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6 Effects of Temperature, Concentration, and Catalysts on Reaction Rates

6 Effects of Temperature, Concentration, and Catalysts on Reaction Rates

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Chemical reactions: energy, rates, and equilibrium


One way to increase reaction rate is to add energy to the reactants by raising the temperature. With more energy in the system, the reactants move faster, so the frequency of collisions increases. Furthermore, the force with which collisions occur increases, making

them more likely to overcome the activation barrier. As a rule of thumb, a 283 K rise in

temperature causes a reaction rate to double.

Increase in frequency

of collisions

Increase in


Increase in

reaction rate

Increase in forcefulness

of collisions


Concentration A measure of the

amount of a given substance in a


A second way to speed up a reaction is to increase the concentrations of the reactants.

As the concentration increases, reactants are crowded together, and collisions between

reactant molecules become more frequent. As the frequency of collisions increases,

reactions between molecules become more likely. Flammable materials burn more rapidly in pure oxygen than in air, for instance, because the concentration of O2 molecules

is higher (air is approximately 21% oxygen). Hospitals must therefore take extraordinary

precautions to ensure that no flames are used near patients receiving oxygen. Although

different reactions respond differently to concentration changes, doubling or tripling a

reactant concentration often doubles or triples the reaction rate.

Increase in


Increase in frequency

of collisions

Increase in

reaction rate


Catalyst A substance that speeds up

the rate of a chemical reaction but is

itself unchanged.

A third way to speed up a reaction is to add a catalyst—a substance that accelerates a

chemical reaction but is itself unchanged in the process. For example, metals such as

nickel, palladium, and platinum catalyze the addition of hydrogen to the carbon–carbon

double bonds in vegetable oils to yield semisolid margarine. Without the metal catalyst,

the reaction does not occur.





A double bond in

vegetable oil



Ni, Pd, or Pt








A single bond

in margarine

A catalyst does not affect the energy level of either reactants or products. Rather,

it increases reaction rate either by letting a reaction take place by an alternative set of

reaction steps with a lower activation energy or by orienting the reacting molecules

appropriately. In a reaction energy diagram, the catalyzed reaction has a lower activation energy (Figure 7.4). A catalyzed reaction releases (or absorbs) the same amount of

energy as an uncatalyzed reaction; it simply occurs more rapidly.

In addition to their widespread use in industry, we also rely on catalysts to reduce

the air pollution created by exhaust from automobile engines. The catalytic converters in most automobiles are tubes packed with catalysts of two types (Figure 7.5). One

catalyst accelerates the complete combustion of hydrocarbons and CO in the exhaust to

give CO2 and H2O and the other decomposes NO to N2 and O2.


effects of temperature, Concentration, and Catalysts on reaction rates


Free energy

Catalyst present



The thousands of biochemical

reactions continually taking place in

our bodies are catalyzed by large protein molecules called enzymes, which

promote reactions by controlling the

orientation of the reacting molecules.

Since almost every reaction is catalyzed

by its own specific enzyme, the study of

enzyme structure, activity, and control

is a central part of biochemistry. We will

look more closely at enzymes and how

they work in Chapter 19.








CxHy, CO, NO, O2

Figure 7.4

A reaction energy diagram for a

reaction in the presence (green

curve) and absence (blue curve) of a


The catalyzed reaction has a lower

1Eact 2 because it uses an alternative

pathway (represented by the multiple

bumps in the green line) with a lower

energy barrier.

No catalyst present




CO2, H2O, N2, O2

Figure 7.5

A catalytic converter.

The exhaust gases from an automobile pass through a two-stage catalytic converter. In one stage,

carbon monoxide and unburned hydrocarbons are converted to CO2 and H2O. In the second stage,

NO is converted to N2 and O2.

Table 7.3 summarizes the effects of changing conditions on reaction rates.

table 7.3 Effects of Changes in Reaction Conditions on Reaction Rates




Increase in reactant concentration increases rate.

Decrease in reactant concentration decreases rate.


Increase in temperature increases rate.

Decrease in temperature decreases rate.

Catalyst added

Increases reaction rate.

ProBlEM 7.10

Ammonia is synthesized industrially by reaction of nitrogen and hydrogen

according to the equation 3 H2 1g2 + N2 1g2 ¡ 2 NH3 1g2. The free-energy

change for this reaction is ∆G = - 16 kJ>mol, yet this reaction does not readily

occur at room temperature. List three ways to increase the rate of this reaction.




