NAME____________________________________ PER____________ DATE DUE____________

ACTIVE LEARNING IN CHEMISTRY EDUCATION

"ALICE"

CHAPTER 20

CHEMICAL

EQUILIBRIUM

Keq

20-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS

All rights reserved. No part of this document may be reproduced or transmitted in any form by any means,electronic, mechanical, photocopying, or otherwise, without the prior written permission of the author.

Copies of this document may be made free of charge for use in public or nonprofit private educationalinstitutions provided that permission is obtained from the author . Please indicate the name and addressof the institution where use is anticipated.

© 1997 A.J. Girondi, Ph.D.505 Latshmere DriveHarrisburg, PA 17109

alicechem@geocities.com

Website: www.geocities.com/Athens/Oracle/2041

20-2 ©1997, A.J. Girondi

SECTION 20.1 Introduction to Equilibrium Systems

In most of the chemical reactions we have studied so far, it appears as though all of the reactantsare converted to products before a reaction stops. In truth, however, experiments show that theconversion of reactants into products is often incomplete in chemical reactions. This is the case no matterhow long the reaction is allowed to continue.

As a reaction progresses, the concentrations of the reactants decrease, while the concentrationsof the products increase. Eventually, a state is established in which the concentrations of products andreactants no longer change. This is known as the state of equilibrium. You have already studied manyexamples of equilibrium. The most obvious example was probably encountered during your study ofsolutions. At that time, you performed some activities that involved the use of saturated solutions.

Equilibrium is a balance between two equal and opposing processes or forces. When a systemreaches equilibrium, it does not undergo additional change unless the equilibrium is somehow disturbed.A saturated solution is in equilibrium because there is a balance between the two opposing forces ofdissolving and crystallizing. (See Figure 20.1 below.) This is an example of a physical equilibrium sincedissolving and crystallizing are physical changes. An equation can be written to represent what isoccurring in this saturated solution after equilibrium has been reached:

REACTANTS PRODUCTSCaCl2(s) Ca2+(aq) + 2 Cl1-(aq)

The (aq) subscript refers to "aqueous" which means dissolved in water.

Ca2+

Cl1-Cl1-

Saturated CaCl2 Solution

Figure 20.1Solution Equilibrium

The double arrow indicates that the forward reaction is occurringat the SAME RATE as the reverse reaction. Therefore, nooverall change occurs in the system. To an observer, it appearsas though nothing whatsoever is happening in the saturatedCaCl2 solution. Actually, the change shown at right is occurringconstantly in both directions. Note that the equation aboveshows chlorine as 2 Cl1-, not as Cl2. Chloride ions (Cl1-) do notexist in diatomic form. This is because these ions already have astable octet of electrons. Only chlorine atoms form diatomicmolecules: (Cl2). This is also true for the other diatomic elements.

CaCl2(s) <===> Ca2+(aq) + Cl2(aq)WRONG! ---> <--- WRONG!

CaCl2(s) <===> Ca2+(aq) + 2 Cl1-(aq)

The type of change shown in Figures 20.1 and 20.2 is referred to as dynamic equilibrium because,although no changes can be observed in the system, the two opposing changes are happeningconstantly.

Figure 20.2 illustrates two more examples of physical equilibria. Examine what is in each beaker,and note how a double arrow is used in the equilibrium equation for each situation. Note, too, how thestate (solid, liquid, gas, or aqueous) of each substance is represented using subsscripts. Finally, noticethat coefficients are used to balance the equations properly.

20-3 ©1997, A.J. Girondi

Figure 20.2 Equilibrium Systems in Saturated Solutions

Saturated AlCl3 Solution

Al3+ Cl1-

AlCl3(s) <===> Al3+(aq) + 3 Cl1-(aq)

Cl1-

Cl1-

AlCl3

Pb2+

I1-

Saturated PbI2 Solution

I1-

PbI2 (s) <===> Pb2+(aq) + 2 I1-(aq)

PbI2

Problem 1. The three equations below represent equilibrium systems created by dissolving solids inwater. Complete the equilibrium equations below. Include any charges on ions and the state of the ionswhich in this case is "aqueous" or (aq):

a. NaCl(s) <====> _____________________________

b. Fe2(SO4)3(s) <====> _____________________________

c. BaBr2(s) <====> ______________________________

The formula for water is not included in the examples above because dissolving is not a chemical changeand, therefore, water is not consider a reactant. When substances dissolve in water they do notchemically react with it. They merely come part in the water. Later in this chapter you will study reactionsthat involve chemical equilibrium, since the changes are chemical, rather than physical.

There are several criteria that must be satisfied before an equilibrium can be achieved. Let's take a

closer look at some of these. Compare the two systems shown below (beakers A and B). Which beaker

contains a system at equilibrium?{1}_________ What do you think prevents the system in the other

beaker from forming an equilibrium system? {2}____________________________________________

Figure 20.3Closed and Open Systems

H2O(g)

H2O(l)

BEAKER "A"

H2O(g)

H2O(l)

BEAKER "B"

H2O(l) H2O(g)

20-4 ©1997, A.J. Girondi

Could equilibrium ever be achieved in an open system such as that shown in beaker B? {3}________ Why

or why not? {4}___________________________________________________________________

Hopefully, you have arrived at the conclusion that an equilibrium can be established in a systemsuch as this only if the system is closed. In addition, this system must be held under conditions ofconstant temperature, pressure, and volume. If one or more of these factors is altered, the equilibriumchanges as well. (Not all equilibrium systems are affected by pressure or volume, but all equilibriumsystems are affected by temperature. We will discuss this later in the chapter.)

Equilibrium equations give an overall picture of the types of chemical changes that are occurring.The equations do not, however, provide any information about the actual amounts of reactants andproducts present. The equilibrium equation for a saturated sodium chloride solution:

NaCl(s) <====> Na1+(aq) + Cl1-(aq)

merely indicates the identities of the ions and molecules involved in the equilibrium, and the doublearrows show that equilibrium prevails. There is absolutely no way that you can tell what the concentrationsof any of the products or reactants are just by looking at this equation. The reaction container may have afew crystals of solid NaCl present or it may have a large pile of solid salt.

