Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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Fast

Step 2: Turn off light

Why is it useful to know the individual steps that make up the overall reaction?

Chemists often are interested in more detailed descriptions of chemical reactions,

including how the reactions occur at the molecular level. We know from our pre-

vious discussions that chemical equations contain a wealth of information about

a reaction. We can determine the spontaneity of the process, the enthalpy of the

process, and the stoichiometry of the reactants by examining the chemical equa-

tion. However, the overall chemical equation doesn’t show how the reactants

collide to become products. To address this concern, investigators study a single

reaction to determine exactly how each molecule moves during the course of the

process. The knowledge of exactly how a reaction proceeds is useful in predicting new

reactions, determining the rates of those reactions, discovering new applications of

chemistry, and learning how substances interact with humans and our environment.

Their study is part of the ﬁeld of mechanistic chemistry.

mechanism for a reaction is the set of steps that compounds take as they

proceed from reactant to product. The mechanism of a reaction accounts for the

experimentally determined rate of the reaction and is consistent with the overall

stoichiometry of the reaction. In some cases, a mechanism is only one step. In oth-

ers, there are a multitude of steps. However, we can’t tell this by looking at the

overall chemical equation.

Each of the single steps in a chemical reaction is called an

elementary step.For

example, as shown in Figure 15.15, the overall process of turning off the lights in

a room is made up of two elementary steps: (1) walking to the light switch and

(2) ﬂipping the light switch. One of these steps, walking to the light switch, is

slower. The other, ﬂipping the switch, is so fast that its contribution to the

time required to complete the overall process is negligible. Any calculations of

the time required to turn off the lights, then, can be reduced to the time it takes

to walk to the light switch. In other words, the time it takes to turn off the lights

in the room is nearly equal to the length of time it takes to walk to the light

switch. This slow step is known as the

rate-determining step.

Within each elementary step, reactants come together and undergo successful

collisions to make products. The number of molecules that collide in this process

deﬁnes the

molecularity of the step (Figure 15.16). A unimolecular reaction involves

one molecule as the only reactant. When two molecules collide, the reaction is

said to be a

bimolecular reaction. Reactions that involve the collision of three mol-

ecules simultaneously (which are called

termolecular reactions,or trimolecular

reactions

) are also known, but they are rare because they require that three mole-

cules collide at the same time and in the proper orientation. In such cases, the

decrease in the entropy associated with three molecules coming together at

one time is often prohibitive (see Chapter 14). In general, an elementary step has

only a single bond breakage or formation.

648 Chapter 15 Chemical Kinetics

Slow

Step 1: Walk to light

Unimolecular reaction

Bimolecular reaction

Termolecular reaction

FIGURE 15.16

Molecularity of reactions. The number of

molecules that must collide at one time

to produce the reaction determines the

molecularity.

FIGURE 15.15

The elementary steps completed

in turning off the lights.

Visualization: Decomposition of

Video Lesson: Deﬁning the

Molecularity of a Reaction

If we know the elementary steps in a reaction, we can obtain a wealth of in-

formation about the rate of the reaction. Most important, rate laws can be written

directly from elementary steps. Speciﬁcally, the rate orders for the reactants in an

elementary step are given by the stoichiometric coefﬁcients of the reactants in

that step. Moreover, the rates of the elementary steps in a mechanism need not be

the same—indeed, in most cases they are not. A mechanism with two elementary

steps typically has a fast step and a slow step. The slow step in a mechanism is the

one that will determine the rate of the overall reaction.

The reaction of NO

and F

gas is a useful case study:

2NO

(g ) + F

(g ) → 2NO

F(g)

After careful experimentation, a mechanism involving two elementary steps has

been suggested for the reaction. This ﬁrst step was determined to be slow; the

second was determined to be relatively fast:

(g ) + F

(g ) → NO

F(g ) + F(g) (slow)

F(g) + NO

(g ) → NO

F(g ) (fast)

2NO

(g ) + F

(g ) → 2NO

F(g)

Note that the sum of the elementary steps gives the balanced equation. From

these steps, we can determine the rate law for the overall reaction in a way that is

similar to our discussion of the length of time required to turn off the lights. The

slow elementary step will determine the rate of the reaction, so we can use this

rate-determining step to write the rate law. Judging on the basis of the stoichio-

metric coefﬁcients of the reactants in the slow step, the overall reaction is ﬁrst

order in NO

, ﬁrst order in F

, and second order overall.

