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PRINCIPLES OF
CHEMICAL
EQUILIBRIUM
CONTENTS
15-1 Dynamic Equilibrium
15-2 The Equilibrium Constant
15-3
15-4
15-5
15-6
15-7
➣
Expression
Relationships Involving
Equilibrium Constants
The Magnitude of an
Equilibrium Constant
The Reaction Quotient, Q:
Predicting the Direction
of Net Change
Altering Equilibrium
Conditions: Le Châtelier’s
Principle
Equilibrium Calculations:
Some Illustrative
Examples
FOCUS ON The Nitrogen
Cycle and the Synthesis of
Nitrogen Compounds
There are many
demonstrations that show
equilibrium: roll a marble in a
parabolic clear dish to show the
establishment of physical
equilibrium; pour AgNO3 into
ordinary tap water, and the
chloride ion in the water will
precipitate out leaving the water
cloudy; burn a match, and when it
goes out, equilibrium is
established. Point out that
equilibrium in chemical reactions is
obtained by minimizing the free
energy (see Chapter 19).
Point out that equilibrium
is less common than
nonequilibrium. For
example, your heart and other
biochemical processes are
oscillating back and forth in cycles.
We certainly don’t want our hearts
to come to equilibrium.
622
A vital natural reaction is in progress in the lightning bolt seen here:
N2(g) + O2(g) ∆ 2 NO(g). Usually this reversible reaction does not occur
to any significant extent in the forward direction, but in the high-temperature
lightning bolt it does. At equilibrium at high temperatures, measurable
conversion of N2(g) and O2(g) to NO(g) occurs. In this chapter we study the
equilibrium condition in a reversible reaction and the factors affecting it.
U
ntil now, we have stressed reactions that go to completion
and the concepts of stoichiometry that allow us to calculate
the outcomes of such reactions. We have made occasional
references to situations involving both a forward and a reverse
reaction—reversible reactions—but in this chapter, we will look at
them in a detailed and systematic way.
Our emphasis will be on the equilibrium condition reached when
forward and reverse reactions proceed at the same rate. Our main
tool in dealing with equilibrium will be the equilibrium constant.
We will begin with some key relationships involving equilibrium
constants; then we will make qualitative predictions about the condition of equilibrium; and finally we will do various equilibrium
calculations. As we will discover throughout the remainder of the
text, the equilibrium condition plays a role in numerous natural
phenomena and affects the methods used to produce many important industrial chemicals.
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15-2
15-1
The Equilibrium Constant Expression
Dynamic Equilibrium
animation
DYNAMIC EQUILIBRIUM
Let’s begin by describing three simple physical and one chemical phenomena
that will help us to establish the core attribute of a system at equilibrium—
two opposing processes take place at equal rates.
1. When a liquid vaporizes within a closed container, after a time, vapor
molecules condense to the liquid state at the same rate at which liquid
molecules vaporize. Even though molecules continue to pass back and
forth between liquid and vapor (a dynamic process), the pressure exerted
by the vapor remains constant with time. The vapor pressure of a liquid is a
property resulting from an equilibrium condition.
2. When a solute is added to a solvent, the system may reach a point at which
the rate of dissolution is just matched by the rate at which dissolved solute
crystallizes—that is, the solution is saturated. Even though solute particles
continue to pass back and forth between the saturated solution and the
undissolved solute, the concentration of dissolved solute remains constant.
The solubility of a solute is a property resulting from an equilibrium condition.
3. When an aqueous solution of iodine, I 2 , is shaken with pure carbon tetrachloride, CCl4(l), the I 2 molecules move into the CCl4 layer. As the concentration of I 2 builds up in the CCl4 , the rate of return of I 2 to the water layer
becomes significant. When I 2 molecules pass between the two liquids at
equal rates—a condition of dynamic equilibrium—the concentration of I 2
in each layer remains constant. At this point, the concentration of I 2 in the
CCl4 is about 85 times greater than in the H 2O (Fig. 15-1). The ratio of concentrations of a solute in two immiscible solvents is called the distribution
coefficient. The distribution coefficient, which represents the partitioning of a
solute between two immiscible solvents, is a property resulting from an equilibrium condition.
4. When gaseous phosphorus pentachloride is heated, it decomposes to phosphorus trichloride and chlorine gases: PCl5(g) ¡ PCl3(g) + Cl2(g).
Consider a sample of PCl5(g) initially exerting a pressure of 1.0 atm in a
closed container at 250 °C. The gas pressure in the container first rises
rapidly and then ever more slowly, reaching a maximum, unchanging pressure of 1.7 atm. Because two moles of gas are produced for each mole of
PCl5(g) that decomposes, if the reaction went to completion the final pressure would have been 2.0 atm. We conclude that the decomposition of PCl5 is
a reversible reaction that reaches an equilibrium condition.
The properties in the first three situations just described—vapor pressure,
solubility, and distribution coefficient—are examples of physical equilibria.
The fourth situation is an example of chemical equilibrium. All four are described through a general quantity known as an equilibrium constant, the subject of the next section.
15-2
623
Paiva, João C. M., Victor M. S.
Gil. “The Complexity of
Teaching and Learning Chemical
Equilibrium.” J. Chem. Educ.
2000: 77, 1560 (December 2000).
Silverstein, Todd P.
“Equilibrium: A Teaching/
Learning Activity.” J. Chem. Educ.
2000: 77, 1410 (November 2000).
The two opposing
processes operate at the
microscopic level.
(a)
(b)
▲ FIGURE 15-1
Dynamic equilibrium in a
physical process
(a) A yellow-brown saturated
solution of I 2 in water
(top layer) is brought into
contact with colorless CCl4(l)
(bottom layer). (b) I 2 molecules distribute themselves
between the H 2O and CCl4 .
When equilibrium is reached,
[I 2] in the CCl4 (violet,
bottom layer) is about 85
times greater than in the
water (colorless, top layer).
THE EQUILIBRIUM CONSTANT EXPRESSION
Methanol (methyl alcohol) is synthesized from a carbon monoxide–hydrogen
mixture called synthesis gas. This reaction is likely to become increasingly important as methanol and its mixtures with gasoline find greater use as motor
fuels. Methanol has a high octane rating, and its combustion produces much
less air pollution than does gasoline.
Methanol synthesis is a reversible reaction, which means that at the same
time CH 3OH(g) is being formed,
CO(g) + 2 H 2(g) ¡ CH 3OH(g)
(15.1)
it decomposes in the reverse reaction
CH 3OH(g) ¡ CO(g) + 2 H 2(g)
(15.2)
▲ Methanol is actively being
considered as an alternative
fuel to gasoline.
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Principles of Chemical Equilibrium
In anticipation of entropy,
point out that a mixing
process and a dissolution
process lead to an increase in
disorder.
Are You Wondering . . .
How we know that an equilibrium is dynamic—that forward and reverse
reactions continue even after equilibrium is reached?
Suppose we have an equilibrium mixture of AgI(s) and its saturated aqueous solution.
AgI(s) ∆ AgI(satd aq)
Now let’s add to this mixture some saturated solution of AgI made from AgI
containing radioactive iodine-131 as iodide ion, as illustrated in Figure 15-2. If
both the forward and reverse processes stopped at equilibrium, radioactivity
would be confined to the solution. What we find, though, is that radioactivity
shows up in the solid in contact with the saturated solution. Over time, the radioactive “hot” spots distribute themselves throughout the solution and undissolved solid. The only way this can happen is if the dissolving of the solid solute
and its crystallization from the saturated solution continue indefinitely. The equilibrium condition is dynamic.
Saturated solution only
added to beaker
(a)
(b)
▲ FIGURE 15-2
Dynamic equilibrium illustrated
(a) A saturated solution of radioactive AgI is added to a saturated solution
of AgI. (b) The radioactive iodide ions distribute themselves throughout the
solution and the solid AgI, showing that the equilibrium is dynamic.
Remind students that the
rates of reactions (forward
and reverse) are affected by
the concentration of reactants.
Initially, only the forward reaction (15.1) occurs, but as soon as some
CH 3OH forms, the reverse reaction (15.2) begins. With passing time, the forward reaction slows because of the decreasing concentrations of CO and H 2
and the reverse reaction speeds up as more CH 3OH accumulates. Eventually,
the forward and reverse reactions proceed at equal rates, and the reaction mixture reaches a condition of dynamic equilibrium, which we can represent with
a double arrow ∆ .
CO(g) + 2 H 2(g) ∆ CH 3OH(g)
(15.3)
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15-2
The Equilibrium Constant Expression
625
TABLE 15.1 Three Approaches to Equilibrium in the Reactiona
CO(g) 2 H2(g) ∆ CH3OH(g)
CO(g)
H2(g)
CH3OH(g)
1.000
0.911
0.0911
1.000
0.822
0.0822
0.000
0.0892
0.00892
0.000
0.753
0.0753
0.000
1.506
0.151
1.000
0.247
0.0247
1.000
1.380
0.138
1.000
1.760
0.176
1.000
0.620
0.0620
Point out to students that
the equilibrium state can be
obtained by starting with
any combination of reactants
and/or products.
Experiment 1
Initial amounts, mol
Equilibrium amounts, mol
Equilibrium concentrations, mol/L
Experiment 2
Initial amounts, mol
Equilibrium amounts, mol
Equilibrium concentrations, mol/L
Experiment 3
Initial amounts, mol
Equilibrium amounts, mol
Equilibrium concentrations, mol/L
The concentrations printed in blue are used in the calculations in Table 15.2.
a
Reaction carried out in a 10.0-L flask at 483 K.
1.80
1.80
1.60
1.40
1.20
te
1.00
0.80
0.60
0.40
0.20
2.00
Jordan, A. D. “Liquid-Liquid
Equilibrium: Verification of
the Lever Rule.” J. Chem. Educ.
2000: 77, 395 (March 2000).
Harrison, J. A., Buckley, P. D.
“Simulating Dynamic
Equilibria: A Class Experiment.”
J. Chem. Educ. 2000: 77, 1013
(August 2000).
1.80
te
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
Time
Experiment 1
Silverstein, Todd P. “The Real
Reason Why Oil and Water
Don’t Mix.” J. Chem. Educ. 1998:
75, 116 (January 1998).
te
1.60
1.40
Chemical Equilibrium
activity
1.20
1.00
0.80
0.60
0.40
0.20
Time
Experiment 2
te time for equilibrium to be reached
mol CO
mol H2
mol CH3OH
▲
2.00
Moles of reactants and products
2.00
Moles of reactants and products
Moles of reactants and products
One consequence of the equilibrium condition is that the amounts of the
reactants and products remain constant with time. These equilibrium amounts,
however, depend on the quantities of reactants and products present initially.
For example, Table 15.1 lists data for three hypothetical experiments. All three
experiments are conducted in a 10.0-L flask at 483 K. In the first experiment,
only CO and H 2 are present initially; in the second, only CH 3OH; and in the
third, CO, H 2 , and CH 3OH. The data from Table 15.1 are plotted in Figure 15-3,
and from these graphs we see that
• in no case is any reacting species completely consumed;
• in all three cases the equilibrium amounts of reactants and products
appear to have nothing in common.
Time
Experiment 3
FIGURE 15-3
Three approaches to equilibrium
in the reaction
CO(g) 2 H2(g) ∆ CH3OH(g)
The initial and equilibrium amounts
for each of these three cases are
listed in Table 15.1. te = time for
equilibrium to be reached.
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Principles of Chemical Equilibrium
TABLE 15.2
[CH3OH]
[CO][H2]
Trial 2:
[CH3OH]
[CO](2 : [H2])
Trial 3:
[CH3OH]
Expt
Trial 1:
1
0.00892
= 1.19
0.0911 * 0.0822
0.00892
= 0.596
0.0911 * 12 * 0.08222
2
0.0247
= 2.17
0.0753 * 0.151
0.0247
= 1.09
0.0753 * 12 * 0.1512
0.0753 * 10.15122
3
0.0620
= 2.55
0.138 * 0.176
0.0620
= 1.280
0.138 * 12 * 0.1762
0.138 * 10.17622
[CO][H2]2
0.00892
0.0911 * 10.082222
0.0247
0.0620
= 14.5
= 14.4
= 14.5
Equilibrium concentration data are from Table 15.1. In Trial 1, the equilibrium concentration of CH 3OH is placed in the numerator
and the product of the equilibrium concentrations, [CO][H 2], in the denominator. In Trial 2, each concentration is multiplied by
its stoichiometric coefficient. In Trial 3, each concentration is raised to a power equal to its stoichiometric coefficient. Trial 3 has
essentially the same value for each experiment. This value is the equilibrium constant Kc .
In the three experiments, point
out that Table 15.1 is consistent
with the equilibrium
concentrations of the two reactants
and the one product in Figure 15-3.
Put the equilibrium concentrations
into the equilibrium expression
(15.4) to show that this equation
is satisfied.
The representation of the
equilibrium expression in
terms of concentrations is
only valid at low concentrations,
usually less than a few moles
per liter.
Stress that “equilibrium”
means equal forward
and reverse reactions,
not equal concentrations of
reactants and products.
Although it is not obvious from a cursory inspection of the data, a particular ratio involving equilibrium concentrations of product and reactants has a
constant value, independent of how the equilibrium is reached. This ratio,
which is central to the study of chemical equilibrium, can be derived theoretically using concepts presented later in the text, but it can also be established
empirically, that is, by trial and error. Three reasonable attempts at formulating the desired ratio for reaction (15.3) are outlined in Table 15.2, and the ratio
that works is identified.
For the methanol synthesis reaction, the ratio of equilibrium concentrations
in the following equation has a constant value of 14.5 at 483 K.
K = ¢
[CH 3OH]
[CO]([H 2])2
≤
= 14.5
(15.4)
eq
This ratio is called the equilibrium constant expression and its numerical
value is the equilibrium constant.
THE EQUILIBRIUM CONSTANT AND ACTIVITIES
Let’s think about expression (15.4) for a moment. We have shown that we obtain a numerical value of 14.5 by substituting equilibrium concentrations from
Table 15.2 (Trial 3) into the equilibrium constant expression. But what happened to the units? The numerator in expression (15.4) has the unit mol L-1 and
the denominator, mol3 L-3. The resultant unit to accompany the numerical
value looks like it should be L2 mol -2. In later chapters we will encounter several expressions that require the logarithm of an equilibrium constant, but we
can only take the logarithms of dimensionless numbers, not of numbers that
carry units. We took the easy way out in equation (15.4) by just “dropping” the
troublesome units. But there is a more satisfactory way to eliminate the units.
We need to replace equation (15.4) by the following expression
K = ¢
aCH3OH
aCO(aH2)2
≤
= 14.5
(15.5)
eq
where a represents the activity of each reactant or product denoted through a
subscript formula. Activity, a thermodynamic concept introduced by G. N.
Lewis,* is the dimensionless ratio [X]>c°, where [X] represents a particular
concentration and c° corresponds to the concentration in a chosen reference
state. Our usual choice of reference state for a substance in solution is a
concentration of one mole per liter (1 mol L-1). For gases, activity can also be
expressed as the dimensionless ratio, P>P°, where P is a particular partial
*G. N. Lewis and M. Randall, Thermodynamics, McGraw Hill, New York, 1923.
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15-2
The Equilibrium Constant Expression
pressure and P° is the partial pressure in the reference state; and our usual
choice of P° is 1 bar (essentially equal to 1 atm). Pure solids and pure liquids
are assigned activities of 1.
Now, by returning to the methanol-synthesis example, let us establish the relationship between an equilibrium constant expressed in activities and the corresponding one expressed in concentrations. We begin by writing the activity
of each species, using the [ ] symbol for the equilibrium concentration and c 0
for the concentration in the reference state.
627
Point out to students that
the equilibrium constant
“14.5” calculated for the
methanol synthesis reaction is
specific to a 483 K reaction
temperature.
[CH 3OH]
c°
[CO]
=
c°
[H 2]
=
c°
aCH3OH =
aCO
aH2
K = §
[CH 3OH]
c°
[CO] [H 2]
¢
≤
c°
c°
2
= 1c°2 ¢
2
¥
[CH 3OH]
[CO]([H 2])2
≤
▲
Then we choose the value c° = 1 mol L-1, substitute these relationships
into the equilibrium constant expression (15.5), and we have
(15.6)
= 14.5
eq
KEEP IN MIND
that any concentrations
substituted into an
equilibrium constant
expression, or obtained
from it, must be equilibrium
concentrations.
eq
Notice that we have arrived at exactly the expression in equation (15.4)! In this
case, however, the unwanted units have been properly cancelled rather than
just conveniently dropped; that is, L2 mol -2 inside the large parentheses is
cancelled by mol2 L-2 on the outside.
Here is another reason for basing equilibrium constants on activities: Under
conditions where gases do not obey the ideal gas law (Section 6-9) or solutions
depart from ideal behavior (Section 13-3), equilibrium constant values may
vary with total concentration or pressure. This problem is eliminated when
activities are used. As we learned in Section 13-9, activities are “effective” or
“active” concentrations. In this text we will generally assume that systems are
ideal, and that activities can be replaced by concentrations or partial pressures.
