Contributed by:
The key highlights are:
1. The concept of equilibrium
2. A system at equilibrium
3. The equilibrium constant
4. Relationship between Kc and Kp
5. What does the value of k mean/
6. Manipulating equilibrium constants
7. Heterogeneous equilibrium
8. Equilibrium calculations
9. The reaction quotient
10. Le-Chatelier's Principle
11. Haber's Process
12. The effect of changes in temperature
1.
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and
Bruce E. Bursten
Chapter 15
Chemical Equilibrium
John D. Bookstaver
St. Charles Community College
Equilibrium
St. Peters, MO
2006, Prentice Hall
2.
The Concept of Equilibrium
play the movie
Chemical equilibrium occurs when a
reaction and its reverse reaction proceed at Equilibrium
the same rate.
3.
The Concept of Equilibrium
• As a system
approaches equilibrium,
both the forward and
reverse reactions are
occurring.
• At equilibrium, the
forward and reverse
reactions are
proceeding at the same
rate.
Equilibrium
4.
A System at Equilibrium
Once equilibrium is
achieved, the
amount of each
reactant and product
remains constant.
Equilibrium
5.
Depicting Equilibrium
In a system at equilibrium, both the
forward and reverse reactions are being
carried out; as a result, we write its
equation with a double arrow
N2O4 (g) 2 NO2 (g)
Equilibrium
6.
The
Constant
Equilibrium
7.
The Equilibrium Constant
• Forward reaction:
N2O4 (g) 2 NO2 (g)
• Rate law:
Rate = kf [N2O4]
Equilibrium
8.
The Equilibrium Constant
• Reverse reaction:
2 NO2 (g) N2O4 (g)
• Rate law:
Rate = kr [NO2]2
Equilibrium
9.
The Equilibrium Constant
• Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
• Rewriting this, it becomes
kf [NO2]2
=
kr [N2O4]
Equilibrium
10.
The Equilibrium Constant
The ratio of the rate constants is a
constant at that temperature, and the
expression becomes
kf [NO2]2
Keq = =
kr [N2O4]
Equilibrium
11.
The Equilibrium Constant
• To generalize this expression, consider
the reaction
aA + bB cC + dD
• The equilibrium expression for this
reaction would be
[C]c[D]d
Kc =
[A]a[B]b Equilibrium
12.
What Are the Equilibrium
Expressions for These Equilibria?
Equilibrium
13.
SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions
Write the equilibrium expression for Kc for the following reactions:
Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each.
Plan: Using the law of mass action, we write each expression as a quotient having the product concentration
terms in the numerator and the reactant concentration terms in the denominator. Each term is raised to the
power of its coefficient in the balanced chemical equation.
PRACTICE EXERCISE
Write the equilibrium-constant expression, Kc for
Equilibrium
14.
The Equilibrium Constant
Because pressure is proportional to
concentration for gases in a closed
system, the equilibrium expression can
also be written
(PC)c (PD)d
Kp =
(PA)a (PB)b
Equilibrium
15.
Relationship between Kc and Kp
• From the ideal gas law we know that
PV = nRT
• Rearranging it, we get
n
P= RT
V
Equilibrium
16.
Relationship between Kc and Kp
Plugging this into the expression for Kp
for each substance, the relationship
between Kc and Kp becomes
Kp = Kc (RT)n
Where
n = (moles of gaseous product) − (moles of gaseous reactant)
Equilibrium
17.
SAMPLE EXERCISE 15.2 Converting Between Kc and Kp
In the synthesis of ammonia from nitrogen and hydrogen,
K c = 9.60 at 300°C. Calculate Kp for this reaction at this temperature.
Analyze: We are given Kc for a reaction and asked to calculate Kp.
Plan: The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must
determine n by comparing the number of moles of product with the number of moles of reactants (Equation
Solve: There are two moles of gaseous products (2 NH3) and four moles of gaseous reactants (1 N2 + 3 H2).
Therefore, n = 2 – 4 = –2. (Remember that functions are always based on products minus reactants.) The
temperature, T, is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.0821 L-atm/mol-K. Using
Kc = 9.60, we therefore have
PRACTICE EXERCISE
For the equilibrium is 4.08 10–3 at 1000 K. Calculate the value for Kp.
Answer: 0.335
Equilibrium
18.
