Download 20. Chemical Equilibrium

NAME____________________________________ PER____________ DATE DUE____________
ACTIVE LEARNING I N C HEMISTRY E DUCATION
"ALICE"
CHAPTER 20
CHEMICAL
EQUILIBRIUM
Keq
20-1
©1997, A.J. Girondi
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© 1997 A.J. Girondi, Ph.D.
505 Latshmere Drive
Harrisburg, PA 17109
[email protected]
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 reactants
are converted to products before a reaction stops. In truth, however, experiments show that the
conversion of reactants into products is often incomplete in chemical reactions. This is the case no matter
how long the reaction is allowed to continue.
As a reaction progresses, the concentrations of the reactants decrease, while the concentrations
of the products increase. Eventually, a state is established in which the concentrations of products and
reactants no longer change. This is known as the state of equilibrium. You have already studied many
examples of equilibrium. The most obvious example was probably encountered during your study of
solutions. 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 system
reaches 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 of
dissolving and crystallizing. (See Figure 20.1 below.) This is an example of a physical equilibrium since
dissolving and crystallizing are physical changes. An equation can be written to represent what is
occurring in this saturated solution after equilibrium has been reached:
CaCl2(s)
Ca2+(aq) + 2 Cl1-(aq)
REACTANTS
PRODUCTS
The (aq) subscript refers to "aqueous"
which means dissolved in water.
The double arrow indicates that the forward reaction is occurring
at the SAME RATE as the reverse reaction. Therefore, no
overall change occurs in the system. To an observer, it appears
as though nothing whatsoever is happening in the saturated
CaCl2 solution. Actually, the change shown at right is occurring
constantly in both directions. Note that the equation above
shows chlorine as 2 Cl1-, not as Cl2. Chloride ions (Cl1-) do not
exist in diatomic form. This is because these ions already have a
stable octet of electrons. Only chlorine atoms form diatomic
molecules: (Cl2). This is also true for the other diatomic elements.
WRONG! ---> CaCl2(s) <===> Ca2+(aq) + Cl2(aq) <--- WRONG!
Saturated CaCl2 Solution
Ca2+
Cl1-
Cl1-
Figure 20.1
Solution Equilibrium
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 happening
constantly.
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 the
state (solid, liquid, gas, or aqueous) of each substance is represented using subsscripts. Finally, notice
that coefficients are used to balance the equations properly.
20-3
©1997, A.J. Girondi
Saturated PbI 2 Solution
Saturated AlCl3 Solution
Pb 2+
Al3+
I 1-
I 1-
Cl1-
Cl1Cl1-
PbI 2
PbI 2 (s) <===>
AlCl3
Pb2+(aq)
Figure 20.2
+
2 I 1-(aq)
AlCl3(s) <===> Al3+(aq) + 3 Cl1-(aq)
Equilibrium Systems in Saturated Solutions
Problem 1. The three equations below represent equilibrium systems created by dissolving solids in
water. Complete the equilibrium equations below. Include any charges on ions and the state of the ions
which 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 change
and, therefore, water is not consider a reactant. When substances dissolve in water they do not
chemically react with it. They merely come part in the water. Later in this chapter you will study reactions
that 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}____________________________________________
H2O(g)
H2O(g)
H2O(l)
H2O(l)
BEAKER "A"
BEAKER "B"
Figure 20.3
Closed and Open Systems
H2O(l)
H2O(g)
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©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 system
such as this only if the system is closed. In addition, this system must be held under conditions of
constant temperature, pressure, and volume. If one or more of these factors is altered, the equilibrium
changes as well. (Not all equilibrium systems are affected by pressure or volume, but all equilibrium
systems 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 and
products 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 double
arrows show that equilibrium prevails. There is absolutely no way that you can tell what the concentrations
of any of the products or reactants are just by looking at this equation. The reaction container may have a
few 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
[Ag 1+], it means "molar concentration of silver ions." This law of chemical equilibrium states that if you find
the product of the concentrations of the products, [C] [D], and divide that result by the product of the
concentrations 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 the
RATE of the reverse (<---) reaction. From the equation for the equilibrium system, we can write what is
called the equilibrium expression for this reaction:
[NH3 ] 4 [O 2 ]5
equilibrium expression =
[NO]4 [H 2 O]6
You should notice two things about the expression above. First, note that the numerator contains the
product of the concentrations of the products, and the denominator is the product of the concentrations
of the reactants. Second, you should see that the coefficients in the equation become exponents in the
equilibrium expression. The reasons why we turn coefficients into exponents can be explained using
kinetic–molecular theory and collision theory. However, we will simply accept this, and save the
explanation for a more advanced course. If you know the concentrations of all the reactants and products
in this equilibrium system, you can plug those values into this expression and solve it. The value you will
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©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 the
expression will maintain the same value. Since it is a constant value, it is given the symbol K. Since it
involves equilibrium, it is often called Keq. (Other authors may designate it as K, Kc, etc.) So,
K eq =
[NH 3 ]4 [O 2 ]5
[NO]4 [H 2 O]6
Problem 2. Write the equilibrium expressions for each of the chemical situations given below.