EXAMPLE 15-1
Relating Equilibrium Concentrations of Reactants and Products. These equilibrium concentrations are measured in
reaction (15.3) at 483 K: [CO] = 1.03 M and [CH 3OH] = 1.56 M. What is the equilibrium concentration of H 2?
Solution
Write the equilibrium constant expression in terms of activities
K = ¢
Assume that the reaction conditions are such that the activities can
be replaced by their concentration values, allowing concentration
units to be canceled as in expression (15.6).
K = ¢
Substitute the known equilibrium concentrations into the equilibrium constant expression (15.6).
K =
Solve for the unknown concentration, [H 2]. (An implicit calculation to restore the concentration unit is [H 2] = aH2 * c° =
0.322 * 1.00 M = 0.322 M.)
aCH3OH
aCO(aH2)2
≤
= 14.5
eq
[CH 3OH]
[CO]([H 2])2
[CH 3OH]
2
[CO][H 2]
≤
= 14.5
eq
1.56
=
1.03[H 2]2
= 14.5
1.56
= 0.104
1.03 * 14.5
[H 2] = 20.104 = 0.322 M
[H 2]2 =
Practice Example A: In another experiment, equal concentrations of CH 3OH and CO are found at equilibrium in
reaction (15.3). What must be the equilibrium concentration of H 2 ?
Practice Example B: At a certain temperature, K = 1.8 * 104 for the reaction N2(g) + 3 H 2(g) ∆ 2 NH 3(g).
If the equilibrium concentrations of N2 and NH 3 are 0.015 M and 2.00 M, respectively, what is the equilibrium concentration of H 2 ?
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Principles of Chemical Equilibrium
A GENERAL EXPRESSION FOR K
Remind students not to
confuse equilibrium
constant expressions
with rate equations (Chapter 14).
The former contains stoichiometric
coefficients whereas the latter
does not. However, a similarity is
that both the rate constant and the
equilibrium constant vary with
temperature.
Before proceeding to other matters, let us emphasize that the equilibrium
constant expression for the methanol synthesis reaction summarized through
expression (15.6) is just a specific example of a more general case. For the
hypothetical, generalized reaction
aA + bB Á ∆ gG + hH Á
The equilibrium constant expression has the form
g
(aG) (aH)h Á
[G]g[H]h Á
K =
=
(aA)a(aB)b Á
[A]a[B]b Á
(15.7)
The numerator of an equilibrium constant expression is the product of the
activities of the species on the right side of the equation (aG , aH , Á ), with each
activity raised to a power given by the stoichiometric coefficient (g, h, Á ). The
denominator is the product of the activities of the species on the left side of the
equation (aA , aB , Á ), and again, with each activity raised to a power given by
the stoichiometric coefficient (a, b, Á ). As previously noted, where equilibrium systems are sufficiently close to ideal in their behavior, equilibrium concentrations are acceptable approximations to true activities.
The numerical value of an equilibrium constant, K, depends on the particular reaction and on the temperature. We will explore the significance of these
numerical values in Section 15-4.
CONCEPT ASSESSMENT
✓
Consider a hypothetical reaction in which one molecule, A, is converted to its isomer,
B, that is, the reversible reaction A ∆ B. Start with a flask containing 54 molecules
of A, represented by open circles. Convert the appropriate number of open circles to
filled circles to represent the isomer B and portray the equilibrium condition if
K = 0.02. Repeat the process for K = 0.5 and then for K = 1.
Are You Wondering . . .
If there is a relationship between the equilibrium constant and rate
constants?
Given the requirement that the rates of the forward and reverse reactions become
equal at equilibrium, it seems that there should be a relationship. The rate laws for
the forward and reverse reactions include numerical constants (rate constants) and
concentration terms raised to powers. Moreover, if we set the rate equations equal
to one another at equilibrium, we should be able to derive an expression involving
a ratio of concentration terms (raised to powers) and a ratio of constants. A major
difficulty, however, is that the exponents of the concentration terms in the equilibrium constant expression must be the same as the coefficients in the balanced equation, whereas the exponents in rate laws are generally not the same as these
coefficients. The trick to discovering the relationship between rate constants and an
equilibrium constant is to work with the detailed mechanism for the reaction in the
manner outlined in Exercise 83. Still, it is generally easier to obtain K directly from
measurements on equilibrium conditions than to attempt a calculation based on
rate constants. Moreover, in Chapters 19 and 20 we will learn about much more
direct measurements and calculations leading to values of equilibrium constants.
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15-3
15-3
Relationships Involving Equilibrium Constants
629
RELATIONSHIPS INVOLVING EQUILIBRIUM
CONSTANTS
Before assessing an equilibrium situation, it may be necessary to make some
preliminary calculations or decisions to get the appropriate equilibrium constant expression. This section presents some useful ideas in working with
equilibrium constants.
RELATIONSHIP OF K TO THE BALANCED CHEMICAL EQUATION
We must always make certain that the expression for K matches the corresponding balanced equation. In doing so, the following hold true.
• When we reverse an equation, we invert the value of K.
• When we multiply the coefficients in a balanced equation by a common
factor (2, 3, Á ), we raise the equilibrium constant to the corresponding
power (2, 3, Á ).
• When we divide the coefficients in a balanced equation by a common factor (2, 3, Á ), we take the corresponding root of the equilibrium constant
(square root, cube root, Á ).
Suppose that in discussing the synthesis of CH 3OH from CO and H 2 , we
had written the reverse of equation (15.3)—that is,
CH 3OH(g) ∆ CO(g) + 2 H 2(g)
K¿ = ?
Now, according to the generalized equilibrium constant expression (15.7), we
should write
K¿ =
[CO][H 2]2
1
1
1
=
=
=
= 0.0690
[CH 3OH]
[CH 3OH]
K
14.5
[CO][H 2]2
In the preceding expression, the terms printed in blue are the equilibrium constant expression and K value originally written as expression (15.4). We see
that K¿ = 1>K.
Suppose that for a certain application we want an equation based on synthesizing two moles of CH 3OH(g).
2 CO(g) + 4 H 2(g) ∆ 2 CH 3OH(g)
K– = ?
2
Here, K– = K . That is,
K– =
[CH 3OH]2
2
4
[CO] [H 2]
= ¢
[CH 3OH]
2
[CO][H 2]
2
≤ = (K)2 = (14.5)2 = 2.10 * 102
EXAMPLE 15-2
Relating K to the Balanced Chemical Equation. The following K value is given at 298 K for the synthesis of NH 3(g)
from its elements.
N2(g) + 3 H 2(g) ∆ 2 NH 3(g)
K = 3.6 * 108
What is the value of K at 298 K for the following reaction?
NH 3(g) ∆
1
3
N (g) + H 2(g)
2 2
2
K = ?
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Solution
First, reverse the given equation. This puts NH 3(g) on the
left side of the equation, where we need it.
2 NH 3(g) ∆ N2(g) + 3 H 2(g)
The equilibrium constant K¿ becomes
K¿ = 1>13.6 * 1082 = 2.8 * 10-9
Then, to base the equation on 1 mol NH 3(g), divide all coefficients by 2.
NH 3(g) ∆
This requires the square root of K¿.
K = 32.8 * 10-9 = 5.3 * 10-5
1
3
N2(g) + H 2(g)
2
2
Practice Example A: Use data from Example 15-2 to determine the value of K at 298 K for the reaction
2
1
N (g) + H 2(g) ∆ NH 3(g)
3 2
3
1
Practice Example B: For the reaction NO(g) + O2(g) ∆ NO2(g) at 184 ° C, K = 7.5 * 102. What is the value
2
of K at 184 °C for the reaction 2 NO2(g) ∆ 2 NO(g) + O2(g)?
CONCEPT ASSESSMENT
✓
Can you conclude whether the numerical value of K for the reaction 2 ICl(g) ∆
I 2(g) + Cl2(g) is greater or less than the numerical value of K for the reaction
1
1
ICl(g) ∆ I 2(g) + Cl2(g)? Explain.
2
2
COMBINING EQUILIBRIUM CONSTANT EXPRESSIONS
In Section 7-7, through Hess’s law, we showed how to combine a series of
equations into a single overall equation. The enthalpy change of the overall reaction was obtained by adding together the enthalpy changes of the individual reactions. A similar procedure can be used with equilibrium constants, but
with this important difference:
When individual equations are combined (that is, added), their equilibrium constants
are multiplied to obtain the equilibrium constant for the overall reaction.
Suppose we seek the equilibrium constant for the reaction
N2O(g) +
1
O (g) ∆ 2 NO(g)
2 2
K = ?
(15.8)
and know the K values of these two equilibria.
N2(g) +
1
O (g) ∆ N2O(g)
2 2
K = 2.7 * 10-18
N2(g) + O2(g) ∆ 2 NO(g)
K = 4.7 * 10-31
(15.9)
(15.10)
Equation (15.8) is obtained by reversing equation (15.9) and adding it to (15.10).
This requires that we also take the reciprocal of the K value of equation (15.9).
N2O(g) ∆ N2(g) +
(a)
1
O (g)
2 2
K(a) = 1>(2.7 * 10-18)
= 3.7 * 1017
(b)
Overall:
N2(g) + O2(g) ∆ 2 NO(g)
N2O(g) +
1
O (g) ∆ 2 NO(g)
2 2
K(b) = 4.7 * 10-31
K(overall) = ?
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Relationships Involving Equilibrium Constants
631
The overall equation is expression (15.8), and according to the general
expression (15.7),
K(overall) =
[NO]2
=
[N2O][O 2]1>2
[N2][O2]1>2
[NO]2
*
= K(a) * K(b)
[N2O]
[N2][O2]
3
3
K(a)
K(b)
= 3.7 * 1017 * 4.7 * 10-31 = 1.7 * 10-13
✓
CONCEPT ASSESSMENT
Point out that the reaction
of sulfur dioxide with
oxygen gas to form sulfur
trioxide is a reaction that takes
place in the atmosphere. This
results in the formation of sulfuric
acid which contributes to acid rain.
You wish to calculate K for the reaction
CH 4(g) + 2 H 2O(g) ∆ CO2(g) + 4 H 2(g)
and you have available a K value for the reaction
CO2(g) + H 2(g) ∆ CO(g) + H 2O(g)
What additional K value do you need, assuming that all K values are at the same
temperature?
EQUILIBRIA INVOLVING GASES: THE EQUILIBRIUM CONSTANTS,
Kc AND Kp
Mixtures of gases are as much solutions as are mixtures in a liquid solvent.
Thus, concentrations in a gaseous mixture can be expressed on a mole-perliter basis, and this is what we have done in writing K expressions using concentration units for our reference state. In order to describe gaseous reactants
and products through their partial pressures in atmospheres, which is commonly done, we need to switch our reference state. We will do this next.
A key step in the manufacture of sulfuric acid is the following reversible
reaction.
(15.11)
2 SO 2(g) + O2(g) ∆ 2 SO 3(g)
Since all of the reaction species are in the gas phase it seems reasonable to use
a partial-pressure reference state. We begin by writing the equilibrium constant expression in terms of activities
K = ¢
(aSO3)2
(aSO2)2aO2
≤
(15.12)
eq
where the activities are
aSO2 =
PSO2
P°
;
aO2 =
PO2
P°
;
aSO3 =
PSO3
P°
The reference-state partial pressure is P° = 1 bar, which we will take to be
essentially the same as 1 atm. Substituting these relationships into equation
(15.12), we obtain
¢
Kp = §
¢
PSO3
P°
PSO2
P°
≤
≤
2
2 PO
= P° ¢
¥
2
P°
(PSO3)2
(PSO2)2PO2
≤
(15.13)
eq
eq
As expected, the equilibrium constant is dimensionless. To designate that a
partial-pressure reference state was used we add the subscript “p” to the equilibrium constant, that is, by writing Kp .
To establish the equilibrium constant based on concentrations, we first use
the ideal gas law, PV = nRT, to relate gas concentrations and partial pressures
[SO2] =
PSO2
n
=
;
V
RT
[O2] =
PO2
n
=
;
V
RT
[SO3] =
PSO3
n
=
V
RT
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and then express activities in terms of those concentrations, where the reference state is c° = 1 mol L-1.
PSO2
aSO2
PSO3
PO2
[SO 2]
RT
=
;
=
c°
c°
aO2
[O2]
RT
=
;
=
c°
c°
aSO3
[SO3]
RT
=
=
c°
c°
Substituting into the equilibrium constant expression (15.12), we obtain
Kp = P° ¢
(PSO3)2
(PSO2)2PO2
≤
eq
= P° ¢
([SO3]RT)2
([SO2]RT)2[O2]RT
≤
=
eq
[SO3]2
P°
¢
≤
RT [SO2]2[O2] eq
(15.14)
As expected the final expression is dimensionless. Because we have designated that the reference state is based on c° = 1 mol L-1, we can attach the
subscript “c” to the equilibrium constant, Kc, which is the ratio of concentrations in the final parenthetical term of expression (15.14). Finally, observe that
(15.14) has a different value than (15.13). This result is not wrong. It simply
stems from a difference in the reference state in the two cases. In working with
equilibrium constant expressions, therefore, it is important to state the experimental conditions and the reference state clearly. Choosing and defining a
reference state is an important concept that we will see again in our discussion
of thermodynamics in Chapter 19.
Eliminating the reference-state terms having a value of 1, the relationship
between Kp and Kc for reaction (15.11) is
Kp = Kc1RT2-1
(15.15)
If we carried out a similar derivation for the general reaction,
aA(g) + bB(g) + Á ∆ gG(g) + hH(g) + Á
the result would be
▲
KEEP IN MIND
that Kp = Kc only if ¢ngas = 0.
That is, Kp = Kc1RT20 = Kc
because any number raised to
the zero power equals one.
Kp = Kc1RT2¢ngas
(15.16)
where ¢ngas is the difference in the stoichiometric coefficients of gaseous products and reactants; that is, ¢ngas = 1g + h + Á2 - 1a + b + Á2. In reaction
(15.11), ¢ngas = 2 - 12 + 12 = -1, which is what we used in equation (15.15).
In summary, although we do not use units for the equilibrium constants, we
do need to use the correct dimensionless values of concentrations or pressures
for terms within equilibrium constant expressions. These are molarity for Kc
expressions and pressures in atmospheres for Kp expressions. These choices then
require that we use a value of R = 0.08206 L atm mol -1 K -1 in equation (15.16).
EXAMPLE 15-3
Illustrating the Dependence of K on the Reference State. Complete the calculation of Kp for reaction (15.11) knowing that Kc = 2.8 * 102 (at 1000 K).
Solution
Write the equation relating the two equilibrium constants with
different reference states.
Kc = RT * Kp
Rearrange the expression to obtain the quantity desired, Kp .
Kp =
Kc
RT
Substitute the given data and solve.
Kp =
2.8 * 102
= 3.4
0.08206 * 1000
Practice Example A: For the reaction 2 NH 3(g) ∆ N2(g) + 3 H 2(g) at 298 K, Kc = 2.8 * 10-9. What is the value
of Kp for this reaction?
Practice Example B: At 1065 °C, for the reaction 2 H 2S(g) ∆ 2 H 2(g) + S 2(g), Kp = 1.2 * 10-2. What is the
value of Kc for the reaction H 2(g) +
1
S (g) ∆ H 2S(g) at 1065 °C?
2 2
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Relationships Involving Equilibrium Constants
633
EQUILIBRIA INVOLVING PURE LIQUIDS AND SOLIDS
Up to this point in the chapter, all our examples have involved gaseous reactions. In subsequent chapters, we will emphasize equilibria in aqueous solutions. Gas-phase reactions and reactions in aqueous solution are homogeneous
reactions: They occur within a single phase. Let’s extend our coverage now
to include reactions involving one or more condensed phases—solids and
liquids—in contact with a gas or solution phase. These are called heterogeneous
reactions. One of the most important ideas about heterogeneous reactions is that
Equilibrium constant expressions do not contain concentration terms for solid or
liquid phases of a single component (that is, for pure solids and liquids).
C(s) + H 2O(g) ∆ CO(g) + H 2(g)
Although solid carbon must be present for the reaction to occur, the equilibrium constant expression contains terms only for the species in the homogeneous gas phase: H 2O, CO, and H 2 .
Kc =
aCOaH2
aC(s)aH2O
=
[CO][H 2]
[H 2O]
The activity of solid carbon is aC(s) = 1, and we have implicitly divided through
each of the remaining concentrations by the reference-state concentration, c 0 = 1
mol L-1, to obtain a dimensionless Kc .
The decomposition of calcium carbonate (limestone) is also a heterogeneous
reaction. The equilibrium constant expression contains just a single term.
CaCO3(s) ∆ CaO(s) + CO2(g)
(15.17)
Kc = [CO2]
We can write Kp for reaction (15.17) by using equation (15.16), with ¢ngas = 1.
Kp = PCO2
and
(15.18)
Kp = Kc(RT)
▲
Equation (15.18) indicates that the equilibrium pressure of CO2(g) in contact
with CaCO3(s) and CaO(s) is a constant equal to Kp . Its value is independent of
the quantities of CaCO3 and CaO (as long as both solids are present). Figure 15-4
offers a conceptualization of this decomposition reaction.
Still another way to think
about solids and liquids is
through their densities.
Density, the mass per unit
volume of a substance, can be
expressed in moles per liter
by converting the unit
volume from milliliter to liter
and dividing the mass in
grams by the molar mass.
