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2: Atomic Masses: Counting Atoms by Weighing

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



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