SECTION 20.2 Solving Problems Involving Equilibrium Systems

There is a simple relationship between the concentrations of the products and the reactants.Through much experimental study, scientists were able to come up with a law of chemical equilibrium.Consider the hypothetical equilibrium system: A + B <===> C + D

Brackets such as, [ ], are used to denote molar (M) concentration. So when you see something like[Ag1+], it means "molar concentration of silver ions." This law of chemical equilibrium states that if you findthe product of the concentrations of the products, [C] [D], and divide that result by the product of theconcentrations of the reactants, [A] [B], the result will be a constant value:

constant value =

[C] [D]

[A] [B]

Now, let's look at a real example:

4 NO(g) + 6 H2O(g) <====> 4 NH3(g) + 5 O2(g)

Since this system is at equilibrium, this means that the RATE of the forward reaction (--->) is equal to theRATE of the reverse (<---) reaction. From the equation for the equilibrium system, we can write what iscalled the equilibrium expression for this reaction:

equilibrium expression =

[NH3 ]4 [O2 ]5

[NO]4 [H2O]6

You should notice two things about the expression above. First, note that the numerator contains theproduct of the concentrations of the products, and the denominator is the product of the concentrationsof the reactants. Second, you should see that the coefficients in the equation become exponents in theequilibrium expression. The reasons why we turn coefficients into exponents can be explained usingkinetic–molecular theory and collision theory. However, we will simply accept this, and save theexplanation for a more advanced course. If you know the concentrations of all the reactants and productsin this equilibrium system, you can plug those values into this expression and solve it. The value you will

20-5 ©1997, A.J. Girondi

get is a constant. Why? Well, if you somehow manage to change the concentration of NH3, for example,the concentrations of the other substances in the system will change, too. The net effect will be that theexpression will maintain the same value. Since it is a constant value, it is given the symbol K. Since itinvolves equilibrium, it is often called Keq. (Other authors may designate it as K, Kc, etc.) So,

Keq =

[NH3 ]4 [O 2 ]5

[NO]4 [H2O]6

Problem 2. Write the equilibrium expressions for each of the chemical situations given below.

a. 2 NO2(g) <===> N2O4(g) Keq =

b. N2(g) + 3 H2(g) <===> 2 NH3(g) Keq =

c. Ag1+(aq) + 2 NH3(aq) <===> Ag(NH3)21+(aq) Keq =

d. 2 NO(g) + 2 H2(g) <===> N2(g) + 2 H2O(g) Keq =

Setting up an equilibrium expression is an extremely useful skill after you become aware of whatinformation it can provide. In other words, what good is Keq?

One way to determine the value of Keq for a particular reaction is to allow the reaction to proceed ata given temperature in such a way as to allow the products to accumulate in the reaction container. After aperiod of time, the reaction will reach equilibrium. At this point, it may be possible to experimentallydetermine the concentrations of the reactants and products in the container. The concentration valuesare then substituted into the equilibrium expression which is then solved for Keq.

Let's consider an equilibrium system involving only gases and assume that some method isavailable to determine the concentrations of reactants and products at equilibrium. Equilibrium equationsare frequently accompanied by the temperature at which the reaction was allowed to achieve equilibrium.Since temperature always affects equilibrium, it also affects the value of Keq. For example:

H2(g) + I2(g) <===> 2 HI(g) (at 250oC)

Let's look at how the value of Keq for this system can be determined (at this temperature).

1. Start by writing the correct formula for the Keq of this reaction: Keq =

[HI]2

[H2 ] [I2 ]

20-6 ©1997, A.J. Girondi

2. Below are the experimentally determined equilibrium concentrations of each reactant and product forthe system above. The brackets mean molarity (moles / liter of sol'n).

Temp. [H2] [I2] [HI]

Experiment 1: 250oC 0.00560 0.000590 0.01270

Experiment 2: 250oC 0.00460 0.000970 0.01476

Notice that two experiments were probably performed under slightly different conditions, producingdifferent equilibrium concentrations. Temperature was constant.

3. Substituting this data into the equilibrium expression:

Experiment 1: Keq =

(0.0127)2

(0.0056) (0.00059) = 48.8

Experiment 2: Keq =

(0.01476)2

(0.0046) (0.00097) = 48.8

Note that while the concentrations of reactants and products changed from experiment 1 to experiment 2,the value of Keq did not change.

Complete the following problems. First, write the expression for Keq for each reaction, and thenuse the given data to calculate the value of Keq. Show all work.

Problem 3. Reaction: N2O4(g) <===> 2 NO2(g) (at 520oC)Equilibrium Concentrations: [N2O4] = 0.014; [NO2] = 0.072

Keq = ____________

Problem 4. Reaction: N2(g) + 3 H2(g) <===> 2 NH3(g) (at 583oC)Equilibrium Concentrations: [N2] = 0.20; [H2] = 0.20; [NH3] = 0.016

Keq = ____________

20-7 ©1997, A.J. Girondi

Problem 5. Reaction: 2 NO(g) + O2(g) <===> 2 NO2(g) (at 500oC)Equilibrium Concentrations: [NO] = 3.49 X 10-4; [O2] = 0.80; [NO2] = 0.25

Keq = ____________

Problem 6. Reaction: NCl3(g) + Cl2(g) <===> NCl5(g) (at 520oC) Equilibrium is established whenthere is 0.0350 mole of NCl3, 0.0200 mole Cl2, and 0.0110 mole of NCl5 in a volume of 2.50 liters. (Hint:

you must first calculate the concentrations in moles per one liter of sol'n.)

Keq = _____________

SECTION 20.3 LeChatelier's Principle

Earlier in this chapter it was mentioned that certain factors can disrupt an equilibrium. If one ormore of these factors is altered, the equilibrium is momentarily upset. These factors are changes in:

1. temperature2. pressure3. volume of the reaction container4. concentration of a reactant and or a product

The exact effect of any change in reaction conditions on equilibrium was studied extensively by HenriLeChatelier, a French chemist. LeChatelier was mainly interested in what occurred when changes in thefour factors listed above were made in a system at equilibrium.