Rate = k[NO

][F

]

Not all mechanisms are so easy to work with. For instance, the reaction of NO

with O

to make NO

is much more complicated than our ﬁrst glance suggests.

From our previous discussion in Section 15.4, we know the overall equation for

the reaction:

2NO(g ) + O

(g ) → 2NO

(g )

We saw that the experimentally determined rate equation for the reaction is

Rate = k[NO]

]

This implies that the reaction is a termolecular process, which is unlikely because

it would require three molecules to collide simultaneously. The mechanism of

this reaction is known, and it has the elementary steps shown below.

2NO(g )

−1

(g )

(

fast)

(g ) + O

(g )

−→ 2NO

(g )

(slow)

What is the meaning of the double arrows in the fast equation? These arrows in-

dicate that the reaction can proceed in both the forward direction and the reverse

direction at the same time. Moreover, each of these directions has a rate constant

associated with it (k

and k

−1

). The slow step in our mechanism is the reaction of

(g) with oxygen, and we could write the rate equation on the basis of this in-

formation. Note that we’ve speciﬁed the rate constants for both reactions:

Rate = k

[NO]

for the fast step (k

)

Rate = k

][O

]

for the slow step (k

)

It follows that the rate law for the overall reaction should be

Rate = k

][O

]

for the overall reaction

15.6 Reaction Mechanisms 649





Video Lesson: Determining the

Rate Laws of Elementary

Reactions

Video Lesson: Calculating the

Rate Laws of Multistep

Reactions

However, this doesn’t agree with the experimentally determined rate law that we

noted above:

Rate = k[NO]

]

This difference arises because N

(g) is an intermediate in the reaction. An

intermediate is a compound that is formed and consumed during the course of a

reaction. If we examine the overall reaction, we don’t see the intermediate N

as one of the reactants. Measuring the concentration of this species, then, could

be difﬁcult. We need to rewrite the rate equation so that the rate reﬂects only the

compounds in the overall reaction. To deal with this, we will assume that the fast

reaction reaches equilibrium, an assumption that greatly simpliﬁes our determi-

nation of the overall rate law for the reaction. What does this assumption mean?

Within a reaction that is at equilibrium, the rate of the forward reaction is equal

to the rate of the reverse reaction.

2NO(g )

−1

(g ) (fast)

[NO]

= k

−1

]

Then we can rearrange our equation to solve for [N

] =

[NO]

−1

= k



[NO]

where the new rate constant



is equal to

−1

We can now substitute for the concentration of the intermediate, [N

], in

our original rate equation and simplify:

Rate = k

][O

]

] = k



[NO]

Rate = k



[NO]

] = k



[NO]

]

where the new rate constant



is equal to



This agrees with our experimentally determined rate equation. The mathe-

matics used to convert our rate law containing the intermediate into the experi-

mentally observed rate law are based on the assumption that the fast reaction has

reached equilibrium. This assumption requires that the rate constant for the re-

verse of the fast reaction be much larger than the rate constant for the slow step.

EXERCISE 15.8 Rate Laws

Write the overall reaction, identify any intermediates, and write the rate law for the

following proposed mechanism for the decomposition of IBr(g) to I

(g) and Br

(g).

IBr(g) n I(g) + Br(g) (slow)

IBr(g) + Br(g) n I(g) + Br

(g) (fast)

I(g) + I(g) n I

(g) (fast)

650 Chapter 15 Chemical Kinetics





* Note: Alternatively, we could rearrange this equation to place the rate constants on one side and the con-

centrations of compounds on the other side:

−1

]

[NO]

This gives rise to something that looks remarkably similar to Q, the reaction quotient from Chapter 14.

We’ll explore this in much greater detail in Chapter 16.

Solution

The overall reaction is the sum of the elementary steps in the mechanism.

2IBr(g) n I

(g) + Br

(g)

The intermediates are produced and consumed in the reaction. They are I(g) and

Br(g). Because the ﬁrst step in the mechanism is the slow step, it is rate determining.

Therefore, the rate law for the reaction can be written directly from this step:

rate =k [IBr]

It is ﬁrst order in IBr(g) and ﬁrst order overall.

PRACTICE 15.8

Write the overall reaction, identify any intermediates, and write the rate law for the

following proposed mechanism for the production of nitrogen dioxide (NO

NO(g) + O

(g) NO

(g) (fast)

(g) + NO(g) n 2NO

(g) (slow)

See Problems 85 and 86.

