Chapter 14 : Chemical Equilibrium

Terms in this set (15)

Have you ever wondered how a baby in the womb gets oxygen? Unlike you and me, a
fetus does not breathe air. Yet, like you and me, a fetus needs oxygen. After we are born,
we inhale air into our lungs; that air diffuses into capillaries, where it comes into contact with our blood. Within our red blood cells, a protein called hemoglobin (Hb) reacts with
oxygen:
Hb + O2 L HbO2
The double arrows in this equation mean that the reaction can occur in both the forward
and reverse directions and can reach chemical equilibrium. We have encountered the term
equlibrium in Chapter 11 and 12, and we define it more carefully in the next section. For
now, understand that the concentrations of the reactants and products in a reaction at
equilibrium are described by the equilibrium constant, K. A large value of K indicates
that the reaction lies far to the right at equilibrium—a high concentration of products and
a low concentration of reactants. A small value of K indicates that the reaction lies far to
the left at equilibrium—a high concentration of reactants and a low concentration of
products. In other words, the value of K is a measure of how far a reaction proceeds—the
larger the value of K, the more the reaction favors the products.
The equilibrium constant for the reaction between hemoglobin and oxygen is such
that hemoglobin efficiently binds oxygen at typical lung oxygen concentrations, but can
also release oxygen under the appropriate conditions. Any system at equilibrium, including the hemoglobin-oxygen system, acts to maintain that equilibrium. If any of the concentrations of the reactants or products change, the reaction shifts to counteract that
change. For the hemoglobin system, as blood flows through the lungs where oxygen
concentrations are high, the equilibrium shifts to the right—hemoglobin binds oxygen.
In our bodies, as blood flows out of the lungs and into muscles and organs where oxygen
concentrations have been depleted (because muscles and organs use oxygen), the equilibrium shifts to the left—hemoglobin releases oxygen.
In other words, to maintain equilibrium, hemoglobin binds oxygen when the surrounding
oxygen concentration is high, but releases oxygen when the surrounding oxygen concentration is low. In this way, hemoglobin transports oxygen from the lungs to all parts of the
body that use oxygen.
A fetus has its own circulatory system. The mother's blood never flows into the
fetus's body, and the fetus cannot get any air in the womb. How, then, does the fetus get
oxygen? The answer lies in the properties of fetal hemoglobin (HbF), which is slightly
different from adult hemoglobin. Like adult hemoglobin, fetal hemoglobin is in equilibrium with oxygen:
HbF + O2 L HbFO2
However, the equilibrium constant for fetal hemoglobin is larger than the equilibrium
constant for adult hemoglobin, meaning that the reaction tends to go farther in the direction of the product. Consequently, fetal hemoglobin loads oxygen at a lower oxygen concentration than does adult hemoglobin. In the placenta, fetal blood flows in close
proximity to maternal blood, without the two ever mixing. Because of the different equilibrium constants, the maternal hemoglobin releases oxygen, which the fetal hemoglobin then binds and carries into its own circulatory system (FigurE 14.1▲). Nature has thus
evolved a chemical system through which the mother's hemoglobin can in effect hand off
oxygen to the hemoglobin of the fetus.
Recall from Chapter 13 that reaction rates generally increase with increasing concentration of the reactants (unless the reaction is zero order) and decrease with decreasing
concentration of the reactants. With this in mind, consider the reaction between hydrogen
and iodine:
H2(g) + I2(g) L 2 HI(g)
In this reaction, H2 and I2 react to form 2 HI molecules, but the 2 HI molecules can also
react to re-form H2 and I2. A reaction such as this one—that can proceed in both the forward and reverse directions—is said to be reversible. Suppose we begin with only H2
and I2 in a container (FigurE 14.2(a)▶ on the next page). What happens initially? H2 and I2
begin to react to form HI (FigurE 14.2(b)▶). However, as H2 and I2 react, their concentrations decrease, which in turn decreases the rate of the forward reaction. At the same time,
HI begins to form. As the concentration of HI increases, the reverse reaction occurs at a
faster and faster rate. Eventually the rate of the reverse reaction (which has been increasing) equals the rate of the forward reaction (which has been decreasing). At that point,
dynamic equilibrium is reached.
Dynamic equilibrium for a chemical reaction is the condition in which the
rate of the forward reaction equals the rate of the reverse reaction
Dynamic equilibrium is "dynamic" because the forward and reverse reactions still occur;
however, they occur at the same rate. When dynamic equilibrium is reached, the concentrations of H2, I2, and HI no longer change. They remain constant because the reactants
and products form at the same rate that they are depleted. However, the constancy of the
reactant and product concentrations at equilibrium does not imply that the concentrations
are equal at equilibrium. Some reactions reach equilibrium only after most of the reactants have formed products. Others reach equilibrium when only a small fraction of the
reactants have formed products. It depends on the reaction.
