534 chapTer 13 properties of Solutions · A solution that is in equilibrium with undissolved solute...

534 chapTer 13 properties of Solutions

We can imagine the solution process as having three components, each with an associated enthalpy change: A cluster of n solute particles must separate from one another 1∆Hsolute2, a cluster of m solvent particles separate from one another 1∆Hsolvent2, and these solute and solvent particles mix 1∆Hmix2.

1. 1solute2n ∆ n solute ∆Hsolute

2. 1solvent2m ∆ m solvent ∆Hsolvent

3. n solute + m solvent ∆ solution ∆Hmix

4. 1solute2n + 1solvent2m ∆ solution ∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix

As seen above, the overall enthalpy change, ∆Hsoln, is the sum of the three steps:

∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix [13.1]

Separation of the solute particles from one another always requires an input of energy to overcome their attractive interactions. The process is therefore endothermic 1∆Hsolute 7 02. Likewise, separation of solvent molecules to accommodate the solute always requires energy 1∆Hsolvent 7 02. The third component, which arises from the attractive interactions between solute particles and solvent particles, is always exother-mic 1∆Hmix 6 02.

The three enthalpy terms in Equation 13.1 can be added together to give either a negative or a positive sum, depending on the actual numbers for the system being con-sidered (▶ Figure 13.4). Thus, the formation of a solution can be either exothermic or endothermic. For example, when magnesium sulfate 1MgSO42 is added to water, the solution process is exothermic: ∆Hsoln = -91.2 kJ>mol. In contrast, the dissolution of ammonium nitrate 1NH4NO32 is endothermic: ∆Hsoln = 26.4 kJ>mol. These particu-lar salts are the main components in the instant heat packs and ice packs used to treat

▲ Figure 13.3 dissolution of the ionic solid naCl in water.

Go FIGureHow does the orientation of H2O molecules around Na+ differ from that around Cl-?

Crystal of NaCl in water Ions hydrated in solution

Hydrated Na+ ionHydrated Cl− ion

Solvent–solute interactionsbetween water molecules andNaCl allow solid to dissolve

SecTion 13.1 The Solution process 535

athletic injuries (▶ Figure 13.5). The packs consist of a pouch of water and the solid salt sealed off from the water—MgSO41s2 for hot packs and NH4NO31s2 for cold packs. When the pack is squeezed, the seal separating the solid from the water is broken and a solution forms, either increasing or decreasing the temperature.

The enthalpy change for a process can provide insight into the extent to which the process occurs. (Section 5.4) Exothermic processes tend to proceed spontane-ously. On the other hand, if ∆Hsoln is too endothermic, the solute might not dissolve to any significant extent in the chosen solvent. Thus, for solutions to form, the solvent–solute interaction must be strong enough to make ∆Hmix comparable in magnitude to ∆Hsolute + ∆Hsolvent. This fact further explains why ionic solutes do not dissolve in nonpolar solvents. The nonpolar solvent molecules experience only weak attractive interactions with the ions, and these interactions do not compensate for the energies required to separate the ions from one another.

By similar reasoning, a polar liquid solute, such as water, does not dissolve in a nonpolar liquid solvent, such as octane 1C8H182. The water molecules experience strong hydrogen-bonding interactions with one another (Section 11.2)—attractive forces that must be overcome if the water molecules are to be dispersed throughout the octane solvent. The energy required to separate the H2O molecules from one another is not recovered in the form of attractive interactions between the H2O and C8H18 molecules.

Give it some thoughtLabel the following processes as exothermic or endothermic:(a) breaking solvent–solvent interactions to form separated particles(b) forming solvent–solute interactions from separated particles

Solution Formation and Chemical ReactionsIn discussing solutions, we must be careful to distinguish the physical process of so-lution formation from chemical reactions that lead to a solution. For example, nickel metal dissolves on contact with an aqueous hydrochloric acid solution because the fol-lowing reaction occurs:

Ni1s2 + 2 HCl1aq2 ¡ NiCl21aq2 + H21g2 [13.2]

▲ Figure 13.4 enthalpy changes accompanying the solution process.

Go FIGureHow does the magnitude of ∆Hmix compare with the magnitude of ∆Hsolvent + ∆Hsolute for exothermic solution processes?

Solution

Solution

∆H

soln

< 0

∆H

mix

Hinitial

H�nal

Exothermic solution process

Solventaggregated

Solventseparated

Soluteseparated

Soluteaggregated

En

thal

py,

H

∆H

solv

ent

∆H

solu

te Solventaggregated

Solventseparated

Soluteseparated

Soluteaggregated

∆H

solu

te +

∆H

solv

ent

∆H

soln

> 0

∆H

mix

Hinitial

H�nal

Endothermic solution processE

nth

alp

y, H

∆H

solv

ent

∆H

solu

te

∆H

solu

te +

∆H

solv

ent

▲ Figure 13.5 Magnesium sulfate instant hot pack.


In this instance, one of the resulting solutes is not Ni metal but rather its salt NiCl2. If the solution is evaporated to dryness, NiCl2 # 6 H2O1s2 is recovered (▲ Figure 13.6). Compounds such as NiCl2 # 6 H2O1s2 that have a defined number of water molecules in the crystal lattice are known as hydrates. When NaCl(s) is dissolved in water, on the other hand, no chemical reaction occurs. If the solution is evaporated to dryness, NaCl is recovered. Our focus throughout this chapter is on solutions from which the solute can be recovered unchanged from the solution.

13.2 | Saturated Solutions and Solubility

As a solid solute begins to dissolve in a solvent, the concentration of solute particles in solution increases, increasing the chances that some solute particles will collide with the surface of the solid and reattach. This process, which is the opposite of the solution process, is called crystallization. Thus, two opposing processes occur in a solution in contact with undissolved solute. This situation is represented in this chemical equation:

Solute + solvent ∆dissolve

crystallizesolution [13.3]

When the rates of these opposing processes become equal, a dynamic equilibrium is established, and there is no further increase in the amount of solute in solution.

(Section 4.1)A solution that is in equilibrium with undissolved solute is saturated. Additional

solute will not dissolve if added to a saturated solution. The amount of solute needed to form a saturated solution in a given quantity of solvent is known as the solubility

▲ Figure 13.6 the reaction between nickel metal and hydrochloric acid is not a simple dissolution. The product is NiCl2 # 6 H2O1s2, nickel(II) chloride hexahydrate, which has exactly 6 waters of hydration in the crystal lattice for every nickel ion.

Nickel metal and hydrochloric acid Nickel reacts with hydrochloric acid,forming NiCl2(aq) and H2(g). Thesolution is of NiCl2, not Ni metal

NiCl2·6 H2O(s) remainswhen solvent evaporated

Go FIGureWhat is the molar mass of nickel chloride hexahydrate, NiCl2 # 6 H2O1s2?

SecTion 13.2 Saturated Solutions and Solubility 537

of that solute. That is, the solubility of a particular solute in a particular solvent is the maximum amount of the solute that can dissolve in a given amount of the solvent at a specified temperature, assuming that excess solute is present. For example, the solubility of NaCl in water at 0 °C is 35.7 g per 100 mL of water. This is the maximum amount of NaCl that can be dissolved in water to give a stable equilibrium solution at that temperature.

If we dissolve less solute than the amount needed to form a saturated solution, the solution is unsaturated. Thus, a solution containing 10.0 g of NaCl per 100 mL of water at 0 °C is unsaturated because it has the capacity to dissolve more solute.

Under suitable conditions, it is possible to form solutions that contain a greater amount of solute than needed to form a saturated solution. Such solutions are super-saturated. For example, when a saturated solution of sodium acetate is made at a high temperature and then slowly cooled, all of the solute may remain dissolved even though its solubility decreases as the temperature decreases. Because the solute in a supersaturated solution is present in a concentration higher than the equilibrium concentration, supersaturated solutions are unstable. For crystallization to occur, however, the solute particles must arrange themselves properly to form crystals. The addition of a small crystal of the solute (a seed crystal) provides a template for crys-tallization of the excess solute, leading to a saturated solution in contact with excess solid (▼ Figure 13.7).

Give it some thoughtWhat happens if a solute is added to a saturated solution?

▲ Figure 13.7 Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving about 170 g of the salt in 100 mL of water at 100 °C and then slowly cooling it to 20 °C. Because the solubility of sodium acetate in water at 20 °C is 46 g per 100 mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.

3

Amount of sodium acetatedissolved is greater than itssolubility at this temperature

Seed crystal of sodium acetate added to supersaturated solution

1 Excess sodium acetatecrystallizes from solution

2 Solution arrives at saturation

Go FIGureWhat is the evidence that the solution in the left photograph is supersaturated?


13.3 | Factors affecting SolubilityThe extent to which one substance dissolves in another depends on the nature of both substances. (Section 13.1) It also depends on temperature and, at least for gases, on pressure.

Solute–Solvent InteractionsThe natural tendency of substances to mix and the various interactions among solute and solvent particles are all involved in determining solubilities. Nevertheless, insight into variations in solubility can often be gained by focusing on the interaction between the solute and solvent. The data in ◀ table 13.1 show that the solubilities of various gases in water increase with increasing molecular mass. The attractive forces between the gas molecules and solvent molecules are mainly dispersion forces, which increase with increasing size and molecular mass. (Section 11.2) Thus, the data indicate that the solubilities of gases in water increase as the attraction between solute (gas) and sol-vent (water) increases. In general, when other factors are comparable, the stronger the attractions between solute and solvent molecules, the greater the solubility of the solute in that solvent.

Because of favorable dipole–dipole attractions between solvent molecules and solute molecules, polar liquids tend to dissolve in polar solvents. Water is both polar and able to form hydrogen bonds. (Section 11.2) Thus, polar molecules, espe-cially those that can form hydrogen bonds with water molecules, tend to be soluble in water. For example, acetone, a polar molecule with the structural formula shown below, mixes in all proportions with water. Acetone has a strongly polar C “ O bond and pairs of nonbonding electrons on the O atom that can form hydrogen bonds with water.

Acetone

O

CH3CCH3

Liquids that mix in all proportions, such as acetone and water, are miscible, whereas those that do not dissolve in one another are immiscible. Gasoline, which is a mixture of hydrocarbons, is immiscible with water. Hydrocarbons are nonpolar sub-stances because of several factors: The C ¬ C bonds are nonpolar, the C ¬ H bonds are nearly nonpolar, and the molecules are symmetrical enough to cancel much of the weak C ¬ H bond dipoles. The attraction between the polar water molecules and the nonpo-lar hydrocarbon molecules is not sufficiently strong to allow the formation of a solution. Nonpolar liquids tend to be insoluble in polar liquids, as ◀ Figure 13.8 shows for hexane 1C6H142 and water.

Many organic compounds have polar groups attached to a nonpolar framework of carbon and hydrogen atoms. For example, the series of organic compounds in ▶ table 13.2 all contain the polar OH group. Organic compounds with this molecu-lar feature are called alcohols. The O ¬ H bond is able to form hydrogen bonds. For example, ethanol 1CH3CH2OH2 molecules can form hydrogen bonds with water molecules as well as with each other (▶ Figure 13.9). As a result, the solute– solute, solvent–solvent, and solute–solvent forces are not greatly different in a mixture of CH3CH2OH and H2O. No major change occurs in the environments of the molecules as they are mixed. Therefore, the increased entropy when the compo-nents mix plays a significant role in solution formation, and ethanol is completely miscible with water.

table 13.1 Solubilities of Gases in Water at 20 °C, with 1 atm Gas Pressure

Molar Gas Mass 1g>mol2

Solubility (M)

N2 28.0 0.69 * 10-3

O2 32.0 1.38 * 10-3

Ar 39.9 1.50 * 10-3

Kr 83.8 2.79 * 10-3

▲ Figure 13.8 Hexane is immiscible with water. Hexane is the top layer because it is less dense than water.

Hexane

Water

SecTion 13.3 Factors affecting Solubility 539

Notice in Table 13.2 that the number of carbon atoms in an alcohol affects its solu-bility in water. As this number increases, the polar OH group becomes an even smaller part of the molecule, and the molecule behaves more like a hydrocarbon. The solubility of the alcohol in water decreases correspondingly. On the other hand, the solubility of alcohols in a nonpolar solvent like hexane 1C6H142 increases as the nonpolar hydrocar-bon chain lengthens.

One way to enhance the solubility of a substance in water is to increase the number of polar groups the substance contains. For example, increasing the number of OH groups in a solute increases the extent of hydrogen bonding between that solute and water, thereby increasing solubility. Glucose (C6H12O6, ▶ Figure 13.10) has five OH groups on a six-carbon framework, which makes the molecule very soluble in water: 830 g dissolves in 1.00 L of water at 17.5 °C. In contrast, cyclohexane 1C6H122, which has a similar structure to glucose but with all of the OH groups replaced by H, is essentially insoluble in water (only 55 mg of cyclohexane can dissolve in 1.00 L of water at 25 °C).

table 13.2 Solubilities of Some alcohols in Water and in hexane*

alcohol

Solubility in h2o

Solubility in C6h14

CH3OH 1methanol2 ∞ 0.12

CH3CH2OH 1ethanol2 ∞ ∞

CH3CH2CH2OH 1propanol2 ∞ ∞

CH3CH2CH2CH2OH 1butanol2 0.11 ∞

CH3CH2CH2CH2CH2OH 1pentanol2 0.030 ∞

CH3CH2CH2CH2CH2CH2OH 1hexanol2 0.0058 ∞

*Expressed in mol alcohol/100 g solvent at 20 °C. The infinity symbol 1∞2 indicates that the alcohol is completely miscible with the solvent.

Hydrogen bond betweenethanol molecule andwater molecule

Hydrogen bond betweentwo ethanol molecules

▲ Figure 13.9 Hydrogen bonding involving oH groups.

C

CC

C

CC

CC

C

OH groups enhance the aqueoussolubility because of their abilityto hydrogen bond with H2O.

Cyclohexane, C6H12, which has no polar OH groups, is essentially insoluble in water

CH2OHH

H

HH

H

H HH

HH

H

HH

HO

H

Hydrogen-bondingsites

HO

HOOH

OH

HH

CC

▲ Figure 13.10 the correlation of molecular structure with solubility.

Chemistry and Life

Fat-soluble and Water-soluble vitamins

Vitamins have unique chemical structures that affect their solubilities in different parts of the human body. Vitamin C and the B vitamins are soluble in water, for example, whereas vitamins A, D, E, and K are soluble in nonpolar solvents and in fatty tissue (which is nonpolar). Because of their water solubility, vitamins B and C are not stored to any appreciable extent in the body, and so foods containing these vitamins should be included in the daily diet. In contrast, the fat-soluble vitamins are stored in sufficient quantities to keep vitamin-deficiency

diseases from appearing even after a person has subsisted for a long period on a vitamin-deficient diet.

That some vitamins are soluble in water and others are not can be ex-plained in terms of their structures. Notice in Figure 13.11 that vitamin A (retinol) is an alcohol with a very long carbon chain. Because the OH group is such a small part of the molecule, the molecule resembles the long-chain alcohols listed in Table 13.2. This vitamin is nearly nonpolar. In contrast, the vitamin C molecule is smaller and has several OH groups that can form hydrogen bonds with water, somewhat like glucose.

Related Exercises: 13.7, 13.48


Over years of study, examination of different solvent–solute combinations has led to an important generalization: Substances with similar intermolecular attractive forces tend to be soluble in one another. This generalization is often simply stated as “like dis-solves like.” Nonpolar substances are more likely to be soluble in nonpolar solvents; ionic and polar solutes are more likely to be soluble in polar solvents. Network solids such as diamond and quartz are not soluble in either polar or nonpolar solvents because of the strong bonding within the solid.

Give it some thoughtSuppose the hydrogens on the OH groups in glucose (Figure 13.10) were replaced with methyl groups, CH3. Would you expect the water solubility of the resulting molecule to be higher than, lower than, or about the same as glucose?

▲ Figure 13.11 the molecular structures of vitamins A and C.

Vitamin CVitamin A

CH3

CH3CH3H3C

H2C

H2C

C

C

OH

CH2

HH

H

HO

O CC

H

H

CCCCCC

HH CH3

H

CCCC

HHH CC

CC

H O

H O

HO

HO

Most of molecule nonpolar

Only one polar groupto interact with water Many polar groups to

interact with water

Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride 1CCl42 or in water: C7H16, Na2SO4, HCl, and I2.

SoLutIonAnalyze We are given two solvents, one that is nonpolar 1CCl42 and the other that is polar 1H2O2, and asked to determine which will be the better solvent for each solute listed.Plan By examining the formulas of the solutes, we can predict whether they are ionic or molecu-lar. For those that are molecular, we can predict whether they are polar or nonpolar. We can then apply the idea that the nonpolar solvent will be better for the nonpolar solutes, whereas the polar solvent will be better for the ionic and polar solutes.Solve C7H16 is a hydrocarbon, so it is molecular and nonpolar. Na2SO4, a compound containing a metal and nonmetals, is ionic. HCl, a diatomic molecule containing two nonmetals that dif-fer in electronegativity, is polar. I2, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C7H16 and I2 (the nonpolar solutes) would be more soluble in the nonpolar CCl4 than in polar H2O, whereas water would be the better solvent for Na2SO4 and HCl (the ionic and polar covalent solutes).

SaMPLe exerCISe 13.1 Predicting Solubility Patterns


Pressure EffectsThe solubilities of solids and liquids are not appreciably affected by pressure, whereas the solubility of a gas in any solvent is increased as the partial pressure of the gas above the solvent increases. We can understand the effect of pressure on gas solubility by considering Figure 13.12, which shows carbon dioxide gas dis-tributed between the gas and solution phases. When equilibrium is established, the rate at which gas molecules enter the solution equals the rate at which solute molecules escape from the solution to enter the gas phase. The equal number of up and down arrows in the left container in Figure 13.12 represent these opposing processes.

Now suppose we exert greater pressure on the piston and compress the gas above the solution, as shown in the middle container in Figure 13.12. If we reduce the gas volume to half its original value, the pressure of the gas increases to about twice its original value. As a result of this pressure increase, the rate at which gas molecules strike the liquid surface and enter the solution phase increases. Thus, the solubility of the gas in the solution increases until equilibrium is again established; that is, solubility increases until the rate at which gas molecules enter the solution equals the rate at which they escape from the solution. Thus,

Practice exercise 1Which of the following solvents will best dissolve wax, which is a complex mixture of com-pounds that mostly are CH3-CH2-CH2-CH2-CH2 ¬ ?

(a) Hexane

(c) Acetone (d) Carbon tetrachloride

(b) Benzene

C

H3C

H3CCH3C

H H

C

H H

H H H H

C CC

C

C

C C

CH3C

H

H

H

H

H

H

(e) Water

H HO

O

C

Cl

ClCl

Cl

Practice exercise 2Arrange the following substances in order of increasing solubility in water:

H C C C C C H

H

H

H

H

H

H

H

H

H

H

HO C C C C C OH

H

H

H

H

H

H

H

H

H

H

H C C C C C

H

H

H

H

H

H

H

H

H

H

HOH C C C C C Cl

H

H

H

H

H

H

H

H

H

H

H C C C C C H

H

H

H

H

H

H

H

H

H

H

HO C C C C C OH

H

H

H

H

H

H

H

H

H

H

H C C C C C

H

H

H

H

H

H

H

H

H

H

HOH C C C C C Cl

H

H

H

H

H

H

H

H

H

H

(a)

(b)

(c)

(d)


the solubility of a gas in a liquid solvent increases in direct proportion to the partial pressure of the gas above the solution (◀ Figure 13.13).

The relationship between pressure and gas solubility is expressed by Henry’s law:

Sg = kPg [13.4]

Here, Sg is the solubility of the gas in the solvent (usually expressed as molarity), Pg is the partial pressure of the gas over the solution, and k is a proportionality constant known as the Henry’s law constant. The value of this constant depends on the solute, solvent, and temperature. As an example, the solubility of N2 gas in water at 25 °C and 0.78 atm pressure is 4.75 * 10-4 M. The Henry’s law constant for N2 in 25 °C water is thus 14.75 * 10-4 mol>L2>0.78 atm = 6.1 * 10-4 mol>L@atm. If the partial pressure of N2 is doubled, Henry’s law predicts that the solubility in water at 25 °C also doubles to 9.50 * 10-4 M.

Bottlers use the effect of pressure on solubility in producing carbonated bever-ages, which are bottled under a carbon dioxide pressure greater than 1 atm. When the bottles are opened to the air, the partial pressure of CO2 above the solution decreases. Hence, the solubility of CO2 decreases, and CO21g2 escapes from the solution as bub-bles (▶ Figure 13.14).

▲ Figure 13.12 effect of pressure on gas solubility.

Equilibrium Pressure is increased.More CO2 dissolves

Equilibrium restored

Go FIGureIf the partial pressure of a gas over a solution is doubled, how has the concentration of gas in the solution changed after equilibrium is restored?

▲ Figure 13.13 the solubility of a gas in water is directly proportional to the partial pressure of the gas. The solubilities are in millimoles of gas per liter of solution.

0

0.50

0.50 1.00

Solu

bilit

y (m

M)

1.00

Partial pressure (atm)

O2

N2

CO

He

Go FIGureHow do the slopes of the lines vary with the molecular weight of the gas? Explain the trend.

Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25 °C. The Henry’s law constant for CO2 in water at this tempera-ture is 3.4 * 10-2 mol>L@atm.

SoLutIonAnalyze We are given the partial pressure of CO2, PCO2

, and the Henry’s law constant, k, and asked to calculate the concentration of CO2 in the solution.Plan With the information given, we can use Henry’s law, Equation 13.4, to calculate the solubility, SCO2

.

SaMPLe exerCISe 13.2 a henry’s Law Calculation


▲ Figure 13.14 Gas solubility decreases as pressure decreases. CO2 bubbles out of solution when a carbonated beverage is opened because the CO2 partial pressure above the solution is reduced.

Solve SCO2= kPCO2

= 13.4 * 10-2 mol>[email protected] atm2 = 0.14 mol>L = 0.14 MCheck The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO2 and the value of Henry’s constant.

Practice exercise 1You double the partial pressure of a gas over a liquid at constant temperature. Which of these statements is then true?(a) The Henry’s law constant is doubled.(b) The Henry’s law constant is decreased by half.(c) There are half as many gas molecules in the liquid.(d) There are twice as many gas molecules in the liquid.(e) There is no change in the number of gas molecules in the liquid.

Practice exercise 2Calculate the concentration of CO2 in a soft drink after the bottle is opened and the solution equilibrates at 25 °C under a CO2 partial pressure of 3.0 * 10-4 atm.

Temperature EffectsThe solubility of most solid solutes in water increases as the solution temperature increases, as ▼ Figure 13.15 shows. There are exceptions to this rule, however, as seen for Ce21SO423, whose solubility curve slopes downward with increasing temperature.

In contrast to solid solutes, the solubility of gases in water decreases with increasing temperature (▼ Figure 13.16). If a glass of cold tap water is warmed, you can see bub-bles on the inside of the glass because some of the dissolved air comes out of solution.

Similarly, as carbonated beverages are allowed to warm, the solubility of CO2 decreases, and CO21g2 escapes from the solution.

Give it some thoughtWhy do bubbles form on the inside wall of a cooking pot when water is heated on the stove, even though the water temperature is well below the boiling point of water?

▲ Figure 13.15 solubilities of some ionic compounds in water as a function of temperature.

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

70

80

90

100

Temperature (°C)

Solu

bilit

y (g

of s

alt i

n 10

0 g

H2O

)

Ce2(SO4)3

KClO 3

K 2Cr 2O

7

KN

O3

Pb(NO 3) 2

CaCl 2

NaNO 3

KCl

NaCl

Go FIGureHow does the solubility of KCl at 80 °C compare with that of NaCl at the same temperature?

▲ Figure 13.16 solubilities of four gases in water as a function of temperature. The solubilities are in millimoles per liter of solution, for a constant total pressure of 1 atm in the gas phase.

0 10 20 30 40 50

1.0

2.0

CH4

O2

He

Temperature (°C)

Solu

bilit

y (m

M)

CO

Go FIGureWhere would you expect N2 to fit on this graph?


13.4 | expressing Solution Concentration

The concentration of a solution can be expressed either qualitatively or quantitatively. The terms dilute and concentrated are used to describe a solution qualitatively. A solu-tion with a relatively small concentration of solute is said to be dilute; one with a large concentration is said to be concentrated. Chemists use various ways to express concen-tration quantitatively, and we examine several of these next.

Mass Percentage, ppm, and ppbOne of the simplest quantitative expressions of concentration is the mass percentage of a component in a solution, given by

Mass % of component =mass of component in soln

total mass of soln* 100 [13.5]

Because percent means “per hundred,” a solution of hydrochloric acid that is 36% HCl by mass contains 36 g of HCl for each 100 g of solution.

We often express the concentration of very dilute solutions in parts per million (ppm) or parts per billion (ppb). These quantities are similar to mass percentage but use 106 (a million) or 109 (a billion), respectively, in place of 100, as a multiplier for the ratio of the mass of solute to the mass of solution. Thus, parts per million is defined as

ppm of component =mass of component in soln

total mass of soln* 106 [13.6]

A solution whose solute concentration is 1 ppm contains 1 g of solute for each million 11062 grams of solution or, equivalently, 1 mg of solute per kilogram of solution. Because the density of water is 1 g>mL, 1 kg of a dilute aqueous solution has a volume

Chemistry and Life

Blood Gases and deep-sea diving

Because gas solubility increases with increasing pressure, divers who breathe compressed air (▶ Figure 13.17) must be concerned about the solubility of gases in their blood. Although the gases are not very soluble at sea level, their solubilities can be appreciable at deep lev-els where their partial pressures are greater. Thus, divers must ascend slowly to prevent dissolved gases from being released rapidly from so-lution and forming bubbles in the blood and other fluids in the body. These bubbles affect nerve impulses and cause decompression sick-ness, or “the bends,” which is a painful and potentially fatal condition. Nitrogen is the main problem because it is the most abundant gas in air and because it can be removed from the body only through the respiratory system. Oxygen, in contrast, is consumed in metabolism.

Deep-sea divers sometimes substitute helium for nitrogen in the air they breathe because helium has a much lower solubility in bio-logical fluids than N2. For example, divers working at a depth of 100 ft experience a pressure of about 4 atm. At this pressure, a mixture of 95% helium and 5% oxygen gives an oxygen partial pressure of about 0.2 atm, which is the partial pressure of oxygen in normal air at 1 atm.

If the oxygen partial pressure becomes too great, the urge to breathe is reduced, CO2 is not removed from the body, and CO2 poisoning occurs. At excessive concentrations in the body, carbon dioxide acts as a neurotoxin, interfering with nerve conduction and transmission.

Related Exercises: 13.59, 13.60, 13.107

▲ Figure 13.17 Gas solubility increases as pressure increases. Divers who use compressed gases must be concerned about the solubility of the gases in their blood.

SecTion 13.4 expressing Solution concentration 545

very close to 1 L. Thus, 1 ppm also corresponds to 1 mg of solute per liter of aqueous solution.

The acceptable maximum concentrations of toxic or carcinogenic substances in the environment are often expressed in ppm or ppb. For example, the maximum allowable concentration of arsenic in drinking water in the United States is 0.010 ppm; that is, 0.010 mg of arsenic per liter of water. This concentration corresponds to 10 ppb.

Give it some thoughtA solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What is the concentration of SO2 in ppm? In ppb?

(a) A solution is made by dissolving 13.5 g of glucose 1C6H12O62 in 0.100 kg of water. What is the mass percentage of solute in this solution?

(b) A 2.5-g sample of groundwater was found to contain 5.4 mg of Zn2+. What is the concentration of Zn2+ in parts per million?

SoLutIon(a) Analyze We are given the number of grams of solute (13.5 g) and the number of grams of

solvent 10.100 kg = 100 g2. From this, we must calculate the mass percentage of solute.Plan We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water).

Solve Mass % of glucose =mass glucose

mass soln* 100 =

13.5 g13.5 g + 100 g

* 100 = 11.9%

Comment The mass percentage of water in this solution is 1100 - 11.92% = 88.1%.

(b) Analyze In this case we are given the number of micrograms of solute. Because 1 mg is 1 * 10-6 g, 5.4 mg = 5.4 * 10-6 g.

Plan We calculate the parts per million using Equation 13.6.

Solve ppm =mass of solutemass of soln

* 106 =5.4 * 10-6 g

2.5 g* 106 = 2.2 ppm

Practice exercise 1Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water.(a) 0.0291%, (b) 0.0300%, (c) 0.0513%, (d) 2.91%, (e) 3.00%.

Practice exercise 2A commercial bleaching solution contains 3.62% by mass of sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution?

SaMPLe exerCISe 13.3 Calculation of Mass-related Concentrations

Mole Fraction, Molarity, and MolalityConcentration expressions are often based on the number of moles of one or more components of the solution. Recall from Section 10.6 that the mole fraction of a compo-nent of a solution is given by

Mole fraction of component =moles of component

total moles of all components [13.7]

The symbol X is commonly used for mole fraction, with a subscript to indicate the compo-nent of interest. For example, the mole fraction of HCl in a hydrochloric acid solution is rep-resented as XHCl. Thus, if a solution contains 1.00 mol of HCl (36.5 g) and 8.00 mol of water (144 g), the mole fraction of HCl is XHCl = 11.00 mol2>11.00 mol + 8.00 mol2 = 0.111.


Mole fractions have no units because the units in the numerator and the denominator can-cel. The sum of the mole fractions of all components of a solution must equal 1. Thus, in the aqueous HCl solution, XH2O = 1.000 - 0.111 = 0.889. Mole fractions are very use-ful when dealing with gases, as we saw in Section 10.6, but have limited use when dealing with liquid solutions.

Recall from Section 4.5 that the molarity (M) of a solute in a solution is defined as

Molarity =moles of solute

liters of soln [13.8]

For example, if you dissolve 0.500 mol of Na2CO3 in enough water to form 0.250 L of solution, the molarity of Na2CO3 in the solution is 10.500 mol2>10.250 L2 = 2.00 M. Molarity is especially useful for relating the volume of a solution to the quantity of solute contained in that volume, as we saw in our discussions of titrations.

(Section 4.6)The molality of a solution, denoted m, is a concentration unit that is also based

on moles of solute. Molality equals the number of moles of solute per kilogram of solvent:

Molality =moles of solute

kilograms of solvent [13.9]

Thus, if you form a solution by mixing 0.200 mol of NaOH (8.00 g) and 0.500 kg of water (500 g), the concentration of the solution is 10.200 mol2>10.500 kg2 = 0.400 m (that is, 0.400 molal) in NaOH.

The definitions of molarity and molality are similar enough that they can be easily confused. Molarity depends on the volume of solution, whereas molality depends on the mass of solvent. When water is the solvent, the molality and molarity of dilute solutions are numerically about the same because 1 kg of solvent is nearly the same as 1 kg of solution, and 1 kg of the solution has a volume of about 1 L.

The molality of a given solution does not vary with temperature because masses do not vary with temperature. The molarity of the solution does change with temperature, however, because the volume of the solution expands or contracts with temperature. Thus, molality is often the concentration unit of choice when a solution is to be used over a range of temperatures.

Give it some thoughtIf an aqueous solution is very dilute, will its molality be greater than its molarity, nearly the same as its molarity, or smaller than its molarity?

SaMPLe exerCISe 13.4 Calculation of Molality

Finally, use Equation 13.9 to obtain the molality:

A solution is made by dissolving 4.35 g of glucose 1C6H12O62 in 25.0 mL of water at 25 °C. Calculate the molality of glucose in the solution. Water has a density of 1.00 g>mL.

SoLutIonAnalyze We are asked to calculate a solution concentration in units of molality. To do this, we must determine the number of moles of solute (glucose) and the number of kilograms of solvent (water).

Plan We use the molar mass of C6H12O6 to convert grams of glucose to moles. We use the density of water to convert milliliters of water to kilograms. The molality equals the number of moles of solute (glucose) divided by the number of kilograms of solvent (water).

Solve Use the molar mass of glucose, 180.2 g>mol, to convert grams to moles: Mol C6H12O6 = 14.35 g C6H12O62a

1 mol C6H12O6

180.2 g C6H12O6b = 0.0241 mol C6H12O6

Because water has a density of 1.00 g>mL, the mass of the solvent is 125.0 mL211.00 g>mL2 = 25.0 g = 0.0250 kg

Molality of C6H12O6 =0.0241 mol C6H12O6

0.0250 kg H2O= 0.964 m

SecTion 13.4 expressing Solution concentration 547

Converting Concentration UnitsIf you follow the dimensional analysis techniques you learned in Chapter 1, you can convert between concentration units, as shown in Sample Exercise 13.5. To convert between molality and molarity, the density of the solution will be needed, as in Sample Exercise 13.6.

Practice exercise 1Suppose you take a solution and add more solvent, so that the original mass of solvent is doubled. You take this new solution and add more solute, so that the original mass of the solute is doubled. What happens to the molality of the final solution, compared to the original molality?(a) It is doubled.(b) It is decreased by half.

(c) It is unchanged.(d) It will increase or decrease depending on the molar mass of the

solute.(e) There is no way to tell without knowing the molar mass of the

solute.

Practice exercise 2What is the molality of a solution made by dissolving 36.5 g of naphthalene 1C10H82 in 425 g of toluene 1C7H82?

SaMPLe exerCISe 13.5 Calculation of Mole Fraction and Molality

An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.

SoLutIon

Analyze We are asked to calculate the concentration of the solute, HCl, in two related concentration units, given only the percentage by mass of the solute in the solution.

Plan In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let’s assume that there is exactly 100 g of solution. Because the solution is 36% HCl, it contains 36 g of HCl and 1100 - 362 g = 64 g of H2O. We must convert grams of solute (HCl) to moles to calculate either mole fraction or molality. We must convert grams of solvent 1H2O2 to moles to calculate mole fractions and to kilograms to calculate molality.

Solve (a) To calculate the mole fraction of HCl, we convert the masses of HCl and H2O to moles and then use Equation 13.7: Moles HCl = 136 g HCl2a 1 mol HCl

36.5 g HClb = 0.99 mol HCl

Moles H2O = 164 g H2O2a 1 mol H2O18 g H2O

b = 3.6 mol H2O

XHCl =moles HCl

moles H2O + moles HCl=

0.993.6 + 0.99

=0.994.6

= 0.22

Practice exercise 1The solubility of oxygen gas in water at 40 °C is 1.0 mmol per liter of solution. What is this concentration in units of mole fraction?(a) 1.00 * 10-6, (b) 1.80 * 10-5, (c) 1.00 * 10-2, (d) 1.80 * 10-2, (e) 5.55 * 10-2.

Practice exercise 2A commercial bleach solution contains 3.62% by mass of NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution.

Notice that we can’t readily calculate the molarity of the solution because we don’t know the volume of the 100 g of solution.

(b) To calculate the molality of HCl in the solution, we use Equation 13.9. We calculated the number of moles of HCl in part (a), and the mass of solvent is 64 g = 0.064 kg: Molality of HCl =

0.99 mol HCl0.064 kg H2O

= 15 m


13.5 | Colligative PropertiesSome physical properties of solutions differ in important ways from those of the pure solvent. For example, pure water freezes at 0 °C, but aqueous solutions freeze at lower temperatures. We apply this behavior when we add ethylene glycol antifreeze to a car’s radiator to lower the freezing point of the solution. The added solute also raises the boiling point of the solution above that of pure water, making it possible to operate the engine at a higher temperature.

Lowering of the freezing point and raising of the boiling point are physical prop-erties of solutions that depend on the quantity (concentration) but not on the kind or identity of the solute particles. Such properties are called colligative properties. (Colligative means “depending on the collection”; colligative properties depend on the collective effect of the number of solute particles.)

In addition to freezing-point lowering and boiling-point raising, vapor-pressure lowering and osmotic pressure are also colligative properties. As we examine each one, notice how solute concentration quantitatively affects the property.

Vapor-Pressure LoweringA liquid in a closed container establishes equilibrium with its vapor. (Section 11.5) The vapor pressure is the pressure exerted by the vapor when it is at equilibrium with the liquid (that is, when the rate of vaporization equals the rate of condensation). A substance that has no measurable vapor pressure is nonvolatile, whereas one that exhibits a vapor pressure is volatile.

A solution consisting of a volatile liquid solvent and a nonvolatile solute forms spontaneously because of the increase in entropy that accompanies their mixing. In

Molarity is moles of solute per liter of solution:

Analyze Our goal is to calculate the molarity of a solution, given the masses of solute (5.0 g) and solvent (225 g) and the density of the so-lution 10.876 g>mL2.

Plan The molarity of a solution is the number of moles of solute di-vided by the number of liters of solution (Equation 13.8). The number of moles of solute 1C7H82 is calculated from the number of grams of solute and its molar mass. The volume of the solution is obtained from the mass of the solution (mass of solution = mass of solute +mass of solvent = 5.0 g + 225 g = 230 g) and its density.

Solve The number of moles of solute is: Moles C7H8 = 15.0 g C7H82a1 mol C7H8

92 g C7H8b = 0.054 mol

The density of the solution is used to convert the mass of the solution to its volume: Milliliters soln = 1230 g2a 1 mL

0.876 gb = 263 mL

Molarity = amoles C7H8

liter solnb = a 0.054 mol C7H8

263 mL solnb a 1000 mL soln

1 L solnb = 0.21 M

SoLutIon

Check The magnitude of our answer is reasonable. Rounding moles to 0.05 and liters to 0.25 gives a molarity of 10.05 mol2>10.25 L2 = 0.2 M.The units for our answer 1mol>L2 are correct, and the answer, 0.21, has two significant figures, corresponding to the number of significant figures in the mass of solute (2).Comment Because the mass of the solvent (0.225 kg) and the volume of the solution (0.263) are similar in magnitude, the molarity and molality are also similar in magnitude: 10.054 mol C7H82>10.225 kg solvent2 = 0.24 m.

Practice exercise 1Maple syrup has a density of 1.325 g>mL, and 100.00 g of maple syrup contains 67 mg of calcium in the form of Ca2+ ions. What is the molarity of calcium in maple syrup?(a) 0.017 M, (b) 0.022 M, (c) 0.89 M, (d) 12.6 M, (e) 45.4 M

Practice exercise 2A solution containing equal masses of glycerol 1C3H8O32 and water has a density of 1.10 g>mL. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution.

SaMPLe exerCISe 13.6 Calculation of Molarity using the density of the Solution

A solution with a density of 0.876 g>mL contains 5.0 g of toluene 1C7H82 and 225 g of benzene. Calculate the molarity of the solution.

SecTion 13.5 colligative properties 549

effect, the solvent molecules are stabilized in their liquid state by this process and thus have a lower tendency to escape into the vapor state. Therefore, when a nonvolatile sol-ute is present, the vapor pressure of the solvent is lower than the vapor pressure of the pure solvent, as illustrated in ▲ Figure 13.18.

Ideally, the vapor pressure of a volatile solvent above a solution containing a nonvol-atile solute is proportional to the solvent’s concentration in the solution. This relation-ship is expressed quantitatively by Raoult’s law, which states that the partial pressure exerted by solvent vapor above the solution, Psolution, equals the product of the mole frac-tion of the solvent, Xsolvent, times the vapor pressure of the pure solvent, P °solvent: Psolution = XsolventP°solvent [13.10]

For example, the vapor pressure of pure water at 20 °C is P°H2O = 17.5 torr. Imag-ine holding the temperature constant while adding glucose 1C6H12O62 to the water so that the mole fractions in the resulting solution are XH2O = 0.800 and XC6H12O6

= 0.200. According to Equation 13.10, the vapor pressure of the water above this solution is 80.0% of that of pure water:

Psolution = 10.8002117.5 torr2 = 14.0 torrThe presence of the nonvolatile solute lowers the vapor pressure of the volatile solvent by 17.5 torr - 14.0 torr = 3.5 torr.

The vapor-pressure lowering, ∆P, is directly proportional to the mole fraction of the solute, Xsolute : ∆P = Xsolute P°solvent [13.11]Thus, for the example of the solution of glucose in water, we have

∆P = XC6H12O6 P°H2O = 10.2002117.5 torr2 = 3.50 torr

The vapor-pressure lowering caused by adding a nonvolatile solute depends on the total concentration of solute particles, regardless of whether they are molecules or ions. Remember that vapor-pressure lowering is a colligative property, so its value for any solu-tion depends on the concentration of solute particles and not on their kind or identity.

Give it some thoughtAdding 1 mol of NaCl to 1 kg of water lowers the vapor pressure of water more than adding 1 mol of C6H12O6. Explain.

▲ Figure 13.18 vapor-pressure lowering. The presence of nonvolatile solute particles in a liquid solvent results in a reduction of the vapor pressure above the liquid.

Equilibrium

Volatile solvent particles

Nonvolatile solute particles

Rate of vaporizationreduced by presenceof nonvolatile solute

Addnonvolatilesolute

Equilibrium reestablishedwith fewer molecules ingas phase

SaMPLe exerCISe 13.7 Calculation of Vapor Pressure of a Solution

Glycerin 1C3H8O32 is a nonvolatile nonelectrolyte with a density of 1.26 g>mL at 25 °C. Calculate the vapor pressure at 25 °C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 °C is 23.8 torr (Appendix B), and its density is 1.00 g>mL.


Consider a mixture of 1.0 mol of benzene 1C6H62 and 2.0 mol of toluene 1C7H82 1Xben = 0.33, Xtol = 0.672. At 20 °C, the vapor pres-sures of the pure substances are P°ben = 75 torr and P°tol = 22 torr. Thus, the partial pressures above the solution are

Pben = 10.332175 torr2 = 25 torrPtol = 10.672122 torr2 = 15 torr

and the total vapor pressure above the liquid is

Ptotal = Pben + Ptol = 25 torr + 15 torr = 40 torr

Note that the vapor is richer in benzene, the more volatile component.The mole fraction of benzene in the vapor is given by the

ratio of its vapor pressure to the total pressure (Equations 10.14 and 10.15):

Xben in vapor =Pben

Ptol=

25 torr40 torr

= 0.63

Although benzene constitutes only 33% of the molecules in the solu-tion, it makes up 63% of the molecules in the vapor.

When an ideal liquid solution containing two volatile components is in equilibrium with its vapor, the more volatile com-ponent will be relatively richer in the vapor. This fact forms the basis of distillation, a technique used to separate (or partially sepa-rate) mixtures containing volatile components. (Section 1.3) Distillation is a way of purifying liquids, and is the procedure by which petrochemical plants achieve the separation of crude petro-leum into gasoline, diesel fuel, lubricating oil, and other products (◀ Figure 13.19). Distillation is also used routinely on a small scale in the laboratory.

Related Exercises: 13.67, 13.68

a Closer Look

ideal solutions with two or More volatile Components

Solutions sometimes have two or more volatile components. Gasoline, for example, is a solution of several volatile liquids. To gain some un-derstanding of such mixtures, consider an ideal solution of two vola-tile liquids, A and B. (For our purposes here, it does not matter which we call the solute and which the solvent.) The partial pressures above the solution are given by Raoult’s law:

PA = XAP°A and PB = XBP°Band the total vapor pressure above the solution is

Ptotal = PA + PB = XAP°A + XBP°B

▲ Figure 13.19 The volatile components of organic mixtures can be separated on an industrial scale in these distillation towers.

Analyze Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute.

Plan We can use Raoult’s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, Xsolvent, is the ratio of the number of moles of solvent 1H2O2to total moles of solution 1moles C3H8O3 + moles H2O2.

Solve To calculate the mole fraction of water in the solution, we must determine the number of moles of C3H8O3 and H2O:

Moles C3H8O3 = 150.0 mL C3H8O32a1.26 g C3H8O3 1 mL C3H8O3

b a 1 mol C3H8O3

92.1 g C3H8O3b = 0.684 mol

Moles H2O = 1500.0 mL H2O2a 1.00 g H2O1 mL H2O

b a 1 mol H2O18.0 g H2O

b = 27.8 mol

XH2O =mol H2O

mol H2O + mol C3H8O3=

27.827.8 + 0.684

= 0.976

We now use Raoult’s law to calculate the vapor pres-sure of water for the solution: PH2O = XH2O P°H2O = 10.9762123.8 torr2 = 23.2 torr

SoLutIon

Comment The vapor pressure of the solution has been lowered by 23.8 torr - 23.2 torr = 0.6 torr relative to that of pure water. The vapor-pressure lowering can be calculated directly using Equa-tion 13.11 together with the mole fraction of the solute, C3H8O3:

∆P = XC3H8O3P °H2O = 10.0242123.8 torr2 = 0.57 torr. Notice that the

use of Equation 13.11 gives one more significant figure than the num-ber obtained by subtracting the vapor pressure of the solution from that of the pure solvent.

Practice exercise 1The vapor pressure of benzene, C6H6, is 100.0 torr at 26.1 °C. Assum-ing Raoult’s law is obeyed, how many moles of a nonvolatile solute must be added to 100.0 mL of benzene to decrease its vapor pressure by 10.0% at 26.1 °C? The density of benzene is 0.8765 g>cm3. (a) 0.011237, (b) 0.11237, (c) 0.1248, (d) 0.1282, (e) 8.765.

Practice exercise 2The vapor pressure of pure water at 110 °C is 1070 torr. A solu-tion of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?


An ideal gas is defined as one that obeys the ideal-gas equation (Section 10.4), and an ideal solution is defined as one that obeys Raoult’s law. Whereas ideality for a gas arises from a complete lack of intermolecular interaction, ideality for a solution implies total uniformity of interaction. The molecules in an ideal solution all influence one another in the same way—in other words, solute–solute, solvent–solvent, and sol-ute–solvent interactions are indistinguishable from one another. Real solutions best approximate ideal behavior when the solute concentration is low and solute and solvent have similar molecular sizes and take part in similar types of intermolecular attractions.

Many solutions do not obey Raoult’s law exactly and so are not ideal. If, for instance, the solvent–solute interactions in a solution are weaker than either the sol-vent–solvent or solute–solute interactions, the vapor pressure tends to be greater than that predicted by Raoult’s law. When the solute–solvent interactions in a solution are exceptionally strong, as might be the case when hydrogen bonding exists, the vapor pressure is lower than that predicted by Raoult’s law. Although you should be aware that these departures from ideality occur, we will ignore them for the remainder of this chapter.

Boiling-Point ElevationIn Sections 11.5 and 11.6, we examined the vapor pressures of pure substances and how to use them to construct phase diagrams. How does the phase diagram of a solution and, hence, its boiling and freezing points differ from that of the pure solvent? The addition of a nonvolatile solute lowers the vapor pressure of the solution. Thus, in ▼ Figure 13.20 the vapor-pressure curve of the solution is shifted downward relative to the vapor-pres-sure curve of the pure solvent.

Recall from Section 11.5 that the normal boiling point of a liquid is the tempera-ture at which its vapor pressure equals 1 atm. Because the solution has a lower vapor pressure than the pure solvent, a higher temperature is required for the solution to achieve a vapor pressure of 1 atm. As a result, the boiling point of the solution is higher than that of the pure solvent. This effect is seen in Figure 13.20. We find the normal boiling point of the pure solvent on the graph by locating the point where the 1-atm pressure horizontal line intersects the black vapor-pressure curve and then tracing this point down to the temperature axis. For the solution, the 1-atm line intersects the blue vapor-pressure curve at a higher temperature, indicating that the solution has a higher boiling point than the pure solvent.

The increase in the boiling point of a solution, relative to the pure solvent, depends on the molality of the solute. But it is important to remember that boiling-point

▲ Figure 13.20 Phase diagram illustrating boiling-point elevation. The black lines show the pure solvent’s phase equilibria curves, and the blue lines show the solution’s phase equilibria curves.

Temperature

1 atm

Liquid

Gas

Solid

Vapor pressure ofpure liquid solvent

Solutionboiling point

Pure solventboiling point

Vaporpressureof solution

∆Tb

Pres

sure


elevation is proportional to the total concentration of solute particles, regardless of whether the particles are molecules or ions. When NaCl dissolves in water, 2 mol of solute particles (1 mol of Na+ and 1 mol of Cl-) are formed for each mole of NaCl that dissolves. We take this into account by defining i, the van’t Hoff factor, as the number of fragments that a solute breaks up into for a particular solvent. The change in boiling point for a solution compared to the pure solvent is:

∆Tb = Tb(solution) - Tb(solvent) = iKbm [13.12]

In this equation, Tb1solution2 is the boiling point of the solution, Tb1solvent2 is the boiling point of the pure solvent, m is the molality of the solute, Kb is the molal boiling-point-elevation constant for the solvent (which is a proportionality constant that is experimentally determined for each solvent), and i is the van’t Hoff factor. For a non-electrolyte, we can always assume i = 1; for an electrolyte, i will depend on how the sub-stance ionizes in that solvent. For instance, i = 2 for NaCl in water, assuming complete dissociation of ions. As a result, we expect the boiling-point elevation of a 1 m aqueous solution of NaCl to be twice as large as the boiling-point elevation of a 1 m solution of a nonelectrolyte such as sucrose. Thus, to properly predict the effect of a particular solute on boiling-point elevation (or any other colligative property), it is important to know whether the solute is an electrolyte or a nonelectrolyte. (Sections 4.1 and 4.3)

Give it some thoughtA solute dissolved in water causes the boiling point to increase by 0.51 °C. Does this necessarily mean that the concentration of the solute is 1.0 m (▶ table 13.3)?

Freezing-Point DepressionThe vapor-pressure curves for the liquid and solid phases meet at the triple point.

(Section 11.6) In ▼ Figure 13.21 we see that the triple-point temperature of the solution is lower than the triple-point temperature of pure liquid because the solution has a lower vapor pressure than the pure liquid.

The freezing point of a solution is the temperature at which the first crystals of pure solvent form in equilibrium with the solution. Recall from Section 11.6 that the line representing the solid–liquid equilibrium rises nearly vertically from the triple point. It is easy to see in Figure 13.21 that the triple-point temperature of the solution is lower than that of the pure liquid, but it is also true for all points along the solid–liquid equilibrium curve: the freezing point of the solution is lower than that of the pure liquid.

▲ Figure 13.21 Phase diagram illustrating freezing-point depression. The black lines show the pure solvent’s phase equilibria curves, and the blue lines show the solution’s phase equilibria curves.

Temperature

1 atm

Liquid

Gas

Solid

Solution triplepoint

Vapor pressureof pure liquidsolvent

Vapor pressureof pure solidsolvent

Pure solventtriple point

Solution freezingpoint

Pure solventfreezing point

Vapor pressureof solution

∆Tf

Pres

sure


Like the boiling-point elevation, the change in freezing point ∆Tf is directly pro-portional to solute molality, taking into account the van’t Hoff factor i:

∆Tf = Tf 1solution2 - Tf 1solvent2 = - iKf m [13.13]

The proportionality constant Kf is the molal freezing-point-depression constant, analogous to Kb for boiling point elevation. Note that because the solution freezes at a lower temperature than does the pure solvent, the value of ∆Tf is negative.

Some typical values of Kb and Kf for several common solvents are given in ▲ table 13.3. For water, the table shows Kb = 0.51 °C>m, which means that the boiling point of any aqueous solution that is 1 m in nonvolatile solute particles is 0.51 °C higher than the boiling point of pure water. Because solutions generally do not behave ideally, the constants listed in Table 13.3 serve well only for solutions that are rather dilute.

For water, Kf is 1.86 °C>m. Therefore, any aqueous solution that is 1 m in nonvola-tile solute particles (such as 1 m C6H12O6 or 0.5 m NaCl) freezes at the temperature that is 1.86 °C lower than the freezing point of pure water.