Equilibrium Can Be Reached from
Either Direction
As you can see, the ratio of [NO2]2 to [N2O4]
remains constant at this temperature no matter
what the initial concentrations of NO2 and N2O4 are.
Equilibrium
19.
Equilibrium Can Be Reached from
Either Direction
This is the data from
the last two trials from
the table on the
previous slide.
Equilibrium
20.
Equilibrium Can Be Reached from
Either Direction
It does not matter whether we start with N2
and H2 or whether we start with NH3. We will
have the same proportions of all three
substances at equilibrium.
Equilibrium
21.
What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product predominates
at equilibrium.
Equilibrium
22.
What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product predominates
at equilibrium.
• If K << 1, the reaction is
reactant-favored;
reactant predominates
at equilibrium.
Equilibrium
23.
SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant
The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen:
The value for the equilibrium constant for this reaction at 25°C is Kc = 1 10–30. Describe the feasibility of
fixing nitrogen by forming NO at 25°C.
Analyze: We are asked to comment on the utility of a reaction based on the magnitude of its equilibrium
Plan: We consider the magnitude of the equilibrium constant to determine whether this reaction is feasible for
the production of NO.
Solve: Because Kc is so small, very little NO will form at 25°C. The equilibrium lies to the left, favoring the
reactants. Consequently, this reaction is an extremely poor choice for nitrogen fixation, at least at 25°C.
PRACTICE EXERCISE
For the reaction at 298 K and Kp = 54 at 700 K. Is the formation of HI
favored more at the higher or lower temperature?
Answer: The formation of product, HI, is favored at the lower temperature because Kp is larger at the lower
Equilibrium
24.
Manipulating Equilibrium Constants
The equilibrium constant of a reaction in
the reverse reaction is the reciprocal of
the equilibrium constant of the forward
reaction.
[NO2]2
N2O4 (g) 2 NO2 (g) Kc = = 0.212 at 100C
[N2O4]
[N2O4] 1
2 NO2 (g) N2O4 (g) K c = =
[NO2]2 0.212
= 4.72 at 100C
Equilibrium
25.
Manipulating Equilibrium Constants
The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power that is equal to that
number.
[NO2]2
N2O4 (g) 2 NO2 (g) Kc = = 0.212 at 100C
[N2O4]
[NO2]4
2 N2O4 (g) 4 NO2 (g) Kc = = (0.212)2 at 100C
[N2O4]2
Equilibrium
26.
Manipulating Equilibrium Constants
The equilibrium constant for a net
reaction made up of two or more steps
is the product of the equilibrium
constants for the individual steps.
Equilibrium
27.
SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions
Given the following information,
determine the value of Kc for the reaction
Analyze: We are given two equilibrium equations and the corresponding equilibrium constants and are asked
to determine the equilibrium constant for a third equation, which is related to the first two.
Plan: We cannot simply add the first two equations to get the third. Instead, we need to determine how to
manipulate the equations in order to come up with the steps that will add to give us the desired equation.
Solve: If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant
(raising to the power 2), we get
Reversing the second equation and again making the corresponding change to its equilibrium constant (taking
the reciprocal) gives
Now we have two equations that sum to give the net equation, and we can multiply the individual Kc values to
get the desired equilibrium constant.
Equilibrium
28.
SAMPLE EXERCISE 15.5 continued
PRACTICE EXERCISE
Given that, at 700 K, Kp = 54.0 for the reaction for the
reaction determine the value of Kp for the reaction
Equilibrium
29.
Equilibrium
Equilibrium
30.
The Concentrations of Solids and
Liquids Are Essentially Constant
Both can be obtained by dividing the
density of the substance by its molar
mass—and both of these are constants
at constant temperature.
Equilibrium
31.
The Concentrations of Solids and
Liquids Are Essentially Constant
Therefore, the concentrations of solids
and liquids do not appear in the
equilibrium expression
PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)
Kc = [Pb2+] [Cl−]2
Equilibrium
32.
CaCO3 (s) CO2 (g) + CaO(s)
As long as some CaCO3 or CaO remain in the
system, the amount of CO2 above the solid
will remain the same.
Equilibrium
33.