Transition State Theory

The destruction of ozone by atomic oxygen is one of the ways in which stratos-

pheric ozone can be depleted:

(g ) + O(g) → 2O

(g )

By examining this reaction in the laboratory, we can determine the

change in enthalpy (∆H =−392 kJ/mol) and measure the rate of

the reaction. However, it is often helpful to examine a reaction by

consulting a plot of the energies for the reactants, products, and

any intermediates. If we make a graph of the energies as the reac-

tants proceed along a

reaction coordinate (the pathway describing

the changes in each molecule in the reaction, even though they are

happening at different times) to become product, we obtain a

reaction proﬁle like that shown in Figure 15.17. On the basis of our

earlier discussion of collision theory, we know that the reactants

must have enough energy to overcome a barrier known as the

activation energy (E

)—assuming that the reactants collide in the

proper orientation. For the reaction of atomic oxygen and ozone,

the barrier is rather small (E

a(forward)

=19 kJ/mol) compared to the

reverse reaction. The reaction proﬁle illustrates this.

The reaction proﬁle is used to explore

transition state theory

and how it applies to a reaction. This theory describes how the

bonds in the reacting molecules reorganize to represent the bonds in the prod-

ucts. At some point on the reaction coordinate the collision occurs, and the atoms

in the reactants occupy a

transition state. The collection of atoms at the transition

state, called the

activated complex, is very energetic at this point in the reaction—

more so than the reactants, products, or intermediates. The activated complex is

not a separate isolable compound. Rather, it is a snapshot of the reaction at the

point in time when the molecules have collided. From the activated complex, the re-

action could proceed to products, or the complex could dissociate back into the

reactants.

The reaction proﬁle shows the activation energy for the forward reaction, the

activation energy for the backward reaction, and the overall change in the energy

of the reaction (∆E) in a graphical way. This change in energy (∆E) is equal to ∆H





15.6 Reaction Mechanisms 651

Progress of reaction

19 kJ/mol

–392 kJ/mol

411 kJ/mol

Transition state

∆H

(reaction)

a(forward)

a(reverse)

Potential energy

+ O

→

FIGURE 15.17

Reaction proﬁle.

Visualization: Transition States

and Activation Energy

for the reaction at constant pressure and volume. Even for cases where ∆E and ∆H

are not equal (unequal volumes or pressures), the difference is usually very small.

By looking at the reaction proﬁle, we can tell whether a reaction is exothermic

(Figure 15.18) or endothermic (Figure 15.19).

H

reaction

∼

E = E

a(forward)

− E

a(reverse)

To what is the activation energy related? In 1888, Svante Arrhenius (1859–1927),

a Swedish chemist, studied how temperature affected the rate of a reaction. What

came out of this work is a relationship between the activation energy and the rate

of a reaction. His mathematical equation is

k = Ae

−E

/RT

where k is the rate constant, A is the frequency factor that relates how many

successfully oriented collisions occur in a particular reaction, E

is the activation

energy, T is the temperature in kelvins, and R is the universal gas constant

(8.314 J/mol·K). This equation says that the rate constant of a reaction is related

to the size of E

.As E

gets larger, the rate constant gets smaller and the rate of the

reaction decreases.

This equation can be used to determine the rate of a reaction on the basis of

the temperature of the reaction and the amount of energy the reactants require to

make the activated complex. To use it, however, we must know the activation

energy (E

) and the frequency factor (A) for the reaction. Fortunately, by per-

forming the reaction at two different temperatures, we can utilize this equation to

calculate the activation energy without knowing the frequency factor. Alterna-

tively, if we know the activation energy and the rate of reaction at a particular

temperature, we can determine the rate of reaction at any other temperature. The

equation that relates this calculation can be derived by taking the natural loga-

rithm of the Arrhenius equation above, at two temperatures:







−



where k

and k

are the rate constants for the reaction obtained at two different

temperatures, T

and T

652 Chapter 15 Chemical Kinetics

Progress of reaction

Transition state

∆H

(reaction)

endothermic

a(forward)

a(reverse)

Potential energy

FIGURE 15.19

Endothermic reaction proﬁle. The products are more energetic

than the reactants in the endothermic reaction.

a(forward)

is larger than E

a(reverse)

Progress of reaction

Transition state

∆H

(reaction)

exothermic

a(forward)

a(reverse)

Potential energy

FIGURE 15.18

Exothermic reaction proﬁle. The reactants are more energetic

than the products in the endothermic reaction.

a(reverse)

is larger than E

a(forward)