The freezing-point depression caused by solutes has useful applications: it is why antifreeze works in car cooling systems, and why calcium chloride 1CaCl22 promotes the melting of ice on roads during winter.

table 13.3 Molal Boiling-Point-elevation and Freezing-Point-depression Constants

Solventnormal Boiling Point 1°C2 Kb1°C>m2

normal Freezing Point 1°C2 Kf 1°C>m2

Water, H2O 100.0 0.51 0.0 1.86Benzene, C6H6 80.1 2.53 5.5 5.12Ethanol, C2H5OH 78.4 1.22 -114.6 1.99Carbon tetrachloride, CCl4 76.8 5.02 -22.3 29.8Chloroform, CHCl3 61.2 3.63 -63.5 4.68

Automotive antifreeze contains ethylene glycol, CH21OH2CH21OH2, a nonvolatile nonelec-trolyte, in water. Calculate the boiling point and freezing point of a 25.0% by mass solution of ethylene glycol in water.

SoLutIonAnalyze We are given that a solution contains 25.0% by mass of a nonvolatile, nonelectrolyte sol-ute and asked to calculate the boiling and freezing points of the solution. To do this, we need to calculate the boiling-point elevation and freezing-point depression.Plan To calculate the boiling-point elevation and the freezing-point depression using Equations 13.12 and 13.13, we must express the concentration of the solution as molality. Let’s assume for convenience that we have 1000 g of solution. Because the solution is 25.0% by mass ethylene glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively. Using these quantities, we can calculate the molality of the solution, which we use with the molal boiling-point-elevation and freezing-point-depression constants (Table 13.3) to calculate ∆Tb and ∆Tf. We add ∆Tb to the boiling point and ∆Tf to the freezing point of the solvent to obtain the boiling point and freezing point of the solution.Solve The molality of the solution is calculated as follows:

Molality =moles C2H6O2

kilograms H2O= a 250 g C2H6O2

750 g H2Ob a 1 mol C2H6O2

62.1 g C2H6O2b a 1000 g H2O

1 kg H2Ob

= 5.37 m

SaMPLe exerCISe 13.8 Calculation of Boiling-Point elevation

and Freezing-Point depression


OsmosisCertain materials, including many membranes in biological systems and synthetic sub-stances such as cellophane, are semipermeable. When in contact with a solution, these materials allow only ions or small molecules—water molecules, for instance—to pass through their network of tiny pores.

Consider a situation in which only solvent molecules are able to pass through a semipermeable membrane placed between two solutions of different concentrations. The rate at which the solvent molecules pass from the less concentrated solution (lower solute concentration but higher solvent concentration) to the more concentrated solu-tion (higher solute concentration but lower solvent concentration) is greater than the rate in the opposite direction. Thus, there is a net movement of solvent molecules from the solution with a lower solute concentration into the one with a higher solute con-centration. In this process, called osmosis, the net movement of solvent is always toward the solution with the lower solvent (higher solute) concentration, as if the solutions were driven to attain equal concentrations.

▶ Figure 13.22 shows the osmosis that occurs between an aqueous solution and pure water, separated by a semipermeable membrane. The U-tube contains water on the left and an aqueous solution on the right. Initially, there is a net movement of water through the membrane from left to right, leading to unequal liquid levels in the two arms of the U-tube. Eventually, at equilibrium (middle panel of Figure 13.22), the pres-sure difference resulting from the unequal liquid heights becomes so large that the net flow of water ceases. This pressure, which stops osmosis, is the osmotic pressure, Π, of the solution. If an external pressure equal to the osmotic pressure is applied to the solution, the liquid levels in the two arms can be equalized, as shown in the right panel of Figure 13.22.

The osmotic pressure obeys a law similar in form to the ideal-gas law, ΠV = inRTwhere Π is the osmotic pressure, V is the volume of the solution, i is the van’t Hoff factor, n is the number of moles of solute, R is the ideal-gas constant, and T is the absolute temperature. From this equation, we can write

Π = ia nVbRT = iMRT [13.14]

We can now use Equations 13.12 and 13.13 to calculate the changes in the boiling and freezing points:

∆Tb = iKbm = 11210.51 °C>m215.37 m2 = 2.7 °C∆Tf = - iKfm = -11211.86 °C>m215.37 m2 = -10.0 °C

Hence, the boiling and freezing points of the solution are readily calculated: ∆Tb = Tb1solution2 - Tb1solvent2

2.7 °C = Tb1solution2 - 100.0 °C Tb1solution2 = 102.7 °C

∆Tf = Tf 1solution2 - Tf 1solvent2 -10.0 °C = Tf 1solution2 - 0.0 °C

Tf 1solution2 = -10.0 °C

Comment Notice that the solution is a liquid over a larger temperature range than the pure solvent.

Practice exercise 1Which aqueous solution will have the lowest freezing point? (a) 0.050 m CaCl2, (b) 0.15 m NaCl, (c) 0.10 m HCl, (d) 0.050 m CH3COOH, (e) 0.20 m C12H22O11.

Practice exercise 2Referring to Table 13.3, calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol 1C10H18O2, a fragrant substance found in the leaves of euca-lyptus trees.


where M is the molarity of the solution. Because the osmotic pressure for any solution depends on the solution concentration, osmotic pressure is a colligative property.

If two solutions of identical osmotic pressure are separated by a semipermeable membrane, no osmosis will occur. The two solutions are isotonic with respect to each other. If one solution is of lower osmotic pressure, it is hypotonic with respect to the more concentrated solution. The more concentrated solution is hypertonic with respect to the dilute solution.

Give it some thoughtOf two KBr solutions, one 0.50 m and the other 0.20 m, which is hypotonic with respect to the other?

Osmosis plays an important role in living systems. The membranes of red blood cells, for example, are semipermeable. Placing a red blood cell in a solution that is hypertonic relative to the intracellular solution (the solution inside the cells) causes water to move out of the cell (Figure 13.23). This causes the cell to shrivel, a process called crenation. Placing the cell in a solution that is hypotonic relative to the intracellular fluid causes water to move into the cell, which may cause the cell to rupture, a process called hemolysis. People who need body fluids or nutrients replaced but cannot be fed orally are given solutions by intravenous (IV) infusion, which feeds nutrients directly into the veins. To prevent crenation or hemolysis of red blood cells, the IV solutions must be isotonic with the intracellular fluids of the blood cells.

▲ Figure 13.22 osmosis is the process of a solvent moving from one compartment to another, across a semipermeable membrane, toward higher solute concentration. Osmotic pressure is generated at equilibrium due to the different heights of liquid on either side of the membrane and is equivalent to the pressure needed to equalize the fluid levels across the membrane.

Go FIGureIf the pure water in the left arm of the U-tube is replaced by a solution more concentrated than the one in the right arm, what will happen?

Soluteparticle

Semipermeablemembrane

Solventparticle(H2O)

Puresolvent

Solution

Initially At equilibriumPressure applied to equalize �uid levels in both arms equals osmotic pressure, ∏

Net movement of H2O is from pure water side to solution side

At equilibrium, the �ow of H2O is the same in both directions, so there is no net movement of H2O


▲ Figure 13.23 osmosis through red blood cell walls. If water moves out of the red blood cell, it shrivels (crenation); if water moves into the red blood cell, it will swell and may burst (hemolysis).

Go FIGureIf the fluid surrounding a patient’s red blood cells is depleted in electrolytes, is crenation or hemolysis more likely to occur?

The average osmotic pressure of blood is 7.7 atm at 25 °C. What molarity of glucose 1C6H12O62 will be isotonic with blood?

SoLutIonAnalyze We are asked to calculate the concentration of glucose in water that would be isotonic with blood, given that the osmotic pressure of blood at 25 °C is 7.7 atm.Plan Because we are given the osmotic pressure and temperature, we can solve for the concentra-tion, using Equation 13.14. Because glucose is a nonelectrolyte, i = 1.Solve

Π = iMRT

M =Π

iRT=

17.7 atm2

112a0.0821L@atmmol@K

b1298 K2= 0.31 M

Comment In clinical situations, the concentrations of solutions are generally expressed as mass percentages. The mass percentage of a 0.31 M solution of glucose is 5.3%. The concentration of NaCl that is isotonic with blood is 0.16 M, because i = 2 for NaCl in water (a 0.155 M solution of NaCl is 0.310 M in particles). A 0.16 M solution of NaCl is 0.9% mass in NaCl. This kind of solution is known as a physiological saline solution.

Practice exercise 1Which of the following actions will raise the osmotic pressure of a solution? (a) decreasing the solute concentration, (b) decreasing the temperature, (c) adding more solvent, (d) increasing the temperature, (e) none of the above.

Practice exercise 2What is the osmotic pressure, in atm, of a 0.0020 M sucrose 1C12H22O112 solution at 20.0 °C?

SaMPLe exerCISe 13.9 osmotic Pressure Calculations

Crenation of red blood cell placed in hypertonic environment

Red blood cell in isotonic medium neither swells nor shrinks.

Hemolysis of red blood cell placed in hypotonic environment

High soluteconcentration

High soluteconcentration

Low soluteconcentration

Low soluteconcentration

The arrows represent the net movement of water molecules.


There are many interesting biological examples of osmosis. A cucumber placed in concentrated brine loses water via osmosis and shrivels into a pickle. People who eat a lot of salty food retain water in tissue cells and intercellular space because of osmosis. The resultant swelling or puffiness is called edema. Water moves from soil into plant roots partly because of osmosis. Bacteria on salted meat or candied fruit lose water through osmosis, shrivel, and die—thus preserving the food.

Give it some thoughtIs the osmotic pressure of a 0.10 M solution of NaCl greater than, less than, or equal to that of a 0.10 M solution of KBr?

Determination of Molar Mass from Colligative PropertiesThe colligative properties of solutions provide a useful means of determining the molar mass of solutes. Any of the four colligative properties can be used, as shown in Sample Exercises 13.10 and 13.11.

The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 °C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass.

SaMPLe exerCISe 13.11 Molar Mass from osmotic Pressure

SoLutIonAnalyze Our goal is to calculate the molar mass of a high-molecular-mass protein, based on its osmotic pressure and a knowledge of the mass of protein and solution volume. Since the protein will be consid-ered as a nonelectrolyte, i = 1.

Plan The temperature 1T = 25 °C2 and osmotic pressure 1Π = 1.54 torr2 are given, and we know the value of R so we can use Equation 13.14 to calculate the molarity of the solution, M. In doing so, we must convert temperature from °C to K and the osmotic pres-sure from torr to atm. We then use the molarity and the volume of

A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357 °C higher than that of the pure solvent. Calculate the molar mass of the solute.

SaMPLe exerCISe 13.10 Molar Mass from Freezing-Point depression

SoLutIonAnalyze Our goal is to calculate the molar mass of a solute based on knowledge of the boiling-point elevation of its solution in CCl4, ∆Tb = 0.357 °C, and the masses of solute and solvent. Table 13.3 gives Kb for the solvent 1CCl42, Kb = 5.02 °C>m.Plan We can use Equation 13.12, ∆Tb = iKbm, to calculate the mo-lality of the solution. Because the solute is a nonelectrolyte, i = 1. Then we can use molality and the quantity of solvent (40.0 g CCl4) to calculate the number of moles of solute. Finally, the molar mass of the solute equals the number of grams per mole, so we divide the number of grams of solute (0.250 g) by the number of moles we have just calculated.Solve From Equation 13.12, we have

Molality =∆Tb

iKb=

0.357 °C1125.02 °C>m

= 0.0711 m

Thus, the solution contains 0.0711 mol of solute per kilogram of sol-vent. The solution was prepared using 40.0 g = 0.0400 kg of solvent 1CCl42. The number of moles of solute in the solution is therefore

10.0400 kg CCl42a0.0711 mol solute

kg CCl4b = 2.84 * 10-3 mol solute

The molar mass of the solute is the number of grams per mole of the substance:

Molar mass =0.250 g

2.84 * 10-3 mol= 88.0 g>mol

Practice exercise 1A mysterious white powder could be powdered sugar 1C12H22O112, cocaine 1C17H21NO42, codeine 1C18H21NO32, nor-fenefrine 1C8H11NO22, or fructose 1C6H12O62. When 80 mg of the powder is dissolved in 1.50 mL of ethanol (d = 0.789 g>cm3, normal freezing point -114.6 °C, Kf = 1.99 °C>m), the freezing point is lowered to -115.5 °C. What is the identity of the white powder? (a) powdered sugar, (b) cocaine, (c) codeine, (d) norfene-frine, (e) fructose.

Practice exercise 2Camphor 1C10H16O2 melts at 179.8 °C, and it has a particularly large freezing-point-depression constant, Kf = 40.0 °C>m. When 0.186 g of an organic substance of unknown molar mass is dis-solved in 22.01 g of liquid camphor, the freezing point of the mix-ture is found to be 176.7 °C. What is the molar mass of the solute?


the solution (5.00 mL) to determine the number of moles of solute. Finally, we obtain the molar mass by dividing the mass of the solute (3.50 mg) by the number of moles of solute.Solve Solving Equation 13.14 for molarity gives

Molarity =Π

iRT=

11.54 torr2a 1 atm760 torr

b

112a0.0821L@atmmol@K

b1298 K2= 8.28 * 10-5

molL

Because the volume of the solution is 5.00 mL = 5.00 * 10-3 L, the number of moles of protein must beMoles = 18.28 * 10-5 mol>L215.00 * 10-3 L2 = 4.14 * 10-7 mol

The molar mass is the number of grams per mole of the substance. Because we know the sample has a mass of 3.50 mg = 3.50 * 10-3 g, we can calculate the molar mass by dividing the number of grams in the sample by the number of moles we just calculated:

Molar mass =gramsmoles

=3.50 * 10-3 g

4.14 * 10-7 mol= 8.45 * 103 g>mol

Comment Because small pressures can be measured easily and accu-rately, osmotic pressure measurements provide a useful way to deter-mine the molar masses of large molecules.

Practice exercise 1Proteins frequently form complexes in which 2, 3, 4, or even more individual proteins (“monomers”) interact specifically with each other via hydrogen bonds or electrostatic interactions. The entire assembly of proteins can act as one unit in solution, and this assembly is called the “quaternary structure” of the protein. Suppose you discover a new protein whose monomer molar mass is 25,000 g>mol. You measure an osmotic pressure of 0.0916 atm at 37 °C for 7.20 g of the protein in 10.00 mL of an aqueous solution. How many protein monomers form the quaternary protein structure in solution? Treat the protein as a nonelectrolyte. (a) 1, (b) 2, (c) 3, (d) 4 , (e) 8.

Practice exercise 2A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form 0.100 L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25 °C. Calculate the molar mass of the polystyrene.

a Closer Look

the van’t Hoff Factor

The colligative properties of solutions depend on the total concentration of solute particles, regardless of whether the particles are ions or mol-ecules. Thus, we expect a 0.100 m solution of NaCl to have a freezing-point depression of 12210.100 m211.86 °C>m2 = 0.372 °C because it is 0.100 m in Na+1aq2 and 0.100 m in Cl-1aq2. The measured freezing-point depression is only 0.348 °C, however, and the situation is similar for other strong electrolytes. A 0.100 m solution of KCl, for example, freezes at -0.344 °C.

The difference between expected and observed colligative properties for strong electrolytes is due to electrostatic attractions between ions. As the ions move about in solution, ions of oppo-site charge collide and “stick together” for brief moments. While they are together, they behave as a single particle called an ion pair (▶ Figure 13.24). The number of independent particles is thereby reduced, causing a reduction in the freezing-point depression (as well as in boiling-point elevation, vapor-pressure reduction, and osmotic pressure).

We have been assuming that the van’t Hoff factor, i, is equal to the number of ions per formula unit of the electrolyte. The true (mea-sured) value of this factor, however, is given by the ratio of the mea-sured value of a colligative property to the value calculated when the substance is assumed to be a nonelectrolyte. Using the freezing-point depression, for example, we have

i =∆Tf 1measured2

∆Tf 1calculated for nonelectrolyte2 [13.15]

The limiting value of i can be determined for a salt from the number of ions per formula unit. For NaCl, for example, the limiting van’t Hoff factor is 2 because NaCl consists of one Na+ and one Cl- per formula unit; for K2SO4 it is 3 because K2SO4 consists of two K+ and

▲ Figure 13.24 ion pairing and colligative properties. A solution of NaCl contains not only separated Na+1aq2 and Cl-1aq2 ions but ion pairs as well.

Cl−Na+

One ion pair

one SO42- per formula unit. In the absence of any information about

the actual value of i for a solution, we will use the limiting value in calculations.

Two trends are evident in ▶ table 13.4 , which gives measured van’t Hoff factors for several substances at different

SecTion 13.6 colloids 559

13.6 | ColloidsSome substances appear to initially dissolve in a solvent, but over time, the substance separates from the pure solvent. For example, finely divided clay particles dispersed in water eventually settle out because of gravity. Gravity affects the clay particles because they are much larger than most molecules, consisting of thousands or even millions of atoms. In contrast, the dispersed particles in a true solution (ions in a salt solution or glucose molecules in a sugar solution) are small. Between these extremes lie dispersed particles that are larger than typical molecules but not so large that the components of the mixture separate under the influence of gravity. These intermediate types of disper-sions are called either colloidal dispersions or simply colloids. Colloids form the di-viding line between solutions and heterogeneous mixtures. Like solutions, colloids can be gases, liquids, or solids. Examples of each are listed in ▼ table 13.5.

Particle size can be used to classify a mixture as colloid or solution. Colloid par-ticles range in diameter from 5 to 1000 nm; solute particles are smaller than 5 nm in diameter. The nanomaterials we saw in Chapter 12 (Section 12.9), when dispersed in a liquid, are colloids. A colloid particle may even consist of a single giant molecule. The hemoglobin molecule, for example, which carries oxygen in your blood, has molecular dimensions of 6.5 * 5.5 * 5.0 nm and a molar mass of 64,500 g>mol.

Although colloid particles may be so small that the dispersion appears uniform even under a microscope, they are large enough to scatter light. Consequently, most colloids appear cloudy or opaque unless they are very dilute. (For example, homog-enized milk is a colloid of fat and protein molecules dispersed in water.) Furthermore, because they scatter light, a light beam can be seen as it passes through a colloidal

dilutions. First, dilution affects the value of i for electrolytes; the more dilute the solution, the more closely i approaches the ex-pected value based on the number of ions in the formula unit. Thus, we conclude that the extent of ion pairing in electrolyte solu-tions decreases upon dilution. Second, the lower the charges on the ions, the less i departs from the expected value because the extent of ion pairing decreases as the ionic charges decrease. Both trends are consistent with simple electrostatics: The force of interaction between charged particles decreases as their separation increases and as their charges decrease.

Related Exercises: 13.83, 13.84, 13.103, 13.105

table 13.4 Measured and expected van’t hoff Factors for Several Substances at 25 °C

Concentration

Compound 0.100 m 0.0100 m 0.00100 mexpected

Value

Sucrose 1.00 1.00 1.00 1.00NaCl 1.87 1.94 1.97 2.00K2SO4 2.32 2.70 2.84 3.00MgSO4 1.21 1.53 1.82 2.00

table 13.5 types of Colloids

Phase of Colloid

dispersing (solvent-like) Substance

dispersed (solute-like) Substance Colloid type example

Gas Gas Gas — None (all are solutions)Gas Gas Liquid Aerosol FogGas Gas Solid Aerosol SmokeLiquid Liquid Gas Foam Whipped creamLiquid Liquid Liquid Emulsion MilkLiquid Liquid Solid Sol PaintSolid Solid Gas Solid foam MarshmallowSolid Solid Liquid Solid emulsion ButterSolid Solid Solid Solid sol Ruby glass


dispersion (▲ Figure 13.25). This scattering of light by colloidal particles, known as the Tyndall effect, makes it possible to see the light beam of an automobile on a dusty dirt road, or the sunlight streaming through trees or clouds. Not all wavelengths are scattered to the same extent. Colors at the blue end of the visible spectrum are scattered more than those at the red end by the molecules and small dust particles in the atmo-sphere. As a result, our sky appears blue. At sunset, light from the sun travels through more of the atmosphere; blue light is scattered even more, allowing the reds and yel-lows to pass through and be seen.

Hydrophilic and Hydrophobic ColloidsThe most important colloids are those in which the dispersing medium is water. These colloids may be hydrophilic (“water loving”) or hydrophobic (“water fear-ing”). Hydrophilic colloids are most like the solutions that we have previously examined. In the human body, the extremely large protein molecules such as en-zymes and antibodies are kept in suspension by interaction with surrounding water molecules. A hydrophilic molecule folds in such a way that its hydrophobic groups are away from the water molecules, on the inside of the folded molecule, while its hydrophilic, polar groups are on the surface, interacting with the water molecules. The hydrophilic groups generally contain oxygen or nitrogen and often carry a charge (◀ Figure 13.26).

Give it some thoughtSome proteins reside in the hydrophobic lipid bilayer of the cell membrane. Would hydrophilic groups of these proteins still be facing the lipid “solvent”?

Hydrophobic colloids can be dispersed in water only if they are stabilized in some way. Otherwise, their natural lack of affinity for water causes them to separate from the water. One method of stabilization involves adsorbing ions on the surface of the hydrophobic particles (▶ Figure 13.27). (Adsorption means to adhere to a surface. It differs from absorption, which means to pass into the interior, as when a

▲ Figure 13.25 tyndall effect in the laboratory. The glass on the right contains a colloidal dispersion; that on the left contains a solution.

OH

N

C

−

−

+

Hydrophilic polar and charged groups onmolecule surface help molecule remaindispersed in water and other polar solvents

▲ Figure 13.26 Hydrophilic colloidal particle. Examples of the hydrophilic groups that help to keep a giant molecule (macromolecule) suspended in water.

Go FIGureWhat is the chemical composition of the groups that carry a negative charge?


sponge absorbs water.) The adsorbed ions can interact with water, thereby stabiliz-ing the colloid. At the same time, the electrostatic repulsion between adsorbed ions on neighboring colloid particles keeps the particles from sticking together rather than dispersing in the water.

Hydrophobic colloids can also be stabilized by hydrophilic groups on their sur-faces. Oil drops are hydrophobic, for example, and they do not remain suspended in water. Instead, they aggregate, forming an oil slick on the water surface. Sodium stea-rate (▼ Figure 13.28), or any similar substance having one end that is hydrophilic (either polar or charged) and one end that is hydrophobic (nonpolar), will stabilize a suspension of oil in water. Stabilization results from the interaction of the hydrophobic ends of the stearate ions with the oil drops and the hydrophilic ends with the water.

Give it some thoughtWhy don’t oil drops stabilized by sodium stearate coagulate to form larger oil drops?

▲ Figure 13.27 Hydrophobic colloids stabilized in water by adsorbed anions.

Adsorbed anionscan interactwith water

Cations insolution Water

RepulsionHydrophobic

particleHydrophobic

particle

Like charges keep particles from colliding, sticking together, and growing large enough to settle out of solution

−

−

−

−

−

−

−

−

−

−−

−

− −

−

−

−

−

−+

+

+

++

+

+

++ +

+

+

▲ Figure 13.28 stabilization of an emulsion of oil in water by stearate ions.

Go FIGureWhich kind of intermolecular force attracts the stearate ion to the oil drop?

Stearateion

Oildroplet

Water

Hydrophobicend

Hydrophilicend

Na+

−

−−

−

−

−

−

−

−−

−−

++

+

+

+

+

+

+

+

+


Colloid stabilization has an interesting application in the human digestive system. When fats in our diet reach the small intestine, a hormone causes the gallbladder to excrete a fluid called bile. Among the components of bile are compounds that have chemical structures similar to sodium stearate; that is, they have a hydrophilic (polar) end and a hydrophobic (nonpolar) end. These compounds emulsify the fats in the intes-tine and thus permit digestion and absorption of fat-soluble vitamins through the intes-tinal wall. The term emulsify means “to form an emulsion,” a suspension of one liquid in another, with milk being one example (Table 13.5). A substance that aids in the forma-tion of an emulsion is called an emulsifying agent. If you read the labels on foods and other materials, you will find that a variety of chemicals are used as emulsifying agents. These chemicals typically have a hydrophilic end and a hydrophobic end.

Chemistry and Life

sickle-Cell Anemia

Our blood contains the complex protein hemoglobin, which carries oxygen from the lungs to other parts of the body. In the genetic dis-ease sickle-cell anemia, hemoglobin molecules are abnormal and have a lower solubility in water, especially in their unoxygenated form. Consequently, as much as 85% of the hemoglobin in red blood cells crystallizes out of solution.

The cause of the insolubility is a structural change in one part of an amino acid. Normal hemoglobin molecules contain an amino acid that has a ¬ CH2CH2COOH group:

O

C OHCH2CH2

Normal Normal

The polarity of the ¬ COOH group contributes to the solubility of the hemoglobin molecule in water. In the hemoglobin molecules of sickle-cell anemia patients, the ¬ CH2CH2COOH chain is absent and in its place is the nonpolar (hydrophobic) ¬ CH1CH322 group:

CH3CH

CH3

Abnormal Abnormal

This change leads to the aggregation of the defective form of hemoglobin into particles too large to remain suspended in bio-logical fluids. It also causes the cells to distort into the sickle shape shown in ▼ Figure 13.29. The sickled cells tend to clog capillar-ies, causing severe pain, weakness, and the gradual deterioration of vital organs. The disease is hereditary, and if both parents carry the defective genes, it is likely that their children will possess only abnormal hemoglobin.

You might wonder how it is that a life-threatening disease such as sickle-cell anemia has persisted in humans through evolutionary time. The answer in part is that people with the disease are far less susceptible to malaria. Thus, in tropical climates rife with malaria, those with sickle-cell disease have lower incidence of this debilitating disease.

▲ Figure 13.29 A scanning electron micrograph of normal (round) and sickle (crescent-shaped) red blood cells. Normal red blood cells are about 6 * 10-3 mm in diameter.

Colloidal Motion in LiquidsWe learned in Chapter 10 that gas molecules move at some average speed that depends inversely on their molar mass, in a straight line, until they collide with some-thing. The mean free path is the average distance molecules travel between collisions.

(Section 10.8) Recall also that the kinetic-molecular theory of gases assumes that gas molecules are in continuous, random motion. (Section 10.7)


table 13.6 Calculated Mean Free Path, after one hour, for uncharged Colloidal Spheres in Water at 20 °C

radius of sphere, nm Mean Free Path, mm

1 1.2310 0.390100 0.1231000 0.039

A 0.100-L solution is made by dissolving 0.441 g of CaCl21s2 in water. (a) Calculate the osmotic pressure of this solution at 27 °C, assuming that it is completely dissociated into its component ions. (b) The measured osmotic pressure of this solution is 2.56 atm at 27 °C. Explain why it is less than the value calculated in (a), and calculate the van’t Hoff factor, i, for the solute in this solution. (c) The enthalpy of solution for CaCl2 is ∆H = -81.3 kJ>mol. If the final temperature of the solution is 27 °C, what was its initial temperature? (Assume that the density of the solu-tion is 1.00 g>mL, that its specific heat is 4.18 J>g@K, and that the solution loses no heat to its surroundings.)

SoLutIon(a) The osmotic pressure is given by Equation 13.14, Π = iMRT. We know the temperature,

T = 27 °C = 300 K, and the gas constant, R = 0.0821 L@atm/mol@K. We can calculate the molarity of the solution from the mass of CaCl2 and the volume of the solution:

Molarity = a 0.441 g CaCl2

0.100 Lb a 1 mol CaCl2

110 g CaCl2b = 0.0397 mol CaCl2>L

Soluble ionic compounds are strong electrolytes. (Sections 4.1 and 4.3) Thus, CaCl2 consists of metal cations 1Ca2+2 and nonmetal anions 1Cl-2. When completely dissociated, each CaCl2 unit forms three ions (one Ca2+ and two Cl-). Hence, the calculated osmotic pressure is

Π = iMRT = 13210.0397 mol>L210.0821 L@atm>mol@K21300 K2 = 2.93 atm(b) The actual values of colligative properties of electrolytes are less than those calculated be-

cause the electrostatic interactions between ions limit their independent movements. In this case, the van’t Hoff factor, which measures the extent to which electrolytes actually dissoci-ate into ions, is given by

i =Π1measured2

Π1calculated for nonelectrolyte2

=2.56 atm

10.0397 mol>L210.0821 L@atm>mol@K21300 K2 = 2.62

Thus, the solution behaves as if the CaCl2 has dissociated into 2.62 particles instead of the ideal 3.

(c) If the solution is 0.0397 M in CaCl2 and has a total volume of 0.100 L, the number of moles of solute is 10.100 L210.0397 mol>L2 = 0.00397 mol. Hence, the quantity of heat gener-ated in forming the solution is 10.00397 mol21-81.3 kJ>mol2 = -0.323 kJ. The solution

SaMPLe InteGratIVe exerCISe Putting Concepts together

Colloidal particles in a solution undergo random motion as a result of collisions with solvent molecules. Because the colloidal particles are massive in comparison with solvent molecules, their movements from any one collision are very tiny. How-ever, there are many such collisions, and they cause a random motion of the entire colloidal particle, called Brownian motion. In 1905, Einstein developed an equation for the average square of the displacement of a colloidal particle, a historically very important development. As you might expect, the larger the colloidal particle, the shorter its mean free path in a given liquid (▼ table 13.6). Today, the understanding of Brownian motion is applied to diverse problems in everything from cheese-making to medical imaging.


Chapter Summary and Key termstHe solution ProCess (seCtion 13.1) Solutions form when one substance disperses uniformly throughout another. The attractive interaction of solvent molecules with solute is called solvation. When the solvent is water, the interaction is called hydration. The dissolution of ionic substances in water is promoted by hydration of the separated ions by the polar water molecules. The overall enthalpy change upon solution formation may be either positive or negative. Solution forma-tion is favored both by a positive entropy change, corresponding to an increased dispersal of the components of the solution, and by a nega-tive enthalpy change, indicating an exothermic process.

sAturAted solutions And soluBility (seCtion 13.2) The equilibrium between a saturated solution and undissolved solute is dynamic; the process of solution and the reverse process, crystallization, occur simultaneously. In a solution in equilibrium with undissolved sol-ute, the two processes occur at equal rates, giving a saturated solution. If there is less solute present than is needed to saturate the solution, the solution is unsaturated. When solute concentration is greater than the equilibrium concentration value, the solution is supersaturated. This is an unstable condition, and separation of some solute from the solution will occur if the process is initiated with a solute seed crystal. The amount of solute needed to form a saturated solution at any particular temperature is the solubility of that solute at that temperature.

FACtors AFFeCtinG soluBility (seCtion 13.3) The solubility of one substance in another depends on the tendency of systems to become more random, by becoming more dispersed in space, and on the relative intermolecular solute–solute and solvent–solvent energies compared with solute–solvent interactions. Polar and ionic solutes tend to dissolve in polar solvents, and nonpolar solutes tend to dis-solve in nonpolar solvents (“like dissolves like”). Liquids that mix in all proportions are miscible; those that do not dissolve significantly in one another are immiscible. Hydrogen-bonding interactions between solute and solvent often play an important role in determining solu-bility; for example, ethanol and water, whose molecules form hydro-gen bonds with each other, are miscible. The solubilities of gases in a liquid are generally proportional to the pressure of the gas over the solution, as expressed by Henry’s law: Sg = kPg. The solubilities of most solid solutes in water increase as the temperature of the solu-tion increases. In contrast, the solubilities of gases in water generally decrease with increasing temperature.

exPressinG solution ConCentrAtions (seCtion 13.4) Concentrations of solutions can be expressed quantitatively by sev-eral different measures, including mass percentage [(mass solute/mass solution) * 100] parts per million (ppm), parts per billion (ppb), and mole fraction. Molarity, M, is defined as moles of solute per liter of solution; molality, m, is defined as moles of solute per kilogram of sol-vent. Molarity can be converted to these other concentration units if the density of the solution is known.

ColliGAtive ProPerties (seCtion 13.5) A physical property of a solution that depends on the concentration of solute particles present, regardless of the nature of the solute, is a colligative property. Colligative properties include vapor-pressure lowering, freezing-point lowering, boiling-point elevation, and osmotic pressure. raoult’s law expresses the lowering of vapor pressure. An ideal solution obeys Raoult’s law. Differences in solvent–solute as compared with solvent–solvent and solute–solute intermolecular forces cause many solutions to depart from ideal behavior.

A solution containing a nonvolatile solute possesses a higher boiling point than the pure solvent. The molal boiling-point-elevation constant, Kb, represents the increase in boiling point for a 1 m solu-tion of solute particles as compared with the pure solvent. Similarly, the molal freezing-point-depression constant, Kf, measures the lowering of the freezing point of a solution for a 1 m solution of solute particles. The temperature changes are given by the equations ∆Tb = iKbm and ∆Tf = - iKf m where i is the van’t Hoff factor, which represents how many particles the solute breaks up into in the solvent. When NaCl dissolves in water, two moles of solute particles are formed for each mole of dissolved salt. The boiling point or freezing point is thus ele-vated or depressed, respectively, approximately twice as much as that of a nonelectrolyte solution of the same concentration. Similar consid-erations apply to other strong electrolytes.

osmosis is the movement of solvent molecules through a semiperme-able membrane from a less concentrated to a more concentrated solution. This net movement of solvent generates an osmotic pressure, Π, which can be measured in units of gas pressure, such as atm. The osmotic pres-sure of a solution is proportional to the solution molarity: Π = iMRT. Osmosis is a very important process in living systems, in which cell walls act as semipermeable membranes, permitting the passage of water but restricting the passage of ionic and macromolecular components.

absorbs this heat, causing its temperature to increase. The relationship between temperature change and heat is given by Equation 5.22:

q = 1specific heat21grams21∆T2The heat absorbed by the solution is q = +0.323 kJ = 323 J. The mass of the 0.100 L of solution is 1100 mL211.00 g>mL2 = 100 g (to three significant figures). Thus, the tempera-ture change is

∆T =q

1specific heat of solution21grams of solution2

=323 J

14.18 J>g@K21100 g2 = 0.773 K

A kelvin has the same size as a degree Celsius. (Section 1.4) Because the solution tem-perature increases by 0.773 °C, the initial temperature was 27.0 °C - 0.773 °C = 26.2 °C.

Key equations 565

Colloids (seCtion 13.6) Particles that are large on the molecular scale but still small enough to remain suspended indefinitely in a sol-vent system form colloids, or colloidal dispersions. Colloids, which are intermediate between solutions and heterogeneous mixtures, have many practical applications. One useful physical property of colloids, the scat-tering of visible light, is referred to as the tyndall effect. Aqueous col-loids are classified as hydrophilic or hydrophobic. Hydrophilic colloids

are common in living organisms, in which large molecular aggregates (enzymes, antibodies) remain suspended because they have many polar, or charged, atomic groups on their surfaces that interact with water. Hydrophobic colloids, such as small droplets of oil, may remain in sus-pension through adsorption of charged particles on their surfaces.

Colloids undergo Brownian motion in liquids, analogous to the random three-dimensional motion of gas molecules.

Learning outcomes after studying this chapter, you should be able to:

• Describe how enthalpy and entropy changes affect solution formation. (Section 13.1)

• Describe the relationship between intermolecular forces and solubility, including use of the “like dissolves like” rule. (Sections 13.1 and 13.3)

• Describe the role of equilibrium in the solution process and its relationship to the solubility of a solute. (Section 13.2)

• Describe the effect of temperature on the solubility of solids and gases in liquids. (Section 13.3)

• Describe the relationship between the partial pressure of a gas and its solubility. (Section 13.3)

• Calculate the concentration of a solution in terms of molarity, molality, mole fraction, percent composition, and parts per million and be able to interconvert between them. (Section 13.4)

• Describe what a colligative property is and explain the difference between the effects of nonelectrolytes and electrolytes on colliga-tive properties. (Section 13.5)

• Calculate the vapor pressure of a solvent over a solution. (Section 13.5)

• Calculate the boiling-point elevation and freezing-point depression of a solution. (Section 13.5)

• Calculate the osmotic pressure of a solution. (Section 13.5)

• Explain the difference between a solution and a colloid. (Section 13.6)

• Describe the similarities between the motions of gas molecules and the motions of colloids in a liquid. (Section 13.6)

Key equations • Sg = kPg [13.4] Henry’s law, which relates gas solubility to partial pressure

• Mass % of component =mass of component in soln

total mass of soln* 100 [13.5] Concentration in terms of mass percent

• ppm of component =mass of component in soln

total mass of soln* 106 [13.6] Concentration in terms of parts per million (ppm)

• Mole fraction of component =moles of component

total moles of all components [13.7] Concentration in terms of mole fraction

• Molarity =moles of solute

liters of soln [13.8] Concentration in terms of molarity

• Molality =moles of solute

kilograms of solvent [13.9] Concentration in terms of molality

• Psolution = XsolventP °solvent [13.10] Raoult’s law, calculating vapor pressure of solvent above a solution

• ∆Tb = iKbm [13.12] Calculating the boiling-point elevation of a solution

• ∆Tf = - iKfm [13.13] Calculating the freezing-point depression of a solution

• Π = ia nVbRT = iMRT [13.14] Calculating the osmotic pressure of a solution


13.6 The solubility of Xe in water at 1 atm pressure and 20 °C is approximately 5 * 10-3 M. (a) Compare this with the solu-bilities of Ar and Kr in water (Table 13.1). (b) What properties of the rare gas atoms account for the variation in solubility? [Section 13.3]

13.7 The structures of vitamins E and B6 are shown below. Predict which is more water soluble and which is more fat soluble. Explain. [Section 13.3]

Vitamin B6 Vitamin E

13.8 You take a sample of water that is at room temperature and in contact with air and put it under a vacuum. Right away, you see bubbles leave the water, but after a little while, the bubbles stop. As you keep applying the vacuum, more bubbles appear. A friend tells you that the first bubbles were water vapor, and the low pressure had reduced the boiling point of water, caus-ing the water to boil. Another friend tells you that the first bubbles were gas molecules from the air (oxygen, nitrogen, and so forth) that were dissolved in the water. Which friend is mostly likely to be correct? What, then, is responsible for the second batch of bubbles? [Section 13.4]

13.9 The figure shows two identical volumetric flasks containing the same solution at two temperatures.(a) Does the molarity of the solution change with the change

in temperature? Explain.(b) Does the molality of the solution change with the change

in temperature? Explain. [Section 13.4]

25 °C 55 °C

Visualizing Concepts

13.1 Rank the contents of the following containers in order of increasing entropy: [Section 13.1]

(a) (b) (c)

13.2 This figure shows the interaction of a cation with surround-ing water molecules.

+

(a) Which atom of water is associated with the cation? Explain.(b) Which of the following explanations accounts for the

fact that the ion-solvent interaction is greater for Li+ than for K+?

a. Li+ is of lower mass than K+. b. The ionization energy of Li is higher than that for K. c. Li+ has a smaller ionic radius than K+. d. Li has a lower density than K. e. Li reacts with water more slowly than K. [Section 13.1]

13.3 Consider two ionic solids, both composed of singly-charged ions, that have different lattice energies. (a) Will the solids have the same solubility in water? (b) If not, which solid will be more soluble in water, the one with the larger lattice energy or the one with the smaller lattice energy? Assume that solute-solvent interactions are the same for both solids. [Section 13.1]

13.4 Are gases always miscible with each other? Explain. [Section 13.1] 13.5 Which of the following is the best representation of a

saturated solution? Explain your reasoning. [Section 13.2]

(a) (c)(b)

exercises

exercises 567

(b) In making a solution, the enthalpy of mixing is always a positive number.

(c) An increase in entropy favors mixing. 13.14 Indicate whether each statement is true or false: (a) NaCl dis-

solves in water but not in benzene 1C6H62 because benzene is denser than water. (b) NaCl dissolves in water but not in ben-zene because water has a large dipole moment and benzene has zero dipole moment. (c) NaCl dissolves in water but not in benzene because the water–ion interactions are stronger than benzene–ion interactions.

13.15 Indicate the type of solute–solvent interaction (Section 11.2) that should be most important in each of the following solu-tions: (a) CCl4 in benzene 1C6H62, (b) methanol 1CH3OH2 in water, (c) KBr in water, (d) HCl in acetonitrile 1CH3CN2.

13.16 Indicate the principal type of solute–solvent interaction in each of the following solutions and rank the solutions from weakest to strongest solute–solvent interaction: (a) KCl in wa-ter, (b) CH2Cl2 in benzene 1C6H62, (c) methanol 1CH3OH2 in water.

13.17 An ionic compound has a very negative ∆Hsoln in water. (a) Would you expect it to be very soluble or nearly insoluble in water? (b) Which term would you expect to be the largest negative number: ∆Hsolvent, ∆Hsolute, or ∆Hmix?

13.18 When ammonium chloride dissolves in water, the solution becomes colder. (a) Is the solution process exothermic or en-dothermic? (b) Why does the solution form?

13.19 (a) In Equation 13.1, which of the enthalpy terms for dissolv-ing an ionic solid would correspond to the lattice energy? (b) Which energy term in this equation is always exothermic?

13.20 For the dissolution of LiCl in water, ∆Hsoln = -37 kJ>mol. Which term would you expect to be the largest negative num-ber: ∆Hsolvent, ∆Hsolute, or ∆Hmix?

13.21 Two nonpolar organic liquids, hexane 1C6H142 and heptane 1C7H162, are mixed. (a) Do you expect ∆Hsoln to be a large positive number, a large negative number, or close to zero? Explain. (b) Hexane and heptane are miscible with each other in all proportions. In making a solution of them, is the entropy of the system increased, decreased, or close to zero, compared to the separate pure liquids?

13.22 The enthalpy of solution of KBr in water is about +198 kJ>mol. Nevertheless, the solubility of KBr in water is relatively high. Why does the solution process occur even though it is endothermic?

Saturated Solutions; Factors affecting Solubility (Sections 13.2 and 13.3)

13.23 The solubility of Cr1NO323 # 9 H2O in water is 208 g per 100 g of water at 15 °C. A solution of Cr1NO323 # 9 H2O in water at 35 °C is formed by dissolving 324 g in 100 g of water. When this solution is slowly cooled to 15 °C, no precipitate forms. (a) What term describes this solution? (b) What action might you take to initiate crystallization? Use molecular-level pro-cesses to explain how your suggested procedure works.

13.24 The solubility of MnSO4 # H2O in water at 20 °C is 70 g per 100 mL of water. (a) Is a 1.22 M solution of MnSO4 # H2O in water at 20 °C saturated, supersaturated, or unsaturated? (b) Given a solution of MnSO4 # H2O of unknown concen-tration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

13.10 This portion of a phase diagram shows the vapor-pressure curves of a volatile solvent and of a solution of that solvent containing a nonvolatile solute. (a) Which line represents the solution? (b) What are the normal boiling points of the sol-vent and the solution? [Section 13.5]

40

1.0

50 60 70T (°C)

P (a

tm)

13.11 Suppose you had a balloon made of some highly flexible semi-permeable membrane. The balloon is filled completely with a 0.2 M solution of some solute and is submerged in a 0.1 M solution of the same solute:

0.1 M

0.2 M

Initially, the volume of solution in the balloon is 0.25 L. Assuming the volume outside the semipermeable membrane is large, as the illustration shows, what would you expect for the solution volume inside the balloon once the system has come to equilibrium through osmosis? [Section 13.5]

13.12 The molecule n-octylglucoside, shown here, is widely used in biochemical research as a nonionic detergent for “solubilizing” large hydrophobic protein molecules. What characteristics of this molecule are important for its use in this way? [Section 13.6]

the Solution Process (Section 13.1)

13.13 Indicate whether each statement is true or false:(a) A solute will dissolve in a solvent if solute–solute interac-

tions are stronger than solute-solvent interactions.


13.25 By referring to Figure 13.15, determine whether the addition of 40.0 g of each of the following ionic solids to 100 g of water at 40 °C will lead to a saturated solution: (a) NaNO3, (b) KCl, (c) K2Cr2O7, (d) Pb1NO322.

13.26 By referring to Figure 13.15, determine the mass of each of the following salts required to form a saturated solution in 250 g of water at 30 °C: (a) KClO3, (b) Pb1NO322, (c) Ce21SO423.

13.27 Consider water and glycerol, CH21OH2CH1OH2CH2OH. (a) Would you expect them to be miscible in all proportions? Explain. (b) List the intermolecular attractions that occur be-tween a water molecule and a glycerol molecule.

13.28 Oil and water are immiscible. Which is the most likely reason? (a) Oil molecules are denser than water. (b) Oil molecules are composed mostly of carbon and hydrogen. (c) Oil molecules have higher molar masses than water. (d) Oil molecules have higher vapor pressures than water. (e) Oil molecules have higher boiling points than water.

13.29 Common laboratory solvents include acetone 1CH3COCH32, methanol 1CH3OH2, toluene 1C6H5CH32, and water. Which of these is the best solvent for nonpolar solutes?

13.30 Would you expect alanine (an amino acid) to be more soluble in water or in hexane? Explain.

Alanine

13.31 (a) Would you expect stearic acid, CH31CH2216COOH, to be more soluble in water or in carbon tetrachloride? Explain.

(b) Which would you expect to be more soluble in water, cyclohexane or dioxane? Explain.

H2C

H2C

CH2

CH2

O

ODioxane

H2C

H2C

CH2

CH2

CH2

CH2

Cyclohexane

13.32 Ibuprofen, widely used as a pain reliever, has a limited solubil-ity in water, less than 1 mg>mL. Which part of the molecule’s structure (gray, white, red) contributes to its water solubility? Which part of the molecule (gray, white, red) contributes to its water insolubility?

Ibuprofen

13.33 Which of the following in each pair is likely to be more soluble in hexane, C6H14: (a) CCl4 or CaCl2, (b) benzene

1C6H62 or glycerol, CH21OH2CH1OH2CH2OH, (c) octanoic acid, CH3CH2CH2CH2CH2CH2CH2COOH, or acetic acid, CH3COOH? Explain your answer in each case.

13.34 Which of the following in each pair is likely to be more solu-ble in water: (a) cyclohexane 1C6H122 or glucose 1C6H12O62, (b) propionic acid 1CH3CH2COOH2 or sodium propionate 1CH3CH2COONa2, (c) HCl or ethyl chloride 1CH3CH2Cl2? Explain in each case.

13.35 (a) Explain why carbonated beverages must be stored in sealed containers. (b) Once the beverage has been opened, why does it maintain more carbonation when refrigerated than at room temperature?

13.36 Explain why pressure substantially affects the solubility of O2 in water but has little effect on the solubility of NaCl in water.

13.37 The Henry’s law constant for helium gas in water at 30 °C is 3.7 * 10-4 M>atm and the constant for N2 at 30 °C is 6.0 * 10-4 M>atm. If the two gases are each present at 1.5 atm pressure, calculate the solubility of each gas.

13.38 The partial pressure of O2 in air at sea level is 0.21 atm. Using the data in Table 13.1, together with Henry’s law, calculate the molar concentration of O2 in the surface water of a mountain lake saturated with air at 20 °C and an atmospheric pressure of 650 torr.

Concentrations of Solutions (Section 13.4)

13.39 (a) Calculate the mass percentage of Na2SO4 in a solution containing 10.6 g of Na2SO4 in 483 g of water. (b) An ore con-tains 2.86 g of silver per ton of ore. What is the concentration of silver in ppm?

13.40 (a) What is the mass percentage of iodine in a solution con-taining 0.035 mol I2 in 125 g of CCl4? (b) Seawater contains 0.0079 g of Sr2+ per kilogram of water. What is the concentra-tion of Sr2+ in ppm?

13.41 A solution is made containing 14.6 g of CH3OH in 184 g of H2O. Calculate (a) the mole fraction of CH3OH, (b) the mass percent of CH3OH, (c) the molality of CH3OH.

13.42 A solution is made containing 20.8 g of phenol 1C6H5OH2 in 425 g of ethanol 1CH3CH2OH2. Calculate (a) the mole frac-tion of phenol, (b) the mass percent of phenol, (c) the molal-ity of phenol.

13.43 Calculate the molarity of the following aqueous solutions: (a) 0.540 g of Mg1NO322 in 250.0 mL of solution, (b) 22.4 g of LiClO4 # 3 H2O in 125 mL of solution, (c) 25.0 mL of 3.50 M HNO3 diluted to 0.250 L.

13.44 What is the molarity of each of the following solutions: (a) 15.0 g of Al21SO423 in 0.250 mL solution, (b) 5.25 g of Mn1NO322 # 2 H2O in 175 mL of solution, (c) 35.0 mL of 9.00 M H2SO4 diluted to 0.500 L?

13.45 Calculate the molality of each of the following solutions: (a) 8.66 g of benzene 1C6H62 dissolved in 23.6 g of carbon tet-rachloride 1CCl42, (b) 4.80 g of NaCl dissolved in 0.350 L of water.

13.46 (a) What is the molality of a solution formed by dissolving 1.12 mol of KCl in 16.0 mol of water? (b) How many grams of sulfur 1S82 must be dissolved in 100.0 g of naphthalene 1C10H82 to make a 0.12 m solution?

13.47 A sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution has a density of 1.329 g>cm3. Calculate

exercises 569

Caffeine

13.59 During a person’s typical breathing cycle, the CO2 concen-tration in the expired air rises to a peak of 4.6% by volume. (a) Calculate the partial pressure of the CO2 in the expired air at its peak, assuming 1 atm pressure and a body tem-perature of 37 °C. (b) What is the molarity of the CO2 in the expired air at its peak, assuming a body temperature of 37 °C?

13.60 Breathing air that contains 4.0% by volume CO2 over time causes rapid breathing, throbbing headache, and nausea, among other symptoms. What is the concentration of CO2 in such air in terms of (a) mol percentage, (b) molarity, assum-ing 1 atm pressure and a body temperature of 37 °C?

Colligative Properties (Section 13.5)

13.61 You make a solution of a nonvolatile solute with a liquid sol-vent. Indicate whether each of the following statements is true or false. (a) The freezing point of the solution is higher than that of the pure solvent. (b) The freezing point of the solution is lower than that of the pure solvent. (c) The boiling point of the solution is higher than that of the pure solvent. (d) The boiling point of the solution is lower than that of the pure solvent.

13.62 You make a solution of a nonvolatile solute with a liquid sol-vent. Indicate if each of the following statements is true or false. (a) The freezing point of the solution is unchanged by addition of the solvent. (b) The solid that forms as the solu-tion freezes is nearly pure solute. (c) The freezing point of the solution is independent of the concentration of the solute. (d) The boiling point of the solution increases in proportion to the concentration of the solute. (e) At any temperature, the vapor pressure of the solvent over the solution is lower than what it would be for the pure solvent.

13.63 Consider two solutions, one formed by adding 10 g of glucose 1C6H12O62 to 1 L of water and the other formed by adding 10 g of sucrose 1C12H22O112 to 1 L of water. Calculate the vapor pressure for each solution at 20 °C; the vapor pressure of pure water at this temperature is 17.5 torr.

13.64 (a) What is an ideal solution? (b) The vapor pressure of pure water at 60 °C is 149 torr. The vapor pressure of water over a solution at 60 °C containing equal numbers of moles of water and ethylene glycol (a nonvolatile solute) is 67 torr. Is the so-lution ideal according to Raoult’s law? Explain.

13.65 (a) Calculate the vapor pressure of water above a solution prepared by adding 22.5 g of lactose 1C12H22O112 to 200.0 g of water at 338 K. (Vapor-pressure data for water are given in Appendix B.) (b) Calculate the mass of propylene glycol 1C3H8O22 that must be added to 0.340 kg of water to reduce the vapor pressure by 2.88 torr at 40 °C.

(a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of H2SO4 in this solution.

13.48 Ascorbic acid 1vitamin C, C6H8O62 is a water-soluble vita-min. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g>mL at 55 °C. Calculate (a) the mass percentage, (b) the mole fraction, (c) the molal-ity, (d) the molarity of ascorbic acid in this solution.

13.49 The density of acetonitrile 1CH3CN2 is 0.786 g>mL and the density of methanol 1CH3OH2 is 0.791 g>mL. A solution is made by dissolving 22.5 mL of CH3OH in 98.7 mL of CH3CN. (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of CH3OH in the solution?