SAMPLE EXERCISE 15.6 Writing Equilibrium-Constant Expressions
for Heterogeneous Reactions
Write the equilibrium-constant expression for Kc for each of the following reactions:
Analyze: We are given two chemical equations, both for heterogeneous equilibria, and asked to write the
corresponding equilibrium-constant expressions.
Plan: We use the law of mass action, remembering to omit any pure solids, pure liquids, and solvents from the
Solve: (a) The equilibrium-constant expression is
Because H2O appears in the reaction as a pure liquid, its concentration does not appear in the equilibrium-
constant expression.
(b) The equilibrium-constant expression is
Because SnO2 and Sn are both pure solids, their concentrations do not appear in the equilibrium-constant
PRACTICE EXERCISE
Write the following equilibrium-constant expressions:
Equilibrium
34.
SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium
Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of
attaining the equilibrium (a) pure CaCO3, (b) CaO and a CO2 pressure greater
than the value of Kp, (c) some CaCO3 and a CO2 pressure greater than the value of Kp, (d) CaCO3 and CaO?
Analyze: We are asked which of several combinations of species can establish an equilibrium between
calcium carbonate and its decomposition products, calcium oxide and carbon dioxide.
Plan: In order for equilibrium to be achieved, it must be possible for both the forward process and the reverse
process to occur. In order for the forward process to occur, there must be some calcium carbonate present. In
order for the reverse process to occur, there must be both calcium oxide and carbon dioxide. In both cases,
either the necessary compounds may be present initially, or they may be formed by reaction of the other
Solve: Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present.
(a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equilibrium pressure of CO2 is attained.
There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium. (b) CO2 continues to
combine with CaO until the partial pressure of the CO2 decreases to the equilibrium value. (c) There is no CaO
present, so equilibrium can’t be attained because there is no way the CO2 pressure can decrease to its
equilibrium value (which would require some of the CO2 to react with CaO). (d) The situation is essentially the
same as in (a): CaCO3 decomposes until equilibrium is attained. The presence of CaO initially makes no
PRACTICE EXERCISE
When added to Fe3O4(s) in a closed container, which one of the following substances—H 2(g), H2O(g), O2(g)
—will allow equilibrium to be established in the reaction Equilibrium
Answers: only H2(g)
36.
Equilibrium Calculations
A closed system initially containing
1.000 x 10−3 M H2 and 2.000 x 10−3 M I2
At 448C is allowed to reach equilibrium. Analysis
of the equilibrium mixture shows that the
concentration of HI is 1.87 x 10−3 M. Calculate Kc at
448C for the reaction taking place, which is
H2 (g) + I2 (g) 2 HI (g)
Equilibrium
37.
What Do We Know?
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
At 1.87 x 10-3
Equilibrium
38.
[HI] Increases by 1.87 x 10-3 M
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change +1.87 x 10-3
At 1.87 x 10-3
Equilibrium
39.
Stoichiometry tells us [H2] and [I2]
decrease by half as much
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At 1.87 x 10-3
Equilibrium
40.
We can now calculate the equilibrium
concentrations of all three compounds…
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
Equilibrium
41.
…and, therefore, the equilibrium constant
[HI]2
Kc =
[H2] [I2]
(1.87 x 10-3)2
=
(6.5 x 10-5)(1.065 x 10-3)
= 51
Equilibrium
42.
SAMPLE EXERCISE 15.8 Calculating K When All Equilibrium Concentrations Are Known
A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472°C. The
equilibrium mixture of gases was analyzed and found to contain 7.38 atm H 2 , 2.46 atm N2 , and 0.166 atm NH3.
From these data, calculate the equilibrium constant Kp for the reaction
Analyze: We are given a balanced equation and equilibrium partial pressures and are asked to calculate the
value of the equilibrium constant.
Plan: Using the balanced equation, we write the equilibrium-constant expression. We then substitute the
equilibrium partial pressures into the expression and solve for Kp.
PRACTICE EXERCISE
An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25°C: [HC 2H3O2] =
1.65 10–2 M; [H+] = 5.44 10–4 M; and [C2H3O2–] = 5.44 10–4 M. Calculate the equilibrium constant Kc for the
ionization of acetic acid at 25°C. The reaction is
Answer: 1.79 10–5 Equilibrium
43.
