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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 13 Chemical Kinetics
IN PREVIOUS CHAPTERS WE STUDIED BASIC DEFINITIONS IN CHEMISTRY, 13.1 THE RATE OF A REACTION AND EXAMINED THE PROPERTIES OF GASES, LIQUIDS, SOLIDS, AND SO- 13.2 THE RATE LAW LUTIONS. 13.3 THE RELATION BETWEEN REACTANT
CONCENTRATION AND TIME WE HAVE DISCUSSED MOLECULAR PROPERTIES AND LOOKED AT SEVERAL TYPES OF REACTIONS IN SOME DETAIL. IN THIS CHAPTER AND 13.4 ACTIVATION ENERGY AND
TEMPERATURE DEPENDENCE OF RATE
CONSTANTS IN SUBSEQUENT CHAPTERS, WE WILL LOOK MORE CLOSELY AT THE RELATIONSHIPS AND LAWS THAT GOVERN CHEMICAL REACTIONS. HOW CAN 13.5 REACTION MECHANISMS
WE PREDICT WHETHER OR NOT A REACTION WILL TAKE PLACE? ONCE
13.6 CATALYSIS STARTED, HOW FAST DOES THE REACTION PROCEED?
REACTION GO BEFORE IT STOPS?
CUSSED IN THE HOW FAR WILL THE LAWS OF THERMODYNAMICS (DIS- CHAPTER 18) HELP US ANSWER THE FIRST QUESTION. CHEMICAL KINETICS, THE SUBJECT OF THIS CHAPTER, PROVIDES ANSWERS TO THE
QUESTION ABOUT THE SPEED OF A REACTION. THE LAST QUESTION IS ONE OF MANY ANSWERED BY THE STUDY OF CHEMICAL EQUILIBRIUM,
WHICH WILL BE CONSIDERED IN CHAPTERS 14, 15, 16, AND 18. 507 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 508 CHEMICAL KINETICS 13.1 THE RATE OF A REACTION Chemical kinetics is the area of chemistry concerned with the speeds, or rates, at which
a chemical reaction occurs. The word “kinetic” suggests movement or change; in
Chapter 5 we defined kinetic energy as the energy available because of the motion of
an object. Here kinetics refers to the rate of a reaction, or the reaction rate, which is
the change in the concentration of a reactant or a product with time (M/s).
There are many reasons for studying the rate of a reaction. To begin with, there
is intrinsic curiosity about why reactions have such vastly different rates. Some
processes, such as the initial steps in vision and photosynthesis and nuclear chain reactions, take place on a time scale as short as 10 12 s to 10 6 s. Others, like the curing of cement and the conversion of graphite to diamond, take years or millions of
years to complete. On a practical level, a knowledge of reaction rates is useful in drug
design, in pollution control, and in food processing. Industrial chemists often place
more emphasis on speeding up the rate of a reaction rather than on maximizing its
We know that any reaction can be represented by the general equation
reactants 88n products This equation tells us that during the course of a reaction, reactants are consumed while
products are formed. As a result, we can follow the progress of a reaction by monitoring either the decrease in concentration of the reactants or the increase in concentration of the products.
Figure 13.1 shows the progress of a simple reaction in which A molecules are
converted to B molecules:
A 88n B The decrease in the number of A molecules and the increase in the number of B molecules with time are shown in Figure 13.2. In general, it is more convenient to express
the reaction rate in terms of the change in concentration with time. Thus, for the reaction A 88n B we can express the rate as
FIGURE 13.1 The progress of
reaction A 88n B at 10-s intervals
over a period of 60 s. Initially,
only A molecules (black dots)
are present. As time progresses,
B molecules (colored dots) are
formed. Back Forward [A]
t or rate [B]
t where [A] and [B] are the changes in concentration (molarity) over a time period
t. Because the concentration of A decreases during the time interval, [A] is a neg- Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.1 509 40 A molecules
Number of molecules FIGURE 13.2 The rate of reaction
A 88n B, represented as the decrease of
A molecules with time and as the increase
of B molecules with time. THE RATE OF A REACTION 30 B molecules
10 0 10 20 30
t (s) 50 60 ative quantity. The rate of a reaction is a positive quantity, so a minus sign is needed
in the rate expression to make the rate positive. On the other hand, the rate of product
formation does not require a minus sign because [B] is a positive quantity (the concentration of B increases with time). These rates are average rates because they are
averaged over a certain time period t.
Our next step is to see how the rate of a reaction is obtained experimentally. By
definition, we know that to determine the rate of a reaction we have to monitor the
concentration of the reactant (or product) as a function of time. For reactions in solution, the concentration of a species can often be measured by spectroscopic means. If
ions are involved, the change in concentration can also be detected by an electrical
conductance measurement. Reactions involving gases are most conveniently followed
by pressure measurements. We will consider two specific reactions for which different
methods are used to measure the reaction rates.
REACTION OF MOLECULAR BROMINE AND FORMIC ACID In aqueous solutions, molecular bromine reacts with formic acid (HCOOH) as follows:
Br2(aq) HCOOH(aq) 88n 2Br (aq) 2H (aq) CO2(g) Molecular bromine is reddish-brown in color. All the other species in the reaction are
colorless. As the reaction progresses, the concentration of Br2 steadily decreases and
its color fades (Figure 13.3). This loss of color and hence concentration can be monitored easily with a spectrometer, which registers the amount of visible light absorbed
by bromine (Figure 13.4).
FIGURE 13.3 The decrease in
bromine concentration as time
elapses shows up as a loss of
color (from left to right). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHEMICAL KINETICS FIGURE 13.4 Plot of absorption of bromine versus wavelength. The maximum absorption
of visible light by bromine occurs
at 393 nm. As the reaction progresses, the absorption, which is
proportional to [Br2], decreases
with time, indicating a depletion
in bromine. Absorption 510 300 400 500 600 Wavelength (nm) Measuring the change (decrease) in bromine concentration at some initial time
and then at some final time allows us to determine the average rate of the reaction during that interval:
t average rate [Br2]final
tinitial Using the data provided in Table 13.1 we can calculate the average rate over the first
50-s time interval as follows:
average rate (0.0101 0.0120) M
50.0 s 3.80 10 5 M/s If we had chosen the first 100 s as our time interval, the average rate would then be
average rate (0.00846 0.0120) M
100.0 s 3.54 10 5 M/s TABLE 13.1 Rates of the Reaction between Molecular
Bromine and Formic Acid at 25°C
TIME (s) Forward RATE (M/s) 0.0
400.0 Back [Br2] (M ) 0.0120
1.04 Main Menu TOC 10
5 Study Guide TOC rate
3 Textbook Website MHHE Website 13.1 FIGURE 13.5 The instantaneous rates of the reaction between molecular bromine and
formic acid at t 100 s, 200 s,
and 300 s are given by the
slopes of the tangents at these
times. THE RATE OF A REACTION 511 0.0120 0.0100 Rate at 100 s:
2.96 × 10 –5 M/s [Br2] (M ) 0.00800 Rate at 200 s:
2.09 × 10 –5 M/s 0.00600 Rate at 300 s:
1.48 × 10 –5 M/s 0.00400 0.00200 0 100 200
t (s) 300 400 These calculations demonstrate that the average rate of the reaction depends on the
time interval we choose.