The resultant molar density
(mol/L) is a concentration
term and, at a fixed
temperature, is a constant
that would be incorporated in
the Kc value.
▲
We can think about this statement in either of two ways: (1) An equilibrium
constant expression includes terms only for reactants and products whose concentrations and/or partial pressures can change during a chemical reaction. The
concentration of the single component within a pure solid or liquid phase
cannot change. (2) Alternatively, recall that the activities of pure liquids and
solids are set equal to 1; thus the effect on the numerical value of the equilibrium constant is the same as not including terms for pure solids and liquids at all.
The water–gas reaction, used to make combustible gases from coal, has reacting species in both gaseous and solid phases.
Of course, the relationship
between concentration and
pressure for gases is based here
on the ideal gas law: PV = ngasRT, or
[gas] = (ngas>V) = P>RT.
Drop two pieces of limestone, one
large and the other small, into a
dilute solution of HCl, seal the
beakers, and leave them until next
class. Some carbon dioxide will be
seen bubbling off. By the next class,
the reaction will have stopped. Point
out that the pressure of CO2 in both
beakers is the same, but one piece of
limestone is still bigger than the other.
That is, the concentration of CO2 does
not depend upon the amount of solid
present. An improvement of the
experiment would be to attach the two
closed vessels to a U-tube manometer
to show the pressures are equal. Relate
this to Figure 15-4.
FIGURE 15-4
Equilibrium in the reaction
CaCO3(s) ∆ CaO(s) CO2(g)
(a)
(b)
(a) Decomposition of CaCO 3(s) upon
heating in a closed vessel yields a few
granules of CaO(s), together with CO 2(g),
which soon exerts its equilibrium partial
pressure. (b) Introduction of additional
CaCO 3(s) and/or more CaO(s) has no
effect on the partial pressure of the
CO 2(g), which remains the same as in (a).
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One of our examples in Section 15-1 was liquid–vapor equilibrium. This is a
physical equilibrium because no chemical reactions are involved. Consider the
liquid–vapor equilibrium for water.
H 2O(l) ∆ H 2O(g)
Kc = [H 2O(g)]
Kp = PH2O
Kp = KcRT
So, equilibrium vapor pressures such as PH2O are just values of Kp . As we have
seen before, these values do not depend on the quantities of liquid or vapor at
equilibrium, as long as some of each is present.
EXAMPLE 15-4
Writing Equilibrium Constant Expressions for Reactions Involving Pure Solids or Liquids. At equilibrium in the
following reaction at 60 °C, the partial pressures of the gases are found to be PHI = 3.65 * 10-3 atm and
PH2S = 9.96 * 10-1 atm. What is the value of Kp for the reaction?
H 2S(g) + I 2(s) ∆ 2 HI(g) + S(s)
Kp = ?
Solution
Write the equilibrium constant expression in terms of
activities. Note that activities for the iodine and sulfur
are not included, since the activity of a pure solid is 1.
K =
Substitute partial pressures for the activities into the
equilibrium constant expression.
Kp =
Substitute the given equilibrium data into the equilibrium constant expression.
Kp =
(aHI)2
(aH2S)
(PHI)2
(PH2S)
13.65 * 10-32
2
9.96 * 10-1
= 1.34 * 10-5
Practice Example A: Teeth are made principally from the mineral hydroxyapatite, Ca 5(PO4)3OH, which can
be dissolved in acidic solution such as that produced by bacteria in the mouth. The reaction that occurs is
Ca 5(PO 4)3OH(s) + 4 H +(aq) ∆ 5 Ca2+(aq) + 3 HPO4 2-(aq) + H 2O(l). Write the equilibrium constant expression
Kc for this reaction.
Practice Example B: The steam–iron process is used to generate H 2(g), mostly for use in hydrogenating oils. Iron
metal and steam [H 2O(g)] react to produce Fe3O4(s) and H 2(g). Write expressions for Kc and Kp for this reversible
reaction. How are the values of Kc and Kp related to each other? Explain.
15-4
Point out the huge range of
values that equilibrium
constants can take.
THE MAGNITUDE OF AN EQUILIBRIUM CONSTANT
In principle, every chemical reaction has an equilibrium constant, but often
the constants are not used. Why is this so? Table 15.3 lists equilibrium constants for several reactions mentioned in this chapter or previously in the text.
The first of these reactions is the synthesis of H 2O from its elements. We have
always assumed that this reaction goes to completion, that is, that the reverse
reaction is negligible and the overall reaction proceeds only in the forward direction. If a reaction goes to completion, one (or more) of the reactants is used
up. A term in the denominator of the equilibrium constant expression approaches zero and makes the value of the equilibrium constant very large. A
very large numerical value of K signifies that the forward reaction, as written,
goes to completion or very nearly so. Because the value of Kp for the water synthesis reaction is 1.4 * 1083, we are entirely justified in saying that the reaction
goes to completion at 298 K.
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15-4
TABLE 15.3
The Magnitude of an Equilibrium Constant
Equilibrium Constants of Some Common Reactions
Reaction
Equilibrium constant, Kp
2 H 2(g) + O2(g) ∆ 2 H 2O(l)
1.4 * 1083 at 298 K
CaCO3(s) ∆ CaO(s) + CO2(g)
1.9 * 10-23 at 298 K
1.0 at about 1200 K
2 SO2(g) + O2(g) ∆ 2 SO3(g)
3.4 at 1000 K
C(s) + H 2O(g) ∆ CO(g) + H 2(g)
1.6 * 10-21 at 298 K
10.0 at about 1100 K
If the equilibrium constant is so large, why is a mixture of hydrogen and
oxygen gases stable at room temperature? The value of the equilibrium constant relates to thermodynamic stability: H 2O(l) is much more thermodynamically stable than a mixture of H 2(g) and O2(g) because it lies at a lower energy
state. As noted in Chapter 14, however, the rate of a chemical reaction is
strongly governed by the activation energy, Ea . Because Ea is very high for the
synthesis of H 2O(l) from H 2(g) and O2(g), the rate of reaction is inconsequential at 298 K. To get the reaction to occur at a measurable rate, we must either
raise the temperature or use a catalyst. A chemist would say that the synthesis
of H 2O(l) at 298 K is a kinetically controlled reaction (as opposed to
thermodynamically controlled).
From Table 15.3, we see that Kp for the decomposition of CaCO3(s) (limestone) is very small at 298 K (only 1.9 * 10-23). To account for a very small numerical value of an equilibrium constant, the numerator must be very small
(approaching zero). A very small numerical value of K signifies that the forward reaction, as written, does not occur to any significant extent. Although limestone does not decompose at ordinary temperatures, the partial pressure of
CO2(g) in equilibrium with CaCO3(s) and CaO(s) increases with temperature.
It becomes 1 atm at about 1200 K. An important application of this decomposition reaction is in the commercial production of quicklime (CaO).
The conversion of SO2(g) and O2(g) to SO3(g) at 1000 K has an equilibrium
constant such that we expect significant amounts of both reactants and products to be present at equilibrium (see Table 15.3). Both the forward and reverse
reactions are important. A similar situation exists for the reaction of C(s) and
H 2O(g) at 1100 K, but not at 298 K where the forward reaction does not occur
to any significant extent 1Kp = 1.6 * 10-212.
In light of the several cases from Table 15.3, we can conclude that
A reaction is most likely to reach a state of equilibrium in which significant quantities of both reactants and products are present if the numerical value of K is neither
very large nor very small, as a very rough approximation, in the range of about 10-10
to 1010.
Thus, we see that equilibrium calculations are not required for all reactions.
At times, we can use simple stoichiometric calculations to determine the outcome of a reaction, and in some cases there may be no reaction at all.
CONCEPT ASSESSMENT
✓
Why is having a balanced equation a necessary condition for predicting the outcome
of a chemical reaction, but often not a sufficient condition?
635
Equilibrium Constant
activity
The products formed in
a reaction under kinetic
control are determined by
reaction rates. The products
formed in a reaction under
thermodynamic control will depend
on the stability of the products.
Equilibrium constants
usually vary in value to a
large extent with
temperature.
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15-5
THE REACTION QUOTIENT, Q: PREDICTING THE
DIRECTION OF NET CHANGE
Let’s return briefly to the set of three experiments that we discussed in Section
15-2, involving the reaction
CO(g) + 2 H 2(g) ∆ CH 3OH(g)
Kc = 14.5
Experiment 1 starts with just the reactants, CO and H 2 . An overall, or net,
change has to occur in which some CH 3OH forms. Only in this way can an
equilibrium condition be reached in which all reacting species are present. We
say that a net change occurs in the forward direction (to the right).
Experiment 2 starts with just the product, CH 3OH. Here, some of the
CH 3OH must decompose back to CO and H 2 before equilibrium can be established. We say that a net change occurs in the reverse direction (to the left).
Experiment 3 starts with all the reacting species present: CO, H 2 , and
CH 3OH. In this system, it is not obvious in what direction a net change occurs
to establish equilibrium.
The ability to predict the direction of net change in establishing equilibrium
is important for two reasons.
• At times we do not need detailed equilibrium calculations. We may need
only a qualitative description of the changes that occur in establishing
equilibrium from a given set of initial conditions.
• In some equilibrium calculations, it is helpful to determine the direction
of net change as a first step.
For any set of initial activities in a reaction mixture, we can set up a ratio of
activities having the same form as the equilibrium constant expression. This
ratio is called the reaction quotient and is designated Q. For a hypothetical
generalized reaction, the reaction quotient, first written in terms of activities,
and then as concentrations assuming a concentration reference state, is
g
Q =
g
(ainit) (ainit)h Á
(ainit)a(ainit)b Á
Qc =
[G]init[H]hinit Á
[A]ainit[B]binit Á
(15.19)
If Q = K, a reaction is at equilibrium, but our primary interest in the relationship between Q and K is for a reaction mixture that is not at equilibrium. To
see what this relationship is, let’s turn again to the experiments in Table 15.1.
In Experiment 1, the initial concentrations of CO and H 2 are
1.000 mol>10.0 L = 0.100 M. Initially there is no CH 3OH. The value of Qc is
Qc =
[CH 3OH]init
[CO]init[H 2]2init
0
=
(0.100)(0.100)2
= 0
(15.20)
We know that a net reaction occurs to the right, producing some CH 3OH.
As it does, the numerator in expression (15.20) increases, the denominator
decreases, and the value of Qc increases; eventually Qc = Kc .
If Qc 6 Kc , a net change occurs from left to right (the direction of the forward
reaction).
In Experiment 2, the initial concentration of CH 3OH is 1.000 mol>10.0 L =
0.100 M. Initially, there is no CO or H 2 . The value of Qc is
Qc =
[CH 3OH]init
[CO]init[H 2]2init
=
0.100
= q
0 * 0
(15.21)
We know that a net reaction occurs to the left, producing some CO and H 2 . As
it does, the numerator in expression (15.21) decreases, the denominator
increases, and the value of Qc decreases; eventually Qc = Kc .
If Qc 7 Kc , a net change occurs from right to left (the direction of the reverse
reaction).
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15-5
Initial condition:
(a)
Pure
reactants
The Reaction Quotient, Q: Predicting the Direction of Net Change
(b)
“Left” of
equilibrium
(c)
Equilibrium
(d)
“Right” of
equilibrium
(e)
Pure
products
Kc
Kc
Kc
Reaction quotient, Qc 0
Reaction proceeds
to the right
637
to the left
▲ FIGURE 15-5
Predicting the direction of net change in a reversible reaction
Five possibilities for the relationship of initial and equilibrium conditions are shown.
From Table 15.1 and Figure 15-3, Experiment 1 corresponds to initial condition (a);
Experiment 2 to condition (e); and Experiment 3 to (d). The situation in Example 15-5
also corresponds to condition (d).
Now let us turn to a case where direction of net change is not immediately
obvious. In Experiment 3, the initial concentrations of all three species are
1.000 mol>10.0 L = 0.100 M. The value of Qc is
Qc =
[CH 3OH]init
0.100
[CO]init[H 2]2init
=
(0.100)(0.100)2
= 100
Because Qc 7 Kc (100 compared with 14.5), a net change occurs in the reverse
direction. Note that you can verify this conclusion from Figure 15-3. The
amounts of CO and H 2 at equilibrium are greater than they were initially, and
the amount of CH 3OH is less.
The criteria for predicting the direction of a net chemical change in a reversible reaction are summarized in Figure 15-5 and applied in Example 15-5.
EXAMPLE 15-5
Predicting the Direction of a Net Chemical Change in Establishing Equilibrium. To increase the yield of H 2(g) in
the water–gas reaction—the reaction of C(s) and H 2O(g) to form CO(g) and H 2(g)—a follow-up reaction called the
“water–gas shift reaction” is generally used. In this reaction, some of the CO(g) of the water gas is replaced by H 2(g).
CO(g) + H 2O(g) ∆ CO2(g) + H 2(g)
Kc = 1.00 at about 1100 K. The following amounts of substances are brought together and allowed to react at this
temperature: 1.00 mol CO, 1.00 mol H 2O, 2.00 mol CO2 , and 2.00 mol H 2 . Compared with their initial amounts, which
of the substances will be present in a greater amount and which in a lesser amount when equilibrium is established?
Solution
[CO2][H 2]
[CO][H 2O]
Our task is to determine the direction of net change by evaluating
Qc . Write down the expression for Qc .
Qc =
Substitute concentrations into the expression for Qc , by assuming
an arbitrary volume V (which cancels out in the calculation).
Qc =
Compare Qc to Kc.
4.00 7 1.00
12.00>V212.00>V2
11.00>V211.00>V2
= 4.00
Because Qc 7 Kc (that is, 4.00 7 1.00), a net change occurs to the
left. When equilibrium is established, the amounts of CO and H 2O
will be greater than the initial quantities and the amounts of CO2
and H 2 will be less.
Practice Example A: In Example 15-5, equal masses of CO, H 2O, CO2 , and H 2 are mixed at a temperature of
about 1100 K. When equilibrium is established, which substance(s) will show an increase in quantity and which will
show a decrease compared with the initial quantities?
Practice Example B: For the reaction PCl5(g) ∆ PCl3(g) + Cl2(g), Kc = 0.0454 at 261 °C. If a vessel is filled
with these gases such that the initial partial pressures are PPCl3 = 2.19 atm, PCl2 = 0.88 atm, PPCl5 = 19.7 atm, in which
direction will a net change occur?
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▲
KEEP IN MIND
2:32 PM
that volume terms cancel in
a reaction quotient or
equilibrium constant
expression whenever the sum
of the exponents in the
numerator equals that in the
denominator. This can
simplify problem solving
in some instances.
CONCEPT ASSESSMENT
A mixture of 1.00 mol each of CO(g), H 2O(g), and CO2(g) is placed in a 10.0-L flask at
a temperature at which Kp = 10.0 in the reaction
CO(g) + H 2O(g) ∆ CO2(g) + H 2(g)
When equilibrium is established, (a) the amount of H 2(g) will be 1.00 mol; (b) the
amounts of all reactants and products will be greater than 1.00 mol; (c) the amounts
of all reactants and products will be less than 1.00 mol; (d) the amount of CO2(g) will
be greater than 1.00 mol and the amounts of CO(g), H 2O(g), and H 2(g) will be less
than 1.00 mol; (e) the amounts of reactants and products cannot be predicted and can
only be determined by analyzing the equilibrium mixture.
15-6
A gas at equilibrium is a
good example of Le
Châtelier’s principle. Use
P = [gas]RT and imagine
changing any one of P, [gas], or T.
The ideal gas law must be satisfied,
so each time one parameter is
changed, the others must readjust.
Huddle, Benjamin P.
“Conceptual Questions” on Le
Châtelier’s Principle. J. Chem. Educ.
2000: 75, 1175 (September 1998).
Thomsen, Volker B. E. “Le
Châtelier’s Principle in the
Sciences.” J. Chem. Educ. 2000: 77,
173 (February 2000).
✓
ALTERING EQUILIBRIUM CONDITIONS:
LE CHÂTELIER’S PRINCIPLE
At times, we want only to make qualitative statements about a reversible reaction: the direction of a net change, whether the amount of a substance will
have increased or decreased when equilibrium is reached, and so on. Also, we
may not have the data needed for a quantitative calculation. In these cases, we
can use a statement attributed to the French chemist Henri Le Châtelier (1884).
Le Châtelier’s principle is hard to state unambiguously, but its essential
meaning is stated here.
When an equilibrium system is subjected to a change in temperature, pressure,
or concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.
As we will see in the examples that follow, it is generally not difficult to predict the outcome of changing one or more variables in a system at equilibrium.
EFFECT OF CHANGING THE
ON EQUILIBRIUM
AMOUNTS OF REACTING SPECIES
Let’s return to reaction (15.11)
2 SO2(g) + O2(g) ∆ 2 SO3(g)
▲
2 SO2(g) O2(g) ∆ 2 SO3(g)
Note that there is a time
delay between the
application of a stress and
the establishment of equilibrium.
This is similar to Figure 15-3.
Imagine if the applied stress is very
small. In this case equilibrium is
always maintained because the
change is too small to notice. Use
this to introduce the concept of
reversible processes.