13.50 The density of toluene 1C7H82 is 0.867 g>mL, and the den-sity of thiophene 1C4H4S2 is 1.065 g>mL. A solution is made by dissolving 8.10 g of thiophene in 250.0 mL of toluene. (a) Calculate the mole fraction of thiophene in the solu-tion. (b) Calculate the molality of thiophene in the solution. (c) Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?

13.51 Calculate the number of moles of solute present in each of the following aqueous solutions: (a) 600 mL of 0.250 M SrBr2, (b) 86.4 g of 0.180 m KCl, (c) 124.0 g of a solution that is 6.45% glucose 1C6H12O62 by mass.

13.52 Calculate the number of moles of solute present in each of the fol-lowing solutions: (a) 255 mL of 1.50 M HNO31aq2, (b) 50.0 mg of an aqueous solution that is 1.50 m NaCl, (c) 75.0 g of an aqueous solution that is 1.50% sucrose 1C12H22O112 by mass.

13.53 Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of 1.5 * 10-2 M KBr, (b) 125 g of 0.180 m KBr, (c) 1.85 L of a solution that is 12.0% KBr by mass (the density of the solu-tion is 1.10 g>mL), (d) a 0.150 M solution of KBr that con-tains just enough KBr to precipitate 16.0 g of AgBr from a solution containing 0.480 mol of AgNO3.

13.54 Describe how you would prepare each of the following aque-ous solutions: (a) 1.50 L of 0.110 M 1NH422SO4 solution, starting with solid 1NH422SO4; (b) 225 g of a solution that is 0.65 m in Na2CO3, starting with the solid solute; (c) 1.20 L of a solution that is 15.0% Pb1NO322 by mass (the density of the solution is 1.16 g>mL), starting with solid solute; (d) a 0.50 M solution of HCl that would just neutralize 5.5 g of Ba1OH22 starting with 6.0 M HCl.

13.55 Commercial aqueous nitric acid has a density of 1.42 g>mL and is 16 M. Calculate the percent HNO3 by mass in the solution.

13.56 Commercial concentrated aqueous ammonia is 28% NH3 by mass and has a density of 0.90 g>mL. What is the molarity of this solution?

13.57 Brass is a substitutional alloy consisting of a solution of cop-per and zinc. A particular sample of red brass consisting of 80.0% Cu and 20.0% Zn by mass has a density of 8750 kg>m3. (a) What is the molality of Zn in the solid solution? (b) What is the molarity of Zn in the solution?

13.58 Caffeine 1C8H10N4O22 is a stimulant found in coffee and tea. If a solution of caffeine in the solvent chloroform 1CHCl32 has a concentration of 0.0500 m, calculate (a) the percentage of caffeine by mass, (b) the mole fraction of caffeine in the solution.


point by 0.49 °C. Calculate the approximate molar mass of adrenaline from this data.

Adrenaline

13.80 Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1 °C. What is the molar mass of lauryl alcohol from this data?

13.81 Lysozyme is an enzyme that breaks bacterial cell walls. A so-lution containing 0.150 g of this enzyme in 210 mL of solu-tion has an osmotic pressure of 0.953 torr at 25 °C. What is the molar mass of lysozyme?

13.82 A dilute aqueous solution of an organic compound soluble in water is formed by dissolving 2.35 g of the compound in water to form 0.250 L of solution. The resulting solution has an osmotic pressure of 0.605 atm at 25 °C. Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

[13.83] The osmotic pressure of a 0.010 M aqueous solution of CaCl2 is found to be 0.674 atm at 25 °C. (a) Calculate the van’t Hoff factor, i, for the solution. (b) How would you expect the value of i to change as the solution becomes more concentrated? Explain.

[13.84] Based on the data given in Table 13.4, which solution would give the larger freezing-point lowering, a 0.030 m solution of NaCl or a 0.020 m solution of K2SO4? How do you explain the departure from ideal behavior and the differences observed between the two salts?

Colloids (Section 13.6)

13.85 (a) Do colloids made only of gases exist? Why or why not? (b) In the 1850’s, Michael Faraday prepared ruby-red col-loids of gold nanoparticles in water that are still stable today. These brightly colored colloids look like solutions. What experiment(s) could you do to determine whether a given col-ored preparation is a solution or colloid?

13.86 Choose the best answer: A colloidal dispersion of one liq-uid in another is called (a) a gel, (b) an emulsion, (c) a foam, (d) an aerosol.

13.87 An “emulsifying agent” is a compound that helps stabilize a hydro-phobic colloid in a hydrophilic solvent (or a hydrophilic colloid in a hydrophobic solvent). Which of the following choices is the best emulsifying agent? (a) CH3COOH, (b) CH3CH2CH2COOH, (c) CH31CH2211COOH, (d) CH31CH2211COONa.

13.88 Aerosols are important components of the atmosphere. Does the presence of aerosols in the atmosphere increase or decrease the amount of sunlight that arrives at the Earth’s surface, compared to an “aerosol-free” atmosphere? Explain your reasoning.

[13.89] Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called “salting out”

13.66 (a) Calculate the vapor pressure of water above a solution prepared by dissolving 28.5 g of glycerin 1C3H8O32 in 125 g of water at 343 K. (The vapor pressure of water is given in Appendix B.) (b) Calculate the mass of ethylene glycol 1C2H6O22 that must be added to 1.00 kg of ethanol 1C2H5OH2 to reduce its vapor pressure by 10.0 torr at 35 °C. The vapor pressure of pure ethanol at 35 °C is 1.00 * 102 torr.

[13.67] At 63.5 °C, the vapor pressure of H2O is 175 torr, and that of ethanol 1C2H5OH2 is 400 torr. A solution is made by mixing equal masses of H2O and C2H5OH. (a) What is the mole frac-tion of ethanol in the solution? (b) Assuming ideal-solution behavior, what is the vapor pressure of the solution at 63.5 °C? (c) What is the mole fraction of ethanol in the vapor above the solution?

[13.68] At 20 °C, the vapor pressure of benzene 1C6H62 is 75 torr, and that of toluene 1C7H82 is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composi-tion in mole fraction of a solution that has a vapor pressure of 35 torr at 20 °C? (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

13.69 (a) Does a 0.10 m aqueous solution of NaCl have a higher boiling point, a lower boiling point, or the same boiling point as a 0.10 m aqueous solution of C6H12O6? (b) The experi-mental boiling point of the NaCl solution is lower than that calculated assuming that NaCl is completely dissociated in solution. Why is this the case?

13.70 Arrange the following aqueous solutions, each 10% by mass in solute, in order of increasing boiling point: glucose 1C6H12O62, sucrose 1C12H22O112, sodium nitrate 1NaNO32.

13.71 List the following aqueous solutions in order of increas-ing boiling point: 0.120 m glucose, 0.050 m LiBr, 0.050 m Zn1NO322.

13.72 List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin 1C3H8O32, 0.020 m KBr, 0.030 m phenol 1C6H5OH2.

13.73 Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.22 m glycerol 1C3H8O32 in ethanol, (b) 0.240 mol of naphthalene 1C10H82 in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water, (d) 2.04 g KBr and 4.82 g glucose 1C6H12O62 in 188 g of water.

13.74 Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.25 m glucose in ethanol; (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3; (c) 3.50 g NaOH in 175 g of water, (d) 0.45 mol ethylene glycol and 0.15 mol KBr in 150 g H2O.

13.75 How many grams of ethylene glycol 1C2H6O22 must be added to 1.00 kg of water to produce a solution that freezes at -5.00 °C?

13.76 What is the freezing point of an aqueous solution that boils at 105.0 °C?

13.77 What is the osmotic pressure formed by dissolving 44.2 mg of aspirin 1C9H8O42 in 0.358 L of water at 25 °C?

13.78 Seawater contains 3.4 g of salts for every liter of solution. As-suming that the solute consists entirely of NaCl (in fact, over 90% of the salt is indeed NaCl), calculate the osmotic pressure of seawater at 20 °C.

13.79 Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of 0.64 g of adrenaline in 36.0 g of CCl4 elevates the boiling

additional exercises 571

same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses?

13.90 Explain how (a) a soap such as sodium stearate stabilizes a colloidal dispersion of oil droplets in water; (b) milk curdles upon addition of an acid.

the protein. (a) Do you think that all proteins would be pre-cipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein–protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein’s hydration shell is stripped away, lead-ing to protein precipitation. Another friend of yours in the

that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in 1.0 * 103 gal of seawater.

13.97 The maximum allowable concentration of lead in drinking water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0-ppb solution. (b) How many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 m3 of water?

13.98 Acetonitrile 1CH3CN2 is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 M LiBr solution in acetonitrile is 0.826 g>cm3. Calcu-late the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of CH3CN.

13.99 A “canned heat” product used to warm buffet dishes consists of a homogeneous mixture of ethanol 1C2H5OH2 and paraf-fin, which has an average formula of C24H50. What mass of C2H5OH should be added to 620 kg of the paraffin to produce 8 torr of ethanol vapor pressure at 35 °C? The vapor pressure of pure ethanol at 35 °C is 100 torr.

13.100 A solution contains 0.115 mol H2O and an unknown number of moles of sodium chloride. The vapor pressure of the solu-tion at 30 °C is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

[13.101] Two beakers are placed in a sealed box at 25 °C. One beaker contains 30.0 mL of a 0.050 M aqueous solution of a non-volatile nonelectrolyte. The other beaker contains 30.0 mL of a 0.035 M aqueous solution of NaCl. The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall? (b) What are the volumes in the two beakers when equilib-rium is attained, assuming ideal behavior?

13.102 A car owner who knows no chemistry has to put antifreeze in his car’s radiator. The instructions recommend a mixture of 30% ethylene glycol and 70% water. Thinking he will improve his protection he uses pure ethylene glycol, which is a liquid at room temperature. He is saddened to find that the solu-tion does not provide as much protection as he hoped. The pure ethylene glycol freezes solid in his radiator on a very cold day, while his neighbor, who did use the 30/70 mixture, has no problem. Suggest an explanation.

13.103 Calculate the freezing point of a 0.100 m aqueous solution of K2SO4, (a) ignoring interionic attractions, and (b) taking in-terionic attractions into consideration by using the van’t Hoff factor (Table 13.4).

13.104 Carbon disulfide 1CS22 boils at 46.30 °C and has a density of 1.261 g>mL. (a) When 0.250 mol of a nondissociating solute

13.91 Butylated hydroxytoluene (BHT) has the following molecular structure:

C CH3H3C

CH3

CH3OH

BHT

CH3

C

CH3

CH3

It is widely used as a preservative in a variety of foods, in-cluding dried cereals. Based on its structure, would you ex-pect BHT to be more soluble in water or in hexane 1C6H142? Explain.

13.92 A saturated solution of sucrose 1C12H22O112 is made by dis-solving excess table sugar in a flask of water. There are 50 g of undissolved sucrose crystals at the bottom of the flask in contact with the saturated solution. The flask is stoppered and set aside. A year later a single large crystal of mass 50 g is at the bottom of the flask. Explain how this experiment provides evidence for a dynamic equilibrium between the saturated so-lution and the undissolved solute.

13.93 Most fish need at least 4 ppm dissolved O2 in water for sur-vival. (a) What is this concentration in mol>L? (b) What par-tial pressure of O2 above water is needed to obtain 4 ppm O2 in water at 10 °C? (The Henry’s law constant for O2 at this temperature is 1.71 * 10-3 mol>L@atm.)

13.94 The presence of the radioactive gas radon (Rn) in well wa-ter presents a possible health hazard in parts of the United States. (a) Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 °C is 7.27 * 10-3 M, what is the Henry’s law constant for radon in water at this temperature? (b) A sample consisting of various gases contains 3.5 * 10-6 mole fraction of radon. This gas at a total pressure of 32 atm is shaken with water at 30 °C. Calculate the molar concentration of radon in the water.

13.95 Glucose makes up about 0.10% by mass of human blood. Calculate this concentration in (a) ppm, (b) molality. (c) What further information would you need to determine the molarity of the solution?

13.96 The concentration of gold in seawater has been reported to be between 5 ppt (parts per trillion) and 50 ppt. Assuming

additional exercises


0.036 g per 100 g of water at 25 °C. The osmotic pressure of this solution is found to be 57.1 torr. Assuming that molal-ity and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solu-tion, determine an appropriate value of n in the formula for the salt.

is dissolved in 400.0 mL of CS2, the solution boils at 47.46 °C. What is the molal boiling-point-elevation constant for CS2? (b) When 5.39 g of a nondissociating unknown is dissolved in 50.0 mL of CS2, the solution boils at 47.08 °C. What is the molecular weight of the unknown?

[13.105] A lithium salt used in lubricating grease has the formula LiCnH2n + 1O2. The salt is soluble in water to the extent of

(a) Account for the variations in heats of vaporization for these substances, considering their relative intermolecular forces. (b) How would you expect the solubilities of these sub-stances to vary in hexane as solvent? In ethanol? Use inter-molecular forces, including hydrogen-bonding interactions where applicable, to explain your responses.

[13.109] A textbook on chemical thermodynamics states, “The heat of solution represents the difference between the lattice en-ergy of the crystalline solid and the solvation energy of the gaseous ions.” (a) Draw a simple energy diagram to illustrate this statement. (b) A salt such as NaBr is insoluble in most polar nonaqueous solvents such as acetonitrile 1CH3CN2 or nitromethane 1CH3NO22, but salts of large cations, such as tetramethylammonium bromide 31CH324NBr4, are gener-ally more soluble. Use the thermochemical cycle you drew in part (a) and the factors that determine the lattice energy (Section 8.2) to explain this fact.

13.110 (a) A sample of hydrogen gas is generated in a closed con-tainer by reacting 2.050 g of zinc metal with 15.0 mL of 1.00 M sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is 122 mL. Calcu-late the partial pressure of the hydrogen gas in this volume at 25 °C, ignoring any solubility of the gas in the solution. (c) The Henry’s law constant for hydrogen in water at 25 °C is 7.8 * 10-4 mol>L@atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hy-drogen in part (b)?

[13.111] The following table presents the solubilities of several gases in water at 25 °C under a total pressure of gas and water vapor of 1 atm. (a) What volume of CH41g2 under standard conditions of temperature and pressure is con-tained in 4.0 L of a saturated solution at 25 °C? (b) Ex-plain the variation in solubility among the hydrocarbons listed (the first three compounds), based on their molecu-lar structures and intermolecular forces. (c) Compare the solubilities of O2, N2, and NO, and account for the varia-tions based on molecular structures and intermolecular forces. (d) Account for the much larger values observed for H2S and SO2 as compared with the other gases listed. (e) Find several pairs of substances with the same or nearly the same molecular masses (for example, C2H4 and N2), and use intermolecular interactions to explain the differences in their solubilities.

13.106 Fluorocarbons (compounds that contain both carbon and fluo-rine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at 25 °C, and their sol-ubilities in water at 25 °C and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Explain why the molarity of each of the solutions should be very close numerically to the molality. (c) Based on their molecular structures, account for the differences in solubility of the four fluorocarbons. (d) Calculate the Henry’s law constant at 25 °C for CHClF2, and compare its magnitude to that for N2 16.8 * 10-4 mol>L@atm2. Suggest a reason for the difference in magnitude.

Fluorocarbon Solubility (mass %)

CF4 0.0015CClF3 0.009CCl2F2 0.028CHClF2 0.30

[13.107] At ordinary body temperature 137 °C2, the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g>L. Air is approximately 78 mol % N2. (a) Calculate the number of moles of N2 dissolved per liter of blood, assuming blood is a sim-ple aqueous solution. (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N2 from air in blood at this pressure? (c) If a scuba diver suddenly surfaces from this depth, how many milliliters of N2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

[13.108] Consider the following values for enthalpy of vaporization 1kJ>mol2 of several organic substances:

CH3C H 30.4

O

Acetaldehyde

H2C CH2 28.5

O

Ethylene oxide

CH3CCH3 32.0

O

Acetone

H2C CH2

CH2

24.7

Cyclopropane

Integrative exercises

Design an experiment 573

bonds between acetone and chloroform molecules to explain the deviation from ideal behavior. (c) Based on the behavior of the solution, predict whether the mixing of acetone and chloroform is an exothermic 1∆Hsoln 6 02 or endothermic 1∆Hsoln 7 02 process. (d) Would you expect the same vapor-pressure behavior for acetone and chloromethane (CH3Cl)? Explain.

13.114 Compounds like sodium stearate, called “surfactants” in gen-eral, can form structures known as micelles in water, once the solution concentration reaches the value known as the critical micelle concentration (cmc). Micelles contain dozens to hun-dreds of molecules. The cmc depends on the substance, the solvent, and the temperature.

Surfactanttail

Surfactanthead

Surfactant monomers Micelle

cmc

At and above the cmc, the properties of the solution vary drastically.(a) The turbidity (the amount of light scattering) of solu-tions increases dramatically at the cmc. Suggest an explana-tion. (b) The ionic conductivity of the solution dramatically changes at the cmc. Suggest an explanation. (c) Chemists have developed fluorescent dyes that glow brightly only when the dye molecules are in a hydrophobic environ-ment. Predict how the intensity of such fluorescence would relate to the concentration of sodium stearate as the sodium stearate concentration approaches and then increases past the cmc.

Gas Solubility (mM)

CH4 1methane2 1.3

C2H6 1ethane2 1.8

C2H4 1ethylene2 4.7

N2 0.6O2 1.2NO 1.9H2S 99SO2 1476

13.112 A small cube of lithium 1density = 0.535 g/cm32 measuring 1.0 mm on each edge is added to 0.500 L of water. The follow-ing reaction occurs:

2 Li1s2 + 2 H2O1l2 ¡ 2 LiOH1aq2 + H21g2What is the freezing point of the resultant solution, assuming that the reaction goes to completion?

[13.113] At 35 °C the vapor pressure of acetone, 1CH322CO, is 360 torr, and that of chloroform, CHCl3, is 300 torr. Acetone and chloroform can form very weak hydrogen bonds between one another; the chlorines on the carbon give the carbon a suf-ficient partial positive charge to enable this behavior:

CH3

CH3OH CC

Cl

Cl

Cl

A solution composed of an equal number of moles of acetone and chloroform has a vapor pressure of 250 torr at 35 °C. (a) What would be the vapor pressure of the solution if it exhibited ideal behavior? (b) Use the existence of hydrogen

design an experimentBased on Figure 13.18, you might think that the reason volatile solvent molecules in a solution are less likely to escape to the gas phase, compared to the pure solvent, is because the solute molecules are physically blocking the solvent molecules from leaving at the surface. This is a com-mon misconception. Design an experiment to test the hypothesis that solute blocking of solvent vaporization is not the reason that solutions have lower vapor pressures than pure solvents.

14 Chemical KineticsChemical reactions take time to occur. Some reactions, such as the rusting of iron or the changing of color in leaves, occur relatively slowly, requiring days, months, or years to complete. Others, such as the combustion reaction that generates the thrust for a rocket, as in the chapter-opening photograph, happen much more rapidly. As chemists, we need to be concerned about the speed with which chemical reactions occur as well as the products of the reactions. For example,

14.3 ConCentration and rate Laws We show that the effect of concentration on rate is expressed quantitatively by rate laws and show how rate laws and rate constants are determined experimentally.

14.4 the Change of ConCentration with time We learn that rate equations can be written to express how concentrations change with time and look at several classifications of rate equations: zero-order, first-order, and second-order reactions.

14.1 faCtors that affeCt reaCtion rates We see that four variables affect reaction rates: concentration, physical states of reactants, temperature, and presence of catalysts. These factors can be understood in terms of the collisions among reactant molecules that lead to reaction.

14.2 reaCtion rates We examine how to express reaction rates and how reactant disappearance rates and product appearance rates are related to the reaction stoichiometry.

What’s ahead

▶ Launch of the Juno Spacecraft at Cape Canaveral, Florida, in August 2011. The spacecraft is mounted to an Atlas V rocket, which at launch uses the very rapid combustion of kerosene and liquid oxygen to generate its thrust.

the chemical reactions that govern the metabolism of food, the transport of essential nutrients, and your body’s ability to adjust to temperature changes (see the Chemistry and Life box on the regulation of body temperature in Section 5.5) all require that reactions occur with the appropriate speed. Indeed, considerations of the speeds of reactions are among the most important aspects of designing new chemistry and chemical processes. The area of chemistry concerned with the speeds, or rates, of reactions is chemical kinetics.

So far, we have focused on the beginning and end of chemical reactions: We start with certain reactants and see what products they yield. This view is useful but does not tell us what happens in the middle—that is, which chemical bonds are broken, which are formed, and in what order these events occur. The speed at which a chemical reac-tion occurs is called the reaction rate. Reaction rates can occur over very different time

14.7 CataLysis We end the chapter with a discussion of how catalysts increase reaction rates, including a discussion of biological catalysts called enzymes.

14.5 temperature and rate We explore the effect of temperature on rate. In order to occur, most reactions require a minimum input of energy called the activation energy.

14.6 reaCtion meChanisms We look more closely at reaction mechanisms, the step-by-step molecular pathways leading from reactants to products.

576 chapTer 14 chemical Kinetics

scales (▲ Figure 14.1). To investigate how reactions happen, we must examine the reac-tion rates and the factors that influence them. Experimental information on the rate of a given reaction provides important evidence that helps us formulate a reaction mecha-nism, which is a step-by-step, molecular-level view of the pathway from reactants to products.

Our goal in this chapter is to understand how to determine reaction rates and to consider the factors that control these rates. What factors determine how rapidly food spoils, for instance? How does one design an automotive airbag that fills extremely rap-idly following a car crash? What determines the rate at which steel rusts? How can we remove hazardous pollutants in automobile exhaust before the exhaust leaves the tail-pipe? Although we will not address these specific questions, we will see that the rates of all chemical reactions are subject to the same principles.

14.1 | Factors that affect Reaction Rates

Four factors affect the rate at which any particular reaction occurs:

1. Physical state of the reactants. Reactants must come together to react. The more readily reactant molecules collide with one another, the more rapidly they react. Reactions may broadly classified as being either homogeneous, involving either all gases or all liquids, or as heterogeneous, in which reactants are in different phases. Under heterogeneous conditions, a reaction is limited by the area of contact of the reactants. Thus, heterogeneous reactions that involve solids tend to proceed more rapidly if the surface area of the solid is increased. For example, a medicine in the form of a fine powder dissolves in the stomach and enters the blood more quickly than the same medicine in the form of a tablet.

2. Reactant concentrations. Most chemical reactions proceed more quickly if the con-centration of one or more reactants is increased. For example, steel wool burns only slowly in air, which contains 20% O2, but bursts into flame in pure oxygen (◀ Figure 14.2). As reactant concentration increases, the frequency with which the reactant molecules collide increases, leading to increased rates.

3. Reaction temperature. Reaction rates generally increase as temperature is in-creased. The bacterial reactions that spoil milk, for instance, proceed more rapidly at room temperature than at the lower temperature of a refrigerator. Increasing temperature increases the kinetic energies of molecules. (Section 10.7) As mol-ecules move more rapidly, they collide more frequently and with higher energy, leading to increased reaction rates.

Ene

rgy

10−15 s 1 s 109 s(30 years)

1015 s(30 million years)

Time scale

▲ Figure 14.1 Reaction rates span an enormous range of time scales. The absorption of light by an atom or a molecule is complete within one femtosecond; explosions occur within seconds; corrosion can occur over years; and the weathering of rocks can occur over millions of years.

Steel wool heated in air (about 20% O2) glows red-hot but oxidizes to Fe2O3 slowly

Red-hot steel wool in 100% O2 burns vigorously, forming Fe2O3 quickly

▲ Figure 14.2 Effect of concentration on reaction rate. The difference in behavior is due to the different concentrations of O2 in the two environments.

Go FiGuReIf a heated steel nail were placed in pure O2, would you expect it to burn as readily as the steel wool does?

secTIon 14.2 reaction rates 577

4. The presence of a catalyst. Catalysts are agents that increase reaction rates without themselves being used up. They affect the kinds of collisions (and therefore alter the mechanism) that lead to reaction. Catalysts play many crucial roles in living organisms, including ourselves.

On a molecular level, reaction rates depend on the frequency of collisions between molecules. The greater the frequency of collisions, the higher the reaction rate. For a col-lision to lead to a reaction, however, it must occur with sufficient energy to break bonds and with suitable orientation for new bonds to form in the proper locations. We will consider these factors as we proceed through this chapter.

Give It Some ThoughtIn a reaction involving reactants in the gas state, how does increasing the partial pressures of the gases affect the reaction rate?

14.2 | Reaction RatesThe speed of an event is defined as the change that occurs in a given time interval, which means that whenever we talk about speed, we necessarily bring in the notion of time. For example, the speed of a car is expressed as the change in the car’s position over a certain time interval. In the United States, the speed of cars is usually measured in units of miles per hour—that is, the quantity that is changing (position measured in miles) divided by a time interval (measured in hours).

Similarly, the speed of a chemical reaction—its reaction rate—is the change in the concentration of reactants or products per unit of time. The units for reaction rate are usually molarity per second 1M>s2—that is, the change in concentration measured in molarity divided by a time interval measured in seconds.

Let’s consider the hypothetical reaction A ¡ B, depicted in ▼ Figure 14.3. Each red sphere represents 0.01 mol of A, each blue sphere represents 0.01 mol of B, and the container has a volume of 1.00 L. At the beginning of the reaction, there is 1.00 mol A, so the concentration is 1.00 mol >L = 1.00 M. After 20 s, the concentration of A has fallen to 0.54 M and the concentration of B has risen to 0.46 M. The sum of the con-centrations is still 1.00 M because 1 mol of B is produced for each mole of A that reacts. After 40 s, the concentration of A is 0.30 M and that of B is 0.70 M.

▲ Figure 14.3 Progress of a hypothetical reaction a ¡ B. The volume of the flask is 1.0 L.

Go FiGuReEstimate the number of moles of A in the mixture after 30 s.

40 s20 s0 s

0.54 mol A0.46 mol B

1.00 mol A0 mol B

0.30 mol A0.70 mol B


The rate of this reaction can be expressed either as the rate of disappearance of reactant A or as the rate of appearance of product B. The average rate of appearance of B over a particular time interval is given by the change in concentration of B divided by the change in time:

Average rate of appearance of B =change in concentration of B

change in time

=3B4 at t2 - 3B4 at t1

t2 - t1=

∆3B4∆t

[14.1]

We use brackets around a chemical formula, as in [B], to indicate molarity. The Greek letter delta, ∆, is read “change in” and is always equal to a final value minus an initial value. (Equation 5.4, Section 5.2) The average rate of appearance of B over the 20-s interval from the beginning of the reaction 1t1 = 0 s to t2 = 20 s2 is

Average rate =0.46 M - 0.00 M

20 s - 0 s= 2.3 * 10-2 M>s

We could equally well express the reaction rate in term of the reactant, A. In this case, we would be describing the rate of disappearance of A, which we express as

Average rate of disappearance of A = - change in concentration of A

change in time

= - ∆3A4

∆t [14.2]

Notice the minus sign in this equation, which we use to indicate that the concen-tration of A decreases. By convention, rates are always expressed as positive quantities. Because [A] decreases, ∆3A4 is a negative number. The minus sign we put in the equa-tion converts the negative ∆3A4 to a positive rate of disappearance.

Because one molecule of A is consumed for every molecule of B that forms, the average rate of disappearance of A equals the average rate of appearance of B:

Average rate = - ∆3A4

∆t= -

0.54 M - 1.00 M20 s - 0 s

= 2.3 * 10-2 M>s

From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from 20 s to 40 s.

solutionanalyze We are given the concentration of A at 20 s 10.54 M2 and at 40 s 10.30 M2 and asked to calculate the average rate of reaction over this time interval.plan The average rate is given by the change in concentration, ∆3A4, divided by the change in time, ∆t. Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity.Solve

Average rate = - ∆3A4

∆t= -

0.30 M - 0.54 M40 s - 20 s

= 1.2 * 10-2 M>s

Practice exercise 1If the experiment in Figure 14.3 is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? (i) After 60 s there are 0.84 mol B in the flask. (ii) The decrease in the number of moles of A from t1 = 0 s to t2 = 20 s is greater than that

from t1 = 40 to t2 = 60 s. (iii) The average rate for the reaction from t1 = 40 s to t2 = 60 s is 7.0 * 10-3 M>s.

(a) Only one of the statements is true. (c) Statements (i) and (iii) are true. (e) All three statements are true.

(b) Statements (i) and (ii) are true. (d) Statements (ii) and (iii) are true.

Practice exercise 2Use the data in Figure 14.3 to calculate the average rate of appearance of B over the time interval from 0 s to 40 s.

samPle exeRCise 14.1 Calculating an average Rate of Reaction

secTIon 14.2 reaction rates 579

Change of Rate with TimeNow let’s consider the reaction between butyl chloride 1C4H9Cl2 and water to form butyl alcohol 1C4H9OH2 and hydrochloric acid:

C4H9Cl1aq2 + H2O1l2 ¡ C4H9OH1aq2 + HCl1aq2 [14.3]

Suppose we prepare a 0.1000@M aqueous solution of C4H9Cl and then measure the concentration of C4H9Cl at various times after time zero (which is the instant at which the reactants are mixed, thereby initiating the reaction). We can use the result-ing data, shown in the first two columns of ▲ table 14.1, to calculate the average rate of disappearance of C4H9Cl over various time intervals; these rates are given in the third column. Notice that the average rate decreases over each 50-s interval for the first several measurements and continues to decrease over even larger intervals through the remaining measurements. It is typical for rates to decrease as a reaction proceeds because the concen-tration of reactants decreases. The change in rate as the reaction proceeds is also seen in a graph of 3C4H9Cl4 versus time (▶ Figure 14.4). Notice how the steepness of the curve decreases with time, indicating a decreas-ing reaction rate.

Instantaneous RateGraphs such as Figure 14.4 that show how the concentration of a re-actant or product changes with time allow us to evaluate the instanta-neous rate of a reaction, which is the rate at a particular instant during the reaction. The instantaneous rate is determined from the slope of the curve at a particular point in time. We have drawn two tangent lines in Figure 14.4, a dashed line running through the point at t = 0 s and a solid line running through the point at t = 600 s. The slopes of these tangent lines give the instantaneous rates at these two time points.* To determine the instantaneous rate at 600 s, for example, we construct horizontal and vertical lines to form the blue right triangle in Figure 14.4. The slope of the tangent line is the ratio of the height of the vertical side to the length of the horizontal side:

Instantaneous rate = - ∆3C4H9Cl4

∆t= -

10.017 - 0.0422 M1800 - 4002 s

= 6.3 * 10-5 M>s

table 14.1 Rate data for Reaction of C4h9Cl with Water

time, t 1s 2 3C4h9Cl 4 1M 2 average Rate 1M ,s 2 0.0 0.1000 1.9 * 10-4

50.0 0.0905 1.7 * 10-4

100.0 0.0820 1.6 * 10-4

150.0 0.0741 1.4 * 10-4

200.0 0.0671 1.22 * 10-4

300.0 0.0549 1.01 * 10-4

400.0 0.0448 0.80 * 10-4

500.0 0.0368 0.560 * 10-4

800.0 0.020010,000 0

▲ Figure 14.4 Concentration of butyl chloride 1c4h9cl2 as a function of time.

Go FiGuReHow does the instantaneous rate of reaction change as the reaction proceeds?

100 200 300 400 500 600 700 800 9000

0.010

0.020

0.030

0.040

0.050

0.060

0.070

0.080

0.090

0.100

Time (s)

[C4H

9Cl]

(M)

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Instantaneousrate at t = 0 s(initial rate)

Instantaneousrate at t = 600 s

∆ [C

4H9C

l]

∆t

Instantaneous rate at time t = slope of tangent to the line at time t.

*You may wish to review graphical determination of slopes in Appendix A. If you are familiar with calculus, you may recognize that the average rate approaches the instantaneous rate as the time interval approaches zero. This limit, in the notation of calculus, is the negative of the derivative of the curve at time t, -d3C4H9Cl4>dt.


In discussions that follow, the term rate means instantaneous rate unless indi-cated otherwise. The instantaneous rate at t = 0 is called the initial rate of the reaction. To understand the difference between average and instantaneous rates, imagine you have just driven 98 mi in 2.0 h. Your average speed for the trip is 49 mi>hr, but your instantaneous speed at any moment during the trip is the speedometer reading at that moment.

Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the initial rate).

solutionanalyze We are asked to determine an instantaneous rate from a graph of reactant concentration versus time.plan To obtain the instantaneous rate at t = 0 s, we must determine the slope of the curve at t = 0. The tangent is drawn on the graph as the hypotenuse of the tan triangle. The slope of this straight line equals the change in the vertical axis divided by the corresponding change in the horizontal axis (which, in the case of this example, is the change in molarity over change in time).Solve The tangent line falls from 3C4H9Cl4 = 0.100 M to 0.060 M in the time change from 0 s to 210 s. Thus, the initial rate is

Rate = - ∆3C4H9Cl4

∆t= -

10.060 - 0.1002 M1210 - 02 s

= 1.9 * 10-4 M>s

Practice exercise 1Which of the following could be the instantaneous rate of the reaction in Figure 14.4 at t = 1000 s?(a) 1.2 * 10-4 M>s, (b) 8.8 * 10-5 M>s, (c) 6.3 * 10-5 M>s, (d) 2.7 * 10-5 M>s,(e) More than one of these.

Practice exercise 2Using Figure 14.4, determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s.

samPle exeRCise 14.2 Calculating an instantaneous Rate of Reaction

Give It Some ThoughtIn Figure 14.4, order the following three rates from fastest to slowest: (i) The average rate of the reaction between 0 s and 600 s, (ii) the instantaneous rate at t = 0 s, and (iii) the instantaneous rate at t = 600 s. You should not have to do any calculations.

Reaction Rates and StoichiometryDuring our discussion of the hypothetical reaction A ¡ B, we saw that the stoi-chiometry requires that the rate of disappearance of A equal the rate of appearance of B. Likewise, the stoichiometry of Equation 14.3 indicates that 1 mol of C4H9OH is produced for each mole of C4H9Cl consumed. Therefore, the rate of appearance of C4H9OH equals the rate of disappearance of C4H9Cl:

Rate = - ∆3C4H9Cl4

∆t=

∆3C4H9OH4∆t

What happens when the stoichiometric relationships are not one-to-one? For example, consider the reaction 2 HI1g2 ¡ H21g2 + I21g2. We can measure either the rate of disappearance of HI or the rate of appearance of either H2 or I2. Because 2 mol of HI disappears for each mole of H2 or I2 that forms, the rate of disappearance of HI is twice the rate of appearance of either H2 or I2. How do we decide which number to use for the rate of the reaction? Depending on whether we monitor HI, I2, or H2, the rates can differ by a factor of 2. To fix this problem, we need to take into account the reaction stoichiometry. To arrive at a number for the reaction rate that does not

secTIon 14.3 concentration and rate Laws 581

depend on which component we measured, we must divide the rate of disappearance of HI by 2 (its coefficient in the balanced chemical equation):

Rate = - 12

∆3HI4∆t

=∆3H24

∆t=

∆3I24∆t

In general, for the reactiona A + b B ¡ c C + d D

the rate is given by

Rate = - 1a

∆3A4∆t

= - 1b

∆3B4

∆t=

1c

∆3C4∆t

=1d

∆3D4

∆t [14.4]

When we speak of the rate of a reaction without specifying a particular reactant or product, we utilize the definition in Equation 14.4.*

*Equation 14.4 does not hold true if substances other than C and D are formed in significant amounts. For example, sometimes intermediate substances build in concentration before forming the final products. In that case, the relationship between the rate of disappearance of reactants and the rate of appearance of prod-ucts is not given by Equation 14.4. All reactions whose rates we consider in this chapter obey Equation 14.4.

14.3 | Concentration and Rate lawsOne way of studying the effect of concentration on reaction rate is to determine the way in which the initial rate of a reaction depends on the initial concentrations. For example, we might study the rate of the reaction

NH4+1aq2 + NO2

- 1aq2 ¡ N21g2 + 2 H2O1l2

solutionanalyze We are given a balanced chemical equation and asked to relate the rate of appearance of the product to the rate of disappearance of the reactant.plan We can use the coefficients in the chemical equation as shown in Equation 14.4 to express the relative rates of reactions.Solve

(a) Using the coefficients in the balanced equation and the relationship given by Equation 14.4, we have:

Rate = - 12

∆3O34

∆t=

13

∆3O24

∆t(b) Solving the equation from part (a) for the rate at which

O3 disappears, - ∆3O34>∆t, we have:

- ∆3O34

∆t=

23

∆3O24

∆t=

23

16.0 * 10-5M>s2 = 4.0 * 10-5M>s

check We can apply a stoichiometric factor to convert the O2 formation rate to the O3 disappearance rate:

- ∆3O34

∆t= a6.0 * 10-5

mol O2>Ls

b a 2 mol O3

3 mol O2b = 4.0 * 10-5

mol O3>Ls

= 4.0 * 10-5M>s

samPle exeRCise 14.3 Relating Rates at Which Products appear and Reactants disappear

(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O31g2 ¡ 3 O21g2?

(b) If the rate at which O2 appears, ∆3O24>∆t, is 6.0 * 10-5 M>s at a particular instant, at what rate is O3 disappearing at this same time, - ∆3O34>∆t?

Practice exercise 1At a certain time in a reaction, substance A is disappearing at a rate of 4.0 * 10-2 M>s, substance B is appearing at a rate of 2.0 * 10-2 M>s, and substance C is appearing at a rate of 6.0 * 10-2 M>s. Which of the following could be the stoichiometry for the reaction being studied?(a) 2A + B ¡ 3C (b) A ¡ 2B + 3C (c) 2A ¡ B + 3C (d) 4A ¡ 2B + 3C(e) A + 2B ¡ 3C

Practice exercise 2If the rate of decomposition of N2O5 in the reaction 2 N2O51g2 ¡ 4 NO21g2 + O21g2 at a particular instant is 4.2 * 10 - 7M>s, what is the rate of appearance of (a) NO2 and (b) O2 at that instant?


by measuring the concentration of NH4+ or NO2

- as a function of time or by measur-ing the volume of N2 collected as a function of time. Because the stoichiometric coef-ficients on NH4

+, NO2- , and N2 are the same, all of these rates are the same.

▲ table 14.2 shows that changing the initial concentration of either reactant changes the initial reaction rate. If we double 3NH4

+4 while holding 3NO2-4 constant,

the rate doubles (compare experiments 1 and 2). If we increase 3NH4+4 by a factor of 4

but leave 3NO2-4 unchanged (experiments 1 and 3), the rate changes by a factor of 4, and

so forth. These results indicate that the initial reaction rate is proportional to 3NH4+4.

When 3NO2-4 is similarly varied while 3NH4

+4 is held constant, the rate is affected in the same manner. Thus, the rate is also directly proportional to the concentration of 3NO2

-4.

table 14.2 Rate data for the Reaction of ammonium and nitrite ions in Water at 25 °C

experiment number

initial nh4+

Concentration (M)initial no2

− Concentration (M)

observed initial Rate 1M ,s 2

1 0.0100 0.200 5.4 * 10 - 7

2 0.0200 0.200 10.8 * 10 - 7

3 0.0400 0.200 21.5 * 10 - 7

4 0.200 0.0202 10.8 * 10 - 7

5 0.200 0.0404 21.6 * 10 - 7

6 0.200 0.0808 43.3 * 10 - 7

passes, and c is the molar concentration of the absorbing substance. Thus, the concentration is directly proportional to absorbance. Many chemical and pharmaceutical companies routinely use Beer’s law to calculate the concentration of purified solutions of the compounds that they make. In the laboratory portion of your course, you may very well perform one or more experiments in which you use Beer’s law to relate absorption of light to concentration.Related Exercises: 14.101, 14.102, Design an Experiment

a Closer look

Using Spectroscopic Methods to Measure Reaction Rates: Beer’s Law

A variety of techniques can be used to monitor reactant and product concentration during a reaction, including spectroscopic methods, which rely on the ability of substances to absorb (or emit) light. Spec-troscopic kinetic studies are often performed with the reaction mix-ture in the sample compartment of a spectrometer, an instrument that measures the amount of light transmitted or absorbed by a sample at different wavelengths. For kinetic studies, the spectrometer is set to measure the light absorbed at a wavelength characteristic of one of the reactants or products. In the decomposition of HI(g) into H21g2 and I21g2, for example, both HI and H2 are colorless, whereas I2 is violet. During the reaction, the violet color of the reaction mixture gets more intense as I2 forms. Thus, visible light of appropriate wavelength can be used to monitor the reaction (▶ Figure 14.5).

▶ Figure 14.6 shows the components of a spectrometer. The spectrometer measures the amount of light absorbed by the sample by comparing the intensity of the light emitted from the light source with the intensity of the light transmitted through the sample, for various wavelengths. As the concentration of I2 increases and its color becomes more intense, the amount of light absorbed by the reaction mixture in-creases, as Figure 14.5 shows, causing less light to reach the detector.

How can we relate the amount of light detected by the spectrometer to the concentration of a species? A relationship called Beer’s law gives us a direct route to the information we seek. Beer’s law connects the amount of light absorbed to the concentration of the absorbing substance:

A = ebc [14.5]

In this equation, A is the measured absorbance, e is the extinc-tion coefficient (a characteristic of the substance being monitored at a given wavelength of light), b is the path length through which the light

▲ Figure 14.5 Visible spectra of l2 at different concentrations.

400 450 500 550 600 6500.0

0.2

0.4

Wavelength (nm)

100 mg/L70 mg/L40 mg/L10 mg/L1 mg/L

Abs

orba

nce

Spectrometer measures intensity of purple color as I2 concentration increases


We express the way in which the rate depends on the reactant concentrations by the equation

Rate = k3NH4+43NO2

- 4 [14.6]

An equation such as Equation 14.6, which shows how the rate depends on reactant concentrations, is called a rate law. For the general reaction

a A + b B ¡ c C + d D

the rate law generally has the form

Rate = k3A4m3B4n [14.7]

Notice that only the concentrations of the reactants generally appear in the rate law. The constant k is called the rate constant. The magnitude of k changes with temperature and therefore determines how temperature affects rate, as we will see in Section 14.5. The exponents m and n are typically small whole numbers. As we will learn shortly, if we know m and n for a reaction, we can gain great insight into the individual steps that occur during the reaction.

Give It Some ThoughtHow do reaction rate, rate law, and rate constant differ?

Once we know the rate law for a reaction and the reaction rate for a set of reac-tant concentrations, we can calculate the value of k. For example, using the values for experiment 1 in Table 14.2, we can substitute into Equation 14.6:

5.4 * 10-7 M>s = k10.0100 M210.200 M2

k =5.4 * 10-7 M>s

10.0100 M210.200 M2 = 2.7 * 10-4 M-1s-1

You should verify that this same value of k is obtained using any of the other experi-mental results in Table 14.2.

Once we have both the rate law and the k value for a reaction, we can calcu-late the reaction rate for any set of concentrations. For example, using Equation 14.7 with k = 2.7 * 10-4 M-1 s-1, m = 1, and n = 1, we can calculate the rate for 3NH4

+4 = 0.100 M and 3NO2- 4 = 0.100 M:

Rate = 12.7 * 10-4M-1 s-1210.100 M210.100 M2 = 2.7 * 10-6M>s

Give It Some ThoughtDoes the rate constant have the same units as the rate?

▲ Figure 14.6 Components of a spectrometer.

Source Lenses/slits/collimators

Monochromator(selects wavelength)

Sample Detector Computer


Reaction Orders: The Exponents in the Rate LawThe rate law for most reactions has the form Rate = k3reactant 14m3reactant 24n c [14.8]

The exponents m and n are called reaction orders. For example, consider again the rate law for the reaction of NH4

+ with NO2- :

Rate = k3NH4+43NO2

- 4Because the exponent of 3NH4

+4 is 1, the rate is first order in NH4+. The rate is also

first order in NO2- . (The exponent 1 is not shown in rate laws.) The overall reaction

order is the sum of the orders with respect to each reactant represented in the rate law. Thus, for the NH4

+ - NO2- reaction, the rate law has an overall reaction order of

1 + 1 = 2, and the reaction is second order overall.The exponents in a rate law indicate how the rate is affected by each reactant con-

centration. Because the rate at which NH4+ reacts with NO2

- depends on 3NH4+4 raised

to the first power, the rate doubles when 3NH4+4 doubles, triples when 3NH4

+4 triples, and so forth. Doubling or tripling 3NO2

- 4 likewise doubles or triples the rate. If a rate law is second order with respect to a reactant, 3A42, then doubling the concentration of that substance causes the reaction rate to quadruple because 3242 = 4, whereas tripling the concentration causes the rate to increase ninefold: 3342 = 9.

The following are some additional examples of experimentally determined rate laws:

2 N2O51g2 ¡ 4 NO21g2 + O21g2 Rate = k3N2O54 [14.9]

H21g2 + I21g2 ¡ 2 HI1g2 Rate = k3H243I24 [14.10]

CHCl31g2 + Cl21g2 ¡ CCl41g2 + HCl1g2 Rate = k3CHCl343Cl241>2 [14.11]

Although the exponents in a rate law are sometimes the same as the coefficients in the balanced equation, this is not necessarily the case, as Equations 14.9 and 14.11 show. For any reaction, the rate law must be determined experimentally. In most rate laws, reac-tion orders are 0, 1, or 2. However, we also occasionally encounter rate laws in which the reaction order is fractional (as is the case with Equation 14.11) or even negative.

Give It Some ThoughtThe experimentally determined rate law for the reaction2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2 is rate = k3NO423H24.(a) What are the reaction orders in this rate law?(b) Would the reaction rate increase more if we doubled the concentration of NO

or the concentration of H2?

samPle exeRCise 14.4 Relating a Rate law to the effect of Concentration on Rate

Consider a reaction A + B ¡ C for which rate = k3A43B42. Each of the following boxes repre-sents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mix-tures in order of increasing rate of reaction.

(1) (2) (3)


Magnitudes and Units of Rate ConstantsIf chemists want to compare reactions to evaluate which ones are relatively fast and which ones are relatively slow, the quantity of interest is the rate constant. A good gen-eral rule is that a large value of k (∼109 or higher) means a fast reaction and a small value of k (10 or lower) means a slow reaction.

Give It Some ThoughtSuppose the reactions A ¡ B and X ¡ Y have the same value of k. When 3A4 = 3X4, will the two reactions necessarily have the same rate?

The units of the rate constant depend on the overall reaction order of the rate law. In a reaction that is second order overall, for example, the units of the rate constant must satisfy the equation:

Units of rate = 1units of rate constant21units of concentration22

Hence, in our usual units of molarity for concentration and seconds for time, we have

Units of rate constant =units of rate

1units of concentration22 =M>sM2 = M-1 s-1

solutionanalyze We are given three boxes containing different numbers of spheres representing mixtures containing different reactant con-centrations. We are asked to use the given rate law and the com-positions of the boxes to rank the mixtures in order of increasing reaction rates.plan Because all three boxes have the same volume, we can put the number of spheres of each kind into the rate law and calculate the rate for each box.Solve Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate:

Box 1: Rate = k1521522 = 125k

Box 2 contains 7 red spheres and 3 purple spheres:

Box 2: Rate = k1721322 = 63k

Box 3 contains 3 red spheres and 7 purple spheres:

Box 3: Rate = k1321722 = 147k

The slowest rate is 63k (Box 2), and the highest is 147k (Box 3). Thus, the rates vary in the order 2 6 1 6 3.check Each box contains 10 spheres. The rate law indicates that in this case [B] has a greater influence on rate than [A] because B has a larger reaction order. Hence, the mixture with the highest concentra-tion of B (most purple spheres) should react fastest. This analysis con-firms the order 2 6 1 6 3.

Practice exercise 1Suppose the rate law for the reaction in this Sample Exercise were rate = k3A423B4. What would be the ordering of the rates for the three mixtures shown above, from slowest to fastest?(a) 1 6 2 6 3 (b) 1 6 3 6 2 (c) 3 6 2 6 1 (d) 2 6 1 6 3 (e) 3 6 1 6 2

Practice exercise 2Assuming that rate = k3A43B4, rank the mixtures represented in this Sample Exercise in order of increasing rate.

solutionanalyze We are given two rate laws and asked to express (a) the overall reaction order for each and (b) the units for the rate constant for the first reaction.plan The overall reaction order is the sum of the exponents in the rate law. The units for the rate constant, k, are found by using the normal units for rate 1M>s2 and concentration (M) in the rate law and apply-ing algebra to solve for k.Solve

(a) The rate of the reaction in Equation 14.9 is first order in N2O5 and first order overall. The reaction in Equation 14.11 is first

samPle exeRCise 14.5 determining Reaction orders and units for Rate Constants

(a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.11? (b) What are the units of the rate constant for the rate law in Equation 14.9?

order in CHCl3 and one-half order in Cl2. The overall reaction order is three halves.

(b) For the rate law for Equation 14.9, we have

Units of rate = 1units of rate constant21units of concentration2so

Units of rate constant =units of rate

units of concentration=

M>sM

= s-1

Notice that the units of the rate constant change as the overall order of the reaction changes.


Using Initial Rates to Determine Rate LawsWe have seen that the rate law for most reactions has the general form

Rate = k3reactant 14m3reactant 24n c

Thus, the task of determining the rate law becomes one of determining the reac-tion orders, m and n. In most reactions, the reaction orders are 0, 1, or 2. As noted earlier in this section, we can use the response of the reaction rate to a change in initial concentration to determine the reaction order.

In working with rate laws, it is important to realize that the rate of a reaction depends on concentration but the rate constant does not. As we will see later in this chapter, the rate constants (and hence the reaction rate) are affected by temperature and by the presence of a catalyst.

Practice exercise 1Which of the following are the units of the rate constant for Equation 14.11?(a) M-1�2 s-1 (b) M -1�2 s-1�2 (c) M 1�2 s-1 (d) M-3�2 s-1 (e) M -3�2s-1�2

Practice exercise 2(a) What is the reaction order of the reactant H2 in Equation 14.10? (b) What are the units of the rate constant for Equation 14.10?

The initial rate of a reaction A + B ¡ C was measured for several different starting concentrations of A and B, and the results are as follows:

samPle exeRCise 14.6 determining a Rate law from initial Rate data

plan (a) We assume that the rate law has the following form: Rate = k3A4m3B4n. We will use the given data to deduce the reac-tion orders m and n by determining how changes in the concentra-tion change the rate. (b) Once we know m and n, we can use the

rate law and one of the sets of data to determine the rate constant k. (c) Upon determining both the rate constant and the reaction orders, we can use the rate law with the given concentrations to calculate rate.

Solve

(a) If we compare experiments 1 and 2, we see that [A] is held constant and [B] is doubled. Thus, this pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with

respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0).

In experiments 1 and 3, [B] is held constant, so these data show how [A] affects rate. Holding [B] constant while doubling [A] increases the rate fourfold. This result indicates that rate is proportional to 3A42 (that is, the reaction is second order in A). Hence, the rate law is Rate = k3A423B40 = k3A42

(b) Using the rate law and the data from experiment 1, we have k =rate3A42 =

4.0 * 10-5 M>s10.100 M22 = 4.0 * 10-3 M-1 s-1

(c) Using the rate law from part (a) and the rate constant from part (b), we have Rate = k3A42 = 14.0 * 10-3 M-1s-1210.050 M22 = 1.0 * 10-5 M>s

Because [B] is not part of the rate law, it is irrelevant to the rate if there is at least some B present to react with A.

experiment number [a] (M) [B] (M) initial Rate 1M ,s 21 0.100 0.100 4.0 * 10-5

2 0.100 0.200 4.0 * 10-5

3 0.200 0.100 16.0 * 10-5

Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when 3A4 = 0.050M and 3B4 = 0.100 M.

solutionanalyze We are given a table of data that relates concentrations of reactants with initial rates of reaction and asked to determine (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentra-tions not listed in the table.

secTIon 14.4 The change of concentration with Time 587

14.4 | the Change of Concentration with time

The rate laws we have examined so far enable us to calculate the rate of a reaction from the rate constant and reactant concentrations. In this section, we will show that rate laws can also be converted into equations that show the relationship between concen-trations of reactants or products and time. The mathematics required to accomplish this conversion involves calculus. We do not expect you to be able to perform the calcu-lus operations, but you should be able to use the resulting equations. We will apply this conversion to three of the simplest rate laws: those that are first order overall, those that are second order overall, and those that are zero order overall.