By calculating the average reaction rate over shorter and shorter intervals, we can
obtain the rate for a specific instant in time, which gives us the instantaneous rate of
the reaction at that time. Figure 13.5 shows the plot of [Br2] versus time, based on the
data shown in Table 13.1. Graphically, the instantaneous rate at 100 s after the start of
the reaction, say, is given by the slope of the tangent to the curve at that instant. The
instantaneous rate at any other time can be determined in a similar manner. Note that
the instantaneous rate determined in this way will always have the same value for the
same concentrations of reactants, as long as the temperature is kept constant. We do
not need to be concerned with what time interval to use. Unless otherwise stated, we
will refer to the instantaneous rate merely as “the rate.”
The following travel analogy helps to distinguish between average rate and instantaneous rate. The distance by car from San Francisco to Los Angeles is 512 miles
along a certain route. If it takes a person 11.4 hours to go from one city to the other,
the average speed is 512 miles/11.4 hours or 44.9 mph. But if the car is traveling at
55.3 mph three hours and twenty-six minutes after departure, then the instantaneous
speed of the car is 55.3 mph. Note that the speed of the car in our example can increase or decrease during the trip, but the instantaneous rate of a reaction must always
decrease with time.
The rate of the bromine-formic acid reaction also depends on the concentration
of formic acid. However, by adding a large excess of formic acid to the reaction mixture we can ensure that the concentration of formic acid remains virtually constant
throughout the course of the reaction. Under this condition the change in the amount
of formic acid present in solution has no effect on the measured rate.
Figure 13.6 is a plot of the rate versus Br2 concentration. The fact that this graph
is a straight line shows that the rate is directly proportional to the concentration; the
higher the concentration, the higher the rate:
k[Br2] Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHEMICAL KINETICS FIGURE 13.6 Plot of rate versus molecular bromine concentration for the reaction between molecular bromine and formic acid.
The straight-line relationship
shows that the rate of reaction is
directly proportional to the molecular bromine concentration. 5.00 × 10 –5
4.00 × 10 –5 Rate (M/s) 512 3.00 × 10 –5
2.00 × 10 –5
1.00 × 10 –5 0 0.00200 0.00600
[Br2] (M ) 0.0100 0.0140 The term k is known as the rate constant, a constant of proportionality between the
reaction rate and the concentrations of reactants. Rearranging the equation gives
[Br2] Since reaction rate has the units M/s, and [Br2] is in M, the unit of k is 1/s, or s 1 in
this case. It is important to understand that k is not affected by the concentration of
Br2. To be sure, the rate is greater at a higher concentration and smaller at a lower concentration of Br2, but the ratio of rate/[Br2] remains the same provided the temperature does not change.
From Table 13.1 we can calculate the rate constant for the reaction. Taking the
data for t 50 s, we write
3.52 10 5 M/s
0.0101 M 3.49 10 3 s 1 We can use the data for any t to calculate k. The slight variations in the values of k
listed in Table 13.1 are due to experimental deviations in rate measurements.
DECOMPOSITION OF HYDROGEN PEROXIDE If one of the products or reactants is a gas, we can use a manometer to find the reaction rate. Consider the decomposition of hydrogen peroxide at 20°C:
2H2O2(aq) 88n 2H2O(l) O2(g) In this case, the rate of decomposition can be determined by monitoring the rate of
oxygen evolution with a manometer (Figure 13.7). The oxygen pressure can be readily converted to concentration by using the ideal gas equation:
FIGURE 13.7 The rate of hydrogen peroxide decomposition
can be measured with a
manometer, which shows the
increase in the oxygen gas
pressure with time. Back Forward PV nRT or Main Menu P TOC n
V Study Guide TOC [O2]RT Textbook Website MHHE Website 13.1 FIGURE 13.8 The instantaneous rate for the decomposition
of hydrogen peroxide at 400 min
is given by the slope of the
tangent. THE RATE OF A REACTION 513 120 100 P (mmHg) 80
Slope = 0.12 mmHg/min
60 40 20 0 200 400 600
t (min) 800 1000 1200 where n/V gives the molarity of oxygen gas. Rearranging the equation, we get
RT The reaction rate, which is given by the rate of oxygen production, can now be written as
t rate 1
t Figure 13.8 shows the increase in oxygen pressure with time and the determination of
an instantaneous rate at 400 min. To express the rate in the normal units of M/s, we
convert mmHg/min to atm/s, then multiply the slope of the tangent ( P/ t) by 1/RT,
as shown in the above equation. REACTION RATES AND STOICHIOMETRY We have seen that for stoichiometrically simple reactions of the type A 88n B, the
rate can be either expressed in terms of the decrease in reactant concentration with
[A]/ t, or the increase in product concentration with time, [B]/ t. For more
complex reactions, we must be careful in writing the rate expressions. Consider, for
example, the reaction
2A 88n B Two moles of A disappears for each mole of B that forms — that is, the rate of disappearance of A is twice as fast as the rate of appearance of B. We write the rate as either Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 514 CHEMICAL KINETICS 1
2 rate [A]
t or [B]
t rate In general, for the reaction
aA bB 88n cC dD the rate is given by
a rate [A]
t The following example requires writing the reaction rate in terms of both reactants and
EXAMPLE 13.1 Write the rate expressions for the following reactions in terms of the disappearance
of the reactants and the appearance of the products:
(a) I (aq) OCl (aq) 88n Cl (aq) OI (aq)
(b) 3O2(g) 88n 2O3(g)
(c) 4NH3(g) 5O2(g) 88n 4NO(g) 6H2O(g)
Answer (a) Since each of the stoichiometric coefficients equals 1,
t rate [OCl ]
t [Cl ]
t [OI ]
t (b) Here the coefficients are 3 and 2, so
t (c) In this reaction
4 rate Similar problems: 13.5, 13.6. 1
t PRACTICE EXERCISE Write the rate expression for the following reaction:
CH4(g) 13.2 2O2(g) 88n CO2(g) 2H2O(g) THE RATE LAW So far we have learned that the rate of a reaction is proportional to the concentration
of reactants and that the proportionality constant k is called the rate constant. The rate
law expresses the relationship of the rate of a reaction to the rate constant and the
concentrations of the reactants raised to some powers. For the general reaction
aA bB 88n cC dD the rate law takes the form
rate k[A]x[B]y (13.1) where x and y are numbers that must be determined experimentally. Note that, in gen- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.2 THE RATE LAW 515 TABLE 13.2 Rate Data for the Reaction
between F2 and ClO2
[F2](M ) [ClO2](M ) INITIAL RATE (M/s) 1. 0.10
3. 0.20 0.010
3 eral, x and y are not equal to the stoichiometric coefficients a and b. When we know
the values of x, y, and k, we can use Equation (13.1) to calculate the rate of the reaction, given the concentrations of A and B.