We have just seen that the concentrations of reactants and products are not equal at
equilibrium—rather, it is the rates of the forward and reverse reactions that are equal.
But what about the concentrations? What can we know about them? The concentrations, as you can see by reexamining Figure 14.2, become constant; they don't change once
equilibrium is reached (as long as the temperature is constant). We can quantify the relative concentrations of reactants and products at equilibrium with a quantity called the
equilibrium constant (K). Consider an equation for a generic chemical reaction:
a A + b B L c C + d D
where A and B are reactants, C and D are products, and a, b, c, and d are the respective
stoichiometric coefficients in the chemical equation. The equilibrium constant (K) for
the reaction is the ratio—at equilibrium—of the concentrations of the products raised to
their stoichiometric coefficients divided by the concentrations of the reactants raised to
their stoichiometric coefficients:
In this notation, [A] represents the molar concentration of A. Why is this particular
ratio of concentrations at equilibrium—and not some other ratio—defined as the equilibrium constant? Because this particular ratio is always a constant when the reactants and
products are at equilibrium (at constant temperature). As we can see from the expression,
the equilibrium constant quantifies the relative concentrations of reactants and products
at equilibrium. The relationship between the balanced chemical equation and the expression of the equilibrium constant is the law of mass action.
Expressing Equilibrium Constants for Chemical Reactions
To express an equilibrium constant for a chemical reaction, we examine the balanced
chemical equation and apply the law of mass action. For example, suppose we want to
express the equilibrium constant for the reaction:
2 N2O5(g) L 4 NO2(g) + O2(g)
The equilibrium constant is [NO2] raised to the fourth power multiplied by [O2] raised to
the first power divided by [N2O5] raised to the second power.
K = [NO2]
4
[O2]
[N2O5]
2
Notice that the coefficients in the chemical equation become the exponents in the expression of the equilibrium constant.
ExamPlE 14.1 Expressing Equilibrium Constants for Chemical Equations
Express the equilibrium constant for the chemical equation.
CH3OH(g) L CO(g) + 2 H2(g)
SOLUTION
The equilibrium constant is the concentrations of the products raised to their
stoichiometric coefficients divided by the concentrations of the reactants
raised to their stoichiometric coefficients.
We have now discussed how to express the equilibrium constant, but what does it mean?
What, for example, does a large equilibrium constant (K W 1) imply about a reaction?
A large equilibrium constant indicates that the numerator of the equilibrium constant
(which specifies the concentrations of products at equilibrium) is larger than the denominator (which specifies the concentrations of reactants at equilibrium). Therefore, when
the equilibrium constant is large, the forward reaction is favored. For example, consider
the reaction:
H2(g) + Br2(g) L 2 HBr(g) K = 1.9 * 1019 (at 25 °C)
The equilibrium constant is large, so the equilibrium point for the reaction lies far to the
right—high concentrations of products, low concentrations of reactants (FigurE 14.3▼).
Remember that the equilibrium constant says nothing about how fast a reaction will reach
equilibrium, only how far the reaction has proceeded once equilibrium is reached. A
reaction with a large equilibrium constant may be kinetically very slow, meaning that it
will take a long time to reach equilibrium.
Conversely, what does a small equilibrium constant (K V 1) mean? It indicates
that the reverse reaction is favored; there will be more reactants than products when equilibrium is reached. For example, consider the reaction:
N2(g) + O2(g) L 2 NO(g) K = 4.1 * 10-31 (at 25 °C)
The equilibrium constant is very small, so the equilibrium point for the reaction lies
far to the left—high concentrations of reactants, low concentrations of products
(FigurE 14.4▼). This is fortunate because N2 and O2 are the main components of air. If
this equilibrium constant were large, much of the N2 and O2 in air would react to form
NO, a toxic gas.
N (g) 2(g) +
+
2 NO(g)
Summarizing the Significance of the Equilibrium Constant:
▶ K V 1 Reverse reaction is favored; forward reaction does not proceed very far.
▶ K ≈ 1 Neither direction is favored; forward reaction proceeds about halfway.
▶ K W 1 Forward reaction is favored; forward reaction proceeds essentially to
completion.
So far, we have expressed the equilibrium constant only in terms of the concentrations of
the reactants and products. However, for gaseous reactions, the partial pressure of a particular gas is proportional to its concentration. Therefore, we can also express the equilibrium constant in terms of the partial pressures of the reactants and products. For example,
consider the gaseous reaction:
2 SO3(g) L 2 SO2(g) + O2(g)
From this point on, we designate Kc as the equilibrium constant with respect to concentration in molarity. For this reaction, we can express Kc using the law of mass action.