First-Order ReactionsA first-order reaction is one whose rate depends on the concentration of a single re-actant raised to the first power. If a reaction of the type A ¡ products is first order, the rate law is:

Rate = - ∆3A4

∆t= k3A4

This form of a rate law, which expresses how rate depends on concentration, is called the differential rate law. Using the operation from calculus called integration, this rela-tionship can be transformed into an equation known as the integrated rate law for a

check A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see if we can correctly calculate the rate. Using data from experiment 3, we have

Rate = k3A42 = 14.0 * 10-3 M-1 s-1210.200 M22 = 1.6 * 10-4 M>s

Thus, the rate law correctly reproduces the data, giving both the correct number and the correct units for the rate.

Practice exercise 1A certain reaction X + Y ¡ Z is described as being first order in [X] and third order overall. Which of the following statements is or are true?: (i) The rate law for the reaction is: Rate = k3X43Y42. (ii) If the concentration of X is increased by a factor of 1.5, the rate will increase by a factor

of 2.25. (iii) If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.

(a) Only one of the statements is true. (c) Statements (i) and (iii) are true. (e) All three statements are true.

(b) Statements (i) and (ii) are true. (d) Statements (ii) and (iii) are true.

Practice exercise 2The following data were measured for the reaction of nitric oxide with hydrogen:

2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2

experiment number [no] (M) 3h2 4 (M) initial Rate 1M ,s 21 0.10 0.10 1.23 * 10-3

2 0.10 0.20 2.46 * 10-3

3 0.20 0.10 4.92 * 10-3

(a) Determine the rate law for this reaction. (b) Calculate the rate constant.(c) Calculate the rate when 3NO4 = 0.050 M and 3H24 = 0.150 M.


first-order reaction that relates the initial concentration of A, 3A40, to its concentration at any other time t, 3A4t :

ln3A4t - ln3A40 = -kt or ln 3A4t

3A40= -kt [14.12]

The function “ln” in Equation 14.12 is the natural logarithm (Appendix A.2). Equation 14.12 can also be rearranged to ln3A4t = -kt + ln3A40 [14.13]

Equations 14.12 and 14.13 can be used with any concentration units as long as the units are the same for both 3A4t and 3A40.

For a first-order reaction, Equation 14.12 or 14.13 can be used in several ways. Given any three of the following quantities, we can solve for the fourth: k, t, 3A40, and 3A4t. Thus, you can use these equations to determine (1) the concentration of a reactant remaining at any time after the reaction has started, (2) the time interval required for a given fraction of a sample to react, or (3) the time interval required for a reactant con-centration to fall to a certain level.

The decomposition of a certain insecticide in water at 12 °C follows first-order kinetics with a rate constant of 1.45 yr-1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentra-tion of 5.0 * 10-7 g>cm3. Assume that the temperature of the lake is

constant (so that there are no effects of temperature variation on the rate). (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concen-tration to decrease to 3.0 * 10 - 7 g>cm3?

samPle exeRCise 14.7 using the integrated First-order Rate law

solutionanalyze We are given the rate constant for a reaction that obeys first-order kinetics, as well as information about concentrations and times, and asked to calculate how much reactant (insecticide) remains after 1 yr. We must also determine the time interval needed to reach a par-ticular insecticide concentration. Because the exercise gives time in (a) and asks for time in (b), we will find it most useful to use the inte-grated rate law, Equation 14.13.

plan

(a) We are given k = 1.45 yr-1, t = 1.00 yr, and 3insecticide40 = 5.0 * 10-7 g>cm3, and so Equation 14.13 can be solved for 3insecticide4t.

(b) We have k = 1.45 yr-1, 3insecticide40 = 5.0 * 10-7 g>cm3, and 3insecticide4t = 3.0 * 10-7 g>cm3, and so we can solve Equation 14.13 for time, t.

Solve

(a) Substituting the known quantities into Equation 14.13, we have ln3insecticide4t = 1 yr = -11.45 yr-1211.00 yr2 + ln15.0 * 10-72We use the ln function on a calculator to evaluate the second term on the right [that is, ln15.0 * 10-724, giving ln3insecticide4t = 1 yr = -1.45 + 1-14.512 = -15.96

check In part (a) the concentration remaining after 1.00 yr (that is, 1.2 * 10 - 7 g>cm3) is less than the original concentration 15.0 * 10 - 7 g>cm32, as it should be. In (b) the given concentration

13.0 * 10-7 g>cm32 is greater than that remaining after 1.00 yr, indicat-ing that the time must be less than a year. Thus, t = 0.35 yr is a reason-able answer.

Practice exercise 1At 25 °C, the decomposition of dinitrogen pentoxide, N2O51g2, into NO21g2 and O21g2 follows first-order kinetics with k = 3.4 * 10-5 s-1. A sample of N2O5 with an initial pressure of 760 torr decomposes at 25 °C until its partial pressure is 650 torr. How much time (in seconds) has elapsed?(a) 5.3 * 10-6 (b) 2000 (c) 4600 (d) 34,000 (e) 190,000

Practice exercise 2The decomposition of dimethyl ether, 1CH322O, at 510 °C is a first-order process with a rate constant of 6.8 * 10-4 s-1:

1CH322O1g2 ¡ CH41g2 + H21g2 + CO1g2If the initial pressure of 1CH322O is 135 torr, what is its pressure after 1420 s?

To obtain 3insecticide4t = 1 yr, we use the inverse natural logarithm, or ex, function on the calculator: 3insecticide4t = 1 yr = e - 15.96 = 1.2 * 10 - 7g>cm3

Note that the concentration units for 3A4t and 3A40 must be the same.(b) Again substituting into Equation 14.13, with

3insecticide4t = 3.0 * 10-7 g>cm3, gives ln13.0 * 10-72 = -11.45 yr-121t2 + ln15.0 * 10-72Solving for t gives t = -3ln13.0 * 10-72 - ln15.0 * 10-724>1.45 yr-1

= -1-15.02 + 14.512>1.45 yr-1 = 0.35 yr


Equation 14.13 can be used to verify whether a reaction is first order and to determine its rate constant. This equation has the form of the general equation for a straight line, y = mx + b, in which m is the slope and b is the y-intercept of the line (Appendix A.4):

ln [A]t = −kt

=

+

+

ln [A]0

mx by

For a first-order reaction, therefore, a graph of ln3A4t versus time gives a straight line with a slope of -k and a y-intercept of ln3A40. A reaction that is not first order will not yield a straight line.

As an example, consider the conversion of methyl isonitrile 1CH3NC2 to its isomer acetonitrile 1CH3CN2 (▶ Figure 14.7). Because experiments show that the reaction is first order, we can write the rate equation:

ln3CH3NC4t = -kt + ln3CH3NC40

We run the reaction at a temperature at which methyl isonitrile is a gas 1199 °C2, and ▼ Figure 14.8(a) shows how the pressure of this gas varies with time. We can use pressure as a unit of concentration for a gas because we know from the ideal-gas law the pressure is directly proportional to the number of moles per unit volume. Figure 14.8(b) shows that a plot of the natural logarithm of the pressure versus time is a straight line. The slope of this line is -5.1 * 10-5 s-1. (You should verify this for yourself, remembering that your result may vary slightly from ours because of inaccu-racies associated with reading the graph.) Because the slope of the line equals -k, the rate constant for this reaction equals 5.1 * 10-5 s-1.

Second-Order ReactionsA second-order reaction is one for which the rate depends either on a reactant concen-tration raised to the second power or on the concentrations of two reactants each raised to the first power. For simplicity, let’s consider reactions of the type A ¡ products or A + B ¡ products that are second order in just one reactant, A:

Rate = - ∆3A4

∆t= k3A42

▲ Figure 14.7 The first-order reaction of ch3nc conversion into ch3cn.

Methyl isonitrile

Acetonitrile

10,0000 20,000 30,0000

20406080

100120140160

Time (s)

(a)

Pres

sure

, CH

3NC

(tor

r)

10,0000 20,000 30,0003.43.6

4.03.8

4.24.44.64.85.05.2

Time (s)

(b)

ln p

ress

ure,

CH

3NC

▲ Figure 14.8 Kinetic data for conversion of methyl isonitrile into acetonitrile.

Go FiGuReWhat can you conclude given that the plot of ln P versus t is linear?


With the use of calculus, this differential rate law can be used to derive the integrated rate law for second-order reactions:

1

3A4t= kt +

13A40

[14.14]

This equation, like Equation 14.13, has four variables, k, t, 3A40, and 3A4t, and any one of these can be calculated knowing the other three. Equation 14.14 also has the form of a straight line 1y = mx + b2. If the reaction is second order, a plot of 1>3A4t ver-sus t yields a straight line with slope k and y-intercept 1>3A40. One way to distinguish between first- and second-order rate laws is to graph both ln3A4t and 1>3A4t against t. If the ln3A4t plot is linear, the reaction is first order; if the 1>3A4t plot is linear, the reaction is second order.

solutionanalyze We are given the concentrations of a reactant at various times during a reaction and asked to determine whether the reaction is first or second order.plan We can plot ln3NO24 and 1>3NO24 against time. If one plot or the other is linear, we will know the reaction is either first or second order.Solve To graph ln3NO24 and 1>3NO24 against time, we first make the following calculations from the data given:

samPle exeRCise 14.8 determining Reaction order from the integrated Rate law

The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300 °C, NO21g2 ¡ NO1g2 + 1

2 O21g2. Is the reaction first or second order in NO2?

time (s) 3no2 4 1M 2 0.0 0.01000 50.0 0.00787100.0 0.00649200.0 0.00481300.0 0.00380

time (s) 3no2 4 1M 2 ln 3no2 4 1 , 3no2 4 11 ,M 20.0 0.01000 -4.605 100

50.0 0.00787 -4.845 127100.0 0.00649 -5.037 154200.0 0.00481 -5.337 208300.0 0.00380 -5.573 263

▲ Figure 14.9 Kinetic data for decomposition of no2.

0 100 200 300–5.8

–5.6

–5.4

–5.2

–5.0

–4.8

–4.6

Time (s)

ln[N

O2]

0 100 200 30050

150

250

Time (s)1/

[NO

2] (1

/M

)

0 100 200 300–5.8

–5.6

–5.4

–5.2

–5.0

–4.8

–4.6

Time (s)

ln[N

O2]

0 100 200 30050

150

250

Time (s)

1/[N

O2]

(1/M

)

As ▶ Figure 14.9 shows, only the plot of 1>3NO24 versus time is linear. Thus, the reaction obeys a second-order rate law: Rate = k3NO242. From the slope of this straight-line graph, we determine that k = 0.543 M -1 s-1 for the disappearance of NO2.

Practice exercise 1For a certain reaction A ¡ products, a plot of ln[A] versus time produces a straight line with a slope of -3.0 * 10-2 s-1. Which of the following statements is or are true?: (i) The reaction follows first-order kinetics. (ii) The rate constant for the reaction is 3.0 * 10-2 s-1. (iii) The initial concentration of [A] was 1.0 M.


Zero-Order ReactionsWe have seen that in a first-order reaction the concentration of a reactant A decreases nonlinearly, as shown by the red curve in ▶ Figure 14.10. As [A] declines, the rate at which it disappears declines in proportion. A zero-order reaction is one in which the rate of disappearance of A is independent of [A]. The rate law for a zero-order reaction is

Rate =- ∆3A4

∆t= k

The integrated rate law for a zero-order reaction is

3A4t = -kt + 3A40

where 3A4t is the concentration of A at time t and 3A40 is the initial concentration. This is the equation for a straight line with vertical intercept 3A40 and slope -kt, as shown in the blue curve in Figure 14.10.

The most common type of zero-order reaction occurs when a gas undergoes decomposition on the surface of a solid. If the surface is completely covered by decomposing molecules, the rate of reaction is constant because the number of react-ing surface molecules is constant, so long as there is some gas-phase substance left.

Half-LifeThe half-life of a reaction, t1>2, is the time required for the concentration of a reac-tant to reach half its initial value, 3A4t1>2

= 123A40. Half-life is a convenient way to de-

scribe how fast a reaction occurs, especially if it is a first-order process. A fast reaction has a short half-life.

We can determine the half-life of a first-order reaction by substituting 3A4t1>2= 1

23A40 for 3A4t and t1>2 for t in Equation 14.12:

ln 123A40

3A40= -kt1>2

ln 12 = -kt1>2

t1>2 = -ln 1

2

k=

0.693k

[14.15]

From Equation 14.15, we see that t1>2 for a first-order rate law does not depend on the initial concentration of any reactant. Consequently, the half-life remains constant throughout the reaction. If, for example, the concentration of a reactant is 0.120 M at some instant in the reac-tion, it will be 1210.120 M2 = 0.060 M after one half-life. After one more half-life passes, the concentration will drop to 0.030 M, and so on. Equa-tion 14.15 also indicates that, for a first-order reaction, we can calculate t1>2 if we know k and calculate k if we know t1>2.

The change in concentration over time for the first-order rearrange-ment of gaseous methyl isonitrile at 199 °C is graphed in ▶ Figure 14.11. Because the concentration of this gas is directly proportional to its pres-sure during the reaction, we have chosen to plot pressure rather than concentration in this graph. The first half-life occurs at 13,600 s (3.78 h). At a time 13,600 s later, the methyl isonitrile pressure (and therefore, concentration) has decreased to half of one-half, or one-fourth, of the initial value. In a first-order reaction, the concentration of the reactant decreases by one-half in each of a series of regularly spaced time intervals, each interval equal to t1>2.

(a) Only one of the statements is true.(b) Statements (i) and (ii) are true.(c) Statements (i) and (iii) are true.(d) Statements (ii) and (iii) are true.(e) All three statements are true.

Practice exercise 2The decomposition of NO2 discussed in the Sample Exercise is second order in NO2 with k = 0.543 M -1 s-1. If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the concentration of this reactant after 0.500 h?

▲ Figure 14.11 Kinetic data for the rearrangement of methyl isonitrile to acetonitrile at 199 °c, showing the half-life of the reaction.

Time[A

]

A0

First-order reaction

Zero-order reactionSlope = −kt

▲ Figure 14.10 Comparison of first-order and zero-order reactions for the disappearance of reactant A with time.

Go FiGuReAt which times during the reaction would you have trouble distinguishing a zero-order reaction from a first-order reaction?

t1/2

t1/2

10,000 20,000 30,0000

37.5

75

150

Time (s)

Pres

sure

, CH

3NC

(tor

r)


Give It Some ThoughtIf a solution containing 10.0 g of a substance reacts by first-order kinetics, how many grams remain after three half-lives?

Chemistry Put to Work

Methyl Bromide in the Atmosphere

The compounds known as chlorofluorocarbons (CFCs) are well-known agents responsible for the destruction of Earth’s protective ozone layer. Another simple molecule that has the potential to destroy the strato-spheric ozone layer is methyl bromide, CH3Br (▼ Figure 14.12). Because this substance has a wide range of uses, including antifun-gal treatment of plant seeds, it has been produced in large quantities in the past (about 150 million pounds per year worldwide in 1997, at the height of its production). In the stratosphere, the C ¬ Br bond is

broken through absorption of short-wavelength radiation. The resultant Br atoms then catalyze decomposition of O3.

Methyl bromide is removed from the lower atmosphere by a vari-ety of mechanisms, including a slow reaction with ocean water:

CH3Br1g2 + H2O1l2 ¡ CH3OH1aq2 + HBr1aq2 [14.16]

To determine the potential importance of CH3Br in destruction of the ozone layer, it is important to know how rapidly the reaction in Equation 14.16 and all other reactions remove CH3Br from the lower atmosphere before it can diffuse into the stratosphere.

The average lifetime of CH3Br in Earth’s lower atmosphere is difficult to measure because the conditions that exist in the atmo-sphere are too complex to be simulated in the laboratory. Instead, scientists analyzed nearly 4000 atmospheric samples collected above the Pacific Ocean for the presence of several trace organic substances, including methyl bromide. From these measurements, it was possible to estimate the atmospheric residence time for CH3Br.

The atmospheric residence time is related to the half-life for CH3Br in the lower atmosphere, assuming CH3Br decom-poses by a first-order process. From the experimental data, the half-life for methyl bromide in the lower atmosphere is estimated to be 0.8 { 0.1 yr. That is, a collection of CH3Br molecules present at any given time will, on average, be 50% decomposed after 0.8 yr, 75% decomposed after 1.6 yr, and so on. A half-life of 0.8 yr, while comparatively short, is still suf-ficiently long so that CH3Br contributes significantly to the de-struction of the ozone layer.

In 1997 an international agreement was reached to phase out use of methyl bromide in developed countries by 2005. How-ever, in recent years exemptions for critical agricultural use have been requested and granted. Nevertheless, authorized worldwide production was down to 26 million pounds in 2012, three-fourths of which is used in the United States.

Related Exercise: 14.122

Stratosphere

Diffusion tostratosphere

50% decomposes in 0.8 years

Troposphere

Methyl bromideapplied as antifungaltreatment

Lower atmosphere

▲ Figure 14.12 Distribution and fate of methyl bromide in Earth’s atmosphere.

solutionanalyze We are asked to estimate the half-life of a reaction from a graph of concentration versus time and then to use the half-life to cal-culate the rate constant for the reaction.plan

(a) To estimate a half-life, we can select a concentration and then determine the time required for the concentration to decrease to half of that value.

samPle exeRCise 14.9 determining the half-life of a First-order Reaction

The reaction of C4H9Cl with water is a first-order reaction. (a) Use Figure 14.4 to estimate the half-life for this reaction. (b) Use the half-life from (a) to calculate the rate constant.

(b) Equation 14.15 is used to calculate the rate constant from the half-life.

Solve

(a) From the graph, we see that the initial value of 3C4H9Cl4 is 0.100 M. The half-life for this first-order reaction is the time required for 3C4H9Cl4 to decrease to 0.050 M, which we can read off the graph. This point occurs at approximately 340 s.

secTIon 14.5 Temperature and rate 593

The half-life for second-order and other reactions depends on reactant concentra-tions and therefore changes as the reaction progresses. We obtained Equation 14.15 for the half-life for a first-order reaction by substituting 3A4t1>2

= 123A40 for 3A4t and t1>2

for t in Equation 14.12. We find the half-life of a second-order reaction by making the same substitutions into Equation 14.14:

1123A40

= kt1>2 +1

3A40

23A40

-1

3A40= kt1>2

t1>2 =1

k3A40 [14.17]

In this case, the half-life depends on the initial concentration of reactant—the lower the initial concentration, the longer the half-life.

Give It Some ThoughtWhy can we report the half-life for a first-order reaction without knowing the initial concentration, but not for a second-order reaction?

14.5 | temperature and RateThe rates of most chemical reactions increase as the temperature rises. For example, dough rises faster at room temperature than when refrigerated, and plants grow more rapidly in warm weather than in cold. We can see the effect of temperature on reaction rate by observing a chemiluminescence reaction (one that produces light), such as that in Cyalume® light sticks (▶ Figure 14.13).

How is this experimentally observed temperature effect reflected in the rate law? The faster rate at higher temperature is due to an increase in the rate constant with increasing temperature. For example, let’s reconsider the first-order reaction we saw in Figure 14.7, namely CH3NC ¡ CH3CN. Figure 14.14 shows the rate constant for this reaction as a function of temperature. The rate constant and, hence, the rate of the reaction increase rapidly with temperature, approximately doubling for each 10 °C rise.

The Collision ModelReaction rates are affected both by reactant concentrations and by temperature. The collision model, based on the kinetic-molecular theory (Section 10.7), accounts for both of these effects at the molecular level. The central idea of the collision model is that molecules must collide to react. The greater the number of collisions per second, the greater the reaction rate. As reactant concentration increases, therefore, the number of collisions increases, leading to an increase in reaction rate. According to the kinetic-molecular theory of gases, increasing the temperature increases molecular speeds. As molecules move faster, they collide more forcefully (with more energy) and more fre-quently, increasing reaction rates.

(b) Solving Equation 14.15 for k, we have

k =0.693t1>2

=0.693340 s

= 2.0 * 10-3 s-1

check At the end of the second half-life, which should occur at 680 s, the concentration should have decreased by yet another factor of 2, to 0.025 M. Inspection of the graph shows that this is indeed the case.

Practice exercise 1We noted in an earlier Practice Exercise that at 25 °C the decom-position of N2O51g2 into NO21g2 and O21g2 follows first-order

kinetics with k = 3.4 * 10-5 s-1. How long will it take for a sample originally containing 2.0 atm of N2O5 to reach a partial pressure of 380 torr?(a) 5.7 h (b) 8.2 h (c) 11 h (d) 16 h (e) 32 h

Practice exercise 2

(a) Using Equation 14.15, calculate t1>2 for the decomposition of the insecticide described in Sample Exercise 14.7.(b) How long does it take for the concentration of the insecticide to reach one-quarter of the initial value?

Hot water Cold water

▲ Figure 14.13 Temperature affects the rate of the chemiluminescence reaction in light sticks: The chemiluminescent reaction occurs more rapidly in hot water, and more light is produced.

Go FiGuReWhy does the light stick glow with less light in cold water than in hot water?


For a reaction to occur, though, more is required than simply a collision—it must be the right kind of collision. For most reactions, in fact, only a tiny fraction of collisions leads to a reaction. For example, in a mixture of H2 and I2 at ordinary tem-peratures and pressures, each molecule undergoes about 1010 collisions per second. If every collision between H2 and I2 resulted in the formation of HI, the reaction would

be over in much less than a second. Instead, at room temperature the reaction proceeds very slowly because only about one in every 1013 col-lisions produces a reaction. What keeps the reaction from occurring more rapidly?

The Orientation FactorIn most reactions, collisions between molecules result in a chemical reac-tion only if the molecules are oriented in a certain way when they collide. The relative orientations of the molecules during collision determine whether the atoms are suitably positioned to form new bonds. For ex-ample, consider the reaction

Cl + NOCl ¡ NO + Cl2

which takes place if the collision brings Cl atoms together to form Cl2, as shown in the top panel of ▼ Figure 14.15. In contrast, in the collision shown in the lower panel, the two Cl atoms are not colliding directly with one another, and no products are formed.

Activation EnergyMolecular orientation is not the only factor influencing whether a molecular collision will produce a reaction. In 1888 the Swedish chemist Svante Arrhenius suggested that molecules must possess a certain minimum amount of energy to react. According to the collision model, this energy comes from the kinetic energies of the colliding mol-ecules. Upon collision, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. That is, the kinetic energy is used to change the potential energy of the molecule. If molecules are moving too slowly—in other words, with too little kinetic energy—they merely bounce off one an-other without changing. The minimum energy required to initiate a chemical reaction is called the activation energy, Ea, and its value varies from reaction to reaction.

180 190 200 210 220 230 240 250

1 × 10−3

2 × 10−3

3 × 10−3

Temperature (°C)

k (s

–1)

CH3CNCH3NC

▲ Figure 14.14 Temperature dependence of the rate constant for methyl isonitrile conversion to acetonitrile. The four points indicated are used in Sample Exercise 14.11.

Go FiGuReWould you expect this curve to eventually go back down to lower values? Why or why not?

▲ Figure 14.15 Molecular collisions may or may not lead to a chemical reaction between Cl and NOCl.

Before collision Collision After collision

Before collision Collision After collision

Effective collision, reaction occurs, Cl2 forms

Ineffective collision, no reaction possible, no Cl2


The situation during reactions is analogous to that shown in ▲ Figure 14.16. The golfer hits the ball to make it move over the hill in the direction of the cup. The hill is a barrier between ball and cup. To reach the cup, the player must impart enough kinetic energy with the putter to move the ball to the top of the barrier. If he does not impart enough energy, the ball will roll partway up the hill and then back down toward him. In the same way, molecules require a certain minimum energy to break existing bonds during a chemical reaction. We can think of this minimum energy as an energy barrier. In the rearrangement of methyl isonitrile to acetonitrile, for example, we might imagine the reaction passing through an intermediate state in which the N ‚ C portion of the methyl isonitrile molecule is sideways:

H3C H3CH3CN NN

C CC

▶ Figure 14.17 shows that energy must be supplied to stretch the bond between the H3C group and the N ‚ C group to allow the N ‚ C group to rotate. After the N ‚ C group has twisted sufficiently, the C ¬ C bond begins to form, and the energy of the molecule drops. Thus, the barrier to formation of acetonitrile represents the energy necessary to force the molecule through the relatively unstable inter-mediate state, analogous to forcing the ball in Figure 14.16 over the hill. The difference between the energy of the starting molecule and the highest energy along the reaction pathway is the activation energy, Ea. The molecule having the arrangement of atoms shown at the top of the barrier is called either the activated complex or the transition state.

The conversion of H3C ¬ N ‚ C to H3C ¬ C ‚ N is exother-mic. Figure 14.17 therefore shows the product as having a lower energy than the reactant. The energy change for the reaction, ∆E, has no effect on reaction rate, however. The rate depends on the magnitude of Ea; generally, the lower the value of Ea is, the faster the reaction.

Notice that the reverse reaction is endothermic. The activation energy for the reverse reaction is equal to the energy that must be overcome if approaching the barrier from the right: ∆E + Ea. Thus, to reach the activated complex for the reverse reaction requires more energy than for the forward reaction—for this reaction, there is a larger barrier to overcome going from right to left than from left to right.

▲ Figure 14.16 Energy is needed to overcome a barrier between initial and final states.

Go FiGuReIf the barrier were lower than as shown in the figure, would the golfer have to hit the ball as hard?

H3C N CPote

ntia

l ene

rgy

CN

H3C N

H3C C

Reaction progress

N

H3C C

C

N

H3C. . .

Ea

∆E

Molecule bends, C—N bond begins to break

Activated complex forms

C—C bond forms

▲ Figure 14.17 Energy profile for conversion of methyl isonitrile 1h3cnc2 to its isomer acetonitrile 1h3ccn2.

Go FiGuReHow does the energy needed to overcome the energy barrier compare with the overall change in energy for this reaction?


Give It Some ThoughtSuppose you could measure the rates for both the forward and reverse reactions of the process in Figure 14.17. In which direction would the rate be larger? Why?

Any particular methyl isonitrile molecule acquires sufficient energy to overcome the energy barrier through collisions with other molecules. Recall from the kinetic-molecular theory of gases that, at any instant, gas molecules are distributed in energy over a wide range. (Section 10.7) ▼ Figure 14.18 shows the distribution of kinetic energies for two temperatures, comparing them with the minimum energy needed for reaction, Ea. At the higher temperature a much greater fraction of the molecules have kinetic energy greater than Ea, which leads to a greater rate of reaction.

Give It Some ThoughtSuppose we have two reactions, A ¡ B and B ¡ C. You can isolate B, and it is stable. Is B the transition state for the reaction A ¡ C?

For a collection of molecules in the gas phase, the fraction of molecules that have kinetic energy equal to or greater than Ea is given by the expression

f = e-Ea>RT [14.18]

In this equation, R is the gas constant 18.314 J>mol@K2 and T is the absolute temperature. To get an idea of the magnitude of f, let’s sup-pose that Ea is 100 kJ>mol, a value typical of many reactions, and that T is 300 K. The calculated value of f is 3.9 * 10-18, an extremely small number! At 320 K, f = 4.7 * 10-17. Thus, only a 20° increase in tem-perature produces a more than tenfold increase in the fraction of mol-ecules possessing at least 100 kJ>mol of energy.

The Arrhenius EquationArrhenius noted that for most reactions the increase in rate with increasing temperature is nonlinear (Figure 14.14). He found that most reaction-rate data obeyed an equation based on (a) the frac-

tion of molecules possessing energy Ea or greater, (b) the number of collisions per second, and (c) the fraction of collisions that have the appropriate orientation. These three factors are incorporated into the Arrhenius equation:

k = Ae-Ea>RT [14.19]

In this equation, k is the rate constant, Ea is the activation energy, R is the gas con-stant 18.314 J>mol@K2, and T is the absolute temperature. The frequency factor, A, is constant, or nearly so, as temperature is varied. This factor is related to the frequency of collisions and the probability that the collisions are favorably oriented for reac-tion.* As the magnitude of Ea increases, k decreases because the fraction of molecules that possess the required energy is smaller. Thus, at fixed values of T and A, reaction rates decrease as Ea increases.

▲ Figure 14.18 The effect of temperature on the distribution of kinetic energies of molecules in a sample.

Go FiGuReWhat would the curve look like for a temperature higher than that for the red curve in the figure?

Kinetic energy

Minimum energyneeded for reaction, Ea

Lower T

Frac

tion

of m

olec

ules

Higher T

Larger fraction of molecules reacts at higher temperature

*Because collision frequency increases with temperature, A also has some temperature dependence, but this dependence is much smaller than the exponential term. Therefore, A is considered approximately constant.


Determining the Activation EnergyWe can calculate the activation energy for a reaction by manipulating the Arrhenius equation. Taking the natural log of both sides of Equation 14.19, we obtain

ln k = −Ea–––RT

=

+

+

ln A

mx by

[14.20]

which has the form of the equation for a straight line. A graph of ln k versus 1>T is a line with a slope equal to -Ea>R and a y-intercept equal to ln A. Thus, the activation energy can be determined by measuring k at a series of temperatures, graphing ln k versus 1>T, and calculating Ea from the slope of the resultant line.

We can also use Equation 14.20 to evaluate Ea in a nongraphical way if we know the rate constant of a reaction at two or more temperatures. For example, suppose that at two different temperatures T1 and T2 a reaction has rate constants k1 and k2. For each condition, we have

ln k1 = -Ea

RT1+ ln A and ln k2 = -

Ea

RT2+ ln A

Subtracting ln k2 from ln k1 gives

ln k1 - ln k2 = a- Ea

RT1+ ln Ab - a-

Ea

RT2+ ln Ab

solutionThe lower the activation energy, the faster the reaction. The value of ∆E does not affect the rate. Hence, the order from slowest reaction to fastest is 2 6 3 6 1.

Practice exercise 1Which of the following statements is or are true? (i) The activation energies for the forward and reverse directions

of a reaction can be different. (ii) Assuming that A is constant, if both Ea and T increase, then k

will increase.

samPle exeRCise 14.10 activation energies and speeds of Reaction

Consider a series of reactions having these energy profiles:

(iii) For two different reactions, the one with the smaller value of Ea will necessarily have the larger value for k.

(a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true.

Practice exercise 2Rank the reverse reactions from slowest to fastest.

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Reaction progress

15 kJ/mol

−10 kJ/mol

Reaction progress

25 kJ/mol

−15 kJ/mol

Reaction progress

20 kJ/mol

5 kJ/mol

1 2 3

Rank the reactions from slowest to fastest assuming that they have nearly the same value for the frequency factor A.


Simplifying this equation and rearranging gives

ln k1

k2=

Ea

Ra 1

T2-

1T1b [14.21]

Equation 14.21 provides a convenient way to calculate a rate constant k1 at some tem-perature T1 when we know the activation energy and the rate constant k2 at some other temperature T2.

Solve

(a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each tem-perature, 1>T, and the natural log of each rate constant, ln k. This gives us the table shown at the right:

A graph of ln k versus 1>T is a straight line (▼ Figure 14.19).

The slope of the line is obtained by choosing any two well-separated points and using the coordinates of each:

Slope =∆y∆x

=-6.6 - 1-10.42

0.00195 - 0.00215= -1.9 * 104

Because logarithms have no units, the numerator in this equa-tion is dimensionless. The denominator has the units of 1>T, namely, K - 1. Thus, the overall units for the slope are K. The slope equals -Ea>R. We use the value for the gas constant R in units of J>mol@K (Table 10.2). We thus obtain

Slope = - Ea

R Ea = -1slope21R2 = -1-1.9 * 104 K2a8.314

Jmol@K

b a 1 kJ1000 J

b = 1.6 * 102 kJ>mol = 160 kJ>mol

We report the activation energy to only two significant figures because we are limited by the precision with which we can read the graph in Figure 14.19.

samPle exeRCise 14.11 determining the activation energy

The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures (these are the data points in Figure 14.14):

temperature 1 °C 2 k 1s−1 2189.7 2.52 * 10-5

198.9 5.25 * 10-5

230.3 6.30 * 10-4

251.2 3.16 * 10-3

(a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at 430.0 K?

solutionanalyze We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, Ea, and the rate constant, k, at a particular temperature.plan We can obtain Ea from the slope of a graph of ln k versus 1>T. Once we know Ea, we can use Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K.

T 1K 2 1 ,T 1K −1 2 ln k

462.9 2.160 * 10-3 -10.589

472.1 2.118 * 10-3 -9.855

503.5 1.986 * 10-3 -7.370

524.4 1.907 * 10-3 -5.757

▲ Figure 14.19 Graphical determination of activation energy Ea.

0.0019 0.0020 0.0021 0.0022−11

−10

−9

−8

−7

−6

−5

1/T (K−1)

ln k

Slope =−Ea

R

secTIon 14.6 reaction Mechanisms 599

14.6 | Reaction mechanismsA balanced equation for a chemical reaction indicates the substances present at the start of the reaction and those present at the end of the reaction. It provides no informa-tion, however, about the detailed steps that occur at the molecular level as the reactants are turned into products. The steps by which a reaction occurs is called the reaction mechanism. At the most sophisticated level, a reaction mechanism describes the order in which bonds are broken and formed and the changes in relative positions of the at-oms in the course of the reaction.

Elementary ReactionsWe have seen that reactions take place because of collisions between reacting mole-cules. For example, the collisions between molecules of methyl isonitrile 1CH3NC2 can provide the energy to allow the CH3NC to rearrange to acetonitrile:

H3C H3CH3CN NN

C CC

Similarly, the reaction of NO and O3 to form NO2 and O2 appears to occur as a result of a single collision involving suitably oriented and sufficiently energetic NO and O3 molecules:

NO1g2 + O31g2 ¡ NO21g2 + O21g2 [14.22]

Both reactions occur in a single event or step and are called elementary reactions.The number of molecules that participate as reactants in an elementary reaction

defines the molecularity of the reaction. If a single molecule is involved, the reaction is unimolecular. The rearrangement of methyl isonitrile is a unimolecular process. Ele-mentary reactions involving the collision of two reactant molecules are bimolecular. The reaction between NO and O3 is bimolecular. Elementary reactions involving the simultaneous collision of three molecules are termolecular. Termolecular reactions are far less probable than unimolecular or bimolecular processes and are extremely rare. The chance that four or more molecules will collide simultaneously with any regularity is even more remote; consequently, such collisions are never proposed as part of a reac-tion mechanism. Thus, nearly all reaction mechanisms contain only unimolecular and bimolecular elementary reactions.

(b) To determine the rate constant, k1, at T1 = 430.0 K, we can use Equation 14.21 with Ea = 160 kJ>mol and one of the rate constants and temperatures from the given data, such as k2 = 2.52 * 10-5 s-1 and T2 = 462.9 K:

ln a k1

2.52 * 10 - 5 s- 1 b =

a 160 kJ>mol8.314 J>mol@K

b a 1462.9 K

-1

430.0 Kb a 1000 J

1 kJb = -3.18

Thus,k1

2.52 * 10 - 5 s- 1 = e - 3.18 = 4.15 * 10 - 2

k1 = 14.15 * 10 - 2212.52 * 10 - 5 s- 12 = 1.0 * 10 - 6 s- 1

Note that the units of k1 are the same as those of k2.

Practice exercise 1Using the data in Sample Exercise 14.11, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C?(a) 8.1 * 10-15 s-1 (b) 2.2 * 10-13 s-1 (c) 2.7 * 10-9 s-1 (d) 2.3 * 10-1 s-1 (e) 9.2 * 103 s-1

Practice exercise 2To one significant figure, what is the value for the frequency factor A for the data presented in Sample Exercise 14.11.


Give It Some ThoughtWhat is the molecularity of the elementary reaction?

NO1g2 + Cl21g2 ¡ NOCl1g2 + Cl1g2

Multistep MechanismsThe net change represented by a balanced chemical equation often occurs by a multi-step mechanism consisting of a sequence of elementary reactions. For example, below 225 °C, the reaction

NO21g2 + CO1g2 ¡ NO1g2 + CO21g2 [14.23]

appears to proceed in two elementary reactions (or two elementary steps), each of which is bimolecular. First, two NO2 molecules collide, and an oxygen atom is transferred from one to the other. The resultant NO3 then collides with a CO molecule and trans-fers an oxygen atom to it:

NO21g2 + NO21g2 ¡ NO31g2 + NO1g2NO31g2 + CO1g2 ¡ NO21g2 + CO21g2

Thus, we say that the reaction occurs by a two-step mechanism.The chemical equations for the elementary reactions in a multistep mech-

anism must always add to give the chemical equation of the overall process. In the present example, the sum of the two elementary reactions is

2 NO21g2 + NO31g2 + CO1g2 ¡ NO21g2 + NO31g2 + NO1g2 + CO21g2Simplifying this equation by eliminating substances that appear on both sides gives Equation 14.23, the net equation for the process.

Because NO3 is neither a reactant nor a product of the reaction—it is formed in one elementary reaction and consumed in the next—it is called an intermediate. Multistep mechanisms involve one or more intermediates. Intermediates are not the same as transition states, as shown in ◀ Figure 14.20. Intermediates can be stable and can therefore sometimes be identified and even

isolated. Transition states, on the other hand, are always inherently unstable and as such can never be isolated. Nevertheless, the use of advanced “ultrafast” techniques sometimes allows us to characterize them.

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Reactants

Intermediate

Products

Transitionstate

Transitionstate

▲ Figure 14.20 The energy profile of a reaction, showing transition states and an intermediate.

Go FiGuReFor this profile, is it easier for a molecule of the intermediate to convert to reactants or products?

solutionanalyze We are given a two-step mechanism and asked for (a) the molecularities of each of the two elementary reactions, (b) the equa-tion for the overall process, and (c) the intermediate.plan The molecularity of each elementary reaction depends on the number of reactant molecules in the equation for that reaction. The overall equation is the sum of the equations for the elementary re-actions. The intermediate is a substance formed in one step of the

samPle exeRCise 14.12 determining molecularity and identifying intermediates

It has been proposed that the conversion of ozone into O2 proceeds by a two-step mechanism:

O31g2 ¡ O21g2 + O1g2O31g2 + O1g2 ¡ 2 O21g2

(a) Describe the molecularity of each elementary reaction in this mechanism.(b) Write the equation for the overall reaction. (c) Identify the intermediate(s).

mechanism and used in another and therefore not part of the equa-tion for the overall reaction.

Solve

(a) The first elementary reaction involves a single reactant and is consequently unimolecular. The second reaction, which involves two reactant molecules, is bimolecular.


Rate Laws for Elementary ReactionsIn Section 14.3, we stressed that rate laws must be determined experimentally; they cannot be predicted from the coefficients of balanced chemical equations. We are now in a position to understand why this is so. Every reaction is made up of a series of one or more elemen-tary steps, and the rate laws and relative speeds of these steps dictate the overall rate law for the reaction. Indeed, the rate law for a reaction can be determined from its mechanism, as we will see shortly, and compared with the experimental rate law. Thus, our next challenge in kinetics is to arrive at reaction mechanisms that lead to rate laws consistent with those observed experimentally. We start by examining the rate laws of elementary reactions.

Elementary reactions are significant in a very important way: If a reaction is ele-mentary, its rate law is based directly on its molecularity. For example, consider the uni-molecular reaction

A ¡ products

As the number of A molecules increases, the number that reacts in a given time interval increases proportionally. Thus, the rate of a unimolecular process is first order:

Rate = k3A4For bimolecular elementary steps, the rate law is second order, as in the reaction

A + B ¡ products Rate = k3A43B4The second-order rate law follows directly from collision theory. If we double the con-centration of A, the number of collisions between the molecules of A and B doubles; likewise, if we double [B], the number of collisions between A and B doubles. There-fore, the rate law is first order in both [A] and [B] and second order overall.

The rate laws for all feasible elementary reactions are given in ▼ table 14.3. Notice how each rate law follows directly from the molecularity of the reaction. It is important

(b) Adding the two elementary reactions gives2 O31g2 + O1g2 ¡ 3 O21g2 + O1g2

Because O(g) appears in equal amounts on both sides of the equation, it can be eliminated to give the net equation for the chemical process:

2 O31g2 ¡ 3 O21g2(c) The intermediate is O(g). It is neither an original reactant nor a

final product but is formed in the first step of the mechanism and consumed in the second.

Practice exercise 1Consider the following two-step reaction mechanism:

A1g2 + B1g2 ¡ X1g2 + Y1g2X1g2 + C1g2 ¡ Y1g2 + Z1g2

Which of the following statements about this mechanism is or are true? (i) Both of the steps in this mechanism are bimolecular.(ii) The overall reaction is A1g2 + B1g2 + C1g2 ¡ Y1g2 + Z1g2.

(iii) The substance X(g) is an intermediate in this mechanism.

(a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true.

Practice exercise 2For the reaction

Mo1CO26 + P1CH323 ¡ Mo1CO25P1CH323 + COthe proposed mechanism is

Mo1CO26 ¡ Mo1CO25 + COMo1CO25 + P1CH323 ¡ Mo1CO25P1CH323

(a) Is the proposed mechanism consistent with the equation for the overall reaction? (b) What is the molecularity of each step of the mechanism? (c) Identify the intermediate(s).

table 14.3 elementary Reactions and their Rate laws

molecularity elementary Reaction Rate law

Unimolecular A ¡ products Rate = k3A4Bimolecular A + A ¡ products Rate = k3A42

Bimolecular A + B ¡ products Rate = k3A43B4Termolecular A + A + A ¡ products Rate = k3A43

Termolecular A + A + B ¡ products Rate = k3A423B4Termolecular A + B + C ¡ products Rate = k3A43B43C4


to remember, however, that we cannot tell by merely looking at a balanced, overall chemical equation whether the reaction involves one or several elementary steps.

solutionanalyze We are given the equation and asked for its rate law, assum-ing that it is an elementary process.plan Because we are assuming that the reaction occurs as a single elementary reaction, we are able to write the rate law using the coef-ficients for the reactants in the equation as the reaction orders.Solve The reaction is bimolecular, involving one molecule of H2 and one molecule of Br2. Thus, the rate law is first order in each reactant and second order overall:

Rate = k3H243Br24comment Experimental studies of this reaction show that the reaction actually has a very different rate law:

Rate = k3H243Br241>2

Because the experimental rate law differs from the one obtained by as-suming a single elementary reaction, we can conclude that the mecha-nism cannot occur by a single elementary step. It must, therefore, involve two or more elementary steps.

samPle exeRCise 14.13 Predicting the Rate law for an elementary Reaction

If the following reaction occurs in a single elementary reaction, predict its rate law:

H21g2 + Br21g2 ¡ 2 HBr1g2

Practice exercise 1Consider the following reaction: 2 A + B ¡ X + 2 Y. You are told that the first step in the mechanism of this reaction has the following rate law: Rate = k3A43B4. Which of the following could be the first step in the reaction mechanism (note that substance Z is an intermediate)?(a) A + A ¡ Y + Z(b) A ¡ X + Z(c) A + A + B ¡ X + Y + Y(d) B ¡ X + Y(e) A + B ¡ X + Z

Practice exercise 2Consider the following reaction: 2 NO1g2 + Br21g2 ¡ 2 NOBr1g2.(a) Write the rate law for the reaction, assuming it involves a single elementary reaction. (b) Is a single-step mechanism likely for this reaction?

The Rate-Determining Step for a Multistep MechanismAs with the reaction in Sample Exercise 14.13, most reactions occur by mechanisms that involve two or more elementary reactions. Each step of the mechanism has its own rate constant and activation energy. Often one step is much slower than the others, and the overall rate of a reaction cannot exceed the rate of the slowest elementary step. Be-cause the slow step limits the overall reaction rate, it is called the rate-determining step (or rate-limiting step).

To understand the concept of the rate-determining step for a reaction, consider a toll road with two toll plazas (▶ Figure 14.21). Cars enter the toll road at point 1 and pass through toll plaza A. They then pass an intermediate point 2 before passing through toll plaza B and arriving at point 3. We can envision this trip along the toll road as occurring in two elementary steps:

Step 1:Step 2:

Overall:

Point 1 ¡ Point 2Point 2 ¡ Point 3Point 1 ¡ Point 3

1through toll plaza A21through toll plaza B21through both toll plazas2

Now suppose that one or more gates at toll plaza A are malfunctioning, so that traffic backs up behind the gates, as depicted in Figure 14.21(a). The rate at which cars can get to point 3 is limited by the rate at which they can get through the traffic jam at plaza A. Thus, step 1 is the rate-determining step of the journey along the toll road. If, however, all gates at A are functioning but one or more at B are not, traffic flows quickly through A but gets backed up at B, as depicted in Figure 14.21(b). In this case, step 2 is the rate-determining step.

In the same way, the slowest step in a multistep reaction determines the overall rate. By analogy to Figure 14.21(a), the rate of a fast step following the rate-determining step does not speed up the overall rate. If the slow step is not the first one, as is the case in


(a) Cars slowed at toll plaza A, rate-determining step is passage through A

(b) Cars slowed at toll plaza B, rate-determining step is passage through B

Toll plaza A Toll plaza B

Toll plaza A Toll plaza B

1 2 3

1 2 3

Figure 14.21(b), the faster preceding steps produce intermediate products that accu-mulate before being consumed in the slow step. In either case, the rate-determining step governs the rate law for the overall reaction.

Give It Some ThoughtWhy can’t the rate law for a reaction generally be deduced from the balanced equation for the reaction?

Mechanisms with a Slow Initial StepWe can most easily see the relationship between the slow step in a mechanism and the rate law for the overall reaction by considering an example in which the first step in a multistep mechanism is the rate-determining step. Consider the reaction of NO2 and CO to produce NO and CO2 (Equation 14.23). Below 225 °C, it is found experi-mentally that the rate law for this reaction is second order in NO2 and zero order in CO: Rate = k3NO242. Can we propose a reaction mechanism consistent with this rate law? Consider the two-step mechanism:*

Step 1: NO21g2 + NO21g2 ¡k1 NO31g2 + NO1g2 1slow2Step 2: NO31g2 + CO1g2 ¡k2 NO21g2 + CO21g2 1fast2Overall: NO21g2 + CO1g2 ¡ NO1g2 + CO21g2

Step 2 is much faster than step 1; that is, k2 77 k1, telling us that the intermediate NO31g2 is slowly produced in step 1 and immediately consumed in step 2.

Because step 1 is slow and step 2 is fast, step 1 is the rate-determining step. Thus, the rate of the overall reaction depends on the rate of step 1, and the rate law of the

▲ Figure 14.21 Rate-determining steps in traffic flow on a toll road.

Go FiGuReFor which of the two scenarios in the figure will one get from point 1 to point 3 most rapidly?

*Note the rate constants k1 and k2 written above the reaction arrows. The subscript on each rate constant identifies the elementary step involved. Thus, k1 is the rate constant for step 1, and k2 is the rate constant for step 2. A negative subscript refers to the rate constant for the reverse of an elementary step. For example, k - 1 is the rate constant for the reverse of the first step.


overall reaction equals the rate law of step 1. Step 1 is a bimolecular process that has the rate law

Rate = k13NO242

Thus, the rate law predicted by this mechanism agrees with the one observed experi-mentally. The reactant CO is absent from the rate law because it reacts in a step that follows the rate-determining step.

A scientist would not, at this point, say that we have “proved” that this mecha-nism is correct. All we can say is that the rate law predicted by the mechanism is consistent with experiment. We can often envision a different sequence of steps that leads to the same rate law. If, however, the predicted rate law of the proposed mecha-nism disagrees with experiment, we know for certain that the mechanism cannot be correct.

samPle exeRCise 14.14 determining the Rate law for a multistep mechanism

The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism:

N2O1g2 ¡ N21g2 + O1g2 1slow2N2O1g2 + O1g2 ¡ N21g2 + O21g2 1fast2

(a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction.

are told that the rate of this reaction is second order overall and second order in [C]. Could any of the following be a rate-deter-mining first step in a reaction mechanism that is consistent with the observed rate law for the reaction (note that substance Z is an intermediate)?(a) C + C ¡ K + Z (b) C ¡ J + Z (c) C + D ¡ J + Z(d) D ¡ J + K (e) None of these are consistent with the observed rate law.

Practice exercise 2Ozone reacts with nitrogen dioxide to produce dinitrogen pentox-ide and oxygen:

O31g2 + 2 NO21g2 ¡ N2O51g2 + O21g2The reaction is believed to occur in two steps:

O31g2 + NO21g2 ¡ NO31g2 + O21g2NO31g2 + NO21g2 ¡ N2O51g2

The experimental rate law is rate = k3O343NO24. What can you say about the relative rates of the two steps of the mechanism?

solutionanalyze Given a multistep mechanism with the relative speeds of the steps, we are asked to write the overall reaction and the rate law for that overall reaction.plan (a) Find the overall reaction by adding the elementary steps and eliminating the intermediates. (b) The rate law for the overall reaction will be that of the slow, rate-determining step.Solve

(a) Adding the two elementary reactions gives2 N2O1g2 + O1g2 ¡ 2 N21g2 + 2 O21g2 + O1g2

Omitting the intermediate, O(g), which occurs on both sides of the equation, gives the overall reaction:

2 N2O1g2 ¡ 2 N21g2 + O21g2(b) The rate law for the overall reaction is just the rate law for the

slow, rate-determining elementary reaction. Because that slow step is a unimolecular elementary reaction, the rate law is first order:

Rate = k3N2O4

Practice exercise 1Let’s consider a hypothetical reaction similar to that in Practice Exercise 1 of Sample Exercise 14.13: 2 C + D ¡ J + 2 K. You

Mechanisms with a Fast Initial StepIt is possible, but not particularly straightforward, to derive the rate law for a mecha-nism in which an intermediate is a reactant in the rate-determining step. This situation arises in multistep mechanisms when the first step is fast and therefore not the rate-determining step. Let’s consider one example: the gas-phase reaction of nitric oxide (NO) with bromine 1Br22:

2 NO1g2 + Br21g2 ¡ 2 NOBr1g2 [14.24]


The experimentally determined rate law for this reaction is second order in NO and first order in Br2:

Rate = k3NO423Br24 [14.25]

We seek a reaction mechanism that is consistent with this rate law. One possibility is that the reaction occurs in a single termolecular step:

NO1g2 + NO1g2 + Br21g2 ¡ 2 NOBr1g2 Rate = k3NO423Br24 [14.26]

As noted in Practice Exercise 2 of Exercise 14.13, this does not seem likely because ter-molecular processes are so rare.

Give It Some ThoughtWhy are termolecular elementary steps rare in gas-phase reactions?

Let’s consider an alternative mechanism that does not involve a termolecular step:

Step 1: NO1g2 + Br21g2 ∆k1

k - 1NOBr21g2 1fast2

Step 2: NOBr21g2 + NO1g2 ¡k2 2 NOBr1g2 1slow2 [14.27]

In this mechanism, step 1 involves two processes: a forward reaction and its reverse.Because step 2 is the rate-determining step, the rate law for that step governs the

rate of the overall reaction:

Rate = k23NOBr243NO4 [14.28]

Note that NOBr2 is an intermediate generated in the forward reaction of step 1. Inter-mediates are usually unstable and have a low, unknown concentration. Thus, the rate law of Equation 14.28 depends on the unknown concentration of an intermediate, which isn’t desirable. We want instead to express the rate law for a reaction in terms of the reactants, or the products if necessary, of the reaction.