SAMPLE EXERCISE 15.9 Calculating K from Initial and Equilibrium Concentrations
A closed system initially containing 1.000 10–3 M H2 and 2.000T10–3 M I2 at 448°C is allowed to reach
equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 10–3 M. Calculate
Kc at 448°C for the reaction taking place, which is
Analyze: We are given the initial concentrations of H2 and I2 and the equilibrium concentration of HI. We
are asked to calculate the equilibrium constant Kc for the reaction
Plan: We construct a table to find equilibrium concentrations of all species and then use the equilibrium
concentrations to calculate the equilibrium constant.
Solve: First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also
provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the
chemical equation as the heading for the table.
Equilibrium
44.
SAMPLE EXERCISE 15.9 continued
Second, we calculate the change in concentration of HI, which is the difference between the equilibrium values
and the initial values:
Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H 2] and
Fourth, we calculate the equilibrium concentrations of H2 and I2 , using the initial concentrations and the
changes. The equilibrium concentration equals the initial concentration minus that consumed:
The completed table now looks like this (with equilibrium concentrations in blue for emphasis):
Equilibrium
45.
SAMPLE EXERCISE 15.9 continued
Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is
Finally, now that we know the equilibrium concentration of each reactant and product, we can use the
equilibrium-constant expression to calculate the equilibrium constant.
Comment: The same method can be applied to gaseous equilibrium problems, in which case partial
pressures are used as table entries in place of molar concentrations.
PRACTICE EXERCISE
Sulfur trioxide decomposes at high temperature in a sealed container:
Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3
partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.
Answer: 0.338
Equilibrium
46.
The Reaction Quotient (Q)
• To calculate Q, one substitutes the
initial concentrations on reactants and
products into the equilibrium
expression.
• Q gives the same ratio the equilibrium
expression gives, but for a system that
is not at equilibrium.
Equilibrium
47.
If Q = K,
the system is at equilibrium.
Equilibrium
48.
If Q > K,
there is too much product and the
equilibrium shifts to the left.
Equilibrium
49.
If Q < K,
there is too much reactant, and the
equilibrium shifts to the right.
Equilibrium
50.
SAMPLE EXERCISE 15.10 Predicting the Direction of Approach to Equilibrium
At 448°C the equilibrium constant Kc for the reaction
is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448°C if we start with 2.0
10–2 mol of HI, 1.0 10–2 mol of H2, and 3.0 10–2 mol of I2 in a 2.00-L container.
Analyze: We are given a volume and initial molar amounts of the species in a reaction and asked to
determine in which direction the reaction must proceed to achieve equilibrium.
Plan: We can determine the starting concentration of each species in the reaction mixture. We can then
substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient,
Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us
in which direction the reaction will proceed.
Solve: The initial concentrations are
The reaction quotient is therefore
Equilibrium
51.
SAMPLE EXERCISE 15.10 continued
Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to
reach equilibrium; the reaction will proceed from left to right as it moves toward equilibrium.
PRACTICE EXERCISE
At 1000 K the value of Kp for the reaction is 0.338. Calculate the value for
Qp , and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures
Answer: Qp = 16; Qp > Kp, and so the reaction will proceed from right to left, forming more SO3.
Equilibrium
52.
SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations
For the Haber process, at 500°C. In an equilibrium
mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is
the partial pressure of NH3 in this equilibrium mixture?
Analyze: We are given an equilibrium constant, Kp , and the equilibrium partial pressures of two of the three
substances in the equation (N2 and H2), and we are asked to calculate the equilibrium partial pressure for the
third substance (NH3).
Plan: We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that
are known. Then we can solve for the only unknown in the equation.
Solve: We tabulate the equilibrium pressures as follows:
Because we do not know the equilibrium pressure of NH3 , we represent it with a variable, x. At equilibrium the
pressures must satisfy the equilibrium-constant expression:
We now rearrange the equation to solve for x:
Equilibrium
53.
SAMPLE EXERCISE 15.11 continued
Comment: We can always check our answer by using it to recalculate the value of the equilibrium constant:
PRACTICE EXERCISE
At 500 K the reaction has Kp = 0.497. In an equilibrium mixture at 500 K, the
partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the
equilibrium mixture?
Answer: 1.22 atm
Equilibrium
54.
SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrations
from Initial Concentrations
A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium
constant Kc for the reaction
at 448°C is 50.5. What are the equilibrium concentrations of H2 , I2 , and HI in moles per liter?