The exponents x and y specify the relationships between the concentrations of reactants A and B and the reaction rate. Added together, they give us the overall reaction order, defined as the sum of the powers to which all reactant concentrations appearing in the rate law are raised. For Equation (13.1) the overall reaction order is
x y. Alternatively, we can say that the reaction is xth order in A, yth order in B, and
(x y)th order overall.
To see how to determine the rate law of a reaction, let us consider the reaction
between fluorine and chlorine dioxide:
F2(g) 2ClO2(g) 88n 2FClO2(g) One way to study the effect of reactant concentration on reaction rate is to determine
how the initial rate depends on the starting concentrations. It is preferable to measure
the initial rates because as the reaction proceeds, the concentrations of the reactants
decrease and it may become difficult to measure the changes accurately. Also, there
may be a reverse reaction of the type
products 88n reactants which would introduce error into the rate measurement. Both of these complications
are virtually absent during the early stages of the reaction.
Table 13.2 shows three rate measurements for the formation of FClO2. Looking
at entries 1 and 3, we see that as we double [F2] while holding [ClO2] constant, the
reaction rate doubles. Thus the rate is directly proportional to [F2]. Similarly, the data
in entries 1 and 2 show that as we quadruple [ClO2] at constant [F2], the rate increases
by four times, so that the rate is also directly proportional to [ClO2]. We can summarize our observations by writing the rate law as
rate k[F2][ClO2] Because both [F2] and [ClO2] are raised to the first power, the reaction is first order
in F2, first order in ClO2, and (1 1) or second order overall. Note that [ClO2] is raised
to the power of 1 whereas its stoichiometric coefficient in the overall equation is 2.
The equality of reaction order (first) and stoichiometric coefficient (1) for F2 is coincidental in this case.
From the reactant concentrations and the initial rate, we can also calculate the rate
constant. Using the first entry of data in Table 13.2, we can write
1.2 10 3 M/s
(0.10 M)(0.010 M)
1.2/M s Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 516 CHEMICAL KINETICS Reaction order enables us to understand how the reaction depends on reactant concentrations. Suppose, for example, that for the general reaction aA bB 88n cC
dD we have x 1 and y 2. The rate law for the reaction is [see Equation (13.1)]
rate k[A][B]2 This reaction is first order in A, second order in B, and third order overall (1 2
3). Let us assume that initially [A] 1.0 M and [B] 1.0 M. The rate law tells us that
if we double the concentration of A from 1.0 M to 2.0 M at constant [B], we also double the reaction rate:
for [A] 1.0 M rate1 k(1.0 M )(1.0 M )2
k(1.0 M3) for [A] 2.0 M rate2 k(2.0 M )(1.0 M )2
k(2.0 M3) Hence rate2 2(rate1) On the other hand, if we double the concentration of B from 1.0 M to 2.0 M at constant
[A], the rate will increase by a factor of 4 because of the power 2 in the exponent:
for [B] 1.0 M rate1 k(1.0 M )(1.0 M )2
k(1.0 M3) for [B] 2.0 M rate2 k(1.0 M )(2.0 M )2
k(4.0 M3) Hence rate2 If, for a certain reaction, x 0 and y
rate 4(rate1) 1, then the rate law is
k[B] This reaction is zero order in A, first order in B, and first order overall. The exponent
zero tells us that the rate of this reaction is independent of the concentration of A. Note
that reaction order can also be a fraction.
The following points summarize our discussion of the rate law:
Rate laws are always determined experimentally. From the concentrations of reactants and the initial reaction rates we can determine the reaction order and then the
rate constant of the reaction.
• Reaction order is always defined in terms of reactant (not product) concentrations.
• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the overall balanced equation.
• The following example illustrates the procedure for determining the rate law of a
reaction. EXAMPLE 13.2 The reaction of nitric oxide with hydrogen at 1280°C is
2NO(g) Back Forward Main Menu TOC 2H2(g) 88n N2(g) Study Guide TOC 2H2O(g) Textbook Website MHHE Website 13.3 RELATION BETWEEN REACTANT CONCENTRATION AND TIME 517 From the following data collected at this temperature, determine the rate law and
calculate the rate constant.
EXPERIMENT [NO] 1
Answer [H2] 5.0
10.0 3 10
10 INITIAL RATE (M/s) 2.0
10 3 1.3
5 We assume that the rate law takes the form
k[NO]x[H2]y rate Experiments 1 and 2 show that when we double the concentration of NO at constant concentration of H2, the rate quadruples. Thus the reaction is second order in
NO. Experiments 2 and 3 indicate that doubling [H2] at constant [NO] doubles the
rate; the reaction is first order in H2. The rate law is given by
k[NO]2[H2] rate which shows that it is a (1 2) or third-order reaction overall.
The rate constant k can be calculated using the data from any one of the experiments. Since
[NO]2[H2] k data from experiment 2 gives us
2.5 10 5 M/s 102/M2 s Note that the reaction is first order in H2 whereas the stoichiometric coefficient for H2 in the balanced equation is 2.
Comment Similar problem: 13.17. PRACTICE EXERCISE The reaction of peroxydisulfate ion (S2O2 ) with iodide ion (I ) is
8 3I (aq) 88n 2SO2 (aq)
4 I3 (aq) From the following data collected at a certain temperature, determine the rate law
and calculate the rate constant.
