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8 Altering an Equilibrium Mixture: Changes in Pressure and Volume

# 8 Altering an Equilibrium Mixture: Changes in Pressure and Volume

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13.8 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN PRESSURE AND VOLUME

To see why Le Châtelier’s principle works for pressure (volume) changes, let’s

look again at the reaction quotient for the equilibrium mixture of 0.50 M N2, 3.00 M

H2, and 1.98 M NH3 at 700 K:

Qc =

[NH 3]t 2

[N2]t[H 2]t 3

=

(1.98)2

(0.50)(3.00)3

= 0.29 = Kc

If we disturb the equilibrium by reducing the volume by a factor of 2, we not only

double the total pressure, we also double the partial pressure and thus the molar concentration of each reactant and product (because molarity = n/V = P/RT). Because

the balanced equation has more moles of gaseous reactants than gaseous products,

the increase in the denominator of the equilibrium constant expression is greater

than the increase in the numerator and the new value of Qc is less than the equilibrium constant Kc:

Qc =

[NH 3]t 2

[N2]t[H 2]t 3

=

(3.96)2

(1.00)(6.00)3

= 0.0726 6 Kc

For the system to move to a new state of equilibrium, Qc must increase, which

means that the net reaction must go from reactants to products, as predicted by Le

Châtelier’s principle (Figure 13.11). In practice, the yield of ammonia in the Haber

process is increased by running the reaction at high pressure, typically 130–300 atm.

(a) A mixture of gaseous N2,

(b) When the pressure is increased

by decreasing the volume, the

mixture is no longer at

equilibrium (Qc < Kc).

H2, and NH3 at equilibrium

(Qc = Kc).

(c) Net reaction occurs from reactants

to products, decreasing the total

number of gaseous molecules until

equilibrium is re-established

(Qc = Kc).

= N2

P increases as

Net reaction

V decreases

to form products

Figure 13.11

Qualitative effect of pressure and volume on the equilibrium

N2(g) ؉ 3 H2(g) Δ 2 NH3(g).

The composition of an equilibrium mixture is unaffected by a change in pressure

if the reaction involves no change in the number of moles of gas. For example, the

reaction of hydrogen with gaseous iodine has 2 mol of gas on both sides of the balanced equation:

H 2(g) + I 2(g) Δ 2 HI(g)

If we double the pressure by halving the volume, the numerator and denominator of

the reaction quotient change by the same factor and Qc remains unchanged:

Qc =

[HI]t 2

[H 2]t[I 2]t

= H2

= NH3

517

518

Chapter 13 CHEMICAL EQUILIBRIUM

In applying Le Châtelier’s principle to a heterogeneous equilibrium, the effect of

pressure changes on solids and liquids can be ignored because the volume (and concentration) of a solid or a liquid is nearly independent of pressure. Consider, for

example, the high-temperature reaction of carbon with steam, the first step in converting coal to gaseous fuels:

C(s) + H 2O(g) Δ CO(g) + H 2(g)

Ignoring the carbon because it’s a solid, we predict that a decrease in volume

(increase in pressure) will shift the equilibrium from products to reactants because

the reverse reaction decreases the amount of gas from 2 mol to 1 mol.

Throughout this section, we’ve been careful to limit the application of Le

Châtelier’s principle to pressure changes that result from a change in volume. What

happens, though, if we keep the volume constant but increase the total pressure by

adding a gas that is not involved in the reaction—say, an inert gas such as argon? In

that case, the equilibrium remains undisturbed because adding an inert gas at constant volume does not change the partial pressures or the molar concentrations of the

substances involved in the reaction. Only if the added gas is a reactant or product

does the reaction quotient change.

WORKED EXAMPLE 13.13

APPLYING LE CHÂTELIER’S PRINCIPLE TO PRESSURE

AND VOLUME CHANGES

Does the number of moles of reaction products increase, decrease, or remain the same

when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a) PCl5(g) Δ PCl3(g) + Cl2(g)

(b) CaO(s) + CO2(g) Δ CaCO3(s)

(c) 3 Fe(s) + 4 H 2O(g) Δ Fe3O4(s) + 4 H 2(g)

STRATEGY

According to Le Châtelier’s principle, the stress of a decrease in pressure is relieved by

net reaction in the direction that increases the number of moles of gas.