Kc = 2.8 * 102 at 1000 K
Suppose we start with certain equilibrium amounts of SO2 , O2 , and SO3 , as
suggested by Figure 15-6(a). Now let’s create a disturbance in the equilibrium
mixture by forcing an additional 1.00 mol SO3 into the 10.0-L flask (Fig. 15-6b).
How will the amounts of the reacting species change to reestablish equilibrium?
According to Le Châtelier’s principle, if the system is to partially offset an
action that increases the equilibrium concentration of one of the reacting
species, it must do so by favoring the reaction in which that species is consumed. In this case, this is the reverse reaction—conversion of some of the
added SO3 to SO2 and O2 . In the new equilibrium, there are greater amounts
of all the substances than in the original equilibrium, but the additional
amount of SO3 is less than the 1.00 mol that was added.
Another way to look at the matter is to evaluate the reaction quotient immediately after adding the SO3 .
Original equilibrium
Qc =
[SO3]
2
[SO2] [O2]
= Kc
Immediately following disturbance
Qc =
[SO3]
[SO2]2[O2]
7 Kc
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639
Add:
1.00 mol SO3
(b)
1.46 mol SO3
0.68 mol SO3
0.32 mol SO2
0.16 mol O2
10.0 L
10.0 L
(a)
(c)
0.54 mol SO2
0.27 mol O2
▲ FIGURE 15-6
Changing equilibrium conditions by changing the amount of a reactant
2 SO2(g) O2(g) ∆ 2 SO3(g), Kc 2.8 : 102 at 1000 K
(a) The original equilibrium condition. (b) Disturbance caused by adding 1.00 mol
SO 3 . (c) The new equilibrium condition. The amount of SO 3 in the new equilibrium
mixture, 1.46 mol, is greater than the original 0.68 mol but it is not as great as
immediately after the addition of 1.00 mol SO 3 . The effect of adding SO 3 to an
equilibrium mixture is partially offset when equilibrium is restored.
Adding any quantity of SO3 to a constant-volume equilibrium mixture
makes Qc larger than Kc . A net change occurs in the direction that reduces
[SO3], that is, to the left, or in the reverse direction. Notice that reaction
in the reverse direction increases [SO2] and [O2], further decreasing the
value of Qc .
EXAMPLE 15-6
Applying Le Châtelier’s Principle: Effect of Adding More of a Reactant to an
Equilibrium Mixture. Predict the effect of adding more H 2(g) to a constant-volume
equilibrium mixture of N2 , H 2 , and NH 3 .
N2(g) + 3 H 2(g) ∆ 2 NH 3(g)
Solution
Increasing [H 2] stimulates the forward reaction and a shift in the equilibrium condition to the right. However, only a portion of the added H 2 is consumed in this reaction. When equilibrium is reestablished, there will be more H 2 than was present
originally, and also more NH 3 , but the amount of N2 will be smaller. Some of the
original N2 must be consumed in converting some of the added H 2 to NH 3 .
Practice Example A: Given the reaction 2 CO(g) + O2(g) ∆ 2 CO2(g), what
is the effect of adding O2(g) to a constant-volume equilibrium mixture?
Practice Example B: Calcination of limestone (decomposition by heating),
CaCO3(s) ∆ CaO(s) + CO2(g), is the commercial source of quicklime, CaO(s).
After this equilibrium has been established in a constant-temperature, constantvolume container, what is the effect on the equilibrium amounts of materials caused
by adding some (a) CaO(s); (b) CO2(g); (c) CaCO3(s)?
Le Châtelier’s Principle
movie
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Principles of Chemical Equilibrium
EFFECT OF CHANGES IN PRESSURE OR VOLUME ON EQUILIBRIUM
There are three ways to change the pressure of a constant-temperature equilibrium mixture.
1. Add or remove a gaseous reactant or product. The effect of these actions
on the equilibrium condition is simply that due to adding or removing a
reaction component, as described previously.
2. Add an inert gas to the constant-volume reaction mixture. This has the
effect of increasing the total pressure, but the partial pressures of the reacting species are all unchanged. An inert gas added to a constant-volume
equilibrium mixture has no effect on the equilibrium condition.
3. Change the pressure by changing the volume of the system. Decreasing
the volume of the system increases the pressure, and increasing the system
volume decreases the pressure. Thus, the effect of this type of pressure
change is simply that of a volume change.
Let’s explore the third situation first. Consider, again, the formation of
SO3(g) from SO2(g) and O2(g).
Kc = 2.8 * 102 at 1000 K
2 SO2(g) + O2(g) ∆ 2 SO3(g)
The equilibrium mixture in Figure 15-7(a) has its volume reduced to one-tenth
of its original value by increasing the external pressure. To see how the equilibrium amounts of the gases change, let’s first rearrange the equilibrium constant expression to the form
Kc =
[SO 3]2
[SO 2]2[O2]
=
(nSO3>V)2
(nSO2>V)2(nO2>V)
=
(nSO3)2
(nSO2)2(nO2)
* V = 2.8 * 102 (15.22)
From equation (15.22), we see that if V is reduced by a factor of 10, the ratio
(nSO3)2
(nSO2)2(nO2)
0.68 mol SO3
0.32 mol SO2
0.16 mol O2
0.83 mol SO3
10.0 L
1.00 L
0.17 mol SO2
(a)
(b)
0.085 mol O2
▲ FIGURE 15-7
Effect of pressure change on equilibrium in the reaction
2 SO2(g) O2(g) ∆ 2 SO3(g)
An increase in external pressure causes a decrease in the reaction volume and a
shift in equilibrium “to the right.” (See Exercise 65 for a calculation of the new
equilibrium amounts.)
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641
must increase by a factor of 10. In this way, the value of Kc is restored, as it must
be to restore equilibrium. There is only one way in which the ratio of moles of
gases will increase in value: The number of moles of SO3 must increase, and the
numbers of moles of SO2 and O2 must decrease. The equilibrium shifts in
the direction producing more SO3—to the right.
Notice that three moles of gas on the left produce two moles of gas on the
right in reaction (15.11). When compared at the same temperature and pressure, two moles of SO3(g) occupy a smaller volume than does a mixture of two
moles of SO2(g) and one mole of O2(g). Given this fact and the observation
from equation (15.22) that a decrease in volume favors the production of additional SO3 , we can formulate a statement that is especially easy to apply.
NO2 ¬ N2O4
Equilibrium animation
Figure 15-7 suggests a way of decreasing the volume of gaseous mixture at
equilibrium—by increasing the external pressure. One way to increase the volume is to lower the external pressure. Another way is to transfer the equilibrium mixture from its original container to one of larger volume. A third
method is to add an inert gas at constant pressure; the volume of the mixture
must increase to make room for the added gas. The effect on the equilibrium,
however, is the same for all three methods: Equilibrium shifts in the direction
of the reaction producing the greater number of moles of gas.
Equilibria between condensed phases are not affected much by changes in
external pressure because solids and liquids are not easily compressible. Also,
we cannot assess whether the forward or reverse reaction is favored by these
changes by examining only the chemical equation.
▲
When the volume of an equilibrium mixture of gases is reduced, a net change occurs
in the direction that produces fewer moles of gas. When the volume is increased, a net
change occurs in the direction that produces more moles of gas.
KEEP IN MIND
that an inert gas has no effect
on an equilibrium condition
if the gas is added to a system
maintained at constant volume, but it can have an effect
if added at constant pressure.
EXAMPLE 15-7
Applying Le Châtelier’s Principle: The Effect of Changing Volume. An equilibrium
mixture of N2(g), H 2(g), and NH 3(g) is transferred from a 1.50-L flask to a 5.00-L flask.
In which direction does a net change occur to restore equilibrium?
N2(g) + 3 H 2(g) ∆ 2 NH 3(g)
Solution
When the gaseous mixture is transferred to the larger flask, the partial pressure of
each gas and the total pressure drop. Whether we think in terms of a decrease in
pressure or an increase in volume, we reach the same conclusion. Equilibrium shifts
in such a way as to produce a larger number of moles of gas. Some of the NH 3 originally present decomposes back to N2 and H 2 . A net change occurs in the direction
of the reverse reaction—to the left—in restoring equilibrium.
Practice Example A: The reaction N2O4(g) ∆ 2 NO2(g) is at equilibrium in
a 3.00-L cylinder. What would be the effect on the concentrations of N2O4(g) and
NO2(g) if the pressure were doubled (that is, cylinder volume decreased to 1.50 L)?
Practice Example B: How is the equilibrium amount of H 2(g) produced in the
water–gas shift reaction affected by changing the total gas pressure or the system
volume? Explain.
CO(g) + H 2O(g) ∆ CO2(g) + H 2(g)
Remind students that the
pressure of a particular gas
inside a container remains
unchanged even if an inert gas is
added to the container as long as
the volume of the container is not
changed. Review Dalton's Law of
Partial Pressures.
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Principles of Chemical Equilibrium
CONCEPT ASSESSMENT
✓
The following reaction is brought to equilibrium at 700 °C.
2 H 2S(g) + CH 4(g) ∆ CS 2(g) + 4 H 2(g)
Indicate whether each of the following statements is true, false, or not possible to
evaluate from the information given.
(a) If the equilibrium mixture is allowed to expand into an evacuated larger container, the mole fraction of H 2 will increase.
(b) If several moles of Ar(g) are forced into the reaction container, the amounts of
H 2S and CH 4 will increase.
(c) If the equilibrium mixture is cooled to 100 °C, the mole fractions of the four
gases will likely change.
(d) If the equilibrium mixture is forced into a slightly smaller container, the partial
pressures of the four gases will all increase.
EFFECT OF TEMPERATURE ON EQUILIBRIUM
Temperature Dependence
of Equilibrium animation
We can think of changing the temperature of an equilibrium mixture in terms
of adding heat (raising the temperature) or removing heat (lowering the temperature). According to Le Châtelier’s principle, adding heat favors the reaction in which heat is absorbed (endothermic reaction). Removing heat favors
the reaction in which heat is evolved (exothermic reaction). Stated in terms of
changing temperature,
Raising the temperature of an equilibrium mixture shifts the equilibrium condition in
the direction of the endothermic reaction. Lowering the temperature causes a shift in the
direction of the exothermic reaction.
Note that an exothermic
reaction is aided by
lowering the temperature
and an endothermic reaction is
aided by raising the temperature.
You can then point out the
difference between the system and
the surroundings.
Remind students not to
confuse shifts in
equilibrium with changes
in reaction rates that result from
temperature changes. Equilibria
of exothermic and endothermic
reactions will shift differently
when temperatures are increased,
but the rates of exothermic and
endothermic reactions both increase
with increasing temperature.
Stress to students that
changing the temperature
of a reaction will result in
a shift of the equilibrium and a
new equilibrium constant.
The principal effect of temperature on equilibrium is in changing the value
of the equilibrium constant. In Chapter 19, we will learn how to calculate equilibrium constants as a function of temperature. For now, we will limit ourselves to making qualitative predictions.
EXAMPLE 15-8
Applying Le Châtelier’s Principle: Effect of Temperature on Equilibrium.
Consider the reaction
2 SO2(g) + O2(g) ∆ 2 SO 3(g)
¢H° = -197.8 kJ
Will the amount of SO3(g) formed from given amounts of SO2(g) and O2(g) be
greater at high or low temperatures?
Solution
Raising the temperature favors the endothermic reaction, the reverse reaction. Lowering the temperature favors the forward (exothermic) reaction. Therefore, an equilibrium mixture would have a higher concentration of SO3 at lower temperatures.
The conversion of SO2 to SO3 is favored at low temperatures.
Practice Example A: The reaction N2O4(g) ∆ 2 NO2(g) has ¢H° = +57.2 kJ.
Will the amount of NO2(g) formed from N2O4(g) be greater at high or low
temperatures?
Practice Example B: The enthalpy of formation of NH 3 is ¢H f°[NH 3(g)] =
-46.11 kJ>mol NH 3 . Will the concentration of NH 3 in an equilibrium mixture with
its elements be greater at 100 or at 300 °C? Explain.
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Altering Equilibrium Conditions: Le Châtelier’s Principle
▲ Sulfuric acid is produced from SO3
SO3(g) H2O(l) ∆ H2SO4(aq)
The catalyst used to speed up the conversion of SO2 to SO3 in the commercial
production of sulfuric acid is V2O5(s).
EFFECT OF A CATALYST ON EQUILIBRIUM
Adding a catalyst to a reaction mixture speeds up both the forward and reverse reactions. Equilibrium is achieved more rapidly, but the equilibrium
amounts are unchanged by the catalyst. Consider again reaction (15.11)
2 SO2(g) + O2(g) ∆ 2 SO3(g)
Kc = 2.8 * 102 at 1000 K
For a given set of reaction conditions, the equilibrium amounts of SO2 , O2 , and
SO3 have fixed values. This is true whether the reaction is carried out by a slow
homogeneous reaction, catalyzed in the gas phase, or conducted as a heterogeneous reaction on the surface of a catalyst. Stated another way, the presence of a
catalyst does not change the numerical value of the equilibrium constant.
We now have two thoughts about a catalyst to reconcile: one from the preceding chapter and one from this discussion.
• A catalyst changes the mechanism of a reaction to one with a lower acti-
vation energy.
• A catalyst has no effect on the condition of equilibrium in a reversible
reaction.
Taken together, these two statements must mean that an equilibrium condition is independent of the reaction mechanism. Thus, even though equilibrium
has been described in terms of opposing reactions occurring at equal rates, we
do not have to be concerned with the kinetics of chemical reactions to work
with the equilibrium concept. This observation is still another indication that
the equilibrium constant is a thermodynamic quantity, as we shall describe
more fully in Chapter 19.
CONCEPT ASSESSMENT
✓
Two students are performing the same experiment in which an endothermic reaction
rapidly attains a condition of equilibrium. Student A does the reaction in a beaker
resting on the surface of the lab bench while student B holds the beaker in which the
reaction occurs. Assuming that all other environmental variables are the same, which
student should end up with more product? Explain.
643
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Principles of Chemical Equilibrium
Good Web sites for
equilibrium calculations:
http://www.chem.vt.edu/
RVGS/ACT/notes/chem-eqm.html
(noncommercial);
http://stokes.lance.colcostate.edu/
tools/equil.html (noncommercial).
Nitrogen Dioxide
and Dinitrogen
Tetroxide movie
Leenson, I. A. “Approaching
Equilibrium in the
N2O4 ¬ NO2 System: A Common
Mistake in Textbooks.” J. Chem.
Educ. 2000: 77, 1652
(December 2000).
15-7
EQUILIBRIUM CALCULATIONS: SOME
ILLUSTRATIVE EXAMPLES
We are now ready to tackle the problem of describing, in quantitative terms,
the condition of equilibrium in a reversible reaction. Part of the approach we
use may seem unfamiliar at first—it has an algebraic look to it. But as you adjust to this “new look,” do not lose sight of the fact that we continue to use
some familiar and important ideas—molar masses, molarities, and stoichiometric factors from the balanced equation, for example.
The five numerical examples that follow apply the general equilibrium
principles described earlier in the chapter. The first four involve gases, while
the fifth deals with equilibrium in an aqueous solution. (The study of equilibria in aqueous solutions is the principal topic of the next three chapters.) Each
example includes a brief section labeled “comments,” which is printed on a
yellow background. Think of the comments as the basic methodology of equilibrium calculations. You may want to refer back to these comments from time
to time while you are studying later chapters.
Example 15-9 is relatively straightforward. It demonstrates how to determine the equilibrium constant of a reaction when the equilibrium concentrations of the reactants and products are known.
EXAMPLE 15-9
Determining a Value of Kc from the Equilibrium Quantities of Substances. Dinitrogen tetroxide, N2O4(l), is an
important component of rocket fuels—for example, as an oxidizer of liquid hydrazine in the Titan rocket. At 25 °C,
N2O4 is a colorless gas that partially dissociates into NO2 , a red-brown gas. The color of an equilibrium mixture of
these two gases depends on their relative proportions, which in turn depends on the temperature (Fig. 15-8).
Equilibrium is established in the reaction N2O4(g) ∆ 2 NO2(g) at 25 °C. The quantities of the two gases present
in a 3.00-L vessel are 7.64 g N2O4 and 1.56 g NO2 . What is the value of Kc for this reaction?
N
N
O
O
O
O
N
N
O
(a)
▲ FIGURE 15-8
O
O
O
(b)
The equilibrium N2O4(g) ∆ 2 NO2(g)
(a) At dry ice temperatures, N2O4 exists as a solid.
The gas in equilibrium with the solid is mostly
colorless N2O4 , with only a trace of brown NO2 .
(b) When warmed to room temperature and above,
the N2O4 melts and vaporizes. The proportion of
NO2(g) at equilibrium increases over that at low
temperatures, and the equilibrium mixture of
N2O4(g) and NO2(g) has a red-brown color.
▲ The Lewis structures of N2O4 and NO2(g)
Nitrogen dioxide is a free radical that combines
exothermically to form dinitrogen tetroxide.
Use these examples to point
out the importance of
significant figures.
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Equilibrium Calculations: Some Illustrative Examples
645
Solution
Convert the mass of N2O4 to moles.
mol N2O4 = 7.64 g N2O4 *
Convert moles of N2O4 to mol>L.