Chemical reactions: energy, rates, and equilibrium

7.7 Reversible Reactions and Chemical Equilibrium

Learning Objective:

• Define chemical equilibrium for reversible reactions.

Many chemical reactions result in the complete conversion of reactants into products.

When sodium metal reacts with chlorine gas, for example, both are entirely consumed.

The sodium chloride product is so much more stable than the reactants that, once

started, the reaction keeps going until it is complete.

What happens, though, when the reactants and products are of approximately equal

stability? This is the case, for example, in the reaction of acetic acid (the main organic

constituent of vinegar) with ethanol to yield ethyl acetate, a solvent used in nail-polish

remover and glue.




Acetic acid

Reversible reaction A reaction

that can go in either direction, from

products to reactants or reactants to


Chemical equilibrium A state in

which the rates of forward and reverse

reactions are the same.


This direction?

Or this direction?


Ethyl acetate


Imagine the situation if you mix acetic acid and ethanol. The two begin to form

ethyl acetate and water. But as soon as ethyl acetate and water form, they begin to go

back to acetic acid and ethanol. Such a reaction, which easily goes in either direction,

is a reversible reaction and is indicated by a double arrow 1 H 2 in equations. The

reaction read from left to right as written is referred to as the forward reaction, and the

reaction from right to left is the reverse reaction.

Now, suppose you mix some ethyl acetate and water. The same thing occurs: as

soon as small quantities of acetic acid and ethanol form, the reaction in the other direction begins to take place. No matter which pair of reactants is mixed together, both

reactions occur until ultimately the concentrations of reactants and products reach constant values and undergo no further change. At this point, the reaction vessel contains

all four substances—acetic acid, ethyl acetate, ethanol, and water—and the reaction is

in a state of chemical equilibrium.

Since the reactant and product concentrations undergo no further change once equilibrium is reached, you might conclude that the forward and reverse reactions have

stopped. That is not the case, however. The forward reaction takes place rapidly at the

beginning of the reaction but then slows down as reactant concentrations decrease. At

the same time, the reverse reaction takes place slowly at the beginning but then speeds

up as product concentrations increase (Figure 7.6). Ultimately, the forward and reverse

rates become equal and change no further.

Chemical equilibrium is an active, dynamic condition. All substances present are

continuously being made and unmade at the same rate, so their concentrations are

constant at equilibrium. As an analogy, think of two floors of a building connected by

up and down escalators. If the number of people moving up is the same as the number of people moving down, the numbers of people on each floor remain constant.

Individual people are continuously changing from one floor to the other, but the total

populations of the two floors are in equilibrium. In complex biological systems, many

reactions may be linked together to establish an equilibrium, called homeostasis, in

which certain conditions such as body temperature or pH of blood are maintained at

optimal levels.

Note that it is not necessary for the concentrations of reactants and products at equilibrium to be equal (just as it is not necessary for the numbers of people on two floors

connected by escalators to be equal). Equilibrium can be reached at any point between

pure products and pure reactants. The extent to which the forward or reverse reaction

is favored over the other is a characteristic property of a given reaction under given




Reaction rate

Rate of forward


equilibrium equations and equilibrium Constants



rates equal

Rate of reverse reaction


Figure 7.6

Reaction rates in an equilibrium reaction.

The forward rate is large initially but decreases as the concentrations of reactants drop. The reverse

rate is small initially but increases as the concentrations of products increase. At equilibrium, the

forward and reverse reaction rates are equal.

7.8 Equilibrium Equations and Equilibrium Constants

Learning Objective:

• Define the equilibrium constant (K), and use the value of K to predict the extent of


Remember that the rate of a reaction depends on the number of collisions between molecules (Section 7.5), and that the number of collisions in turn depends on concentration,

that is, the number of molecules in a given volume (Section 7.6). For a reversible reaction, then, the rates of both the forward and the reverse reactions must depend on the

concentration of reactants and products, respectively. When a reaction reaches equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of

reactants and products remain constant. We can use this fact to obtain useful information about a reaction.

Let us look at the details of a specific equilibrium reaction. Suppose that you allow

various mixtures of sulfur dioxide and oxygen to come to equilibrium with sulfur trioxide at a temperature of 1000 K and then measure the concentrations of all three gases

in the mixtures.

