2: Atomic Masses: Counting Atoms by Weighing
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As solid copper(II) sulfate
pentahydrate, CuSO4и5H2O, is
heated, it loses H2O, eventually
forming white CuSO4.
© Cengage Learning
558 Chapter 17 Equilibrium
b. The reaction is
CuSO4 и 5H2O(s) 4
3 CuSO4(s) ϩ 5H2O(g)
The two solids are not included. The equilibrium expression is
K ϭ [H2O]5
Self-Check EXERCISE 17.2 Write the equilibrium expression for each of the following reactions.
a. 2KClO3(s) 4
3 2KCl(s) ϩ 3O2(g)
This reaction is often used to produce oxygen gas in the laboratory.
b. NH4NO3(s) 4
3 N2O(g) ϩ 2H2O(g)
c. CO2(g) ϩ MgO(s) 4
3 MgCO3(s)
d. SO3(g) ϩ H2O(l) 4
3 H2SO4(l)
See Problems 17.25 through 17.28. ■
17.7 Le Châtelier’s Principle
OBJECTIVE:
To learn to predict the changes that occur when a system at equilibrium is
disturbed.
It is important to understand the factors that control the position of a chemical equilibrium. For example, when a chemical is manufactured, the
chemists and chemical engineers in charge of production want to choose
conditions that favor the desired product as much as possible. That is, they
want the equilibrium to lie far to the right (toward products). When the
process for the synthesis of ammonia was being developed, extensive studies were carried out to determine how the equilibrium concentration of ammonia depended on the conditions of temperature and pressure.
17.7 Le Châtelier’s Principle
559
In this section we will explore how various changes in conditions affect
the equilibrium position of a reaction system. We can predict the effects of
changes in concentration, pressure, and temperature on a system at equilibrium by using Le Châtelier’s principle, which states that when a change
is imposed on a system at equilibrium, the position of the equilibrium shifts in a
direction that tends to reduce the effect of that change.
▲
The Effect of a Change in Concentration
Let us consider the ammonia synthesis reaction. Suppose there is an equilibrium position described by these concentrations:
[N2] ϭ 0.399 M
[H2] ϭ 1.197 M
[NH3] ϭ 0.203 M
What will happen if 1.000 mol/L of N2 is suddenly injected into the system?
We can begin to answer this question by remembering that for the system at
equilibrium, the rates of the forward and reverse reactions exactly balance,
N2(g) ϩ 3H2(g) 4
3 2NH3(g)
as indicated here by arrows of the same length. When N2 is added, there are
suddenly more collisions between N2 and H2 molecules. This increases the
rate of the forward reaction (shown here by the greater length of the arrow
pointing in that direction),
N2(g) ϩ 3H2(g) 3z 2NH3(g)
and the reaction produces more NH3. As the concentration of NH3 increases,
the reverse reaction also speeds up (as more collisions between NH3 molecules occur) and the system again comes to equilibrium. However, the new
equilibrium position has more NH3 than was present in the original position. We say that the equilibrium has shifted to the right—toward the products. The original and new equilibrium positions are shown below.
Equilibrium Position I
Equilibrium Position II
[N2] ϭ 0.399 M
[H2] ϭ 1.197 M
[NH3] ϭ 0.203 M
[N2] ϭ 1.348 M
[H2] ϭ 1.044 M
[NH3] ϭ 0.304 M
1.000 mol/L
of N2 added
Note that the equilibrium does, in fact, shift to the right; the concentration
of H2 decreases (from 1.197 M to 1.044 M), the concentration of NH3 increases (from 0.203 M to 0.304 M), and, of course, because nitrogen was
added, the concentration of N2 shows an increase relative to the original
amount present.
It is important to note at this point that, although the equilibrium
shifted to a new position, the value of K did not change. We can demonstrate
this by inserting the equilibrium concentrations from positions I and II into
the equilibrium expression.
• Position I: K ϭ
• Position II: K ϭ
[NH3]2
[N2][H2]3
[NH3]2
[N2][H2]3
ϭ
ϭ
(0.203)2
(0.399)(1.197)3
(0.304)2
(1.348)(1.044)3
ϭ 0.0602
ϭ 0.0602
These values of K are the same. Therefore, although the equilibrium position
shifted when we added more N2, the equilibrium constant K remained the
same.
Could we have predicted this shift by using Le Châtelier’s principle? Because the change in this case was to add nitrogen, Le Châtelier’s principle
560 Chapter 17 Equilibrium
N2
added
N2
H2
NH3
a
Figure 17.9
A system at equilibrium shifts in
the direction that compensates
for any imposed change.