Video Lesson: The Arrhenius

Equation

Video Lesson: Using the

Arrhenius Equation

This equation provides a quick method to determine the energy of activation

for a reaction, but the resulting answer can contain signiﬁcant error, because its

value is based on only two reactions. Alternatively, we can determine the value of

graphically by using a series of rate constants calculated at different tempera-

tures. To see how this is done, let’s examine the original equation produced by

Arrhenius:

k = Ae

−E

/RT

If we take the natural log of both sides of the equation, we get

ln k = lnA −

Rearranging this equation as shown enables us to determine the activation en-

ergy by plotting the rate constant of a reaction at several temperatures.

ln k =−





+ ln A

y = mx + b

If we construct an “Arrhenius plot” of ln k versus 1/T for a series of reactions,

we obtain a straight line whose slope is equal to –E

/R and whose intercept is ln A,

as shown in Figure 15.20. This method not only enables us to determine the en-

ergy of activation quite accurately but also provides a convenient way to deter-

mine the frequency factor A.

EXERCISE 15.9 Energy Barrier

The reaction of NO, a component in smog, with ozone has been extensively studied.

Data for the temperature dependence are tabulated below. What is the activation

energy for this reaction?

NO(g ) + O

(g ) → NO

(g ) + O

(g )

Temperature (K) k (1/M

195 1.08 × 10

298 12.0 × 10

15.6 Reaction Mechanisms 653

ln k

1/T

–8

–7

–6

–5

–4

0.00315 0.00325 0.00335 0.00345 0.00355 0.00365

y = –6517.3x + 16.089

FIGURE 15.20

A plot of ln k versus 1/T gives a straight line whose slope

is −E

/R. Each data point on this plot corresponds to the

measurement of the temperature and rate constant in

separate reactions. The slope of the line of best ﬁt

(−6517.3) is equal to −E

/R. Therefore, E

= 54.2 kJ/mol

for this reaction. If the axes went to zero, the intercept

(16.089) corresponding to ln A would be found. The

frequency factor is 9,710,000.

Solution

The activation energy can be calculated by substituting the values from the table

into the equation:







−





12.0 ×10

1.08 ×10



8.314 J·mol

−1

·K

−1



195 K

−

298 K



2.408 =

8.314 J·mol

−1

·K

−1



1.773 ×10

−3

−1



20.02 J·mol

−1

= E



1.773 ×10

−3



= 11292 J·mol

−1

= 11.3kJ·mol

−1

PRACTICE 15.9

More data on the reaction of NO and O

are shown in the table below. Calculate E

for each pair of reactions. Are the values the same? Explain.

See Problems 83, 84, 87, 88, and 101.

15.7 Applications of Catalysts

Environmental chemists have comprehensively explored the rate of decomposi-

tion of atrazine in aqueous solutions. Because the rate in water is so slow

(

1/2

= 400

days),they’ve spent time considering how the decomposition could be

accelerated to clean up the environment more quickly. The exercise in the previ-

ous section showed that the rate of a reaction increases with an increase in

temperature. That’s one way to speed up a reaction. However, raising the temper-

ature of groundwater, rivers, and lakes is not a feasible way to increase the rate

of decomposition of herbicides and pesticides. Researchers at the University of

Wisconsin have found a better way to enhance the rate of the decomposition.They

have discovered that ﬁltering atrazine-contaminated water through a container

full of titanium(IV) oxide in the presence of ultraviolet light greatly increases the

rate of decomposition (to t

1/2

=15 min). What does the titanium(IV) oxide do?

The titanium(IV) oxide is used as a

catalyst in the decomposition of atrazine.

A catalyst is a compound that, when added to a reaction mixture, changes the

mechanism of the reaction to a new pathway with a lower activation energy.

Rather than lowering the activation energy of the existing mechanism, it creates

a new set of elementary steps whose rate-determining step has a lower energy of

activation than the reaction would have without the catalyst. Because E

is lower,

the new mechanism is faster, so we often say that a catalyst increases the rate of the

reaction. Another useful aspect of catalysts is that they can be recovered from

the reaction; catalysts are not consumed in a reaction. Because of this, they don’t

appear in the net chemical equation. However, because they are involved in the

reaction, they do appear in the mechanism. To show that a particular reaction in-

volves a catalyst, we typically place it over or under the arrow in the overall equa-

tion. There are two types of catalysts, homogeneous catalysts and heterogeneous

654 Chapter 15 Chemical Kinetics

Temperature (K) k (1/M·s)