SOLUTION

(a) Because the forward reaction converts 1 mol of gas to 2 mol of gas, net reaction will

go from reactants to products, thus increasing the number of moles of PCl3 and Cl2.

(b) Because there is 1 mol of gas on the reactant side of the balanced equation and none

on the product side, the stress of a decrease in pressure is relieved by net reaction

from products to reactants. The number of moles of CaCO3 therefore decreases.

(c) Because there are 4 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by a change in pressure. The number

of moles of Fe3O4 and H2 remains the same.

Ī PROBLEM 13.17 Does the number of moles of products increase, decrease, or remain

the same when each of the following equilibria is subjected to an increase in pressure by

decreasing the volume?

(a) CO(g) + H 2O(g) Δ CO 2(g) + H 2(g)

(b) 2 CO(g) Δ C(s) + CO2(g)

(c) N2O4(g) Δ 2 NO 2(g)

CONCEPTUAL PROBLEM 13.18 The following picture represents the equilibrium mixture for the gas-phase reaction A2 Δ 2 A:

Draw a picture that shows how the concentrations change when the pressure is increased

by reducing the volume.

13.9 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN TEMPERATURE

519

13.9 ALTERING AN EQUILIBRIUM MIXTURE:

CHANGES IN TEMPERATURE

When an equilibrium is disturbed by a change in concentration, pressure, or volume,

the composition of the equilibrium mixture changes because the reaction quotient Qc

no longer equals the equilibrium constant Kc. As long as the temperature remains

constant, however, concentration, pressure, or volume changes don’t change the

value of the equilibrium constant.

By contrast, a change in temperature nearly always changes the value of the equilibrium constant. For the Haber synthesis of ammonia, which is an exothermic

reaction, the equilibrium constant Kc decreases by a factor of 1011 over the temperature range 300–1000 K (Figure 13.12).

N2(g) + 3 H 2(g) Δ 2 NH 3(g) + 92.2 kJ

¢H° = -92.2 kJ

Figure 13.12

Temperature dependence of the

equilibrium constant for the reaction

N2(g) ؉ 3 H2(g) Δ 2 NH3(g).

Kc is plotted on a

logarithmic scale.

Equilibrium constant, Kc

10

Temp

(K)

8

Kc

300

2.6 × 10 8

400

3.9 × 10 4

500

1.7 × 10 2

600

4.2

700

2.9 × 10−1

1

800

3.9 × 10−2

10−2

900

8.1 × 10−3

1000

2.3 × 10−3

10 6

Kc decreases by a factor

of 1011 on raising T from

300 K to 1000 K.

10 4

10 2

10−4

200

400

600

800

N2O4(g)

2 NO2(g); ΔH > 0

1000

Temperature (K)

At low temperatures, the equilibrium mixture is rich in NH3 because Kc is large.

At high temperatures, the equilibrium shifts in the direction of N2 and H2.

In general, the temperature dependence of an equilibrium constant depends on

the sign of ¢H° for the reaction.

• The equilibrium constant for an exothermic reaction (negative ÂH) decreases

as the temperature increases.

The equilibrium constant for an endothermic reaction (positive ¢H°) increases as

the temperature increases.

You can predict the way in which Kc depends on temperature by using Le

Châtelier’s principle. Take the endothermic decomposition of N2O4, for example:

N2O4(g) + 55.3 kJ Δ 2 NO2(g)

Colorless

¢H° = +55.3 kJ

Brown

Le Châtelier’s principle says that if heat is added to an equilibrium mixture, thus

increasing its temperature, net reaction occurs in the direction that relieves the stress

of the added heat. For an endothermic reaction, such as the decomposition of N2O4,

heat is absorbed by reaction in the forward direction. The equilibrium therefore shifts

to the product side at the higher temperature, which means that Kc increases with

increasing temperature.