The initial
equilibrium mixture
of N2 , H2 , and NH3 .
b
Addition of N2 .
c
The new equilibrium position for
the system containing more N2
(because of the addition of N2 ),
less H2 , and NH3 than in a.
predicts that the system will shift in a direction that consumes nitrogen.
This tends to offset the original change—the addition of N2. Therefore, Le
Châtelier’s principle correctly predicts that adding nitrogen will cause the
equilibrium to shift to the right (Figure 17.9) as some of the added nitrogen
is consumed.
If ammonia had been added instead of nitrogen, the system would have
shifted to the left, consuming ammonia. Another way of stating Le Châtelier’s principle, then, is to say that when a reactant or product is added to a system at equilibrium, the system shifts away from the added component. On the
other hand, if a reactant or product is removed, the system shifts toward the removed component. For example, if we had removed nitrogen, the system
would have shifted to the left and the amount of ammonia present would
have been reduced.
A real-life example that shows the importance of Le Châtelier’s principle is the effect of high elevations on the oxygen supply to the body. If you
have ever traveled to the mountains on vacation, you may have noticed that
you felt “light-headed” and especially tired during the first few days of your
visit. These feelings resulted from a decreased supply of oxygen to your body
because of the lower air pressure that exists at higher elevations. For example, the oxygen supply in Leadville, Colorado (elevation ϳ 10,000 ft), is only
about two-thirds that found at sea level. We can understand the effects of diminished oxygen supply in terms of the following equilibrium:
Hb(aq) ϩ 4O2(g) 4
3 Hb(O2)4(aq)
where Hb represents hemoglobin, the iron-containing protein that transports O2 from your lungs to your tissues, where it is used to support metabolism. The coefficient 4 in the equation signifies that each hemoglobin molecule picks up four O2 molecules in the lungs. Note by Le Châtelier’s
principle that a lower oxygen pressure will cause this equilibrium to shift to
the left, away from oxygenated hemoglobin. This leads to an inadequate
oxygen supply at the tissues, which in turn results in fatigue and a “woozy”
feeling.
This problem can be solved in extreme cases, such as when climbing
Mt. Everest or flying in a plane at high altitudes, by supplying extra oxygen
from a tank. This extra oxygen pushes the equilibrium to its normal position. However, lugging around an oxygen tank would not be very practical
for people who live in the mountains. In fact, nature solves this problem
in a very interesting way. The body adapts to living at high elevations by
17.7 Le Châtelier’s Principle
561
producing additional hemoglobin—the other way to shift this equilibrium
to the right. Thus, people who live at high elevations have significantly
higher hemoglobin levels than those living at sea level. For example, the
Sherpas who live in Nepal can function in the rarefied air at the top of Mt.
Everest without an auxiliary oxygen supply.
EXAMPLE 17.4
Using Le Châtelier’s Principle: Changes in Concentration
Arsenic, As4, is obtained from nature by first reacting its ore with oxygen
(called roasting) to form solid As4O6. (As4O6, a toxic compound fatal in doses
of 0.1 g or more, is the “arsenic” made famous in detective stories.) The
As4O6 is then reduced using carbon:
As4O6(s) ϩ 6C(s) 4
3 As4(g) ϩ 6CO(g)
Predict the direction of the shift in the equilibrium position for this reaction
that occurs in response to each of the following changes in conditions.
a. Addition of carbon monoxide
b. Addition or removal of C(s) or As4O6(s)
c. Removal of As4(g)
SOLUTION
a. Le Châtelier’s principle predicts a shift away from the substance
whose concentration is increased. The equilibrium position will shift
to the left when carbon monoxide is added.
b. Because the amount of a pure solid has no effect on the equilibrium
position, changing the amount of carbon or tetraarsenic hexoxide
will have no effect.
c. When gaseous arsenic is removed, the equilibrium position will shift
to the right to form more products. In industrial processes, the
desired product is often continuously removed from the reaction
system to increase the yield.
Self-Check EXERCISE 17.3 Novelty devices for predicting rain contain cobalt(II) chloride and are based
on the following equilibrium:
CoCl2(s) ϩ 6H2O(g) 4
3 CoCl2 и 6H2O(s)
Blue
Pink
© Cengage Learning
What color will this indicator be when rain is likely due to increased water
vapor in the air?
See Problems 17.33 through 17.36. ■
▲
When blue anhydrous CoCl2
reacts with water, pink
CoCl2и6H2O is formed.