3 13.3 [S2O2 ]
8 [I ] INITIAL RATE (M/s) 0.080
4 RELATION BETWEEN REACTANT CONCENTRATION AND TIME Rate law expressions enable us to calculate the rate of a reaction from the rate constant and reactant concentrations. The rate laws can also be used to determine the con- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 518 CHEMICAL KINETICS centrations of reactants at any time during the course of a reaction. We will illustrate
this application by considering two of the simplest kinds of rate laws — those applying to reactions that are first order overall and those applying to reactions that are second order overall. FIRST-ORDER REACTIONS A first-order reaction is a reaction whose rate depends on the reactant concentration
raised to the first power. In a first-order reaction of the type
A 88n product the rate is
For a first-order reaction,
doubling the concentration of
the reactant doubles the rate. From the rate law we also know that
rate k[A] To obtain the units of k for this rate law, we write
M 1 1/s or s Combining the first two equations for the rate we get
t k[A] (13.2) Using calculus, we can show from Equation (13.2) that
See Appendix 4 for a discussion
of natural logarithms and of
logarithms and significant figures. [A]0
[A] kt (13.3) where ln is the natural logarithm, and [A]0 and [A] are the concentrations of A at times
t 0 and t t, respectively. It should be understood that t 0 need not correspond
to the beginning of the experiment; it can be any time when we choose to start monitoring the change in the concentration of A.
Equation (13.3) can be rearranged as follows:
ln [A]0 ln [A] kt or
ln [A] kt ln [A]0 Equation (13.4) has the form of the linear equation y
of the line that is the graph of the equation: mx b, in which m is the slope ↔ ln [A]0 ↔ y ( k)(t)
↔ ↔ ln [A] (13.4) mx b Thus a plot of ln [A] versus t (or y versus x) gives a straight line with a slope of k(or
m). This graphical analysis allows us to calculate the rate constant k. Figure 13.9 shows
the characteristics of a first-order reaction. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website [A] FIGURE 13.9 First-order reaction characteristics: (a) Decrease
of reactant concentration with
time; (b) plot of the straight-line
relationship to obtain the rate
constant. The slope of the line is
equal to k. RELATION BETWEEN REACTANT CONCENTRATION AND TIME 519 ln [A] 13.3 t t (a) (b) There are many first-order reactions. An example is the decomposition of ethane
(C2H6) into highly reactive fragments called methyl radicals (CH3):
C2H6 88n 2CH3 The decomposition of N2O5 is also a first-order reaction
2N2O5(g) 88n 4NO2(g) O2(g) In the following example we apply Equation (13.3) to an organic reaction.
EXAMPLE 13.3 The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 10 4 s 1 at 500°C.
CH2 O CH2
cyclopropane CH3 O CH P CH2
propene (a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? (b) How long will it take for the concentration of cyclopropane
to decrease from 0.25 M to 0.15 M? (c) How long will it take to convert 74 percent
of the starting material?
Answer (a) Applying Equation (13.3),
ln Because k is given in units of
s 1, we must convert 8.8 min
to seconds. ln [A]0
[A] kt 0.25 M
[A] (6.7 10 4 s 1) 8.8 min 60 s
1 min Solving the equation, we obtain
[A] 0.354 0.25 M
[A] ln e0.354 [A] 1.42 0.18 M (b) Again using Equation (13.3), we have Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 520 CHEMICAL KINETICS ln 0.25 M
0.15 M 4 (6.7
7.6 t 10 s 1)t 102 s 13 min (c) In a calculation of this type, we do not need to know the actual concentration
of the starting material. If 74 percent of the starting material has reacted, then the
amount left after time t is (100% 74%), or 26%. Thus [A]/[A]0 26%/100%, or
0.26. From Equation (13.3), we write
10 2.0 1.0
0.26 103 s 4 s 1 ln 33 min Similar problems: 13.27, 13.28.
PRACTICE EXERCISE The reaction 2A 88n B is first order in A with a rate constant of 2.8 10 2 s 1
at 80°C. How long (in seconds) will it take for A to decrease from 0.88 M to 0.14 M?
Now let us determine graphically the order and rate constant of the decomposition of nitrogen pentoxide in carbon tetrachloride (CCl4) solvent at 45°C:
2N2O5(CCl4) 88n 4NO2(g) O2(g) The following table shows the variation of N2O5 concentration with time, and the corresponding ln [N2O5] values
t (s) N2O5 decomposes to give NO2
(brown color). [N2O5] 0
0.16 ln [N2O5] 0.094
1.83 Applying Equation (13.4) we plot ln [N2O5] versus t, as shown in Figure 13.10. The
fact that the points lie on a straight line shows that the rate law is first order. Next we
determine the rate constant from the slope. We select two points far apart on the line
and subtract their y and x values as follows:
slope (m) y
5.7 ( 0.34)
10 4 s 1 Since m
k, we get k 5.7 10 4 s 1.
For gas-phase reactions we can replace the concentration terms in Equation (13.3)
with the pressures of the gaseous reactant. Consider the first-order reaction
A(g) 88n product Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.3 FIGURE 13.10 Plot of
ln [N2O5] versus time. The rate
constant can be determined from
the slope of the straight line. RELATION BETWEEN REACTANT CONCENTRATION AND TIME 521 0 ln [N2O5 ] –0.50 –1.00 –1.50 –2.00 0 500 1000 1500 2000 2500 3000 3500 t (s) Using the ideal gas equation we write
Substituting [A] nART
RT P/RT in Equation (13.3) we get
[A] ln P0 /RT
P/RT ln P0
P kt The equation corresponding to Equation (13.4) now becomes
ln P kt ln P0 The following example shows the use of pressure measurements to study the kinetics of a first-order reaction.
EXAMPLE 13.4 The rate of decomposition of azomethane is studied by monitoring the partial pressure of the reactant as a function of time:
CH3ONPNOCH3(g) 88n N2(g) C2H6(g) The data obtained at 300°C are shown in the following table:
TIME (s) 0
250 Back Forward Main Menu TOC PARTIAL PRESSURE OF
AZOMETHANE (mmHg) 284
150 Study Guide TOC Textbook Website MHHE Website 522 CHEMICAL KINETICS Are these values consistent with first-order kinetics? If so, determine the rate constant.