LeChatelier found that changes in these factors could put a "stress" on the system at equilibrium,causing it to move away from the state of equilibrium. This stress causes a change in the rate of either theforward or reverse reaction. After the system moves away from equilibrium, he observed that the amountsof reactants and products adjusted in order to restore the system to equilibrium. The results of hisobservations allowed him to formulate the following law, or principle, known as

LeChatelier's Principle: When a stress is placed on a system at equilibrium, thesystem will adjust to relieve the stress and to restore equilibrium in the system.

The stress changes the RATE of either the forward or the reverse reaction. LeChatelier's principle statesthat when a stress is applied, the RATES of the reactions will adjust so that the RATES once againbecome equal.

20-8 ©1997, A.J. Girondi

As an example, let's look at the equilibrium system: 2 NO(g) + O2(g) <===> 2 NO2(g)

Assume we mix some O2 molecules with an excess (more than needed) of NO molecules:

original equilibrium concentrations

NO O2 NO2

+

(Excess NO present)

length of arrows indicates relativerates of forward and reverse reactions

LeChatelier's principle enables us to predict the direction in which the equilibrium would shift when theconcentration of one or more of the products or reactants is changed. Suppose more oxygen wereadded to the reaction container. We could ask, "What effect would this change have on the rates of theforward and reverse reactions, and would this concentration increase affect the concentrations of theother components in the system?"

You are putting a "stress" on a system in equilibrium whenever you change the concentration ofany of the products or reactants. This means that, as a result of the change, the system is no longer inequilibrium. LeChatelier's principle can be explained using the collision theory. Part of this theory statesthat a collision must occur before a chemical reaction can take place between reactants. Furthermore, thecollisions must produce enough energy and the particles must often collide in just the right way. Not allcollisions result in a reaction. When more oxygen is added, there will be more collisions between the NOand O2 molecules. This will increase the rate of the forward (---->) reaction:

NO O2 NO2

+

(Excess NO present)

some oxygen is added rate of forward reaction is now greater

The system will try to remove the stress by getting rid of the excess oxygen. In this way, an equilibriumcondition can be restored. The only way for the system to get rid of the extra oxygen is to have it react withNO to produce more product, NO2. So, momentarily the system "shifts to the right" and produces moreproduct:

NO O2

+

NO2

(Excess NO present)

more product is now produced

Looking at it another way, the added oxygen molecules temporarily increase the rate of the forward (--->)reaction. However, since there are now more product molecules (NO2), they collide more often whichincreases the rate of the reverse reaction (<-----), and eventually a new equilibrium is established.However, this equilibrium situation is different from the original equilibrium situation because theconcentrations of the reactants and the products are not the same as they were originally. The rates of the

20-9 ©1997, A.J. Girondi

reactions are again equal, but they are not the same as the original rates. Because of the "shift to theright" the system now has more product than it did before more oxygen was added. Notice, too, that theshift caused the concentration of a reactant (NO) to decrease. The stress was relieved in the sense that atleast some of the added oxygen has been removed.

NO O2

+

NO2

(Excess NO present)

new equilibrium rates

new equilibrium concentrations

Since oxygen was added, how do you think the rates of the forward and reverse reactions in the newequilibrium compare to those in the old equilibrium? {5}______________________________________

Using these principles, predict the direction of the shift (forward ---->) or (reverse <----) inequilibrium when the concentration of Cl2 is increased in the system below, and how this will affect theconcentrations of each of the other products and reactants.

4 HCl(g) + O2(g) <===> 2 H2O(g) + 2 Cl2(g)

When the concentration of Cl2 is increased, the equilibrium will shift to the {6}_______________. The

concentration of HCl will {7}____________________ . As the system begins to shift after the addition of

Cl2, the concentration of Cl2 will {8}_________________. The concentration of O2 will

{9}_________________. The concentration of H2O will {10}_________________.

Problem 7. Complete Table 20.1 below. The first blank has been completed as an example.

Table 20.1Direction of Equilibrium Shift

Equation Added Equilibrium Shift Substance to Right or Left

a. CO(g) + 2 H2(g) <===> CH3OH(g) CO ------>

b. PCl5(g) <===> PCl3(g) + Cl2(g) PCl3 ________________

c. N2(g) + 3 H2(g) <===> 2 NH3(g) H2 ________________

d. CO(g) + H2O(g) <===> CO2(g) + H2(g) H2 ________________

e. 4 NO(g) + 6 H2O(g) <===> 4 NH3(g) + 5 O2(g) H2O ________________

f. 2 SO2(g) + O2(g) <===> 2 SO3(g) SO3 ________________

g. 2 NCl3(g) <===> N2(g) + 3 Cl2(g) Cl2 ________________

h. CH3COOH(aq) <===> H1+(aq) + CH3COO1-(aq) CH3COOH ________________

i. C2H6(g) <===> H2(g) + C2H4(g) H2 ________________

20-10 ©1997, A.J. Girondi

ACTIVITY 20.4 Lechatelier's Principle and Changes in Concentration

Now that you are able to predict the direction of an equilibrium shift, you will perform anexperiment that will allow you to apply this skill. The shifting of an equilibrium will be examinedexperimentally by using the reaction for the formation of a "complex ion" with the formula Fe(SCN)2+ bythe combination of Fe3+ ions with an SCN1- ion:

Fe3+(aq) + SCN1-(aq) <===> Fe(SCN)2+(aq)

(pale yellow) (colorless) (reddish)

This is a good example to use because a solution of Fe3+ ions has a faint yellow color, while a solution ofSCN1- ions has no color. When these two solutions are mixed, a deep-red color results due to theformation of the Fe(SCN)2+ ion. In this activity you will examine what happens when certain stresses areplaced on the equilibrium that exists between these three ions. Wear safety glasses.

