CHAPTER 14
Chemical Equilibrium
CHAPTER TERMS AND DEFINITIONS
Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred
terms are italicized in the text. Where a term does not fall directly under a text section heading,
additional information is given for you to locate it.
reversible* describes chemical reactions in which products formed can themselves react, giving
back the original reactants (chapter introduction)
catalytic methanation* conversion of carbon monoxide and hydrogen to methane and water in the
presence of a catalyst (chapter introduction)
steam re-forming*
preparing carbon monoxide and hydrogen by reacting hydrocarbons with steam
(chapter introduction)
dynamic equilibrium* state in which the reactants and products of a reversible reaction or process
are being formed at the same rate, such that there is no apparent change in the system (14.1)
chemical equilibrium state reached by a reaction mixture when the rates of forward and reverse
reactions have become equal so that net change no longer occurs (14.1)
equilibrium constant* quantity relating equilibrium compositions for a particular reaction at a
given temperature (14.2)
equilibrium-constant expression arrangement of symbols showing multiplication of the
concentrations of reaction products and division by the concentrations of reactants, the concentration of
each raised to a power equal to its coefficient in the chemical equation (14.2)
equilibrium constant (Kc) value obtained for the equilibrium-constant expression when equilibrium
concentrations are substituted (14.2)
equilibrium constant (Kp)
pressures (14.2)
equilibrium constant for a gaseous reaction expressed in terms of partial
law of mass action relation stating that the values of the equilibrium-constant expression Kc are
constant for a particular reaction at a given temperature whatever equilibrium concentrations are
substituted (14.2)
activities* dimensionless quantities defining the equilibrium constant; for an ideal mixture, the
activity of a substance is the ratio of its concentration (or partial pressure if a gas) to a standard
concentration of 1 M (or partial pressure of 1 atm) so that units cancel (14.2, marginal note)
homogeneous equilibrium
heterogeneous equilibrium
(14.3)
equilibrium that involves reactants and products in a single phase (14.3)
equilibrium that involves reactants and products in more than one phase
oscillating reaction* a reaction that cycles back and forth over time (A Chemist Looks at: Slime
Molds and Leopard’s Spots)
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Chapter 14: Chemical Equilibrium
313
reaction quotient (Qc) expression identical to the equilibrium-constant expression but with
concentrations not necessarily those at equilibrium (14.5)
quadratic formula*
solutions to a quadratic equation of the form ax2 + bx + c = 0;
x =   b  b 2 - 4ac  / 2a (14.6, marginal note)