a. 2 NO2(g) <===> N2O4(g)
K eq =
b. N2(g) + 3 H2(g) <===> 2 NH3(g)
K eq =
c. Ag1+(aq) + 2 NH3(aq) <===> Ag(NH3)21+(aq)
K eq =
d. 2 NO(g) + 2 H2(g) <===> N2(g) + 2 H2O(g)
K eq =
Setting up an equilibrium expression is an extremely useful skill after you become aware of what
information 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 at
a given temperature in such a way as to allow the products to accumulate in the reaction container. After a
period of time, the reaction will reach equilibrium. At this point, it may be possible to experimentally
determine the concentrations of the reactants and products in the container. The concentration values
are 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 is
available to determine the concentrations of reactants and products at equilibrium. Equilibrium equations
are 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) + I 2(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:
20-6
K eq =
[HI] 2
[H 2 ] [I 2 ]
©1997, A.J. Girondi
2. Below are the experimentally determined equilibrium concentrations of each reactant and product for
the system above. The brackets mean molarity (moles / liter of sol'n).
Temp.
[H2]
[I 2]
[HI]
Experiment 1:
250 oC
0.00560
0.000590
0.01270
Experiment 2:
250 oC
0.00460
0.000970
0.01476
Notice that two experiments were probably performed under slightly different conditions, producing
different equilibrium concentrations. Temperature was constant.
3. Substituting this data into the equilibrium expression:
Experiment 1:
K eq =
(0.0127) 2
(0.0056) (0.00059)
=
48.8
Experiment 2:
K eq =
(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 then
use 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
K eq = ____________
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©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
K eq = ____________
Problem 6. Reaction: NCl3(g) + Cl2(g) <===> NCl5(g) (at 520oC) Equilibrium is established when
there 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.)
K eq = _____________
SECTION 20.3
LeChatelier's Principle
Earlier in this chapter it was mentioned that certain factors can disrupt an equilibrium. If one or
more of these factors is altered, the equilibrium is momentarily upset. These factors are changes in:
1.
2.
3.
4.
temperature
pressure
volume of the reaction container
concentration of a reactant and or a product
The exact effect of any change in reaction conditions on equilibrium was studied extensively by Henri
LeChatelier, a French chemist. LeChatelier was mainly interested in what occurred when changes in the
four 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 the
forward or reverse reaction. After the system moves away from equilibrium, he observed that the amounts
of reactants and products adjusted in order to restore the system to equilibrium. The results of his
observations allowed him to formulate the following law, or principle, known as
LeChatelier's Principle: When a stress is placed on a system at equilibrium, the
system 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 states
that when a stress is applied, the RATES of the reactions will adjust so that the RATES once again
become equal.
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©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 relative
rates of forward and reverse reactions
LeChatelier's principle enables us to predict the direction in which the equilibrium would shift when the
concentration of one or more of the products or reactants is changed. Suppose more oxygen were
added to the reaction container. We could ask, "What effect would this change have on the rates of the
forward and reverse reactions, and would this concentration increase affect the concentrations of the
other components in the system?"
