca2 +(aq) + SO42 -(aq)

122. At 25�C, 1.5 � 10�2 mol of Ag2SO4 dissolves in 1.0 Lof water.

(a) How many moles of Ag� are present?(b) How many moles of SO4

2� are present?(c) Calculate the equilibrium constant for the reaction.

123. The following equilibrium concentrations were found forthe reaction between NO and O2 to form NO2 at 230�C:[NO] � 0.0542 M, [O2] � 0.127 M, [NO2] � 15.5 M.

(a) What does it mean to say that a reaction has cometo equilibrium? Must all reactions eventually cometo equilibrium?

2 NO(g) + O2(g) uv 2 NO2(g)

Ag2SO4(s) uv 2 Ag+(aq) + SO2 -

4 (aq)


CaSO4 (s)

1.0 L

Ca2+

SO 2−4

CaSO4 (s)

0.50 L

Ca2+ SO 2−4

120. The equilibrium constant for the following reaction is1.6 � 10�5 at 35�C.

If 1.0 mol of NOCl is placed into an empty1.0-L flask, what will be the equilibrium con-centrations of all species? State all assumptionsand show all work. Provide a justification forany assumptions.

121. The rate constant for the following reaction in the for-ward direction, kf, is 2.10 � 10�7 s�1, and that in thereverse direction, kr, is 1.65 � 10�7 s�1.

cis-2-butene(g) uv trans-2-butene(g)

2 NOCl(g) uv 2 NO(g) + Cl2(g)

cis

trans

t(a)

conc

(M

) cis

trans

t(b)

conc

(M

)

cis

trans

t(c)

conc

(M

)

cis

trans

t(d)

conc

(M

)

cis

trans

t(e)

conc

(M

)

Which of the following graphs could represent thechange in concentrations with time? More than onegraph could be correct. Explain your reasoning.


(b) What is the equilibrium constant, Kc, for the reac-tion?

(c) If sufficient O2 and NO2 are added to increase [O2]and [NO2] to 1.127 and 16.5 M, respectively, whilekeeping [NO] at 0.0542 M, in which direction willthe reaction proceed?

124. At a certain temperature, the following reaction has anequilibrium constant of 5.0 � 10�9.

(a) If 1.0 mol of N2F4 is placed in a 1.0-L flask with noNF2 present, what will be the equilibrium concen-trations of NF2 and N2F4?

(b) If 1.0 mol of NF2 is placed in a 1.0-L flask with noN2F4 present, what will be approximately the equi-librium concentration of N2F4? No detailed equi-librium calculations are necessary.

125. The following equation represents a system at equilib-rium.

Which of the following could be a valid representationof this system? ( � Cl2 and � Cl).

Cl2(g) uv 2 Cl(g)

N2F4(g) uv 2 NF2(g)

126. A student devises the following qualitative analysisscheme for a solution containing the ions Ni2�, Mg2�,and Ba2�. First add a solution of NaF, filter, then add asolution of Na2CO3, filter, and finally add a solution ofNa2SO4. Will this scheme succeed? If not, explain howthe scheme could be modified to selectively separatethe ions.

PROBLEMS 467

(a) (b)

(c) (d)

(e) all of the above


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Chapter Ten - UNAMdepa.fquim.unam.mx/amyd/archivero/Unidad-ChemicalEquilibrium_3… · Chapter Ten THE CONNECTION BETWEEN KINETICS AND EQUILIBRIUM 10.1 Reactions That DonÕt Go to