Le Châtelier’s principle when a system in chemical equilibrium is disturbed by a change of
temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that
tends to counteract this change of variable (14.7)
contact process*
dioxide (14.9)
industrial method of preparing sulfuric acid by the catalytic oxidation of sulfur
catalyst* substance that speeds up the attainment of equilibrium, is not consumed by the reaction,
and has no effect on the equilibrium composition of the reaction mixture (14.9)
acid rain* rain with increased acidity owing to the presence of sulfuric and nitric acids (14.9,
marginal note)
Ostwald process*
(14.9)
industrial method of preparing nitric acid by the catalytic oxidation of ammonia
CHAPTER DIAGNOSTIC TEST
1.
Write equilibrium-constant expressions for the following equilibria in terms of Kc.
a.
2HCl(g) +½O2(g)
b.
2NO(g) + Br2(g)
2.
2NOBr(g)
+
c.
Ag (aq) + 2NH3(aq)
d.
HCN(aq) + H2O(l)
e.
4NH3(g) + 3O2(g)
f.
H2O(g) + Cl2(g)
–
I3 (aq) + H2O(l)
Ag(NH3)2+(aq)
H3O+(aq) + CN–(aq)
2N2(g) + 6H2O(g)
HOI(aq) + 2I–(aq) + H+(aq)
The equilibrium concentrations for the decomposition of PCl5(g) at 433 K,
PCl5(g)
PCl3(g) + Cl2(g)
are [PCl5] = 0.865 mol/L, [PCl3] = [Cl2] = 0.135 mol/L. Calculate Kc.
3.
What effect would an increase in pressure have on the equilibrium of the system in Problem 2?
4.
A system containing nitrogen, hydrogen, and ammonia is allowed to come to equilibrium. The
total equilibrium pressure is 5 atm. The partial pressures are PN2 = 1 atm, PH 2 = 2 atm, and PNH3 =
2 atm. Calculate Kp for the reaction
N2(g) + 3H2(g)
5.
2NH3(g)
Using the data from Problem 4, calculate Kp for
1
3
N2(g) + H2(g)
2
2
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NH3(g)
314
6.
Chapter 14: Chemical Equilibrium
If 1.00 mol CO2 and 1.00 mol H2 are placed in a 1.00-L flask at 825 K and react according to
CO2(g) + H2(g)
CO(g) + H2O(g)
analysis of the equilibrium mixture shows 0.27 mol CO present. Determine Kc at this temperature.
7.
Consider the following equilibrium:
2H2S(g)
2H2(g) + S2(g)
For a 5.00-L vessel containing the following amounts of gases, determine whether the initial
concentrations of these gases will remain fixed or change (and if they change, indicate which
gases will show an increase in concentration):
0.0131 mol H2, 0.00650 mol S2, 0.0383 mol H2S. Kc equals 2.3  104.
8.
Consider the following reaction:
4HCl(g) + O2(g)
2H2O(g) + 2Cl2(g)
ΔH = +28 kcal
Describe what happens to the composition of the equilibrium mixture and to the equilibrium
constant K with each of the following changes to the system at equilibrium.
9.
a.
Addition of oxygen gas
b.
An increase in temperature
c.
Reduction of the volume of the reaction container
d.
Addition of a catalyst
e.
Removal of HCl(g) from the reaction vessel
I2 vapor is a deep purple color. The dissociation of HI(g) into H2(g) and I2(g) in a closed vessel
can be followed qualitatively by observing changes in the relative intensity of the purple color of
I2 vapor. When H2 gas is added to the reaction at equilibrium, the vapor slowly takes on a less
intense purple color. Explain this observation in terms of Le Châtelier’s principle.
10. Consider the reaction
2SiO(g)
2Si(l) + O2(g)
Kc = 9.62  101
If 1.00 mol SiO is placed into a 1.00-L container, what are the equilibrium concentrations of SiO
and O2?
11. Once the equilibrium in Problem 10 is reached, how would adding 10.0 g of Si affect that
equilibrium?
12. Consider the reaction
CO2(g) + H2(g)
CO(g) + H2O(g); Kc = 0.137
If 5.0 mol each of CO2 and H2 are placed in a 10.0-L flask, what are the equilibrium
concentrations?
13. Consider the following equilibrium:
2HI(g)
H2(g) + I2(g)
At equilibrium, a 2.00-L vessel contains 1.25 mol I2, 1.25 mol H2, and an unknown amount of HI.
Kc for this equilibrium is 0.0183. Calculate the equilibrium concentration of HI.
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Chapter 14: Chemical Equilibrium
315
ANSWERS TO CHAPTER DIAGNOSTIC TEST
If you missed an answer, study the text section and problem-solving skill given in parentheses after the
answer.
1.
a.
[H 2 O][Cl 2 ]
[HCl] 2 [O 2 ]1/2
(Note that [H2O] is included because the reaction is in the gas phase and [H2O] is not a
constant.)
b.
c.
d.
e.
f.
[NOBr] 2
[NO] 2 [Br 2 ]
[Ag(NH 3 ) 2 ]
[Ag  ][NH 3 ] 2
[H 3 O  ][CN]
[HCN]
[N 2 ] 2 [H 2 O] 6
[NH 3 ] 4 [O 2 ] 3
[HOI][I - ] 2 [H  ]
[I -3 ]
(14.2 , 14.3, PS Sk. 2)
2.
Kc = 0.0211 (14.2, PS Sk. 3)
3.
An increase in pressure would cause a shift to the side of the reaction that has the smaller number
of moles of gaseous materials, which would reduce the increased pressure. Therefore, the rate of
the reverse reaction would increase, and we say that we would observe a shift to the left. (14.8, PS
Sk. 7)
4.
Kp = 0.5 (14.2, PS Sk. 2, 3)
5.
Kp = 0.7 (14.2, PS Sk. 2, 3)
6.
Kc = 0.14 (14.1, 14.2, PS Sk. 1, 3)
7.
Qc = 1.52  104, which is < Kc. Therefore, initial concentrations will change, and the
concentrations of H2 and S2 gases will increase as the equilibrium shifts to the right. (14.4, 14.5,
PS Sk. 4)
8.
a.
Shift right to consume the added oxygen; no change in K. (14.7, PS Sk. 7)
b.
Favors endothermic reaction, so it shifts right; K increases. (14.8, PS Sk. 7)
c.
This increases the pressure, so reaction shifts right, toward fewer moles of gas; no change in
K. (14.8, PS Sk. 7)
d.
No effect on the equilibrium composition; no change in K. (14.9, PS Sk. 7)
e.
Shift left to replace the HCl lost; no change in K. (14.7, 14.8, PS Sk. 7)
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316
9.
Chapter 14: Chemical Equilibrium
The equilibrium may be expressed
2HI(g)
H2(g) + I2(g)
When H2 is added to the reaction at equilibrium, the rate of the reverse reaction increases. We say
that the equilibrium shifts to decrease the concentration of H2 and thus to form more HI. I2 vapor
is consumed in this equilibrium shift to the left. A loss of I2 vapor results in a less intense purple
color. (14.5, 14.7, PS Sk. 7)
10. [SiO] = 0.51 mol/L, [O2] = 0.247 mol/L (14.6, PS Sk. 6)
11. It would not. The Si formed is a pure liquid with fixed density and thus fixed concentration.
Adding more of it will not change the Si concentration. This constant value is, in effect, included
in the value of Kc for the reaction and is not part of the equilibrium-constant expression. (14.3,
14.7, PS Sk. 7)
12. [CO2] = [H2] = 0.37 mol/L
[CO] = [H2O] = 0.14 mol/L (14.6, PS Sk. 6)
13. [HI] = 4.62 mol/L (14.6, PS Sk. 5)
SUMMARY OF CHAPTER TOPICS
Chapter 14 is the first of four chapters devoted to the study of chemical equilibrium. Most of the
problems in these chapters, although involving different species, are worked in essentially the same
manner. The key to performing these calculations is to set up the table shown in Example 14.1 in the
text and used in the solutions in this study guide. You may balk at using this table format because it
takes extra time to set it up. Take this extra time. It is extremely worthwhile.
In preparing the table, there usually will be an unknown quantity, which we designatex. For each
problem, be sure to define, and write down, what you are letting x be. This is another crucial step to
ensure that you work the problem correctly.
Pay close attention to the coefficients in the reaction when working with unknowns. If two molecules of
a product substance are formed when one molecule reacts, then be sure your amount of substance that
reacts is x and the amount of product formed is 2x.
Make sure that you then have the correct form of the equilibrium-constant expression. Many students
forget exponents. Always go back and check that they are there if necessary. And remember that the
form of the expression is always “products over reactants.”
We will not follow our typical “Wanted, Given, etc.,” format too closely in solving the exercises in
these chapters because the table gives an equally useful structure to problem solving. You should be
quite familiar with the necessary steps of problem solving by now.
14.1 Chemical Equilibrium—A Dynamic Equilibrium
Learning Objectives

Define dynamic equilibrium and chemical equilibrium.