Because N2O4 is colorless and NO2 has a brown color, the effect of temperature on

the N2O4 -NO2 equilibrium is readily apparent from the color of the mixture (Figure

13.13). For an exothermic reaction, such as the Haber synthesis of NH3, heat is absorbed

by net reaction in the reverse direction, so Kc decreases with increasing temperature.

Kc increases as

T increases

The darker brown color of the sample

at the highest temperature indicates that

the equilibrium N2O4(g)

2 NO2(g)

shifts from reactants to products with

increasing temperature, as expected for

an endothermic reaction.

Figure 13.13

Sample tubes containing an equilibrium

mixture of N2O4 and NO2 immersed in

ice water (left), at room temperature

(center), and immersed in hot water

(right).

520

Chapter 13 CHEMICAL EQUILIBRIUM

WORKED EXAMPLE 13.14

APPLYING LE CHÂTELIER’S PRINCIPLE TO TEMPERATURE CHANGES

In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction

4 NH 3(g) + 5 O 2(g) Δ 4 NO(g) + 6 H 2O(g)

¢H° = -901.2 kJ

How does the equilibrium amount of NO vary with an increase in temperature?

STRATEGY

Le Châtelier’s principle predicts that the stress of added heat when the temperature is

increased will be relieved by net reaction in the direction that absorbs the heat. It’s

helpful to include the heat in the balanced equation—on the reactant side if the reaction

is endothermic, or on the product side if the reaction is exothermic.

SOLUTION

Because the oxidation of ammonia is exothermic, we include the heat (901.2 kJ) on the

product side:

4 NH 3(g) + 5 O 2(g) Δ 4 NO(g) + 6 H 2O(g) + 901.2 kJ

The stress of added heat when the temperature is increased will be relieved by net reaction from products to reactants, which absorbs the added heat. The equilibrium will

therefore shift to the reactant side (Kc will decrease) with an increase in temperature.

Consequently, the equilibrium mixture will contain less NO at higher temperatures.

Ī PROBLEM 13.19 When air is heated at very high temperatures in an automobile

engine, the air pollutant nitric oxide is produced by the reaction

N2(g) + O 2(g) Δ 2 NO(g)

¢H° = +182.6 kJ

How does the equilibrium amount of NO vary with an increase in temperature?

Ī PROBLEM 13.20 Ethyl acetate, a solvent used in many fingernail-polish removers, is

made by the reaction of acetic acid with ethanol:

CH 3CO2H(soln) + CH 3CH 2OH(soln) Δ CH 3CO2CH 2CH 3(soln) + H 2O(soln)

Acetic acid

Ethanol

¢H° = -2.9 kJ

Ethyl acetate

Does the amount of ethyl acetate in an equilibrium mixture increase or decrease when the

temperature is increased? How does Kc change when the temperature is decreased?

Justify your answers using Le Châtelier’s principle.

WORKED CONCEPTUAL EXAMPLE 13.15

APPLYING LE CHÂTELIER’S PRINCIPLE TO TEMPERATURE CHANGES

᭡ Fingernail polish can be removed

by dissolving it in ethyl acetate.

The following pictures represent the composition of the equilibrium mixture at 600 K

and 650 K for the combination of two A molecules, 2 A(g) Δ A2(g):

T = 600 K

T = 650 K

Is the reaction endothermic or exothermic? Explain using Le Châtelier’s principle.

STRATEGY

We can determine the direction of net reaction on raising the temperature by counting

the number of A and A2 molecules at each temperature. According to Le Châtelier’s

principle, if the net reaction converts reactants to products on raising the temperature,

heat is on the reactant side of the chemical equation and the reaction is endothermic.

Conversely, if the net reaction converts products to reactants on raising the temperature, heat is on the product side and the reaction is exothermic.