Kc = [SO2]
2
[O2]
[SO3]
2
We now designate Kp as the equilibrium constant with respect to partial pressures in
atmospheres. The expression for Kp takes the same form as the expression for Kc, except
that we use the partial pressure of each gas in place of its concentration. For the SO3
reaction, we write Kp:
Kp = (PSO2
)
2
PO2
(PSO3
)
2
where PA is the partial pressure of gas A in units of atmospheres.
Since the partial pressure of a gas in atmospheres is not the same as its concentration
in molarity, the value of Kp for a reaction is not necessarily equal to the value of Kc.
However, as long as the gases behave ideally, we can derive a relationship between the
two constants. The concentration of an ideal gas A is the number of moles of A (nA)
divided by its volume (V) in liters:
[A] = nA
V
From the ideal gas law, we can relate the quantity nA>V to the partial pressure of A
PAV = nART
PA = nA
V
RT
Since [A] = nA>V, we can write:
PA = [A]RT or [A] = PA
RT
Now consider the general equilibrium chemical equation:
a A + b B L c C + d D
Now consider the general equilibrium chemical equation:
a A + b B L c C + d D
According to the law of mass action, we write Kc:
Kc = [C]c
[D]d
[A]a
[B]b
Substituting [X] = PX>RT for each concentration term, we get:
Kc =
a
PC
RT b
c
a
PD
RT b
d
a
PA
RT b
a
a
PB
RT b
b =
Pc
CPd
Da 1
RT b
c+d
Pa
APb
Ba 1
RT b
a+b = Pc
CPd
D
Pa
APb
B
a 1
RT b
c+d-(a+b)
= Kp a 1
RT b
c+d-(a+b)
Rearranging, we arrive at:
Kp = Kc(RT)
c+d-(a+b)
Finally, if we let ∆n = c + d - (a + b), which is the sum of the stoichiometric coefficients of the gaseous products minus the sum of the stoichiometric coefficients of the
gaseous reactants, we arrive at the following general result:
Kp = Kc(RT)
∆n
Notice that if the sum of the stoichiometric coefficients for the reactants equals the sum
for the products, then ∆n = 0, and Kp is equal to Kc.
Throughout this book, we express concentrations and partial pressures within the equilibrium constant in units of molarity and atmospheres, respectively. When expressing the value
of the equilibrium constant, however, we have not included the units. Formally, the values
of concentration or partial pressure that we substitute into the equilibrium constant expression are ratios of the concentration or pressure to a reference concentration (exactly 1 M) or
a reference pressure (exactly 1 atm). For example, within the equilibrium constant expression, a pressure of 1.5 atm becomes:
1.5 atm
1 atm = 1.5
Similarly, a concentration of 1.5 M becomes:
1.5 M
1 M = 1.5
As long as concentration units are expressed in molarity for Kc and pressure units are
expressed in atmospheres for Kp, we can skip this formality and simply enter the quantities directly into the equilibrium expression, dropping their corresponding units.
ExamPlE 14.3 Relating Kp and Kc
Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere according to the equation shown here
2 NO(g) + O2(g) L 2 NO2(g)
Kp = 2.2 * 1012 at 25 °C
Find Kc for this reaction
SORT
You are given Kp for the reaction and asked to find Kc
STRATEGIZE
Use Equation 14.2 to relate Kp and Kc
SOLVE
Solve the equation for Kc.
Calculate ∆n.
Substitute the required quantities to calculate Kc. The temperature must be in kelvins. The units are dropped when reporting
Kc as described in the text.
The easiest way to check this answer is to substitute it back into Equation 14.2 and confirm that you get the original value for Kp.
Kp = Kc(RT)
∆n
= 5.4 * 1013 a0.08206
L # atm
mol # K
* 298 Kb
-1
= 2.2 * 1012
Many chemical reactions involve pure solids or pure liquids as reactants or products.
Consider the reaction:
2 CO(g) L CO2(g) + C(s)
We might expect the expression for the equilibrium constant to be:
Kc = [CO2][C]
[CO]2 (incorrect)
However, since carbon is a solid, its concentration is constant—its concentration does not
change no matter what amount is present. The concentration of a solid does not change
because a solid does not expand to fill its container. A solid's concentration, therefore,
depends only on its density, which is constant as long as some solid is present (FigurE 14.5▼).
Consequently, pure solids—those reactants or products labeled in the chemical equation
with an (s)—are not included in the equilibrium expression. The correct equilibrium expression for this reaction is therefore:
Kc = [CO2]
[CO]2 (correct)
Similarly, the concentration of a pure liquid does not change. So, pure liquids—those
reactants or products labeled in the chemical equation with an (/)—are also excluded
from the equilibrium expression. For example, what is the equilibrium expression for the
following reaction?