Apply stoichiometry to an equilibrium mixture. (Example 14.1)
Problem-Solving Skill
1.
Applying stoichiometry to an equilibrium mixture. Given the starting amounts of reactants and
the amount of one substance at equilibrium, find the equilibrium composition (Example 14.1).
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Chapter 14: Chemical Equilibrium
317
Exercise 14.1
Synthesis gas (a mixture of CO and H2) is increased in concentration of hydrogen by passing it with
steam over a catalyst. This is the so-called water–gas shift reaction. Some of the CO is converted to
CO2, which can be removed:
CO(g) + H2O(g)
CO2(g) + H2(g)
Suppose that you start with a gaseous mixture containing 1.00 mol CO and 1.00 mol H2O. When
equilibrium is reached at 1000C, the mixture contains 0.43 mol H2. What is the molar composition of
the equilibrium mixture?
Solution: Set up the table under the equation; let x = moles CO that react.
Amounts (mol): CO(g) + H2O(g)
Starting
Change
Equilibrium
1.00
x
1.00  x
CO2(g) + H2(g)
1.00
x
1.00  x
0
+x
x
0
+x
0.43
For every mole of CO that reacts, 1 mol of H2 is produced. Therefore, x = 0.43 mol.
Equilibrium amount of CO = 1.00 – 0.43 = 0.57 mol CO
Equilibrium amount of H2O = 1.00 – 0.43 = 0.57 mol H2O
Equilibrium amount of CO2 = x = 0.43 mol
Equilibrium amount of H2 = 0.43 mol (as given)
14.2 The Equilibrium Constant
Learning Objectives

Define equilibrium-constant expression and equilibrium constant.

State the law of mass action.

Write equilibrium-constant expressions. (Example 14.2)

Describe the kinetics argument for the approach to chemical equilibrium.

Obtain an equilibrium constant from reaction composition. (Example 14.3)

Describe the equilibrium constant Kp; indicate how Kp and Kc are related.

Obtain Kc for a reaction that can be written as a sum of other reactions of known Kc values.
Problem-Solving Skills
2.
Writing equilibrium-constant expressions. Given the chemical equation, write the equilibriumconstant expression (Example 14.2). (See also Section 14.3)
3.
Obtaining an equilibrium constant from reaction composition. Given the equilibrium
composition, find Kc (Example 14.3).
The equilibrium constant for a given reaction and equation is constant for that equation as long as the
temperature remains unchanged. No matter what the starting mixture or how the equilibrium system is
perturbed, the value of this constant will be the same as long as the temperature is not changed. This
concept is called the law of mass action.
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318
Chapter 14: Chemical Equilibrium
Exercise 14.2
a.
Write the equilibrium-constant expression Kc for the equation
2NO2(g) + 7H2(g)
b.
2NH3(g) + 4H2O(g)
Write the equilibrium-constant expression Kc when this reaction is written
NO2(g) +
7
H2(g)
2
NH3(g) + 2H2O(g)
Known: The equilibrium-constant expression is equal to product concentrations over reactant
concentrations, each to the coefficient power.
Solution: For (a):
Kc =
[NH 3 ] 2 [H 2 O] 4
[NO 2 ] 2 [H 2 ] 7
For (b):
Kc =
[NH 3 ][H 2 O] 2
[NO 2 ][H 2 ] 7/2
Exercise 14.3
When 1.00 mol each of carbon monoxide and water reach equilibrium at 1000C in a 10.0-L vessel, the
equilibrium mixture contains 0.57molCO, 0.57 mol H2O, 0.43 mol CO2, and 0.43 mol H2. Write the
chemical equation for the equilibrium. What is the value of Kc?
Solution: The equation is
CO(g) + H2O(g)
CO2(g) + H2(g)
The equilibrium-constant expression is
Kc =
[CO 2 ][H 2 ]
[CO][H 2 O]
Each concentration must be calculated at equilibrium and then placed in this expression:
[CO] = [H2O] =
[CO2] = [H2] =
Kc =
0.57 mol
= 0.057 M
10.0 L
0.43 mol
= 0.043 M
10.0 L
(0.043 )( 0.043 )
= 0.57
(0.057 )( 0.057 )
In this case, the units cancel. It is conventional, however, that even when they do not cancel, we
do not write units for an equilibrium constant. The reason for this is discussed in text Section 18.6.
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Chapter 14: Chemical Equilibrium
319
Exercise 14.4
Hydrogen sulfide, a colorless gas with a foul odor, dissociates on heating:
2H2S(g)
2H2(g) + S2(g)
When 0.100 mol H2S was put into a 10.0-L vessel and heated to 1132C, it gave an equilibrium
mixture containing 0.0285 mol H2. What is the value of Kc at this temperature?
Solution: First, change amounts to concentrations:
Starting concentration of H2S =
0.100 mol
= 0.0100 M
10.0 L
Equilibrium concentration of H2 =
0.0285 mol
= 0.00285 M
10.0 L
Second, set up a table under the equation, letting x = mol/L of H2S that react.
Concentration (M) 2H2S(g)
2H2(g) + S2(g)
Starting
0.0100
0
0
Change
+2x
+x
2x
Equilibrium 0.0100  2x 0.00285 = 2x
x
Third, calculate equilibrium concentrations from the bottom line in the table:
[H2S] = (0.0100 – 0.00285) M = 0.00715 M
[H2] = 0.00285 M (as given)
[S2] =
0.00285 M
= 0.00143 M
2
Finally, write the equilibrium-constant expression from the equation and substitute in the
calculated molarities:
Kc =
[H 2 ] 2 [S 2 ]
[H 2 S]
2
=
[0.00285 ] 2 [0.00143 ]
[0.00715]
2
=
[1.153  10 8 ]
[5.11  10
5
= 2.3  104
]
Exercise 14.5
Phosphorus pentachloride dissociates on heating:
PCl5(g)
PCl3(g) + Cl2(g)
If Kc equals 3.26  102 at 191C, what is Kp at this temperature?
Known: Kp = Kc(RT) n, where n is the sum of gaseous product coefficients minus gaseous
reactant coefficients, and R = 0.0821(L ∙ atm)/(K ∙ mol).
Solution:
n = 2 – 1 = 1
T = 191 + 273 = 464 K
Kp = Kc(RT) = 3.26  102  0.0821  464 = 1.24
Note that we do not use units because Kc had no units. However, R must be expressed as (L ∙
atm)/(K ∙ mol) and T in kelvins.
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320
Chapter 14: Chemical Equilibrium
14.3 Heterogeneous Equilibria; Solvents in Homogeneous
Equilibria
Learning Objectives