[N2O4] =
Convert the mass of NO2 to moles.
mol NO 2 = 1.56 g NO 2 *
Convert moles of NO 2 to mol>L.
[NO 2] =
Write the equilibrium constant expression, substitute the equilibrium concentrations, and solve
for Kc .
Kc =
1 mol N2O4
= 8.303 * 10-2 mol
92.01 g N2O4
8.303 * 102 mol
= 0.0277 M
3.00 L
1 mol NO 2
= 3.391 * 10-2 mol
46.01 g NO 2
3.391 * 10-2
= 0.0113 M
3.00 L
(0.0113)2
[NO2]2
=
= 4.61 * 10-3
[N2O4]
(0.0277)
Practice
Example A: Equilibrium is established in a 3.00-L flask at 1405 K for the reaction
2 H 2S(g) ∆ 2 H 2(g) + S 2(g). At equilibrium, there is 0.11 mol S 2(g), 0.22 mol H 2(g), and 2.78 mol H 2S(g). What is
the value of Kc for this reaction?
Practice Example B: Equilibrium is established at 25 °C in the reaction N2O4(g) ∆ 2 NO2(g), Kc = 4.61 * 10-3.
If [NO 2] = 0.0236 M in a 2.26-L flask, how many grams of N2O4 are also present?
Comments
1. The quantities required in an equilibrium constant expression, Kc , are equilibrium concentrations in moles per
liter, not simply equilibrium amounts in moles or masses in grams. You will find it helpful to organize all the
equilibrium data and carefully label each item.
Example 15-10 is somewhat more involved than Example 15-9. We are still
interested in determining the equilibrium constant for a reaction, but we do not
have the same sort of information as in Example 15-9. We are given the initial
concentrations of all the reactants and products, but the equilibrium concentration of only one substance. This case requires a little algebra and some careful
bookkeeping. We will introduce a tabular system, sometimes called an ICE table,
for keeping track of changing concentrations of reactants and products. The table
contains the initial, change in, and equilibrium concentration of each species. It is a
helpful device that we will use throughout the next three chapters.
EXAMPLE 15-10
Determining a Value of Kp from Initial and Equilibrium Amounts of Substances:
Relating Kc and Kp The equilibrium condition for SO2(g), O2(g), and SO3(g) is important in sulfuric acid production. When a 0.0200-mol sample of SO3 is introduced
into an evacuated 1.52-L vessel at 900 K, 0.0142 mol SO3 is present at equilibrium.
What is the value of Kp for the dissociation of SO3(g) at 900 K?
2 SO3(g) ∆ 2 SO 2(g) + O2(g)
Kp = ?
Solution
Let’s first determine Kc and then convert to Kp by using equation (15.16). In the ICE
table on page 646, the key term leading to the other data is the change in amount of
SO3 : In progressing from 0.0200 mol SO3 to 0.0142 mol SO3 , 0.0058 mol SO3 is dissociated. The negative sign (-0.0058 mol) indicates that this amount of SO3 is consumed
in establishing equilibrium. In the row labeled “changes,” the changes in amounts of
SO2 and O2 must be related to the change in amount of SO3 . For this, we use the stoichiometric coefficients from the balanced equation: 2, 2, and 1. That is, two moles of
SO2 and one mole of O2 are produced for every two moles of SO3 that dissociate.
Silverstein, Todd P.
“Graphing Calculator
Strategies for Solving Chemical
Equilibrium Problems.” J. Chem.
Educ. 2000: 77, 1120.
(September 2000).
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Principles of Chemical Equilibrium
The reaction:
2 SO3(g)
initial amounts:
changes:
equil amounts:
0.0200 mol
-0.0058 mol
0.0142 mol
equil concns:
[SO3] =
2 SO2(g)
∆
O2(g)
0.00 mol
+0.0058 mol
0.0058 mol
0.0142 mol
;
1.52 L
[SO2] =
[SO3] = 9.34 * 10-3 M;
Kc =
0.00 mol
+0.0029 mol
0.0029 mol
0.0058 mol
;
1.52 L
[O2] =
[O2] = 1.9 * 10-3 M
[SO2] = 3.8 * 10-3 M;
[SO2]2[O2]
[SO 3]2
0.0029 mol
1.52 L
13.8 * 10-32 11.9 * 10-32
2
=
19.34 * 10-32
2
= 3.1 * 10-4
Kp = Kc1RT2¢ngas = 3.1 * 10-4 10.0821 * 900212 + 12 - 2
= 3.1 * 10-4 10.0821 * 90021 = 2.3 * 10-2
Practice Example A: A 5.00-L evacuated flask is filled with 1.86 mol NOBr. At
equilibrium at 25 °C, there is 0.082 mol of Br2 present. Determine Kc and Kp for the
reaction 2 NOBr(g) ∆ 2 NO(g) + Br2(g).
Practice Example B: 0.100 mol SO2 and 0.100 mol O2 are introduced into an
evacuated 1.52-L flask at 900 K. When equilibrium is reached, the amount of SO3
found is 0.0916 mol. Use these data to determine Kp for the reaction
2 SO3(g) ∆ 2 SO 2(g) + O2(g).
Comments
2. The chemical equation for a reversible reaction serves both to establish the
form of the equilibrium constant expression and to provide the conversion
factors (stoichiometric factors) to relate the equilibrium quantity of one
species to equilibrium quantities of the others.
3. For equilibria involving gases, you can use either Kc or Kp . In general, if the
data given involve amounts of substances and volumes, it is easier to work
with Kc . If data are given as partial pressures, then work with Kp . Whether
working with Kc or Kp or the relationship between them, you must always
base these expressions on the given chemical equation, not on equations
you may have used in other situations.
The methods used in Examples 15-9 and 15-10 are summarized in Figure
15-9. Example 15-11 demonstrates that we can often determine several pieces
of useful information about an equilibrium system from just the equilibrium
constant and the reaction equation.
EXAMPLE 15-11
Determining Equilibrium Partial and Total Pressures from a Value of Kp . Ammonium hydrogen sulfide,
NH 4HS(s), used as a photographic developer, is unstable and dissociates at room temperature.
NH 4HS(s) ∆ NH 3(g) + H 2S(g)
Kp = 0.108 at 25 °C
A sample of NH 4HS(s) is introduced into an evacuated flask at 25 °C. What is the total gas pressure at equilibrium?
Solution
Write out the equilibrium constant expression in
terms of pressures.
Kp = (PNH3)(PH2S) = 0.108
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Are the experimental data
exclusively equilibrium data?
No
Yes
Convert the data
to concentrations or
partial pressures
Relate changes in initial amounts,
concentrations, or partial pressures to
a known equilibrium amount,
concentration, or partial pressure
Establish all the equilibrium
concentrations or partial
pressures
Substitute concentrations
into Kc or partial pressures
into Kp expressions
▲ FIGURE 15-9
Determining Kc or Kp from experimental data
Kp for this reaction is just the product of the equilibrium partial pressures of NH 3(g) and H 2S(g),
each stated in atmospheres. (There is no term for
NH 4HS because it is a solid.) Because these gases
are produced in equimolar amounts, PNH3 = PH2S.
Kp = (PNH3)(PH2S) = (PNH3)(PNH3) = (PNH3)2 = 0.108
Find PNH3 . (Note that the unit atm appears
because in the equilibrium expression the reference pressure P° was implicitly included.)
PNH3 = 20.108 = 0.329 atm
The total pressure is
Ptot = PNH3 + PH2S = 0.329 atm + 0.329 atm = 0.658 atm
PH2S = PNH3 = 0.329 atm
Practice Example A: Sodium hydrogen carbonate (baking soda) decomposes at elevated temperatures and is one
of the sources of CO2(g) when this compound is used in baking.
2 NaHCO3(s) ∆ Na 2CO3(s) + H 2O(g) + CO2(g)
Kp = 0.231 at 100 °C
What is the partial pressure of CO2(g) when this equilibrium is established starting with NaHCO3(s)?
Practice Example B: If enough additional NH 3(g) is added to the flask in Example 15-11 to raise its partial pressure to 0.500 atm at equilibrium, what will be the total gas pressure when equilibrium is reestablished?
Comments
4. When using Kp expressions, look for relationships among partial pressures of the reactants. If you need to
relate the total pressure to the partial pressures of the reactants, you should be able to do this with some equations presented in Chapter 6 (for example, equations 6.15, 6.16, and 6.17).
Example 15-12 brings back the ICE format, but with a twist. This time the
known values include the equilibrium constant and an initial amount of the reactant, but no information is given about the equilibrium amount of the reactant
or the product. That means that we do not know how much the initial value will
change. We show this by using an “x” in that part of the table. The setup will be
quite algebraic; in fact, we must use the quadratic formula to obtain a solution.
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Principles of Chemical Equilibrium
EXAMPLE 15-12
Calculating Equilibrium Concentrations from Initial Conditions. A 0.0240-mol
sample of N2O4(g) is allowed to come to equilibrium with NO2(g) in a 0.372-L flask
at 25 °C. Calculate the amount of N2O4 present at equilibrium (Fig. 15-10).
N2O4(g) ∆ 2 NO 2(g)
(a)
Kc = 4.61 * 103 at 25 °C
(b)
N2O4
NO2
▲ FIGURE 15-10
Equilibrium in the reaction
N2O4(g) ∆ 2 NO2(g)
at 25 °C—Example 15-12 illustrated
Each “molecule” illustrated represents 0.001 mol. (a) Initially, the bulb contains
0.024 mol N 2O 4 , represented by 24 “molecules.” (b) At equilibrium, some
“molecules” of N 2O 4 have dissociated to NO 2 . The 21 “molecules” of N 2O 4 and 6
of NO 2 correspond to 0.021 mol N 2O 4 and 0.006 mol NO 2 at equilibrium.
Most students can solve a
quadratic equation, but few
of us can solve polynomials
greater than a quadratic. Suggest to
the students that if they get an
equation that is a cubic or higher
degree equation, it is likely that it
will simplify by using an
approximation.
Solution
We need to determine the amount of N2O4 that dissociates to establish equilibrium.
For the first time, we introduce an algebraic unknown, x. Suppose we let x = the
number of moles of N2O4 that dissociate. In the following ICE table, we enter the
value -x into the row labeled “changes.” The amount of NO2 produced is +2x because the stoichiometric coefficient of NO2 is 2 and that of N2O4 is 1.
2 NO2(g)
The reaction:
N2O4(g)
∆
initial amounts:
0.0240 mol
0.00 mol
-x mol
+2x mol
changes:
10.0240 - x2 mol
equil amounts:
2x mol
[N2O4] = 10.0240 - x mol2>0.372 L [NO 2] = 2x mol>0.372 L
equil concns:
2x 2
b
[NO2]
0.372
4x 2
Kc =
=
= 4.61 * 10-3
=
[N2O4]
0.37210.0240 - x2
0.0240 - x
a
b
0.372
2
4x 2 = 4.12 * 10-5 - 11.71 * 10-32x
▲
For a discussion of
quadratic equations,
see Appendix A.
x 2 + 14.28 * 10-42x - 1.03 * 10-5 = 0
-4.28 * 10-4 ; 414.28 * 10-42 + 4 * 1.03 * 10-5
2
2
▲
The symbol ; signifies
that there are two possible
roots. In this problem, x must
be a positive quantity smaller
than 0.0240.
a
x =
=
-4.28 * 10-4 ; 411.83 * 10-72 + 4.12 * 10-5
2
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15-7
x =
Equilibrium Calculations: Some Illustrative Examples
649
-4.28 * 10-4 ; 44.14 * 10-5
2
=
-4.28 * 10-4 ; 6.43 * 10-3
2
=
-4.28 * 10-4 + 6.43 * 10-3
6.00 * 10-3
=
2
2
= 3.00 * 10-3 mol N2O4
The amount of N2O4 at equilibrium is 10.0240 - x2 = 10.0240 - 0.00302 =
0.0210 mol N2O4.
Practice Example A: If 0.150 mol H 2(g) and 0.200 mol I 2(g) are introduced into
a 15.0-L flask at 445 °C and allowed to come to equilibrium, how many moles of
HI(g) will be present?
H 2(g) + I 2(g) ∆ 2 HI(g)
Kc = 50.2 at 445 °C
Practice Example B: Suppose the equilibrium mixture of Example 15-12 is
transferred to a 10.0-L flask. (a) Will the equilibrium amount of N2O4 increase or
decrease? Explain. (b) Calculate the number of moles of N2O4 in the new equilibrium condition.
Determine the direction of
net change by comparing
Q and K
Comments
Our final example is similar to the previous one, but with this slight complication: Initially, we don’t know whether a net change occurs to the right or
to the left to establish equilibrium. We can find out, though, by using the reaction quotient, Qc , and proceeding in the manner suggested in Figure 15-11.
Also, because the reactants and products are in solution, we can work exclusively with concentrations in formulating the Kc expression.
EXAMPLE 15-13
Using the Reaction Quotient, Qc , in an Equilibrium Calculation. Solid silver is
added to a solution with these initial concentrations: [Ag +] = 0.200 M,
[Fe 2+] = 0.100 M, and [Fe 3+] = 0.300 M. The following reversible reaction occurs.
Ag +(aq) + Fe 2+(aq) ∆ Ag(s) + Fe 3+(aq)
Kc = 2.98
What are the ion concentrations when equilibrium is established?
Let x change in amount,
concentration, or partial
pressure of one reactant or
product to reach equilibrium
Relate changes in amounts,
concentrations, or partial
pressures of other reactants
or products to the chosen x
Express equilibrium
amounts, concentrations,
partial pressures
in terms of x
Substitute equilibrium
concentrations or
partial pressures into Kc or Kp
expression; solve for x;
substitute the value of x
into any equation in which
x appeared to find the
desired quantities
▲ FIGURE 15-11
Determining equilibrium
concentrations and partial
pressures
Tabulate these data in an ICE table
5. When you need to introduce an algebraic unknown, x, into an equilibrium
calculation, follow these steps.
• Introduce x into the ICE setup in the row labeled “changes.”
• Decide which change to label as x, that is, the amount of a reactant consumed or of a product formed. Usually, we base this on the species
that has the smallest stoichiometric coefficient in the balanced chemical
equation.
• Use stoichiometric factors to relate the other changes to x (that is, 2x,
3x, Á ).
• Consider that equilibrium amounts = intial amounts + “changes.– (If
you have assigned the correct signs to the changes, equilibrium amounts
will also be correct.)
• After substitutions have been made into the equilibrium constant expression, the equation will often be a quadratic equation in x, which you
can solve by the quadratic formula. Occasionally you may encounter a
higher-degree equation. Appendix A-3 outlines a straightforward method
of dealing with these.
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Principles of Chemical Equilibrium
Solution
Because all reactants and products are present initially, we need to use the reaction
quotient Qc to determine the direction in which a net change occurs.
Qc =
[Fe 3+]
2+
[Ag ][Fe ]
+
=
0.300
= 15.0
(0.200)(0.100)
Because Qc (15.0) is larger than Kc (2.98), a net change must occur in the direction of
the reverse reaction, to the left. Let’s define x as the change in molarity of Fe 3+. Because the net change occurs to the left, we designate the changes for the species on
the left side of the equation as positive and those on the right side as negative. The
relevant data are tabulated as follows.
Fe 2+(aq)
Ag(s) Fe 3(aq)
The reaction:
Ag +(aq)
∆
Initial concns:
0.200 M
0.100 M
0.300 M
+x M
+x M
-x M
changes:
10.200 + x2 M
10.100 + x2 M
10.300 - x2 M
equil concns:
Kc =
[Fe 3+]
=
2+
[Ag ][Fe ]
+
(0.300 - x)
= 2.98
(0.200 + x)(0.100 + x)
This equation, which is solved in Appendix A-3, is a quadratic equation for which
the acceptable root is x = 0.11. To obtain the equilibrium concentrations, we substitute this value of x into the terms shown in the table of data.
[Ag +]equil = 0.200 + 0.11 = 0.31 M
[Fe 2+]equil = 0.100 + 0.11 = 0.21 M
[Fe 3+]equil = 0.300 - 0.11 = 0.19 M
CHECK: If we have done the calculation correctly, we should obtain a value very
close to that given for Kc when we substitute the calculated equilibrium concentrations into the reaction quotient, Qc . We do.
After making an
approximation, it is always
necessary to check its
validity by substituting the
approximate value back into the
equation and verifying the 5% rule.
Qc =
[Fe 3+]
2+
[Ag ][Fe ]
+
=
(0.19)
= 2.9
(0.31)(0.21)
(Kc = 2.98)
Practice Example A: Excess Ag(s) is added to 1.20 M Fe 3+(aq). Given that
Ag +(aq) + Fe 2+(aq) ∆ Ag(s) + Fe 3+(aq)
Kc = 2.98
what are the equilibrium concentrations of the species in solution?
Practice Example B: A solution is prepared with [V 3+] = [Cr 2+] = 0.0100 M
and [V 2+] = [Cr 3+] = 0.150 M. The following reaction occurs.
▲
KEEP IN MIND
that if one or more of the
substances appearing in Qc
(or Qp) is not present initially,
a net change must occur to
produce some of the
substance(s). You need to
compare Qc (or Qp) with Kc
(or Kp) only if all the
substances appearing in
these expressions are
present initially.