230 2.95 × 10

260 5.42 × 10

369 35.5 × 10

Video Lesson: Catalysts and

Types of Catalysts

catalysts. The type of catalyst, as well as the reaction proﬁle that results, depends

on how the catalyst is mixed with the reaction.

homogeneous catalyst is part of a reaction that is catalyzed in a homogeneous

mixture (that is, the catalyst is intimately mixed with the reactants). For example,

the destruction of ozone by chlorine radicals (a homogeneous catalyst) is illus-

trated by the net reaction of ozone with oxygen atoms:

+ O−−−−−−−→2O

The mechanism of this reaction consists of two elementary steps:

Step 1: The ozone molecule reacts with elemental chlorine (the cata-

lyst) to make a molecule of ClO and a molecule of oxygen.

Step 2: The ClO reacts with an oxygen atom (the other reactant in the

net equation) to make another molecule of oxygen and regenerate

elemental chlorine.

The chlorine atom appears ﬁrst as a reactant and later as a product; chlorine be-

gins and escapes the reaction without being changed and therefore is not con-

sumed. This is exactly what catalysts do. Is ClO a catalyst too? No. It is formed in

the course of the reaction and then consumed before products are made. The ClO

never escapes the reaction. As we saw earlier, this is exactly what happens to an

intermediate.

The presence of chlorine atoms causes a tremendous increase in the rate of

the decomposition of ozone. How does a catalyst cause a reaction’s rate to in-

crease? Consider the reaction proﬁle, shown in Figure 15.21, for the ﬁrst-order

decomposition of hydrogen peroxide by a homogeneous catalyst. In the noncat-

alyzed reaction, the reaction follows a unimolecular mechanism with a clearly

deﬁned high-energy barrier (the activation energy) separating the reactants from

the products. If a catalyst such as iodide is added to this reaction, the mechanism

of the reaction changes. The iodide is the catalyst, and it

is shown over the reaction arrow to identify it as such.

What is the role of OI

–

in the reaction? The reaction makes

an intermediate, which is more stable than the activated

complex. The solid line in the reaction proﬁle in Figure

15.21 illustrates the new reaction.

What is the outcome of adding a catalyst? Remember

that our catalyst has produced a new mechanism for

the reaction. If we plot the reaction proﬁle for this new

mechanism, we can see that the activation energy is

lower. This means that more molecules will have the

15.7 Applications of Catalysts 655

Application

HEMICAL ENCOUNTERS:

Destruction of Ozone

Step 1

Step 2

Intermediate Catalyst

ClO++

ClClO + +

Step 1

Step 2

Net reaction

OOI

–

OOI

–

O+

Progress of reaction

Reactants

Products

Catalyzed E

Uncatalyzed E

Potential energy

FIGURE 15.21

Decomposition of H

with and without

a homogeneous catalyst. The dotted line

represents the reaction proﬁle without

the catalyst; the solid line indicates the

effect of a homogeneous catalyst.

Video Lesson: CIA

Demonstration: Elephant Snot

Visualization: Homogeneous

Catalysis

Progress of reaction

Reactants

Products

a(uncatalyzed)

a(catalyzed)

a(adsorb)

a(desorb)

Potential energy

FIGURE 15.22

Heterogeneous catalysis of NO.

requisite energy to become activated complexes in the mechanism. Although the

overall reaction enthalpy (∆H) hasn’t changed, the activation energy has been re-

duced dramatically. A reduction in the activation energy causes the rate of the

reaction to increase.

In systems with a

heterogeneous catalyst, the catalyst and the reactants are in

different physical states. The catalyst is typically a metal in solid form, whereas the

reactants are typically in gaseous, aqueous, or liquid form. The TiO

decomposi-

tion of atrazine is an example of this type of catalysis. The catalytic converter

(usually made up of platinum, palladium, and/or rhodium metal) in your auto-

mobile is another. One of the reactions in the catalytic converter cuts down on

smog-forming NO by reducing it to N

. The general unbalanced reaction is

NO(g)

→

(g)

The heterogeneous catalyst works in a three-step process. In the ﬁrst step

shown in Figure 15.22, the reactant molecules

adsorb to the catalyst—that is, they

sticks to the catalyst’s surface. Note that the word adsorb is different than the word

absorb, which means being taken up or mixed into a substance.A small activation

energy barrier must be crossed to achieve the surface binding of the reactant.