Before you begin note that: There are two concentrations of Fe(NO3)3 solutions of the materials shelf forthis activity. Be sure to use 0.1M Fe(NO3)3 in step 1 and 0.2M Fe(NO3)3 in step 5, part b.

Procedure:

1. Look at the bottle of 0.1M Fe(NO3)3 solution. This solution contains Fe3+ ions and NO31- ions. Thecolor of this solution is due to the Fe3+ ions. What's the color of the aqueous Fe3+ ions? {11}__________(The NO31- ions are only spectator ions here, so we will ignore them.)

2. Look at the bottle of 0.1M KSCN solution. The KSCN solution contains K1+ and SCN1- ions. Based onthe appearance of this solution, what can you conclude about the color of the K1+ and SCN1- ions?___________________________ (The K1+ ions are only spectator ions here, so we will ignore them.)

3. In a test tube, mix 1 or 2 mL of the Fe(NO3)3 solution with 1 or 2 mL of the KSCN solution. The resultingcolor is due to the product, FeSCN2+. What's the color of the FeSCN2+ ion?{12}________________Look again at the equilibrium equation for this reaction. Note that FeSCN2+ is the product of the reactionbetween Fe3+ and SCN1-.

4. To a 100 or 150 mL beaker, add 1 mL of 0.1M Fe(NO3)3 solution. (Be sure it is the 0.1 M solution.) Next,add 2 mL of 0.1M KSCN solution. Finally, add 75 mL of distilled water. Stir well. This solution containssome reactants (Fe3+ and SCN1-) and some product (FeSCN2+). The reddish color of the product(FeSCN2+) is not very evident at this point, since it is diluted.

Fe3+(aq) + SCN1-(aq) <===> Fe(SCN)2+(aq)

(pale yellow) (colorless) (reddish)

5. Now place four 4-mL portions of the solution formed in step 4 into separate clean standard-sized (150mm) test tubes. (Make sure the test tubes all have the same internal diameter.) Label the tubes A throughD and perform the following tests:

a.) Do nothing to tube A. Use it as a control in order to compare its color to the other tubes.

b.) To tube B, add 15 drops of 0.2M Fe(NO3)3 solution and compare the color of test tube B to test tube A.

Has the color gotten lighter or darker?{13}____________________ Does this color change indicate an

increase or a decrease in the concentration of Fe(SCN)2+.{14}_________________________

20-11 ©1997, A.J. Girondi

When you added Fe(NO3)3, you were adding Fe3+ ions to the equilibrium system. This resulted in an

increase in the concentration of Fe3+ ions, [Fe3+], in the system. Explain the effect (on the equilibrium) of

increasing the Fe3+ concentration in the system. {15}_______________________________________

______________________________________________________________________________

c.) To tube C, add several drops of 6M NaOH (dangerous). Handle this solution with great care. If you get

any on you, wash with lots of water. Be sure to wear safety glasses! Mix well. Does the solution's color get

lighter or darker?{16}______________ Does this change indicate an increase or a decrease in the

concentration of Fe(SCN)2+?{17}___________________________

Thus, adding NaOH has the overall effect of DECREASING the concentration of Fe3+ ions in theequilibrium system:

Fe3+(aq) + SCN1-(aq) <===> Fe(SCN)2+(aq)

(pale yellow) (colorless) (reddish)

Explain the effect on the equilibrium system which resulted from the addition of NaOH: {18}___________

______________________________________________________________________________

d.) To tube D, add about 1 mL (20 drops) of 0.1M AgNO3 solution. (Avoid getting this solution on yourhands. After several hours, a dark stain can result.) Is the color of this mixture lighter or darker than that oftube A? {19}________________ Does this change indicate an increase or a decrease in theconcentration of Fe(SCN)2+? {20}___________________ When you add AgNO3 to the equilibriumsystem, some of the SCN1- ions react with it. The SCN1- that reacted this way is removed from theequilibrium system. So, adding AgNO3 has the effect of DECREASING the concentration of SCN1- ionsin the equilibrium system. What is the effect of this change on the equilibrium system?{21}________________________________

Problem 8. Using your observations from the activities above, complete Table 20.2 for the Fe(SCN)2+

equilibrium system.

Table 20.2Shifts in the Fe(SCN)2+ Equilibrium System

Test Substance Added Effect Shift Forward (--->) Toward Products, or Shift Reverse (<---) Toward Reactants

a. nothing none none

b. Fe(NO3)3 increases Fe3+ ___________________

c. NaOH decreases Fe3+ ___________________

d. AgNO3 decreases SCN1- ___________________

SECTION 20.5 LeChatelier's Principle and Changes in Pressure

Now let's "shift" our discussion to the effects of pressure changes on equilibrium systems. Forreactions which occur in liquid solutions, a change in pressure will not affect the equilibrium to any great

20-12 ©1997, A.J. Girondi

extent because the volume of the liquid solution will not change very much even if extreme pressure isplaced on the system. However, when a gas is involved, changing pressure will have an effect on theequilibrium. Consider the equilibrium system below:

N2O4(g) <===> 2 NO2(g)

If a container full of these two gases is at equilibrium and the pressure is changed, a stress is placed on thesystem. If we increase the pressure on the system, it will adjust itself to reduce the pressure. This is inaccord with LeChatelier's principle. Since the pressure of the system is directly proportional to thenumber of gas molecules present, the only way to reduce the pressure (at constant temperature) is toreduce the total number of molecules in the system. This can occur if two NO2 molecules combine to formone N2O4 molecule. Therefore, to relieve the strain caused by increasing the pressure, the equilibrium willshift to the left toward the reactant which is N2O4.

Increasing the pressure on a gaseous system atequilibrium causes the equilibrium to shift to the side withthe fewest number of molecules.

On the other hand, if the pressure is decreased, more gas molecules must be formed to bring thesystem's pressure back to equilibrium. In this case, some N2O4 molecules will decompose into two NO2

molecules. This will increase the number of gas molecules present, thereby increasing the pressure inthe system.