V 3+(aq) + Cr 2+(aq) ∆ V 2+(aq) + Cr 3+(aq)
Kc = 7.2 * 102
What are the ion concentrations when equilibrium is established?
[Hint: The algebra can be greatly simplified by extracting the square root of both
sides of an equation at the appropriate point.]
Comments
6. It is sometimes helpful to compare the reaction quotient Q to the equilibrium constant K to determine the direction of the net change.
7. In many equilibrium calculations—often those in aqueous solutions—you
can work with molarities directly, without having to work with moles of reactants and solution volumes.
8. Where possible, check your calculation, for instance, by substituting
calculated equilibrium concentrations into the reaction quotient Q to see if its
numerical value is close to that of K.
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Integrative Example
651
Summary
15-1 Dynamic Equilibrium—Equilibrium is the condition
in which the forward and reverse reaction rates of reversible processes are equal. Chemical and physical
processes in equilibrium are dynamic by nature.
15-2 The Equilibrium Constant Expression—This condition of dynamic equilibrium is described through an
equilibrium constant expression. The form of the equilibrium constant expression is established from the balanced
chemical equation using activities to express the “effective” concentrations (equation 15.7). The numerical value
obtained from the equilibrium constant expression is referred to as the equilibrium constant, K. Equilibrium constants are unitless.
15-3 Relationships Involving Equilibrium Constants—
When the equation for a reversible reaction is written in
the reverse order, the equilibrium constant expression and
the value of K are both inverted from their original form.
When two or more reactions are coupled together, the
equilibrium constant for the overall reaction is the product
of the K values of the individual reactions. The equilibrium constant of a reaction can have different values depending on the reference state used. For Kc a
concentration reference state is used, while for Kp , a pressure reference state is used. The relationship between Kc
and Kp is given by equation (15.16).
15-4 The Magnitude of an Equilibrium Constant—The
magnitude of the equilibrium constant can be used to
determine the outcome of a reaction. For large values of K
the reaction goes to completion, with all reactants converted to products. A very small equilibrium constant,
for example, a large negative power of ten, indicates that
practically none of the reactants have been converted to
products. Finally, equilibrium constants of an intermedi-
ate value, for example, between 10 -10 and 1010, indicate
that some of the reactants have been converted to
products.
15-5 The Reaction Quotient, Q: Predicting the Direction
of Net Change—The reaction quotient, Q (equation
15.19), has the same form as the equilibrium constant
expression; however, its numerical value is determined
using the initial reaction activities. A comparison of the
reaction quotient with the equilibrium constant makes it
possible to predict the direction of net change leading to
equilibrium (Fig. 15-5). If Q 6 K, the forward reaction is
favored, meaning that when equilibrium is established the
amounts of products will have increased and the amounts
of reactants will have decreased. If Q 7 K, the reverse reaction is favored until equilibrium is established. If Q = K,
neither the forward nor reverse reaction is favored. The
initial conditions are in fact equilibrium conditions.
15-6 Altering Equilibrium Conditions: Le Châtelier’s
Principle—Le Châtelier’s principle is used to make qualitative predictions of the effects of different variables on an
equilibrium condition. This principle describes how an
equilibrium condition is modified, or “shifts,” in response
to the addition or removal of reactants or changes in reaction volume, external pressure, or temperature. Catalysts,
by speeding up the forward and reverse reactions equally,
have no effect on an equilibrium condition.
15-7 Equilibrium
Calculations:
Some
Illustrative
Examples—For quantitative equilibrium calculations, a
few basic principles and algebraic techniques are
required. A useful method employs a tabular system,
called an ICE table, for keeping track of the initial concentrations of the reactants and products, changes in these
concentrations, and the equilibrium concentrations.
Integrative Example
In the manufacture of ammonia, the chief source of hydrogen gas is the following reaction for the reforming of methane
at high temperatures.
(15.23)
CH 4(g) + 2 H 2O(g) ∆ CO2(g) + 4 H 2(g)
The following data are also given.
(a) CO(g) + H 2O(g) ∆ CO2(g) + H 2(g)
(b) CO(g) + 3 H 2(g) ∆ H 2O(g) + CH 4(g)
¢H° = -40 kJ; Kc = 1.4 at 1000 K
¢H° = -230 kJ; Kc = 190 at 1000 K
At 1000 K, 1.00 mol each of CH 4 and H 2O are allowed to come to equilibrium in a 10.0-L vessel. Calculate the number of
moles of H 2 present at equilibrium. Would the yield of H 2 increase if the temperature were raised above 1000 K?
Strategy
First, we should assemble the data needed to solve this problem. The amounts of substances and a reaction volume are
given, so we should be able to work with a Kc expression. However, because the Kc value for the reaction of interest is not
given, we will have to derive this value by combining the two equations for which data are given. This will yield values
of both Kc and ¢H for the reaction of interest.
To calculate the number of moles of H 2 at equilibrium we can use the ICE method, and to assess the effect of temperature on the equilibrium yield of H 2 we can apply Le Châtelier’s principle.
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THE NITROGEN CYCLE AND THE SYNTHESIS OF
NITROGEN COMPOUNDS
▲ Use of liquid ammonia as a fertilizer by direct injection into
the soil.
Nitrogen is an element essential both in living things and
in industry. The atmosphere contains a vast pool of elemental nitrogen, but most organisms cannot use nitrogen
in that form. Instead, they require certain nitrogencontaining compounds. The conversion of elemental atmospheric nitrogen to nitrogen compounds is called nitrogen
fixation, and in nature, this process is carried out only by
certain types of bacteria. Nitrogen fixation is part of the nitrogen cycle, which is the path of nitrogen through the
environment and a variety of living organisms and back
into the environment. A simplified version of the nitrogen
cycle is depicted in Figure 15-12, which shows that elemental nitrogen from the atmosphere is fixed, becomes part of
plants, animals, and other organisms, and is then returned
to the atmosphere. The numbers in parentheses refer to the
various parts of the cycle described below.
A few leguminous plants, such as beans, peas, and alfalfa, have bacteria residing on their roots that convert elemental nitrogen from the air into compounds used to
make plant protein (1). Plants also convert nitrates in the
soil to proteins (5). Animals obtain their protein-building
nitrogen by feeding on plants and/or other animals (2).
The decay of plant and animal proteins produces ammonia (3). Through a series of bacterial actions, ammonia is
converted to nitrites and nitrates (4). Denitrifying bacteria
decompose nitrites and nitrates, returning N2O and N2 to
the atmosphere, thus completing the cycle (6). Some atmospheric nitrogen is converted to nitrates during electrical storms (1a).
Reversible chemical reactions play an important role in
the nitrogen cycle. One example is the reaction of N2(g)
and O2(g) to form NO(g).
N2(g) + O2(g) ∆ 2 NO(g)
Kp = 4.7 * 10-31 at 298 K
= 1.3 * 10-4 at 1800 K
The reaction does not occur to any measurable extent at
298 K, but at 1800 K the situation is a little different. An
equilibrium mixture of N2(g) and O2(g) at 1800 K
Atmospheric
nitrogen
(1) Bacteria
(6) Bacteria
(1a) Lightning
Nitrates
(5)
Plant
protein
(6) Bacteria
(2)
(4) Bacteria
Nitrites
Animal
protein
(3) Decay
(4) Bacteria
Ammonia
▲ FIGURE 15-12
The nitrogen cycle
652
(3) Decay
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▲ During electrical storms, N2(g) and O2(g) combine to produce
small quantities of NO(g) in a reversible chemical reaction.
contains about 1 or 2% NO(g). NO(g) is introduced into
the nitrogen cycle through this high-temperature reaction
occurring naturally in lightning flashes during thunderstorms and artificially in high-temperature combustion
processes, as in internal combustion engines. Further reactions account for the formation of HNO 3(aq) in rainstorms
and the introduction of nitrates into the soil.
of NH 3 is greatest at low temperatures. Thus, the optimum conditions for the equilibrium production of NH 3 is
greatest at low temperatures. Thus, the optimum conditions for the equilibrium production of NH 3 are high pressures and low temperatures. However, these “optimum”
conditions do not take into account the rate of reaction.
Although the equilibrium production of NH 3 is favored at low temperatures, equilibrium is achieved so
slowly that the synthesis is not feasible at those temperatures. One way to speed up the reaction is to raise the temperature, even though doing so decreases the equilibrium
concentration of NH 3 . Another way is to use a catalyst.
The usual industrial operating conditions are a temperature of about 550 °C, pressures ranging from 150 to 350
atm, and a catalyst—usually iron in the presence of Al2O3 ,
MgO, CaO, and K 2O. Figure 15-13 suggests the dramatic
difference between the theoretical optimum conditions
and the actual operating conditions.
Another essential feature of the Haber–Bosch method
is to remove NH 3 as it forms. This is done by liquefying
the NH 3(g). In fact, the mixture need not be brought to
equilibrium at all, and nearly 100% conversion of N2 and
H 2 to NH 3 is achieved.
“Theoretical”
conditions
100
Ammonia at equilibrium, %
(Ratio H2/N2 3:1)
2 NO(g) + O2(g) ¡ 2 NO2(g)
3 NO2(g) + H 2O(l) ¡ 2 HNO3(aq) + NO(g)
Ammonia is an important industrial chemical. Its synthesis is achieved by the following reaction.
N2(g) + 3 H 2(g) ∆ 2 NH 3(g)
Kp = 6.2 * 105 at 298 K
80
70
60
50
40
30
20
10
100
300
era
tur
e, C
100
mp
Since ammonia can be converted by soil bacteria to nitrites and nitrates that can be used by plants, one of the
principal uses of artificial NH 3 is as a fertilizer that is injected directly into the soil. NH 3 is also used in the production of other nitrogen compounds, such as urea, hydrazine,
ammonium sulfate, ammonium nitrate, ammonium dihydrogen phosphate, and ammonium hydrogen phosphate.
Several of these compounds are used as fertilizers, and
others are used in the manufacture of explosives, pharmaceuticals, and plastics. So much nitrogen is now being
fixed artificially that fixed nitrogen is accumulating in the
environment somewhat faster than it is being returned to
the atmosphere. This causes environmental problems such
as the buildup of nitrates in groundwater.
Let us look at the commercial synthesis of ammonia,
called the Haber–Bosch process, from the standpoint of
chemical equilibrium and kinetics. In the synthesis reaction, two moles of gaseous product are formed for every
four moles of gaseous reactants. Carrying out the reaction
at high pressure favors the production of NH 3 . Because
the forward reaction is exothermic, the equilibrium yield
Te
¢H° = -92.22 kJ
90
500
700
900
Pressure, atm
200
300
400
500
600
700
Commercial
operating conditions
▲ FIGURE 15-13
Equilibrium conversion of N2(g) and H2(g) to NH3(g)
as a function of temperature and pressure
653
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Principles of Chemical Equilibrium
Solution
We combine equations (a) and (b) to obtain the data needed in this problem.
(a)
(b)
CO(g) + H 2O(g) ∆ CO 2(g) + H 2(g)
CH 4(g) + H 2O(g) ∆ CO(g) + 3 H 2(g)
¢H = -40 kJ
¢H = 230 kJ
Kc = 1.4
Kc = 1>190
Overall:
CH 4(g) + 2 H 2O(g) ∆ CO 2(g) + 4 H 2(g)
¢H = 190 kJ
Kc = 1.4>190 = 7.4 * 10 -3
Next we set up an ICE table in which x represents the number of moles of CH 4 consumed in reaching equilibrium.
The reaction:
initial amounts:
changes:
equil amounts:
equil concns, M:
CH4(g)
1.00 mol
-x mol
11.00 - x2 mol
11.00 - x2>10.0
Now we set up Kc and make substitutions into the
expression.
2 H2O(g)
1.00 mol
-2x mol
11.00 - 2x2 mol
11.00 - 2x2>10.0
Kc =
=
∆
CO2(g)
0.00 mol
x mol
x mol
x>10.0
4 H2(g)
0.00 mol
4x mol
4x mol
4x>10.0
[CO2][H 2]4
[CH 4][H 2O]2
1x>10.0214x>10.024
311.00 - x2/10.04311.00 - 2x2>10.042
x14x24
=
10011.00 - x211.00 - 2x22
= 7.4 * 10-3
The above equation reduces to
256x 5 = 0.74311.00 - x211.00 - 2x224
and then to
256x 5 - 0.74311.00 - x211.00 - 2x224 = 0
(15.24)
The solution to this equation is x = 0.23 mol. The number
of moles of H 2 at equilibrium is 4x = 0.92 mol.
Because the reaction is endothermic 1¢H = 190 kJ2,
the forward reaction is favored at higher temperatures.
The equilibrium yield of H 2 will increase if the temperature is raised above 1000 K.
Assessment
Equation (15.24) looks impossibly difficult to solve, but it is not. It can be solved for x rather simply by the method of successive approximations. This is done in Appendix A-3, equation (A.2). An important clue as to the possible range of values for x can be found in the ICE table. Note that the equilibrium amount of H 2O(g) is 1.00 - 2x, meaning that x 6 0.50,
or else all of the H 2O(g) would be consumed. This marks a good place to start the approximations.
Exercises
Writing Equilibrium Constant Expressions
1. Based on these descriptions, write a balanced equation and the corresponding Kc expression for each reversible reaction.
(a) Carbonyl fluoride, COF2(g), decomposes into
gaseous carbon dioxide and gaseous carbon tetrafluoride.
(b) Copper metal displaces silver(I) ion from aqueous
solution, producing silver metal and an aqueous solution of copper(II) ion.
(c) Peroxodisulfate ion, S 2O8 2-, oxidizes iron(II) ion
to iron(III) ion in aqueous solution and is itself reduced to sulfate ion.
2. Based on these descriptions, write a balanced equation and the corresponding Kp expression for each reversible reaction.
(a) Oxygen gas oxidizes gaseous ammonia to
gaseous nitrogen and water vapor.
(b) Hydrogen gas reduces gaseous nitrogen dioxide
to gaseous ammonia and water vapor.
(c) Nitrogen gas reacts with the solid sodium carbonate and carbon to produce solid sodium cyanide and
carbon monoxide gas.
3. Write equilibrium constant expressions, Kc , for the
reactions
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Exercises
4.
5.
6.
7.
8.
9.
(a) 2 NO(g) + O2(g) ∆ 2 NO2(g)
(b) Zn(s) + 2 Ag +(aq) ∆ Zn2+(aq) + 2 Ag(s)
(c) Mg(OH)2(s) + CO3 2-(aq) ∆
MgCO3(s) + 2 OH -(aq)
Write equilibrium constant expressions, Kp , for the
reactions
(a) CS 2(g) + 4 H 2(g) ∆ CH 4(g) + 2 H 2S(g)
1
(b) Ag2O(s) ∆ 2 Ag(s) + O2(g)
2
(c) 2 NaHCO3(s) ∆
Na 2CO3(s) + CO2(g) + H 2O(g)
Write an equilibrium constant, Kc , for the formation
from its gaseous elements of (a) 1 mol HF(g);
(b) 2 mol NH 3(g); (c) 2 mol N2O(g) (d) 1 mol ClF3(l).
Write an equilibrium constant, Kp , for the formation
from its gaseous elements of (a) 1 mol NOCl(g);
(b) 2 mol ClNO 2(g); (c) 1 mol N2H 4(g); (d) 1 mol
NH 4Cl(s).
Determine values of Kc from the Kp values given.
(a) SO2Cl2(g) ∆ SO2(g) + Cl2(g)
Kp = 2.9 * 10-2 at 303 K
(b) 2 NO(g) + O2(g) ∆ 2 NO2(g)
Kp = 1.48 * 104 at 184 °C
(c) Sb2S 3(s) + 3 H 2(g) ∆ 2 Sb(s) + 3 H 2S(g)
Kp = 0.429 at 713 K
Determine the values of Kp from the Kc values given.
(a) N2O4(g) ∆ 2 NO2(g)
Kc = 4.61 * 10-3 at 25 °C
(b) 2 CH 4(g) ∆ C2H 2(g) + 3 H 2(g)
Kc = 0.154 at 2000 K
(c) 2 H 2S(g) + CH 4(g) ∆ 4 H 2(g) + CS 2(g)
Kc = 5.27 * 10-8 at 973 K
The vapor pressure of water at 25 °C is 23.8 mmHg.
Write Kp for the vaporization of water, with pressures
in atmospheres. What is the value of Kc for the vaporization process?
655
10. If Kc = 5.12 * 10-3 for the equilibrium established
between liquid benzene and its vapor at 25 °C, what is
the vapor pressure of C6H 6 at 25 °C, expressed in millimeters of mercury?
11. Determine Kc for the reaction
1
1
1
N (g) + O2(g) + Br2(g) ∆ NOBr(g)
2 2
2
2
from the following information (at 298 K).