The Effect of a Change in Volume
When the volume of a gas is decreased (when a gas is compressed), the pressure increases. This occurs because the molecules present are now contained
in a smaller space and they hit the walls of their container more often, giving a greater pressure. Therefore, when the volume of a gaseous reaction system at equilibrium is suddenly reduced, leading to a sudden increase in pressure, by Le Châtelier’s principle the system will shift in the direction that
reduces the pressure.
562 Chapter 17 Equilibrium
For example, consider the reaction
CaCO3(s) 4
3 CaO(s) ϩ CO2(g)
in a container with a movable piston (Figure 17.10). If the volume is suddenly
decreased by pushing in the piston, the pressure of the CO2 gas initially increases. How can the system offset this pressure increase? By shifting to the
left—the direction that reduces the amount of gas present. That is, a shift to
the left will use up CO2 molecules, thus lowering the pressure. (There will then
be fewer molecules present to hit the walls, because more of the CO2 molecules
have combined with CaO and thus have become part of the solid CaCO3.)
Therefore, when the volume of a gaseous reaction system at equilibrium is decreased (thus increasing the pressure), the system shifts in the direction that gives the smaller number of gas molecules. So a decrease in the system
volume leads to a shift that decreases the total number of gaseous molecules
in the system.
Suppose we are running the reaction
N2(g) ϩ 3H2(g) 4
3 2NH3(g)
and we have a mixture of the gases nitrogen, hydrogen, and ammonia at
equilibrium (Figure 17.11a). If we suddenly reduce the volume, what will
happen to the equilibrium position? Because the decrease in volume initially
increases the pressure, the system moves in the direction that lowers its pressure. The reaction system can reduce its pressure by reducing the number of
gas molecules present. This means that the reaction
N2(g) ϩ 3H2(g)
4
3
4 gaseous molecules
2NH3(g)
2 gaseous molecules
shifts to the right, because in this direction four molecules (one of nitrogen
and three of hydrogen) react to produce two molecules (of ammonia), thus
reducing the total number of gaseous molecules present. The equilibrium position
shifts to the right—toward the side of the reaction that involves the smaller
number of gaseous molecules in the balanced equation.
The opposite is also true. When the container volume is increased
(which lowers the pressure of the system), the system shifts so as to increase
its pressure. An increase in volume in the ammonia synthesis system produces a shift to the left to increase the total number of gaseous molecules
present (to increase the pressure).
CO2
CaCO3
CaO
a
The system is initially at equilibrium.
Figure 17.10
The reaction system
CaCO3(s) 4
3 CaO(s) ϩ CO2(g).
b
Then the piston is pushed in,
decreasing the volume and
increasing the pressure. The
system shifts in the direction
that consumes CO2 molecules,
thus lowering the pressure
again.
563
17.7 Le Châtelier’s Principle
N2
H2
NH3
a
c
b
A mixture of NH3(g),
N2(g), and H2(g) at
equilibrium.
The volume is
suddenly decreased.
The new equilibrium
position for the
system containing
more NH3 and less N2
and H2. The reaction
N2(g) ؉ 3H2(g) I
2NH3(g) shifts to the
right (toward the side
with fewer molecules)
when the container
volume is decreased.
Figure 17.11
EXAMPLE 17.5
Using Le Châtelier’s Principle: Changes in Volume
Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.
a. The preparation of liquid phosphorus trichloride by the reaction
P4(s) ϩ 6Cl2(g)
4
3
6 gaseous molecules
4PCl3(l)
0 gaseous molecules
SOLUTION a
P4 and PCl3 are a pure solid and a pure liquid, respectively, so we need to consider only the effect on Cl2. If the volume is decreased, the Cl2 pressure will
initially increase, so the position of the equilibrium will shift to the right,
consuming gaseous Cl2 and lowering the pressure (to counteract the original
change).
b. The preparation of gaseous phosphorus pentachloride according to
the equation
PCl3(g) ϩ Cl2(g)
2 gaseous molecules
4
3
PCl5(g)
1 gaseous molecule
SOLUTION b
Decreasing the volume (increasing the pressure) will shift this equilibrium to
the right, because the product side contains only one gaseous molecule
while the reactant side has two. That is, the system will respond to the decreased volume (increased pressure) by lowering the number of molecules
present.
c. The reaction of phosphorus trichloride with ammonia:
PCl3(g) ϩ 3NH3(g) 4
3 P(NH2)3(g) ϩ 3HCl(g)
564 Chapter 17 Equilibrium
SOLUTION c
Both sides of the balanced reaction equation have four gaseous molecules. A
change in volume will have no effect on the equilibrium position. There is
no shift in this case, because the system cannot change the number of molecules present by shifting in either direction.