CO2(g) + H2O(/) L H+(aq) + HCO3
-(aq)
Since H2O(/) is pure liquid, it is omitted from the equilibrium expression. The correct equilibrium expression for this reaction is therefore:
Kc = [H+][HCO3
-]
[CO2]
ExamPlE 14.4 writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid
Write an expression for the equilibrium constant (Kc) for the chemical equation.
CaCO3(s) L CaO(s) + CO2(g)
SOLUTION
Since CaCO3(s) and CaO(s) are both solids, you omit them from the equilibrium expression.
The most direct way to obtain an experimental value for the equilibrium constant of a
reaction is to measure the concentrations of the reactants and products in a reaction mixture at equilibrium. Consider the reaction:
H2(g) + I2(g) L 2 HI(g)
Suppose a mixture of H2 and I2 is allowed to come to equilibrium at 445 °C. The measured equilibrium concentrations are [H2] = 0.11 M, [I2] = 0.11 M, and [HI] = 0.78 M.
What is the value of the equilibrium constant at this temperature? We can write the
expression Kc from the balanced equation:
Kc = [HI]2
[H2][I2]
To calculate the value of Kc, we substitute the correct equilibrium concentrations into the
expression for Kc :
Kc = [HI]2
[H2][I2]
= 0.782
(0.11)(0.11)
= 5.0 * 101
The concentrations within Kc should always be written in moles per liter (M); however, as
noted in Section 14.4, we do not normally include units when expressing the value of the
equilibrium constant, so Kc is unitless.
For any reaction, the equilibrium concentrations of the reactants and products
depend on the initial concentrations and, in general, vary from one set of initial concentrations to another. However, the equilibrium constant is always the same at a given temperature, regardless of the initial concentrations. Table 14.1 lists several different
equilibrium concentrations of H2, I2, and HI, each from a different set of initial
concentrations. Notice that the equilibrium constant is always the same, regardless of the
initial concentrations. Notice also that, whether we start with only reactants or only products, the reaction reaches equilibrium concentrations in which the equilibrium constant is
the same. In other words, no matter what the initial concentrations are, the reaction
always goes in a direction so that the equilibrium concentrations—when substituted into
the equilibrium constant expression—give the same constant, K
In each entry in Table 14.1, we calculated the equilibrium constant from values of
the equilibrium concentrations of all the reactants and products. In most cases, however,
we need only know the initial concentrations of the reactant(s) and the equilibrium concentration of any one reactant or product. We can deduce the other equilibrium concentrations from the stoichiometry of the reaction. For example, consider the simple
reaction:
A(g) L 2 B(g)
Suppose that we have a reaction mixture in which the initial concentration of A is 1.00 M
and the initial concentration of B is 0.00 M. When equilibrium is reached, the concentration
of A is 0.75 M. Since [A] has changed by -0.25 M, we can deduce (based on the stoichiometry) that [B] must have changed by 2 * (+0.25 M) or +0.50 M. We summarize the initial
conditions, the changes, and the equilibrium conditions in the following table:
This type of table is often referred to as an ICE table (I = initial, C = change,
E = equilibrium). To calculate the equilibrium constant, we use the balanced equation to
write an expression for the equilibrium constant and then substitute the equilibrium concentrations from the ICE table.
Kc = [B]2
[A] = (0.50)2
(0.75) = 0.33
In examples 14.5 and 14.6, we show the general procedure for solving these kinds of
equilibrium problems in the left column and two examples exemplifying the procedure in
the center and right columns.
When the reactants of a chemical reaction are mixed, they generally react to form
products—the reaction proceeds to the right (toward the products). The amount of products that forms when the reaction reaches equilibrium depends on the magnitude of the
equilibrium constant, as we have seen. However, what if a reaction mixture that is not at
equilibrium contains both reactants and products? Can we predict the direction of change
for such a mixture?
To gauge the progress of a reaction relative to equilibrium, we use a quantity called
the reaction quotient. The definition of the reaction quotient takes the same form as the
definition of the equilibrium constant, except that the reaction need not be at equilibrium.
For the general reaction:
a A + b B L c C + d D
we define the reaction quotient (Qc) as the ratio—at any point in the reaction—of the
concentrations of the products raised to their stoichiometric coefficients divided by the
concentrations of the reactants raised to their stoichiometric coefficients. For gases with
quantities measured in atmospheres, the reaction quotient uses the partial pressures in
place of concentrations and is called Qp.