2 NO(g) ∆ N 2(g) + O 2(g) Kc = 2.1 * 1030
NO(g) +
1
Br (g) ∆ NOBr(g)
2 2
Kc = 1.4
12. Given the equilibrium constant values
1
N 2(g) + O 2(g) ∆ N 2O(g) Kc = 2.7 * 10 -18
2
N 2O 4(g) ∆ 2 NO 2(g) Kc = 4.6 * 10 -3
1
N (g) + O 2(g) ∆ NO 2(g)
2 2
Kc = 4.1 * 10 -9
Determine a value of Kc for the reaction
2 N2O(g) + 3 O2(g) ∆ 2 N2O4(g)
13. Use the following data to estimate a value of Kp at 1200 K
for the reaction 2 H 2(g) + O2(g) ∆ 2 H 2O(g)
C(graphite) + CO2(g) ∆ 2 CO(g)
Kc = 0.64
CO2(g) + H 2(g) ∆ CO(g) + H 2O(g) Kc = 1.4
C(graphite) +
1
O2(g) ∆ CO(g)
2
Kc = 1 * 108
14. Determine Kc for the reaction N2(g) + O2(g) +
Cl2(g) ∆ 2 NOCl(g), given the following data at
298 K.
1
N2(g) + O2(g) ∆ NO2(g)
Kp = 1.0 * 10-9
2
1
NOCl(g) + O2(g) ∆ NO2Cl(g)
Kp = 1.1 * 102
2
1
NO2(g) + Cl2(g) ∆ NO2Cl(g)
Kp = 0.3
2
Experimental Determination of Equilibrium Constants
15. 1.00 * 10-3 mol PCl5 is introduced into a 250.0-mL
flask, and equilibrium is established at 284 °C:
PCl5(g) ∆ PCl3(g) + Cl2(g). The quantity of Cl2(g)
present at equilibrium is found to be 9.65 * 10-4 mol.
What is the value of Kc for the dissociation reaction
at 284 °C?
16. A mixture of 1.00 g H 2 and 1.06 g H 2S in a 0.500-L flask
comes to equilibrium at 1670 K: 2 H 2(g) + S 2(g) ∆
2 H 2S(g). The equilibrium amount of S 2(g) found is
8.00 * 10-6 mol. Determine the value of Kp at 1670 K.
17. The two common chlorides of phosphorus, PCl3 and
PCl5 , both important in the production of other phos-
phorus compounds, coexist in equilibrium through
the reaction
PCl3(g) + Cl2(g) ∆ PCl5(g)
At 250 °C, an equilibrium mixture in a 2.50-L flask
contains 0.105 g PCl5 , 0.220 g PCl3 , and 2.12 g Cl2 .
What are the values of (a) Kc and (b) Kp for this reaction at 250 °C?
18. A 0.682-g sample of ICl(g) is placed in a 625-mL reaction
vessel at 682 K. When equilibrium is reached between
the ICl(g) and I 2(g) and Cl2(g) formed by its dissociation, 0.0383 g I 2 is present. What is Kc for this reaction?
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Equilibrium Relationships
19. Equilibrium is established at 1000 K, where Kc = 281
for the reaction 2 SO 2(g) + O2(g) ∆ 2 SO 3(g). The
equilibrium amount of O2(g) in a 0.185-L flask is
0.00247 mol. What is the ratio of [SO2] to [SO3] in this
equilibrium mixture?
20. For the dissociation of I 2(g) at about 1200 °C,
I 2(g) ∆ 2 I(g), Kc = 1.1 * 10-2. What volume flask
should we use if we want 0.37 mol I to be present for
every 1.00 mol I 2 at equilibrium?
21. In the Ostwald process for oxidizing ammonia, a variety of products is possible—N2 , N2O, NO, and NO2—
depending on the conditions. One possibility is
NH 3(g) +
5
3
O2(g) ∆ NO(g) + H 2O(g)
4
2
19
Kp = 2.11 * 10 at 700 K
For the decomposition of NO2 at 700 K,
1
NO2(g) ∆ NO(g) + O2(g) Kp = 0.524
2
(a) Write a chemical equation for the oxidation of
NH 3(g) to NO2(g).
(b) Determine Kp for the chemical equation you have
written.
22. At 2000 K, Kc = 0.154 for the reaction 2 CH 4(g) ∆
C2H 2(g) + 3 H 2(g). If a 1.00-L equilibrium mixture at
2000 K contains 0.10 mol each of CH 4(g) and H 2(g),
(a) what is the mole fraction of C2H 2(g) present?
(b) Is the conversion of CH 4(g) to C2H 2(g) favored at
high or low pressures?
(c) If the equilibrium mixture at 2000 K is transferred
from a 1.00-L flask to a 2.00-L flask, will the number
of moles of C2H 2(g) increase, decrease, or remain
unchanged?
23. An equilibrium mixture at 1000 K contains
0.276 mol H 2 , 0.276 mol CO 2 , 0.224 mol CO, and
0.224 mol H 2O.
CO2(g) + H 2(g) ∆ CO(g) + H 2O(g)
(a) Show that for this reaction, Kc is independent of
the reaction volume, V.
(b) Determine the value of Kc and Kp .
24. For the reaction CO(g) + H 2O(g) ∆ CO2(g) +
H 2(g), Kp = 23.2 at 600 K. Explain which of the following situations might be found at equilibrium:
(a) PCO = PH2O = PCO2 = PH2 ; (b) PH2>PH2O =
(PCO2)(PH2) = (PCO)(PH2O);
PCO2>PCO ;
(c)
(d) PCO2>PH2O = PH2>PCO .
Direction and Extent of Chemical Change
25. Can a mixture of 2.2 mol O2 , 3.6 mol SO 2 , and
1.8 mol SO3 be maintained indefinitely in a 7.2-L flask
at a temperature at which Kc = 100 in this reaction?
Explain.
2 SO 2(g) + O2(g) ∆ 2 SO 3(g)
26. Is a mixture of 0.0205 mol NO2(g) and 0.750 mol
N2O4(g) in a 5.25-L flask at 25 °C at equilibrium? If
not, in which direction will the reaction proceed—
toward products or reactants?
N2O4(g) ∆ 2 NO2(g)
Kc = 4.61 * 10-3 at 25 °C
27. In the reaction 2 SO2(g) + O2(g) ∆ 2 SO3(g), 0.455
mol SO2 , 0.183 mol O2 , and 0.568 mol SO3 are introduced simultaneously into a 1.90-L vessel at 1000 K.
(a) Is this mixture at equilibrium?
(b) If not, in which direction will a net change occur?
28. In the reaction CO(g) + H 2O(g) ∆ CO2(g) +
H 2(g), K = 31.4 at 588 K. Equal masses of each reactant and product are brought together in a reaction
vessel at 588 K.
(a) Can this mixture be at equilibrium?
(b) If not, in which direction will a net change occur?
29. A mixture consisting of 0.150 mol H 2 and 0.150 mol I 2
is brought to equilibrium at 445 °C in a 3.25-L flask.
What are the equilibrium amounts of H 2 , I 2 , and HI?
H 2(g) + I 2(g) ∆ 2 HI (g)
Kc = 50.2 at 445 °C
30. Starting with 0.280 mol SbCl3 and 0.160 mol Cl2 , how
many moles of SbCl5 , SbCl3 , and Cl2 are present when
equilibrium is established at 248 °C in a 2.50-L flask?
SbCl5(g) ∆ SbCl3(g) + Cl2(g)
Kc = 2.5 * 10-2 at 248 °C
31. Starting with 0.3500 mol CO(g) and 0.05500 mol
COCl2(g) in a 3.050-L flask at 668 K, how many moles
of Cl2(g) will be present at equilibrium?
CO(g) + Cl2(g) ∆ COCl2(g)
Kc = 1.2 * 103 at 668 K
32. 1.00 g each of CO, H 2O, and H 2 are sealed in a 1.41-L
vessel and brought to equilibrium at 600 K. How
many grams of CO2 will be present in the equilibrium
mixture?
CO(g) + H 2O(g) ∆ CO2(g) + H 2(g)
Kc = 23.2
33. Equilibrium is established in a 2.50-L flask at 250 °C
for the reaction
PCl5(g) ∆ PCl3(g) + Cl2(g)
Kc = 3.8 * 10-2
How many moles of PCl5 , PCl3 , and Cl2 are present at
equilibrium, if
(a) 0.550 mol each of PCl5 and PCl3 are initially introduced into the flask?
(b) 0.610 mol PCl5 alone is introduced into the flask?
34. For the following reaction, Kc = 2.00 at 1000 °C.
2 COF2(g) ∆ CO2(g) + CF4(g)
If a 5.00-L mixture contains 0.145 mol COF2 , 0.262 mol
CO2 , and 0.074 mol CF4 at a temperature of 1000 °C,
(a) Will the mixture be at equilibrium?
(b) If the gases are not at equilibrium, in what direction will a net change occur?
(c) How many moles of each gas will be present at
equilibrium?
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Exercises
35. In the following reaction, Kc = 4.0.
C2H 5OH + CH 3COOH ∆ CH 3COOC2H 5 + H 2O
A reaction is allowed to occur in a mixture of 17.2 g
C2H 5OH, 23.8 g CH 3COOH, 48.6 g CH 3COOC2H 5 ,
and 71.2 g H 2O.
(a) In what direction will a net change occur?
(b) How many grams of each substance will be present at equilibrium?
36. The N2O4 –NO2 equilibrium mixture in the flask on
the left in the figure is allowed to expand into the
evacuated flask on the right. What is the composition
of the gaseous mixture when equilibrium is reestablished in the system consisting of the two flasks?
N2O4(g) ∆ 2 NO2(g)
657
41. One sketch below represents an initial nonequilibrium mixture in the reversible reaction
SO2(g) + Cl2(g) ∆ SO2Cl2(g)
Kc = 4.0
Which of the other three sketches best represents an
equilibrium mixture? Explain.
Kc = 4.61 * 10-3 at 25 °C
0.971 mol N2O4
0.0580 mol NO2
Initial mixture
(a)
(b)
(c)
2.25 L
25 C
0.750 L
25 C
37. Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at
high temperatures.
HCONH 2(g) ∆ NH 3(g) + CO(g)
Kc = 4.84 at 400 K
If 0.186 mol HCONH 2(g) dissociates in a 2.16-L flask at
400 K, what will be the total pressure at equilibrium?
38. A mixture of 1.00 mol NaHCO3(s) and 1.00 mol
Na 2CO3(s) is introduced into a 2.50-L flask in which
the partial pressure of CO2 is 2.10 atm and that of
H 2O(g) is 715 mmHg. When equilibrium is established at 100 °C, will the partial pressures of CO2(g)
and H 2O(g) be greater or less than their initial partial
pressures? Explain.
42. One sketch below represents an initial nonequilibrium mixture in the reversible reaction
2 NO(g) + Br2(g) ∆ 2 NOBr(g)
Kc = 3.0
Which of the other three sketches best represents an
equilibrium mixture? Explain.
2 NaHCO3(s) ∆ Na 2CO3(s) + CO2(g) + H 2O(g)
Kp = 0.23 at 100 °C
39. Cadmium metal is added to 0.350 L of an aqueous solution in which [Cr 3+] = 1.00 M. What are the concentrations of the different ionic species at equilibrium?
What is the minimum mass of cadmium metal required to establish this equilibrium?
Initial mixture
(a)
(b)
(c)
2 Cr 3+(aq) + Cd(s) ∆ 2 Cr 2+(aq) + Cd2+(aq)
Kc = 0.288
3+
40. Lead metal is added to 0.100 M Cr (aq). What are
[Pb 2+], [Cr 2+], and [Cr 3+] when equilibrium is established in the reaction?
Pb(s) + 2 Cr 3+(aq) ∆ Pb 2+(aq) + 2 Cr 2+(aq)
Kc = 3.2 * 10-10
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Partial Pressure Equilibrium Constant, Kp
43. Refer to Example 15-4. H 2S(g) at 747.6 mmHg pressure and a 1.85-g sample of I 2(s) are introduced into a
725-mL flask at 60 °C. What will be the total pressure
in the flask at equilibrium?
H 2S(g) + I 2(s) ∆ 2 HI(g) + S(s)
Kp = 1.34 * 105 at 60 °C
44. A sample of NH 4HS(s) is placed in a 2.58-L flask containing 0.100 mol NH 3(g). What will be the total gas
pressure when equilibrium is established at 25 °C?
NH 4HS(s) ∆ NH 3(g) + H 2S(g)
Kp = 0.108 at 25 °C
45. The following reaction is used in some self-contained
breathing devices as a source of O2(g).
4 KO2(s) + 2 CO2(g) ∆ 2 K 2CO3(s) + 3 O2(g)
Kp = 28.5 at 25 °C
Suppose that a sample of CO2(g) is added to an evacuated flask containing KO2(s) and equilibrium is established. If the equilibrium partial pressure of CO2(g) is
found to be 0.0721 atm, what are the equilibrium partial pressure of O2(g) and the total gas pressure?
46. Concerning the reaction in Exercise 45, if KO2(s)
and K 2CO3(s) are maintained in contact with air at
1.00 atm and 25 °C, in which direction will a net
change occur to establish equilibrium? Explain.
[Hint: Recall equation (6.17) and the composition of
air (Table 6.3.)]
47. 1.00 mol each of CO and Cl2 are introduced into an
evacuated 1.75-L flask, and the following equilibrium
is established at 668 K.
CO(g) + Cl2(g) ¡ COCl2(g)
Kp = 22.5
For this equilibrium, calculate (a) the partial pressure
of COCl2(g); (b) the total gas pressure.
48. For the reaction 2 NO2(g) ∆ 2 NO(g) + O2(g),
Kc = 1.8 * 10-6 at 184 °C. What is the value of Kp for
this reaction at 184 °C?
NO(g) +
1
O (g) ∆ NO2(g)
2 2
Le Châtelier’s Principle
49. Continuous removal of one of the products of a chemical reaction has the effect of causing the reaction to go
to completion. Explain this fact in terms of Le Châtelier’s principle.
50. We can represent the freezing of H 2O(l) at 0 °C as H 2O
(l, d = 1.00 g>cm3) ∆ H 2O(s, d = 0.92 g>cm3). Explain why increasing the pressure on ice causes it to
melt. Is this the behavior you expect for solids in
general? Explain.
51. Explain how each of the following affects the amount
of H 2 present in an equilibrium mixture in the reaction
3 Fe(s) + 4 H 2O(g) ∆ Fe3O4(s) + 4 H 2(g)
¢H° = -150 kJ
(a) Raising the temperature of the mixture; (b) introducing more H 2O(g); (c) doubling the volume of the
container holding the mixture; (d) adding an appropriate catalyst.
52. In the gas phase, iodine reacts with cyclopentene
(C5H 8) by a free radical mechanism to produce cyclopentadiene (C5H 6) and hydrogen iodide. Explain
how each of the following affects the amount of HI(g)
present in the equilibrium mixture in the reaction
I 2(g) + C5H 8(g) ∆ C5H 6(g) + 2 HI(g)
¢H° = 92.5 kJ
(a) Raising the temperature of the mixture; (b) introducing more C5H 6(g); (c) doubling the volume of the
container holding the mixture; (d) adding an appropriate catalyst; (e) adding an inert gas such as He to a
constant-volume reaction mixture.
53. The reaction N2(g) + O2(g) ∆ 2 NO(g), ¢H° =
+181 kJ, occurs in high-temperature combustion
processes carried out in air. Oxides of nitrogen produced from the nitrogen and oxygen in air are intimately
involved in the production of photochemical smog.
What effect does increasing the temperature have on
(a) the equilibrium production of NO(g); (b) the rate
of this reaction?
54. Use data from Appendix D to determine whether the
forward reaction is favored by high temperatures or
low temperatures.
(a) PCl3(g) + Cl2(g) ∆ PCl5(g)
(b) SO2(g) + 2 H 2S(g) ∆ 2 H 2O(g) + 3 S(s)
(c) 2 N2(g) + 3 O2(g) + 4 HCl(g) ∆
4 NOCl(g) + 2 H 2O(g)
55. If the volume of an equilibrium mixture of N2(g),
H 2(g), and NH 3(g) is reduced by doubling the pressure, will PN2 have increased, decreased, or remained
the same when equilibrium is reestablished? Explain.
N2(g) + 3 H 2(g) ∆ 2 NH 3(g)
56. For the reaction
A(s) ∆ B(s) + 2 C(g) +
1
D(g)
2
¢H° = 0
(a) Will Kp increase, decrease, or remain constant
with temperature? Explain.
(b) If a constant-volume mixture at equilibrium at 298
K is heated to 400 K and equilibrium reestablished,
will the number of moles of D(g) increase, decrease, or
remain constant? Explain.
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Integrative and Advanced Exercises
57. What effect does increasing the volume of the system
have on the equilibrium condition in each of the following reactions?
(a) C(s) + H 2O(g) ∆ CO(g) + H 2(g)
(b) Ca(OH)2(s) + CO2(g) ∆ CaCO 3(s) + H 2O(g)
(c) 4 NH 3(g) + 5 O2(g) ∆ 4 NO(g) + 6 H 2O(g)
58. For which of the following reactions would you expect the extent of the forward reaction to increase
with increasing temperatures? Explain.
1
1
N2(g) + O2(g)
2
2
1
(b)
SO 3(g) ∆ SO2(g) + O2(g)
2
(c) N2H 4(g) ∆ N2(g) + 2 H 2(g)
(a)
NO(g) ∆
(d) COCl2(g) ∆ CO(g) + Cl2(g)
659
¢H° = -90.2 kJ
¢H° = +98.9 kJ
¢H° = -95.4 kJ
¢H° = +108.3 kJ
Integrative and Advanced Exercises
59. Explain why the percent of molecules that dissociate
into atoms in reactions of the type I 2(g) ∆ 2 I(g)
always increases with an increase in temperature.