Decreasing the pressure on a gaseous system atequilibrium causes the equilibrium to shift to the side withthe larger number of molecules.

Consider a container in which the reactionshown at right has come to equilibrium: N2(g) + 3 H2(g) <===> 2 NH3(g)

In which direction will the equilibrium shift if the pressure on the system above is

decreased? {22}_____________ Explain: {23}____________________________________________

______________________________________________________________________________

How should the pressure be changed on the system above in order to produce a larger amount of

ammonia, NH3? {24}____________________ Explain: {25}_________________________________

______________________________________________________________________________

Shown at right is another reaction which you studiedpreviously involving the formation of HI gas by the reaction:

H2(g) + I2(g) <===> 2 HI(g)

How will the equilibrium in the reaction above be affected by an increase in pressure? {26}_____________

Explain: {27}_____________________________________________________________________

In part a of Problem 9 below, the equation shows 1 molecule of N2O4 on the left side and 2molecules of NO2 on the right side. Therefore, a decrease in pressure would cause a shift of theequilibrium to the right (toward more molecules). In part b of Problem 9, the equation reveals 2 moleculeson the left side of the arrow and only 1 molecule on the right side. Indicate the direction of shift when thepressure decreases in the space provided. Then, complete Problem 9.

20-13 ©1997, A.J. Girondi

Problem 9. Complete Table 20.3 below. Indicate the direction in which the equilibrium will shift if thepressure is changed in the manner indicated.

Table 20.3Pressure Changes and Equilibrium Shifts

Equilibrium Equations Pressure Shifts

a. N2O4(g) <===> 2 NO2(g) decreased ---->b. PCl3(g) + Cl2(g) <===> PCl5(g) decreased _____c. 2 SO3(g) <===> 2 SO2(g) + O2(g) decreased _____d. 2 CO(g) + O2(g) <===> 2 CO2(g) decreased _____e. N2(g) + O2(g) <===> 2 NO(g) increased _____f. 2 H2(g) + O2(g) <===> 2 H2O(g) decreased _____g. C3H8(g) + 5 O2(g) <===> 3 CO2(g) + 4 H2O(g) decreased _____h. 2 N2O(g) <===> 2 N2(g) + O2(g) increased _____i. 2 HBr(g) <===> H2(g) + Br2(g) increased _____j. CH4(g) + 2 O2(g) <===> CO2(g) + 2 H2O(g) increased _____

SECTION 20.6 LeChatelier's Principle and Changes in Volume

LeChatelier's principle can also be used to predict the effect of changes in the volume of thereaction container on equilibrium. Equilibrium shifts caused by volume changes are similar to thosecaused by pressure changes. When the volume of a particular reaction container is reduced, themolecules get crowded together and collide more frequently. This stress can be relieved by decreasingthe number of molecules present. Look again at the nitrogen dioxide equilibrium:

2 NO(g) + O2(g) <===> 2 NO2(g)

A decrease in the volume of the reaction container can be compensated for by forming fewer molecules.The result is that the equilibrium above shifts to the right to form more molecules of NO2. In the process,three molecules of reactants will become two molecules of product. In general terms, this relationship canbe stated as follows:

For reactions in which there is a change in the number ofgas molecules, a decrease in the volume favors the reactionthat produces fewer molecules. An increase in volumefavors the reaction that produces the larger number ofmolecules.

If the forward and reverse reactions of an equilibrium systemproduce the same number of molecules, then changes in volumehave no effect on the system. Consider the system shown at right:

H2(g) + Cl2(g) <===> 2 HCl(g)

Note that the forward reaction (--->) produces two molecules of HCl, while the reverse reaction (<---) alsoproduces two molecules – one H2 and one Cl2. (This is also true for pressure changes.)

20-14 ©1997, A.J. Girondi

Problem 10. Based on this generalization, predict whether equilibrium shifts toward the products (--->)or the reactants (<---) in each example below when the volume of the reaction container is decreased:

a. PCl5(g) <===> PCl3(g) + Cl2(g) _____________________________

b. N2(g) + 3 H2(g) <===> 2 NH3(g) _____________________________

c. 2 CO(g) + O2(g) <===> 2 CO2(g) _____________________________

Problem 11. In which direction will the systems below shift if the volume of the reaction container isincreased:

a. H2(g) + I2(g) <===> 2 HI(g) _____________________________

b. CO(g) + 2 H2(g) <===> CH3OH(g) _____________________________

c. C3H8(g) + 5 O2(g) <===> 3 CO2(g) + 4 H2O(g) _____________________________

Problem 12. Study the equations in Table 20.4. Determine the direction of equilibrium shift. Answerby writing either "forward" or "reverse," or by using arrows: ---> or <---. These equations are a little morecomplicated, because they involve liquids and solids in addition to gases. Pressure changes do not havemuch, if any, effect on liquids and solids, so you should only consider molecules of gases in an equilibriumsystem when effects of pressure changes are being evaluated. Therefore, substances which are liquidsor solids with subscripts (l) or (s) in the equations should be ignored in Table 20.4.

Remember: As you complete Table 20.4, ignore any substances below which are solids or pure liquids.

Table 20.4Equilibrium Shifts and Pressure Changes

Pressure ShiftEquilibrium Equation Change Direction

a. CO(g) + H2O(l) <===> CO2(g) + H2(g) increase ___________

b. 4 FeS(s) + 7 O2(g) <===> 2 Fe2O3(s) + 4 SO2(g) increase ___________

c. 4 NH3(g) + 5 O2(g) <===> 4 NO(g) + 6 H2O(l) decrease ___________

d. C(s) + O2(g) <===> CO2(g) decrease ___________

When an equilibrium system involving gases is present in a closed container, changes in volumecause changes in pressure. If the size of the container is decreased (volume is decreased) the pressuregoes up. For example, imagine that you are squeezing a balloon which contains gases into a smallervolume. To relieve the increased pressure, the system will shift in the direction of fewer gas atoms ormolecules. If the size of the container is increased, the system will shift in the direction which will providemore gas atoms or molecules which can occupy the added volume of space. Changes in volume – likechanges in pressure – do not affect solids or pure liquids.