Qc = [C]c
[D]d
[A]a
[B]b Qp = Pc
C Pd
D
Pa
A Pb
B
The difference between the reaction quotient and the equilibrium constant is that, at a
given temperature, the equilibrium constant has only one value and it specifies the relative amounts of reactants and products at equilibrium. The reaction quotient, however,
depends on the current state of the reaction and has many different values as the reaction
proceeds. For example, in a reaction mixture containing only reactants, the reaction quotient is zero (Qc = 0):
Qc = [0]c
[0]d
[A]a
[B]b = 0
In a reaction mixture containing only products, the reaction quotient is infinite (Qc = ∞):
Qc = [C]c
[D]d
[0]a
[0]b = ∞
In a reaction mixture containing both reactants and products, each at a concentration of 1
M, the reaction quotient is one (Qc = 1):
Qc = (1)c
(1)d
(1)a
(1)b = 1
The reaction quotient is useful because the value of Q relative to K is a measure of the progress of the reaction toward equilibrium. At equilibrium, the reaction quotient is equal to the
equilibrium constant. FigurE 14.6▶ on the next page is a plot of Q as a function of the concentrations of A and B for the simple reaction A(g) L B(g), which has an equilibrium constant
of K = 1.45. The following points are representative of three possible conditions:
Summarizing Direction of Change Based on Q and K:
The reaction quotient (Q) is a measure of the progress of a reaction toward equilibrium.
▶ Q 6 K Reaction goes to the right (toward products).
▶ Q 7 K Reaction goes to the left (toward reactants).
▶ Q = K Reaction is at equilibrium—the reaction does not proceed in either direction.
ExamPlE 14.7 Predicting the Direction of a Reaction by Comparing Q and K
Consider the reaction and its equilibrium constant:
2(g) + Cl2(g) L 2 ICl(g) Kp = 81.9
A reaction mixture contains PI2 = 0.114 atm, PCl2 = 0.102 atm, and PICl = 0.355 atm. Is the reaction mixture at equilibrium?
If not, in which direction will the reaction proceed?
SOLUTION
To determine the progress of the reaction relative to the equilibrium
state, calculate Q.
Compare Q to K.
In Section 14.6, we demonstrated how to calculate an equilibrium constant given the
equilibrium concentrations of the reactants and products. Just as commonly, we want to
calculate equilibrium concentrations of reactants or products given the equilibrium constant. These kinds of calculations are important because they allow us to calculate the
amount of a reactant or product at equilibrium. For example, in a synthesis reaction, we
might want to know how much of the product forms when the reaction reaches equilibrium. Or for the hemoglobin-oxygen equilibrium discussed in Section 14.1, we might
want to know the concentration of oxygenated hemoglobin present under certain oxygen
concentrations within the lungs or muscles.
We can divide these types of problems into two categories: (1) finding equilibrium
concentrations when we are given the equilibrium constant and all but one of the equilibrium concentrations of the reactants and products; and (2) finding equilibrium concentrations when we are given the equilibrium constant and only initial concentrations. The
second category of problem is more difficult than the first. We examine each separately.
Finding Equilibrium Concentrations when we are given the
Equilibrium Constant and all but One of the Equilibrium
Concentrations of the Reactants and Products
We can use the equilibrium constant to calculate the equilibrium concentration of one of
the reactants or products, given the equilibrium concentrations of the others. To solve
this type of problem, we follow our general problem-solving procedure, as shown in
Example 14.8.
ExamPlE 14.8 Finding Equilibrium Concentrations when You are given the Equilibrium Constant and
all but One of the Equilibrium Concentrations of the Reactants and Products
Consider the reaction:
2 COF2(g) L CO2(g) + CF4(g) Kc = 2.00 at 1000 °C
In an equilibrium mixture, the concentration of COF2 is 0.255 M and the concentration of CF4 is 0.118 M. What is the
equilibrium concentration of CO2?
SORT
You are given the equilibrium constant of a chemical
reaction, together with the equilibrium concentrations of the
reactant and one product. You are asked to find the equilibrium concentration of the other product.
STRATEGIZE
Calculate the concentration of the product by
using the given quantities and the expression for Kc.
SOLVE
Solve the equilibrium expression for [CO2] and then
substitute in the appropriate values to calculate it.
More commonly, we know the equilibrium constant and only initial concentrations of
reactants, and we need to find the equilibrium concentrations of the reactants or products.