The partial pressure of azomethane at any time is directly proportional to
its concentration (in mol/L). Therefore we substitute partial pressures for concentration in Equation (13.4): Answer ln P kt ln P0 where P0 and P are the partial pressures of azomethane at time t 0 and t t.
Figure 13.11, which is based on the data given below, shows that a plot of ln P versus t yields a straight line, so the reaction is indeed first order.
t (s) ln P 0
4.883 The slope of the line is given by
Similar problems: 13.21, 13.22. 5.05
33) s 2.55 3 10 According to Equation (13.4), the slope is equal to s 1 k, so k 2.55 10 3 s 1. PRACTICE EXERCISE Azomethane (C2H6N2) decomposes at a certain temperature in the gas phase as follows:
C2H6N2(g) 88n C2H6(g) N2(g) From the following data determine the order of the reaction and the rate constant.
TIME (min) [C2H6N2] (M) 0
0.13 As a reaction proceeds, the concentration of the reactant(s) decreases. Another
measure of the rate of a reaction, relating concentration to time, is the half-life, t ,
which is the time required for the concentration of a reactant to decrease to half of its
initial concentration. We can obtain an expression for t for a first-order reaction as
follows. From Equation (13.3) we write
2 t By the definition of half-life, when t Back Forward Main Menu TOC 1
[A] t , [A]
2 Study Guide TOC [A]0 /2, so Textbook Website MHHE Website 13.3 FIGURE 13.11 Plot of ln P versus time for the decomposition of
azomethane. RELATION BETWEEN REACTANT CONCENTRATION AND TIME 523 5.80
5.60 1n P 5.40
4.80 0 100 200 300 t (s) t 1
[A]0 /2 1
k (13.5) Equation (13.5) tells us that the half-life of a first-order reaction is independent of the
initial concentration of the reactant. Thus, it takes the same time for the concentration
of the reactant to decrease from 1.0 M to 0.50 M, say, as it does for a decrease in concentration from 0.10 M to 0.050 M (Figure 13.12). Measuring the half-life of a reaction is one way to determine the rate constant of a first-order reaction.
The following analogy may be helpful for understanding Equation (13.5). The duration of a college undergraduate’s career, assuming the student does not take any time
off, is four years. Thus the half-life of his or her stay at the college is two years. This
half-life is not affected by how many other students are present. Similarly, the half-life
of a first-order reaction is concentration independent.
The usefulness of t is that it gives us an estimate of the magnitude of the rate
constant — the shorter the half-life, the larger the k. Consider, for example, two ra1
2 Concentration FIGURE 13.12 Change in
concentration of a reactant with
number of half-lives for a firstorder reaction. 0 Back Forward Main Menu 1
Number of half–lives elapsed TOC Study Guide TOC Textbook Website MHHE Website 524 CHEMICAL KINETICS dioactive isotopes used in nuclear medicine: 24Na (t
14.7 hr) and 60Co (t
It is obvious that the Na isotope decays faster because it has a shorter half-life. If we
started with 1 g each of the isotopes, most of the 24Na would be gone in a week while
the 60Co sample would be mostly intact.
In Example 13.5 we calculate the half-life of a first-order reaction.
2 EXAMPLE 13.5 The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with
a rate constant of 5.36 10 4 s 1 at 700°C:
C2H6(g) 88n 2CH3(g) Calculate the half-life of the reaction in minutes.
Answer Applying Equation (13.5)
10 4 s 5.36 1 3 1.29 10 s 21.5 min Similar problem: 13.27.
PRACTICE EXERCISE Calculate the half-life of the decomposition of N2O5, discussed on p. 520. SECOND-ORDER REACTIONS A second-order reaction is a reaction whose rate depends on the concentration of one
reactant raised to the second power or on the concentrations of two different reactants,
each raised to the first power. The simpler type involves only one kind of reactant
A 88n product where
t rate From the rate law
k[A]2 rate As before, we can determine the units of k by writing
M2 1/M s Another type of second-order reaction is
A B 88n product and the rate law is given by
rate Back Forward Main Menu TOC k[A][B] Study Guide TOC Textbook Website MHHE Website 13.3 RELATION BETWEEN REACTANT CONCENTRATION AND TIME 525 The reaction is first order in A and first order in B, so it has an overall reaction order
Using calculus, we can obtain the following expressions for “A 88n product” second-order reactions:
[A]0 kt (13.6) (The corresponding equation for “A B 88n product” reactions is too complex for
We can obtain an equation for the half-life of a second-order reaction by setting
[A] [A]0 /2 in Equation (13.6).
[A]0 /2 1
[A]0 kt Solving for t we obtain
2 t 1
2 (13.7) Note that the half-life of a second-order reaction is inversely proportional to the initial
reactant concentration. This result makes sense because the half life should be shorter
in the early stage of the reaction when more reactant molecules are present to collide
with each other. Measuring the half-lives at different initial concentrations is one way
to distinguish between a first-order and a second-order reaction.
The kinetic analysis of a second-order reaction is shown in the following example.
EXAMPLE 13.6 Iodine atoms combine to form molecular iodine in the gas phase
I(g) I(g) 88n I2(g) This reaction follows second-order kinetics and has the high rate constant 7.0
109/M s at 23°C. (a) If the initial concentration of I was 0.086 M, calculate the
concentration after 2.0 min. (b) Calculate the half-life of the reaction if the initial
concentration of I is 0.60 M and if it is 0.42 M.
Answer (a) We begin with Equation (13.6):
0.086 M kt
(7.0 where [A] is the concentration at t
[A] 109/M s) 2.0 min 60 s
1 min 2.0 min. Solving the equation, we get
1.2 10 12 M This is such a low concentration that it is virtually undetectable. The very large rate
constant for the reaction means that practically all the I atoms combine after only
2.0 min of reaction time.
(b) We need Equation (13.7) for this part.
For [I]0 Back Forward Main Menu TOC 0.60 M Study Guide TOC Textbook Website MHHE Website 526 CHEMICAL KINETICS t 1
(7.0 109/M s)(0.60 M )
2.4 10 10 s For [I]0 0.42 M
109/M s)(0.42 M)
10 10 s These results confirm that the half-life of a second-order reaction is not
a constant but depends on the initial concentration of the reactant(s). Comment Similar problems: 13.29, 13.30. PRACTICE EXERCISE The reaction 2A 88n B is second order with a rate constant of 51/M min at 24°C.
(a) Starting with [A] 0.0092 M, how long will it take for [A] 3.7 10 3 M?