Define homogeneous equilibrium and heterogeneous equilibrium.

Write Kc for a reaction with pure solids or liquids. (Example 14.4)
Problem-Solving Skill
2.
Writing equilibrium-constant expressions. Given the chemical equation, write the equilibriumconstant expression (Example 14.4).*
In working problems with gas reactions that may include heterogeneous equilibria, pay careful attention
to the states of the substances in the equations. Only the gases are included in an equilibrium-constant
expression.
Exercise 14.6
The Mond process for purifying nickel involves the formation of nickel tetracarbonyl, Ni(CO) 4, a
volatile liquid, from nickel metal and carbon monoxide. Carbon monoxide is passed over impure
nickel to form nickel carbonyl vapor, which, when heated, decomposes and deposits pure nickel.
Ni(s) + 4CO(g)
Ni(CO)4(g)
Write the expression for Kc for this reaction.
Solution: Kc =
[Ni(CO) 4 ]
[CO] 4
*Note that Problem-Solving Skill 2 is used again in this section. Here, as well as in later chapters,
problem-solving skills are repeated as needed.
14.4 Qualitatively Interpreting the Equilibrium Constant
Learning Objective

Give a qualitative interpretation of the equilibrium constant based on its value.
Exercise 14.7
The equilibrium constant Kc for the reaction
2NO(g) + O2(g)
2NO2(g)
equals 4.0  1013 at 25C. Does the equilibrium mixture contain predominantly reactants or products? If
[NO] = [O2] = 2.0  10–6 M at equilibrium, what is the equilibrium concentration of NO2?
Solution: Since Kc is large, the equilibrium mixture contains mostly products. Determine [NO2]
at equilibrium by substituting values in the equilibrium-constant expression and solving for
[NO2]:
Kc =
[NO 2 ] 2
2
[NO] [O 2 ]
=
[NO 2 ] 2
(2.0  10
6 2
) (2.0  10
6
)
= 4.0  1013
[NO2] = 1.8  10–2 M
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Chapter 14: Chemical Equilibrium
321
14.5 Predicting the Direction of Reaction
Learning Objective

Use the reaction quotient Q.

Describe the direction of reaction after comparing Q with Kc.

Use the reaction quotient. (Example 14.5)
Problem-Solving Skill
4.
Using the reaction quotient. Given the concentrations of substances in a reaction mixture,
predict the direction of reaction (Example 14.5).
Exercise 14.8
A 10.0-L vessel contains 0.0015 mol CO2 and 0.10 mol CO. If a small amount of carbon is added to
this vessel and the temperature raised to 1000C, will more CO form? The reaction is
CO2(g) + C(s)
2CO(g)
The value of Kc for this reaction is 1.17 at 1000C. Assume that the volume of gas in the vessel is 10.0
L.
Wanted: Will more CO form?
Given:
10.0-L vessel, 0.0015 mol CO2, 0.10 mol CO; Kc at 1000C = 1.17
Known:
Calculate the value of the reaction quotient Qc and compare with Kc.
Qc =
[CO] i2
[CO 2 ] i
Solution: Calculate concentrations:
[CO2]I =
0.0015 mol
= 0.00015 M CO2
10.0 L
[CO] I =
0.10 mol
= 0.010 M CO
10.0 L
Qc =
[0.010 ] 2
= 0.667
[0.00015 ]
Qc is less than Kc. Thus, yes, the rate of the forward reaction will increase to produce more CO.
14.6 Calculating Equilibrium Concentrations
Learning Objectives

Obtain one equilibrium concentration given the others. (Example 14.6)

Solve an equilibrium problem (involving a linear equation in x). (Example 14.7)