2 SO2 1g2 + O2 1g2 H 2 SO3 1g2

In one experiment, we start with only 1.00 mol of SO2 and 1.00 mol of O2 in a 1.00 L

container. In other words, the initial concentrations of reactants are 1.00 mol>L. When

the reaction reaches equilibrium, we have 0.0620 mol>L of SO2, 0.538 mol>L of O2, and

0.938 mol>L of SO3. In another experiment, we start with 1.00 mol>L of SO3. When

this reaction reaches equilibrium, we have 0.150 mol> L of SO2, 0.0751 mol>L of O2,

and 0.850 mol>L of SO3. In both cases, we see that there is substantially more product

1SO3 2 than reactants when the reaction reaches equilibrium, regardless of the starting

conditions. Is it possible to predict what the equilibrium conditions will be for any given


As it turns out, the answer is YES! No matter what the original concentrations

were, and no matter what concentrations remain at equilibrium, we find that a constant

When the number of people moving up is the same as the number of

people moving down, the number of

people on each floor remains constant, and the two populations are in





Chemical reactions: energy, rates, and equilibrium

numerical value is obtained if the equilibrium concentrations are substituted into the


3SO3 4 2

3SO2 4 2 3O2 4

= constant at a given temperature

The square brackets in this expression indicate the concentration of each substance

expressed as moles per liter. Using the equilibrium concentrations for each of the

experiments previously described, we can calculate the value and verify that it is constant:

Experiment 1.

Experiment 2.

3SO3 4 2

3SO2 4 2 3O2 4

3SO3 4 2

3SO2 4 2 3O2 4



10.938 mol>L2 2

10.0620 mol>L2 2 10.538 mol>L2

10.850 mol>L2 2

10.150 mol>L2 2 10.0751 mol>L2

= 425

= 428

At a temperature of 1000 K, the actual value of the constant is 429. Within experimental error, the ratios of product and reactant concentrations for the two experiments

at equilibrium yield the same result. Numerous experiments like those just described

have led to a general equation that is valid for any reaction. Consider a general reversible reaction:

aA + bB + c H mM + nN + c

Equilibrium constant (K) Value

obtained at a given temperature from

the ratio of the concentrations of

products and reactants, each raised to

a power equal to its coefficient in the

balanced equation.

where A, B, c are reactants; M, N, c are products; and a, b, c, m, n, c are coefficients in the balanced equation. At equilibrium, the composition of the reaction mixture obeys the following equilibrium equation, where K is the equilibrium constant.

Equilibrium equation

[M] m[N] n . . .

K =

[A] a[B] b . . .

Product concentrations

Reactant concentrations

Equilibrium constant

The equilibrium constant K is the number obtained by multiplying the equilibrium

concentrations of the products and dividing by the equilibrium concentrations of the

reactants, with the concentration of each substance raised to a power equal to its coefficient in the balanced equation. If we take another look at the reaction between sulfur

dioxide and oxygen, we can now see how the equilibrium constant was obtained:

2 SO2(g) + O2(g)


The practice of omitting pure

substances in the equilibrium constant

expression will be utilized in Chapter 10

when we discuss equilibria involving

acids and bases.

2 SO3(g)


[SO2]2 [O2]

Note that if there is no coefficient for a reactant or product in the reaction equation, it is assumed to be 1. The value of K varies with temperature, but a temperature of

298 K is assumed unless otherwise specified—and units are usually omitted.

For reactions that involve pure solids or liquids, these pure substances are omitted

when writing the equilibrium constant expression. To explain why, consider the decomposition of limestone:

CaCO3 1s2 ¡ CaO1s2 + CO2 1g2

Writing the equilibrium constant expression for this reaction as the concentration of

products over the concentration of reactions would yield

K =

3CaO4 3CO2 4

3CaCO3 4


Consider the solids CaO and CaCO3. Their concentrations (in mol > L) can be calculated from their molar masses and densities at a given temperature. For example, the

concentration of CaO at 298 K can be calculated as


a 3.25

equilibrium equations and equilibrium Constants

1000 cm3



mol CaO


= 58.0

g CaO



mol CaO

g CaO


The ratio of products over reactants would change if CO2 was added to or removed

from the reaction. The concentration of CaO, however, is the same whether we have

10 g or 500 g. Adding solid CaO will not change the ratio of products over reactants.

Since the concentration of solids is independent of the amount of solid present, these

concentrations are omitted and the expression for K becomes

K =

3CaO4 3CO2 4

= 3CO2 4

3CaCO3 4

The value of the equilibrium constant indicates the position of a reaction at equilibrium. If the forward reaction is favored, the product term 3M4 m 3N4 n (numerator) is larger than the reactant term 3A4 a 3B4 b 1denominator2, and the value of K is

larger than one. If instead the reverse reaction is favored, 3M4 m 3N4 n is smaller than

3A4 a 3B4 b at equilibrium, and the value of K is smaller than one.