13.10 THE EFFECT OF A CATALYST ON EQUILIBRIUM

521

SOLUTION

Two A and five A2 molecules are present at 600 K, and six A and three A2 molecules are

present at 650 K. On raising the temperature, the net reaction converts products to reactants, and so heat is on the product side of the chemical equation:

2 A(g) Δ A2(g) + heat

The reaction is therefore exothermic, as expected for a reaction in which a chemical

bond is formed. Note that Le Châtelier’s principle predicts that net reaction will occur

in the direction that uses up the added heat.

CONCEPTUAL PROBLEM 13.21 The following pictures represent the composition of

the equilibrium mixture for the reaction A(g) + B(s) Δ AB(g) at 400 K and 500 K:

T = 400 K

T = 500 K

Is the reaction endothermic or exothermic? Explain using Le Châtelier’s principle.

13.10 THE EFFECT OF A CATALYST

ON EQUILIBRIUM

Recall from Section 12.14 that a catalyst increases the rate of a chemical reaction by

making available a new, lower-energy pathway for the conversion of reactants to

products. Because the forward and reverse reactions pass through the same transition state, a catalyst lowers the activation energy for the forward and reverse

reactions by exactly the same amount. As a result, the rates of the forward and

reverse reactions increase by the same factor (Figure 13.14).

Remember...

A catalyst is a substance that increases the

rate of a transformation without being consumed in the process. (Section 12.14)

The activation energy for the catalyzed pathway (red curve) is lower than that for the

uncatalyzed pathway (blue curve) by an amount ΔEa.

Potential energy

ΔEa

Ea (forward)

without

catalyst

Ea (reverse)

without

catalyst

Ea (forward)

with

catalyst

Ea (reverse)

with

catalyst

Reactants

Products

Reaction progress

Because the forward and reverse reactions pass through the same transition state, the

catalyst lowers the activation energy barrier for the forward and reverse reactions by the

same amount. The catalyst therefore accelerates the forward and reverse reactions by

the same factor, and the composition of the equilibrium mixture is unchanged.

Figure 13.14

Potential energy profiles for a reaction

whose activation energy is lowered by

the presence of a catalyst.

522

Chapter 13 CHEMICAL EQUILIBRIUM

If a reaction mixture is at equilibrium in the absence of a catalyst (that is, the forward and reverse rates are equal), it will still be at equilibrium after a catalyst is

added because the forward and reverse rates, though faster, remain equal. If a reaction mixture is not at equilibrium, a catalyst accelerates the rate at which equilibrium

is reached, but it does not affect the composition of the equilibrium mixture. Because

a catalyst has no effect on the equilibrium concentrations, it does not appear in the

balanced chemical equation or in the equilibrium constant expression.

Even though a catalyst doesn’t change the position of an equilibrium, it can

nevertheless significantly influence the choice of optimum conditions for a

reaction. Look again at the Haber synthesis of ammonia. Because the reaction

N2(g) + 3 H 2(g) Δ 2 NH 3(g) is exothermic, its equilibrium constant decreases

with increasing temperature, and optimum yields of NH3 are obtained at low temperatures. At those low temperatures, however, the rate at which equilibrium is

reached is too slow for the reaction to be practical. We thus have what appears to be

a no-win situation: Low temperatures give good yields but slow rates, whereas high

temperatures give satisfactory rates but poor yields. The answer to the dilemma is to

find a catalyst.

In the early 1900s, the German chemist Fritz Haber discovered that a catalyst consisting of iron mixed with certain metal oxides causes the reaction to occur at a

satisfactory rate at temperatures where the equilibrium concentration of NH3 is reasonably favorable. The yield of NH3 can be improved further by running the reaction

at high pressures. Typical reaction conditions for the industrial synthesis of ammonia

are 400–500 °C and 130–300 atm.

Ī PROBLEM 13.22 A platinum catalyst is used in automobile catalytic converters to

hasten the oxidation of carbon monoxide:

Pt

2 CO(g) + O 2(g) ERF 2 CO2(g)

¢H° = -566 kJ

Suppose that you have a reaction vessel containing an equilibrium mixture of CO(g),

O 2(g), and CO2(g). Under the following conditions, will the amount of CO increase,

decrease, or remain the same?

(a) A platinum catalyst is added.

(b) The temperature is increased.