Solve an equilibrium problem (involving a quadratic equation in x). (Example 14.8)
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322
Chapter 14: Chemical Equilibrium
Problem-Solving Skills
5.
Obtaining one equilibrium concentration given the others. Given Kc and all concentrations of
substances but one in an equilibrium mixture, calculate the concentration of this one substance
(Example 14.6).
6.
Solving equilibrium problems. Given the starting composition and Kc of a reaction mixture,
calculate the equilibrium composition (Examples 14.7 and 14.8).
Exercise 14.9
Phosphorus pentachloride gives an equilibrium mixture of PCl5, PCl3, and Cl2 when heated.
PCl5(g)
PCl3(g) + Cl2(g)
A 1.00-L vessel contains an unknown amount of PCl5 and 0.020 mol each of PCl3 and Cl2 at
equilibrium at 250C. How many moles of PCl5 are in the vessel if Kc for this reaction is 0.0415 at
250C?
Wanted: moles PCl5
Given:
V = 1.00 L; 0.020 mol PCl3, 0.020 mol Cl2; Kc = 0.0415
Known:
Kc =
[PCl 3 ][Cl 2 ]
[PCl 5 ]
Solution: Solve the preceding for [PCl5] and substitute in known values:
[PCl5] =
[PCl 3 ][Cl 2 ]
[0.020 / 1.00 ][ 0.020 / 1.00 ]
=
0.0415
Kc
= 0.0096
Moles PCl5 = 0.0096
Note that in substituting values into the equilibrium-constant expression in Exercise 14.9, we did not
write [0.020/1.00]2. We wrote the number twice. The reason for doing this, when you are in a hurry,
such as when taking a test, is that you will not forget to square the value. If you write it twice, you will
avoid making that error.
Exercise 14.10
What is the equilibrium composition of a reaction mixture if you start with 0.500 mol each of H2 and I2
in a 1.0-L vessel? The reaction is
H2(g) + I2(g)
2HI(g)
Kc = 49.7 at 458C
Solution: First, set up a table of concentrations. Since it is a 1-L vessel, the concentrations are the
mole amounts to two significant figures.
Concentrations (M): H2(g) + I2(g)
Starting
Change
Equilibrium
0.500
x
0.500  x
0.500
x
0.500  x
2HI(g)
0
2x
2x
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Chapter 14: Chemical Equilibrium
323
Second, substitute values into the equilibrium-constant expression:
Kc =
(2 x)( 2 x)
( 2 x) 2
[HI] 2
=
=
= 49.7
(0.500  x)( 0.500  x)
[H 2 ][I 2 ]
(0.500  x) 2
Third, take the square root of both sides and solve for x:
2x
= ±7.050 (only the positive value is possible)
(0.500  x)
2x = 3.52 – 7.050x
9.050x = 3.52
x=
3.52
= 0.389 M
9.050
Thus equilibrium concentrations are
[H2] = [I2] = 0.500  0.389 = 0.11 M
[HI] = 2(0.389) = 0.78 M
When calculating equilibrium concentrations given only starting amounts, be sure to look for ways to
simplify your work so that you do not have to deal with x2. First, check to see whether the expression
with the x is a perfect square. If it appears not to be, check to see that you’ve written the equilibriumconstant expression with the proper exponents. If you still do not have a perfect square, then see if you
can eliminate the x that is subtracted from a beginning concentration. You can ignore this change when
it is very small compared with the other number. This often will be the case when the equilibrium
constant divided by the initial concentration is 103 or less. If you cannot simplify the work, then you
will have to use the quadratic formula to solve for x. It is very useful but takes a long time to solve, and
when you use it, you must take great care to avoid errors.
Exercise 14.11
Phosphorus pentachloride, PCl5, decomposes when heated.
PCl5(g)
PCl3(g) + Cl2(g)
If the initial concentration of PCl5 is 1.00 mol/L, what is the equilibrium composition of the gaseous
mixture at 160C? The equilibrium constant Kc at 160C is 0.0211.
Solution: Set up the table of concentrations.
Concentrations (M): PCl5(g)
PCl3(g) + Cl2(g)
Starting
Change
1.00
x
0
+x
0
+x
Equilibrium
1.00  x
x
x
Putting these values into the equilibrium-constant expression, we have
Kc =
[PCl 3 ][Cl 2 ]
( x)( x)
=
= 0.0211
[PCl 5 ]
(1.00  x)
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Chapter 14: Chemical Equilibrium
Since this is not a perfect square, and Kc divided by the initial concentration of PCl5 is too large to
ignore x, we must use the quadratic formula:
b  b 2  4ac
2a
Put the Kc expression into the proper form:
x2 = (0.0211)(1.00 – x) = 0.0211 – 0.0211x
We rearrange and assign values as follows:
x2
+ 0.0211x
– 0.0211
a=1
b
c
=0
Substitute values into the quadratic formula:
x=
 0.0211  (0.0211)
2
 4(1)( 0.0211)
2
=
0.0211  0.2913
2
Since a negative value for x is impossible, the solution is
x = 0.1351
Referring to the bottom line of the table, we find that the equilibrium compositions are
[PCl5] = 1.00 – 0.1351 = 0.86 M
[PCl3] = [Cl2] = 0.135 M
14.7 Removing Products or Adding Reactants
Learning Objectives

State Le Châtelier’s principle.

State what happens to an equilibrium when a reactant or product is added or removed.