With the aid of some assumptions, we can express the concentration of the inter-mediate NOBr2 in terms of the concentrations of the starting reactants NO and Br2. We first assume that NOBr2 is unstable and does not accumulate to any significant extent in the reaction mixture. Once formed, NOBr2 can be consumed either by reacting with NO to form NOBr or by falling back apart into NO and Br2. The first of these possibili-ties is step 2 of our alternative mechanism, a slow process. The second is the reverse of step 1, a unimolecular process:

NOBr21g2 ¡k-1 NO1g2 + Br21g2Because step 2 is slow, we assume that most of the NOBr2 falls apart according to this reaction. Thus, we have both the forward and reverse reactions of step 1 occurring much faster than step 2. Because they occur rapidly relative to step 2, the forward and reverse reactions of step 1 establish an equilibrium. As in any other dynamic equilib-rium, the rate of the forward reaction equals that of the reverse reaction:

k3NO43Br24 = k-13NOBr24

Rate of forward reaction Rate of reverse reaction

Solving for 3NOBr24, we have

3NOBr24 =k1

k-1 3NO43Br24

Substituting this relationship into Equation 14.28, we have

Rate = k2k1

k-13NO43Br243NO4 = k3NO423Br24

where the experimental rate constant k equals k2k1>k-1. This expression is consistent with the experimental rate law (Equation 14.25). Thus, our alternative mechanism


(Equation 14.27), which involves two steps but only unimolecular and bimolecu-lar processes, is far more probable than the single-step termolecular mechanism of Equation 14.26.

In general, whenever a fast step precedes a slow one, we can solve for the concentra-tion of an intermediate by assuming that an equilibrium is established in the fast step.

solutionanalyze We are given a mechanism with a fast initial step and asked to write the rate law for the overall reaction.plan The rate law of the slow elementary step in a mechanism deter-mines the rate law for the overall reaction. Thus, we first write the rate law based on the molecularity of the slow step. In this case, the slow step involves the intermediate N2O2 as a reactant. Experimental rate laws, however, do not contain the concentrations of intermediates; instead they are expressed in terms of the concentrations of starting substances. Thus, we must relate the concentration of N2O2 to the concentration of NO by assuming that an equilibrium is established in the first step.Solve The second step is rate determining, so the overall rate is

Rate = k23N2O243Br24We solve for the concentration of the intermediate N2O2 by assuming that an equilibrium is established in step 1; thus, the rates of the for-ward and reverse reactions in step 1 are equal:

k13NO42 = k-13N2O24Solving for the concentration of the intermediate, N2O2, gives

3N2O24 =k1

k-13NO42

Substituting this expression into the rate expression gives

Rate = k2k1

k-13NO423Br24 = k3NO423Br24

samPle exeRCise 14.15 deriving the Rate law for a mechanism with a Fast initial step

Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the experimentally observed one:

Step 1: NO1g2 + NO1g2 ∆k1

k -1N2O21g2 1fast, equilibrium2

Step 2: N2O21g2 + Br21g2 ¡k2 2 NOBr1g2 1slow2

Thus, this mechanism also yields a rate law consistent with the experi-mental one. Remember: There may be more than one mechanism that leads to an observed experimental rate law!

Practice exercise 1Consider the following hypothetical reaction: 2 P + Q ¡ 2 R + S. The following mechanism is proposed for this reaction:

P + P ∆ T (fast)

Q + T ¡ R + U 1slow2

U ¡ R + S 1fast2Substances T and U are unstable intermediates. What rate law is predicted by this mechanism?(a) Rate = k3P42 (b) Rate = k3P43Q4 (c) Rate = k3P423Q4(d) Rate = k3P43Q42 (e) Rate = k3U4Practice exercise 2The first step of a mechanism involving the reaction of bromine is

Br21g2 ∆k1

k-12 Br1g2 1fast, equilibrium2

What is the expression relating the concentration of Br(g) to that of Br21g2?

So far we have considered only three reaction mechanisms: one for a reaction that occurs in a single elementary step and two for simple multistep reactions where there is one rate-determining step. There are other more complex mechanisms, however. If you take a biochemistry class, for example, you will learn about cases in which the concen-tration of an intermediate cannot be neglected in deriving the rate law. Furthermore, some mechanisms require a large number of steps, sometimes 35 or more, to arrive at a rate law that agrees with experimental data!

14.7 | CatalysisA catalyst is a substance that changes the speed of a chemical reaction without under-going a permanent chemical change itself. Most reactions in the body, the atmosphere, and the oceans occur with the help of catalysts. Much industrial chemical research is

secTIon 14.7 catalysis 607

devoted to the search for more effective catalysts for reactions of commercial impor-tance. Extensive research efforts also are devoted to finding means of inhibiting or re-moving certain catalysts that promote undesirable reactions, such as those that corrode metals, age our bodies, and cause tooth decay.

Homogeneous CatalysisA catalyst that is present in the same phase as the reactants in a reaction mixture is called a homogeneous catalyst. Examples abound both in solution and in the gas phase. Consider, for example, the decomposition of aqueous hydrogen peroxide, H2O21aq2, into water and oxygen:

2 H2O21aq2 ¡ 2 H2O1l2 + O21g2 [14.29]

In the absence of a catalyst, this reaction occurs extremely slowly. Many substances are capa-ble of catalyzing the reaction, however, including bromide ion, which reacts with hydrogen peroxide in acidic solution, forming aqueous bromine and water (▼ Figure 14.22).

2 Br-1aq2 + H2O21aq2 + 2 H+ ¡ Br21aq2 + 2 H2O1l2 [14.30]

2 Br−(aq) + H2O2(aq) + 2 H+(aq)

Br2(aq) + 2 H2O(l)

brown

Br2(aq) + H2O2(aq)

2 Br−(aq) + 2 H+(aq) + O2(g)

colorless bubbles

NaBr catalyst about to be added to reaction mixture

H2O2(aq) in acidic solution

▲ Figure 14.22 Homogeneous catalysis. Effect of catalyst on the speed of hydrogen peroxide decomposition to water and oxygen gas.

Go FiGuReWhy does the solution in the middle cylinder have a brownish color?


If this were the complete reaction, bromide ion would not be a catalyst because it undergoes chemical change during the reaction. However, hydrogen peroxide also reacts with the Br21aq2 generated in Equation 14.30:

Br21aq2 + H2O21aq2 ¡ 2 Br - 1aq2 + 2 H+1aq2 + O21g2 [14.31]

The sum of Equations 14.30 and 14.31 is just Equation 14.29, a result which you can check for yourself.

When the H2O2 has been completely decomposed, we are left with a colorless solution of Br - 1aq2, which means that this ion is indeed a catalyst of the reac-tion because it speeds up the reaction without itself undergoing any net change. In contrast, Br2 is an intermediate because it is first formed (Equation 14.30) and then consumed (Equation 14.31). Neither the catalyst nor the intermediate appears in the equation for the overall reaction. Notice, however, that the catalyst is there at the start of the reaction, whereas the intermediate is formed during the course of the reaction.

How does a catalyst work? If we think about the general form of rate laws (Equation 14.7, rate = k3A4m3B4n), we must conclude that the catalyst must affect the numerical value of k, the rate constant. On the basis of the Arrhenius equation (Equation 14.19, k = Ae-Ea>RT), k is determined by the activa-tion energy 1Ea2 and the frequency factor (A). A catalyst may affect the rate of reaction by altering the value of either Ea or A. We can envision this happening in two ways: The catalyst could provide a new mechanism for the reaction that has an Ea value lower than the Ea value for the uncatalyzed reaction, or the cat-alyst could assist in the orientation of reactants and so increase A. The most dramatic catalytic effects come from lowering Ea. As a general rule, a catalyst lowers the overall activation energy for a chemical reaction.

A catalyst can lower the activation energy for a reaction by providing a different mechanism for the reaction. In the decomposition of hydrogen peroxide, for example, two succes-

sive reactions of H2O2, first with bromide and then with bromine, take place. Because these two reactions together serve as a catalytic pathway for hydrogen peroxide decom-position, both of them must have significantly lower activation energies than the uncat-alyzed decomposition (◀ Figure 14.23).

Heterogeneous CatalysisA heterogeneous catalyst is one that exists in a phase different from the phase of the reactant molecules, usually as a solid in contact with either gaseous reactants or reac-tants in a liquid solution. Many industrially important reactions are catalyzed by the surfaces of solids. For example, raw petroleum is transformed into smaller hydrocar-bon molecules by using what are called “cracking” catalysts. Heterogeneous catalysts are often composed of metals or metal oxides.

The initial step in heterogeneous catalysis is usually adsorption of reactants. Adsorption refers to the binding of molecules to a surface, whereas absorption refers to the uptake of molecules into the interior of a substance. (Section 13.6) Adsorp-tion occurs because the atoms or ions at the surface of a solid are extremely reactive. Because the catalyzed reaction occurs on the surface, special methods are often used to prepare catalysts so that they have very large surface areas. Unlike their counter-parts in the interior of the substance, surface atoms and ions have unused bonding capacity that can be used to bond molecules from the gas or solution phase to the surface of the solid.

Reaction progress

2 H2O2

2 H2O + O2

2 H2O + O2 +2 Br− + 2 H+

2 H2O2 + 2 Br− + 2 H+

Uncatalyzed reactionCatalyzedreaction

Pote

ntia

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rgy

▲ Figure 14.23 Energy profiles for the uncatalyzed and bromide-catalyzed decomposition of h2o2.

Go FiGuReWhere are the intermediates and transition states in this diagram?


The reaction of hydrogen gas with ethylene gas to form ethane gas provides an example of heterogeneous catalysis:

C2H41g2 + H21g2 ¡ C2H61g2 ∆H ° = -137 kJ>mol [14.32]Ethylene Ethane

Even though this reaction is exothermic, it occurs very slowly in the absence of a cat-alyst. In the presence of a finely powdered metal, however, such as nickel, palladium, or platinum, the reaction occurs easily at room temperature via the mechanism diagrammed in ▲ Figure 14.24. Both ethylene and hydrogen are adsorbed on the metal surface. Upon adsorption, the H ¬ H bond of H2 breaks, leaving two H atoms initially bonded to the metal surface but relatively free to move. When a hydrogen encounters an adsorbed ethylene molecule, it can form a s bond to one of the carbon atoms, effectively destroying the C ¬ C p bond and leaving an ethyl group 1C2H52 bonded to the surface via a metal-to-carbon s bond. This s bond is relatively weak, so when the other carbon atom also encounters a hydrogen atom, a sixth C ¬ H s bond is readily formed, and an ethane molecule 1C2H62 is released from the metal surface.

Give It Some ThoughtHow does a homogeneous catalyst compare with a heterogeneous one regarding the ease of recovery of the catalyst from the reaction mixture?

EnzymesThe human body is characterized by an extremely complex system of interrelated chemical reactions, all of which must occur at carefully controlled rates to maintain life. A large number of marvelously efficient biological catalysts known as enzymes are necessary for many of these reactions to occur at suitable rates. Most enzymes

▲ Figure 14.24 Heterogeneous catalysis. Mechanism for reaction of ethylene with hydrogen on a catalytic surface.

Second free H about to attach to C2H5 intermediate to form C2H6

One free H has attached to C2H4 to form C2H5 (ethyl group) intermediate

Hydrogen

Carbon

H2 and C2H4 adsorbed on metal surface

1 After H—H bond breaks, H atoms migrate along metal surface

2

Ethane, C2H6, desorbs from metal surface

5

3

4


are large protein molecules with molecular weights ranging from about 10,000 to about 1 million amu. They are very selective in the reactions they catalyze, and some are absolutely specific, operating for only one sub-stance in only one reaction. The decomposition of hydrogen peroxide, for example, is an important biological process. Because hydrogen peroxide is strongly oxidizing, it can be physiologically harmful. For this reason, the blood and liver of mammals contain an enzyme, catalase, that catalyzes the decomposition of hydrogen peroxide into water and oxygen (Equation 14.29). ◀ Figure 14.26 shows the dramatic acceleration of this chemical reaction by the catalase in beef liver.

The reaction any given enzyme catalyzes takes place at a specific loca-tion in the enzyme called the active site. The substances that react at this site are called substrates. The lock-and-key model provides a simple expla-nation for the specificity of an enzyme (▶ Figure 14.27). The substrate is pictured as fitting neatly into the active site, much like a key fits into a lock.

Lysozyme is an enzyme that is important to the functioning of our immune system because it accelerates reactions that damage (or “lyse”) bacterial cell walls. ▶ Figure 14.28 shows a model of the enzyme lyso-zyme without and with a bound substrate molecule.


Catalytic Converters

Heterogeneous catalysis plays a major role in the fight against urban air pollution. Two components of automobile exhausts that help form pho-tochemical smog are nitrogen oxides and unburned hydrocarbons. In ad-dition, automobile exhaust may contain considerable quantities of carbon monoxide. Even with the most careful attention to engine design, it is im-possible under normal driving conditions to reduce the quantity of these pollutants to an acceptable level in the exhaust gases. It is therefore neces-sary to remove them from the exhaust before they are vented to the air. This removal is accomplished in the catalytic converter.

The catalytic converter, which is part of an automobile’s exhaust system, must perform two functions: (1) oxidation of CO and un-burned hydrocarbons 1CxHy2 to carbon dioxide and water, and (2) reduction of nitrogen oxides to nitrogen gas:

CO, CxHy ¡O2

CO2 + H2O

NO, NO2 ¡ N2

These two functions require different catalysts, so the development of a successful catalyst system is a difficult challenge. The catalysts must be effective over a wide range of operating temperatures. They must continue to be active despite the fact that various components of the exhaust can block the active sites of the catalyst. And the catalysts must be sufficiently rugged to withstand exhaust gas turbulence and the mechanical shocks of driving under various conditions for thousands of miles.

Catalysts that promote the combustion of CO and hydrocarbons are, in general, the transition-metal oxides and the noble metals. These mate-rials are supported on a structure (▶ Figure 14.25) that allows the best possible contact between the flowing exhaust gas and the catalyst surface. A honeycomb structure made from alumina 1Al2O32 and impregnated with the catalyst is employed. Such catalysts operate by first adsorbing ox-ygen gas present in the exhaust gas. This adsorption weakens the O ¬ O bond in O2, so that oxygen atoms are available for reaction with adsorbed CO to form CO2. Hydrocarbon oxidation probably proceeds somewhat similarly, with the hydrocarbons first being adsorbed followed by rupture of a C ¬ H bond.

Transition-metal oxides and noble metals are also the most effective catalysts for reduction of NO to N2 and O2. The catalysts that are most effective in one reaction, however, are usually much less effective in the other. It is therefore necessary to have two catalytic components.

Catalytic converters contain remarkably efficient heterogeneous catalysts. The automotive exhaust gases are in contact with the catalyst for only 100 to 400 ms, but in this very short time, 96% of the hydro-carbons and CO is converted to CO2 and H2O, and the emission of nitrogen oxides is reduced by 76%.

There are costs as well as benefits associated with the use of cata-lytic converters, one being that some of the metals are very expensive. Catalytic converters currently account for about 35% of the platinum, 65% of the palladium, and 95% of the rhodium used annually. All of these metals, which come mainly from Russia and South Africa, can be far more expensive than gold.

Related Exercises: 14.62, 14.81, 14.82, 14.124

Metal catalyst impregnated on high surface area alumina “honeycomb”

▲ Figure 14.25 Cross section of a catalytic converter.

Catalase present in beef liver rapidly converts H2O2 to water and O2

Ground beef liver

O2 gas

H2O2andH2O

▲ Figure 14.26 Enzymes speed up reactions.

Go FiGuReWhy is the reaction faster when the liver is ground up?


The combination of enzyme and substrate is called the enzyme–substrate complex. Although Figure 14.27 shows both the active site and its substrate as having a fixed shape, the active site is often fairly flexible and so may change shape as it binds the sub-strate. The binding between substrate and active site involves dipole–dipole attractions, hydrogen bonds, and dispersion forces. (Section 11.2)

As substrate molecules enter the active site, they are somehow activated so that they are capable of reacting rapidly. This activation process may occur, for example, by the withdrawal or donation of electron density from a particular bond or group of atoms in the enzyme’s active site. In addition, the substrate may become distorted in the process of fitting into the active site and made more reactive. Once the reaction occurs, the products depart from the active site, allowing another substrate molecule to enter.

The activity of an enzyme is destroyed if some molecule other than the substrate specific to that enzyme binds to the active site and blocks entry of the substrate. Such substances are called enzyme inhibitors. Nerve poisons and certain toxic metal ions, such as lead and mercury, are believed to act in this way to inhibit enzyme activity. Some other poisons act by attaching elsewhere on the enzyme, thereby distorting the active site so that the substrate no longer fits.

Enzymes are enormously more efficient than nonbiochemical catalysts. The num-ber of individual catalyzed reaction events occurring at a particular active site, called the turnover number, is generally in the range of 103 to 107 per second. Such large turn-over numbers correspond to very low activation energies. Compared with a simple chemical catalyst, enzymes can increase the rate constant for a given reaction by a mil-lionfold or more.

Give It Some ThoughtIs it reasonable to say that enzymes lower the energy of the transition state for a reaction?

▲ Figure 14.28 Lysozyme was one of the first enzymes for which a structure–function relationship was described. This model shows how the substrate (yellow) “fits” into the active site of the enzyme.

Enzyme Enzyme

ProductsSubstrate

Enzyme–substratecomplex

▲ Figure 14.27 Lock-and-key model for enzyme action.

Go FiGuReWhich molecules must bind more tightly to the active site, substrates or products?


Chemistry and life

Nitrogen Fixation and Nitrogenase

Nitrogen is one of the most essential elements in living organisms, found in many compounds vital to life, including proteins, nucleic acids, vitamins, and hormones. Nitrogen is continually cycling through the biosphere in

various forms, as shown in ◀ Figure 14.29. For example, certain mi-croorganisms convert the nitrogen in animal waste and dead plants and animals into N21g2, which then returns to the atmosphere. For the food chain to be sustained, there must be a means of converting atmospheric

N21g2 into a form plants can use. For this reason, if a chemist were asked to name the most important chemical reaction in the world,

she might easily say nitrogen fixation, the process by which at-mospheric N21g2 is converted into compounds suitable for

plant use. Some fixed nitrogen results from the action of lightning on the atmosphere, and some is produced

industrially using a process we will discuss in Chapter 15. About 60% of fixed nitrogen, however, is a consequence of the action of the remarkable and complex enzyme nitrogenase. This enzyme is not present in humans or other animals; rather, it is found in bacteria that live in the root nodules of certain plants, such as the legumes clover and alfalfa.

Nitrogenase converts N2 into NH3, a process that, in the absence of a catalyst, has a very large activation energy. This process is a reduction reaction in which the oxidation state of N is reduced from 0 in N2 to -3 in NH3. The mechanism by which nitrogenase reduces N2 is not fully understood. Like many other enzymes, including catalase, the active site of nitrogenase contains transition-metal atoms; such enzymes are called metalloen-zymes. Because transition metals can readily change oxidation state, metalloenzymes are

especially useful for effecting transformations in which substrates are either oxidized or reduced.It has been known for nearly 40 years that a portion of

nitrogenase contains iron and molybdenum atoms. This portion, called the FeMo-cofactor, is thought to serve as the active site of the

enzyme. The FeMo-cofactor of nitrogenase is a cluster of seven Fe atoms and one Mo atom, all linked by sulfur atoms (▼ Figure 14.30).

It is one of the wonders of life that sim-ple bacteria can contain beautifully complex and vitally important enzymes such as nitro-genase. Because of this enzyme, nitrogen is continually cycled between its comparatively inert role in the atmosphere and its critical role in living organisms. Without nitrogenase, life as we know it could not exist on Earth.


◀ Figure 14.30 The FeMo-cofactor of nitrogenase. Nitrogenase is found in nodules in the roots of certain plants, such as the white clover roots shown at the left. The cofactor, which is thought to be the active site of the enzyme, contains seven Fe atoms and one Mo atom, linked by sulfur atoms. The molecules on the outside of the cofactor connect it to the rest of the protein.

S

Fe

Mo

Nitrogen inatmosphere(N2)

Nitrogen�xationcaused bylightning

Groundwater Nitrate (NO3−)

Nitrite(NO2

−)

Ammonia(NH3)

Plant andanimal waste

▲ Figure 14.29 Simplified picture of the nitrogen cycle.


samPle inteGRative exeRCise Putting Concepts together

▲ Figure 14.31 Variation in pressure of HCOOH(g) as a function of time at 838 K.

0 500 1000 1500

75

150

225

300

Time (s)

Pres

sure

, HC

OO

H (t

orr)

HCOOH(g) alone

HCOOH(g)with ZnO(s)

Formic acid (HCOOH) decomposes in the gas phase at elevated temperatures as follows:HCOOH1g2 ¡ CO21g2 + H21g2

The uncatalyzed decomposition reaction is determined to be first order. A graph of the partial pressure of HCOOH versus time for decomposition at 838 K is shown as the red curve in ▶ Figure 14.31. When a small amount of solid ZnO is added to the reaction chamber, the partial pressure of acid versus time varies as shown by the blue curve in Figure 14.31.(a) Estimate the half-life and first-order rate constant for formic acid decomposition.(b) What can you conclude from the effect of added ZnO on the decomposition of formic acid?(c) The progress of the reaction was followed by measuring the partial pressure of formic acid

vapor at selected times. Suppose that, instead, we had plotted the concentration of formic acid in units of mol>L. What effect would this have had on the calculated value of k?

(d) The pressure of formic acid vapor at the start of the reaction is 3.00 * 102 torr. Assuming constant temperature and ideal-gas behavior, what is the pressure in the system at the end of the reaction? If the volume of the reaction chamber is 436 cm3, how many moles of gas occupy the reaction chamber at the end of the reaction?

(e) The standard heat of formation of formic acid vapor is ∆Hf° = -378.6 kJ>mol. Calculate ∆H° for the overall reaction. If the activation energy 1Ea2 for the reaction is 184 kJ>mol, sketch an approximate energy profile for the reaction, and label Ea, ∆H °, and the transition state.

solution(a) The initial pressure of HCOOH is 3.00 * 102 torr. On the graph we move to the level at which the

partial pressure of HCOOH is 1.50 * 102 torr, half the initial value. This corresponds to a time of about 6.60 * 102 s, which is therefore the half-life. The first-order rate constant is given by Equa-tion 14.15: k = 0.693>t1>2 = 0.693>660 s = 1.05 * 10-3 s-1.

(b) The reaction proceeds much more rapidly in the presence of solid ZnO, so the surface of the oxide must be acting as a catalyst for the decomposition of the acid. This is an example of heterogeneous catalysis.

(c) If we had graphed the concentration of formic acid in units of moles per liter, we would still have determined that the half-life for decomposition is 660 s, and we would have computed the same value for k. Because the units for k are s-1, the value for k is independent of the units used for concentration.

(d) According to the stoichiometry of the reaction, two moles of product are formed for each mole of reactant. When reaction is completed, therefore, the pressure will be 600 torr, just twice the initial pressure, assuming ideal-gas behavior. (Because we are working at quite high tempera-ture and fairly low gas pressure, assuming ideal-gas behavior is reasonable.) The number of moles of gas present can be calculated using the ideal-gas equation (Section 10.4):

n =PVRT

=1600>760 atm210.436 L2

10.08206 L@atm>mol@K21838 K2 = 5.00 * 10-3 mol

(e) We first calculate the overall change in energy, ∆H° (Section 5.7 and Appendix C), as in ∆H° = ∆Hf°1CO21g22 + ∆Hf°1H21g22 - ∆Hf°1HCOOH1g22

= -393.5 kJ>mol + 0 - 1-378.6 kJ>mol2 = -14.9 kJ>mol

From this and the given value for Ea, we can draw an approximate energy profile for the reaction, in analogy to Figure 14.17.

∆H°

Ea

Transition state

Reaction progress

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Chapter summary and Key termsINTROdUCTION TO KINETICS (INTROdUCTION ANd SECTION 14.1) Chemical kinetics is the area of chemistry in which reaction rates are studied. Factors that affect reaction rate are the physical state of the reactants; concentration; temperature; and the presence of catalysts.

REACTION RATES (SECTION 14.2) Reaction rates are usually expressed as changes in concentration per unit time: Typically, for reactions in solu-tion, rates are given in units of molarity per second 1M>s2. For most reac-tions, a plot of molarity versus time shows that the rate slows down as the reaction proceeds. The instantaneous rate is the slope of a line drawn tangent to the concentration-versus-time curve at a specific time. Rates can be written in terms of the appearance of products or the disappear-ance of reactants; the stoichiometry of the reaction dictates the relationship between rates of appearance and disappearance.

RATES ANd CONCENTRATION (SECTION 14.3) The quantitative relationship between rate and concentration is expressed by a rate law, which usually has the following form:

Rate = k3reactant 14m3reactant 24n cThe constant k in the rate law is called the rate constant; the exponents m, n, and so forth are called reaction orders for the reactants. The sum of the reaction orders gives the overall reaction order. Reaction orders must be determined experimentally. The units of the rate constant depend on the overall reaction order. For a reaction in which the over-all reaction order is 1, k has units of s-1; for one in which the overall reaction order is 2, k has units of M -1 s-1.

Spectroscopy is one technique that can be used to monitor the course of a reaction. According to Beer’s law, the absorption of elec-tromagnetic radiation by a substance at a particular wavelength is directly proportional to its concentration.

CONCENTRATION ANd TIME (SECTION 14.4) Rate laws can be used to determine the concentrations of reactants or products at any time dur-ing a reaction. In a first-order reaction the rate is proportional to the con-centration of a single reactant raised to the first power: Rate = k3A4. In such cases the integrated form of the rate law is ln3A4t = -kt + ln3A40, where 3A4t is the concentration of reactant A at time t, k is the rate con-stant, and 3A40 is the initial concentration of A. Thus, for a first-order reaction, a graph of ln[A] versus time yields a straight line of slope -k.

A second-order reaction is one for which the overall reaction order is 2. If a second-order rate law depends on the concentration of only one reactant, then rate = k3A42, and the time dependence of [A] is given by the integrated form of the rate law: 1>3A4t = 1>3A40 + kt. In this case, a graph of 1>3A4t versus time yields a straight line. A zero-order reaction is one for which the overall reaction order is 0. Rate = k if the reaction is zero order.

The half-life of a reaction, t1>2, is the time required for the con-centration of a reactant to drop to one-half of its original value. For a first-order reaction, the half-life depends only on the rate constant and not on the initial concentration: t1>2 = 0.693>k. The half-life of a second-order reaction depends on both the rate constant and the initial concentration of A: t1>2 = 1>k3A40.

THE EFFECT OF TEMPERATURE ON RATES (SECTION 14.5) The collision model, which assumes that reactions occur as a result of collisions between molecules, helps explain why the magnitudes of rate constants

increase with increasing temperature. The greater the kinetic energy of the colliding molecules, the greater is the energy of collision. The minimum energy required for a reaction to occur is called the activation energy, Ea. A collision with energy Ea or greater can cause the atoms of the colliding molecules to reach the activated complex (or transition state), which is the highest energy arrangement in the pathway from reactants to products. Even if a collision is energetic enough, it may not lead to reaction; the reac-tants must also be correctly oriented relative to one another in order for a collision to be effective.

Because the kinetic energy of molecules depends on tempera-ture, the rate constant of a reaction is very dependent on temperature. The relationship between k and temperature is given by the Arrhe-nius equation: k = Ae-Ea>RT. The term A is called the frequency factor; it relates to the number of collisions that are favorably oriented for reaction. The Arrhenius equation is often used in logarithmic form: ln k = ln A - Ea>RT. Thus, a graph of ln k versus 1>T yields a straight line with slope -Ea>R.

REACTION MECHANISMS (SECTION 14.6) A reaction mechanism details the individual steps that occur in the course of a reaction. Each of these steps, called elementary reactions, has a well-defined rate law that depends on the number of molecules (the molecularity) of the step. Elemen-tary reactions are defined as either unimolecular, bimolecular, or termolecular, depending on whether one, two, or three reactant molecules are involved, respectively. Termolecular elementary reactions are very rare. Unimolecu-lar, bimolecular, and termolecular reactions follow rate laws that are first order overall, second order overall, and third order overall, respectively.

Many reactions occur by a multistep mechanism, involving two or more elementary reactions, or steps. An intermediate is produced in one elementary step, is consumed in a later elementary step, and therefore does not appear in the overall equation for the reaction. When a mechanism has several elementary steps, the overall rate is limited by the slowest elementary step, called the rate-determining step. A fast elementary step that follows the rate-determining step will have no effect on the rate law of the reaction. A fast step that precedes the rate-determining step often creates an equilibrium that involves an intermediate. For a mechanism to be valid, the rate law predicted by the mechanism must be the same as that observed experimentally.

CATALySTS (SECTION 14.7) A catalyst is a substance that increases the rate of a reaction without undergoing a net chemical change itself. It does so by providing a different mechanism for the reaction, one that has a lower activation energy. A homogeneous catalyst is one that is in the same phase as the reactants. A heterogeneous catalyst has a different phase from the reactants. Finely divided metals are often used as hetero-geneous catalysts for solution- and gas-phase reactions. Reacting mol-ecules can undergo binding, or adsorption, at the surface of the catalyst. The adsorption of a reactant at specific sites on the surface makes bond breaking easier, lowering the activation energy. Catalysis in living organ-isms is achieved by enzymes, large protein molecules that usually cata-lyze a very specific reaction. The specific reactant molecules involved in an enzymatic reaction are called substrates. The site of the enzyme where the catalysis occurs is called the active site. In the lock-and-key model for enzyme catalysis, substrate molecules bind very specifically to the active site of the enzyme, after which they can undergo reaction.

learning outcomes after studying this chapter, you should be able to:

• List the factors that affect the rate of chemical reactions. (Section 14.1)

• Determine the rate of a reaction given time and concentration. (Section 14.2)

• Relate the rate of formation of products and the rate of disappear-ance of reactants given the balanced chemical equation for the reaction. (Section 14.2)

exercises 615

• Explain the form and meaning of a rate law, including the ideas of reaction order and rate constant. (Section 14.3)

• Determine the rate law and rate constant for a reaction from a se-ries of experiments given the measured rates for various concen-trations of reactants. (Section 14.3)

• Apply the integrated form of a rate law to determine the concentra-tion of a reactant at a given time. (Section 14.4)

• Apply the relationship between the rate constant of a first-order reaction and its half-life. (Section 14.4)

• Explain how the activation energy affects a rate and be able to use the Arrhenius equation. (Section 14.5)

• Predict a rate law for a reaction having a multistep mechanism given the individual steps in the mechanism. (Section 14.6)

• Explain the principles underlying catalysis. (Section 14.7)

Key equations

Rate = - 1a

∆3A4

∆t= -

1b

∆3B4

∆t=

1c

∆3C4

∆t=

1d

∆3D4

∆t [14.4] Definition of reaction rate in terms of the components of the bal-

anced chemical equation a A + b B ¡ c C + d D

Rate = k3A4m3B4n [14.7] General form of a rate law for the reaction A + B ¡ products

ln3A4t - ln3A40 = -kt or ln 3A4t

3A40= -kt [14.12] The integrated form of a first-order rate law for the reaction

A ¡ products

13A4t

= kt +1

3A40 [14.14] The integrated form of the second-order rate law for the reaction

A ¡ products

t1>2 =0.693

k [14.15] Relating the half-life and rate constant for a first-order reaction

k = Ae-Ea>RT [14.19] The Arrhenius equation, which expresses how the rate constant depends on temperature

ln k = - Ea

RT+ ln A [14.20] Logarithmic form of the Arrhenius equation

exercisesvisualizing Concepts

14.1 An automotive fuel injector dispenses a fine spray of gasoline into the automobile cylinder, as shown in the bottom draw-ing here. When an injector gets clogged, as shown in the top drawing, the spray is not as fine or even and the performance of the car declines. How is this observation related to chemi-cal kinetics? [Section 14.1]

14.2 Consider the following graph of the concentration of a sub-stance X over time. Is each of the following statements true or false? (a) X is a product of the reaction. (b) The rate of the reaction remains the same as time progresses. (c) The average rate between points 1 and 2 is greater than the av-erage rate between points 1 and 3. (d) As time progresses, the curve will eventually turn downward toward the x-axis. [Section 14.2]

Time

1

2

3

[X]


14.3 You study the rate of a reaction, measuring both the concen-tration of the reactant and the concentration of the product as a function of time, and obtain the following results:

TimeB

ACon

centration

(a) Which chemical equation is consistent with these data: (i) A ¡ B, (ii) B ¡ A, (iii) A ¡ 2 B,

(iv) B ¡ 2 A?(b) Write equivalent expressions for the rate of the reaction

in terms of the appearance or disappearance of the two substances. [Section 14.2]

14.4 Suppose that for the reaction K + L ¡ M, you monitor the production of M over time, and then plot the following graph from your data:

5 10 15 20 25 300t, min

[M]

(a) Is the reaction occurring at a constant rate from t = 0 to t = 15 min? (b) Is the reaction completed at t = 15 min? (c) Suppose the reaction as plotted here were started with 0.20 mol K and 0.40 mol L. After 30 min, an additional 0.20 mol K are added to the reaction mixture. Which of the following correctly describes how the plot would look from t = 30 min to t = 60 min? (i) [M] would remain at the same constant value it has at t = 30 min, (ii) [M] would increase with the same slope as t = 0 to 15 min, un-til t = 45 min at which point the plot becomes horizontal again, or (iii) [M] decreases and reaches 0 at t = 45 min. [Section 14.2]

14.5 The following diagrams represent mixtures of NO(g) and O21g2. These two substances react as follows:

2 NO1g2 + O21g2 ¡ 2 NO21g2 It has been determined experimentally that the rate is second

order in NO and first order in O2. Based on this fact, which

of the following mixtures will have the fastest initial rate? [Section 14.3]

(1) (2) (3)

14.6 A friend studies a first-order reaction and obtains the follow-ing three graphs for experiments done at two different tem-peratures. (a) Which two graphs represent experiments done at the same temperature? What accounts for the difference in these two graphs? In what way are they the same? (b) Which two graphs represent experiments done with the same start-ing concentration but at different temperatures? Which graph probably represents the lower temperature? How do you know? [Section 14.4]

Time

1

2

3ln

[A]

14.7 (a) Given the following diagrams at t = 0 min and t = 30 min, what is the half-life of the reaction if it follows first-order kinetics?

t = 0 min t = 30 min

(b) After four half-life periods for a first-order reaction, what fraction of reactant remains? [Section 14.4]

14.8 Which of the following linear plots do you expect for a reaction A ¡ products if the kinetics are (a) zero order, (b) first order, or (c) second order? [Section 14.4]

exercises 617

Time Time

Time Time

Time Time

(i) (ii)

(iii) (iv)

(v) (vi)

ln[A

]1/

[A]

1/[A

]

[A]

[A]

ln[A

]

14.9 The following diagram shows a reaction profile. Label the components indicated by the boxes. [Section 14.5]

(4)

Reaction progress

Ene

rgy

(3)

(1)

(2)

14.10 The accompanying graph shows plots of ln k versus 1>T for two different reactions. The plots have been extrapolated to the y-intercepts. Which reaction (red or blue) has (a) the larger value for Ea, and (b) the larger value for the frequency factor, A? [Section 14.5]

1/T

ln k

14.11 The following graph shows two different reaction pathways for the same overall reaction at the same temperature. Is each of the following statements true or false? (a) The rate is faster for the red path than for the blue path. (b) For both paths, the rate of the reverse reaction is slower than the rate of the for-ward reaction. (c) The energy change ∆E is the same for both paths. [Section 14.6]

Reaction progress

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14.12 Consider the diagram that follows, which represents two steps in an overall reaction. The red spheres are oxy-gen, the blue ones nitrogen, and the green ones fluorine. (a) Write the chemical equation for each step in the reaction. (b) Write the equation for the overall reaction. (c) Identify the intermediate in the mechanism. (d) Write the rate law for the overall reaction if the first step is the slow, rate-deter-mining step. [Section 14.6]

+ +

+

14.13 Based on the following reaction profile, how many intermedi-ates are formed in the reaction A ¡ C? How many tran-sition states are there? Which step, A ¡ B or B ¡ C, is the faster? For the reaction A ¡ C, is ∆E positive, negative, or zero? [Section 14.6]

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A

B

C


14.14 Draw a possible transition state for the bimolecular reaction de-picted here. (The blue spheres are nitrogen atoms, and the red ones are oxygen atoms.) Use dashed lines to represent the bonds that are in the process of being broken or made in the transition state. [Section 14.6]

++

14.15 The following diagram represents an imaginary two-step mechanism. Let the red spheres represent element A, the green ones element B, and the blue ones element C. (a) Write the equation for the net reaction that is occurring. (b) Identify the intermediate. (c) Identify the catalyst. [Sections 14.6 and 14.7]

14.16 Draw a graph showing the reaction pathway for an overall exothermic reaction with two intermediates that are pro-duced at different rates. On your graph indicate the reactants, products, intermediates, transition states, and activation en-ergies. [Sections 14.6 and 14.7]

Reaction Rates (sections 14.1 and 14.2)

14.17 (a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

14.18 (a) What are the units usually used to express the rates of re-actions occurring in solution? (b) From your everyday experi-ence, give two examples of the effects of temperature on the rates of reactions. (c) What is the difference between average rate and instantaneous rate?

14.19 Consider the following hypothetical aqueous reaction: A1aq2 S B1aq2. A flask is charged with 0.065 mol of A in a total volume of 100.0 mL. The following data are collected:

Time (min) 0 10 20 30 40Moles of A 0.065 0.051 0.042 0.036 0.031

(a) Calculate the number of moles of B at each time in the table, assuming that there are no molecules of B at time zero, and that A cleanly converts to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 10-min interval in units of M>s. (c) Between t = 10 min and t = 30 min, what is the average rate of appearance of B in units of M>s? Assume that the volume of the solution is constant.

14.20 A flask is charged with 0.100 mol of A and allowed to react to form B according to the hypothetical gas-phase reaction A1g2 ¡ B1g2. The following data are collected:

Time (s) 0 40 80 120 160Moles of A 0.100 0.067 0.045 0.030 0.020

(a) Calculate the number of moles of B at each time in the table, assuming that A is cleanly converted to B with no in-termediates. (b) Calculate the average rate of disappearance of A for each 40-s interval in units of mol>s. (c) Which of the following would be needed to calculate the rate in units of concentration per time: (i) the pressure of the gas at each time, (ii) the volume of the reaction flask, (iii) the tempera-ture, or (iv) the molecular weight of A?

14.21 The isomerization of methyl isonitrile 1CH3NC2 to acetoni-trile 1CH3CN2 was studied in the gas phase at 215 °C, and the following data were obtained:

time (s) 3Ch3nC 4 1M 20 0.0165

2000 0.01105000 0.005918000 0.00314

12,000 0.0013715,000 0.00074

(a) Calculate the average rate of reaction, in M>s, for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from t = 0 to t = 15,000 s. (c) Which is greater, the average rate between t = 2000 and t = 12,000 s, or between t = 8000 and t = 15,000 s? (d) Graph 3CH3NC4 versus time and determine the instantaneous rates in M>s at t = 5000 s and t = 8000 s.

14.22 The rate of disappearance of HCl was measured for the fol-lowing reaction:

CH3OH1aq2 + HCl1aq2 ¡ CH3Cl1aq2 + H2O1l2 The following data were collected:

time (min) [hCl] (M)

0.0 1.8554.0 1.58107.0 1.36215.0 1.02430.0 0.580

(a) Calculate the average rate of reaction, in M>s, for the time interval between each measurement. (b) Calculate the average rate of reaction for the entire time for the data from t = 0.0 min to t = 430.0 min. (c) Which is greater, the aver-age rate between t = 54.0 and t = 215.0 min, or between t = 107.0 and t = 430.0 min? (d) Graph [HCl] versus time and determine the instantaneous rates in M>min and M>s at t = 75.0 min and t = 250 min.

14.23 For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product:(a) H2O21g2 ¡ H21g2 + O21g2(b) 2 N2O1g2 ¡ 2 N21g2 + O21g2(c) N21g2 + 3 H21g2 ¡ 2 NH31g2(d) C2H5NH21g2 ¡ C2H41g2 + NH31g2

exercises 619

14.24 For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant:(a) 2 H2O1g2 ¡ 2 H21g2 + O21g2(b) 2 SO21g2 + O21g2 ¡ 2 SO31g2(c) 2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2(d) N21g2 + 2 H21g2 ¡ N2H41g2

14.25 ( a ) C o n s i d e r t h e c o m b u s t i o n o f H21g2: 2 H21g2 +O21g2 ¡ 2 H2O1g2. If hydrogen is burning at the rate of 0.48 mol>s, what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction 2 NO1g2 + Cl21g2 ¡ 2 NOCl1g2 is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of 56 torr/min, what is the rate of change of the total pressure of the vessel?

14.26 (a) Consider the combustion of ethylene, C2H41g2 +3 O21g2 ¡ 2 CO21g2 + 2 H2O1g2. If the concentration of C2H4 is decreasing at the rate of 0.036 M>s, what are the rates of change in the concentrations of CO2 and H2O? (b) The rate of decrease in N2H4 partial pressure in a closed reaction ves-sel from the reaction N2H41g2 + H21g2 ¡ 2 NH31g2 is 74 torr per hour. What are the rates of change of NH3 partial pressure and total pressure in the vessel?

Rate laws (section 14.3)

14.27 A reaction A + B ¡ C obeys the following rate law: Rate = k3B42. (a) If [A] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction or-ders for A and B? What is the overall reaction order? (c) What are the units of the rate constant?

14.28 Consider a hypothetical reaction between A, B, and C that is first order in A, zero order in B, and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when [A] is doubled and the other reactant concentra-tions are held constant? (c) How does the rate change when [B] is tripled and the other reactant concentrations are held con-stant? (d) How does the rate change when [C] is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled? (f) By what factor does the rate change when the concentrations of all three reactants are cut in half?

14.29 The decomposition reaction of N2O5 in carbon tetrachloride is 2 N2O5 ¡ 4 NO2 + O2. The rate law is first order in N2O5. At 64 °C the rate constant is 4.82 * 10-3 s-1. (a) Write the rate law for the reaction. (b) What is the rate of reaction when 3N2O54 = 0.0240 M? (c) What happens to the rate when the concentration of N2O5 is doubled to 0.0480 M? (d) What happens to the rate when the concentration of N2O5 is halved to 0.0120 M?

14.30 Consider the following reaction:

2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2 (a) The rate law for this reaction is first order in H2 and second

order in NO. Write the rate law. (b) If the rate constant for this reaction at 1000 K is 6.0 * 104 M -2 s-1, what is the re-action rate when 3NO4 = 0.035 M and 3H24 = 0.015 M? (c) What is the reaction rate at 1000 K when the concentration of NO is increased to 0.10 M, while the concentration of H2 is 0.010 M? (d) What is the reaction rate at 1000 K if [NO] is decreased to 0.010 M and 3H24 is increased to 0.030 M?

14.31 Consider the following reaction:

CH3Br1aq2 + OH-1aq2 ¡ CH3OH1aq2 + Br-1aq2 The rate law for this reaction is first order in CH3Br and first

order in OH- . When 3CH3Br4 is 5.0 * 10-3 M and 3OH- 4 is 0.050 M, the reaction rate at 298 K is 0.0432 M>s. (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the con-centration of OH- were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

14.32 The reaction between ethyl bromide 1C2H5Br2 and hydroxide ion in ethyl alcohol at 330 K, C2H5Br1alc2 + OH- 1alc2 ¡ C2H5OH1l2 + Br - 1alc2, is first order each in ethyl bromide and hydroxide ion. When 3C2H5Br4 is 0.0477 M and 3OH- 4 is 0.100 M, the rate of disappearance of ethyl bromide is 1.7 * 10-7 M>s. (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the so-lution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

14.33 The iodide ion reacts with hypochlorite ion (the ac-tive ingredient in chlorine bleaches) in the following way: OCl- + I - ¡ OI - + Cl- . This rapid reaction gives the following rate data:

3oCl−4 1M 2 3i−4 1M 2 initial Rate 1M ,s 21.5 * 10-3 1.5 * 10-3 1.36 * 10-4

3.0 * 10-3 1.5 * 10-3 2.72 * 10-4

1.5 * 10-3 3.0 * 10-3 2.72 * 10-4

(a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when 3OCl-4 = 2.0 * 10-3 M and 3I - 4 = 5.0 * 10 - 4 M.

14.34 The reaction 2 ClO21aq2 + 2 OH- 1aq2 ¡ ClO3- 1aq2 +

ClO2- 1aq2 + H2O1l2 was studied with the following results:

experiment 1Clo2 2 1M 2 3oh−4 1M 2 initial Rate 1M ,s 21 0.060 0.030 0.02482 0.020 0.030 0.002763 0.020 0.090 0.00828

(a) Determine the rate law for the reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when 3ClO24 = 0.100 M and 3OH- 4 = 0.050 M.

14.35 The following data were measured for the reaction BF31g2 + NH31g2 ¡ F3BNH31g2:

experiment 3BF3 4 1m 2 3nh3 4 1M 2 initial Rate 1M ,s 21 0.250 0.250 0.21302 0.250 0.125 0.10653 0.200 0.100 0.06824 0.350 0.100 0.11935 0.175 0.100 0.0596

(a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when 3BF34 = 0.100 M and 3NH34 = 0.500 M?


14.36 The following data were collected for the rate of disappearance of NO in the reaction 2 NO1g2 + O21g2 ¡ 2 NO21g2:

experiment 3no 4 1M 2 3o2 4 1M 2 initial Rate 1M ,s 21 0.0126 0.0125 1.41 * 10-2

2 0.0252 0.0125 5.64 * 10-2

3 0.0252 0.0250 1.13 * 10-1

(a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets? (d) What is the rate of disappearance of NO when 3NO4 = 0.0750 M and 3O24 = 0.0100 M ? (e) What is the rate of disappearance of O2 at the concentrations given in part (d)?

[14.37] Consider the gas-phase reaction between nitric oxide and bromine at 273 °C: 2 NO1g2 + Br21g2 ¡ 2 NOBr1g2. The following data for the initial rate of appearance of NOBr were obtained:

experiment 3no 4 1m 2 3Br2 4 1M 2 initial Rate 1M ,s 21 0.10 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735

(a) Determine the rate law. (b) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets. (c) How is the rate of appearance of NOBr related to the rate of disappearance of Br2? (d) What is the rate of disappearance of Br2 when 3NO4 = 0.075 M and 3Br24 = 0.25 M ?

[14.38] Consider the reaction of peroxydisulfate ion 1S2O82-2 with

iodide ion 1I - 2 in aqueous solution:

S2O82 - 1aq2 + 3 I - 1aq2 ¡ 2 SO4

2 - 1aq2 + I3- 1aq2

At a particular temperature the initial rate of disappearance of S2O8

2 - varies with reactant concentrations in the following manner:

experiment 3s2o82−4 1M 2 3i−4 1M 2 initial Rate 1M ,s 2

1 0.018 0.036 2.6 * 10-6

2 0.027 0.036 3.9 * 10-6

3 0.036 0.054 7.8 * 10-6

4 0.050 0.072 1.4 * 10-5

(a) Determine the rate law for the reaction and state the units of the rate constant. (b) What is the average value of the rate constant for the disappearance of S2O8

2 - based on the four sets of data? (c) How is the rate of disappearance of S2O8

2 - related to the rate of disappearance of I - ? (d) What is the rate of disappearance of I - when 3S2O8

2-4 = 0.025 M and 3I - 4 = 0.050 M ?

Change of Concentration with time (section 14.4)

14.39 (a) Define the following symbols that are encountered in rate equations for the generic reaction A S B: 3A40, t1>2 3A4t, k. (b) What quantity, when graphed versus time, will yield a

straight line for a first-order reaction? (c) How can you calcu-late the rate constant for a first-order reaction from the graph you made in part (b)?

14.40 (a) For a generic second-order reaction A ¡ B, what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

14.41 (a) The gas-phase decomposition of SO2Cl2, SO2Cl21g2 ¡SO21g2 + Cl21g2, is first order in SO2Cl2. At 600 K the half-life for this process is 2.3 * 105 s. What is the rate con-stant at this temperature? (b) At 320 °C the rate constant is 2.2 * 10-5 s-1. What is the half-life at this temperature?

14.42 Molecular iodine, I21g2, dissociates into iodine atoms at 625 K with a first-order rate constant of 0.271 s-1. (a) What is the half-life for this reaction? (b) If you start with 0.050 M I2 at this temperature, how much will remain after 5.12 s assum-ing that the iodine atoms do not recombine to form I2?

14.43 As described in Exercise 14.41, the decomposition of sulfuryl chloride 1SO2Cl22 is a first-order process. The rate constant for the decomposition at 660 K is 4.5 * 10-2 s-1. (a) If we begin with an initial SO2Cl2 pressure of 450 torr, what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of SO2Cl2 decline to one-tenth its initial value?

14.44 The first-order rate constant for the decomposition of N2O5, 2 N2O51g2 ¡ 4 NO21g2 + O21g2, at 70 °C is 6.82 * 10-3 s-1. Suppose we start with 0.0250 mol of N2O51g2 in a volume of 2.0 L. (a) How many moles of N2O5 will remain after 5.0 min? (b) How many minutes will it take for the quantity of N2O5 to drop to 0.010 mol? (c) What is the half-life of N2O5 at 70 °C ?

14.45 The reaction SO2Cl21g2 ¡ SO21g2 + Cl21g2 is first order in SO2Cl2. Using the following kinetic data, determine the magnitude and units of the first-order rate constant:

time (s) Pressure so2Cl2 (atm)

0 1.0002500 0.9475000 0.8957500 0.848

10,000 0.803

14.46 From the following data for the first-order gas-phase isomeri-zation of CH3NC at 215 °C, calculate the first-order rate con-stant and half-life for the reaction:

time (s) Pressure Ch3nC (torr)

0 5022000 3355000 1808000 95.5

12,000 41.715,000 22.4

14.47 Consider the data presented in Exercise 14.19. (a) By using appropriate graphs, determine whether the reaction is first order or second order. (b) What is the rate constant for the reaction? (c) What is the half-life for the reaction?

exercises 621

14.48 Consider the data presented in Exercise 14.20. (a) Determine whether the reaction is first order or second order. (b) What is the rate constant? (c) What is the half-life?

14.49 The gas-phase decomposition of NO2, 2 NO21g2 ¡2 NO1g2 + O21g2, is studied at 383 °C, giving the following data:

time (s) 3no2 4 (M)

0.0 0.1005.0 0.017

10.0 0.009015.0 0.006220.0 0.0047

(a) Is the reaction first order or second order with respect to the concentration of NO2? (b) What is the rate constant? (c) Predict the reaction rates at the beginning of the reac-tion for initial concentrations of 0.200 M, 0.100 M, and 0.050 M NO2.

14.50 Sucrose 1C12H22O112, commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H22O6. At 23 °C and in 0.5 M HCl, the following data were obtained for the disappearance of sucrose:

time (min) 3C12h22o11 4 1M 20 0.316

39 0.27480 0.238

140 0.190210 0.146

(a) Is the reaction first order or second order with respect to 3C12H22O114? (b) What is the rate constant? (c) Using this rate constant, calculate the concentration of sucrose at 39, 80, 140, and 210 min if the initial sucrose concentration was 0.316 M and the reaction were zero order in sucrose.

temperature and Rate (section 14.5)

14.51 (a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) According to the collision model, why does temperature affect the value of the rate constant? (c) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature?

14.52 (a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction: NO + O ¡ NO2 or H + Cl ¡ HCl? (b) How does the kinetic-molecular theory help us understand the tempera-ture dependence of chemical reactions?