(b) Calculate the half-life of the reaction.
First- and second-order reactions are the most common reaction types. Reactions
whose order is zero are rare. For a zero-order reaction
A 88n product the rate law is given by
k[A]0 rate k Thus the rate of a zero-order reaction is a constant, independent of reactant concentration. Third-order and higher order reactions are quite complex; they are not presented in this book. Table 13.3 summarizes the kinetics for first-order and second-order reactions of the type A 88n product.
The Chemistry in Action essay on p. 527 describes the application of chemical
kinetics to estimating the ages of objects. 13.4 ACTIVATION ENERGY AND TEMPERATURE DEPENDENCE OF RATE
CONSTANTS With very few exceptions, reaction rates increase with increasing temperature. For example, the time required to hard-boil an egg in water is much shorter if the “reaction”
is carried out at 100°C (about 10 min) than at 80°C (about 30 min). Conversely, an efTABLE 13.3
ORDER Summary of the Kinetics of First-order and Second-order
RATE LAW CONCENTRATION-TIME EQUATION [A]0
[A] 1 Forward Main Menu k[A] ln 2 Back Rate
Rate k[A]2 1
[A] TOC Study Guide TOC HALF-LIFE kt
[A]0 kt Textbook Website 0.693
k[A]0 MHHE Website 13.4 ACTIVATION ENERGY AND TEMPERATURE DEPENDENCE OF RATE CONSTANTS 527 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Determining the Age of the Shroud of Turin
How do scientists determine the ages of artifacts from
archaeological excavations? If someone tried to sell
you a manuscript supposedly dating from 1000 B.C.,
how could you be certain of its authenticity? Is a
mummy found in an Egyptian pyramid really three
thousand years old? Is the so-called Shroud of Turin
truly the burial cloth of Jesus Christ? The answers to
these and other similar questions can usually be found
by applying chemical kinetics and the radiocarbon
Earth’s atmosphere is constantly being bombarded by cosmic rays of extremely high penetrating
power. These rays, which originate in outer space,
consist of electrons, neutrons, and atomic nuclei. One
of the important reactions between the atmosphere
and cosmic rays is the capture of neutrons by atmospheric nitrogen (nitrogen-14 isotope) to produce the
radioactive carbon-14 isotope and hydrogen. The unstable carbon atoms eventually form 14CO2, which
mixes with the ordinary carbon dioxide (12CO2) in
the air. As the carbon-14 isotope decays, it emits
particles (electrons). The rate of decay (as measured
by the number of electrons emitted per second) obeys
first-order kinetics. It is customary in the study of radioactive decay to write the rate law as
rate kN where k is the first-order rate constant and N the number of 14C nuclei present. The half-life of the decay,
t , is 5.73 103 yr, so that from Equation (13.7) we
2 k 0.693
5.73 103 yr 1.21 10 4 yr 1 The carbon-14 isotopes enter the biosphere when
carbon dioxide is taken up in plant photosynthesis.
Plants are eaten by animals, which exhale carbon-14
in CO2. Eventually, carbon-14 participates in many aspects of the carbon cycle. The 14C lost by radioactive
decay is constantly replenished by the production of
new isotopes in the atmosphere. In this decay-replenishment process, a dynamic equilibrium is established
whereby the ratio of 14C to 12C remains constant in
living matter. But when an individual plant or an ani- Forward Main Menu TOC The Shroud of Turin. For generations there was controversy
about whether the shroud, a piece of linen bearing the image of a man, was the burial cloth of Jesus Christ. The age
of the shroud has been determined by radiocarbon dating. mal dies, the carbon-14 isotope in it is no longer replenished, so the ratio decreases as 14C decays. This
same change occurs when carbon atoms are trapped
in coal, petroleum, or wood preserved underground,
and, of course, in Egyptian mummies. After a number
of years, there are proportionately fewer 14C nuclei
in, say, a mummy than in a living person.
In 1955, Willard F. Libby† suggested that this fact
could be used to estimate the length of time the
carbon-14 isotope in a particular specimen has been
decaying without replenishment. Using Equation
(13.3), we can write
N kt † Willard Frank Libby (1908–1980). American chemist. Libby received the
Nobel Prize in Chemistry in 1960 for his work on radiocarbon dating. Study Guide TOC Textbook Website MHHE Website 528 CHEMICAL KINETICS t 1
ln decay rate at t
decay rate at t 0
t decay rate of fresh sample
decay rate of old sample Knowing k and the decay rates for the fresh sample
and the old sample, we can calculate t, which is the
age of the old sample. This ingenious technique is
based on a remarkably simple idea. Its success de- pends on how accurately we can measure the rate of
decay. In fresh samples, the ratio 14C/12C is about
1/1012, so the equipment used to monitor the radioactive decay must be very sensitive. Precision is
more difficult with older samples because they contain
even fewer 14C nuclei. Nevertheless, radiocarbon dating has become an extremely valuable tool for estimating the age of archaeological artifacts, paintings,
and other objects dating back 1000 to 50,000 years.
A recent major application of radiocarbon dating was the determination of the age of the Shroud
of Turin. In 1988 three laboratories in Europe and the
United States, working on samples of less than 50 mg
of the shroud, independently showed by carbon-14
dating that the shroud dates from between A.D. 1260
and A.D. 1390. Thus the shroud could not have been
the burial cloth of Christ. fective way to preserve foods is to store them at subzero temperatures, thereby slowing the rate of bacterial decay. Figure 13.13 shows a typical example of the relationship between the rate constant of a reaction and temperature. In order to explain this
behavior, we must ask how reactions get started in the first place. Rate constant Chemistry in Action Chemistry in Action Chemistry in where N0 and N are the number of 14C nuclei present at t 0 and t t, respectively. Since the rate of
decay is directly proportional to the number of 14C
nuclei present, the above equation can be rewritten
as THE COLLISION THEORY OF CHEMICAL KINETICS Temperature
FIGURE 13.13 Dependence of
rate constant on temperature. The
rate constants of most reactions
increase with increasing temperature. The kinetic molecular theory of gases (p. 178) postulates that gas molecules frequently
collide with one another. Therefore, it seems logical to assume — and it is generally
true — that chemical reactions occur as a result of collisions between reacting molecules. In terms of the collision theory of chemical kinetics, then, we expect the rate of
a reaction to be directly proportional to the number of molecular collisions per second,
or to the frequency of molecular collisions:
rate number of collisions
s This simple relationship explains the dependence of reaction rate on concentration.