Apply Le Châtelier’s principle when a concentration is altered. (Example 14.9)
Problem Solving Skill
7.
Applying Le Châtelier’s principle. Given a reaction, use Le Châtelier’s principle to decide the
effect of adding or removing a substance (Example 14.9).
It is important that you understand that Le Châtelier’s principle is an observation, not an explanation.
Recall from Chapter 13 that collision theory supposes that reaction occurs because of molecular
collisions. Thus it is easy to see why more product could form when the concentration of reactant is
increased—because there is more chance for collision of reactants to form a product. If product is
removed, there are fewer collisions to convert product back to reactant. Thus Le Châtelier’s principle
supports collision theory.
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Chapter 14: Chemical Equilibrium
325
Exercise 14.12
Consider each of the following equilibria that are disturbed in the manner indicated. Predict the
direction of reaction.
a.
The equilibrium
CaCO3(s)
CaO(s) + CO2(g)
is disturbed by increasing the pressure (that is, concentration) of carbon dioxide.
b.
The equilibrium
2Fe(s) + 3H2O(g)
Fe2O3(s) + 3H2(g)
is disturbed by increasing the concentration of hydrogen.
Solution:
a.
Increasing the pressure of carbon dioxide in the reaction mixture is the same as increasing
the concentration of CO2. This would cause a shift left or cause the rate of the reverse
reaction to increase, which would reduce the added CO2 concentration.
b.
Increasing the concentration of hydrogen again increases the rate of the reverse reaction,
which lowers the added H2 concentration.
Remember that once disturbed and then left alone, a reaction system will again reach equilibrium;
even though the concentrations will be different, the value of Kc will be the same as long as the
temperature is not changed.
14.8 Changing the Pressure and Temperature
Learning Objectives

Describe the effect of a pressure change on chemical equilibrium.

Apply Le Châtelier’s principle when the pressure is altered. (Example 14.10)

Describe the effect of a temperature change on chemical equilibrium.

Apply Le Châtelier’s principle when the temperature is altered. (Example 14.11)

Describe how the optimum conditions for a reaction are chosen.
Problem-Solving Skill
7.
Applying Le Châtelier’s principle. Given a reaction, use Le Châtelier’s principle to decide the
effect of adding or removing a substance (Example 14.9), changing the pressure (Example 14.10),
or changing the temperature (Example 14.11).
Looking again at an equilibrium system from the viewpoint of molecular collisions, if we increase the
pressure, we push molecules closer together. The side of the equation having more gas molecules will
have more collisions. Thus the reaction will go toward the other side of the equation. On the other hand,
the side with more gas molecules has to have more collisions for reaction to occur and thus is affected
more by a decrease in pressure than is the side with fewer gas molecules. Thus, when the pressure is
decreased, reaction is toward the side with more molecules. Again, Le Châtelier’s principle describes
what we expect.
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326
Chapter 14: Chemical Equilibrium
Exercise 14.13
Can you increase the amount of product in each of the following reactions by increasing the pressure?
Explain.
a.
CO2(g) + H2(g)
b.
4CuO(s)
c.
2SO2(g) + O2(g)
CO(g) + H2O(g)
2Cu2O(s) + O2(g)
2SO3(g)
Solution:
a.
No. According to Le Châtelier’s principle, if the pressure is increased, reaction favors the
direction to make fewer gas molecules. Both sides have the same number of gas molecules,
so the rates of both forward and reverse reactions will be similarly affected, with no shift.
b.
No. Increasing the pressure will favor the reverse reaction because there are no gas
molecules on the left side.
c.
Yes. There are fewer gas molecules on the right—two versus three on the left—so an
increase in pressure will yield more product.
In the next exercise, to understand Le Châtelier’s principle in relation to how temperature affects
reactions, we must consider the reaction energetics. If the temperature is raised, the reaction that uses
heat will be favored. If the temperature is lowered, the reaction that needs heat will be slowed down,
and the opposite reaction then will go more quickly. In these instances, the value of the equilibrium
constant will change. When equilibrium is again established after the heat perturbation, the value of K
will not be the same as it was before the change. The value of K changes consistently with the rate
change. If the rate of the forward reaction is increased, K will increase; if the rate of the reverse reaction
is increased, K will decrease. This will be discussed further in text Chapter 17.
Exercise 14.14
Consider the possibility of converting carbon dioxide to carbon monoxide by the endothermic reaction
CO2(g) + H2(g)
CO(g) + H2O(g)
Is a high or a low temperature more favorable to the production of carbon monoxide? Explain.
Solution: Endothermic reactions will be favored with an increase in temperature. So a high
temperature would be favorable.
Exercise 14.15
Consider the reaction
2CO2(g)
2CO(g) + O2(g)
H = 566 kJ
Discuss the temperature and pressure conditions that would give the best yield of carbon
monoxide.
Solution: High temperatures would favor the formation of products in this endothermic reaction.
Low pressure would favor the products because there are more gas molecules on the product side.
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Chapter 14: Chemical Equilibrium
327
14.9 Effect of a Catalyst
Learning Objectives

Define catalyst.

Compare the effect of a catalyst on rate of reaction with its effect on equilibrium.