14.53 Calculate the fraction of atoms in a sample of argon gas at 400 K that has an energy of 10.0 kJ or greater.

14.54 (a) The activation energy for the isomerization of methyl iso-nitrile (Figure 14.7) is 160 kJ>mol. Calculate the fraction of methyl isonitrile molecules that has an energy of 160.0 kJ or greater at 500 K. (b) Calculate this fraction for a temperature of 520 K. What is the ratio of the fraction at 520 K to that at 500 K?

14.55 The gas-phase reaction Cl1g2 + HBr1g2 ¡ HCl1g2 + Br1g2 has an overall energy change of -66 kJ. The activation energy for the reaction is 7 kJ. (a) Sketch the energy profile for the re-action, and label Ea and ∆E. (b) What is the activation energy for the reverse reaction?

14.56 For the elementary process N2O51g2 ¡ NO21g2 + NO31g2 the activation energy 1Ea2 and overall ∆E are 154 kJ>mol and 136 kJ>mol, respectively. (a) Sketch the energy profile for this reaction, and label Ea and ∆E. (b) What is the activation en-ergy for the reverse reaction?

14.57 Indicate whether each statement is true or false.(a) If you compare two reactions with similar collision factors,

the one with the larger activation energy will be faster.(b) A reaction that has a small rate constant must have a

small frequency factor.(c) Increasing the reaction temperature increases the frac-

tion of successful collisions between reactants. 14.58 Indicate whether each statement is true or false.

(a) If you measure the rate constant for a reaction at differ-ent temperatures, you can calculate the overall enthalpy change for the reaction.

(b) Exothermic reactions are faster than endothermic reactions.

(c) If you double the temperature for a reaction, you cut the activation energy in half.

14.59 Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest?(a) Ea = 45 kJ>mol; ∆E = -25 kJ>mol(b) Ea = 35 kJ>mol; ∆E = -10 kJ>mol(c) Ea = 55 kJ>mol; ∆E = 10 kJ>mol

14.60 Which of the reactions in Exercise 14.61 will be fastest in the reverse direction? Which will be slowest?

14.61 (a) A certain first-order reaction has a rate constant of 2.75 * 10-2 s-1 at 20 °C. What is the value of k at 60 °C if Ea = 75.5 kJ>mol ? (b) Another first-order reaction also has a rate constant of 2.75 * 10-2 s-1 at 20 °C What is the value of k at 60 °C if Ea = 125 kJ>mol ? (c) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

14.62 Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in auto-mobile engines. The decomposition of nitric oxide (NO) to N2 and O2 is second order with a rate constant of 0.0796 M -1s-1 at 737 °C and 0.0815 M -1s-1 at 947 °C. Calculate the activa-tion energy for the reaction.

14.63 The rate of the reaction

CH3COOC2H51aq2 + OH- 1aq2 ¡ CH3COO - 1aq2 + C2H5OH1aq2 was measured at several temperatures, and the following data

were collected:

temperature 1 °C 2 k 1M−1 s−1215 0.052125 0.10135 0.18445 0.332


Calculate the value of Ea by constructing an appropriate graph.

14.64 The temperature dependence of the rate constant for a reac-tion is tabulated as follows:

temperature (K) k 1M −1 s−1 2600 0.028650 0.22700 1.3750 6.0800 23

Calculate Ea and A.

Reaction mechanisms (section 14.6)

14.65 (a) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism? (d) What is meant by the term rate-determin-ing step?

14.66 (a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an in-termediate in a mechanism? (d) What are the differences be-tween an intermediate and a transition state?

14.67 What is the molecularity of each of the following elementary reactions? Write the rate law for each.(a) Cl21g2 ¡ 2 Cl1g2(b) OCl- 1aq2 + H2O1l2 ¡ HOCl1aq2 + OH- 1aq2(c) NO1g2 + Cl21g2 ¡ NOCl21g2

14.68 What is the molecularity of each of the following elementary reactions? Write the rate law for each.(a) 2 NO1g2 ¡ N2O21g2 CH2

(b) H2C ¬ CH21g2 ¡ CH2 “ CH ¬ CH31g2 (c) SO31g2 ¡ SO21g2 + O1g2

14.69 (a) Based on the following reaction profile, how many inter-mediates are formed in the reaction A ¡ D? (b) How many transition states are there? (c) Which step is the fast-est? (d) For the reaction A ¡ D, is ∆E positive, nega-tive, or zero?

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14.70 Consider the following energy profile.

Reaction progess

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(a) How many elementary reactions are in the reaction mechanism? (b) How many intermediates are formed in the reaction? (c) Which step is rate limiting? (d) For the overall reaction, is ∆E positive, negative, or zero?

14.71 The following mechanism has been proposed for the gas-phase reaction of H2 with ICl:

H21g2 + ICl1g2 ¡ HI1g2 + HCl1g2HI1g2 + ICl1g2 ¡ I21g2 + HCl1g2

(a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

14.72 The decomposition of hydrogen peroxide is catalyzed by io-dide ion. The catalyzed reaction is thought to proceed by a two-step mechanism:

H2O21aq2 + I - 1aq2 ¡ H2O1l2 + IO - 1aq2 1slow2 IO - 1aq2 + H2O21aq2 ¡ H2O1l2 + O21g2 + I - 1aq2 1fast2 (a) Write the chemical equation for the overall process. (b)

Identify the intermediate, if any, in the mechanism. (c) As-suming that the first step of the mechanism is rate determin-ing, predict the rate law for the overall process.

[14.73] The reaction 2 NO1g2 + Cl21g2 ¡ 2 NOCl1g2 was per-formed and the following data obtained under conditions of constant 3Cl24:

Time

(under conditionsof constant [Cl2])

1/[N

O]

(a) Is the following mechanism consistent with the data?

NO1g2 + Cl21g2 ¡ NOCl21g2 1fast2 NOCl21g2 + NO1g2 ¡ 2 NOCl1g2 1slow2

(b) Does the linear plot guarantee that the overall rate law is second order?

14.74 You have studied the gas-phase oxidation of HBr by O2:

4 HBr1g2 + O21g2 ¡ 2 H2O1g2 + 2 Br21g2

exercises 623

You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism:

HBr1g2 + O21g2 ¡ HOOBr1g2HOOBr1g2 + HBr1g2 ¡ 2 HOBr1g2

HOBr1g2 + HBr1g2 ¡ H2O1g2 + Br21g2 (a) Confirm that the elementary reactions add to give the

overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to de-tect HOBr or HOOBr among the products, does this disprove your mechanism?

Catalysis (section 14.7)

14.75 (a) What is a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst? (c) Do catalysts affect the overall enthalpy change for a reaction, the activa-tion energy, or both?

14.76 (a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a hetero-geneous catalyst?

14.77 In Figure 14.22, we saw that Br - 1aq2 catalyzes the decompo-sition of H2O21aq2 into H2O1l2 and O21g2. Suppose that some KBr1s2 is added to an aqueous solution of hydrogen peroxide. Make a sketch of 3Br - 1aq24 versus time from the addition of the solid to the end of the reaction.

14.78 In solution, chemical species as simple as H+ and OH- can serve as catalysts for reactions. Imagine you could measure the 3H+4 of a solution containing an acid-catalyzed reaction as it occurs. Assume the reactants and products themselves are nei-ther acids nor bases. Sketch the 3H+4 concentration profile you would measure as a function of time for the reaction, assuming t = 0 is when you add a drop of acid to the reaction.

14.79 The oxidation of SO2 to SO3 is accelerated by NO2. The reac-tion proceeds according to:

NO21g2 + SO21g2 ¡ NO1g2 + SO31g22 NO1g2 + O21g2 ¡ 2 NO21g2

(a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of SO2 by O2 to give SO3. (b) Do we consider NO2 a catalyst or an intermedi-ate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

14.80 The addition of NO accelerates the decomposition of N2O, possibly by the following mechanism:

NO1g2 + N2O1g2 ¡ N21g2 + NO21g22 NO21g2 ¡ 2 NO1g2 + O21g2

(a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equa-tion. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposi-tion of N2O, NO2 does not accumulate in measurable quanti-ties, does this rule out the proposed mechanism?

14.81 Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina 1Al2O32 or silica 1SiO22. (a) Why is this an effective way of utilizing the cata-lyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

14.82 (a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advan-tage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

14.83 When D2 reacts with ethylene 1C2H42 in the presence of a finely divided catalyst, ethane with two deuteriums, CH2D ¬ CH2D, is formed. (Deuterium, D, is an isotope of hy-drogen of mass 2). Very little ethane forms in which two deu-teriums are bound to one carbon (for example, CH3 ¬ CHD2). Use the sequence of steps involved in the reaction (Figure 14.24) to explain why this is so.

14.84 Heterogeneous catalysts that perform hydrogenation reac-tions, as illustrated in Figure 14.24, are subject to “poisoning,” which shuts down their catalytic ability. Compounds of sulfur are often poisons. Suggest a mechanism by which such com-pounds might act as poisons.

[14.85] The enzyme carbonic anhydrase catalyzes the reaction CO21g2 + H2O1l2 ¡ HCO3

- 1aq2 + H+1aq2. In water, without the enzyme, the reaction proceeds with a rate constant of 0.039 s-1 at 25 °C. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 1.0 * 106 s-1 at 25 °C. Assuming the collision factor is the same for both situ-ations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

[14.86] The enzyme urease catalyzes the reaction of urea, 1NH2CONH22, with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of 4.15 * 10-5 s-1 at 100 °C. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 3.4 * 104 s-1 at 21 °C. (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at 100 °C as it is at 21 °C, what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you ex-pect for the rate of the catalyzed reaction at 100 °C as compared to that at 21 °C? (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

[14.87] The activation energy of an uncatalyzed reaction is 95 kJ>mol. The addition of a catalyst lowers the activation energy to 55 kJ>mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) 25 °C, (b) 125 °C?

[14.88] Suppose that a certain biologically important reaction is quite slow at physiological temperature 137 °C2 in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a 1 * 105@fold increase in the reac-tion rate?


additional exercises 14.89 Consider the reaction A + B ¡ C + D. Is each of the fol-

lowing statements true or false? (a) The rate law for the reaction must be Rate = k3A43B4. (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction.

14.90 Hydrogen sulfide 1H2S2 is a common and troublesome pol-lutant in industrial wastewaters. One way to remove H2S is to treat the water with chlorine, in which case the following reaction occurs:

H2S1aq2 + Cl21aq2 ¡ S1s2 + 2 H+1aq2 + 2 Cl- 1aq2 The rate of this reaction is first order in each reactant.

The rate constant for the disappearance of H2S at 28 °C is 3.5 * 10-2 M -1 s-1. If at a given time the concentration of H2S is 2.0 * 10 - 4 M and that of Cl2 s 0.025 M, what is the rate of formation of Cl- ?

14.91 The reaction 2 NO1g2 + O21g2 ¡ 2 NO21g2 is second or-der in NO and first order in O2. When 3NO4 = 0.040 M, and 3O24 = 0.035 M, the observed rate of disappearance of NO is 9.3 * 10-5 M>s. (a) What is the rate of disappearance of O2 at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8?

14.92 You perform a series of experiments for the reaction A ¡ B + C and find that the rate law has the form rate = k3A4x. Determine the value of x in each of the follow-ing cases: (a) There is no rate change when 3A40 is tripled. (b) The rate increases by a factor of 9 when 3A40 is tripled. (c) When 3A40 is doubled, the rate increases by a factor of 8.

14.93 Consider the following reaction between mercury(II) chloride and oxalate ion:

2 HgCl21aq2 + C2O42 - 1aq2 ¡ 2 Cl- 1aq2 + 2 CO21g2 + Hg2Cl21s2

The initial rate of this reaction was determined for several concentrations of HgCl2 and C2O4

2 - , and the following rate data were obtained for the rate of disappearance of C2O4

2 - :

experiment 3hgCl2 4 1M 2 3C2o42 −4 1M 2 Rate 1m ,s 2

1 0.164 0.15 3.2 * 10-5

2 0.164 0.45 2.9 * 10-4

3 0.082 0.45 1.4 * 10-4

4 0.246 0.15 4.8 * 10-5

(a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of HgCl2 is 0.100 M and that of 1C2O4

2-2 is 0.25 M if the tempera-ture is the same as that used to obtain the data shown?

14.94 The following kinetic data are collected for the initial rates of a reaction 2 X + Z ¡ products:

experiment 3x 40 1M 2 3Z 40 1M 2 Rate 1M ,s 21 0.25 0.25 4.0 * 101

2 0.50 0.50 3.2 * 102

3 0.50 0.75 7.2 * 102

(a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M?

14.95 The reaction 2 NO2 ¡ 2 NO + O2 has the rate constant k = 0.63 M - 1s- 1. (a) Based on the units for k, is the reac-tion first or second order in NO2? (b) If the initial con-centration of NO2 is 0.100 M, how would you determine how long it would take for the concentration to decrease to 0.025 M?

14.96 Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reac-tion proceeds. What can you conclude about the rate laws of these reactions from these observations?

[14.97] A first-order reaction A ¡ B has the rate constant k = 3.2 * 10-3 s-1. If the initial concentration of A is 2.5 * 10-2 M, what is the rate of the reaction at t = 660 s?

14.98 (a) The reaction H2O21aq2 ¡ H2O1l2 + 12O21g2 is first

order. Near room temperature, the rate constant equals 7.0 * 10-4 s-1. Calculate the half-life at this tempera-ture. (b) At 415 °C, 1CH222O decomposes in the gas phase, 1CH222O1g2 ¡ CH41g2 + CO1g2. If the reaction is first order with a half-life of 56.3 min at this temperature, calculate the rate constant in s-1.

14.99 Americium-241 is used in smoke detectors. It has a first–order rate constant for radioactive decay of k = 1.6 * 10-3 yr-1. By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of k = 0.011 day-1. (a) What are the half-lives of these two iso-topes? (b) Which one decays at a faster rate? (c) How much of a 1.00-mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00-mg sample of each isotope remains after 4 days?

14.100 Urea 1NH2CONH22 is the end product in protein metabolism in animals. The decomposition of urea in 0.1 M HCl occurs according to the reaction

NH2CONH21aq2 + H +1aq2 + 2 H2O1l2 ¡ 2 NH4+1aq2 + HCO -

3 1aq2 The reaction is first order in urea and first order overall.

When 3NH2CONH24 = 0.200 M, the rate at 61.05 °C is 8.56 * 10-5 M>s. (a) What is the rate constant, k? (b) What is the concentration of urea in this solution after 4.00 * 103 s if the starting concentration is 0.500 M? (c) What is the half-life for this reaction at 61.05 °C ?

[14.101] The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 * 103 M -1 cm-1 at 520 nm. (a) Calculate the initial con-centration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at 30.0 min. Calculate the rate constant in units of s- 1. (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to 0.100?

[14.102] A colored dye compound decomposes to give a colorless product. The original dye absorbs at 608 nm and has an ex-tinction coefficient of 4.7 * 104 M -1cm-1 at that wavelength. You perform the decomposition reaction in a 1-cm cuvette in a spectrometer and obtain the following data:


time (min) absorbance at 608 nm

0 1.25430 0.94160 0.75290 0.672

120 0.545

From these data, determine the rate law for the reaction “dye ¡ product” and determine the rate constant.

14.103 Cyclopentadiene 1C5H62 reacts with itself to form dicyclopen-tadiene 1C10H122. A 0.0400 M solution of C5H6 was monitored as a function of time as the reaction 2 C5H6 ¡ C10H12 proceeded. The following data were collected:

time (s) 3C5h6 4 1M 2 0.0 0.0400 50.0 0.0300100.0 0.0240150.0 0.0240200.0 0.0174

Plot 3C5H64 versus time, ln3C5H64 versus time, and 1>3C5H64 versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

14.104 The first-order rate constant for reaction of a particular or-ganic compound with water varies with temperature as follows:

temperature (K) Rate Constant 1s−1 2300 3.2 * 10-11

320 1.0 * 10-9

340 3.0 * 10-8

355 2.4 * 10-7

From these data, calculate the activation energy in units of kJ>mol.

14.105 At 28 °C, raw milk sours in 4.0 h but takes 48 h to sour in a refrigerator at 5 °C. Estimate the activation energy in kJ>mol for the reaction that leads to the souring of milk?

[14.106] The following is a quote from an article in the August 18, 1998, issue of The New York Times about the breakdown of cellulose and starch: “A drop of 18 degrees Fahrenheit [from 77 °F to 59 °F4 lowers the reaction rate six times; a 36-degree drop [from 77 °F to 41 °F] produces a fortyfold decrease in the rate.” (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of Ea calculated from the 36° drop and that the rate of breakdown is first order with a half-life at 25 °C of 2.7 yr, calculate the half-life for breakdown at a temperature of -15 °C.

14.107 The following mechanism has been proposed for the reaction of NO with H2 to form N2O and H2O:

NO1g2 + NO1g2 ¡ N2O21g2N2O21g2 + H21g2 ¡ N2O1g2 + H2O1g2

(a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reac-tion. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate = k3NO423H24. If the pro-posed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

14.108 Ozone in the upper atmosphere can be destroyed by the fol-lowing two-step mechanism:

Cl1g2 + O31g2 ¡ ClO1g2 + O21g2ClO1g2 + O1g2 ¡ Cl1g2 + O21g2

(a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?

14.109 Using Figure 14.23 as your basis, draw the energy profile for the bromide-catalyzed decomposition of hydrogen peroxide. (a) Label the curve with the activation energies for reactions [14.30] and [14.31]. (b) Notice from Figure 14.22 that when Br - 1aq2 is first added, Br2 accumulates to some extent dur-ing the reaction and the solution turns brown. What does this tell us about the relative rates of the reactions represented by Equations 14.30 and 14.31?

[14.110] The following mechanism has been proposed for the gas-phase reaction of chloroform 1CHCl32 and chlorine:

Step 1: Cl21g2 ∆k1

k - 1 2 Cl1g2 1fast2

Step 2: Cl1g2 + CHCl31g2 ¡k2 HCl1g2 + CCl31g2 1slow2

Step 3: Cl1g2 + CCl31g2 ¡k3 CCl4 1fast2 (a) What is the overall reaction? (b) What are the inter-

mediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

[14.111] Consider the hypothetical reaction 2 A + B ¡ 2 C + D. The following two-step mechanism is proposed for the reaction:

Step 1: A + B ¡ C + X

Step 2: A + X ¡ C + D

X is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction or-der in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

[14.112] In a hydrocarbon solution, the gold compound 1CH323AuPH3 decomposes into ethane 1C2H62 and a different gold com-pound, 1CH32AuPH3. The following mechanism has been proposed for the decomposition of 1CH323AuPH3:

Step 1: 1CH323 AuPH3 ∆k1

k -1 1CH323Au + PH3 1fast2

Step 2: 1CH323 Au ¡k2 C2H6 + 1CH32Au 1slow2

Step 3: 1CH32Au + PH3 ¡k3 1CH32AuPH3 1fast2


(a) What is the overall reaction? (b) What are the intermedi-ates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding PH3 to the solution of 1CH323AuPH3?

[14.113] Platinum nanoparticles of diameter ∼2 nm are important catalysts in carbon monoxide oxidation to carbon diox-ide. Platinum crystallizes in a face-centered cubic arrange-ment with an edge length of 3.924 Å. (a) Estimate how many platinum atoms would fit into a 2.0-nm sphere; the volume of a sphere is 14>32pr3. Recall that 1 A° = 1 * 10-10 m and 1 nm = 1 * 10-9 m. (b) Estimate how many platinum at-oms are on the surface of a 2.0-nm Pt sphere, using the sur-face area of a sphere 14pr22 and assuming that the “footprint” of one Pt atom can be estimated from its atomic diameter of 2.8 A° . (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0-nm nanoparticle. (d) Repeat these calculations for a 5.0-nm plati-num nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

14.114 One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself

rapidly enough of the CO2 accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to 107 CO2 molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

14.115 Suppose that, in the absence of a catalyst, a certain biochemi-cal reaction occurs x times per second at normal body tem-perature 137 °C2. In order to be physiologically useful, the reaction needs to occur 5000 times faster than when it is un-catalyzed. By how many kJ>mol must an enzyme lower the activation energy of the reaction to make it useful?

14.116 Enzymes are often described as following the two-step mechanism:

E + S ∆ ES 1fast2ES ¡ E + P 1slow2

where E = enzyme, S = substrate, ES = enzyme9substrate complex, and P = product.

(a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor.

integrative exercises [14.117] Dinitrogen pentoxide 1N2O52 decomposes in chloroform as a sol-

vent to yield NO2 and O2. The decomposition is first order with a rate constant at 45 °C of 1.0 * 10-5 s-1. Calculate the partial pressure of O2 produced from 1.00 L of 0.600 M N2O5 solution at 45 °C over a period of 20.0 h if the gas is collected in a 10.0-L con-tainer. (Assume that the products do not dissolve in chloroform.)

[14.118] The reaction between ethyl iodide and hydroxide ion in etha-nol 1C2H5OH2 solution, C2H5I1alc2 + OH- 1alc2 ¡C2H5OH1l2 + I - 1alc2, has an activation energy of 86.8 kJ>mol and a frequency factor of 2.10 * 1011 M -1 s-1. (a) Predict the rate constant for the reaction at 35 °C. (b) A solution of KOH in ethanol is made up by dissolving 0.335 g KOH in ethanol to form 250.0 mL of solution. Similarly, 1.453 g of C2H5I is dissolved in ethanol to form 250.0 mL of solution. Equal vol-umes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at 35 °C? (c) Which reagent in the reaction is limiting, assuming the re-action proceeds to completion? (d) Assuming the frequency factor and activation energy do not change as a function of tem-perature, calculate the rate constant for the reaction at 50 °C.

[14.119] You obtain kinetic data for a reaction at a set of different tempera-tures. You plot ln k versus 1>T and obtain the following graph:

1/T

ln k

Suggest a molecular-level interpretation of these unusual data.

14.120 The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kJ>mol. and a frequency fac-tor of A = 6.0 * 108 M -1 s-1. The reaction is believed to be bimolecular:

NO1g2 + F21g2 ¡ NOF1g2 + F1g2 (a) Calculate the rate constant at 100 °C. (b) Draw the Lewis

structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a pos-sible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

14.121 The mechanism for the oxidation of HBr by O2 to form 2 H2O and Br2 is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process. (b) HBr does not react with O2 at a measurable rate at room temperature under ordinary conditions. What can you infer from this about the magnitude of the activation energy for the rate-determining step? (c) Draw a plausible Lewis structure for the intermediate HOOBr. To what familiar compound of hydrogen and oxygen does it appear similar?

[ 14.122] The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to pro-duce methyl chloride and hydrogen chloride:

Reaction 1: CH41g2 + Cl21g2 ¡ CH3Cl1g2 + HCl1g2


This reaction is very slow in the absence of light. However, Cl21g2 can absorb light to form Cl atoms:

Reaction 2: Cl21g2 + hv ¡ 2 Cl1g2 Once the Cl atoms are generated, they can catalyze the re-

action of CH4 and Cl2, according to the following proposed mechanism:Reaction 3: CH41g2 + Cl1g2 ¡ CH31g2 + HCl1g2

Reaction 4: CH31g2 + Cl21g2 ¡ CH3Cl1g2 + Cl1g2 The enthalpy changes and activation energies for these two

reactions are tabulated as follows:

Reaction ∆h° 1kJ ,mol 2 Ea 1kJ ,mol 23 +4 174 -109 4

(a) By using the bond enthalpy for Cl2 (Table 8.4), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the elec-tromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, CH41g2 + Cl21g2, should be placed on your diagram in part (b). Use this result to estimate the value of Ea for the reaction CH41g2 + Cl21g2 ¡ CH31g2 + HCl1g2 + Cl1g2. (d) The species Cl(g) and CH31g2 in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of CH3, and verify that it is a radical. (e) The sequence of reactions 3 and 4 comprises a

radical chain mechanism. Why do you think this is called a “chain reaction”? Propose a reaction that will terminate the chain reaction.

[14.123] Many primary amines, RNH2, where R is a carbon-containing fragment such as CH3, CH3CH2, and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid or-bital picture to visualize the bonding at the nitrogen in a primary amine (just use a C atom for “R”). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

[14.124] The NOx waste stream from automobile exhaust includes spe-cies such as NO and NO2. Catalysts that convert these spe-cies to N2. are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of NO, NO2, and N2. (b) Using a resource such as Table 8.4, look up the energies of the bonds in these molecules. In what region of the electro-magnetic spectrum are these energies? (c) Design a spectro-scopic experiment to monitor the conversion of NOx into N2, describing what wavelengths of light need to be monitored as a function of time.

[14.125] As shown in Figure 14.24, the first step in the heteroge-neous hydrogenation of ethylene is adsorption of the ethylene molecule on a metal surface. One proposed expla-nation for the “sticking” of ethylene to a metal surface is the interaction of the electrons in the C ¬ C p bond with vacant orbitals on the metal surface. (a) If this notion is correct, would ethane be expected to adsorb to a metal sur-face, and, if so, how strongly would ethane bind compared to ethylene? (b) Based on its Lewis structure, would you expect ammonia to adsorb to a metal surface using a simi-lar explanation as for ethylene?

design an experiment Let’s explore the chemical kinetics of our favorite hypothetical reaction: a A + b B ¡ c C + d D. We shall assume that all the substances are soluble in water and that we carry out the reaction in aque-ous solution. Substances A and C both absorb visible light, and the absorption maxima are 510 nm for A and for 640 nm for C. Substances B and D are colorless. You are provided with pure samples of all four substances and you know their chemical formulas. You are also provided appropriate in-strumentation to obtain visible absorption spectra (see the Closer Look box on using spectroscopic methods in Section 14.3). Let’s design an experiment to ascertain the kinetics of our reaction. (a) What experiments could you design to determine the rate law and the rate constant for the reaction at room temperature? Would you need to know the values of the stoichiometric constants a and c in order to find the rate law? (b) Design an experiment to determine the activation energy for the reaction. What challenges might you face in actually carrying out this experiment? (c) You now want to test whether a particular water-soluble substance Q is a homogeneous catalyst for the reaction. What experiments can you carry out to test this notion? (d) If Q does indeed catalyze the reaction, what follow-up experi-ments might you undertake to learn more about the reaction profile for the reaction?

15 Chemical EquilibriumTo be in equilibrium is to be in a state of balance. A tug of war in which the two sides pull with equal force so that the rope does not move is an example of a static equilibrium, one in which an object is at rest. Equilibria can also be dynamic, whereby a forward process and the reverse process take place at the same rate so that no net change occurs.

15.3 Understanding and Working With eqUilibriUm Constants We learn to interpret the magnitude of an equilibrium constant and how its value depends on the way the corresponding chemical equation is expressed.

15.4 heterogeneoUs eqUilibria We learn how to write equilibrium-constant expressions for heterogeneous reactions.

15.5 CalCUlating eqUilibriUm Constants We see that the value of an equilibrium constant can be calculated from equilibrium concentrations of reactants and products.

15.1 the ConCept of eqUilibriUm We begin by examining reversible reactions and the concept of equilibrium.

15.2 the eqUilibriUm Constant We define the equilibrium constant based on rates of forward and reverse reactions, and learn how to write equilibrium-constant expressions for homogeneous reactions.

WhaT’s ahEad

▶ SALT PILLARS in the Dead Sea. These pillars are formed in shallow bays where evaporation can lead to salt concentrations that exceed equilibrium values.

A saturated solution (Section 13.2) of an ionic compound in contact with undis-solved crystals of the same compound is a good example of dynamic equilibrium. The rate at which ions leave the surface of the crystals and enter the solution (dissolution) is equal to the rate at which ions leave the solution and become part of the solid (crystal-lization). Hence the concentration of ions in solution and the amount of undissolved solid do not change with time.

If some of the solvent is lost due to evaporation, the solution becomes more concen-trated, which increases the rate of crystallization. This change leads to a net migration of ions from the solution into the solid until the solution concentration is reduced so that the rate of crystallization and dissolution are once again equal and equilibrium is rees-tablished. A striking example of this effect is the formation of intricate salt formations in the Dead Sea, as illustrated in the chapter-opening photograph. The Dead Sea, which borders Jordan and Israel, is the lowest point on the surface of the Earth, and has a salt concentration that is almost nine times higher than the ocean. An extended period of hot weather leads to extensive evaporation of the Dead Sea, resulting in an increase in the salt concentration and subsequent crystallization and growth of intricate salt formations.

A saturated solution is one of many instances of dynamic equilibrium that we have already encountered. Vapor pressure (Section 11.5) is another example of dynamic equilibrium. The pressure of a vapor above a liquid in a closed container reaches equi-librium with the liquid phase, and therefore stops changing when the rate at which molecules escape from the liquid into the gas phase equals the rate at which molecules

15.7 le Châtelier’s prinCiple We discuss Le Châtelier’s principle, which predicts how a system at equilibrium responds to changes in concentration, volume, pressure, and temperature.

15.6 appliCations of eqUilibriUm Constants We learn that equilibrium constants can be used to predict equilibrium concentrations of reactants and products and to determine the direction in which a reaction mixture must proceed to achieve equilibrium.

630 chapter 15 chemical equilibrium

in the gas phase become part of the liquid. Henry’s law (Section 13.3), which gov-erns the solubility of gases in a solvent, is yet another example of dynamic equilibrium.

In this chapter, we consider dynamic equilibria in chemical reactions. Chemical equilibrium occurs when opposing reactions proceed at equal rates: The rate at which the products form from the reactants equals the rate at which the reactants form from the products. As a result, concentrations cease to change, making the reaction appear to be stopped. In this and the next two chapters, we will explore chemical equilibrium in some detail. Later, in Chapter 19, we will learn how to relate chemical equilibria to thermodynamics. Here, we learn how to express the equilibrium state of a reaction in quantitative terms and study the factors that determine the relative concentrations of reactants and products in equilibrium mixtures.

15.1 | The Concept of EquilibriumLet’s examine a simple chemical reaction to see how it reaches an equilibrium state—a mixture of reactants and products whose concentrations no longer change with time. We begin with N2O4, a colorless substance that dissociates to form brown NO2. ▼ Figure 15.1 shows a sample of frozen N2O4 inside a sealed tube. The solid N2O4 becomes a gas as it is warmed above its boiling point 121.2 °C2, and the gas turns darker as the colorless N2O4 gas dissociates into brown NO2 gas. Eventually, even though there is still N2O4 in the tube, the color stops getting darker because the system reaches equilibrium. We are left with an

N2O4NO2

Frozen N2O4 sample is nearly colorless

On warming, the N2O4 becomes a gas and partially dissociates to form brown NO2(g)

Colors stop changing, equilibrium reached: rate of reactionN2O4(g) 2 NO2(g) = rate of reaction 2 NO2(g) N2O4(g)

▲ Figure 15.1 The equilibrium between NO2 and N2O4.

Go FiGurEIf you were to let the tube on the right sit overnight and then take another picture would the brown color look darker, lighter, or the same?

section 15.1 the concept of equilibrium 631

equilibrium mixture of N2O4 and NO2 in which the concentrations of the gases no longer change as time passes. Because the reaction is in a closed system, where no gases can es-cape, equilibrium will eventually be reached.

The equilibrium mixture results because the reaction is reversible: N2O4 can form NO2, and NO2 can form N2O4. Dynamic equilibrium is represented by writ-ing the equation for the reaction with two half arrows pointing in opposite directions

(Section 4.1):

N2O41g2 ∆ 2 NO21g2 [15.1] Colorless Brown

We can analyze this equilibrium using our knowledge of kinetics. Let’s call the decomposition of N2O4 the forward reaction and the formation of N2O4 the reverse reaction. In this case, both the forward reaction and the reverse reaction are elementary reactions. As we learned in Section 14.6, the rate laws for elementary reactions can be written from their chemical equations:

Forward reaction: N2O41g2 ¡ 2 NO21g2 Ratef = kf3N2O44 [15.2]

Reverse reaction: 2 NO21g2 ¡ N2O41g2 Rater = kr3NO242 [15.3]

At equilibrium, the rate at which NO2 forms in the forward reaction equals the rate at which N2O4 forms in the reverse reaction:

kf 3N2O44 = kr3NO242 [15.4]Forward reaction Reverse reaction

Rearranging this equation gives

3NO242

3N2O44 =kf

kr= a constant [15.5]

From Equation 15.5, we see that the quotient of two rate constants is another constant. We also see that, at equilibrium, the ratio of the concentration terms equals this same constant. (We consider this constant, called the equilibrium constant, in Section 15.2.) It makes no difference whether we start with N2O4 or with NO2, or even with some mixture of the two. At equilibrium, at a given temperature, the ratio equals a specific value. Thus, there is an important constraint on the proportions of N2O4 and NO2 at equilibrium.

Once equilibrium is established, the concentrations of N2O4 and NO2 no longer change, as shown in ▼ Figure 15.2(a). However, the fact that the composition of the equilibrium mixture remains constant with time does not mean that N2O4 and NO2 stop reacting. On the contrary, the equilibrium is dynamic—which means some N2O4 is always converting to NO2 and some NO2 is always converting to N2O4. At equilibrium, however, the two processes occur at the same rate, as shown in Figure 15.2(b).

Time

(a)

0

Con

cent

rati

on

Equilibriumachieved

Equilibriumachieved

(rates are equal)

Time

(b)

0

Rat

e

kf [N2O4]

kr[NO2]2

NO2

N2O4

▲ Figure 15.2 Achieving chemical equilibrium in the N2O4 1g 2 ∆ 2 NO2 1g 2 reaction. Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction.

Go FiGurEAt equilibrium, is the ratio 3NO24>3N2O44 less than, greater to, or equal to 1?


We learn several important lessons about equilibrium from this example: • At equilibrium, the concentrations of reactants and products no longer change

with time. • For equilibrium to occur, neither reactants nor products can escape from the

system. • At equilibrium, a particular ratio of concentration terms equals a constant.

Give It Some Thought(a) Which quantities are equal in a dynamic equilibrium?(b) If the rate constant for the forward reaction in Equation 15.1 is larger than

the rate constant for the reverse reaction, will the constant in Equation 15.5 be greater than 1 or smaller than 1?

15.2 | The Equilibrium ConstantA reaction in which reactants convert to products and products convert to reactants in the same reaction vessel naturally leads to an equilibrium, regardless of how compli-cated the reaction is and regardless of the nature of the kinetic processes for the forward and reverse reactions. Consider the synthesis of ammonia from nitrogen and hydrogen:

N21g2 + 3 H21g2 ∆ 2 NH31g2 [15.6]

This reaction is the basis for the Haber process, which is critical for the production of fertilizers and therefore critical to the world’s food supply. In the Haber process, N2 and H2 react at high pressure and temperature in the presence of a catalyst to form ammonia. In a closed system, however, the reaction does not lead to complete consumption of the N2 and H2. Rather, at some point the reaction appears to stop with all three compo-nents of the reaction mixture present at the same time.

How the concentrations of H2, N2, and NH3 vary with time is shown in ▼ Figure 15.3. Notice that an equilibrium mixture is obtained regardless of whether we begin with N2 and H2 or with NH3. The equilibrium condition is reached from either direction.

▲ Figure 15.3 The same equilibrium is reached whether we start with only reactants (N2 and H2) or with only product 1NH3 2 .

Time

Con

cent

rati

onC

once

ntra

tion

H2

NH3N2

Equilibrium achieved

Time

H2NH3N2


Starting with only reactants

Starting with only products

Time

Con

cent

rati

onC

once

ntra

tion

H2

NH3N2


Time

H2NH3N2


Starting with only reactants

Starting with only products

Give It Some ThoughtHow do we know when equilibrium has been reached in a chemical reaction?

An expression similar to Equation 15.5 governs the concentrations of N2, H2,and NH3 at equilibrium. If we were to systematically change the relative amounts of the three gases in the starting mixture and then analyze each equilibrium mixture, we could determine the relationship among the equilibrium concentrations.

section 15.2 the equilibrium constant 633

Chemists carried out studies of this kind on other chemical systems in the nine-teenth century before Haber’s work. In 1864, Cato Maximilian Guldberg (1836–1902) and Peter Waage (1833–1900) postulated their law of mass action, which expresses, for any reaction, the relationship between the concentrations of the reactants and products present at equilibrium. Suppose we have the general equilibrium equation

a A + b B ∆ d D + e E [15.7]

where A, B, D, and E are the chemical species involved and a, b, d, and e are their coef-ficients in the balanced chemical equation. According to the law of mass action, the equilibrium condition is described by the expression

Kc =3D4d3E4e

3A4a3B4b — products— reactants

[15.8]

We call this relationship the equilibrium-constant expression (or merely the equilibrium expression) for the reaction. The constant Kc, the equilibrium constant, is the numerical value obtained when we substitute molar equilibrium concentrations into the equilibrium-constant expression. The subscript c on the K indicates that concentrations expressed in molarity are used to evaluate the constant.


The Haber Process

The quantity of food required to feed the ever-increasing human population far exceeds that provided by nitrogen-fixing plants.

(Section 14.7, “Nitrogen Fixation and Nitrogenase”) Therefore, human agriculture requires substantial amounts of ammonia-based fertilizers for croplands. Of all the chemical reactions that humans have learned to control for their own purposes, the synthesis of ammonia from hydrogen and atmospheric nitrogen is one of the most important.

In 1912, the German chemist Fritz Haber (1868–1934) developed the Haber process (Equation 15.6). The process is sometimes also called the Haber–Bosch process to honor Karl Bosch, the engineer who developed the industrial process on a large scale (▶ Figure 15.4). The engineering needed to implement the Haber process requires the use of temperatures and pressures (approximately 500 °C and 200 to 600 atm) that were difficult to achieve at that time.

The Haber process provides a historically interesting example of the complex impact of chemistry on our lives. At the start of World War I, in 1914, Germany depended on nitrate deposits in Chile for the nitrogen-containing compounds needed to manufacture explosives. During the war, the Allied naval blockade of South America cut off this supply. How-ever, by using the Haber reaction to fix nitrogen from air, Germany was able to continue to produce explosives. Experts have estimated that World War I would have ended before 1918 had it not been for the Haber process.

From these unhappy beginnings as a major factor in international warfare, the Haber process has become the world’s principal source of fixed nitrogen. The same process that prolonged World War I has enabled the manufacture of fertilizers that have increased crop yields, thereby saving millions of people from starvation. About 40 billion pounds of ammonia are manufactured annually in the United States, mostly by the Haber process. The ammonia can be applied directly to the soil, or it can be converted into ammonium salts that are also used as fertilizers.

Haber was a patriotic German who gave enthusiastic support to his nation’s war effort. He served as chief of Germany’s Chemical Warfare Service during World War I and developed the use of chlo-rine as a poison-gas weapon. Consequently, the decision to award him the Nobel Prize in Chemistry in 1918 was the subject of considerable controversy and criticism. The ultimate irony, however, came in 1933 when Haber was expelled from Germany because he was Jewish.


▲ Figure 15.4 A high-pressure steel reactor used in the Haber process is on display at Karlsruhe Institute of Technology in Germany where the Haber process was developed.


The numerator of the equilibrium-constant expression is the product of the con-centrations of all substances on the product side of the equilibrium equation, each raised to a power equal to its coefficient in the balanced equation. The denominator is similarly derived from the reactant side of the equilibrium equation. Thus, for the Haber process, N21g2 + 3 H21g2 ∆ 2 NH31g2, the equilibrium-constant expression is

Kc =3NH342

3N243H243 [15.9]

Once we know the balanced chemical equation for a reaction that reaches equi-librium, we can write the equilibrium-constant expression even if we do not know the reaction mechanism. The equilibrium-constant expression depends only on the stoichi-ometry of the reaction, not on its mechanism.

The value of the equilibrium constant at any given temperature does not depend on the initial amounts of reactants and products. It also does not matter whether other substances are present, as long as they do not react with a reactant or a product. The value of Kc depends only on the particular reaction and on the temperature.

soluTionAnalyze We are given three equations and are asked to write an equilibrium-constant expression for each.

Plan Using the law of mass action, we write each expression as a quo-tient having the product concentration terms in the numerator and the reactant concentration terms in the denominator. Each concen-tration term is raised to the power of its coefficient in the balanced chemical equation.

Solve

(a) Kc =3O243

3O342 (b) Kc =3NOCl42

3NO423Cl24 (c) Kc =

3Ag1NH322+4

3Ag+43NH342

samPlE ExErCisE 15.1 Writing Equilibrium-Constant Expressions

Write the equilibrium expression for Kc for the following reactions:(a) 2 O31g2 ∆ 3 O21g2(b) 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2(c) Ag+1aq2 + 2 NH31aq2 ∆ Ag1NH322

+1aq2

Practice Exercise 1For the reaction 2 SO21g2 + O21g2 ∆ 2 SO31g2 which of the following is the correct equilibrium-constant expression?

(a) KC =3SO2423O243SO342 (b) KC =

23SO243O2423SO34

(c) KC =3SO342

3SO2423O24 (d) KC =

23SO3423SO243O24

Practice Exercise 2Write the equilibrium-constant expression Kc for (a) H21g2 + I21g2 ∆ 2 HI1g2, (b) Cd2+1aq2 + 4 Br-1aq2 ∆ CdBr4

2-1aq2.

Evaluating Kc

We can illustrate how the law of mass action was discovered empirically and demon-strate that the equilibrium constant is independent of starting concentrations by exam-ining a series of experiments involving dinitrogen tetroxide and nitrogen dioxide:

N2O41g2 ∆ 2 NO21g2 Kc =3NO242

3N2O44 [15.10]

We start with several sealed tubes containing different concentrations of NO2 and N2O4. The tubes are kept at 100 °C until equilibrium is reached. We then analyze the mixtures and determine the equilibrium concentrations of NO2 and N2O4, which are shown in ▶ Table 15.1.

To evaluate Kc, we insert the equilibrium concentrations into the equilibrium-constant expression. For example, using Experiment 1 data, 3NO24 = 0.0172 M and 3N2O44 = 0.00140 M, we find

Kc =3NO242

3N2O44 =30.017242

0.00140= 0.211

section 15.2 the equilibrium constant 635

Proceeding in the same way, the values of Kc for the other samples are calculated. Note from Table 15.1 that the value for Kc is constant (within the limits of experimental error) even though the initial concentrations vary. Furthermore, Experiment 4 shows that equilibrium can be achieved beginning with N2O4 rather than with NO2. That is, equilibrium can be approached from either direction. ▶ Figure 15.5 shows how Experi-ments 3 and 4 result in the same equilibrium mixture even though the two experiments start with very different NO2 concentrations.

Notice that no units are given for Kc either in Table 15.1 or in the calculation we just did using Experiment 1 data. It is common practice to write equilibrium constants without units for reasons that we address later in this section.

Give It Some ThoughtHow does the value of Kc in Equation 15.10 depend on the starting concentrations of NO2 and N2O4?

Equilibrium Constants in Terms of Pressure, Kp

When the reactants and products in a chemical reaction are gases, we can formulate the equilibrium-constant expression in terms of partial pressures. When partial pres-sures in atmospheres are used in the expression, we denote the equilibrium constant Kp (where the subscript p stands for pressure). For the general reaction in Equation 15.7, we have

Kp =1PD2d1PE2e

1PA2a1PB2b [15.11]

where PA is the partial pressure of A in atmospheres, PB is the partial pressure of B in atmospheres, and so forth. For example, for our N2O4>NO2 reaction, we have

Kp =1PNO2

22

PN2O4

Give It Some ThoughtWhat is the difference between the equilibrium constant Kc and the equilibrium constant Kp?

For a given reaction, the numerical value of Kc is generally different from the numerical value of Kp. We must therefore take care to indicate, via subscript c or p, which constant we are using. It is possible, however, to calculate one from the other using the ideal-gas equation (Section 10.4):

PV = nRT, so P =nV

RT [15.12]

The usual units for n>V are mol>L, which equals molarity, M. For substance A in our generic reaction, we therefore see that

PA =nA

VRT = 3A4RT [15.13]

Table 15.1 initial and Equilibrium Concentrations of n2o4 1g 2 and no2 1g 2 at 100 °C

Experimentinitial 3n2o4 4 1M 2

initial 3no2 4 1M 2

Equilibrium 3n2o4 4 1M 2

Equilibrium 3no2 4 1M 2 Kc

1 0.0 0.0200 0.00140 0.0172 0.2112 0.0 0.0300 0.00280 0.0243 0.2113 0.0 0.0400 0.00452 0.0310 0.2134 0.0200 0.0 0.00452 0.0310 0.213

▲ Figure 15.5 The same equilibrium mixture is produced regardless of the initial NO2 concentration. The concentration of NO2 either increases or decreases until equilibrium is reached.

0

0.0100

0.0200

0.0300

0.0400

[NO

2] (M

)

Time

Experiment 4

Experiment 3


When we substitute Equation 15.13 and like expressions for the other gaseous com-ponents of the reaction into Equation 15.11, we obtain a general expression relating Kp and Kc :

Kp = Kc1RT2∆n [15.14]

The quantity ∆n is the change in the number of moles of gas in the balanced chemical equation. It equals the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants:

∆n = 1moles of gaseous product2 - 1moles of gaseous reactant2 [15.15]

For example, in the N2O41g2 ∆ 2 NO21g2 reaction, there are two moles of product NO2 and one mole of reactant N2O4. Therefore, ∆n = 2 - 1 = 1, and Kp = Kc1RT2, for this reaction.

Give It Some ThoughtIs it possible to have a reaction where Kc = Kp? If so, under what conditions would this relationship hold?

For the Haber process,

N21g2 + 3 H21g2 ∆ 2 NH31g2Kc = 9.60 at 300 °C. Calculate Kp for this reaction at this temperature.

soluTionAnalyze We are given Kc for a reaction and asked to calculate Kp.Plan The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must determine ∆n by comparing the number of moles of product with the number of moles of reactants (Equation 15.15).Solve With 2 mol of gaseous products 12 NH32 and 4 mol of gaseous reactants 11 N2 + 3 H22, ∆n = 2 - 4 = -2. (Remember that ∆ functions are always based on products minus re-actants.) The temperature is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.08206 L@atm>mol@K. Using Kc = 9.60, we therefore have

Kp = Kc1RT2∆n = 19.60210.08206 * 5732-2 =19.602

10.08206 * 57322 = 4.34 * 10-3

Practice Exercise 1For which of the following reactions is the ratio Kp>Kc largest at 300 K? (a) N21g2 + O21g2 ∆ 2 NO1g2 (b) CaCO31s2 ∆ CaO1s2 + CO21g2 (c) Ni1CO241g2 ∆ Ni1s2 + 4 CO1g2 (d) C1s2 + 2 H21g2 ∆ CH41g2Practice Exercise 2For the equilibrium 2 SO31g2 ∆ 2 SO21g2 + O21g2, Kc is 4.08 * 10-3 at 1000 K. Calcu-late the value for Kp.

samPlE ExErCisE 15.2 Converting between Kc and Kp

Equilibrium Constants and UnitsYou may wonder why equilibrium constants are reported without units. The equilib-rium constant is related to the kinetics of a reaction as well as to the thermodynamics. (We explore this latter connection in Chapter 19.) Equilibrium constants derived from thermodynamic measurements are defined in terms of activities rather than concentra-tions or partial pressures.

section 15.3 Understanding and Working with equilibrium constants 637

The activity of any substance in an ideal mixture is the ratio of the concentration or pressure of the substance either to a reference concentration 11 M2 or to a refer-ence pressure (1 atm). For example, if the concentration of a substance in an equilib-rium mixture is 0.010 M, its activity is 0.010 M>1 M = 0.010. The units of such ratios always cancel and, consequently, activities have no units. Furthermore, the numeri-cal value of the activity equals the concentration. For pure solids and pure liquids, the situation is even simpler because the activities then merely equal 1 (again with no units).

In real systems, activities are also ratios that have no units. Even though these activities may not be exactly numerically equal to concentrations, we will ignore the differences. All we need to know at this point is that activities have no units. As a result, the thermodynamic equilibrium constants derived from them also have no units. It is therefore common practice to write all types of equilibrium constants without units, a practice that we adhere to in this text. In more advanced chemis-try courses, you may make more rigorous distinctions between concentrations and activities.

Give It Some ThoughtIf the concentration of N2O4 in an equilibrium mixture is 0.00140 M, what is its activity? (Assume the solution is ideal.)

15.3 | understanding and Working with Equilibrium Constants

Before doing calculations with equilibrium constants, it is valuable to understand what the magnitude of an equilibrium constant can tell us about the relative concentrations of reactants and products in an equilibrium mixture. It is also useful to consider how the magnitude of any equilibrium constant depends on how the chemical equation is expressed.

The Magnitude of Equilibrium ConstantsThe magnitude of the equilibrium constant for a reaction gives us important informa-tion about the composition of the equilibrium mixture. For example, consider the ex-perimental data for the reaction of carbon monoxide gas and chlorine gas at 100 °C to form phosgene 1COCl22, a toxic gas used in the manufacture of certain polymers and insecticides:

CO1g2 + Cl21g2 ∆ COCl21g2 Kc =3COCl243CO43Cl24 = 4.56 * 109

For the equilibrium constant to be so large, the numerator of the equilibrium-constant expression must be approximately a billion 11092 times larger than the denominator. Thus, the equilibrium concentration of COCl2 must be much greater than that of CO or Cl2, and in fact, this is just what we find experimentally. We say that this equilibrium lies to the right (that is, toward the product side). Likewise, a very small equilibrium constant indicates that the equilibrium mixture contains mostly reactants. We then say that the equilibrium lies to the left. In general,

If K W 1 (large K): Equilibrium lies to right, products predominate

If K V 1 (small K): Equilibrium lies to left, reactants predominate

These situations are summarized in ▶ Figure 15.6. Remember, it is forward and reverse reaction rates, not reactant and product concentrations, that are equal at equilibrium.

▲ Figure 15.6 Relationship between magnitude of K and composition of an equilibrium mixture.

Go FiGurEWhat would this figure look like for a reaction in which K ≈ 1?

Reactants Products

Reactants Products

K >> 1, equilibrium “lies to the right”

K << 1, equilibrium “lies to the left”


soluTionAnalyze We are asked to judge the relative magnitudes of three equilibrium constants and then to calculate them.Plan (a) The more product present at equilibrium, relative to reactant, the larger the equilibrium constant. (b) The equilibrium constant is given by Equation 15.8.Solve

(a) Each box contains 10 spheres. The amount of product in each varies as follows: (i) 6, (ii) 1, (iii) 8. Therefore, the equilibrium constant varies in the order (ii) 6 (i) 6 (iii), from smallest (most reactant) to largest (most products).

(b) In (i), we have 0.60 mol>L product and 0.40 mol>L reactant, giving Kc = 0.60>0.40 = 1.5. (You will get the same result by merely dividing the number of spheres of each kind: 6 spheres>4 spheres = 1.5.) In (ii), we have 0.10 mol>L product and 0.90 mol>L reactant, giving Kc = 0.10>0.90 = 0.11 (or 1 sphere>9 spheres = 0.11). In (iii), we have 0.80 mol>L product and 0.20 mol>L reactant, giving Kc = 0.80>0.20 = 4.0 (or 8 spheres>2 spheres = 4.0). These calculations verify the order in (a).

Comment Imagine a drawing that represents a reaction with a very small or very large value of Kc. For example, what would the drawing look like if Kc = 1 * 10-5? In that case there would need to be 100,000 reactant molecules for only 1 product molecule. But then, that would be impractical to draw.

Practice Exercise 1The equilibrium constant for the reaction N2O41g2 ∆ 2 NO21g2 at 2 °C is Kc = 2.0. If each yellow sphere represents 1 mol of N2O4 and each brown sphere 1 mol of NO2 which of the following 1.0 L containers represents the equilibrium mixture at 2 °C?

(ii)(i) (iii)

(b)(a) (c) (d) (e)

2

Practice Exercise 2For the reaction H21g2 + I21g2 ∆ 2 HI1g2, Kp = 794 at 298 K and Kp = 55 at 700 K. Is the formation of HI favored more at the higher or lower temperature?