20-15 ©1997, A.J. Girondi

Problem 13. Complete Table 20.5 below. Reminder: Ignore any substances which are solids (s) orpure liquids (l).

Table 20.5Equilibrium Shifts and Volume Changes

Volume ShiftEquilibrium Equation Change Direction

a. CO(g) + H2O(l) <===> CO2(g) + H2(g) increase ___________

b. 4 FeS(s) + 7 O2(g) <===> 2 Fe2O3(s) + 4 SO2(g) increase ___________

c. 4 NH3(g) + 5 O2(g) <===> 4 NO(g) + 6 H2O(l) decrease ___________

d. C(s) + O2(g) <===> CO2(g) decrease ___________

SECTION 20.7 Equilibrium Systems Involving Solids and/or Liquids

Many of the equilibrium equations that you have seen so far in this chapter have reactants andproducts which are either gases (g) or water (aqueous) solutions (aq). Changes in volume or pressurehave an effect on gases. Changes in concentration have an effect on both gases or aqueous solutions.However, changes in volume, pressure, or concentration in equilibrium systems do not affect solids orpure liquids. They are not variables that play a role in determining the value of Keq. We will save an in-depth discussion of the reasons for this for a future chemistry course! Solids and pure liquids are neverincluded in Keq expressions. Examine the following examples.

CaO(s) + CO 2(g) <===> CaCO 3(s) Keq =

1

[CO2 ]

H2O(l) + HF(g) <===> H 3O 1+

(aq) + F 1-(aq) Keq =

[H3O1+ ] [F1- ]

[HF]

Problem 14. Write the equilibrium (Keq) expressions for the following systems. Do not includesubstances which are solids (s) or pure liquids (l).

a. BaCO3(s) <===> BaO(s) + CO2(g) Keq =

b. HCN(aq) + H2O(l) <===> H3O1+(aq) + CN1-(aq) Keq =

c. CuSO4•3H2O(s) + 2 H2O(g) <===> CuSO4•5H2O(s) Keq =

20-16 ©1997, A.J. Girondi

SECTION 20.8 LeChatelier's Principle and Changes in Temperature

Changes in temperature will also create a stress on a system at equilibrium. In this case, the effectis more complicated than that for stresses caused by concentration and pressure changes. This isbecause Keq is temperature dependent, meaning that its value will be numerically different at differenttemperatures for a given reaction. Thus, heating or cooling an equilibrium system will result in a shifting ofthe equilibrium forward (to the right) or reverse (to the left), depending on which of the two reactions isexothermic and which is endothermic.

Let's first consider a reaction that is exothermic (as most reactions tend to be). Such a reactioncan be written as: reactants <===> products + heat energy

Suppose the temperature of the system is increased. This involves adding heat energy to the system.

This is like increasing the amount of a product of the reaction. (Since heat appears on the right side of the

equation, it can be considered a product.) You might think of this as placing a stress on the right side of

the equation. To relieve this stress, the equilibrium will shift to the left. The concentrations of products will

then {28}_____________ , and the concentrations of reactants will {29}_______________. .

Lowering the temperature of an exothermic reaction that has come to equilibrium should have the

opposite effect. Lowering the temperature amounts to removing heat energy from the system. This

stress will result in the equilibrium shifting to the right. This will {30}______________the concentration of

products and {31}________________ the concentration of reactants.

Next, let's consider an endothermic reaction that can be written as:

reactants + heat energy ----> products

The heat energy can be regarded as part of the reaction, in this case as one of the reactants. Adding heat

energy by increasing the temperature of this system at equilibrium amounts to placing a stress on the left

side of the equation. To relieve the stress, the concentration of reactants will decrease by forming more

{32}___________. (The system shifts to the right.) Decreasing the temperature of this endothermic

reaction will cause the equilibrium to shift to the {33}_____________.

In summary, you know that in an equilibrium system one of the reactions is exothermic and the

other is {34}______________. Raising the temperature (which amounts to adding heat) will cause an

increase in the rate of both reactions. In terms of collision theory, why would this be true? {35}_________

______________________________________________________________________________

However, when heat is added, the rate of the endothermic reaction (the reaction which uses heat) will

generally increase more than the rate of the exothermic reaction (which gives off heat). In other words,

adding heat causes a shift in favor of the endothermic reaction. Removing heat, causes a shift in favor of

the exothermic reaction.

20-17 ©1997, A.J. Girondi

Problem 15. Complete Table 20.6 by indicating the direction of the equilibrium shift when thetemperature is changed as indicated.

Table 20.6The Effect of Temperature Changes on Equilibrium Systems

Temperature Shift Equilibrium Equation Change Direction

a. H2(g) + Cl2(g) <===> 2 HCl(g) + 44184 kJ decrease __________b. 50.2 kJ + H2(g) + I2(g) <===> 2 HI(g) decrease __________c. CH4(g) + 2 O2(g) <===> CO(g) + 2 H2O(l) + 887 kJ increase __________d. C(s) + O2(g) <===> CO2(g) + 393 kJ decrease __________e. N2(g) + 3 H2(g) <===> 2 NH3(g) + 46 kJ increase __________f. 2376 kJ + 8 SO2(g) <===> S8(s) + 16 O2(g) decrease __________g. 75.3 kJ + CH4(g) <===> C(s) + 2 H2(g) increase __________

ACTIVITY 20.9 Testing LeChatelier's Principle With Cobalt Ions

This next activity will allow you to study the effects of concentration and temperature changes onan equilibrium system that exists between two different cobalt complexes. In water solutions, the Co2+ ionis pink. The pink color is due to the existence of a complex ion with the formula: Co(H2O)62+. This is theform in which cobalt normally exists in water. When Cl1- ions are also present in high concentrations, theCo(H2O)62+ is converted to Co(H2O)4Cl2, which is blue:

Co(H2O)62+(aq) + 2 Cl1-(aq) <===> Co(H2O)4Cl2(aq) + 2 H2O(l)

pink blue

The two colored Co2+ species can be converted to one another by appropriate changes in theconcentration of Cl1- ion or of water and by changes in temperature. Follow the procedure below, and besure to wear your glasses.