Often, there is only a very small barrier for binding of a reactant to the surface of

a catalyst. Then the reactants migrate around on the surface until they collide to

make the product. A larger, yet still low, activation energy barrier exists for this

step. In the ﬁnal step, the products are

desorbed, or released from the surface of

the catalyst (the reverse of being adsorbed). The resulting reaction proﬁle dia-

gram has a characteristic shape.

656 Chapter 15 Chemical Kinetics

All of the different-colored heteroge-

neous catalysts in this collection work in

a similar manner. The reactants must ad-

sorb to the surface of the catalyst before

the reaction can be catalyzed. After reac-

tion, the products desorb from the cata-

lyst. The reaction shown here illustrates

the catalytic hydrogenation of ethylene

to make ethane.

(a) (b) (c) (d)

Hydrogen

)

Ethene

)

Ethane

)

Catalyst

surface

Hydrogen and ethene bond to the surface of the catalyst (a). The hy-

drogen atoms then migrate to the ethene molecule in steps (b) and

(c). Finally, the product molecule is released from catalyst (d).

Visualization: Heterogeneous

Catalysis

Video Lesson: CIA

Demonstration: The Cobalt(II)-

Catalyzed Reaction of

Potassium Sodium Tartrate

Video Lesson: CIA

Demonstration: The Copper-

Catalyzed Decomposition of

Acetone

15.7 Applications of Catalysts 657

The modern confectionery industry operates a booming

business, helping to satisfy that sweet tooth in most of

us. Fanciful desserts require sucrose as a sweetener. For-

tunately, the American farmer can meet the large de-

mand for sweeteners. For instance, sugar beet produc-

tion in Colorado, Montana, Nebraska, and Wyoming

yields 4.5 to 6 million tons of sucrose per year. Sugar

cane production, primarily in Hawaii, Louisiana, and

Florida, adds another 6 million tons to the total. But

even more sugar is needed. One of the ways to meet the

public’s demand for sweeteners involves corn starch and

a biological molecule known as

D-glucose isomerase,

shown in Figure 15.23. The product of these molecules is

sweeter than sucrose alone. It is known as high-fructose

corn syrup, and its use has surpassed that of sucrose in

the confectionery industry.

Like other enzymes, D-glucose isomerase acts as a

catalyst that speeds up a reaction. The enzymes work by

binding selectively to a particular molecule, forcing it

into just the right shape, and then assisting in the

NanoWorld / MacroWorld

Big effects of the very small:

Enzymes—nature’s catalysts

FIGURE 15.24

Diagram of an enzyme-catalyzed reaction. The substrate binds to the

active site on an enzyme, where the reaction is catalyzed.

reaction that makes the product. They increase the value

of A (the frequency factor from the Arrhenius equation)

and decrease the energy of activation (E

) at the same

time. This activity arises because the backbone of the

amino acid polymer weaves the enzyme into a structure

similar to a catcher’s mitt. Along the inside of the

catcher’s mitt (the active site of the enzyme) lie portions

of the enzyme that are polar and portions that are non-

polar. The arrangements of the polar and nonpolar

groups provide a template that exactly matches that of

the molecule they bind (the reactant or substrate). When

the substrate binds, the enzyme bends it into a confor-

mation similar to that of the product of the reaction.

Then, when the conformation is just right, the reaction

takes place. After releasing the product, the enzyme re-

turns to its original shape, ready to accept another sub-

strate (Figure 15.24). The net result is an increase in the

reaction rate without an increase in temperature, which

is particularly useful in the food industry.

For example, lactase (an enzyme that converts lactose

into glucose and galactose) is used in the dairy industry

to make digestible milk products. Because many of these

products spoil at temperatures warmer than those found

in a refrigerator, the use of enzymes to speed the reaction

without a temperature increase is quite helpful. After the

lactase has been added to milk, lactose-intolerant people

can drink all of the milk they want. And they owe their

settled stomach to one of nature’s catalysts.

FIGURE 15.23

High-fructose corn syrup is made from corn starch using

an enzyme.

D-Glucose isomerase catalyzes the reaction

that converts glucose into fructose. The reaction pro-

ceeds much more rapidly with the enzyme than without

it. The enzyme is shown here as a series of ribbons that

represent the atoms that make up the strands of protein

polymers. The strands loop and wind their way together

to make a pocket that can catalyze the reaction of glu-

cose to make fructose.

Enzyme

Substrate

Active site

Enzyme–substrate

complex

Products

Enzyme