The following diagrams represent three systems at equilibrium, all in the same-size containers. (a) Without doing any calculations, rank the systems in order of increasing Kc. (b) If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate Kc for each system.

samPlE ExErCisE 15.3 interpreting the magnitude of an

Equilibrium Constant

section 15.3 Understanding and Working with equilibrium constants 639

The Direction of the Chemical Equation and KWe have seen that we can represent the N2O4>NO2 equilibrium as

N2O41g2 ∆ 2 NO21g2 Kc =3NO242

3N2O44 = 0.212 1at 100 °C2 [15.16]

We could equally well consider this equilibrium in terms of the reverse reaction:

2 NO21g2 ∆ N2O41g2The equilibrium expression is then

Kc =3N2O443NO242 =

10.212

= 4.72 1at 100 °C2 [15.17]

Equation 15.17 is the reciprocal of the expression in Equation 15.16. The equilib-rium-constant expression for a reaction written in one direction is the reciprocal of the expression for the reaction written in the reverse direction. Consequently, the numerical value of the equilibrium constant for the reaction written in one direc-tion is the reciprocal of that for the reverse reaction. Both expressions are equally valid, but it is meaningless to say that the equilibrium constant for the equilibrium between NO2 and N2O4 is “0.212” or “4.72” unless we indicate how the equilibrium reaction is written and specify the temperature. Therefore, whenever you are using an equilibrium constant, you should always write the associated balanced chemi-cal equation.

Give It Some ThoughtFor the reaction PCl51g2 ∆ PCl31g2 + Cl21g2, the equilibrium constant KC = 1.1 * 10-2 at 400 K. What is the equilibrium constant for the reaction PCl31g2 + Cl21g2 ∆ PCl51g2 at 400 K?

Relating Chemical Equation Stoichiometry and Equilibrium ConstantsThere are many ways to write a balanced chemical equation for a given reaction. For example, if we multiply Equation 15.1, N2O41g2 ∆ 2 NO21g2 by 2, we have

2 N2O41g2 ∆ 4 NO21g2This chemical equation is balanced and might be written this way in some contexts. Therefore, the equilibrium-constant expression for this equation is

Kc =3NO244

3N2O442

which is the square of the equilibrium-constant expression given in Equation 15.10 for the reaction as written in Equation 15.1: 3NO242>3N2O44. Because the new equilibrium-constant expression equals the original expression squared, the new equi-librium constant Kc equals the original constant squared: 0.2122 = 0.0449 1at 100 °C2. Once again, it is important to remember that you must relate each equilibrium con-stant you work with to a specific balanced chemical equation. The concentrations of the substances in the equilibrium mixture will be the same no matter how you write the chemical equation, but the value of Kc you calculate depends completely on how you write the reaction.

Give It Some ThoughtHow does the magnitude of Kp for the reaction 2 HI1g2 ∆ H21g2 + I21g2 change if the equilibrium is written 6 HI1g2 ∆ 3 H21g2 + 3 I21g2?


It is also possible to calculate the equilibrium constant for a reaction if we know the equilibrium constants for other reactions that add up to give us the one we want, similar to Hess’s law. (Section 5.6) For example, consider the following two reac-tions, their equilibrium-constant expressions, and their equilibrium constants at 100 °C:

1. 2 NOBr1g2 ∆ 2 NO1g2 + Br21g2 Kc1 =3NO423Br243NOBr42 = 0.014

2. Br21g2 + Cl21g2 ∆ 2 BrCl1g2 Kc2 =3BrCl42

3Br243Cl24 = 7.2

The net sum of these two equations is: 3. 2 NOBr1g2 + Cl21g2 ∆ 2 NO1g2 + 2 BrCl1g2You can prove algebraically that the equilibrium-constant expression for reaction 3 is the product of the expressions for reactions 1 and 2:

Kc3 =3NO423BrCl42

3NOBr423Cl24=

3NO423Br243NOBr42 *

3BrCl42

3Br243Cl24Thus,

Kc3 = 1Kc121Kc22 = 10.014217.22 = 0.10

To summarize:

1. The equilibrium constant of a reaction in the reverse direction is the inverse (or reciprocal) of the equilibrium constant of the reaction in the forward direction:

A + B ∆ C + D K1

C + D ∆ A + B K = 1>K1

2. The equilibrium constant of a reaction that has been multiplied by a number is equal to the original equilibrium constant raised to a power equal to that number.

A + B ∆ C + D K1

nA + nB ∆ nC + nD K = K1n

3. The equilibrium constant for a net reaction made up of two or more reactions is the product of the equilibrium constants for the individual reactions:

A + B ∆ C + D K1

C + F ∆ G + A K2

B + F ∆ D + G K3 = 1K121K22

samPlE ExErCisE 15.4 Combining Equilibrium Expressions

Given the reactions HF1aq2 ∆ H+1aq2 + F-1aq2 Kc = 6.8 * 10-4

H2C2O41aq2 ∆ 2 H+1aq2 + C2O42 - 1aq2 Kc = 3.8 * 10-6

determine the value of Kc for the reaction2 HF1aq2 + C2O4

2 - 1aq2 ∆ 2 F-1aq2 + H2C2O41aq2soluTionAnalyze We are given two equilibrium equations and the correspond-ing equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two.

Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate these equations to come up with equations that we can add to give us the desired equation.

section 15.4 heterogeneous equilibria 641

2 H+1aq2 + C2O42 - 1aq2 ∆ H2C2O41aq2

15.4 | heterogeneous EquilibriaMany equilibria involve substances that are all in the same phase. Such equilibria are called homogeneous equilibria. In some cases, however, the substances in equilibrium are in different phases, giving rise to heterogeneous equilibria. As an example of the lat-ter, consider the equilibrium that occurs when solid lead(II) chloride dissolves in water to form a saturated solution:

PbCl21s2 ∆ Pb2+1aq2 + 2 Cl-1aq2 [15.18]

This system consists of a solid in equilibrium with two aqueous species. If we want to write the equilibrium-constant expression for this process, we encounter a problem we have not encountered previously: How do we express the concentration of a solid? If we were to carry out experiments starting with varying amounts of products and reactants we would find that the equilibrium-constant expression for the reaction of Equation 15.18 is

Kc = 3Pb2+43Cl-42 [15.19]

Thus, our problem of how to express the concentration of a solid is not relevant in the end, because PbCl21s2 does not show up in the equilibrium-constant expression. More generally, we can state that whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression.

The fact that pure solids and pure liquids are excluded from equilibrium-constant expressions can be explained in two ways. First, the concentration of a pure solid or liq-uid has a constant value. If the mass of a solid is doubled, its volume also doubles. Thus, its concentration, which relates to the ratio of mass to volume, stays the same. Because

Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get 2 HF1aq2 ∆ 2 H+1aq2 + 2 F-1aq2 Kc = 16.8 * 10-422 = 4.6 * 10-7

Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives Kc =

13.8 * 10-6 = 2.6 * 105

Now, we have two equations that sum to give the net equation, and we can multiply the individual Kc values to get the desired equilibrium constant.

2 HF1aq2 ∆ 2 H+1aq2 + 2 F-1aq22 H+1aq2 + C2O4

2 - 1aq2 ∆ H2C2O41aq22 HF1aq2 + C2O4

2 - 1aq2 ∆ 2 F-1aq2 + H2C2O41aq2

Kc = 4.6 * 10-7

Kc = 2.5 * 105

Kc = 14.6 * 10-7212.6 * 1052 = 0.12

Practice Exercise 1Given the equilibrium constants for the following two reactions in aqueous solution at 25 °C,

HNO21aq2 ∆ H+1aq2 + NO2-1aq2 Kc = 4.5 * 10-4

H2SO31aq2 ∆ 2 H+1aq2 + SO3-1aq2 Kc = 1.1 * 10-9

what is the value of Kc for the reaction?

2 HNO21aq2 + SO 2 -3 1aq2 ∆ H2SO31aq2 + 2 NO2

-1aq2(a) 4.9 * 10-13, (b) 4.1 * 105, (c) 8.2 * 105, (d) 1.8 * 102, (e) 5.4 * 10-3.

Practice Exercise 2Given that, at 700 K, Kp = 54.0 for the reaction H21g2 + I21g2 ∆ 2 HI1g2 and Kp = 1.04 * 10-4 for the reaction N21g2 + 3 H21g2 ∆ 2 NH31g2, determine the value of Kp for the reaction 2 NH31g2 + 3 I21g2 ∆ 6 HI1g2 + N21g2 at 700 K.


equilibrium-constant expressions include terms only for reactants and products whose concentrations can change during a chemical reaction, the concentrations of pure sol-ids and pure liquids are omitted.

The omission can also be rationalized in a second way. Recall from Section 15.2 that what is substituted into a thermodynamic equilibrium expression is the activity of each substance, which is a ratio of the concentration to a reference value. For a pure substance, the reference value is the concentration of the pure substance, so that the activity of any pure solid or liquid is always 1.

Give It Some ThoughtWrite the equilibrium-constant expression for the evaporation of water, H2O1l2 ∆ H2O1g2, in terms of partial pressures.

Decomposition of calcium carbonate is another example of a heterogeneous reaction:

CaCO31s2 ∆ CaO1s2 + CO21g2

Omitting the concentrations of the solids from the equilibrium-constant expression gives

Kc = 3CO24 and Kp = PCO2

These equations tell us that at a given temperature, an equilibrium among CaCO3, CaO, and CO2 always leads to the same CO2 partial pressure as long as all three components are pres-ent. As shown in ▼ Figure 15.7, we have the same CO2 pressure regardless of the relative amounts of CaO and CaCO3.

CaO CaOCaCO3 CaCO3

CO2 (g)CO2 (g) CO2 (g)CO2 (g)

CaCO3(s) CaO(s) + CO2(g)

Large amount of CaCO3, small amount of CaO, gas pressure P

Small amount of CaCO3, large amount of CaO, gas pressure still P

▲ Figure 15.7 At a given temperature, the equilibrium pressure of CO2 in the bell jars is the same no matter how much of each solid is present.

Go FiGurEIf some of the CO21g2 were released from the upper bell jar and the seal then restored and the system allowed to return to equilibrium, would the amount of CaCO31s2 increase, decrease or remain the same?

section 15.4 heterogeneous equilibria 643

soluTionAnalyze We are given two chemical equations, both for heterogeneous equilibria, and asked to write the corresponding equilibrium-constant expressions.Plan We use the law of mass action, remembering to omit any pure solids and pure liquids from the expressions.Solve

(a) The equilibrium-constant expression is

Kc =3CO4

3CO243H24Because H2O appears in the reaction as a liquid, its concentration does not appear in the equilibrium-constant expression.(b) The equilibrium-constant expression is

Kc =3CO242

3CO42

samPlE ExErCisE 15.5 Writing Equilibrium-Constant Expressions for heterogeneous reactions

Write the equilibrium-constant expression Kc for(a) CO21g2 + H21g2 ∆ CO1g2 + H2O1l2(b) SnO21s2 + 2 CO1g2 ∆ Sn1s2 + 2 CO21g2

Because SnO2 and Sn are pure solids, their concentrations do not ap-pear in the equilibrium-constant expression.

Practice Exercise 1Consider the equilibrium that is established in a saturated solution of silver chloride, Ag+1aq2 + Cl-1aq2 ∆ AgCl1s2. If solid AgCl is added to this solution, what will happen to the concentra-tion of Ag+ and Cl- ions in solution? (a) 3Ag+4 and 3Cl-4 will both increase, (b) 3Ag+4 and 3Cl-4 will both decrease, (c) 3Ag+4 will increase and 3Cl-4 will decrease, (d) 3Ag+4 will decrease and 3Cl-4 will increase, (e) neither 3Ag+4 nor 3Cl-4 will change.

Practice Exercise 2Write the following equilibrium-constant expressions:(a) Kc for Cr1s2 + 3 Ag+1aq2 ∆ Cr3+1aq2 + 3 Ag1s2,(b) Kp for 3 Fe1s2 + 4 H2O1g2 ∆ Fe3O41s2 + 4 H21g2.

soluTionAnalyze We are asked which of several combinations of species can establish an equilibrium between calcium carbonate and its decompo-sition products, calcium oxide and carbon dioxide.Plan For equilibrium to be achieved, it must be possible for both the forward process and the reverse process to occur. For the forward process to occur, some calcium carbonate must be present. For the reverse process to occur, both calcium oxide and carbon dioxide must be present. In both cases, either the necessary compounds may be present initially or they may be formed by reaction of the other species.Solve Equilibrium can be reached in all cases except (c) as long as suf-ficient quantities of solids are present. (a) CaCO3 simply decomposes, forming CaO(s) and CO21g2 until the equilibrium pressure of CO2 is attained. There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium. (b) CO2 continues to combine with CaO until the partial pressure of the CO2 decreases to the equilibrium value. (c) Because there is no CaO present, equilibrium cannot be attained; there is no way the CO2 pressure can decrease to its equilib-rium value (which would require some CO2 to react with CaO). (d) The situation is essentially the same as in (a): CaCO3 decomposes

samPlE ExErCisE 15.6 analyzing a heterogeneous Equilibrium

Each of these mixtures was placed in a closed container and allowed to stand:(a) CaCO31s2(b) CaO(s) and CO21g2 at a pressure greater than the value of Kp

(c) CaCO31s2 and CO21g2 at a pressure greater than the value of Kp

(d) CaCO31s2 and CaO1s2Determine whether or not each mixture can attain the equilibrium

CaCO31s2 ∆ CaO1s2 + CO21g2

until equilibrium is attained. The presence of CaO initially makes no difference.

Practice Exercise 1If 8.0 g of NH4HS1s2 is placed in a sealed vessel with a volume of 1.0 L and heated to 200 °C the reaction NH4HS1s2 ∆ NH31g2 + H2S1g2 will occur. When the sys-tem comes to equilibrium, some NH4HS1s2 is still present. Which of the following changes will lead to a reduction in the amount of NH4HS1s2 that is present, assuming in all cases that equilibrium is re-established following the change? (a) Adding more NH31g2 to the vessel, (b) Adding more H2S1g2 to the ves-sel, (c) Adding more NH4HS1s2 to the vessel, (d) Increasing the volume of the vessel, (e) decreasing the volume of the vessel.

Practice Exercise 2When added to Fe3O41s2 in a closed container, which one of the following substances—H21g2, H2O1g2, O21g2—allows equilibrium to be established in the reaction 3 Fe1s2 + 4 H2O1g2 ∆ Fe3O41s2 + 4 H21g2?


When a solvent is a reactant or product in an equilibrium, its concentration is omit-ted from the equilibrium-constant expression, provided the concentrations of reactants and products are low, so that the solvent is essentially a pure substance. Applying this guideline to an equilibrium involving water as a solvent,

H2O1l2 + CO3 2-1aq2 ∆ OH-1aq2 + HCO3

-1aq2 [15.20]

gives an equilibrium-constant expression that does not contain 3H2O4:

Kc =3OH-43HCO3

-43CO3

2-4 [15.21]

Give It Some ThoughtWrite the equilibrium-constant expression for the reaction

NH31aq2 + H2O1l 2 ∆ NH4+1aq2 + OH-1aq2

15.5 | Calculating Equilibrium Constants

If we can measure the equilibrium concentrations of all the reactants and products in a chemical reaction, as we did with the data in Table 15.1, calculating the value of the equilibrium constant is straightforward. We simply insert all the equilibrium concen-trations into the equilibrium-constant expression for the reaction.

soluTionAnalyze We are given a balanced equation and equilibrium partial pressures and are asked to calculate the value of the equilibrium constant.Plan Using the balanced equation, we write the equilibrium-constant expression. We then substitute the equilibrium partial pressures into the expression and solve for Kp.Solve

Kp =1PNH3

22

PN21PH2

23 =10.16622

12.46217.3823 = 2.79 * 10-5

Practice Exercise 1A mixture of gaseous sulfur dioxide and oxygen are added to a reaction vessel and heated to 1000 K where they react to form SO31g2. If the vessel contains 0.669 atm SO21g2, 0.395 atm O21g2, and 0.0851 atm SO31g2 after the system has reached

samPlE ExErCisE 15.7 Calculating K When all Equilibrium Concentrations are Known

After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium at 472 °C, it is found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate the equilibrium constant Kp for the reaction

N21g2 + 3 H21g2 ∆ 2 NH31g2

equilibrium, what is the equilibrium constant Kp for the reaction 2 SO21g2 + O21g2 ∆ 2 SO31g2? (a) 0.0410, (b) 0.322, (c) 24.4, (d) 3.36, (e) 3.11.

Practice Exercise 2An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 °C: 3CH3COOH4 = 1.65 * 10-2 M; 3H+4 = 5.44 * 10-4 M; and 3CH3COO-4 = 5.44 * 10-4 M. Calculate the equilibrium con-stant Kc for the ionization of acetic acid at 25 °C. The reaction is

CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2

Often, we do not know the equilibrium concentrations of all species in an equilib-rium mixture. If we know the initial concentrations and the equilibrium concentration of at least one species, however, we can generally use the stoichiometry of the reaction

section 15.5 calculating equilibrium constants 645

to deduce the equilibrium concentrations of the others. The following steps outline the procedure:

1. Tabulate all known initial and equilibrium concentrations of the species that ap-pear in the equilibrium-constant expression.

2. For those species for which initial and equilibrium concentrations are known, calculate the change in concentration that occurs as the system reaches equilibrium.

3. Use the stoichiometry of the reaction (that is, the coefficients in the balanced chemical equation) to calculate the changes in concentration for all other species in the equilibrium-constant expression.

4. Use initial concentrations from step 1 and changes in concentration from step 3 to calculate any equilibrium concentrations not tabulated in step 1.

5. Determine the value of the equilibrium constant.

The best way to illustrate how to do this type of calculation is by example, as we do in Sample Exercise 15.8.

soluTionAnalyze We are given the initial concentrations of H2 and I2 and the equilibrium concentration of HI. We are asked to calculate the equi-librium constant Kc for H21g2 + I21g2 ∆ 2 HI1g2.

samPlE ExErCisE 15.8 Calculating K from initial and Equilibrium Concentrations

A closed system initially containing 1.000 * 10-3 M H2 and 2.000 * 10-3 M I2 at 448 °C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 * 10-3 M. Calculate Kc at 448 °C for the reaction taking place, which is

H21g2 + I21g2 ∆ 2 HI1g2

Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant.

Solve

(1) We tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemi-cal equation as the heading for the table.

H21g2 + I21g2 ∆ 2 HI1g2Initial concentration (M) 1.000 * 10-3 2.000 * 10-3 0

Change in concentration (M)Equilibrium concentration (M) 1.87 * 10-3

(2) We calculate the change in HI concentra-tion, which is the difference between the equilibrium and initial values: Change in 3HI4 = 1.87 * 10-3 M - 0 = 1.87 * 10-3 M

(3) We use the coefficients in the balanced equation to relate the change in [HI] to the changes in 3H24 and 3I24:

a1.87 * 10-3 mol HI

Lb a 1 mol H2

2 mol HIb = 0.935 * 10-3

mol H2

L

a1.87 * 10-3 mol HI

Lb a 1 mol I2

2 mol HIb = 0.935 * 10-3

mol I2

L

(4) We calculate the equilibrium concentra-tions of H2 and I2, using initial concentra-tions and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed:

3H24 = 1.000 * 10-3 M - 0.935 * 10-3 M = 0.065 * 10-3 M

3I24 = 2.000 * 10-3 M - 0.935 * 10-3 M = 1.065 * 10-3 M

(5) Our table now is complete (with equilibrium concentrations in blue for emphasis):

H21g2 + I21g2 ∆ 2 HI1g2

Initial concentration (M) 1.000 * 10-3 2.000 * 10-3 0

Change in concentration (M) -0.935 * 10-3 -0.935 * 10-3 +1.87 * 10-3

Equilibrium concentration (M) 0.065 * 10-3 1.065 * 10-3 1.87 * 10-3


15.6 | applications of Equilibrium Constants

We have seen that the magnitude of K indicates the extent to which a reaction pro-ceeds. If K is very large, the equilibrium mixture contains mostly substances on the product side of the equation for the reaction. That is, the reaction proceeds far to the right. If K is very small (that is, much less than 1), the equilibrium mixture contains mainly substances on the reactant side of the equation. The equilibrium constant also allows us to (1) predict the direction in which a reaction mixture achieves equilibrium and (2) calculate equilibrium concentrations of reactants and products.

Predicting the Direction of ReactionFor the formation of NH3 from N2 and H2 (Equation 15.6), Kc = 0.105 at 472 °C. Sup-pose we place 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of NH3 in a 1.00-L container at 472 °C. How will the mixture react to reach equilibrium? Will N2 and H2 react to form more NH3, or will NH3 decompose to N2 and H2?

To answer this question, we substitute the starting concentrations of N2, H2, and NH3 into the equilibrium-constant expression and compare its value to the equilibrium constant:

3NH342

3N243H243 =12.0022

11.00212.0023 = 0.500 whereas Kc = 0.105 [15.22]

To reach equilibrium, the quotient 3NH342>3N243H243 must decrease from the start-ing value of 0.500 to the equilibrium value of 0.105. Because the system is closed, this change can happen only if 3NH34 decreases and 3N24 and 3H24 increase. Thus, the reac-tion proceeds toward equilibrium by forming N2 and H2 from NH3; that is, the reaction as written in Equation 15.6 proceeds from right to left.

This approach can be formalized by defining a quantity called the reaction quo-tient. The reaction quotient, Q, is a number obtained by substituting reactant and product concentrations or partial pressures at any point during a reaction into an equi-librium-constant expression. Therefore, for the general reaction

a A + b B ∆ d D + e E

Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed. Finally, we use the equilibrium-constant expression to calculate the equilibrium constant:

Kc =3HI42

3H243I24 =11.87 * 10-322

10.065 * 10-3211.065 * 10-32 = 51

Comment The same method can be applied to gaseous equilibrium problems to calculate Kp, in which case partial pressures are used as table entries in place of molar concentrations. Your instructor may refer to this kind of table as an ICE chart, where ICE stands for Initial – Change – Equilibrium.

Practice Exercise 1In Section 15.1, we discussed the equilibrium between N2O41g2 and NO21g2. Let’s return to that equation in a quantitative example. When 9.2 g of frozen N2O4 is added to a 0.50 L reaction ves-sel and the vessel is heated to 400 K and allowed to come to equilbirum, the concentration of N2O4 is determined to be 0.057 M. Given this information, what is the value of Kc for the reac-tion N2O41g2 ¡ 2 NO21g2 at 400 K? (a) 0.23, (b) 0.36, (c) 0.13, (d) 1.4, (e) 2.5.

Practice Exercise 2The gaseous compound BrCl decomposes at high temperature in a sealed container: 2 BrCl1g2 ∆ Br21g2 + Cl21g2. Initially, the vessel is charged at 500 K with BrCl(g) at a partial pressure of 0.500 atm. At equilibrium, the BrCl(g) partial pressure is 0.040 atm. Calculate the value of Kp at 500 K.

section 15.6 applications of equilibrium constants 647

the reaction quotient in terms of molar concentrations is

Qc =3D4d3E4e

3A4a3B4b [15.23]

(A related quantity Qp can be written for any reaction that involves gases by using par-tial pressures instead of concentrations.)

Although we use what looks like the equilibrium-constant expression to calculate the reaction quotient, the concentrations we use may or may not be the equilibrium concentrations. For example, when we substituted the starting concentrations into the equilibrium-constant expression of Equation 15.22, we obtained Qc = 0.500 whereas Kc = 0.105. The equilibrium constant has only one value at each temperature. The reaction quotient, however, varies as the reaction proceeds.

Of what use is Q? One practical thing we can do with Q is tell whether our reaction really is at equilibrium, which is an especially valuable option when a reaction is very slow. We can take samples of our reaction mixture as the reaction proceeds, separate the components, and measure their concentrations. Then we insert these numbers into Equation 15.23 for our reaction. To determine whether we are at equilibrium, or in which direction the reaction proceeds to achieve equilibrium, we compare the values of Qc and Kc or Qp and Kp. Three possible situations arise: • Q 6 K: The concentration of products is too small and that of reactants too large.

The reaction achieves equilibrium by forming more products; it proceeds from left to right.

• Q = K: The reaction quotient equals the equilibrium constant only if the system is at equilibrium.

• Q 7 K: The concentration of products is too large and that of reactants too small. The reaction achieves equilibrium by forming more reactants; it proceeds from right to left.

These relationships are summarized in ▶ Figure 15.8.

▲ Figure 15.8 Predicting the direction of a reaction by comparing Q and K at a given temperature.

Q

At equilibrium

Q < K

Q = K

Q > K

K

K

Q

Q

K

Equilibrium

Reaction proceeds to form more products

Reaction proceeds to form more reactants

soluTionAnalyze We are given a volume and initial molar amounts of the spe-cies in a reaction and asked to determine in which direction the reac-tion must proceed to achieve equilibrium.Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concen-trations into the equilibrium-constant expression to calculate the reaction quotient, Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed.Solve The initial concentrations are

3HI4 = 2.0 * 10-2 mol>2.00 L = 1.0 * 10-2 M 3H24 = 1.0 * 10-2 mol>2.00 L = 5.0 * 10-3 M 3I24 = 3.0 * 10-2 mol>2.00 L = 1.5 * 10-2 M

The reaction quotient is therefore

Qc =3HI42

3H243I24=

11.0 * 10-222

15.0 * 10-3211.5 * 10-22 = 1.3

samPlE ExErCisE 15.9 Predicting the direction of approach to Equilibrium

At 448 °C, the equilibrium constant Kc for the reaction

H21g2 + I21g2 ∆ 2 HI1g2is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 * 10-2 mol of HI, 1.0 * 10-2 mol of H2, and 3.0 * 10-2 mol of I2 in a 2.00-L container.

Because Qc 6 Kc, the concentration of HI must increase and the con-centrations of H2 and I2 must decrease to reach equilibrium; the reac-tion as written proceeds left to right to attain equilibrium.

Practice Exercise 1Which of the following statements accurately describes what would happen to the direction of the reaction described in the sample exercise above, if the size of the container were different from 2.00 L? (a) The reaction would proceed in the opposite direc-tion (from right to left) if the container volume were reduced suffi-ciently. (b) The reaction would proceed in the opposite direction if the container volume were expanded sufficiently. (c) The direction of this reaction does not depend on the volume of the container.

Practice Exercise 2At 1000 K, the value of Kp for the reaction 2 SO31g2 ∆ 2 SO21g2 + O21g2 is 0.338. Calculate the value for Qp, and predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are PSO3

= 0.16 atm; PSO2

= 0.41 atm; PO2= 2.5 atm.


Calculating Equilibrium ConcentrationsChemists frequently need to calculate the amounts of reactants and products present at equilibrium in a reaction for which they know the equilibrium constant. The approach in solving problems of this type is similar to the one we used for evaluating equilibrium constants: We tabulate initial concentrations or partial pressures, changes in those con-centrations or pressures, and final equilibrium concentrations or partial pressures. Usu-ally, we end up using the equilibrium-constant expression to derive an equation that must be solved for an unknown quantity, as demonstrated in Sample Exercise 15.10.

soluTionAnalyze We are given an equilibrium constant, Kp, and the equilib-rium partial pressures of two of the three substances in the equation 1N2 and H22, and we are asked to calculate the equilibrium partial pressure for the third substance 1NH32.

Plan We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that we know. Then we can solve for the only unknown in the equation.

Solve We tabulate the equilibrium pressures:

N21g2 + 3 H21g2 ∆ 2 NH31g2

Equilibrium pressure 1atm2 0.432 0.928 x

Because we do not know the equilibrium pressure of NH3, we represent it with x. At equilibrium, the pressures must satisfy the equilibrium-constant expression:

Kp =1PNH3

22

PN21PH2

23 =x2

10.432210.92823 = 1.45 * 10-5

samPlE ExErCisE 15.10 Calculating Equilibrium Concentrations

For the Haber process, N21g2 + 3 H21g2 ∆ 2 NH31g2, Kp = 1.45 * 10-5, at 500 °C. In an equi-librium mixture of the three gases at 500 °C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?

We now rearrange the equation to solve for x: x2 = 11.45 * 10-5210.432210.92823 = 5.01 * 10-6

x = 25.01 * 10-6 = 2.24 * 10-3 atm = PNH3

Check We can always check our answer by using it to recalculate the value of the equilibrium constant:

Kp =12.24 * 10-322

10.432210.92823 = 1.45 * 10-5

Practice Exercise 1At 500 K, the reaction 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2 has Kp = 51. In an equilibrium mixture at 500 K, the partial pressure of NO is 0.125 atm and Cl2 is 0.165 atm. What is the partial pres-sure of NOCl in the equilibrium mixture? (a) 0.13 atm, (b) 0.36 atm, (c) 1.0 atm, (d) 5.1 * 10-5 atm, (e) 0.125 atm.

Practice Exercise 2At 500 K, the reaction PCl51g2 ∆ PCl31g2 + Cl21g2 has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pres-sure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture?

In many situations, we know the value of the equilibrium constant and the ini-tial amounts of all species. We must then solve for the equilibrium amounts. Solving this type of problem usually entails treating the change in concentration as a variable. The stoichiometry of the reaction gives us the relationship between the changes in the amounts of all the reactants and products, as illustrated in Sample Exercise 15.11. The calculations frequently involve the quadratic formula, as you will see in this exercise.

soluTionAnalyze We are given the volume of a container, an equilibrium con-stant, and starting amounts of reactants in the container and are asked to calculate the equilibrium concentrations of all species.

samPlE ExErCisE 15.11 Calculating Equilibrium Concentrations from initial Concentrations

A 1.000-L flask is filled with 1.000 mol of H21g2 and 2.000 mol of I21g2 at 448 °C. The value of the equilibrium constant Kc for the reaction

H21g2 + I21g2 ∆ 2 HI1g2at 448 °C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter?

Plan In this case, we are not given any of the equilibrium concentra-tions. We must develop some relationships that relate the initial con-centrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.8, where we calculated an equilibrium constant using initial concentrations.

section 15.6 applications of equilibrium constants 649

Solve

(1) We note the initial concentrations of H2 and I2:

(2) We construct a table in which we tabulate the initial concentrations:

3H24 = 1.000 M and 3I24 = 2.000 M

H21g2 + I21g2 ∆ 2 HI1g2Initial concentration (M) 1.000 2.000 0Change in concentration (M)Equilibrium concentration (M)

(3) We use the stoichiometry of the reaction to determine the changes in concentration that occur as the reaction proceeds to equilib-rium. The H2 and I2 concentrations will de-crease as equilibrium is established and that of HI will increase. Let’s represent the change in concentration of H2 by x. The balanced chemical equation tells us the relationship between the changes in the concentrations of the three gases. For each x mol of H2 that re-acts, x mol of I2 are consumed and 2x mol of HI are produced:

H21g2 + I21g2 ∆ 2 HI1g2Initial concentration (M) 1.000 2.000 0Change in concentration (M) -x -x +2x

Equilibrium concentration (M)

(4) We use initial concentrations and changes in concentrations, as dictated by stoichiometry, to express the equilibrium concentrations. With all our entries, our table now looks like this:

H21g2 + I21g2 ∆ 2 HI1g2Initial concentration (M) 1.000 2.000 0Change in concentration (M) -x -x +2xEquilibrium concentration (M) 1.000 - x 2.000 - x 2x

(5) We substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x:

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

4x2 = 50.51x2 - 3.000x + 2.000246.5x2 - 151.5x + 101.0 = 0

Kc =3HI42

3H243I24=

12x22

11.000 - x212.000 - x2 = 50.5

x =-1-151.52 { 21-151.522 - 4146.521101.02

2146.52 = 2.323 or 0.935

When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find nega-tive concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this solution. We then use x = 0.935 to find the equilibrium concentrations:

3H24 = 1.000 - x = 0.065 M

3I24 = 2.000 - x = 1.065 M

3HI4 = 2x = 1.87 M

Check We can check our solution by putting these numbers into the equilibrium-constant expression to assure that we correctly calculate the equilibrium constant:

Kc =3HI42

3H243I24=

11.8722

10.065211.0652 = 51

Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the so-lutions to the equation will give you a value that leads to negative concentrations and thus is not chemically meaningful. Reject this solution to the quadratic equation.


Practice Exercise 1For the equilibrium Br21g2 + Cl21g2 ∆ 2 BrCl1g2, the equilibrium constant Kp is 7.0 at 400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl? (a) 0.57 atm, (b) 0.22 atm, (c) 0.45 atm, (d) 0.15 atm, (e) 0.31 atm.

Practice Exercise 2For the equilibrium PCl51g2 ∆ PCl31g2 + Cl21g2, the equilibrium constant Kp is 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl51g2 at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature?

15.7 | le Châtelier’s PrincipleMany of the products we use in everyday life are obtained from the chemical in-dustry. Chemists and chemical engineers in industry spend a great deal of time and effort to maximize the yield of valuable products and minimize waste. For example, when Haber developed his process for making ammonia from N2 and H2, he exam-ined how reaction conditions might be varied to increase yield. Using the values of the equilibrium constant at various temperatures, he calculated the equilibrium amounts of NH3 formed under a variety of conditions. Some of Haber’s results are shown in ▼ Figure 15.9.

Notice that the percent of NH3 present at equilibrium decreases with increasing temperature and increases with increasing pressure.

We can understand these effects in terms of a principle first put forward by Henri-Louis Le Châtelier* (1850–1936), a French industrial chemist: If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concen-tration, the system will shift its equilibrium position so as to counteract the effect of the disturbance.

*Pronounced “le-SHOT-lee-ay.”

Percent of NH3 decreases with increasing temperature

Percent of NH3 increases with increasing pressure

60.6%

54.9%

47.8%

38.7%48.8%

42.9%

35.9%

27.4%37.8%

32.2%

26.0%

18.9%20.8%

16.9%

12.9%

8.8%

450

Tem

pera

ture

(°C)

Total pressure (atm)

500

550

400500

400

300

200

Percent NH3produced

▲ Figure 15.9 Effect of temperature and pressure on NH3 yield in the Haber process. Each mixture was produced by starting with a 3 : 1 molar mixture of H2 and N2.

Go FiGurEAt what combination of pressure and temperature should you run the reaction to maximize NH3 yield?

section 15.7 Le châtelier’s principle 651

If a system at equilibrium is disturbed by a change in concentration, pressure, or temperature,the system will shift its equilibrium position so as to counter the effect of the disturbance.

Le Châtelier’s Principle

Concentration: adding or removing a reactant or productIf a substance is added to a system at equilibrium, the system reacts to consume some of the substance. If a substance is removedfrom a system, the system reacts to produce more of substance.

Pressure: changing the pressure by changing the volumeAt constant temperature, reducing the volume of a gaseous equilibriummixture causes the system to shift in the direction that reduces thenumber of moles of gas.

Temperature:

Endothermic

Reaction shifts left

Reaction shifts right

Reaction shifts right

Reaction shifts left

If the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reactionor a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the “excess reactant,” namely heat.

Initial equilibrium Substance added

Substances react

Equilibrium reestablished

Initial volume

PressureSystem shifts to direction of fewer moles of gas

Increasing T

Decreasing T Decreasing T

Increasing T

+ + +

Exothermic

In this section, we use Le Châtelier’s principle to make qualitative predictions about how a system at equilibrium responds to various changes in external conditions. We consider three ways in which a chemical equilibrium can be disturbed: (1) adding or removing a reactant or product, (2) changing the pressure by changing the volume, and (3) changing the temperature.

Change in Reactant or Product ConcentrationA system at dynamic equilibrium is in a state of balance. When the concentrations of species in the reaction are altered, the equilibrium shifts until a new state of bal-ance is attained. What does shift mean? It means that reactant and product concen-trations change over time to accommodate the new situation. Shift does not mean that the equilibrium constant itself is altered; the equilibrium constant remains the same. Le Châtelier’s principle states that the shift is in the direction that minimizes or reduces the effect of the change. Therefore, if a chemical system is already at equi-librium and the concentration of any substance in the mixture is increased (either reactant or product), the system reacts to consume some of that substance. Conversely, if the concentration of a substance is decreased, the system reacts to produce some of that substance.

There is no change in the equilibrium constant when we change the concentra-tions of reactants or products. As an example, consider our familiar equilibrium mix-ture of N2, H2, and NH3:

N21g2 + 3 H21g2 ∆ 2 NH31g2


Adding H2 causes the system to shift so as to reduce the increased concentration of H2 (▲ Figure 15.10). This change can occur only if the reaction consumes H2 and simul-taneously consumes N2 to form more NH3. Adding N2 to the equilibrium mixture like-wise causes the reaction to shift toward forming more NH3. Removing NH3 also causes a shift toward producing more NH3, whereas adding NH3 to the system at equilibrium causes the reaction to shift in the direction that reduces the increased NH3 concentra-tion: Some of the added ammonia decomposes to form N2 and H2. All of these “shifts” are entirely consistent with predictions that we would make by comparing the reaction quotient Q with the equilibrium constant K.

In the Haber reaction, therefore, removing NH3 from an equilibrium mixture of N2, H2, and NH3 causes the reaction to shift right to form more NH3. If the NH3 can be removed continuously as it is produced, the yield can be increased dramatically. In the industrial production of ammonia, the NH3 is continuously removed by selectively liquefying it (▶ Figure 15.11). (The boiling point of NH3, -33 °C, is much higher than those of N2, -196 °C, and H2, -253 °C.) The liquid NH3 is removed, and the N2 and H2 are recycled to form more NH3. As a result of the product being continuously removed, the reaction is driven essentially to completion.

Give It Some ThoughtDoes the equilibrium 2 NO1g2 + O21g2 ∆ 2 NO21g2 shift to the right (more products) or left (more reactants) if(a) O2 is added to the system?(b) NO is removed?

Effects of Volume and Pressure ChangesIf a system containing one or more gases is at equilibrium and its volume is decreased, thereby increasing its total pressure, Le Châtelier’s principle indicates that the system responds by shifting its equilibrium position to reduce the pressure. A system can re-duce its pressure by reducing the total number of gas molecules (fewer molecules of gas

Part

ial p

ress

ure

Time

N2

NH3

H2

Initialequilibrium

Equilibriumreestablished

N2(g) + 3 H2(g) 2 NH3(g)

H2 added

▲ Figure 15.10 Effect of adding H2 to an equilibrium mixture of N2, H2, and NH3. Adding H2 causes the reaction as written to shift to the right, consuming some N2 to produce more NH3.

Go FiGurEWhy does the nitrogen concentration decrease after hydrogen is added?


exert a lower pressure). Thus, at constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas. Increasing the volume causes a shift in the direction that produces more gas molecules (▼ Figure 15.12).

▲ Figure 15.11 Diagram of the industrial production of ammonia. Incoming N21g2 and H21g2 are heated to approximately 500 °C and passed over a catalyst. When the resultant N2, H2, and NH3 mixture is cooled, the NH3 liquefies and is removed from the mixture, shifting the reaction to produce more NH3.

N2 and H2 gases heated to approximately 500 °C

2

Heated gases pass over catalyst, NH3 forms

3

As gas mixture cools,NH3(g) liqui�es

4

Incoming N2 and H2 gases

1

Unreacted N2 and H2 recycled

5

N2

Liquid NH3outlet

Heatexchanger

Hot gases

Heatexchanger

H2

▲ Figure 15.12 Pressure and Le Châtelier’s principle.

Increase volume,decrease pressure

Initial volume

New equilibrium favorsproducts to reduce totalmoles of gas

New equilibrium favorsreactants to increase totalmoles of gas

Decrease volume,increase pressure

2 A(g) B(g)


Give It Some ThoughtWhat happens to the equilibrium 2 SO21g2 + O21g2 ∆ 2 SO31g2, if the volume of the system is increased?

In the reaction N21g2 + 3 H21g2 ∆ 2 NH31g2, four molecules of reactant are consumed for every two molecules of product produced. Consequently, an increase in pressure (caused by a decrease in volume) shifts the reaction in the direction that pro-duces fewer gas molecules, which leads to the formation of more NH3, as indicated in Figure 15.9. In the reaction H21g2 + I21g2 ∆ 2 HI1g2, the number of molecules of gaseous products (two) equals the number of molecules of gaseous reactants; therefore, changing the pressure does not influence the position of equilibrium.

Keep in mind that, as long as temperature remains constant, pressure–volume changes do not change the value of K. Rather, these changes alter the partial pressures of the gaseous substances. In Sample Exercise 15.7, we calculated Kp = 2.79 * 10-5 for the Haber reaction, N21g2 + 3 H21g2 ∆ 2 NH31g2, in an equilibrium mixture at 472 °C containing 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. Consider what hap-pens when we suddenly reduce the volume of the system by one-half. If there were no shift in equilibrium, this volume change would cause the partial pressures of all sub-stances to double, giving PH2

= 14.76 atm, PN2= 4.92 atm, and PNH3

= 0.332 atm. The reaction quotient would then no longer equal the equilibrium constant:

Qp =1PNH3

22

PN21PH2

23 =10.33222

14.922114.7623 = 6.97 * 10-6 ≠ Kp

Because Qp 6 Kp, the system would no longer be at equilibrium. Equilibrium would be rees-tablished by increasing PNH3

and/or decreasing PN2 and PH2

until Qp = Kp = 2.79 * 10-5. Therefore, the equilibrium shifts to the right in the reaction as written, as Le Châtelier’s prin-ciple predicts.

It is possible to change the pressure of a system in which a chemical reaction is running without changing its volume. For example, pressure increases if additional amounts of any reacting components are added to the system. We have already seen how to deal with a change in concentration of a reactant or product. However, the total pressure in the reaction vessel might also be increased by adding a gas that is not involved in the equilibrium. For example, argon might be added to the ammonia equi-librium system. The argon would not alter the partial pressures of any of the reacting components and therefore would not cause a shift in equilibrium.

Effect of Temperature ChangesChanges in concentrations or partial pressures shift equilibria without changing the value of the equilibrium constant. In contrast, almost every equilibrium constant changes as the temperature changes. For example, consider the equilibrium established when cobalt(II) chloride 1CoCl22 is dissolved in hydrochloric acid, HCl(aq), in the endothermic reaction

Co1H2O262+1aq2 + 4 Cl-1aq2 ∆ CoCl42-1aq2 + 6 H2O1l2 ∆H 7 0 [15.24]

Pale pink Deep blue

Because Co1H2O262+ is pink and CoCl4

2- is blue, the position of this equilib-rium is readily apparent from the color of the solution (▶ Figure 15.13). When the solution is heated it turns blue, indicating that the equilibrium has shifted to form more CoCl4

2-. Cooling the solution leads to a pink solution, indicating that the equi-librium has shifted to produce more Co1H2O26

2+. We can monitor this reaction by spectroscopic methods, measuring the concentration of all species at the different temperatures. (Section 14.2) We can then calculate the equilibrium constant at each temperature. How do we explain why the equilibrium constants and therefore the position of equilibrium both depend on temperature?


We can deduce the rules for the relationship between K and temperature from Le Châtelier’s principle. We do this by treating heat as a chemical reagent. In an endo-thermic (heat-absorbing) reaction, we consider heat a reactant, and in an exothermic (heat-releasing) reaction, we consider heat a product:

Endothermic: Reactants + heat ∆ productsExothermic: Reactants ∆ products + heat

When the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reaction or a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the excess reactant (or product), namely heat.

Give It Some ThoughtUse Le Châtelier’s principle to explain why the equilibrium vapor pressure of a liquid increases with increasing temperature.

In an endothermic reaction, such as Equation 15.24, heat is absorbed as reactants are converted to products. Thus, increasing the temperature causes the equilibrium to shift to the right, in the direction of making more products, and K increases. In an exo-thermic reaction, the opposite occurs: Heat is produced as reactants are converted to

▲ Figure 15.13 Temperature and Le Châtelier’s principle. In the molecular level views, only the CoCl4

2 - and Co1H2O262+ ions are shown for clarity.

Cool Heat

Heat + Co(H2O)62+(aq) + 4 Cl−(aq)

∆H > 0, endothermic reaction

Pink Blue

CoCl42−(aq) + 6 H2O(l)

CoCl42+

Solution appears violet because appreciable amounts of both pink Co(H2O)6

2+ and blue CoCl42– are

present.

Solution appears pink because lowering the temperature shifts the equilibrium to favor formation of the pink Co(H2O)6

2+ ion.

Solution appears blue because raising the temperature shifts the equilibrium to favor formation of the blue CoCl4

2– ion.

Co(H2O)62–


products. Thus, increasing the temperature in this case causes the equilibrium to shift to the left, in the direction of making more reactants, and K decreases.

Endothermic: Increasing T results in higher K value

Exothermic: Increasing T results in lower K value

Cooling a reaction has the opposite effect. As we lower the temperature, the equi-librium shifts in the direction that produces heat. Thus, cooling an endothermic reac-tion shifts the equilibrium to the left, decreasing K, as shown in Figure 15.13, and cooling an exothermic reaction shifts the equilibrium to the right, increasing K.

In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the pressure is increased by ad-dition of N21g2, (d) the volume is increased, (e) the temperature is decreased?

soluTionAnalyze We are given a series of changes to be made to a system at equilibrium and are asked to predict what effect each change will have on the position of the equilibrium.Plan Le Châtelier’s principle can be used to determine the effects of each of these changes.Solve

(a) The system will adjust to decrease the concentration of the added N2O4, so the equilibrium shifts to the right, in the direction of product.

(b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the equilibrium shifts to the right.

(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the posi-tion of the equilibrium.

(d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas molecules); thus, the equilibrium shifts to the right.

samPlE ExErCisE 15.12 using le Châtelier’s Principle to Predict shifts in Equilibrium

Consider the equilibrium

N2O41g2 ∆ 2 NO21g2 ∆H ° = 58.0 kJ

(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation. Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of the equilibrium constant, K.

Practice Exercise 1For the reaction4 NH31g2 + 5 O21g2 ∆ 4 NO1g2 + 6 H2O1g2 ∆H ° = -904 kJ

which of the following changes will shift the equilibrium to the right, toward the formation of more products? (a) Adding more water vapor, (b) Increasing the temperature, (c) Increasing the vol-ume of the reaction vessel, (d) Removing O2(g), (e) Adding 1 atm of Ne(g) to the reaction vessel.

Practice Exercise 2For the reaction

PCl51g2 ∆ PCl31g2 + Cl21g2 ∆H° = 87.9 kJin which direction will the equilibrium shift when (a) Cl21g2 is removed, (b) the temperature is decreased, (c) the volume of the reaction system is increased, (d) PCl31g2 is added?

soluTionAnalyze We are asked to determine the standard enthalpy change of a reaction and how the equilibrium constant for the reaction varies with temperature.

Plan

(a) We can use standard enthalpies of formation to calculate ∆H° for the reaction.

(b) We can then use Le Châtelier’s principle to determine what effect temperature will have on the equilibrium constant.

samPlE ExErCisE 15.13 Predicting the Effect of Temperature on K

(a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction

N21g2 + 3 H21g2 ∆ 2 NH31g2(b) Determine how the equilibrium constant for this reaction should change with temperature.

Solve

(a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar enthalpies of formation of the products, each multiplied by its coefficient in the balanced chemical equation, minus the same quantities for the reactants. (Section 5.7) At 25 °C, ∆Hf° for NH31g2 is -46.19 kJ>mol. The ∆Hf° values for H21g2 and N21g2 are zero by definition because the enthalpies of formation of the elements in their normal


The Effect of CatalystsWhat happens if we add a catalyst to a chemical system that is at equilibrium? As shown in ▼ Figure 15.14, a catalyst lowers the activation barrier between reactants and prod-ucts. The activation energies for both the forward and reverse reactions are lowered. The catalyst thereby increases the rates of both forward and reverse reactions. Since K is the ratio of the forward and reverse rate constants for a reaction, you can predict, correctly, that the presence of a catalyst, even though it changes the reaction rate, does not affect the numeric value of K (Figure 15.14). As a result, a catalyst increases the rate at which equilibrium is achieved but does not change the composition of the equilibrium mixture.

The rate at which a reaction approaches equilibrium is an important practi-cal consideration. As an example, let’s again consider the synthesis of ammonia from

states at 25 °C are defined as zero. (Section 5.7) Because 2 mol of NH3 is formed, the total enthalpy change is

12 mol21-46.19 kJ>mol2 - 0 = -92.38 kJ

(b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction. An increase in temperature causes the reaction to shift in the direction of less NH3 and more N2 and H2. This effect is seen in the values for Kp presented in ▼ Table 15.2. Notice that Kp changes markedly

with changes in temperature and that it is larger at lower temperatures.

Comment The fact that Kp for the formation of NH3 from N2 and H2 decreases with increasing temperature is a matter of great practical importance. To form NH3 at a reasonable rate requires higher tem-peratures. At higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to NH3 is smaller. To compensate for this, higher pressures are needed because high pres-sure favors NH3 formation.

Practice Exercise 1The standard enthalpy of formation of HCl(g) is -92.3 kJ>mol. Given only this information, in which direction would you expect the equilibrium for the reaction H21g2 + Cl21g2 ∆ 2 HCl1g2 to shift as the temperature increases: (a) to the left, (b) to the right, (c) no shift in equilibrium?

Practice Exercise 2Using the thermodynamic data in Appendix C, determine the en-thalpy change for the reaction

2 POCl31g2 ∆ 2 PCl31g2 + O21g2Use this result to determine how the equilibrium constant for the reaction should change with temperature.

Table 15.2 Variation in Kp with Temperature for n2 + 3h2 ∆ 2 nh3

Temperature 1 °C 2 Kp

300 4.34 * 10-3

400 1.64 * 10-4

450 4.51 * 10-5

500 1.45 * 10-5

550 5.38 * 10-6

600 2.25 * 10-6

Ene

rgy

Time

[B]eq

[B]

Reaction pathway

Transition states

B

A

Catalyzed reaction has lower activation energy

Catalyzed reaction proceeds more rapidly

The two reactions reach the same equilibrium mixture, but the catalyzed reaction achieves equilibrium faster

▲ Figure 15.14 An energy profile for the reaction A ∆ B (left), and the change in concentration of B as a function of time (right), with and without a catalyst. Green curves show the reaction with a catalyst; black curves show the reaction without a catalyst.

Go FiGurEWhat quantity dictates the speed of a reaction: (a) the energy difference between the initial state and the transition state or (b) the energy difference between the initial state and the final state?


N2 and H2. In designing his process, Haber had to deal with a rapid decrease in the equilibrium constant with increasing temperature (Table 15.2). At temperatures suf-ficiently high to give a satisfactory reaction rate, the amount of ammonia formed was too small. The solution to this dilemma was to develop a catalyst that would produce a reasonably rapid approach to equilibrium at a sufficiently low temperature, so that the equilibrium constant remained reasonably large. The development of a suitable catalyst thus became the focus of Haber’s research efforts.

After trying different substances to see which would be most effective, Carl Bosch settled on iron mixed with metal oxides, and variants of this catalyst formulation are still used today. (Section 15.2, “The Haber Process”) These catalysts make it possible to obtain a reasonably rapid approach to equilibrium at around 400 to 500 °C and 200 to 600 atm. The high pressures are needed to obtain a satisfactory equilibrium amount of NH3. If a catalyst could be found that leads to sufficiently rapid reaction at temperatures lower than 400 °C, it would be possible to obtain the same extent of equilibrium conver-sion at pressures much lower than 200 to 600 atm. This would result in great savings in both the cost of the high-pressure equipment and the energy consumed in the pro-duction of ammonia. It is estimated that the Haber process consumes approximately 1% of the energy generated in the world each year. Not surprisingly chemists and chemical engineers are actively searching for improved catalysts for the Haber process. A break-through in this field would not only increase the supply of ammonia for fertilizers, it would also reduce the global consumption of fossil fuels in a significant way.