1. Mark two 50 mL Erlenmeyer flasks or 50 mL beakers "1" and "2."

2. Prepare the following two solutions:

Solution 1: Dissolve 0.50 g of CoCl2•6H2O in 10 mL of 6M HCl (hydrochloric acid). Handle the HCl withcare. The high concentration of Cl1- in the HCl pushes the equilibrium to the right and most of the cobalt inthis mixture is in the form of Co(H2O)4Cl2. Stir the solution until the solid is completely dissolved. What isthe color of Co(H2O)4Cl2 in solution 1? {36}______________

Solution 2: Dissolve 0.50 g of CoCl2•6H2O in 15 mL of water. The high concentration of H2O in thismixture pushes the equilibrium to the left and most of the cobalt in this mixture is in the form ofCo(H2O)62+. Stir the solution until the solid is completely dissolved. What is the color of Co(H2O)62+ insolution 2? {37}_______________

Adding HCl (and therefore Cl1-) to this system causes it to shift to the {38}____________ and the color

turns more {39}___________. Adding H2O to this system causes it to shift to the {40}____________and

the color turns more {41}________________.

20-18 ©1997, A.J. Girondi

2. Add 5 mL of water (or more if necessary) to solution 1 until a color change occurs. Now what is thecolor of the solution? {42}_______________ Heat the flask of solution 1 on a hotplate or with a burner untila color change occurs. What is the color of the heated solution 1?{43}______________

3. What do you think will happen to the color of the solution if it is cooled? {44}__________________Now place the flask of solution 1 into an ice water bath. Allow the flask to remain in the ice water bath until achange occurs. Was your prediction correct?{45}_________________

4. Keeping in mind that solution 2 contains CoCl2•6H2O, what two things could you do to solution 2 to getit to form more Co(H2O)4Cl2? {46}______________________________________________________Now do both of these two things – simultaneously – to a 15 mL portion of solution 2 in a 50 mL flask orbeaker. Describe the result: {47}______________________________________________________Did you manage to make more Co(H2O)4Cl2 in solution 2? {48}_________________________________

How do you know? {49}_____________________________________________________________

5. Based on your results, is the forward (--->) reaction for this cobalt system endothermic or exothermic?

{50}________________. Explain how you know: {51}_______________________________________

______________________________________________________________________________

If you have a little time left try this. Add some solid CoCl2 to a small volume of water in a test tube and shaketo dissolve. Pour the solution onto a piece of filter paper. Note that it is red in color. Now hold the wetfilter paper with your crucible tongs and warm it gently over the flame of a lab burner. Be careful not toignite the paper as it drys. Note the color change as you evaporate the water out of the system. Wet thepaper withplain water to restore the red color.

SECTION 20.10 A Review of LeChatelier's Principle

1. State LeChatelier's principle: {52}___________________________________________________

______________________________________________________________________________

2. In which direction will an equilibrium system shift if the concentration of one of the products isdecreased (at constant T and P)? {53}__________________________________________________

3. In which direction will an equilibrium shift if a reaction has more gas molecules on the left (reactant side)than on the right (product side) and if the pressure of the system is increased? {54}__________________

4. Suppose for a hypothetical equilibrium system such as A(g) <===> B(g), the forward reaction (from leftto right) is exothermic. In which direction (forward --->) or (<--- reverse) will the equilibrium shift if thetemperature is increased? {55}___________________

5. Will an equilibrium reaction shift forward (--->) or reverse (<---) if the concentration of one of the

reactants is decreased (at constant T and P)? {56}__________________

6. For a reaction involving equal numbers of gas molecules on both sides of the equation, will the

equilibrium shift forward or reverse if the pressure is decreased? {57}____________________________

7. For an endothermic reaction, will the equilibrium shift toward products or reactants if the temperature is

increased? {58}_______________________________

20-19 ©1997, A.J. Girondi

SECTION 20.11 Using Keq Values to Make Predictions

Equilibrium constants (Keq) are quite useful to chemists because they provide a clue about theamount of product that can be produced in a given chemical reaction. Normally, a chemical reaction iscarried out because an experimenter wants to obtain and use the product. Ideally, one would like to get a100% yield, which means that all of the reactants would be converted into products. However, 100%yields are not always possible. Instead, a system may go to equilibrium resulting in less than a 100% yield.

We are able to predict the extent to which reactants will be converted into products based on thesize of Keq. Remember that Keq is related to a ratio involving products over reactants. Look at the Keq

values calculated below:

Keq =

100

2 = 50; Keq =

10

0.001 = 1 X 10 4 ; Keq =

50

0.02 = 2.5 X 103

1. Suppose the three Keq values above represent very similar reactions. Which Keq value represents thereaction which produced the most product?{59}_________________

2. Which Keq value represents the reaction which produced the least product?{60}__________________

3. Explain how Keq values can be used to determine which of a series of similar reactions will produce the

most product? {61}________________________________________________________________

4. For the gaseous reaction A + B <===> C, a chemist is interested in getting as large a yield of the

product C as possible. He varies the reaction temperature which is the one variable that can change the

Keq value of a system. At 300oC he experimentally calculates that Keq for the system is 26.2. At 10oC he

finds that Keq = 0.012. To maximize the amount of C produced, should the chemist heat the reaction

container or cool it?{62}_______________ Explain: {63}_____________________________________

______________________________________________________________________________

Problem 16. For the equilibrium system: 2 SO2(g) + O2(g) <===> 2 SO3(g) the value of K eq at roomtemperature is 30.0. Predict the concentration of SO3 gas in the system when it is at equilibrium if theother concentrations are: [SO2] = 0.20M and [O2] = 0.30M.