Self-Check EXERCISE 17.4 For each of the following reactions, predict the direction the equilibrium will
shift when the volume of the container is increased.
a. H2(g) ϩ F2(g) 4
3 2HF(g)
b. CO(g) ϩ 2H2(g) 4
3 CH3OH(g)
c. 2SO3(g) 4
3 2SO2(g) ϩ O2(g)
See Problems 17.33 through 17.36. ■
▲
The Effect of a Change in Temperature
It is important to remember that although the changes we have just discussed may alter the equilibrium position, they do not alter the equilibrium
constant. For example, the addition of a reactant shifts the equilibrium position to the right but has no effect on the value of the equilibrium constant;
the new equilibrium concentrations satisfy the original equilibrium constant. This was demonstrated earlier in this section for the addition of N2 to
the ammonia synthesis reaction.
The effect of temperature on equilibrium is different, however, because
the value of K changes with temperature. We can use Le Châtelier’s principle to
predict the direction of the change in K.
To do this we need to classify reactions according to whether they produce heat or absorb heat. A reaction that produces heat (heat is a “product”)
is said to be exothermic. A reaction that absorbs heat is called endothermic.
Because heat is needed for an endothermic reaction, energy (heat) can be
regarded as a “reactant” in this case.
In an exothermic reaction, heat is treated as a product. For example, the
synthesis of ammonia from nitrogen and hydrogen is exothermic (produces
heat). We can represent this by treating energy as a product:
N2(g) ϩ 3H2(g) 4
3 2NH3(g) ϩ 92 kJ
Energy
released
Le Châtelier’s principle predicts that when we add energy to this system at
equilibrium by heating it, the shift will be in the direction that consumes
energy—that is, to the left.
On the other hand, for an endothermic reaction (one that absorbs energy), such as the decomposition of calcium carbonate,
CaCO3(s) ϩ 556 kJ 4
3 CaO(s) ϩ CO2(g)
Energy
needed
energy is treated as a reactant. In this case an increase in temperature causes
the equilibrium to shift to the right.
In summary, to use Le Châtelier’s principle to describe the effect of a
temperature change on a system at equilibrium, simply treat energy as a reactant (in an endothermic process) or as a product (in an exothermic process), and
predict the direction of the shift in the same way you would if an actual reactant or product were being added or removed.
17.7 Le Châtelier’s Principle
EXAMPLE 17.6
565
Using Le Châtelier’s Principle: Changes in Temperature
For each of the following reactions, predict how the equilibrium will shift as
the temperature is increased.
a. N2(g) ϩ O2(g) 4
3 2NO( g) (endothermic)
SOLUTION a
This is an endothermic reaction, so energy can be viewed as a reactant.
N2(g) ϩ O2(g) ϩ energy 4
3 2NO(g)
Thus the equilibrium will shift to the right as the temperature is increased
(energy added).
b. 2SO2(g) ϩ O2(g) 4
3 2SO3(g) (exothermic)
SOLUTION b
This is an exothermic reaction, so energy can be regarded as a product.
2SO2(g) ϩ O2(g) 4
3 2SO3(g) ϩ energy
As the temperature is increased, the equilibrium will shift to the left.
Self-Check EXERCISE 17.5 For the exothermic reaction
2SO2(g) ϩ O2(g) 4
3 2SO3(g)
predict the equilibrium shift caused by each of the following changes.
a. SO2 is added.
b. SO3 is removed.
c. The volume is decreased.
d. The temperature is decreased.
See Problems 17.33 through 17.42. ■
We have seen how Le Châtelier’s principle can be used to predict the effects of several types of changes on a system at equilibrium. To summarize
these ideas, Table 17.2 shows how various changes affect the equilibrium position of the endothermic reaction N2O4(g) 4
3 2NO2(g). The effect of a temperature change on this system is depicted in Figure 17.12.
Table 17.2 Shifts in the Equilibrium Position for
3 2NO2(g)
the Reaction N2O4(g) ؉ Energy 4
addition of N2O4(g)
right
addition of NO2(g)
left
removal of N2O4(g)
left
removal of NO2(g)
right
decrease in container volume
left
increase in container volume
right
increase in temperature
right
decrease in temperature
left