Give It Some ThoughtCan a catalyst be used to increase the amount of product produced for a reaction that reaches equilibrium quickly without a catalyst?

samPlE inTEGraTiVE ExErCisE Putting Concepts Together

At temperatures near 800 °C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H2:

C1s2 + H2O1g2 ∆ CO1g2 + H21g2The mixture of gases that results is an important industrial fuel called water gas. (a) At 800 °C the equilibrium constant for this reaction is Kp = 14.1. What are the equilibrium partial pres-sures of H2O, CO, and H2 in the equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H2O in a 1.00-L vessel? (b) What is the minimum amount of carbon required to achieve equilibrium under these conditions? (c) What is the total pressure in the vessel at equilibrium? (d) At 25 °C the value of Kp for this reaction is 1.7 * 10-21. Is the reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H2 at equilibrium, should the pressure of the system be increased or decreased?

soluTion(a) To determine the equilibrium partial pressures, we use the ideal-gas equation, first deter-mining the starting partial pressure of water.

PH2O =nH2ORT

V=

10.100 mol210.08206 L@atm>mol@K211073 K21.00 L

= 8.81 atm

We then construct a table of initial partial pressures and their changes as equilibrium is achieved:C1s2 + H2O1g2 ∆ CO1g2 + H21g2

Initial partial pressure (atm) 8.81 0 0Change in partial pressure (atm) -x +x +xEquilibrium partial pressure (atm) 8.81 - x x x

There are no entries in the table under C(s) because the reactant, being a solid, does not appear in the equilibrium-constant expression. Substituting the equilibrium partial pressures of the other species into the equilibrium-constant expression for the reaction gives

Kp =PCOPH2

PH2O=

1x21x218.81 - x2 = 14.1


Multiplying through by the denominator gives a quadratic equation in x: x2 = 114.1218.81 - x2

x2 + 14.1x - 124.22 = 0

Solving this equation for x using the quadratic formula yields x = 6.14 atm. Hence, the equilibrium partial pressures are PCO = x = 6.14 atm, PH2

= x = 6.14 atm, and PH2O = 18.81 - x2 = 2.67 atm.(b) Part (a) shows that x = 6.14 atm of H2O must react for the system to achieve equilibrium. We can use the ideal-gas equation to convert this partial pressure into a mole amount.

n =PVRT

=16.14 atm211.00 L2

10.08206 L@atm>mol@K211073 K2 = 0.0697 mol

Thus, 0.0697 mol of H2O and the same amount of C must react to achieve equilibrium. As a result, there must be at least 0.0697 mol of C (0.836 g C) present among the reactants at the start of the reaction.(c) The total pressure in the vessel at equilibrium is simply the sum of the equilibrium partial pressures:

Ptotal = PH2O + PCO + PH2= 2.67 atm + 6.14 atm + 6.14 atm = 14.95 atm

(d) In discussing Le Châtelier’s principle, we saw that endothermic reactions exhibit an increase in Kp with increasing temperature. Because the equilibrium constant for this reaction increases as temperature increases, the reaction must be endothermic. From the enthalpies of formation given in Appendix C, we can verify our prediction by calculating the enthalpy change for the reaction,

∆H° = ∆Hf°1CO1g22 + ∆Hf°1H21g22 - ∆Hf°1C(s, graphite2 - ∆Hf°1H2O1g22 = +131.3 kJ

The positive sign for ∆H° indicates that the reaction is endothermic.(e) According to Le Châtelier’s principle, a decrease in the pressure causes a gaseous equilib-rium to shift toward the side of the equation with the greater number of moles of gas. In this case, there are 2 mol of gas on the product side and only one on the reactant side. Therefore, the pressure should be decreased to maximize the yield of the CO and H2.


Controlling Nitric Oxide Emissions

The formation of NO from N2 and O2,

12 N21g2 + 1

2 O21g2 ∆ NO1g2 ∆H ° = 90.4 kJ [15.25]

provides an interesting example of the practical importance of the fact that equilibrium constants and reaction rates change with tempera-ture. By applying Le Châtelier’s principle to this endothermic reaction and treating heat as a reactant, we deduce that an increase in tempera-ture shifts the equilibrium in the direction of more NO. The equilib-rium constant Kp for formation of 1 mol of NO from its elements at 300 K is only about 1 * 10-15 (▶ Figure 15.15). At 2400 K, how-ever, the equilibrium constant is about 0.05, which is 1013 times larger than the 300 K value.

Figure 15.15 helps explain why NO is a pollution problem. In the cylinder of a modern high-compression automobile engine, the tem-perature during the fuel-burning part of the cycle is approximately 2400 K. Also, there is a fairly large excess of air in the cylinder. These conditions favor the formation of NO. After combustion, however, the gases cool quickly. As the temperature drops, the equilibrium in Equation 15.25 shifts to the left (because the reactant heat is being removed). However, the lower temperature also means that the reaction rate decreases, so the NO formed at 2400 K is essentially “trapped” in that form as the gas cools.

The gases exhausting from the cylinder are still quite hot, perhaps 1200 K. At this temperature, as shown in Figure 15.15, the equilibrium constant for formation of NO is about 5 * 10-4, much smaller than the

0 1000 2000

1 × 10−15

1 × 10−10

1 × 10−5

1

Temperature (K)

N2(g) + O2(g) NO(g)12

KpCylinder temperature

during combustionExhaust gastemperature

12

▲ Figure 15.15 Equilibrium and temperature. The equilibrium constant increases with increasing temperature because the reaction is endothermic. It is necessary to use a log scale for Kp because the values vary over such a large range.

Go FiGurEEstimate the value of Kp at 1200 K, the exhaust gas temperature.


Chapter summary and Key TermsTHE CONCEPT Of EquILIBRIum (SECTION 15.1) A chemical reac-tion can achieve a state in which the forward and reverse processes are occurring at the same rate. This condition is called chemical equilibrium, and it results in the formation of an equilibrium mixture of the reactants and products of the reaction. The composition of an equilibrium mixture does not change with time if temperature is held constant.

THE EquILIBRIum CONSTANT (SECTION 15.2) An equilib-rium that is used throughout this chapter is the reaction N21g2 +3 H21g2 ∆ 2 NH31g2. This reaction is the basis of the Haber process for the production of ammonia. The relationship between the concentrations of the reactants and products of a system at equilibrium is given by the law of mass action. For an equilibrium equation of the form a A + b B ∆ d D + e E, the equilibrium- constant expression is written as

Kc =3D4d3E4e

3A4a3B4b

where Kc is a dimensionless constant called the equilibrium constant. When the equilibrium system of interest consists of gases, it is often convenient to express the concentrations of reactants and products in terms of gas pressures:

Kp =1PD2d1PE2e

1PA2a1PB2b

Kc and Kp are related by the expression Kp = Kc1RT2∆n. To do this conversion properly, use R = 0.08206 L@atm>mol@K and temperature in kelvins.

uNDERSTANDING AND WORKING WITH EquILIBRIum CON-STANTS (SECTION 15.3) The value of the equilibrium constant changes with temperature. A large value of Kc indicates that the equilib-rium mixture contains more products than reactants and therefore lies toward the product side of the equation. A small value for the equilib-rium constant means that the equilibrium mixture contains less products than reactants and therefore lies toward the reactant side. The equilib-rium-constant expression and the equilibrium constant of the reverse of a reaction are the reciprocals of those of the forward reaction. If a reac-tion is the sum of two or more reactions, its equilibrium constant will be the product of the equilibrium constants for the individual reactions.

HETEROGENEOuS EquILIBRIA (SECTION 15.4) Equilibria for which all substances are in the same phase are called homogeneous

equilibria; in heterogeneous equilibria, two or more phases are present. Because their activities are exactly 1 the concentrations of pure solids and liquids are left out of the equilibrium-constant expression for a heterogeneous equilibrium.

CALCuLATING EquILIBRIum CONSTANTS (SECTION 15.5) If the concentrations of all species in an equilibrium are known, the equilibrium-constant expression can be used to calculate the equilib-rium constant. The changes in the concentrations of reactants and products on the way to achieving equilibrium are governed by the stoichiometry of the reaction.

APPLICATIONS Of EquILIBRIum CONSTANTS (SECTION 15.6) The reaction quotient, Q, is found by substituting reactant and product concentrations or partial pressures at any point during a reac-tion into the equilibrium-constant expression. If the system is at equi-librium, Q = K. If Q ≠ K, however, the system is not at equilibrium. When Q 6 K, the reaction will move toward equilibrium by form-ing more products (the reaction proceeds from left to right); when Q 7 K, the reaction will move toward equilibrium by forming more reactants (the reaction proceeds from right to left). Knowing the value of K makes it possible to calculate the equilibrium amounts of reac-tants and products, often by the solution of an equation in which the unknown is the change in a partial pressure or concentration.

LE CHâTELIER’S PRINCIPLE (SECTION 15.7) Le Châtelier’s prin-ciple states that if a system at equilibrium is disturbed, the equilibrium will shift to minimize the disturbing influence. Therefore, if a reac-tant or product is added to a system at equilibrium, the equilibrium will shift to consume the added substance. The effects of removing reactants or products and of changing the pressure or volume of a reac-tion can be similarly deduced. For example, if the volume of the system is reduced, the equilibrium will shift in the direction that decreases the num-ber of gas molecules. While changes in concentration or pressure lead to shifts in the equilibrium concentrations they do not change the value of the equilibrium constant, K.

Changes in temperature affect both the equilibrium concentra-tions and the equilibrium constant. We can use the enthalpy change for a reaction to determine how an increase in temperature affects the equi-librium: For an endothermic reaction, an increase in temperature shifts the equilibrium to the right; for an exothermic reaction, a temperature increase shifts the equilibrium to the left. Catalysts affect the speed at which equilibrium is reached but do not affect the magnitude of K.

learning outcomes after studying this chapter, you should be able to:

• Explain what is meant by chemical equilibrium and how it relates to reaction rates. (Section 15.1)

• Write the equilibrium-constant expression for any reaction. (Section 15.2)

• Given the value of Kc convert to Kp and vice versa. (Section 15.2)

• Relate the magnitude of an equilibrium constant to the relative amounts of reactants and products present in an equilibrium mixture. (Section 15.3)

• Manipulate the equilibrium constant to reflect changes in the chemical equation. (Section 15.3)

value at 2400 K. However, the rate of conversion of NO to N2 and O2 is too slow to permit much loss of NO before the gases are cooled further.

As discussed in the “Chemistry Put to Work” box in Section 14.7, one of the goals of automotive catalytic converters is to achieve rapid conversion of NO to N2 and O2 at the temperature of the exhaust gas.

Some catalysts developed for this reaction are reasonably effective under the grueling conditions in automotive exhaust systems. Nev-ertheless, scientists and engineers are continuously searching for new materials that provide even more effective catalysis of the decomposi-tion of nitrogen oxides.

exercises 661

Visualizing Concepts

15.1 (a) Based on the following energy profile, predict whether kf 7 kr or kf 6 kr. (b) Using Equation 15.5, predict whether the equilibrium constant for the process is greater than 1 or less than 1. [Section 15.1]

Reaction progress

Pote

ntia

l ene

rgy

Reactants

Products

15.2 The following diagrams represent a hypothetical reaction A ¡ B, with A represented by red spheres and B rep-resented by blue spheres. The sequence from left to right represents the system as time passes. Does the system reach equilibrium? If so, in which diagram is the system in equilib-rium? [Sections 15.1 and 15.2]

1 2 3 4 5

15.3 The following diagram represents an equilibrium mixture produced for a reaction of the type A + X ∆ AX. Is K

greater or smaller than 1 if the volume is 1 L and each atom/molecule in the diagram represents 1 mol? [Section 15.2]

15.4 The following diagram represents a reaction shown going to completion. Each molecule in the diagram represents 0.1 mol and the volume of the box is 1.0 L. (a) Letting A = red spheres and B = blue spheres, write a balanced equation for the reac-tion. (b) Write the equilibrium-constant expression for the re-action. (c) Calculate the value of Kc. (d) Assuming that all of the molecules are in the gas phase, calculate ∆n, the change in the number of gas molecules that accompanies the reaction. (e) Calculate the value of Kp. [Section 15.2]

15.5 Snapshots of two hypothetical reactions, A1g2 + B1g2 ∆AB1g2 and X1g2 + Y1g2 ∆ XY1g2 at five different times

• Write the equilibrium-constant expression for a heterogeneous re-action. (Section 15.4)

• Calculate an equilibrium constant from concentration measure-ments. (Section 15.5)

• Predict the direction of a reaction given the equilibrium constant and the concentrations of reactants and products. (Section 15.6)

• Calculate equilibrium concentrations given the equilibrium con-stant and all but one equilibrium concentration. (Section 15.6)

• Calculate equilibrium concentrations, given the equilibrium con-stant and the starting concentrations. (Section 15.6)

• Use Le Châtelier’s principle to predict how changing the concentra-tions, volume, or temperature of a system at equilibrium affects the equilibrium position. (Section 15.7)

Key Equations

• Kc =3D4d3E4e

3A4a3B4b [15.8] The equilibrium-constant expression for a general reaction of the type a A + b B ∆ d D + e E, the concentrations are equilib-rium concentrations only

• Kp =1PD2d1PE2e

1PA2a1PB2b [15.11] The equilibrium-constant expression in terms of equilibrium par-tial pressures

• Kp = Kc1RT2∆n [15.14] Relating the equilibrium constant based on pressures to the equi-librium constant based on concentration

• Qc =3D4d3E4e

3A4a3B4b [15.23] The reaction quotient. The concentrations are for any time during a reaction. If the concentrations are equilibrium concentrations, then Qc = Kc.

Exercises


5.0 g PbO2(g)in both vessels

Vessel AV = 50 mL

Vessel BV = 100 mL

15.8 The reaction A2 + B2 ∆ 2 AB has an equilibrium con-stant Kc = 1.5. The following diagrams represent reaction mixtures containing A2 molecules (red), B2 molecules (blue), and AB molecules. (a) Which reaction mixture is at equilib-rium? (b) For those mixtures that are not at equilibrium, how will the reaction proceed to reach equilibrium? [Sections 15.5 and 15.6]

(i) (ii) (iii)

15.9 The reaction A21g2 + B1g2 ∆ A1g2 + AB1g2 has an equilibrium constant of Kp = 2. The accompanying diagram shows a mixture containing A atoms (red), A2 molecules, and AB molecules (red and blue). How many B atoms should be added to the diagram to illustrate an equilibrium mixture? [Section 15.6]

15.10 The diagram shown here represents the equilibrium state for the reaction A21g2 + 2 B1g2 ∆ 2AB1g2. (a) Assuming the volume is 2 L, calculate the equilibrium constant Kc for the reaction. (b) If the volume of the equilibrium mixture is decreased, will the number of AB molecules increase or de-crease? [Sections 15.5 and 15.7]

are shown here. Which reaction has a larger equilibrium con-stant? [Sections 15.1 and 15.2]

t = 10 s 20 s 30 s 40 s 50 s

X(g) + Y(g) XY(g)

t = 10 s 20 s 30 s 40 s 50 s

A(g) + B(g) AB(g)

15.6 Ethene 1C2H42 reacts with halogens 1X22 by the following reaction:

C2H41g2 + X21g2 ∆ C2H4X21g2 The following figures represent the concentrations at equi-

librium at the same temperature when X2 is Cl2 (green), Br2 (brown), and I2 (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3]

(a) (b)

(c)

15.7 When lead (IV) oxide is heated above 300 °C it decom-poses according to the following reaction PbO21s2 ∆PbO1s2 + O21g2. Consider the two sealed vessels of PbO2 shown here. If both vessels are heated to 400 °C and allowed to come to equilibitum which of the following statements is true? (a) There will be less PbO2 remaining in vessel A, (b) There will be less PbO2 remaining in vessel B, (c) The amount of PbO2 remaining in each vessel will be the same. [Section 15.4]

exercises 663

15.15 Write the expression for Kc for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous.(a) 3 NO1g2 ∆ N2O1g2 + NO21g2(b) CH41g2 + 2 H2S1g2 ∆ CS21g2 + 4 H21g2(c) Ni1CO241g2 ∆ Ni1s2 + 4 CO1g2(d) HF1aq2 ∆ H+1aq2 + F-1aq2(e) 2 Ag1s2 + Zn2+1aq2 ∆ 2 Ag+1aq2 + Zn1s2(f) H2O1l2 ∆ H+1aq2 + OH-1aq2(g) 2 H2O1l2 ∆ 2 H+1aq2 + 2 OH-1aq2

15.16 Write the expressions for Kc for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous.(a) 2 O31g2 ∆ 3 O21g2(b) Ti1s2 + 2 Cl21g2 ∆ TiCl41l2(c) 2 C2H41g2 + 2 H2O1g2 ∆ 2 C2H61g2 + O21g2(d) C1s2 + 2 H21g2 ∆ CH41g2(e) 4 HCl1aq2 + O21g2 ∆ 2 H2O1l2 + 2 Cl21g2(f) 2 C8H181l2 + 25 O21g2 ∆ 16 CO21g2 + 18 H2O1g2(g) 2 C8H181l2 + 25 O21g2 ∆ 16 CO21g2 + 18 H2O1l2

15.17 When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products?(a) N21g2 + O21g2 ∆ 2 NO1g2; Kc = 1.5 * 10-10

(b) 2 SO21g2 + O21g2 ∆ 2 SO31g2; Kp = 2.5 * 109

15.18 Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring for-mation of reactants?(a) 2 NO1g2 + O21g2 ∆ 2 NO21g2; Kp = 5.0 * 1012

(b) 2 HBr1g2 ∆ H21g2 + Br21g2; Kc = 5.8 * 10-18

15.19 Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative num-ber. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases the speed at which a reaction reaches equilibrium increases.

15.20 Which of the following statements are true and which are false? (a) For the reaction 2 A1g2 + B1g2 ∆ A2B1g2 Kc and Kp are numerically the same. (b) It is possible to distin-guish Kc from Kp by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of Kc increases with increasing pressure.

15.21 If Kc = 0.042 for PCl31g2 + Cl21g2 ∆ PCl51g2 at 500 K, what is the value of Kp for this reaction at this temperature?

15.22 Calculate Kc at 303 K for SO21g2 + Cl21g2 ∆ SO2Cl21g2 if Kp = 34.5 at this temperature.

15.23 The equilibrium constant for the reaction

2 NO1g2 + Br21g2 ∆ 2 NOBr1g2 is Kc = 1.3 * 10-2 at 1000 K. (a) At this temperature does

the equilibrium favor NO and Br2, or does it favor NOBr? (b) Calculate Kc for 2 NOBr1g2 ∆ 2 NO1g2 + Br21g2. (c) Calculate Kc for NOBr1g2 ¡ NO1g2 + 1

2 Br21g2. 15.24 Consider the following equilibrium:

2 H21g2 + S21g2 ∆ 2 H2S1g2 Kc = 1.08 * 107at 700 °C

15.11 The following diagrams represent equilibrium mixtures for the reaction A2 + B ∆ A + AB at 300 K and 500 K. The A atoms are red, and the B atoms are blue. Is the reaction exo-thermic or endothermic? [Section 15.7]

300 K 500 K

15.12 The following graph represents the yield of the compound AB at equilibrium in the reaction A1g2 + B1g2 ¡ AB1g2 at two different pressures, x and y, as a function of temperature.

Temperature

[AB]

P=x

P=y

(a) Is this reaction exothermic or endothermic? (b) Is P = x greater or smaller than P = y? [Section 15.7]

Equilibrium; The Equilibrium Constant (sections 15.1–15.4)

15.13 Suppose that the gas-phase reactions A ¡ B and B ¡ A are both elementary processes with rate constants of 4.7 * 10-3 s-1 and 5.8 * 10-1 s-1, respectively. (a) What is the value of the equilibrium constant for the equilibrium A1g2 ∆ B1g2? (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

15.14 Consider the reaction A + B ∆ C + D. Assume that both the forward reaction and the reverse reaction are el-ementary processes and that the value of the equilibrium constant is very large. (a) Which species predominate at equi-librium, reactants or products? (b) Which reaction has the larger rate constant, the forward or the reverse?


15.32 Gaseous hydrogen iodide is placed in a closed container at 425 °C, where it partially decomposes to hydrogen and io-dine: 2 HI1g2 ∆ H21g2 + I21g2. At equilibrium it is found that 3HI4 = 3.53 * 10-3 M, 3H24 = 4.79 * 10-4 M, and 3I24 = 4.79 * 10-4 M. What is the value of Kc at this temperature?

15.33 The equilibrium 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2 is es-tablished at 500 K. An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl2, and NOCl, respectively. (a) Calculate Kp for this re-action at 500.0 K. (b) If the vessel has a volume of 5.00 L, cal-culate Kc at this temperature.

15.34 Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl31g2 + Cl21g2 ∆ PCl51g2. A 7.5-L gas vessel is charged with a mixture of PCl31g2 and Cl21g2, which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are PPCl3 = 0.124 atm, PCl2 = 0.157 atm, a n d PPCl5 = 1.30 atm. (a) What is the value of Kp at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate Kc for this reaction at 450 K.

15.35 A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilib-rium is established:

2 NO1g2 + 2 H21g2 ∆ N21g2 + 2 H2O1g2 At equilibrium 3NO4 = 0.062 M. (a) Calculate the equilib-

rium concentrations of H2, N2, and H2O. (b) Calculate Kc. 15.36 A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a

2.00-L vessel at 700 K. These substances react according toH21g2 + Br21g2 ∆ 2 HBr1g2

At equilibrium, the vessel is found to contain 0.566 g of H2. (a) Calculate the equilibrium concentrations of H2, Br2, and HBr. (b) Calculate Kc.

15.37 A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and 0.1600 mol of H2O is placed in a 2.000-L vessel. The following equi-librium is established at 500 K:

CO21g2 + H21g2 ∆ CO1g2 + H2O1g2 (a) Calculate the initial partial pressures of CO2, H2, and H2O.

(b) At equilibrium PH2O = 3.51 atm. Calculate the equilib-rium partial pressures of CO2, H2, and CO. (c) Calculate Kp for the reaction. (d) Calculate Kc for the reaction.

15.38 A flask is charged with 1.500 atm of N2O41g2 and 1.00 atm NO21g2 at 25 °C, and the following equilibrium is achieved:

N2O41g2 ∆ 2 NO21g2 After equilibrium is reached, the partial pressure of NO2 is

0.512 atm. (a) What is the equilibrium partial pressure of N2O4? (b) Calculate the value of Kp for the reaction. (c) Calcu-late Kc for the reaction.

15.39 Two different proteins X and Y are dissolved in aqueous solu-tion at 37 °C. The proteins bind in a 1:1 ratio to form XY. A solution that is initially 1.00 mM in each protein is allowed to reach equilibrium. At equilibrium, 0.20 mM of free X and 0.20 mM of free Y remain. What is Kc for the reaction?

15.40 A chemist at a pharmaceutical company is measuring equi-librium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug

(a) Calculate Kp. (b) Does the equilibrium mixture contain mostly H2 and S2 or mostly H2S? (c) Calculate the value of Kc if you rewrote the equation H21g2 + 1

2 S21g2 ∆ H2S1g2.

15.25 At 1000 K, Kp = 1.85 for the reaction

SO21g2 + 12 O21g2 ∆ SO31g2

(a) What is the value of Kp for the reaction SO31g2 ∆ SO21g2 + 12 O21g2? (b) What is the value of Kp for the reaction

2 SO21g2 + O21g2 ∆ 2 SO31g2? (c) What is the value of Kc for the reaction in part (b)?

15.26 Consider the following equilibrium, for which Kp = 0.0752 at 480 °C:

2 Cl21g2 + 2 H2O1g2 ∆ 4 HCl1g2 + O21g2(a) What is the value of Kp for the reaction

4 HCl1g2 + O21g2 ∆ 2 Cl21g2 + 2 H2O1g2?(b) What is the value of Kp for the reaction

Cl21g2 + H2O1g2 ∆ 2 HCl1g2 + 12 O21g2?

(c) What is the value of Kc for the reaction in part (b)? 15.27 The following equilibria were attained at 823 K:

CoO1s2 + H21g2 ∆ Co1s2 + H2O1g2 Kc = 67

CoO1s2 + CO1g2 ∆ Co1s2 + CO21g2 Kc = 490

Based on these equilibria, calculate the equilibrium constant for H21g2 + CO21g2 ∆ CO1g2 + H2O1g2 at 823 K.

15.28 Consider the equilibrium

N21g2 + O21g2 + Br21g2 ∆ 2 NOBr1g2 Calculate the equilibrium constant Kp for this reaction, given

the following information (at 298 K):2 NO1g2 + Br21g2 ∆ 2 NOBr1g2 Kc = 2.0

2 NO1g2 ∆ N21g2 + O21g2 Kc = 2.1 * 1030

15.29 Mercury(I) oxide decomposes into elemental mercury and e l e m e nt a l ox y ge n : 2 Hg2O1s2 ∆ 4 Hg1l2 + O21g2. (a) Write the equilibrium-constant expression for this reac-tion in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and el-emental oxygen. Rewrite the equilibrium-constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

15.30 Consider the equilibrium Na2O1s2 + SO21g2 ∆ Na2SO31s2. (a) Write the equilibrium-constant expression for this reac-tion in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibrium-constant expression in terms of molarities for the aqueous reaction.

Calculating Equilibrium Constants (section 15.5)

15.31 Methanol 1CH3OH2 is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: CO1g2 + 2 H21g2 ∆ CH3OH1g2. A n e q u i l i b r i u m mixture in a 2.00-L vessel is found to contain 0.0406 mol CH3OH, 0.170 mol CO, and 0.302 mol H2 at 500 K. Calculate Kc at this temperature.

exercises 665

15.47 At 1285 °C, the equilibrium constant for the reaction Br21g2 ∆ 2 Br1g2 is Kc = 1.04 * 10-3. A 0.200-L vessel containing an equilibrium mixture of the gases has 0.245 g Br21g2 in it. What is the mass of Br1g2 in the vessel?

15.48 For the reaction H21g2 + I21g2 ∆ 2 HI1g2, Kc = 55.3 at 700 K. In a 2.00-L flask containing an equilibrium mixture of the three gases, there are 0.056 g H2 and 4.36 g I2. What is the mass of HI in the flask?

15.49 At 800 K, the equilibrium constant for I21g2 ∆ 2 I1g2 is Kc = 3.1 * 10-5. If an equilibrium mixture in a 10.0-L vessel contains 2.67 * 10-2 g of I(g), how many grams of I2 are in the mixture?

15.50 For 2 SO21g2 + O21g2 ∆ 2 SO31g2, Kp = 3.0 * 104 at 700 K. In a 2.00-L vessel, the equilibrium mixture contains 1.17 g of SO3 and 0.105 g of O2. How many grams of SO2 are in the vessel?

15.51 At 2000 °C, the equilibrium constant for the reaction

2 NO1g2 ∆ N21g2 + O21g2 is Kc = 2.4 * 103. If the initial concentration of NO is

0.175 M, what are the equilibrium concentrations of NO, N2, and O2?

15.52 For the equilibrium

Br21g2 + Cl21g2 ∆ 2 BrCl1g2 at 400 K, Kc = 7.0. If 0.25 mol of Br2 and 0.55 mol of Cl2 are

introduced into a 3.0-L container at 400 K, what will be the equilibrium concentrations of Br2, Cl2, and BrCl?

15.53 At 373 K, Kp = 0.416 for the equilibrium

2 NOBr1g2 ∆ 2 NO1g2 + Br21g2 If the pressures of NOBr(g) and NO(g) are equal, what is the

equilibrium pressure of Br21g2? 15.54 At 218 °C, Kc = 1.2 * 10-4 for the equilibrium

NH4SH1s2 ∆ NH31g2 + H2S1g2 Calculate the equilibrium concentrations of NH3 and H2S if a

sample of solid NH4SH is placed in a closed vessel at 218 °C and decomposes until equilibrium is reached.

15.55 Consider the reaction

CaSO41s2 ∆ Ca2+1aq2 + SO42-1aq2

At 25 °C, the equilibrium constant is Kc = 2.4 * 10-5 for this reaction. (a) If excess CaSO41s2 is mixed with water at 25 °C to produce a saturated solution of CaSO4, what are the equilibrium concentrations of Ca2+ and SO4

2-? (b) If the resulting solution has a volume of 1.4 L, what is the mini-mum mass of CaSO41s2 needed to achieve equilibrium?

15.56 At 80 °C, Kc = 1.87 * 10-3 for the reaction

PH3BCl31s2 ∆ PH31g2 + BCl31g2 (a) Calculate the equilibrium concentrations of PH3 and BCl3 if

a solid sample of PH3BCl3 is placed in a closed vessel at 80 °C and decomposes until equilibrium is reached. (b) If the flask has a volume of 0.250 L, what is the minimum mass of PH3BCl31s2 that must be added to the flask to achieve equilibrium?

15.57 For the reaction I2 + Br21g2 ∆ 2 IBr1g2, Kc = 280 at 150 °C. Suppose that 0.500 mol IBr in a 2.00-L flask is allowed to reach equilibrium at 150 °C. What are the equilibrium con-centrations of IBr, I2, and Br2?

molecules bind the protein in a 1:1 ratio to form a drug– protein complex. The protein concentration in aqueous solu-tion at 25 °C is 1.50 * 10-6M. Drug A is introduced into the protein solution at an initial concentration of 2.00 * 10-6M. Drug B is introduced into a separate, identical protein solu-tion at an initial concentration of 2.00 * 10-6M. At equilib-rium, the drug A-protein solution has an A-protein complex concentration of 1.00 * 10-6M, and the drug B solution has a B-protein complex concentration of 1.40 * 10-6M. Calculate the Kc value for the A-protein binding reaction and for the B-protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

applications of Equilibrium Constants (section 15.6)

15.41 (a) If Qc 6 Kc, in which direction will a reaction proceed in order to reach equilibrium? (b) What condition must be satis-fied so that Qc = Kc?

15.42 (a) If Qc 7 Kc, how must the reaction proceed to reach equi-librium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of Qc at this point in the reaction?

15.43 At 100 °C, the equilibrium constant for the reaction COCl21g2 ∆ CO1g2 + Cl21g2 h a s t h e v a l u e Kc =2.19 * 10-10. Are the following mixtures of COCl2, CO, and Cl2 at 100 °C at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium.(a) 3COCl24 = 2.00 * 10-3 M, 3CO4 = 3.3 * 10-6 M,

3Cl24 = 6.62 * 10-6 M(b) 3COCl24 = 4.50 * 10-2 M, 3CO4 = 1.1 * 10-7 M,

3Cl24 = 2.25 * 10-6 M(c) 3COCl24 = 0.0100 M, 3CO4 = 3Cl24 = 1.48 * 10-6 M

15.44 As shown in Table 15.2, Kp for the equilibrium

N21g2 + 3 H21g2 ∆ 2 NH31g2 is 4.51 * 10-5 at 450 °C. For each of the mixtures listed here,

indicate whether the mixture is at equilibrium at 450 °C. If it is not at equilibrium, indicate the direction (toward prod-uct or toward reactants) in which the mixture must shift to achieve equilibrium.(a) 98 atm NH3, 45 atm N2, 55 atm H2

(b) 57 atm NH3, 143 atm N2, no H2

(c) 13 atm NH3, 27 atm N2, 82 atm H2

15.45 At 100 °C, Kc = 0.078 for the reaction

SO2Cl21g2 ∆ SO21g2 + Cl21g2 In an equilibrium mixture of the three gases, the concentra-

tions of SO2Cl2 and SO2 are 0.108 M and 0.052 M, respec-tively. What is the partial pressure of Cl2 in the equilibrium mixture?

15.46 At 900 K, the following reaction has Kp = 0.345:

2 SO21g2 + O21g2 ∆ 2 SO31g2 In an equilibrium mixture the partial pressures of SO2 and O2

are 0.135 atm and 0.455 atm, respectively. What is the equilib-rium partial pressure of SO3 in the mixture?


(d) decrease the volume of the container in which the reac-tion occurs; (e) add a catalyst; (f) increase temperature.

15.63 How do the following changes affect the value of the equi-librium constant for a gas-phase exothermic reaction: (a) re-moval of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

15.64 For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction ves-sel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

15.65 Consider the following equilibrium between oxides of nitrogen

3 NO1g2 ∆ NO21g2 + N2O1g2 (a) Use data in Appendix C to calculate ∆H° for this

reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

15.66 Methanol 1CH3OH2 can be made by the reaction of CO with H2:

CO1g2 + 2 H21g2 ∆ CH3OH1g2 (a) Use thermochemical data in Appendix C to calculate ∆H°

for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

15.67 Ozone, O3, decomposes to molecular oxygen in the strato-sphere according to the reaction 2 O31g2 ¡ 3 O21g2. Would an increase in pressure favor the formation of ozone or of oxygen?

15.68 The water–gas shift reaction CO1g2 + H2O1g2 ∆CO21g2 + H21g2 is used industrially to produce hydrogen. The reaction enthalpy is ∆H° = -41 kJ. (a) To increase the equilibrium yield of hydrogen would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen by controlling the pressure of this reac-tion? If so would high or low pressure favor formation of H21g2?

15.58 At 25 °C, the reaction

CaCrO41s2 ∆ Ca2+1aq2 + CrO42-1aq2

has an equilibrium constant Kc = 7.1 * 10-4. What are the equilibrium concentrations of Ca2 + and CrO4

2- in a satu-rated solution of CaCrO4?

15.59 Methane, CH4, reacts with I2 according to the reaction CH41g2 + l21g2 ∆ CH3l1g2 + HI1g2. At 630 K, Kp for this reaction is 2.26 * 10-4. A reaction was set up at 630 K with initial partial pressures of methane of 105.1 torr and of 7.96 torr for I2. Calculate the pressures, in torr, of all reactants and products at equilibrium.

15.60 The reaction of an organic acid with an alcohol, in organic sol-vent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually H2SO4). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water:

CH3COOH1solv2 + CH3CH2OH1solv2 ∆CH3COOCH2CH31solv2 + H2O1solv2

where “(solv)” indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at 55 °C is 6.68. A pharmaceutical chemist makes up 15.0 L of a solution that is initially 0.275 M in acetic acid and 3.85 M in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

le Châtelier’s Principle (section 15.7)

15.61 Consider the following equilibrium for which ∆H 6 0

2 SO21g2 + O21g2 ∆ 2 SO31g2 How will each of the following changes affect an equilibrium

mixture of the three gases: (a) O21g2 is added to the system; (b) the reaction mixture is heated; (c) the volume of the reac-tion vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) SO31g2 is removed from the system?

15.62 Consider the reaction

4 NH31g2 + 5 O21g2 ∆4 NO1g2 + 6 H2O1g2, ∆H = -904.4 kJ

Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (a) in-crease 3NH34; (b) increase 3H2O4; (c) decrease 3O24;

additional Exercises 15.69 Both the forward reaction and the reverse reaction in the fol-

lowing equilibrium are believed to be elementary steps:

CO1g2 + Cl21g2 ∆ COCl1g2 + Cl1g2 At 25 °C, the rate constants for the forward and reverse reac-

tions are 1.4 * 10-28 M-1 s-1 and 9.3 * 1010 M-1 s-1, respec-tively. (a) What is the value for the equilibrium constant at 25 °C? (b) Are reactants or products more plentiful at equilibrium?

15.70 If Kc = 1 for the equilibrium 2 A1g2 ∆ B1g2, what is the relationship between [A] and [B] at equilibrium?

15.71 A mixture of CH4 and H2O is passed over a nickel cata-lyst at 1000 K. The emerging gas is collected in a 5.00-L

flask and is found to contain 8.62 g of CO, 2.60 g of H2, 43.0 g of CH4, and 48.4 g of H2O. Assuming that equilib-rium has been reached, calculate Kc and Kp for the reaction CH41g2 + H2O1g2 ∆ CO1g2 + 3 H21g2.

15.72 When 2.00 mol of SO2Cl2 is placed in a 2.00-L flask at 303 K, 56% of the SO2Cl2 decomposes to SO2 and Cl2 :

SO2Cl21g2 ∆ SO21g2 + Cl21g2 (a) Calculate Kc for this reaction at this temperature.

(b) Calculate Kp for this reaction at 303 K. (c) According to Le Châtelier’s principle, would the percent of SO2Cl2 that de-composes increase, decrease or stay the same if the mixture


At 700 K, the equilibrium constant Kp for this reaction is 0.26. Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium.(a) PNO = 0.15 atm, PCl2 = 0.31 atm, PNOCl = 0.11 atm(b) PNO = 0.12 atm, PCl2 = 0.10 atm,

PNOCl = 0.050 atm(c) PNO = 0.15 atm, PCl2 = 0.20 atm,

PNOCl = 5.10 * 10-3 atm 15.82 At 900 °C, Kc = 0.0108 for the reaction

CaCO31s2 ∆ CaO1s2 + CO21g2 A mixture of CaCO3, CaO, and CO2 is placed in a 10.0-L ves-

sel at 900 °C. For the following mixtures, will the amount of CaCO3 increase, decrease, or remain the same as the system approaches equilibrium?(a) 15.0 g CaCO3, 15.0 g CaO, and 4.25 g CO2

(b) 2.50 g CaCO3, 25.0 g CaO, and 5.66 g CO2

(c) 30.5 g CaCO3, 25.5 g CaO, and 6.48 g CO2

15.83 When 1.50 mol CO2 and 1.50 mol H2 are placed in a 3.00-L container at 395 °C, the following reaction occurs: CO21g2 + H21g2 ∆ CO1g2 + H2O1g2. I f Kc = 0.802, what are the concentrations of each substance in the equilib-rium mixture?

15.84 The equilibrium constant Kc for C1s2 + CO21g2 ∆ 2 CO1g2 is 1.9 at 1000 K and 0.133 at 298 K. (a) If excess C is allowed to react with 25.0 g of CO2 in a 3.00-L vessel at 1000 K, how many grams of CO are produced? (b) How many grams of C are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

15.85 NiO is to be reduced to nickel metal in an industrial process by use of the reaction

NiO1s2 + CO1g2 ∆ Ni1s2 + CO21g2 At 1600 K, the equilibrium constant for the reaction is

Kp = 6.0 * 102. If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?

15.86 Le Châtelier noted that many industrial processes of his time could be improved by an understanding of chemical equilib-ria. For example, the reaction of iron oxide with carbon mon-oxide was used to produce elemental iron and CO2 according to the reaction

Fe2O31s2 + 3 CO1g2 ∆ 2 Fe1s2 + 3 CO21g2 Even in Le Châtelier’s time, it was noted that a great deal of

CO was wasted, expelled through the chimneys over the fur-naces. Le Châtelier wrote, “Because this incomplete reaction was thought to be due to an insufficiently prolonged contact between carbon monoxide and the iron ore [oxide], the dimen-sions of the furnaces have been increased. In England, they have been made as high as 30 m. But the proportion of carbon monoxide escaping has not diminished, thus demonstrating, by an experiment costing several hundred thousand francs, that the reduction of iron oxide by carbon monoxide is a limited reaction. Acquaintance with the laws of chemical equilibrium would have permitted the same conclusion to be reached more rapidly and far more economically.” What does this anecdote tell us about the equilibrium constant for this reaction?

were transferred to a 15.00-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of SO2Cl2 that decomposes when 2.00 mol of SO2Cl2 is placed in a 15.00-L vessel at 303 K.

15.73 A mixture of H2, S, and H2S is held in a 1.0-L vessel at 90 °C and reacts according to the equation:

H21g2 + S1s2 ∆ H2S1g2 At equilibrium, the mixture contains 0.46 g of H2S and 0.40

g H2. (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of Kc for the reaction at this temperature?

15.74 A sample of nitrosyl bromide (NOBr) decomposes according to the equation

2 NOBr1g2 ∆ 2 NO1g2 + Br21g2 An equilibrium mixture in a 5.00-L vessel at 100 °C contains

3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br2. (a) Calculate Kc. (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of NOBr?

15.75 Consider the hypothetical reaction A1g2 ∆ 2 B1g2. A flask is charged with 0.75 atm of pure A, after which it is al-lowed to reach equilibrium at 0 °C. At equilibrium, the partial pressure of A is 0.36 atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of Kp? (c) What could we do to maximize the yield of B?

15.76 As shown in Table 15.2, the equilibrium constant for the reaction N21g2 + 3 H21g2 ∆ 2 NH31g2 is Kp = 4.34 * 10-3 at 300 °C. Pure NH3 is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are 1.05 g NH3 in the equi-librium mixture. (a) What are the masses of N2 and H2 in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?


2 IBr1g2 ∆ I21g2 + Br21g2 Kp = 8.5 * 10-3 at 150 °C. If 0.025 atm of IBr is placed in a

2.0-L container, what is the partial pressure of all substances after equilibrium is reached?


PH3BCl31s2 ∆ PH31g2 + BCl31g2 Kp = 0.052 at 60 °C. (a) Calculate Kc. (b) After 3.00 g of solid

PH3BCl3 is added to a closed 1.500-L vessel at 60 °C, the vessel is charged with 0.0500 g of BCl31g2. What is the equilibrium concentration of PH3?

[15.79] Solid NH4SH is introduced into an evacuated flask at 24 °C. The following reaction takes place:

NH4SH1s2 ∆ NH31g2 + H2S1g2 At equilibrium, the total pressure (for NH3 and H2S taken to-

gether) is 0.614 atm. What is Kp for this equilibrium at 24 °C? [15.80] A 0.831-g sample of SO3 is placed in a 1.00-L container and

heated to 1100 K. The SO3 decomposes to SO2 and O2:

2 SO31g2 ∆ 2 SO21g2 + O21g2 At equilibrium, the total pressure in the container is 1.300 atm.

Find the values of Kp and Kc for this reaction at 1100 K. 15.81 Nitric oxide (NO) reacts readily with chlorine gas as follows:

2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2


equilibrium concentrations (in terms of x) into the equilibrium- constant expression, derive an equation that can be solved for x. (d) The equation from part (c) is a cubic equation (one that has the form ax3 + bx2 + cx + d = 0). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the al-lowed range of x that you specified in part (b). The point at which the cubic equation crosses the x-axis is the solution. (e) From the plot in part (d), estimate the equilibrium concen-trations of A, B, and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equi-librium expression.)

15.91 At 1200 K, the approximate temperature of automobile ex-haust gases (Figure 15.15), Kp for the reaction

2 CO21g2 ∆ 2 CO1g2 + O21g2 is about 1 * 10-13. Assuming that the exhaust gas (total

pressure 1 atm) contains 0.2% CO, 12% CO2, and 3% O2 by volume, is the system at equilibrium with respect to the CO2 reaction? Based on your conclusion, would the CO concen-tration in the exhaust be decreased or increased by a catalyst that speeds up the CO2 reaction? Recall that at a fixed pres-sure and temperature, volume % = mol %.

15.92 Suppose that you worked at the U.S. Patent Office and a patent application came across your desk claiming that a newly devel-oped catalyst was much superior to the Haber catalyst for ammo-nia synthesis because the catalyst led to much greater equilibrium conversion of N2 and H2 into NH3 than the Haber catalyst under the same conditions. What would be your response?

[15.87] At 700 K, the equilibrium constant for the reaction

CCl41g2 ∆ C1s2 + 2 Cl21g2 is Kp = 0.76. A flask is charged with 2.00 atm of CCl4, which

then reaches equilibrium at 700 K. (a) What fraction of the CCl4 is converted into C and Cl2? (b) What are the partial pressures of CCl4 and Cl2 at equilibrium?

[15.88] The reaction PCl31g2 + Cl21g2 ∆ PCl51g2 has Kp =0.0870 at 300 °C. A flask is charged with 0.50 atm PCl3, 0.50 atm Cl2, and 0.20 atm PCl5 at this temperature. (a) Use the re-action quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of Cl2 in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of Cl2 in the equilibrium mixture?

[15.89] An equilibrium mixture of H2, I2, and HI at 458 °C contains 0.112 mol H2, 0.112 mol I2, and 0.775 mol HI in a 5.00-L vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 mol of HI?

[15.90] Consider the hypothetical reaction A1g2 + 2 B1g2 ∆ 2 C1g2, for which Kc = 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol>L of compound A present at equilibrium. (a) In terms of x, what are the equilibrium concentrations of compounds B and C? (b) What limits must be placed on the value of x so that all concentrations are positive? (c) By putting the

integrative Exercises 15.93 Consider the reaction IO4

-1aq2 + 2 H2O1l2 ∆ H4IO6-1aq2;

Kc = 3.5 * 10-2. If you start with 25.0 mL of a 0.905 M solu-tion of NaIO4, and then dilute it with water to 500.0 mL, what is the concentration of H4IO6

- at equilibrium? [15.94] Silver chloride, AgCl(s), is an “insoluble” strong electro-

lyte. (a) Write the equation for the dissolution of AgCl(s) in H2O1l2. (b) Write the expression for Kc for the reaction in part (a). (c) Based on the thermochemical data in Appendix C and Le Châtelier’s principle, predict whether the solubility of AgCl in H2O increases or decreases with increasing tempera-ture. (d) The equilibrium constant for the dissolution of AgCl in water is 1.6 * 10-10 at 25 °C. In addition, Ag+1aq2 can re-act with Cl-1aq2 according to the reaction

Ag+1aq2 + 2 Cl-1aq2 ∆ AgCl2-1aq2

where Kc = 1.8 * 105 at 25 °C. Although AgCl is “not solu-ble” in water, the complex AgCl2

- is soluble. At 25 °C, is the solubility of AgCl in a 0.100 M NaCl solution greater than the solubility of AgCl in pure water, due to the formation of sol-uble AgCl2

- ions? Or is the AgCl solubility in 0.100 M NaCl less than in pure water because of a Le Châtelier-type argu-ment? Justify your answer with calculations. (Hint: Any form in which silver is in solution counts as “solubility.”)

[15.95] Consider the equilibrium A ∆ B in which both the for-ward and reverse reactions are elementary (single-step) reac-tions. Assume that the only effect of a catalyst on the reaction is to lower the activation energies of the forward and reverse reactions, as shown in Figure 15.14. Using the Arrhenius equation (Section 14.5), prove that the equilibrium constant is the same for the catalyzed reaction as for the uncatalyzed one.

[15.96] The phase diagram for SO2 is shown here. (a) What does this diagram tell you about the enthalpy change in the reaction SO21l2 ¡ SO21g2? (b) Calculate the equilibrium constant for this reaction at 100 °C and at 0 °C. (c) Why is it not pos-sible to calculate an equilibrium constant between the gas and liquid phases in the supercritical region? (d) At which of the three points marked in red does SO21g2 most closely approach ideal-gas behavior? (e) At which of the three red points does SO21g2 behave least ideally?

Gas

Critical point

Supercriticalregion

Liquid

Temperature (°C)

Pres

sure

(atm

)

0 100 200 300

102

10

1

10−1


15.99 Water molecules in the atmosphere can form hydrogen-bonded dimers, 1H2O22. The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSEPR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer for-mation? (c) The Kp for water dimer formation in the gas phase is 0.050 at 300 K and 0.020 at 350 K. Is water dimer formation endothermic or exothermic?

15.100 The protein hemoglobin (Hb) transports O2 in mamma-lian blood. Each Hb can bind 4 O2 molecules. The equilib-rium constant for the O2 binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the P50 value, defined as the partial pressure of oxygen at which 50% of the protein is saturated. Fetal hemoglobin has a P50 value of 19 torr, and adult hemoglobin has a P50 value of 26.8 torr. Use these data to estimate how much larger Kc is for the aqueous reaction 4 O21g2 + Hb1aq2 ¡3Hb1O2241aq24.

[15.97] Write the equilibrium-constant expression for the equilibriumC1s2 + CO21g2 ∆ 2 CO1g2

The table that follows shows the relative mole percentages of CO21g2 and CO(g) at a total pressure of 1 atm for several temperatures. Calculate the value of Kp at each temperature. Is the reaction exothermic or endothermic?

Temperature 1 °C 2 Co2 1mol % 2 Co 1mol % 2850 6.23 93.77950 1.32 98.681050 0.37 99.631200 0.06 99.94

15.98 In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the cor-responding expression for Kp. (b) By using data in Appendix B, give the value of Kp for this reaction at 30 °C. (c) What is the value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

design an ExperimentThe reaction between hydrogen and iodine to form hydrogen iodide was used to illus-trate Beer’s law in Chapter 14 (Figure 14.5). The reaction can be monitored using visi-ble-light spectroscopy because I2 has a violet color while H2 and HI are colorless. At 300 K, the equilibrium constant for the reaction H21g2 + I21g2 ∆ 2 HI1g2 is Kc = 794. To answer the following questions assume you have access to hydrogen, iodine, hydro-gen iodide, a transparent reaction vessel, a visible-light spectrometer, and a means for changing the temperature. (a) Which gas or gases concentration could you readily moni-tor with the spectrometer? (b) To use Beer’s law (Equation 14.5) you need to determine the extinction coefficient, e, for the substance in question. How would you determine e? (c) Describe an experiment for determining the equilibrium constant at 600 K. (d) Use the bond enthalpies in Table 8.4 to estimate the enthalpy of this reaction. (e) Based on your answer to part (d), would you expect Kc at 600 K to be larger or smaller than at 300 K?

16Acid–Base EquilibriaThe acids and bases that you have used so far in the laboratory are probably solutions of relatively simple inorganic substances, such as hydrochloric acid, sulfuric acid, sodium hydroxide, and the like. But acids and bases are important even when we are not in the lab.

Kw = [H3O+][OH−] defines the relationship between H3O

+ and OH− concentrations in aqueous solutions.

16.4 The ph Scale We use the pH scale to describe the acidity or basicity of an aqueous solution. Neutral solutions have a pH = 7, acidic solutions have pH below 7, and basic solutions have pH above 7.

16.5 STrong acidS and BaSeS We categorize acids and bases as being either strong or weak electrolytes. Strong acids and bases are strong electrolytes, ionizing or dissociating completely in aqueous solution. Weak acids and bases are weak electrolytes and ionize only partially.

16.1 acidS and BaSeS: a Brief review We begin by reviewing the Arrhenius definition of acids and bases.

16.2 BrønSTed–lowry acidS and BaSeS We learn that a Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry base is a proton acceptor. Two species that differ by the presence or absence of a proton are known as a conjugate acid–base pair.

16.3 The auToionizaTion of waTer We see that the autoionization of water produces small quantities of H3O

+ and OH− ions. The equilibrium constant for autoionization,

WhAT’s AhEAd

▶ Clusters of grapes and balsamiC vinegar. Grapes contain several acids that contribute to their characteristic flavor. The distinctive flavor of all vinegars is due to acetic acid. Balsamic vinegar is obtained by fermenting grapes.

They are ubiquitous, including in the foods we eat. The characteristic flavor of the grapes shown in the opening photograph is largely due to tartaric acid 1H2C4H4O62 and malic acid 1H2C4H4O52 (Figure 16.1), two closely related (they differ by only one O atom) organic acids that are found in biological systems. Fermentation of the sugars in the grapes ultimately forms vinegar, the tangy, sour flavor of which is due to acetic acid 1CH3COOH2, a substance we discussed in Section 4.3. The sour taste of oranges, lemons, and other citrus fruits is due to citric acid 1H3C6H5O72, and, to a lesser extent, ascorbic acid 1H2C6H6O62, better known as Vitamin C.

Acids and bases are among the most important substances in chemistry, and they affect our daily lives in innumerable ways. Not only are they present in our foods, but acids and bases are also crucial components of living systems, such as the amino acids that are used to synthesize proteins and the nucleic acids that code genetic information. Both citric and malic acids are among several acids involved in the Krebs cycle (also called the citric acid cycle) that is used to generate energy in aerobic organisms. The application of acid–base chemistry has also had critical roles in shaping modern society, including such human-driven activities as indus-trial manufacturing, the creation of advanced pharmaceuticals, and many aspects of the environment.

The impact of acids and bases depends not only on the type of acid or base, but also on how much is present. The time required for a metal object immersed in water to corrode, the ability of an aquatic environment to support fish and plant life, the fate of pollutants washed out of the air by rain, and even the rates of reactions that maintain

534 chapTer 13 properties of Solutions · A solution that is in equilibrium with undissolved solute...

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