(c) The pressure is increased by decreasing the volume.

(d) The pressure is increased by adding argon gas.

(e) The pressure is increased by adding O2 gas.

13.11 THE LINK BETWEEN CHEMICAL

EQUILIBRIUM AND CHEMICAL KINETICS

We emphasized in Section 13.1 that the equilibrium state is a dynamic one in which

reactant and product concentrations remain constant, not because the reaction stops

but because the rates of the forward and reverse reactions are equal. To explore this

idea further, let’s consider the general, reversible reaction

A + B Δ C + D

Remember...

Because an elementary reaction describes

an individual molecular event, its rate law

follows directly from its stoichiometry.

(Sections 12.9 and 12.10)

Let’s assume that the forward and reverse reactions occur in a single bimolecular

step; that is, they are elementary reactions (Section 12.9). We can then write the

following rate laws:

Rate forward = kf[A][B]

Rate reverse = kr[C][D]

13.11 THE LINK BETWEEN CHEMICAL EQUILIBRIUM AND CHEMICAL KINETICS

523

If we begin with a mixture that contains all reactants and no products, the initial rate

of the reverse reaction is zero because [C] = [D] = 0. As A and B are converted to C

and D by the forward reaction, the rate of the forward reaction decreases because [A]

and [B] are getting smaller. At the same time, the rate of the reverse reaction increases

because [C] and [D] are getting larger. Eventually, the decreasing rate of the forward

reaction and the increasing rate of the reverse reaction become equal, and thereafter

the concentrations remain constant; that is, the system is at chemical equilibrium

(Figure 13.2, page 495).

Because the forward and reverse rates are equal at equilibrium, we can write

kf[A][B] = kr[C][D]

at equilibrium

which can be rearranged to give

kf

[C][D]

=

kr

[A][B]

The right side of this equation is the equilibrium constant expression for the forward reaction, which equals the equilibrium constant Kc since the reaction mixture is

at equilibrium.

Kc =

[C][D]

[A][B]

Therefore, the equilibrium constant is simply the ratio of the rate constants for the

forward and reverse reactions:

Kc =

kf

kr

In deriving this equation for Kc, we have assumed a single-step mechanism. For a

multistep mechanism, each step has a characteristic rate constant ratio, kf/kr. When

equilibrium is reached, each step in the mechanism must be at equilibrium, and Kc

for the overall reaction is equal to the product of the rate constant ratios for the individual steps.

The equation relating Kc to kf and kr provides a fundamental link between chemical equilibrium and chemical kinetics: The relative values of the rate constants for

the forward and reverse reactions determine the composition of the equilibrium mixture. When kf is much larger than kr, Kc is very large and the reaction goes almost to

completion. Such a reaction is sometimes said to be irreversible because the reverse

reaction is often too slow to be detected. When kf and kr have comparable values, Kc

has a value near 1, and comparable concentrations of both reactants and products are

present at equilibrium. This is the usual situation for a reversible reaction.

Addition of a catalyst to a reaction mixture increases both rate constants kf and kr

because the reaction takes place by a different, lower-energy mechanism. Because kf

and kr increase by the same factor, though, the ratio kf/kr is unaffected, and the value

of the equilibrium constant Kc = kf/kr remains unchanged. Thus, addition of a catalyst does not alter the composition of an equilibrium mixture.

The equation Kc = kf/kr also helps explain why equilibrium constants depend on

temperature. Recall from Section 12.12 that the rate constant increases as the temperature increases, in accord with the Arrhenius equation k = Ae - Ea/RT. In general, the

forward and reverse reactions have different values of the activation energy, so kf and

kr increase by different amounts as the temperature increases. The ratio kf/kr = Kc is

therefore temperature-dependent. For an exothermic reaction, Ea for the reverse reaction is greater than Ea for the forward reaction. Consequently, as the temperature

increases, kr increases by more than kf increases, and so Kc = kf/kr for an exothermic

reaction decreases as the temperature increases. Conversely, Kc for an endothermic

reaction increases as the temperature increases. These results are in accord with Le

Châtelier’s principle (Section 13.9).