Describe how a catalyst can affect the product formed.
ADDITIONAL PROBLEMS
1.
Calculate the concentrations of all substances present in the equilibrium mixture at a given
temperature if 2.35 mol H2 and 2.35 mol I2 are placed in a 10.0-L flask and allowed to come to
equilibrium, at which time 3.76 mol HI is present. What is the value of Kc? The reaction is
H2(g) + I2(g)
2.
The following concentrations were found for the substances present in a reaction flask: CS2 = 0.48
M, H2 = 0.35 M, CH4 = 0.42 M, and H2S = 0.52 M. Is the reaction at equilibrium? If not, for which
reaction will the rate be greater? Kc at this temperature is 0.28. The reaction is
CS2(g) + 4H2(g)
3.
4.
2HI(g)
CH4(g) + 2H2S(g)
Write the equilibrium-constant expression for each of the following equations.
a.
SbBr3(g)
Sb(s) + 3/2Br2(g)
b.
SO2(g) + ½O2(g)
c.
2CaSO4(s)
d.
2Cl2(g) + 2H2O(g)
4HCl(g) + O2(g)
e.
2Fe(s) + 4H2O(g)
Fe2O4(s) + 4H2(g)
SO3(g)
2CaO(s) + 2SO2(g) + O2(g)
The value of Kp for the equilibrium
B2H6(g) + 4BF3(g)
6HBF2(g)
is 2.94 at 296 K. What is the value of Kc for this equilibrium at 296 K?
5.
The equilibrium constant for the following reaction at a given temperature is 45.0. How many
moles of each component are present in the reaction mixture at this temperature if 0.340 mol H2
and 0.340 mol I2 are placed in a 10.0-L vessel? The reaction is
H2(g) + I2(g)
6.
2HI(g)
The equilibrium constant Kc for the following reaction at 150C is 1.20  102. Calculate the
concentrations of all components in the equilibrium mixture if 5.00 mol I2 and 8.00 mol Br2 are
reacted in a 10.0-L vessel. The reaction is
I2(g) + Br2(g)
2IBr(g)
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328
7.
Chapter 14: Chemical Equilibrium
The decomposition of SO3 proceeds according to the following equation:
2SO3(g)
2SO2(g) + O2(g)
At 298 K, Kp for this reaction is 8.30  1023. If 2.46 mol SO3 is placed in a 1.00-L gas cylinder at
298 K, what will be the partial pressure of each species present at equilibrium?
8.
When a mixture of air and gasoline vapor explodes in the cylinder of a gasoline engine, some of
the N2 and O2 from the air combine to form NO gas:
N2(g) + O2(g)
2NO(g)
The NO is discharged in the exhaust gases and contributes to smog formation. Kc for the reaction
is 1.21  104 at 1800 K. An experimental combustion chamber having a volume of 1.00 L is
filled with air at 1.00 atm pressure at 0C in the presence of a catalyst. It contains 0.0357 mol N2
and 0.00892 mol O2. The chamber is heated to 1800 K and held at this temperature until there is
no observed change in pressure. How many moles of NO are present in the equilibrium mixture?
9.
Anhydrous CaSO4 is used commonly as a drying agent. At 25C, the heterogeneous equilibrium
CaSO4(s) + 2H2O(g)
CaSO4 ∙ 2H2O(s)
has a Kp value of 1.55  103. A 1.50  102 g sample of anhydrous CaSO4 and 1.00  102 g CaSO4 ∙
2H2O are placed in the bottom of a small desiccator at 25C, and the desiccator is closed. What
will be the equilibrium vapor pressure of the solid mixture?
10. Predict the direction of increased reaction resulting from each of the following changes to the
equilibrium system. Explain your answers. Also state any expected change in the value of K in
each case.
2NO2(g) + 7H2(g)
2NH3(g) + 4H2O(g) + 993 kJ
a.
An increase in the volume of the container
b.
An increase in temperature
c.
Removal of NH3
d.
Addition of a catalyst
e.
Addition of 0.15 mol of helium gas
ANSWERS TO ADDITIONAL PROBLEMS
If you missed an answer, study the text section and problem-solving skill given in parentheses after the
answer.
1.
[H2] = [I2] = 0.047 M; [HI] = 0.376 M; Kc = 64 (14.1, PS Sk. 1)
2.
The reaction is not at equilibrium because Qc = 16, which is greater than Kc. The rate of the
reverse reaction will be increased. (14.5, PS Sk. 4)
3.
a.
b.
c.
[Br 2 ] 3/2
[SbBr 3 ]
[SO 3 ]
[SO 2 ][O 2 ]1/2
[SO2]2 [O2]
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Chapter 14: Chemical Equilibrium
d.
e.
329
[HCl] 4 [O 2 ]
[Cl 2 ] 2 [H 2 O] 2
[H 2 ] 4
[H 2 O] 4
(14.2, PS Sk. 2)
4.
1.21  101 (14.2)
5.
mol H2 = mol I2 = 0.078; mol HI = 0.524 (14.6, PS Sk. 6)
6.
[I2] = 0.023 M; [Br2] = 0.32 M; [IBr] = 0.95 M (14.6, PS Sk. 6)
7.
PSO3 = 60.2 atm; PSO2 = 8.44  107 atm; PO2 = 4.22  107 atm (14.2, 14.6, PS Sk. 2, 6)
8.
1.95  104 mol (14.6, PS Sk. 6)
9.
PH 2O = 2.54  102 atm (14.2, 14.6, PS Sk. 6)
10.
a.
An increase in volume would decrease the pressure. This would affect the reverse reaction
(with 6 gas molecules) less than it would the forward reaction (with 9 gas molecules).
Therefore, the rate of the forward reaction will be slowed, and increased reaction will occur
to the left. The value of K will not change because the temperature remains constant.
b.
Because the reaction is exothermic (with the heat of reaction on the product side), the
reverse reaction will be favored with an increase in temperature, and the value of K will
decrease.
c.
Removal of product will decrease the rate of the reverse reaction; thus reaction will increase
to the right, or product, side; K will not change.
d.
The addition of a catalyst will affect both forward and reverse reactions similarly; there will
be no change.
e.
Because helium is not part of the reacting system and will not react with other substances