You are putting a "stress" on a system in equilibrium whenever you change the concentration of
any of the products or reactants. This means that, as a result of the change, the system is no longer in
equilibrium. LeChatelier's principle can be explained using the collision theory. Part of this theory states
that a collision must occur before a chemical reaction can take place between reactants. Furthermore, the
collisions must produce enough energy and the particles must often collide in just the right way. Not all
collisions result in a reaction. When more oxygen is added, there will be more collisions between the NO
and O2 molecules. This will increase the rate of the forward (---->) reaction:
NO
O2
NO2
+
(Excess NO present)
rate of forward reaction is now greater
some oxygen is added
The system will try to remove the stress by getting rid of the excess oxygen. In this way, an equilibrium
condition can be restored. The only way for the system to get rid of the extra oxygen is to have it react with
NO to produce more product, NO2. So, momentarily the system "shifts to the right" and produces more
product:
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 which
increases 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 the
concentrations 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 the
right" the system now has more product than it did before more oxygen was added. Notice, too, that the
shift caused the concentration of a reactant (NO) to decrease. The stress was relieved in the sense that at
least some of the added oxygen has been removed.
new equilibrium concentrations
NO
O2
NO2
+
(Excess NO present)
new equilibrium rates
Since oxygen was added, how do you think the rates of the forward and reverse reactions in the new
equilibrium compare to those in the old equilibrium? {5}______________________________________
Using these principles, predict the direction of the shift (forward ---->) or (reverse <----) in
equilibrium when the concentration of Cl2 is increased in the system below, and how this will affect the
concentrations 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
{9}_________________.
of
Cl 2 will
{8}_________________.
The
concentration
of
O2 will
The concentration of H2O will {10}_________________.
Problem 7. Complete Table 20.1 below. The first blank has been completed as an example.
Table 20.1
Direction of Equilibrium Shift
Equation
Added
Substance
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)
SO 3
________________
g. 2 NCl3(g) <===> N2(g) + 3 Cl2(g)
Cl2
________________
h. CH3COOH(aq) <===> H1+(aq) + CH3COO1-(aq)
i. C2H6(g) <===> H2(g) + C2H4(g)
CH3COOH
H2
20-10
Equilibrium Shift
to Right or Left
------>
________________
________________
©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 an
experiment that will allow you to apply this skill. The shifting of an equilibrium will be examined
experimentally by using the reaction for the formation of a "complex ion" with the formula Fe(SCN)2+ by
the combination of Fe 3+ ions with an SCN1- ion:
Fe3+(aq)
+
(pale yellow)
SCN1-(aq)
(colorless)
<===> Fe(SCN)2+(aq)
(reddish)
This is a good example to use because a solution of Fe3+ ions has a faint yellow color, while a solution of
SCN1- ions has no color. When these two solutions are mixed, a deep-red color results due to the
formation of the Fe(SCN) 2+ ion. In this activity you will examine what happens when certain stresses are
placed 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 for
this 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(NO 3)3 solution. This solution contains Fe 3+ ions and NO31- ions. The
color 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 on
the appearance of this solution, what can you conclude about the color of the K 1+ 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(NO 3)3 solution with 1 or 2 mL of the KSCN solution. The resulting
color 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 reaction
between 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 contains
some 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)
(pale yellow)
+
SCN1-(aq)
(colorless)
<===> Fe(SCN)2+(aq)
(reddish)
5. Now place four 4-mL portions of the solution formed in step 4 into separate clean standard-sized (150
mm) test tubes. (Make sure the test tubes all have the same internal diameter.) Label the tubes A through
D 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}_________________________
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©1997, A.J. Girondi
When you added Fe(NO3)3, you were adding Fe 3+ 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 Fe 3+ 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}______________
concentration of
Does this change indicate an increase or a decrease in the
Fe(SCN) 2+?{17}___________________________
Thus, adding NaOH has the overall effect of DECREASING the concentration of Fe3+ ions in the
equilibrium 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 your
hands. After several hours, a dark stain can result.) Is the color of this mixture lighter or darker than that of
tube A? {19}________________
Does this change indicate an increase or a decrease in the
concentration of Fe(SCN)2+? {20}___________________
When you add AgNO3 to the equilibrium
11system, some of the SCN ions react with it. The SCN that reacted this way is removed from the
equilibrium system. So, adding AgNO3 has the effect of DECREASING the concentration of SCN1- ions
in 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 20.2
Shifts in the Fe(SCN) 2+ Equilibrium System
Test
Substance Added
Effect
a.
nothing
none
b.