Consider the reaction of A molecules with B molecules to form some product.
Suppose that each product molecule is formed by the direct combination of an A molecule and a B molecule. If we doubled the concentration of A, then the number of
A-B collisions would also double, because there would be twice as many A molecules
that could collide with B molecules in any given volume (Figure 13.14). Consequently,
the rate would increase by a factor of 2. Similarly, doubling the concentration of B
molecules would increase the rate twofold. Thus, we can express the rate law as
rate k[A][B] The reaction is first order in both A and B and obeys second-order kinetics.
The collision theory is intuitively appealing, but the relationship between rate and
molecular collisions is more complicated than you might expect. The implication of Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.4 (c)
FIGURE 13.14 Dependence of
number of collisions on concentration. We consider here only
A-B collisions, which can lead to
formation of products. (a) There
are four possible collisions
among two A and two B molecules. (b) Doubling the number of
either type of molecule (but not
both) increases the number of
collisions to eight. (c) Doubling
both the A and B molecules increases the number of collisions
to sixteen. FIGURE 13.15 Potential energy profiles for (a) exothermic
and (b) endothermic reactions.
These plots show the change in
potential energy as reactants A
and B are converted to products
C and D. The activated complex
is a highly unstable species with
a high potential energy. The activation energy is defined for the
forward reaction in both (a) and
(b). Note that the products C and
D are more stable than the reactants in (a) and less stable than
the reactants in (b). Back Forward Main Menu A B 88n C D If the products are more stable than the reactants, then the reaction will be accompanied by a release of heat; that is, the reaction is exothermic [Figure 13.15(a)]. On the
other hand, if the products are less stable than the reactants, then heat will be absorbed
by the reacting mixture from the surroundings and we have an endothermic reaction
[Figure 13.15(b)]. In both cases we plot the potential energy of the reacting system
versus the progress of the reaction. Qualitatively, these plots show the potential energy
changes as reactants are converted to products. Activated
Potential energy (b) the collision theory is that a reaction always occurs when an A and a B molecule collide. However, not all collisions lead to reactions. Calculations based on the kinetic
molecular theory show that, at ordinary pressures (say, 1 atm) and temperatures (say,
298 K), there are about 1 1027 binary collisions (collisions between two molecules)
in 1 mL of volume every second, in the gas phase. Even more collisions per second
occur in liquids. If every binary collision led to a product, then most reactions would
be complete almost instantaneously. In practice, we find that the rates of reactions differ greatly. This means that, in many cases, collisions alone do not guarantee that a reaction will take place.
Any molecule in motion possesses kinetic energy; the faster it moves, the greater
the kinetic energy. But a fast-moving molecule will not break up into fragments on its
own. To react, it must collide with another molecule. When molecules collide, part of
their kinetic energy is converted to vibrational energy. If the initial kinetic energies are
large, then the colliding molecules will vibrate so strongly as to break some of the
chemical bonds. This bond fracture is the first step toward product formation. If the
initial kinetic energies are small, the molecules will merely bounce off each other intact. Energetically speaking, there is some minimum collision energy below which no
reaction occurs. Lacking this energy, the molecules remain intact, and no change results from the collision.
We postulate that in order to react, the colliding molecules must have a total kinetic energy equal to or greater than the activation energy (Ea), which is the minimum
amount of energy required to initiate a chemical reaction. When molecules collide they
form an activated complex, a temporary species formed by the reactant molecules as
a result of the collision before they form the product.
Figure 13.15 shows two different potential energy profiles for the reaction Potential energy (a) 529 ACTIVATION ENERGY AND TEMPERATURE DEPENDENCE OF RATE CONSTANTS Ea A+B C+D
(a) TOC Study Guide TOC Ea C+D A+B
(b) Textbook Website MHHE Website 530 CHEMICAL KINETICS We can think of activation energy as a barrier that prevents less energetic molecules from reacting. Because the number of reactant molecules in an ordinary reaction
is very large, the speeds, and hence also the kinetic energies of the molecules, vary
greatly. Normally, only a small fraction of the colliding molecules — the fastest-moving ones — have enough kinetic energy to exceed the activation energy. These molecules can therefore take part in the reaction. The increase in the rate (or the rate constant) with temperature can now be explained: The speeds of the molecules obey the
Maxwell distributions shown in Figure 5.15. Compare the speed distributions at two
different temperatures. Since more high-energy molecules are present at the higher temperature, the rate of product formation is also greater at the higher temperature. THE ARRHENIUS EQUATION The dependence of the rate constant of a reaction on temperature can be expressed by
the following equation, known as the Arrhenius equation:
k Ae Ea/RT (13.8) where Ea is the activation energy of the reaction (in kJ/mol), R the gas constant
(8.314 J/K mol), T the absolute temperature, and e the base of the natural logarithm
scale (see Appendix 4). The quantity A represents the collision frequency and is called
the frequency factor. It can be treated as a constant for a given reacting system over a
fairly wide temperature range. Equation (13.8) shows that the rate constant is directly
proportional to A and, therefore, to the collision frequency. In addition, because of the
minus sign associated with the exponent Ea/RT, the rate constant decreases with increasing activation energy and increases with increasing temperature. This equation
can be expressed in a more useful form by taking the natural logarithm of both sides:
ln k ln Ae
ln A Ea/RT Ea
RT (13.9) Equation (13.9) can be rearranged to a linear equation: ↔ ln A ↔ 1
T ↔ Ea
R ↔ ln k y m x b Thus, a plot of ln k versus 1/T gives a straight line whose slope m is equal to Ea/R
and whose intercept b with the ordinate (the y-axis) is ln A.
The following example demonstrates a graphical method for determining the activation energy of a reaction.
EXAMPLE 13.7 The rate constants for the decomposition of acetaldehyde
CH3CHO(g) 88n CH4(g) CO(g) were measured at five different temperatures. The data are shown in the table. Plot
ln k versus 1/T, and determine the activation energy (in kJ/mol) for the reaction.
Note that the reaction is “ 3 ” order in CH3CHO, so k has the units of 1/M s.