60. A 1.100-L flask at 25 °C and 1.00 atm pressure contains
CO2(g) in contact with 100.0 mL of a saturated aqueous solution in which [CO2(aq)] = 3.29 * 10-2 M.
(a) What is the value of Kc at 25 °C for the equilibrium CO2(g) ∆ CO2(aq)?
(b) If 0.01000 mol of radioactive 14CO2 is added to the
flask, how many moles of the 14CO2 will be found in
the gas phase and in the aqueous solution when equilibrium is reestablished?
[Hint: The radioactive 14CO2 distributes itself between the two phases in exactly the same manner as
the nonradioactive 12CO2 .]
61. Refer to Example 15-13. Suppose that 0.100 L of the
equilibrium mixture is diluted to 0.250 L with water.
What will be the new concentrations when equilibrium is reestablished?
62. In the equilibrium described in Example 15-12, the
percent dissociation of N2O4 can be expressed as
3.00 * 10-3 mol N2O4
* 100% = 12.5%
0.0240 mol N2O4 initially
What must be the total pressure of the gaseous mixture if N2O4(g) is to be 10.0% dissociated at 298 K?
N2O4 ∆ 2 NO2(g)
Kp = 0.113 at 298 K
63. Starting with SO3(g) at 1.00 atm, what will be the total
pressure when equilibrium is reached in the following
reaction at 700 K?
2 SO 3(g) ∆ 2 SO2(g) + O2(g)
Kp = 1.6 * 10-5
64. A sample of air with a mole ratio of N2 to O2 of 79 : 21
is heated to 2500 K. When equilibrium is established
in a closed container with air initially at 1.00 atm, the
mole percent of NO is found to be 1.8%. Calculate Kp
for the reaction.
N2(g) + O2(g) ∆ 2 NO(g)
65. Derive, by calculation, the equilibrium amounts of
SO2 , O2 , and SO3 listed in (a) Figure 15-6(c); (b)
Figure 15-7(b).
66. The decomposition of salicylic acid to phenol and carbon dioxide was carried out at 200.0 °C, a temperature
at which the reactant and products are all gaseous. A
0.300-g sample of salicylic acid was introduced into a
50.0-mL reaction vessel, and equilibrium was established. The equilibrium mixture was rapidly cooled to
condense salicylic acid and phenol as solids; the
CO2(g) was collected over mercury and its volume
was measured at 20.0 °C and 730 mmHg. In two identical experiments, the volumes of CO2(g) obtained
were 48.2 and 48.5 mL, respectively. Calculate Kp for
this reaction.
OH CO2
OH
COOH
67. One of the key reactions in the gasification of coal is
the methanation reaction, in which methane is produced from synthesis gas—a mixture of CO and H 2 .
CO(g) + 3 H 2(g) ∆ CH 4(g) + H 2O(g)
¢H = -230 kJ; Kc = 190 at 1000 K
(a) Is the equilibrium conversion of synthesis gas to
methane favored at higher or lower temperatures?
Higher or lower pressures?
(b) Assume you have 4.00 mol of synthesis gas with a
3:1 mol ratio of H 2(g) to CO(g) in a 15.0-L flask. What
will be the mole fraction of CH 4(g) at equilibrium at
1000 K?
68. A sample of pure PCl5(g) is introduced into an evacuated flask and allowed to dissociate.
PCl5(g) ∆ PCl3(g) + Cl2(g)
If the fraction of PCl5 molecules that dissociate is denoted by a, and if the total gas pressure is P, show that
a2 P
Kp =
1 - a2
69. Nitrogen dioxide obtained as a cylinder gas is always a
mixture of NO2(g) and N2O4(g). A 5.00-g sample obtained from such a cylinder is sealed in a 0.500-L flask at
298 K. What is the mole fraction of NO2 in this mixture?
N2O4(g) ∆ 2 NO2(g)
Kc = 4.61 * 10-3
70. What is the apparent molar mass of the gaseous mixture that results when COCl2(g) is allowed to dissociate at 395 °C and a total pressure of 3.00 atm?
COCl2(g) ∆ CO(g) + Cl2(g)
Kp = 4.44 * 10-2 at 395 °C
Think of the apparent molar mass as the molar mass
of a hypothetical single gas that is equivalent to the
gaseous mixture.
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Principles of Chemical Equilibrium
71. Show that in terms of mole fractions of gases and total
gas pressure the equilibrium constant expression for
N2(g) + 3 H 2(g) ∆ 2 NH 3(g)
is
Kp =
(xNH3)2
(xN2)(xH2)2
1
*
(Ptot)2
72. For the synthesis of ammonia at 500 K,
N2(g) + 3 H 2(g) ∆ 2 NH 3(g), Kp = 9.06 * 10-2.
Assume that N2 and H 2 are mixed in the mole ratio
1:3 and that the total pressure is maintained at 1.00
atm. What is the mole percent NH 3 at equilibrium?
[Hint: Use the equation from Exercise 71.]
73. A mixture of H 2S(g) and CH 4(g) in the mole ratio 2 : 1
was brought to equilibrium at 700 °C and a total pressure of 1 atm. On analysis, the equilibrium mixture
was found to contain 9.54 * 10-3 mol H 2S. The CS 2
present at equilibrium was converted successively to
H 2SO4 and then to BaSO4 ; 1.42 * 10-3 mol BaSO4
was obtained. Use these data to determine Kp at
700 °C for the reaction
2 H 2S(g) + CH 4(g) ∆ CS 2(g) + 4 H 2(g)
mixture is allowed to come to equilibrium at 1000 K in
reaction (15.23). What will be the equilibrium amount,
in moles, of each gas?
76. Concerning the reaction in Exercise 22 and the situation described in part (c) of that exercise, will the mole
fraction of C2H 2(g) increase, decrease, or remain unchanged when equilibrium is reestablished? Explain.
77. The formation of nitrosyl chloride is given by the following equation: 2 NO(g) + Cl2(g) ∆ 2 NOCl(g);
Kc = 4.6 * 104 at 298 K. In a 1.50-L flask, there are
4.125 mol of NOCl and 0.1125 mol of Cl2 present at
equilibrium (298 K).
(a) Determine the partial pressure of NO at
equilibrium.
(b) What is the total pressure of the system at
equilibrium?
78. At 500 K, a 10.0-L equilibrium mixture contains
0.424 mol N2 , 1.272 mol H 2 , and 1.152 mol NH 3 . The
mixture is quickly chilled to a temperature at which the
NH 3 liquefies, and the NH 3(l) is completely removed.
The 10.0-L gaseous mixture is then returned to 500 K,
and equilibrium is reestablished. How many moles of
NH 3(g) will be present in the new equilibrium mixture?
Kp at 700 °C = ?
74. A solution is prepared having these initial concentrations: [Fe 3+] = [Hg 2 2+] = 0.5000 M; [Fe 2+] = [Hg 2+] =
0.03000 M. The following reaction occurs among the
ions at 25 °C.
N2(g) + 3 H 2(g) ∆ 2 NH 3
Kc = 152 at 500 K
79. Recall the formation of methanol from synthesis gas,
the reversible reaction with which we began our discussion of the equilibrium constant expression (page 623).
CO(g) + 2 H 2(g) ∆ CH 3OH(g)
2 Fe 3+(aq) + Hg2 2+(aq) ∆ 2 Fe 2+(aq) + 2 Hg 2+(aq)
Kc = 14.5 at 483 K
Kc = 9.14 * 10-6
A particular synthesis gas consisting of 35.0 mole percent CO(g) and 65.0 mole percent H 2(g) at a total pressure of 100.0 atm at 483 K is allowed to come to
equilibrium. Determine the partial pressure of
CH 3OH(g) in the equilibrium mixture.
What will be the ion concentrations at equilibrium?
75. Refer to the Integrative Example. A gaseous mixture is
prepared containing 0.100 mol each of CH 4(g),
H 2O(g), CO2(g), and H 2(g) in a 5.00-L flask. Then the
Feature Problems
HI(g) is introduced into five identical 400-cm3 glass
bulbs, and the five bulbs are maintained at 623 K.
Each bulb is opened after a period of time and analyzed for I 2 by titration with 0.0150 M Na 2S 2O3(aq).
80. A classic experiment in equilibrium studies dating
from 1862 involved the reaction in solution of ethanol
(C2H 5OH) and acetic acid (CH 3COOH) to produce
ethyl acetate and water.
C2H 5OH + CH 3COOH ∆ CH 3COOC2H 5 + H 2O
I 2(aq) + 2 Na 2S 2O3(aq) ¡
Na 2S 4O6(aq) + 2 NaI(aq)
The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content.
Data for this experiment are provided in the table
below. What is the value of Kc at 623 K?
2 CH 3COOH(aq) + Ba(OH)2(aq) ∆
Ba(CH 3COO)2(aq) + 2 H 2O(l)
In one experiment, a mixture of 1.000 mol acetic acid
and 0.5000 mol ethanol is brought to equilibrium. A
sample containing exactly one-hundredth of the equilibrium mixture requires 28.85 mL 0.1000 M Ba(OH)2
for its titration. Calculate the equilibrium constant,
Kc , for the ethanol–acetic acid reaction based on this
experiment.
81. The decomposition of HI(g) is represented by the
equation
2 HI(g) ∆ H 2(g) + I 2(g)
Bulb
Number
Initial
Mass of
HI(g), g
Time
Bulb
Opened
(h)
1
2
3
4
5
0.300
0.320
0.315
0.406
0.280
2
4
12
20
40
Volume
0.0150 M Na2S2O3
Required
for Titration
(in mL)
20.96
27.90
32.31
41.50
28.68
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Self-Assessment Exercises
82. In one of Fritz Haber’s experiments to establish the conditions required for the ammonia synthesis reaction,
pure NH 3(g) was passed over an iron catalyst at 901 °C
and 30.0 atm. The gas leaving the reactor was bubbled
through 20.00 mL of a HCl(aq) solution. In this way, the
NH 3(g) present was removed by reaction with HCl.
The remaining gas occupied a volume of 1.82 L at STP.
The 20.00 mL of HCl(aq) through which the gas had
been bubbled required 15.42 mL of 0.0523 M KOH for its
titration. Another 20.00-mL sample of the same HCl(aq)
through which no gas had been bubbled required
18.72 mL of 0.0523 M KOH for its titration. Use these
data to obtain a value of Kp at 901 °C for the reaction
N2(g) + 3 H 2(g) ∆ 2 NH 3(g).
83. The following is an approach to establishing a relationship between the equilibrium constant and rate
constants mentioned in the Are You Wondering feature on page 628.
• Work with the detailed mechanism for the reaction.
• Use the principle of microscopic reversibility, the
idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter 14, we treated some reaction
661
steps as reversible and others as going to completion. However, as noted in Table 15.3, every
reaction has an equilibrium constant even though
a reaction is generally considered to go to completion if its equilibrium constant is very large.)
• Use the idea that when equilibrium is attained in
an overall reaction, it is also attained in each step
of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the
mechanism, similar to what we did with the
steady-state assumption in describing reaction
mechanisms.
• Combine the Kc expressions for the elementary
steps into a Kc expression for the overall reaction.
The numerical value of the overall Kc can thereby
be expressed as a ratio of rate constants, k.
Use this approach to establish the equilibrium constant
expression for the overall reaction,
H 2(g) + I 2(g) ∆ 2 HI(g)
The mechanism of the reaction appears to be the
following:
I 2(g) ∆ 2 I(g)
Fast:
Slow:
2 I(g) + H 2(g) ∆ 2 HI(g)
Self-Assessment Exercises
84. In your own words, define or explain the following
terms or symbols: (a) Kp ; (b) Qc ; (c) ¢ngas .
85. Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a
net chemical change; (c) Le Châtelier’s principle;
(d) effect of a catalyst on equilibrium.
86. Explain the important distinctions between each pair
of terms: (a) reaction that goes to completion and reversible reaction; (b) Kc and Kp ; (c) reaction quotient
(Q) and equilibrium constant expression (K); (d) homogeneous and heterogeneous reaction.
87. In the reversible reaction H 2(g) + I 2(g) ∆ 2 HI(g),
an initial mixture contains 2 mol H 2 and 1 mol I 2 . The
amount of HI expected at equilibrium is (a) 1 mol;
(b) 2 mol; (c) less than 2 mol; (d) more than 2 mol but
less than 4 mol.
88. Equilibrium is established in the reaction 2 SO2(g) +
O2(g) ∆ 2 SO3(g) at a temperature where Kc = 100.
If the number of moles of SO3(g) in the equilibrium
mixture is the same as the number of moles of SO2(g),
(a) the number of moles of O2(g) is also equal to the
number of moles of SO2(g); (b) the number of moles of
O2(g) is half the number of moles of SO2 ; (c) [O 2] may
have any of several values; (d) [O2] = 0.010 M.
89. The volume of the reaction vessel containing an equilibrium mixture in the reaction SO2Cl2(g) ∆
SO2(g) + Cl2(g) is increased. When equilibrium is
reestablished, (a) the amount of Cl2 will have
increased; (b) the amount of SO2 will have decreased;
(c) the amounts of SO2 and Cl2 will have remained the
same; (d) the amount of SO2Cl2 will have increased.
90. For the reaction 2 NO2(g) ∆ 2 NO(g) + O2(g),
Kc = 1.8 * 10-6 at 184 °C. At 184 °C, the value of Kc
for the reaction NO(g) + 12 O2(g) ∆ NO2(g) is
(a) 0.9 * 106; (b) 7.5 * 102; (c) 5.6 * 105;
(d) 2.8 * 105.
91. For the dissociation reaction 2 H 2S(g) ∆ 2 H 2(g) +
S 2(g), Kp = 1.2 * 10-2 at 1065 °C. For this same
reaction at 1000 K, (a) Kc is less than Kp ; (b) Kc is
greater than Kp ; (c) Kc = Kp ; (d) whether Kc is less
than, equal to, or greater than Kp depends on the total
gas pressure.
92. The following data are given at 1000 K: CO(g) +
H 2O(g) ∆ CO2(g) + H 2(g);
¢H° = -42 kJ;
Kc = 0.66. After an initial equilibrium is established
in a 1.00-L container, the equilibrium amount of H 2
can be increased by (a) adding a catalyst; (b) increasing the temperature; (c) transferring the mixture to a
10.0-L container; (d) in some way other than (a), (b),
or (c).
93. Equilibrium is established in the reversible reaction
2 A + B ∆ 2 C. The equilibrium concentrations
are [A] = 0.55 M, [B] = 0.33 M, [C] = 0.43 M. What
is the value of Kc for this reaction?
94. The Deacon process for producing chlorine gas from
hydrogen chloride is used in situations where HCl is
available as a by-product from other chemical
processes.
4 HCl(g) + O2(g) ∆ 2 H 2O(g) + 2 Cl2(g)
¢H° = -114 kJ
A mixture of HCl, O2 , H 2O, and Cl2 is brought to
equilibrium at 400 °C. What is the effect on the equilibrium amount of Cl2(g) if
(a) additional O2(g) is added to the mixture at constant volume;
(b) HCl(g) is removed from the mixture at constant
volume;
(c) the mixture is transferred to a vessel of twice the
volume;
(d) a catalyst is added to the reaction mixture;
(e) the temperature is raised to 500 °C?
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Principles of Chemical Equilibrium
95. An equilibrium mixture of SO2 , SO3 , and O2 gases is
maintained in a 2.05-L flask at a temperature at which
Kc = 35.5 for the reaction
2 SO 2(g) + O2(g) ∆ 2 SO 3(g)
(a) If the numbers of moles of SO2 and SO3 in the
flask are equal, how many moles of O2 are present?
(b) If the number of moles of SO3 in the flask is twice
the number of moles of SO2 , how many moles of O2
are present?
96. Using the method in Appendix E, construct a concept
map of Section 15-6, illustrating the shift in equilibrium due to the various stresses.
eMedia Exercises
97. After viewing the Dynamic Equilibrium animation
of equilibrium through a physical change (Activebook
15-1), describe the influence of temperature on rates
of change.
98. After viewing the Chemical Equilibrium activity
(eBook 15-2), estimate the value of the equilibrium
constant of the reaction depicted. What information
did you use to make this estimate?
99. (a) Describe how the stoichiometry of the reaction
in the Chemical Equilibrium activity (Activebook
15-2 ) influences the values of concentrations at equilibrium. (b) If the initial concentrations of both reactants are doubled in this exercise, does this change the
value of the equilibrium constant? (c) Does this
change, if any, affect the concentration of the product
at equilibrium?
100. Using equal initial concentrations of reactant and
product in the Equilibrium Constant activity
(Activebook 15-4), describe how the value of the equilibrium constant is related to the concentration of the
product once equilibrium has been reached.
101. (a) What would be the change in the results seen
in the NO2 –N2O4 Equilibrium animation (Activebook 15-6) if the pressure were decreased by a factor of
two instead of increased? (b) Given that this forward
reaction is exothermic, what would be the effect on
the ratio of the NO2 and N2O4 concentrations by
increasing the temperature of the system?