These kinds of problems are generally more involved than those we just examined and
require a specific procedure to solve them. The procedure has some similarities to the
procedure demonstrated in Examples 14.5 and 14.6; we must set up an ICE table showing
the initial conditions, the changes, and the equilibrium conditions. However, unlike
Examples 14.5 and 14.6, where the change in concentration of at least one reactant or
product is known, here the changes in concentration are not known and are represented
with the variable x. For example, consider again the simple reaction:
A(g) L 2 B(g)
Suppose that, as before, we have a reaction mixture in which the initial concentration of
A is 1.0 M and the initial concentration of B is 0.00 M. However, now we know the equilibrium constant, K = 0.33, and want to find the equilibrium concentrations. We know
that since Q = 0, the reaction proceeds to the right (toward the products). We set up the
ICE table with the given initial concentrations and then represent the unknown change in
[A] in terms of the variable x:
Notice that, due to the stoichiometry of the reaction, the change in [B] must be +2x. As
before, each equilibrium concentration is the sum of the two entries above it in the ICE
table. In order to find the equilibrium concentrations of A and B, we must find the value
of the variable x. Since we know the value of the equilibrium constant, we can use the
equilibrium expression to set up an equation in which x is the only variable:
K = [B]2
[A] = (2x)
2
1.0 - x = 0.33
4x2
1.0 - x = 0.33
The above equation is a quadratic equation—it contains the variable x raised to the second power. In general, we can solve quadratic equations with the quadratic formula,
which we introduce in Example 14.10. If the quadratic equation is a perfect square, however, we can solve it by simpler means, as shown in Example 14.9. For both of these
examples, we give the general procedure in the left column and apply the procedure to
the example problems in the center and right columns. Later in this section, we will see
that quadratic equations can often be simplified by making some approximations based
on our chemical knowledge.
When the initial conditions are given in terms of partial pressures (instead of concentrations) and the equilibrium constant is given as Kp instead of Kc, use the same procedure, but substitute partial pressures for concentrations, as shown in Example 14.11.
ExamPlE 14.11 Finding Equilibrium Partial Pressures when we are given the Equilibrium Constant
and Initial Partial Pressures
Consider the reaction.
I2(g) + Cl2(g) L 2 ICl(g) Kp = 81.9 (at 25 °C)
A reaction mixture at 25 °C initially contains PI2 = 0.100 atm, PCl2 = 0.100 atm, and PICl = 0.100 atm. Find the equilibrium
partial pressures of I2, Cl2, and ICl at this temperature.
SOLUTION
Follow the procedure outlined in Examples 14.5 and 14.6 (using the pressures in place of the concentrations) to solve the
problem.
1. Using the balanced equation as a guide, prepare a table
with the known initial partial pressures of the reactants
and products.
2. Use the initial partial pressures to calculate the reaction
quotient (Q). Compare Q to K and predict the direction in
which the reaction will proceed.
3. Represent the change in the partial pressure of one of
the reactants or products with the variable x. Define the
changes in the partial pressures of the other reactants or
products in terms of x.
4. Sum each column for each reactant and product to determine the equilibrium partial pressures in terms of the initial partial pressures and the variable x.
5. Substitute the expressions for the equilibrium partial
pressures (from step 4) into the expression for the equilibrium constant. Use the given value of the equilibrium
constant to solve the expression for the variable x.
6. Substitute x into the expressions for the equilibrium partial pressures of the reactants and products (from step 4)
and calculate the partial pressures.
7. Check your answer by substituting the computed equilibrium partial pressures into the equilibrium expression. The
calculated value of K should match the given value of K.
For some equilibrium problems of the type shown in Examples 14.9, 14.10, and 14.11,
there is the possibility of using an approximation that makes solving the problem much
easier without significant loss of accuracy. For example, if the equilibrium constant is
relatively small, the reaction will not proceed very far to the right. If the initial reactant
concentration is relatively large, we can make the assumption that x is small relative to
the initial concentration of reactant. To see how this approximation works, consider again
the simple reaction A L 2 B. Suppose that, as before, we have a reaction mixture in
which the initial concentration of A is 1.0 M and the initial concentration of B is 0.0 M,
and we want to find the equilibrium concentrations. However, suppose that in this case
the equilibrium constant is much smaller, say K = 3.3 * 10-5
. The ICE table is identical to the one we set up previously:
With the exception of the value of K, we end up with the exact quadratic equation that we
had before:
K = [B]2
[A] = (2x)
2
1.0 - x = 3.3 * 10-5
With the exception of the value of K, we end up with the exact quadratic equation that we
had before:
K = [B]2
[A] = (2x)
2
1.0 - x = 3.3 * 10-5
or more simply,
4x2
1.0 - x = 3.3 * 10-5
We can multiply this quadratic equation out and solve it using the quadratic formula.
However, since we know that K is small, we know that the reaction will not proceed very
far toward products; therefore, x will also be small. If x is much smaller than 1.0, then
(1.0 - x) (the quantity in the denominator) can be approximated by (1.0).
4x2
(1.0 - x) = 3.3 * 10-5
This approximation greatly simplifies the equation, which we can then readily solve for x
as follows:
4x2
1.0 = 3.3 * 10-5
4x2 = 3.3 * 10-5
x = C
3.3 * 10-5
4 = 0.0029
We can check the validity of this approximation by comparing the calculated value
of x to the number it was subtracted from. The ratio of x to the number it is subtracted
from should be less than 0.05 (or 5%) for the approximation to be valid. In this case, we
subtracted x from 1.0, and we calculate the ratio of the value of x to 1.0 as follows:
0.0029
1.0
* 100% = 0.29%
The approximation is therefore valid. In Examples 14.12 and 14.13, we present two
nearly identical problems—the only difference is the initial concentration of the reactant.