Fe(NO3)3
increases Fe3+
___________________
c.
NaOH
decreases Fe 3+
___________________
d.
AgNO 3
SECTION 20.5
decreases
Shift Forward (--->) Toward Products, or
Shift Reverse (<---) Toward Reactants
none
SCN1-
___________________
LeChatelier's Principle and Changes in Pressure
Now let's "shift" our discussion to the effects of pressure changes on equilibrium systems. For
reactions 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 is
placed on the system. However, when a gas is involved, changing pressure will have an effect on the
equilibrium. 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 the
system. If we increase the pressure on the system, it will adjust itself to reduce the pressure. This is in
accord with LeChatelier's principle. Since the pressure of the system is directly proportional to the
number of gas molecules present, the only way to reduce the pressure (at constant temperature) is to
reduce the total number of molecules in the system. This can occur if two NO2 molecules combine to form
one N 2O4 molecule. Therefore, to relieve the strain caused by increasing the pressure, the equilibrium will
shift to the left toward the reactant which is N2O4.
Increasing the pressure on a gaseous system at
equilibrium causes the equilibrium to shift to the side with
the fewest number of molecules.
On the other hand, if the pressure is decreased, more gas molecules must be formed to bring the
system'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 in
the system.
Decreasing the pressure on a gaseous system at
equilibrium causes the equilibrium to shift to the side with
the larger number of molecules.
Consider a container in which the reaction
shown at right has come to equilibrium:
In
which
direction
will
the
equilibrium
decreased? {22}_____________ Explain:
N2(g) + 3 H2(g) <===> 2 NH3(g)
shift
if
the
pressure
on
the
system
above
is
{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 studied
previously 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?
Explain:
{26}_____________
{27}_____________________________________________________________________
In part a of Problem 9 below, the equation shows 1 molecule of N2O4 on the left side and 2
molecules of NO2 on the right side. Therefore, a decrease in pressure would cause a shift of the
equilibrium to the right (toward more molecules). In part b of Problem 9, the equation reveals 2 molecules
on the left side of the arrow and only 1 molecule on the right side. Indicate the direction of shift when the
pressure 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 the
pressure is changed in the manner indicated.
Table 20.3
Pressure Changes and Equilibrium Shifts
Equilibrium Equations
Pressure
Shifts
a. N2O4(g) <===> 2 NO2(g)
b. PCl3(g) + Cl2(g) <===> PCl5(g)
c. 2 SO3(g) <===> 2 SO2(g) + O2(g)
d. 2 CO(g) + O2(g) <===> 2 CO2(g)
e. N2(g) + O2(g) <===> 2 NO(g)
f. 2 H2(g) + O2(g) <===> 2 H2O(g)
g. C3H8(g) + 5 O2(g) <===> 3 CO2(g) + 4 H2O(g)
h. 2 N2O(g) <===> 2 N2(g) + O2(g)
i. 2 HBr(g) <===> H2(g) + Br2(g)
j. CH4(g) + 2 O2(g) <===> CO2(g) + 2 H2O(g)
decreased
decreased
decreased
decreased
increased
decreased
decreased
increased
increased
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 the
reaction container on equilibrium. Equilibrium shifts caused by volume changes are similar to those
caused by pressure changes. When the volume of a particular reaction container is reduced, the
molecules get crowded together and collide more frequently. This stress can be relieved by decreasing
the 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 can
be stated as follows:
For reactions in which there is a change in the number of
gas molecules, a decrease in the volume favors the reaction
that produces fewer molecules. An increase in volume
favors the reaction that produces the larger number of
molecules.
If the forward and reverse reactions of an equilibrium system
produce the same number of molecules, then changes in volume H 2(g) + Cl 2(g) <===> 2 HCl (g)
have no effect on the system. Consider the system shown at right:
Note that the forward reaction (--->) produces two molecules of HCl, while the reverse reaction (<---) also
produces two molecules – one H 2 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 is
increased:
a.