For a reaction such as the combination of hydrogen and oxygen to form water vapor,

the equilibrium constant is enormous 13.1 * 1081 2, showing how greatly the formation of water is favored. Equilibrium is effectively nonexistent for such reactions, and

the reaction is described as going to completion.

On the other hand, the equilibrium constant is very small for a reaction such as the

combination of nitrogen and oxygen at 298 K to give NO 14.7 * 10-31 2, showing what

we know from observation—that N2 and O2 in the air do not combine noticeably at

room temperature:

N2 1g2 + O2 1g2 H 2 NO1g2

K =

3NO4 2

= 4.7 * 10-31

3N2 4 3O2 4

When K is close to 1, say between 103 and 10-3, significant amounts of both reactants and products are present at equilibrium. An example is the reaction of acetic acid

with ethanol to give ethyl acetate (Section 7.7). For this reaction, K = 3.4.


3CH3CO2CH2CH3 4 3H2O4

K =

= 3.4


We can summarize the meaning of equilibrium constants in the following way:

K very



Reaction goes

hardly at all

K very



More reactants

than products




More products

than reactants


Reaction goes

to completion

K much smaller than 0.001

Only reactants are present at equilibrium; essentially no

reaction occurs.

K between 0.001 and 1

More reactants than products are present at equilibrium.

K between 1 and 1000

More products than reactants are present at equilibrium.

K much larger than 1000

Only products are present at equilibrium; reaction goes

essentially to completion.





Chemical reactions: energy, rates, and equilibrium

Worked Example 7.7 Writing equilibrium equations

The first step in the industrial synthesis of hydrogen is the reaction of steam with methane to give carbon

monoxide and hydrogen. Write the equilibrium equation for the reaction.

H2O1g2 + CH4 1g2 H CO1g2 + 3 H2 1g2

AnAlySiS The equilibrium constant K is the number obtained by multiplying the equilibrium concentrations

of the products (CO and H2) and dividing by the equilibrium concentrations of the reactants (H2O and CH4),

with the concentration of each substance raised to the power of its coefficient in the balanced equation.


K =

3CO4 3H2 4 3

3H2O4 3CH4 4

Worked Example 7.8 equilibrium equations: Calculating K

In the reaction of Cl2 with PCl3, the concentrations of reactants and products were determined experimentally

at equilibrium and found to be 7.2 mol>L for PCl3, 7.2 mol>L for Cl2, and 0.050 mol>L for PCl5.

PCl3 1g2 + Cl2 1g2 H PCl5 1g2

Write the equilibrium equation, and calculate the equilibrium constant for the reaction. Which reaction is

favored, the forward one or the reverse one?

AnAlySiS All the coefficients in the balanced equation are 1, so the equilibrium constant equals the concen-

tration of the product, PCl5, divided by the product of the concentrations of the two reactants, PCl3 and Cl2.

Insert the values given for each concentration, and calculate the value of K.

BAllPArK EStiMAtE At equilibrium, the concentration of the reactants (7.2 mol>L for each reactant) is higher

than the concentration of the product (0.05 mol>L), so we expect a value of K less than 1.


K =

0.050 mol>L

3PCl5 4


= 9.6 * 10-4

3PCl3 4 3Cl2 4

17.2 mol>L2 17.2 mol>L2

The value of K is less than 1, so the reverse reaction is favored. Note that units for K are omitted.

BAllPArK CHECK Our calculated value of K is just as we predicted: K 6 1.

ProBlEM 7.11

Write equilibrium equations for the following reactions:

(a) N2O4 1g2 H 2 NO2 1g2

(b) 2 H2S1g2 + O2 1g2 H 2 S1s2 + 2 H2O1g2

(c) 2 BrF5 1g2 H Br2 1g2 + 5 F2 1g2

ProBlEM 7.12

Do the following reactions favor reactants or products at equilibrium? Give relative

concentrations at equilibrium.