Analyze: We are given the volume of a container, an equilibrium constant, and starting amounts of reactants
in the container and are asked to calculate the equilibrium concentrations of all species.
Plan: In this case we are not given any of the equilibrium concentrations. We must develop some relationships
that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that
outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial concentrations.
Solve: First, we note the initial concentrations of H2 and I2 in the 1.000-L flask:
Second, we construct a table in which we tabulate the initial concentrations:
Equilibrium
55.
SAMPLE EXERCISE 15.12 continued
Third, we use the stoichiometry of the reaction to determine the changes in concentration that occur as the
reaction proceeds to equilibrium. The concentrations of H2 and I2 will decrease as equilibrium is established,
and that of HI will increase. Let’s represent the change in concentration of H2 by the variable x. The balanced
chemical equation tells us the relationship between the changes in the concentrations of the three gases:
Fourth, we use the initial concentrations and the changes in concentrations, as dictated by stoichiometry, to
express the equilibrium concentrations. With all our entries, our table now looks like this:
Fifth, we substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the
single unknown, x:
Equilibrium
56.
SAMPLE EXERCISE 15.12 continued
If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this
expression to obtain a quadratic equation in x:
Solving the quadratic equation (Appendix A.3) leads to two solutions for x:
When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative
concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this
solution. We then use x = 0.935 to find the equilibrium concentrations:
Check: We can check our solution by putting these numbers into the equilibrium-constant expression:
Equilibrium
57.
SAMPLE EXERCISE 15.12 continued
Comment: Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions will
not be chemically meaningful and should be rejected.
PRACTICE EXERCISE
For the equilibrium the equilibrium constant Kp has the value 0.497 at 500 K.
A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium
pressures of PCl5 , PCl3 , and Cl2 at this temperature?
Equilibrium
58.
Le Châtelier’s
Principle
Equilibrium
59.
Le Châtelier’s Principle
“If a system at equilibrium is disturbed by
a change in temperature, pressure, or the
concentration of one of the components,
the system will shift its equilibrium
position so as to counteract the effect of
the disturbance.”
Equilibrium
60.
What Happens When More of a
Reactant Is Added to a System?
Play insert CD
Equilibrium
61.
The Haber Process
The transformation of nitrogen and hydrogen into
ammonia (NH3) is of tremendous significance in
agriculture, where ammonia-based fertilizers are of
utmost importance.
Equilibrium
62.
The Haber Process
If H2 is added to the
system, N2 will be
consumed and the
two reagents will
form more NH3.
Equilibrium
63.
The Haber Process
This apparatus
helps push the
equilibrium to the
right by removing
the ammonia (NH3)
from the system as
a liquid.
Equilibrium
64.
The Effect of Changes in Pressure
Equilibrium
65.
The Effect of Changes in
Temperature
Co(H2O)62+(aq) + 4 Cl(aq) CoCl4 (aq) + 6 H2O (l)
Equilibrium
66.
The Effect of Changes in
Temperature
Play insert CD
Equilibrium
67.
Catalysts increase the rate of both the
forward and reverse reactions.
Equilibrium
68.
Equilibrium is achieved faster, but the
equilibrium composition remains unaltered.
Equilibrium
69.
SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium
Consider the equilibrium
In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure
is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?
Analyze: We are given a series of changes to be made to a system at equilibrium and are asked to predict
what effect each change will have on the position of the equilibrium.
Plan: Le Châtelier’s principle can be used to determine the effects of each of these changes.
Solve: (a) The system will adjust to decrease the concentration of the added N2O4, so the equilibrium shifts to
the right, in the direction of products.
(b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the
equilibrium shifts to the right.
(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial
pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the position of the equilibrium.
(d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas
molecules); thus, the equilibrium shifts to the right. (This is the opposite of the effect observed in Figure 15.13,
where the volume was decreased.)
(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation.
Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium
shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of the
equilibrium constant, K. Equilibrium
70.
SAMPLE EXERCISE 15.13 continued
PRACTICE EXERCISE
For the reaction
in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the temperature is decreased, (c)
the volume of the reaction system is increased, (d) PCl3(g) is added?
Answers: (a) right, (b) left, (c) right, (d) left
Equilibrium
71.