Remember...

Because the fraction of collisions with sufficient energy for reaction is given by e-Ea /RT,

the Arrhenius equation indicates that the

rate constant decreases as Ea increases and

increases as T increases. (Section 12.12)

524

Chapter 13 CHEMICAL EQUILIBRIUM

WORKED EXAMPLE 13.16

EXPLORING THE LINK BETWEEN EQUILIBRIUM AND KINETICS

The equilibrium constant Kc for the reaction of hydrogen with iodine is 57.0 at 700 K,

and the reaction is endothermic (¢E = 9 kJ).

kf

H 2(g) + I 2(g) ERF 2 HI(g)

kr

Kc = 57.0 at 700 K

(a) Is the rate constant kf for the formation of HI larger or smaller than the rate constant

kr for the decomposition of HI?

(b) The value of kr at 700 K is 1.16 * 10-3 M -1 s -1. What is the value of kf at the same

temperature?

(c) How are the values of kf, kr, and Kc affected by the addition of a catalyst?

(d) How are the values of kf, kr, and Kc affected by an increase in temperature?

STRATEGY

Remember...

The greater the activation energy, the

steeper the slope of an Arrhenius plot (a

graph of ln k versus 1/T) and the greater

the increase in k for a given increase in T.

(Section 12.13)

To answer these questions, make use of the relationship Kc = kf/kr. Also, remember

that a catalyst increases kf and kr by the same factor, and recall that the temperature

dependence of a rate constant increases with increasing value of the activation energy

(Section 12.13).

SOLUTION

(a) Because Kc = kf/kr = 57.0, the rate constant for the formation of HI (forward reaction) is larger than the rate constant for the decomposition of HI (reverse reaction)

by a factor of 57.0.

(b) Because Kc = kf/kr,

kf = (Kc)(kr) = (57.0)(1.16 * 10-3 M -1 s -1) = 6.61 * 10-2 M -1 s -1

(c) A catalyst lowers the activation energy barrier for the forward and reverse reactions

by the same amount, thus increasing the rate constants kf and kr by the same factor.

Because the equilibrium constant Kc equals the ratio of kf to kr, the value of Kc is

unaffected by the addition of a catalyst.

(d) Because the reaction is endothermic, Ea for the forward reaction is greater than Ea

for the reverse reaction. Consequently, as the temperature increases, kf increases by

more than kr increases, and therefore Kc = kf/kr increases, consistent with Le

Châtelier’s principle.

Ī PROBLEM 13.23 Nitric oxide emitted from the engines of supersonic aircraft can

contribute to the destruction of stratospheric ozone:

kf

NO(g) + O 3(g) ERF NO2(g) + O 2(g)

k

r

᭡ Nitric oxide emissions from supersonic

aircraft can contribute to destruction of

the ozone layer.

This reaction is highly exothermic (¢E = -201 kJ), and its equilibrium constant Kc is

3.4 * 1034 at 300 K.

(a) Which rate constant is larger, kf or kr?

(b) The value of kf at 300 K is 8.5 * 106 M -1 s -1. What is the value of kr at the same

temperature?

(c) A typical temperature in the stratosphere is 230 K. Do the values of kf, kr, and Kc

increase or decrease when the temperature is lowered from 300 K to 230 K?

INQUIRY HOW DOES EQUILIBRIUM AFFECT OXYGEN TRANSPORT IN THE BLOODSTREAM?

525

INQUIRY HOW DOES EQUILIBRIUM AFFECT

OXYGEN TRANSPORT IN THE BLOODSTREAM?

Humans, like all animals, need oxygen. The oxygen comes from breathing: About

500 mL of air is drawn into the lungs of an average person with each breath. When

the freshly inspired air travels through the bronchial passages and enters the approximately 150 million alveolar sacs of the lungs, it picks up moisture and mixes with air

remaining from the previous breath. As it mixes, the concentrations of both water

vapor and carbon dioxide increase. These gas concentrations are measured by their

partial pressures, with the partial pressure of oxygen in the lungs usually around

100 mm Hg (Table 13.2). Oxygen then diffuses through the delicate walls of the lung

alveoli and into arterial blood, which transports it to all body tissues.