In Example 14.12, the initial concentration of the reactant is relatively large, the equilibrium constant is small, and the x is small approximation works well. In Example 14.13,
however, the initial concentration of the reactant is much smaller, and even though the
equilibrium constant is the same as it is in Example 14.12, the x is small approximation
does not work (because the initial concentration is also small). In cases like Example
14.13, we have a couple of options to solve the problem. We can either go back and solve
the equation exactly (using the quadratic formula, for example), or we can use the method
of successive approximations. In this method, which is introduced in Example 14.13, we
essentially solve for x as if it were small, and then substitute the value obtained back into
the equation to solve for x again. This can be repeated until the calculated value of x stops
changing with each iteration, which is an indication that we have arrived at an acceptable
value for x.
Note that the x is small approximation does not imply that x is zero. If that were the
case, the reactant and product concentrations would not change from their initial values.
The x is small approximation simply means that when x is added or subtracted to another
number, it does not change that number by very much. For example, let's calculate the
value of the difference 1.0 - x when x = 3.0 * 10-4:
1.0 - x = 1.0 - 3.0 * 10-4 = 0.9997 = 1.0
Since the value of 1.0 is known only to two significant figures, subtracting the small x
does not change the value at all. This situation is similar to weighing yourself on a bathroom scale with and without a penny in your pocket. Unless your scale is unusually precise, removing the penny from your pocket does not change the reading on the scale. This
does not imply that the penny is weightless, only that its weight is small when compared
to your weight. You can neglect the weight of the penny in reading your weight with no
detectable loss in accuracy.
We know that a chemical system not in equilibrium tends to progress toward equilibrium
and that the relative concentrations of the reactants and products at equilibrium are characterized by the equilibrium constant, K. What happens when a chemical system already
at equilibrium is disturbed? Le Châtelier's principle states that the chemical system
responds to minimize the disturbance.
Le Châtelier's principle: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance.
In other words, a system at equilibrium tends to maintain equilibrium—it bounces back
when disturbed. We can disturb a system in chemical equilibrium in several different
ways, including changing the concentration of a reactant or product, changing the volume or pressure, and changing the temperature. Let's consider each of these separately.
The Effect of a Concentration Change on Equilibrium
Consider this reaction in chemical equilibrium:
N2O4(g) L 2 NO2(g)
Suppose we disturb the equilibrium by adding NO2 to the equilibrium mixture and
increasing the concentration of NO2. What happens? According to Le Châtelier's principle, the system will shift in a direction to minimize the disturbance. The reaction goes to
the left (it proceeds in the reverse direction), consuming some of the added NO2 and
bringing its concentration back down, which is shown graphically in FigurE 14.7(a)▶
The reaction shifts to the left because the value of Q changes as follows:
- Before addition of NO2 : Q = K
- Immediately after addition of NO2 : Q 7 K.
- Reaction shifts to left to reestablish equilibrium.
On the other hand, what happens if we add extra N2O4, increasing its concentration? In
this case, the reaction shifts to the right, consuming some of the added N2O4 and bringing
its concentration back down, as shown in FigurE 14.7(b)▶.
The reaction shifts to the right because the value of Q changes as follows:
- Before addition of N2O4 : Q = K.
- Immediately after addition of N2O4 : Q 6 K.
- Reaction shifts to right to reestablish equilibrium
In both cases, the system shifts in a direction that minimizes the disturbance. Conversely,
lowering the concentration of a reactant (which makes Q 7 K) causes the system to shift in
the direction of the reactants to minimize the disturbance. Lowering the concentration of a
product (which makes Q 6 K) causes the system to shift in the direction of products.
ExamPlE 14.14 The Effect of a Concentration Change on Equilibrium
Consider the following reaction at equilibrium:
CaCO3(s) L CaO(s) + CO2(g)
What is the effect of adding additional CO2 to the reaction mixture? What is the effect of adding additional CaCO3?
SOLUTION
Adding additional CO2 increases the concentration of CO2 and causes the reaction to shift to the left. Adding additional
CaCO3, however, does not increase the concentration of CaCO3 because CaCO3 is a solid and therefore has a constant concentration. Thus, adding additional CaCO3 has no effect on the position of the equilibrium. (Recall from Section 14.5 that solids are not included in the equilibrium expression.)