present, it will have no effect on the reaction; there will be no change. (14.7, 14.8, PS Sk. 7)
CHAPTER POST-TEST
1.
Which of the following represents the correct equilibrium-constant expression for the reaction
3D(aq) + 5E(aq)
a.
b.
c.
d.
2
1
F(aq) + G(aq)
2
3
[F] 2 [G] 3/2
[D] 1/3[E] 1/5
[D] 3 [F] 1/2
[E] 5 [G] 2/3
[F] 1/2
[G]
2/3
∙
[E] 5
[D] 3
[F]1/2 [G] 2/3
[E] 5 [D] 3
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330
Chapter 14: Chemical Equilibrium
e.
2.
[F]1/2[G]3/2[D]3[E]5
The equilibrium constant Kc for the dissociation
PCl5(g)
PCl3(g) + Cl2(g)
equals 3.26  102 at 191C. Does the equilibrium mixture contain predominantly reactant or
products? When 0.285 mol PCl5 is introduced into a 10.0-L vessel and heated to 191C, the
measured concentrations of PCl3 and Cl2 sometime later are both found to be [0.00853] or
0.00853 M. At this point has equilibrium been reached?
3.
Determine whether the following statements are true or false. If a statement is false, change it so
that it is true.
a.
For the equilibrium 2H2(g) + O2(g)
2H2O(l), a decrease in pressure will favor the
increased formation of H2O(l). True/False: ______________________________________
_________________________________________________________________________
b.
Changes in either temperature or pressure of a system at equilibrium will alter the
equilibrium constant for the system. True/False: __________________________________
_________________________________________________________________________
4.
5.
Write equilibrium expressions for the following reactions in terms of Kp.
a.
3O2(g)
b.
4CuO(s)
c.
C(s) + S2(g)
2O3(g)
2Cu2O(s) + O2(g)
CS2(g)
Hydrogen sulfide gas, H2S, when heated to 1132C, dissociates to H2 and S2 gases. The equation
is
2H2S(g)
2H2(g) + S2(g)
The equilibrium concentrations of H2 and S2 were measured to be 0.0384 M and 0.0192 M,
respectively. If the Kc value is 2.3  104 at 1132C, what is the equilibrium concentration of H2S
gas?
6.
When 1.000 mol of gaseous HI is sealed in a 1.000-L flask at a specific temperature, it
decomposes to form 0.182 mol each of hydrogen and iodine. Calculate Kc for the reaction
2HI(g)
7.
H2(g) + I2(g)
The reaction
2NO(g) + O2(g)
2NO2(g)
is endothermic. If we increase the temperature, will the Kc value increase or decrease?
8.
If we suddenly decreased the volume of the container for the system in Problem 7, how would this
affect the equilibrium?
9.
If we inject more O2 into the system in Problem 7, how would this affect the equilibrium? How
would it affect the value of the equilibrium constant?
10. At 395 K, the reaction
CO(g) + Cl2(g)
COCl2(g)
has equilibrium partial pressures for each species of 0.128, 0.116, and 0.334 atm, respectively.
Determine Kp.
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Chapter 14: Chemical Equilibrium
331
11. In a 1-L tank, 1 mol MgCO3, 1 mol MgO, and 1 mol CO2 are in equilibrium:
MgCO3(s)
MgO(s) + CO2(g)
If more CO2 is added, will there be a change in the amounts of MgCO3 and MgO present at
equilibrium?
12. Consider the following reaction:
H2(g) + I2(g)
2HI(g)
Kc = 54.7
If 2.00 mol H2 and 2.00 mol I2 are placed in a 1.00-L container, what will be the equilibrium
concentrations?
13. Consider the following reaction:
2HI(g)
H2(g) + I2(g)
Kc = 0.0183
At equilibrium, [H2] = 0.32 mol/L, [I2] = 0.32 mol/L, and [HI] = 2.36 mol/L. Suppose that 0.70
mol/L HI is added, and a shift in equilibrium occurs. Calculate the new equilibrium
concentrations.
ANSWERS TO CHAPTER POST-TEST
If you missed an answer, study the text section and problem-solving skill given in parentheses after the
answer.
1.
d (14.2, PS Sk. 2)
2.
Predominantly reactant; Qc = 3.63  103, hence equilibrium has not been reached. (14.4, 14.5, PS
Sk. 4)
3.
a.
False. The equilibrium will favor the increased formation of H2(g) and O2(g). (14.8, PS Sk.
7)
b.
False. Changes in temperature will alter the equilibrium constant. Changes in pressure will
alter the equilibrium composition but not the equilibrium constant. (14.8, PS Sk. 7)
4.
PO23
a.
Kp =
b.
Kp = PO2
c.
Kp =
PO32
PCS 2
PS2
(14.2, 14.3, PS Sk. 2)
5.
[H2S] = 0.35 mol/L (14.6, PS Sk. 5)
6.
Kc = 0.0819 (14.2, PS Sk. 1, 3)
7.
Increasing the temperature favors a shift to the right and increases Kc, which is temperaturedependent. (14.8, PS Sk. 7)
8.
Decreasing the volume would increase the internal pressure. By Le Châtelier’s principle, a shift
should occur to decrease this pressure. The pressure will decrease if there are fewer moles of
gaseous particles, so the reaction would shift to the right. Kc does not change. (14.8, PS Sk. 7)
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332
9.
Chapter 14: Chemical Equilibrium
With the addition of O2 to this system, the rate of the forward reaction increases, and additional O2
is consumed. Eventually, the product–reactant ratio once again is equal to the original value of the
equilibrium constant. Kc does not change unless the temperature changes. (14.7, PS Sk. 7)
10. Kp = 22.5 (14.2, PS Sk. 3)
11. Yes. If CO2 is added, the reaction shifts left, forming more MgCO3 and consuming CO2 and MgO.
(14.7, PS Sk. 7)
12. [H2] = [I2] = 0.43 mol/L
[HI] = 3.15 mol/L (14.6, PS Sk. 6)
13. [HI] = 2.91 mol/L
[H2] = [I2] = 0.39 mol/L (14.6, 14.7, PS Sk. 6, 7)
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