SAMPLE EXERCISE 15.14 Predicting the Effect of Temperature on K
(a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the
(b) Determine how the equilibrium constant for this reaction should change with temperature.
Analyze: We are asked to determine the standard enthalpy change of a reaction and how the equilibrium
constant for the reaction varies with temperature.
Plan: (a) We can use standard enthalpies of formation to calculate H° for the reaction. (b) We can then use
Le Châtelier’s principle to determine what effect temperature will have on the equilibrium constant.
Solve: (a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar
enthalpies of formation of the products, each multiplied by its coefficient in the balanced chemical equation,
less the same quantities for the reactants. At 25°C, for NH3(g) is The values for
H2(g) and N2(g) are zero by definition because the enthalpies of formation of the elements in their normal states
at 25°C are defined as zero (Section 5.7). Because 2 mol of NH3 is formed, the total enthalpy change is
Equilibrium
72.
SAMPLE EXERCISE 15.14 continued
(b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction.
An increase in temperature causes the reaction to shift in the direction of less NH 3 and more N2 and H2. This
effect is seen in the values for Kp presented in Table 15.2. Notice that Kp changes markedly with changes in
temperature and that it is larger at lower temperatures.
Comment: The fact that Kp for the formation of NH3 from N2 and H2 decreases with increasing temperature
is a matter of great practical importance. To form NH3 at a reasonable rate requires higher temperatures. At
higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to NH 3 is
smaller. To compensate for this, higher pressures are needed because high pressure favors NH 3 formation.
Equilibrium
73.
SAMPLE EXERCISE 15.14 continued
PRACTICE EXERCISE
Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction
Use this result to determine how the equilibrium constant for the reaction should change with temperature.
the equilibrium constant will increase with increasing temperature
Equilibrium
74.
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
At temperatures near 800°C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form
CO and H2:
The mixture of gases that results is an important industrial fuel called water gas. (a) At 800°C the equilibrium
constant for this reaction is Kp = 14.1. What are the equilibrium partial pressures of H2O, CO, and H2 in the
equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H 2O in a 1.00-L vessel?
(b) What is the minimum amount of carbon required to achieve equilibrium under these conditions? (c) What is
the total pressure in the vessel at equilibrium? (d) At 25°C the value of Kp for this reaction is 1.7 10–21. Is the
reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H2 at equilibrium, should
the pressure of the system be increased or decreased?
Solution (a) To determine the equilibrium partial pressures, we use the ideal gas equation, first determining
the starting partial pressure of hydrogen.
We then construct a table of starting partial pressures and their changes as equilibrium is achieved:
Equilibrium
75.
SAMPLE INTEGRATIVE EXERCISE continued
There are no entries in the table under C(s) because the reactant, being a solid, does not appear in the
equilibrium-constant expression. Substituting the equilibrium partial pressures of the other species into the
equilibrium-constant expression for the reaction gives
Multiplying through by the denominator gives a quadratic equation in x:
Solving this equation for x using the quadratic formula yields x = 6.14 atm. Hence, the equilibrium partial
pressures are and
(b) Part (a) shows that x = 6.14atm of H2O must react in order for the system to achieve equilibrium. We can
use the ideal-gas equation to convert this partial pressure into a mole amount.
Thus, 0.0697 mol of H2O and the same amount of C must react to achieve equilibrium. As a result, there
must be at least 0.0697 mol of C (0.836 g C) present among the reactants at the start of the reaction.
(c) The total pressure in the vessel at equilibrium is simply the sum of the equilibrium partial pressures:
Equilibrium
76.
SAMPLE INTEGRATIVE EXERCISE continued
(d) In discussing Le Châtelier’s principle, we saw that endothermic reactions exhibit an increase in Kp with
increasing temperature. Because the equilibrium constant for this reaction increases as temperature increases,
the reaction must be endothermic. From the enthalpies of formation given in Appendix C, we can verify our
prediction by calculating the enthalpy change for the reaction,
The positive sign for H° indicates that the reaction is endothermic.
(e) According to Le Châtelier’s principle, a decrease in the pressure causes a gaseous equilibrium to shift
toward the side of the equation with the greater number of moles of gas. In this case there are two moles of gas
on the product side and only one on the reactant side. Therefore, the pressure should be reduced to maximize
the yield of the CO and H2.
Equilibrium
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