H2(g) + I 2(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. Answer
by writing either "forward" or "reverse," or by using arrows: ---> or <---. These equations are a little more
complicated, because they involve liquids and solids in addition to gases. Pressure changes do not have
much, if any, effect on liquids and solids, so you should only consider molecules of gases in an equilibrium
system when effects of pressure changes are being evaluated. Therefore, substances which are liquids
or 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 20.4
Equilibrium Shifts and Pressure Changes
Equilibrium Equation
Pressure
Change
Shift
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 volume
cause changes in pressure. If the size of the container is decreased (volume is decreased) the pressure
goes up. For example, imagine that you are squeezing a balloon which contains gases into a smaller
volume. To relieve the increased pressure, the system will shift in the direction of fewer gas atoms or
molecules. If the size of the container is increased, the system will shift in the direction which will provide
more gas atoms or molecules which can occupy the added volume of space. Changes in volume – like
changes 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) or
pure liquids (l).
Table 20.5
Equilibrium Shifts and Volume Changes
Equilibrium Equation
Volume
Change
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
Shift
Direction
Equilibrium Systems Involving Solids and / or Liquids
Many of the equilibrium equations that you have seen so far in this chapter have reactants and
products which are either gases (g) or water (aqueous) solutions (aq). Changes in volume or pressure
have 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 or
pure liquids. They are not variables that play a role in determining the value of Keq. We will save an indepth discussion of the reasons for this for a future chemistry course! Solids and pure liquids are never
included in K eq expressions. Examine the following examples.
CaO (s) + CO 2(g)
H 2 O(l) + HF(g)
<===> CaCO 3(s)
<===> H 3O 1+(aq)
+ F 1- (aq)
K eq =
1
[CO 2 ]
K eq =
[H 3 O1+ ] [F 1- ]
[HF]
Problem 14. Write the equilibrium (Keq) expressions for the following systems.
substances which are solids (s) or pure liquids (l).
a.
BaCO3(s) <===> BaO(s) + CO2(g)
K eq =
b.
HCN(aq) + H2O(l) <===> H3O1+(aq) + CN1-(aq)
K eq =
c.
CuSO 4•3H 2O(s) + 2 H2O(g) <===> CuSO4•5H 2O(s)
K eq =
20-16
Do not include
©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 effect
is more complicated than that for stresses caused by concentration and pressure changes. This is
because K eq is temperature dependent, meaning that its value will be numerically different at different
temperatures for a given reaction. Thus, heating or cooling an equilibrium system will result in a shifting of
the equilibrium forward (to the right) or reverse (to the left), depending on which of the two reactions is
exothermic and which is endothermic.
Let's first consider a reaction that is exothermic (as most reactions tend to be). Such a reaction
can 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
products and
{31}________________
{30}______________the
concentration of
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.)
reaction will cause the equilibrium to shift to the
Decreasing the temperature of this endothermic
{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 the
temperature is changed as indicated.
Table 20.6
The Effect of Temperature Changes on Equilibrium Systems
Equilibrium Equation
a. H2(g) + Cl2(g) <===> 2 HCl(g) + 44184 kJ
b. 50.2 kJ + H2(g) + I2(g) <===> 2 HI (g)
c. CH4(g) + 2 O2(g) <===> CO(g) + 2 H2O(l) + 887 kJ
d. C(s) + O2(g) <===> CO2(g) + 393 kJ
e. N2(g) + 3 H2(g) <===> 2 NH3(g) + 46 kJ
f. 2376 kJ + 8 SO2(g) <===> S8(s) + 16 O2(g)
g. 75.3 kJ + CH4(g) <===> C(s) + 2 H2(g)
ACTIVITY 20.9
Temperature
Change
Shift
Direction
decrease
decrease
increase
decrease
increase
decrease
increase
__________
__________
__________
__________
__________
__________
__________
Testing LeChatelier's Principle With Cobalt Ions
This next activity will allow you to study the effects of concentration and temperature changes on
an equilibrium system that exists between two different cobalt complexes. In water solutions, the Co2+ ion
is pink. The pink color is due to the existence of a complex ion with the formula: Co(H 2O)62+. This is the
form in which cobalt normally exists in water. When Cl1- ions are also present in high concentrations, the
Co(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 Co 2+ species can be converted to one another by appropriate changes in the
concentration of Cl 1- ion or of water and by changes in temperature. Follow the procedure below, and be
sure 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•6H 2O in 10 mL of 6M HCl (hydrochloric acid). Handle the HCl with
care. The high concentration of Cl1- in the HCl pushes the equilibrium to the right and most of the cobalt in
this mixture is in the form of Co(H2O)4Cl2. Stir the solution until the solid is completely dissolved. What is
the color of Co(H2O)4Cl2 in solution 1? {36}______________
Solution 2: Dissolve 0.50 g of CoCl2•6H 2O in 15 mL of water. The high concentration of H 2O in this
mixture pushes the equilibrium to the left and most of the cobalt in this mixture is in the form of
Co(H2O)62+. Stir the solution until the solid is completely dissolved. What is the color of Co(H2O)62+ in
solution 2? {37}_______________
Adding HCl (and therefore Cl 1-) to this system causes it to shift to the
turns more
{39}___________.