The Effect of a Volume (or Pressure) Change on Equilibrium
How does a system in chemical equilibrium respond to a volume change? Recall from
Chapter 5 that changing the volume of a gas (or a gas mixture) results in a change in pressure. Remember also that pressure and volume are inversely related: a decrease in volume
causes an increase in pressure, and an increase in volume causes a decrease in pressure.
So, if the volume of a reaction mixture at chemical equilibrium is changed, the pressure
changes and the system shifts in a direction to minimize that change. Consider the following reaction at equilibrium in a cylinder equipped with a moveable piston:
N2(g) + 3 H2(g) L 2 NH3(g)
What happens if we push down on the piston, lowering the volume and raising the pressure (FigurE 14.8(A)▼)? How can the chemical system change to bring the pressure back
down? Look carefully at the reaction coefficients. If the reaction shifts to the right, 4 mol
of gas particles are converted to 2 mol of gas particles. From the ideal gas law
(PV = nRT), we know that decreasing the number of moles of a gas (n) results in a lower
pressure (P). Therefore, the system shifts to the right, decreasing the number of gas molecules and bringing the pressure back down, minimizing the disturbance.
Consider the same reaction mixture at equilibrium again. What happens if, this time,
we pull up on the piston, increasing the volume (FigurE 14.8(b)▼)? The higher volume results in a lower pressure, and the system responds in such a way as to bring the pressure
back up. It does this by shifting to the left, converting 2 mol of gas particles into 4 mol of
gas particles, increasing the pressure and minimizing the disturbance.
Consider again the same reaction mixture at equilibrium. What happens if, this time,
we keep the volume the same, but increase the pressure by adding an inert gas to the
mixture? Although the overall pressure of the mixture increases, the partial pressures of
the reactants and products do not change. Consequently, there is no effect, and the reaction does not shift in either direction.
ExamPlE 14.15 The Effect of a Volume Change on Equilibrium
Consider the following reaction at chemical equilibrium:
2 KClO3(s) L 2 KCl(s) + 3 O2(g)
What is the effect of decreasing the volume of the reaction mixture? Increasing the
volume of the reaction mixture? Adding an inert gas at constant volume?
SOLUTION
The chemical equation has 3 mol of gas on the right and zero moles of gas on the
left. Decreasing the volume of the reaction mixture increases the pressure and
causes the reaction to shift to the left (toward the side with fewer moles of gas particles). Increasing the volume of the reaction mixture decreases the pressure and
causes the reaction to shift to the right (toward the side with more moles of gas particles). Adding an inert gas has no effect.
When a system at equilibrium is disturbed by a change in concentration or a change in
volume, the equilibrium shifts to counter the change, but the equilibrium constant does
not change. In other words, changes in volume or concentration generally change Q, not
K, and the system responds by shifting so that Q becomes equal to K. In contrast, a
change in temperature changes the actual value of the equilibrium constant. Nonetheless,
we can use Le Châtelier's principle to predict the effects of a temperature change. If we
increase the temperature of a reaction mixture at equilibrium, the reaction shifts in the
direction that tends to decrease the temperature and vice versa. Recall from Chapter 6
that an exothermic reaction (negative ∆H) emits heat:
Exothermic reaction: A + B L C + D + heat
We can think of heat as a product in an exothermic reaction. In an endothermic reaction
(positive ∆H), the reaction absorbs heat:
Endothermic reaction: A + B + heat L C + D
We can think of heat as a reactant in an endothermic reaction.
At constant pressure, raising the temperature of an exothermic reaction—think of
this as adding heat—is similar to adding more product, causing the reaction to shift left.
For example, the reaction of nitrogen with hydrogen to form ammonia is exothermic:
Raising the temperature of an equilibrium mixture of these three gases causes the reaction to shift left, absorbing some of the added heat and forming fewer products and more
reactants. Note that, unlike adding additional NH3 to the reaction mixture (which does
not change the value of the equilibrium constant), adding heat does change the value of
the equilibrium constant. The new equilibrium mixture has more reactants and fewer
products and therefore a smaller value of K.
Conversely, lowering the temperature causes the reaction to shift right, releasing heat
and producing more products because the value of K has increased In contrast, for an endothermic reaction, raising the temperature (adding heat) causes the
reaction to shift right, and lowering the temperature (removing heat) causes the reaction
to shift left.
ExamPlE 14.16 The Effect of a Temperature Change on Equilibrium
The following reaction is endothermic:
CaCO3(s) L CaO(s) + CO2(g)
What is the effect of increasing the temperature of the reaction mixture? Decreasing
the temperature?
SOLUTION
Since the reaction is endothermic, we can think of heat as a reactant.
Heat + CaCO3(s) L CaO(s) + CO2(g)
Raising the temperature is equivalent to adding a reactant, causing the reaction to shift
to the right. Lowering the temperature is equivalent to removing a reactant, causing the
reaction to shift to the left.
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