(a) Sucrose1aq2 + H2O1l2 H Glucose1aq2 + Fructose1aq2 K = 1.4 * 105

(b) NH3 1aq2 + H2O1l2 H NH4 + 1aq2 + OH - 1aq2 K = 1.6 * 10 -5

(c) Fe 2O3 1s2 + 3 CO1g2 H 2 Fe1s2 + 3 CO2 1g2 K 1at 1000 K2 = 24.2

ProBlEM 7.13

For the reaction H2 1g2 + I2 1g2 H 2 HI1g2, equilibrium concentrations at 298 K are

3H2 4 = 0.0510 mol>L,3I2 4 = 0.174 mol>L, and 3HI4 = 0.507 mol>L. What is the

value of K at 298 K?



Le Châtelier’s Principle: the effect of Changing Conditions on equilibria

KEy ConCEPt ProBlEM 7.14

The following diagrams represent two similar reactions that have achieved equilibrium:

A2 + B 2

2 AB

A2 + 2B

2 AB

(a) Write the expression for the equilibrium constant for each reaction.

(b) Calculate the value for the equilibrium constant for each reaction.

7.9 Le Châtelier’s Principle: The Effect of Changing Conditions

on Equilibria

Learning Objective:

• Use Le Châtelier’s principle to predict the effect of changes in temperature, pressure,

and concentrations on an equilibrium reaction.

The effect of a change in reaction conditions on chemical equilibrium is predicted by a

general rule called Le Châtelier’s principle.

le Châtelier’s principle When a stress is applied to a system at equilibrium, the equilibrium shifts to relieve the stress.

The word “stress” in this context means any change in concentration, pressure, volume, or temperature that disturbs the original equilibrium and causes the rates of the

forward and reverse reactions to become temporarily unequal.

We saw in Section 7.6 that reaction rates are affected by changes in temperature

and concentration and by addition of a catalyst. But what about equilibria? Are they

similarly affected? The answer is that changes in concentration, temperature, and pressure do affect equilibria, but that addition of a catalyst does not (except to reduce the

time it takes to reach equilibrium). The change caused by a catalyst affects forward and

reverse reactions equally so that equilibrium concentrations are the same in both the

presence and the absence of the catalyst.

Effect of Changes in Concentration

Let us look at the effect of a concentration change by considering the reaction of CO with

H2 to form CH3OH (methanol). Once equilibrium is reached, the concentrations of the

reactants and product are constant, and the forward and reverse reaction rates are equal.

CO1g2 + 2 H2 1g2 H CH3OH1g2

What happens if the concentration of CO is increased? To relieve the stress of added

CO, according to Le Châtelier’s principle, the extra CO must be used up. In other words,

the rate of the forward reaction must increase to consume CO. Think of the CO added

on the left as “pushing” the equilibrium to the right:


CO( g) + 2 H2( g)


CH3OH( g)

Of course, as soon as more CH3OH forms, the reverse reaction also speeds up,

some CH3OH converts back to CO and H2. Ultimately, the forward and reverse reaction

rates adjust until they are again equal, and equilibrium is reestablished. At this new





equilibrium state, the value of 3H2 4 is lower because some of the H2 reacted with the

added CO and the value of 3CH3OH4 is higher because CH3OH formed as the reaction

was driven to the right by the addition of CO. The changes offset each other, however,

so that the value of the equilibrium constant K remains constant.

Chemical reactions: energy, rates, and equilibrium

CO( g) + 2 H2( g)

If this increases . . .

CH3OH( g)

. . . then this decreases . . .

. . . but this remains constant.

K =

. . . and this increases . . .


[CO] [H2]2

What happens if CH3OH is added to the reaction at equilibrium? Some of the

methanol reacts to yield CO and H2, making the values of 3CO4, 3H2 4, and 3CH3OH4

higher when equilibrium is reestablished. As before, the value of K does not change.

If this increases . . .

CO( g) + 2 H2( g)

. . . then this increases . . .

. . . but this remains constant.

K =

CH3OH( g)

. . . and this increases . . .


[CO] [H2]2

Alternatively, we can view chemical equilibrium as a balance between the free energy of the reactants (on the left) and the free energy of the products (on the right).

Adding more reactants tips the balance in favor of the reactants. In order to restore the

balance, reactants must be converted to products, or the reaction must shift to the right.

If, instead, we remove reactants, then the balance is too heavy on the product side and

the reaction must shift left, generating more reactants to restore balance.

▶ Equilibrium represents a balance

between the free energy of reactants

and products. Adding reactants (or

products) to one side upsets the balance, and the reaction will proceed in

a direction to restore the balance.

Adding reactants

to left side...

...will shift the

reaction to the right.

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6 Effects of Temperature, Concentration, and Catalysts on Reaction Rates

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