{38}____________
and the color
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 the
color of the solution? {42}_______________ Heat the flask of solution 1 on a hotplate or with a burner until
a 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 a
change occurs. Was your prediction correct?{45}_________________
4. Keeping in mind that solution 2 contains CoCl2•6H 2O, what two things could you do to solution 2 to get
it 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 or
beaker. 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 shake
to dissolve. Pour the solution onto a piece of filter paper. Note that it is red in color. Now hold the wet
filter paper with your crucible tongs and warm it gently over the flame of a lab burner. Be careful not to
ignite the paper as it drys. Note the color change as you evaporate the water out of the system. Wet the
paper 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 is
decreased (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 left
to right) is exothermic. In which direction (forward --->) or (<--- reverse) will the equilibrium shift if the
temperature 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 K e q Values to Make Predictions
Equilibrium constants (Keq) are quite useful to chemists because they provide a clue about the
amount of product that can be produced in a given chemical reaction. Normally, a chemical reaction is
carried out because an experimenter wants to obtain and use the product. Ideally, one would like to get a
100% 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 the
size of K eq. Remember that Keq is related to a ratio involving products over reactants. Look at the K eq
values calculated below:
K eq =
100
2
= 50;
K eq =
10
= 1 X 10 4 ;
0.001
K eq =
50
0.02
= 2.5 X 10 3
1. Suppose the three Keq values above represent very similar reactions. Which K eq value represents the
reaction 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
K eq value of a system. At 300 oC he experimentally calculates that K eq 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 SO 2(g) + O2(g) <===> 2 SO 3(g) the value of K eq at room
temperature is 30.0. Predict the concentration of SO3 gas in the system when it is at equilibrium if the
other concentrations are: [SO 2] = 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 contains
1.50 moles of CO, 1.00 mole of Cl2, and 4.00 moles of COCl2. (These are all gases.) Calculate the value of
the 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 of
CO, and 0.15 mole of H 2. Calculate the value of Keq for this system. (Note that C is a solid while the other
substances are in the gas phase.)
________________
There is a supplementary discussion of another type of equilibrium constant known as the
solubility product constant, Ksp, in Appendix E of your ALICE materials. Ask your teacher if you should
study 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 very
useful to you in the upcoming chapter on acids and bases. Look over the learning outcomes and make
sure that you have mastered each of them. Check them off once you are satisfied. Arrange to take any
quizzes 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 in
Appendix E.
_____7. Determine the identity of unknown chemicals by testing and comparing them with a set of known
chemicals.
_____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 increase
the pressure; {24} increase the pressure; {25} increasing the pressure will create a stress which the
system 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 not
change pressure; {28} decrease; {29} increase; {30} increase; {31} decrease; {32} products;
{33} left toward reactants; {34} endothermic; {35} more collisions between particles occur at higher
temperatures; {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 the
change in color; {50} endothermic; {51} because adding heat speeds up an endothermic reaction more
than it speeds up an exothermic one; {52} when a stress is placed on a system at equilibrium, the system
will adjust to relieve the stress and to restore equilibrium in the system; {53} shifts to the right toward
products; {54} shifts to the right toward products; {55} shift to the left toward reactants; {56} shifts to the
left (reverse) toward reactants; {57} neither, it will not shift either way; {58} shift to the right toward
products; {59} 1X 10 4; {60} 50; {61} bigger Keq value means more products; {62} heat it; {63} since K eq is
greater at the higher temperature, the forward reaction which forms C is endothermic
Problems:
3.
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)
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
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
1.
2.
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