__________M

20-20 ©1997, A.J. Girondi

SECTION 20.12 Review Problems

Problem 17. A five-liter flask contains the system: CO(g) + Cl2(g) <===> COCl2(g). The flask contains1.50 moles of CO, 1.00 mole of Cl2, and 4.00 moles of COCl2. (These are all gases.) Calculate the value ofthe equilibrium constant for this system. (Remember values used must be in moles per 1.00 liter.)

________________

Problem 18. In a 1-liter flask the following system is at equilibrium: C(s) + H2O(g) <===> CO(g) + H2(g).The amounts of substances present in the 1-liter flask are 0.16 mole of C, 0.58 mole of H2O, 0.15 mole ofCO, and 0.15 mole of H2. Calculate the value of Keq for this system. (Note that C is a solid while the othersubstances are in the gas phase.)

________________

There is a supplementary discussion of another type of equilibrium constant known as thesolubility product constant, Ksp, in Appendix E of your ALICE materials. Ask your teacher if you shouldstudy that Appendix, or if you should end Chapter 20 here.

Go To Appendix E???(Ask the Instructor)

20-21 ©1997, A.J. Girondi

SECTION 20.13 Learning Outcomes

Equilibrium is an extremely important topic in chemistry and in all of the sciences. It will be veryuseful to you in the upcoming chapter on acids and bases. Look over the learning outcomes and makesure that you have mastered each of them. Check them off once you are satisfied. Arrange to take anyquizzes or exams on Chapter 20, and then move on to Chapter 21.

_____1. Define equilibrium and state the general characteristics of a system in equilibrium.

_____2. Write equilibrium expressions for chemical systems involving solids, liquids, and gases.

_____3. Calculate Keq values given the equilibrium concentrations of the products and reactants.

_____4. State LeChatelier's principle in general terms.

_____5. Use LeChatelier's principle to predict the direction an equilibrium will shift if there is a change in pressure, concentrations, temperature, or volume of the reaction container.

_____6. Given the Keq values for two or more similar systems in equilibrium, predict which system contains the greater concentration of products.

The following learning outcomes are to be included only if you studied the material concerning Ksp inAppendix E.

_____7. Determine the identity of unknown chemicals by testing and comparing them with a set of knownchemicals.

_____8. Given the solubility of a substance, calculate its Ksp value.

_____9. Given the Ksp value and the concentration of one ion, calculate the concentration of the other ion.

20-22 ©1997, A.J. Girondi

SECTION 20.14 Answers to Questions and Problems

Questions:

{1} beaker A; {2} molecules are escaping; {3} no; {4} molecules that escape cannot return to liquid phase;{5} they will be greater; {6} left; {7} increase; {8} decrease; {9} increase; {10} decrease; {11} amber(depends on your color vision); {12} deep red (depends on your color vision); {13} darker; {14} increase;{15} causes more collisions between Fe3+ ions and SCN1- ions and shifts system toward right (products);{16} lighter; {17} decrease; {18} system shifted toward the left (toward reactants); {19} lighter;{20} decrease; {21} system shifts toward the left (toward reactants}; {22} shift to left toward reactants;{23} decreased pressure is a stress, so system shifts to the left to form more molecules to help increasethe pressure; {24} increase the pressure; {25} increasing the pressure will create a stress which thesystem will try to relieve by forming fewer molecules (shift to the right) which will help lower the pressure;{26} no effect; {27} since both sides of equation have same number of molecules, shifting would notchange pressure; {28} decrease; {29} increase; {30} increase; {31} decrease; {32} products;{33} left toward reactants; {34} endothermic; {35} more collisions between particles occur at highertemperatures; {36} blue (depends on your color vision); {37} pink (depends on your color vision);{38} right; {39} blue; {40} left; {41} pink; {42} pink; {43} blue; {44} will turn pink; {45} I hope so!{46} heat it and add more HCl; {47} solution should shift toward blue; {48} yes; {49} because of thechange in color; {50} endothermic; {51} because adding heat speeds up an endothermic reaction morethan it speeds up an exothermic one; {52} when a stress is placed on a system at equilibrium, the systemwill adjust to relieve the stress and to restore equilibrium in the system; {53} shifts to the right towardproducts; {54} shifts to the right toward products; {55} shift to the left toward reactants; {56} shifts to theleft (reverse) toward reactants; {57} neither, it will not shift either way; {58} shift to the right towardproducts; {59} 1X 104; {60} 50; {61} bigger Keq value means more products; {62} heat it; {63} since Keq isgreater at the higher temperature, the forward reaction which forms C is endothermic

Problems:

1. a. NaCl(s) <===> Na1+(aq) + Cl1-(aq); b. Fe2(SO4)3(s) <===> 2 Fe3+(aq) + 3 SO42-(aq)

c. BaBr2(s) <====> Ba2+(aq) + 2 Br1-(aq)

2. a. Keq = [N2O4] / [NO2]2; b. Keq = [NH3]2 / [N2] [H2]3; c. Keq = [Ag(NH3)21+] / [Ag1+] [NH3]2

d. Keq = [N2] [H2O]2 / [NO]2[H2]2

3. 0.37

4. 0.16

5. 6.4 X 105

6. 39.3

7. a. --->; b. <---; c. --->; d. <---; e. --->; f. <---; g. <---; h. --->; i. <---

8. a. none; b. --->; c. <---; d. <---

9. a. --->; b. <---; c. --->; d. <---; e. no effect; f. <---; g. --->; h. <---; i. no effect; j. no effect

10. a. <---; b. --->; c. --->

11. a. no effect; b. <---; c. --->

12. a. <---; b. --->; c. <---; d. no effect

13. a. --->; b. <---; c. --->; d. no effect

14. a. Keq = [CO2]; b. Keq = [H3O1+] [CN1-] / [HCN]; c. Keq = 1 / [H2O]2

15. a. --->; b. <---; c. <---; d. --->; e. <---; f. <---; g. --->

20-23 ©1997, A.J. Girondi

16. 0.60

17. 13.3

18. 3.9 X 10-2

20-24 ©1997, A.J. Girondi