Only about 3% of the oxygen in blood is dissolved; the rest is chemically bound

to hemoglobin molecules (Hb), large proteins that contain heme groups embedded in

them. Each hemoglobin molecule contains four heme groups, and each heme group

contains an iron atom that can bind to one O2 molecule. Thus, a single hemoglobin

molecule can bind four molecules of oxygen.

Arterial

blood vessel

PO2 (mm Hg)

Dry air

Alveolar air

Arterial blood

Venous blood

H3C C

Alveolar

wall

Red blood cell

O2

H

C

C

C

N

C

N

C

CH3

Alveoli

CO2

Capillaries

1. Air enters the

lungs and reaches

the alveoli, where

gases are picked up

by capillaries.

CH2CH2CO2–

C

C

C

C

C

CH3

N C

Fe

C

H2C CH C

159

100

95

40

H

O2CCH2CH2

Capillary

Venous

blood vessel

Partial Pressure of

Oxygen in Human

Lungs and Blood

at Sea Level

Source

2. Oxygen in air binds to

hemoglobin molecules in

red blood cells.

Bronchiole

TABLE 13.2

C

N

C

H

C

H

C

C

C

CH

CH3

CH2

Heme — an O2 molecule binds

to the central iron atom.

3. Each hemoglobin

molecule can bind four

molecules of oxygen.

The entire system of oxygen transport and delivery in the body depends on

the pickup and release of O2 by hemoglobin according to the following series of

equilibria:

Hb + O2 Δ Hb(O2)

Hb(O2) + O2 Δ Hb(O2)2

Hb(O2)2 + O2 Δ Hb(O2)3

Hb(O2)3 + O2 Δ Hb(O 2)4

The positions of the different equilibria depend on the partial pressures of O2 (PO2) in

the various tissues. In hard-working, oxygen-starved muscles, where PO2 is low, oxygen is released from hemoglobin as the equilibria shift toward the left, according to

Le Châtelier’s principle. In the lungs, where PO2 is high, oxygen is absorbed by

hemoglobin as the equilibria shift toward the right.

526

Chapter 13 CHEMICAL EQUILIBRIUM

The amount of oxygen carried by hemoglobin at any given value of PO2 is usually

expressed as a percent saturation and can be found from the curve shown in Figure

13.15. The saturation is 97.5% in the lungs, where PO2 = 100 mm Hg, meaning that

each hemoglobin molecule is carrying close to its maximum possible amount of 4 O2

molecules. When PO2 = 26 mm Hg, however, the saturation drops to 50%.

100

% Saturation

80

The saturation is 97.5% in the lungs,

where PO2 = 100 mm Hg, meaning

that each hemoglobin molecule is

carrying close to its maximum

possible amount of 4 O2 molecules.

60

40

When PO2 = 26 mm Hg, however,

the saturation drops to 50%.

20

0

0

20

40

60

80

100

120

PO (mm Hg)

2

Figure 13.15

An oxygen-carrying curve for hemoglobin. The percent saturation

of the oxygen-binding sites on hemoglobin depends on the partial

pressure of oxygen (PO2).

What about people who live at high altitudes? In Leadville, Colorado, for example, where the altitude is 10,156 ft, the partial pressure of O2 in the lungs is only about

68 mm Hg. Hemoglobin is only 90% saturated with O2 at this pressure, so less oxygen is available for delivery to the tissues. People who climb suddenly from sea level

to high altitude thus experience a feeling of oxygen deprivation, or hypoxia, as their

bodies are unable to supply enough oxygen to their tissues. The body soon copes

with the situation, though, by producing more hemoglobin molecules, which both

provide more capacity for O2 transport and also drive the Hb + O2 equilibria to the

right. The time required to adapt to the lower O2 pressures is typically days to weeks,

so athletes and hikers must train at high altitudes for some time.

Ī PROBLEM 13.24 The affinity of hemoglobin (Hb) for CO is greater than its affinity

for O2. Use Le Châtelier’s principle to predict how CO affects the equilibrium

Hb + O 2 Δ Hb(O2). Suggest a reason for the toxicity of CO.

᭡ The bodies of mountain dwellers

produce increased amounts of

hemoglobin to cope with the low O2

pressures at high altitudes.

Ī PROBLEM 13.25

How many O2 molecules are drawn into the lungs of an average

person with each breath? Assume that the ambient air pressure is 1.00 atm and the temperature is 25 °C.

CONCEPTUAL PROBLEMS

527

SUMMARY

Chemical equilibrium is a dynamic state in which the concentrations of reactants and products remain constant because the rates

of the forward and reverse reactions are equal. For the general

reaction a A + b B Δ c C + d D, concentrations in the

equilibrium mixture are related by the equilibrium equation:

Kc =

[C]c[D]d

[A]a[B]b

The quotient on the right side of the equation is called the

equilibrium constant expression. The equilibrium constant Kc is the

number obtained when equilibrium concentrations (in mol/L) are

substituted into the equilibrium constant expression. The value of

Kc varies with temperature and depends on the form of the balanced chemical equation.

The equilibrium constant Kp can be used for gas-phase reactions. It is defined in the same way as Kc except that the

equilibrium constant expression contains partial pressures (in

atmospheres) instead of molar concentrations. The constants Kp

and Kc are related by the equation Kp = Kc(RT)¢n, where

¢n = (c + d) - (a + b).

Homogeneous equilibria are those in which all reactants and

products are in a single phase; heterogeneous equilibria are those

in which reactants and products are present in more than one

phase. The equilibrium equation for a heterogeneous equilibrium

does not include concentrations of pure solids or pure liquids.

The value of the equilibrium constant for a reaction makes it

possible to judge the extent of reaction, predict the direction of

reaction, and calculate equilibrium concentrations (or partial pressures) from initial concentrations (or partial pressures). The farther

the reaction proceeds toward completion, the larger the value of

Kc. The direction of a reaction not at equilibrium depends on the

relative values of Kc and the reaction quotient Qc, which is

defined in the same way as Kc except that the concentrations in the

equilibrium constant expression are not necessarily equilibrium

concentrations. If Qc 6 Kc, the net reaction goes from reactants to

products to attain equilibrium; if Qc 7 Kc, the net reaction goes

from products to reactants; if Qc = Kc, the system is at equilibrium.

The composition of an equilibrium mixture can be altered by

changes in concentration, pressure (volume), or temperature. The

qualitative effect of these changes is predicted by Le Châtelier’s

principle, which says that if a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves

the stress. Temperature changes affect equilibrium concentrations

because Kc is temperature-dependent. As the temperature

increases, Kc for an exothermic reaction decreases and Kc for an

endothermic reaction increases.

A catalyst increases the rate at which chemical equilibrium is

reached, but it does not affect the equilibrium constant or the equilibrium concentrations. The equilibrium constant for a single-step

reaction equals the ratio of the rate constants for the forward and

reverse reactions: Kc = kf/kr.

KEY WORDS

chemical equilibrium 493

equilibrium constant Kc 496

equilibrium constant Kp 499

equilibrium equation 496

equilibrium mixture 493

heterogeneous equilibria 502

homogeneous equilibria 502

Le Châtelier’s principle 512

reaction quotient Qc 505

CONCEPTUAL PROBLEMS

Increasing time

Problems 13.1–13.25 appear within the chapter.

13.26 Consider the interconversion of A molecules (red spheres)

and B molecules (blue spheres) according to the reaction

A Δ B. Each of the series of pictures at the right represents a separate experiment in which time increases from

left to right:

(a) Which of the experiments has resulted in an equilibrium state?

(b) What is the value of the equilibrium constant Kc for the

reaction A Δ B?

(1)

(2)

(c) Explain why you can calculate Kc without knowing the

volume of the reaction vessel.

(3) ### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

8 Altering an Equilibrium Mixture: Changes in Pressure and Volume

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