2 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.4 ACTIVATION ENERGY AND TEMPERATURE DEPENDENCE OF RATE CONSTANTS FIGURE 13.16
sus 1/ T. Plot of ln k ver- 531 0.00 1n k –1.00
1.20 × 10 –3 k (1/M 1
2 s) 1.30 × 10 –3
1/ T (K–1) 1.40 × 10 –3 T (K) 0.011
810 Answer We want to plot ln k on the y-axis versus 1/T on the x-axis. From the given
data we obtain
1/T (K 1) ln k 4.51
3 A plot of these data yields the graph in Figure 13.16. The slope of the line is calculated from two pairs of coordinates:
4.00 slope ( 0.45) 2.09 104 K From the linear form of Equation (13.9)
R 2.09 (8.314 J/K mol)(2.09
1.74 104 K) 105 J/mol 1.74 104 K 102 kJ/mol It is important to note that although the rate constant itself has the units
1/M •s, the quantity ln k has no units (we cannot take the logarithm of a unit). Comment Similar problem: 13.37. 1
2 PRACTICE EXERCISE The second-order rate constant for the decomposition of nitrous oxide (N2O) into
nitrogen molecule and oxygen atom has been measured at different temperatures: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 532 CHEMICAL KINETICS k (1/M s) t (°C) 1.87 10
0.2440 3 600
750 Determine graphically the activation energy for the reaction.
An equation relating the rate constants k1 and k2 at temperatures T1 and T2 can
be used to calculate the activation energy or to find the rate constant at another temperature if the activation energy is known. To derive such an equation we start with
ln k1 ln A Ea
RT1 ln k2 ln A Ea
RT2 Subtracting ln k2 from ln k1 gives
ln k2 ln k1 Ea 1
R T2 1
k2 Ea 1
R T2 1
T1 ln Ea T1 T2
R (13.10) The following example illustrates the use of the equation we have just derived.
EXAMPLE 13.8 The rate constant of a first-order reaction is 3.46 10 2 s 1 at 298 K. What is the
rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol?
Answer The data are
k1 3.46 10 T1 2 s 1 298 K k2 ? T2 350 K Substituting in Equation (13.10),
ln 3.46 10 2 k2 50.2 103 J/mol 298 K 350 K
8.314 J/K mol (298 K)(350 K) Solving the equation gives
ln 10 2 10 3.46 2 3.01 k2
3.46 e k2
Similar problem: 13.40. Back Forward 3.01 0.702 s 0.0493
1 The rate constant is expected to be greater at a higher temperature.
Therefore, the answer is reasonable. Comment Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.5 FIGURE 13.17 Relative orientation of reacting molecules. Only
when the K atom collides directly
with the I atom will the reaction
most likely occur. REACTION MECHANISMS +
K 533 + + CH3I KI + CH3 PRACTICE EXERCISE The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 10 10 s 1
at 25°C. Calculate the rate constant at 40°C if the activation energy is 116 kJ/mol.
For simple reactions (for example, reactions between atoms), we can equate the
frequency factor (A) in the Arrhenius equation with the frequency of collisions between
the reacting species. For more complex reactions, we must also consider the “orientation factor,” that is, how reacting molecules are oriented relative to each other. The reaction between potassium atoms (K) and methyl iodide (CH3I) to form potassium iodide (KI) and a methyl radical (CH3) illustrates this point:
K CH3I 88n KI CH3 This reaction is most favorable when the K atom collides with the I atom in CH3I headon (Figure 13.17). Otherwise, few or no products are formed. The nature of the orientation factor is satisfactorily dealt with in a more advanced treatment of chemical kinetics. 13.5 REACTION MECHANISMS As we mentioned earlier, an overall balanced chemical equation does not tell us much
about how a reaction actually takes place. In many cases, it merely represents the sum
of several elementary steps, or elementary reactions, a series of simple reactions that
represent the progress of the overall reaction at the molecular level. The term for the
sequence of elementary steps that leads to product formation is reaction mechanism.
The reaction mechanism is comparable to the route of travel followed during a trip;
the overall chemical equation specifies only the origin and destination.
As an example of a reaction mechanism, let us consider the reaction between nitric oxide and oxygen:
2NO(g) O2(g) 88n 2NO2(g) We know that the products are not formed directly from the collision of two NO molecules with an O2 molecule because N2O2 is detected during the course of the reaction. Let us assume that the reaction actually takes place via two elementary steps as
2NO(g) 88n N2O2(g)
N2O2(g) O2(g) 88n 2NO2(g) In the first elementary step, two NO molecules collide to form a N2O2 molecule. This
event is followed by the reaction between N2O2 and O2 to give two molecules of NO2.
The net chemical equation, which represents the overall change, is given by the sum
of the elementary steps: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 534 CHEMICAL KINETICS The sum of the elementary
steps must give the overall
balanced equation. Elementary step:
Overall reaction: NO
2NO NO 88n N2O2
O2 88n 2NO2 N2O2 O2 88n N2O2 2NO2 Species such as N2O2 are called intermediates because they appear in the mechanism
of the reaction (that is, the elementary steps) but not in the overall balanced equation.
Keep in mind that an intermediate is always formed in an early elementary step and
consumed in a later elementary step.
The molecularity of a reaction is the number of molecules reacting in an elementary step. These molecules may be of the same or different types. Each of the elementary steps discussed above is called a bimolecular reaction, an elementary step
that involves two molecules. An example of a unimolecular reaction, an elementary
step in which only one reacting molecule participates, is the conversion of cyclopropane
to propene discussed in Example 13.3. Very few termolecular reactions, reactions that
involve the participation of three molecules in one elementary step, are known, because the simultaneous encounter of three molecules is a far less likely event than a
bimolecular collision. RATE LAWS AND ELEMENTARY STEPS Knowing the elementary steps of a reaction enables us to deduce the rate law. Suppose
we have the following elementary reaction:
A 88n products Because there is only one molecule present, this is a unimolecular reaction. It follows
that the larger the number of A molecules present, the faster the rate of product formation. Thus the rate of a unimolecular reaction is directly proportional to the concentration of A, or is first order in A:
rate k[A] For a bimolecular elementary reaction involving A and B molecules
A B 88n product the rate of product formation depends on how frequently A and B collide, which in
turn depends on the concentrations of A and B. Thus we can express the rate as
rate k[A][B] Similarly, for a bimolecular elementary reaction of the type
A or A 88n products
2A 88n products the rate becomes
rate k[A]2 The above examples show that the reaction order for each reactant in an elementary reaction is equal to its stoichiometric coefficient in the chemical equation for that step. In
general we cannot tell by merely looking at a balanced equation whether the reaction
occurs as shown or in a series of steps. This determination is made in the laboratory. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.5 FIGURE 13.18 Sequence of
steps in the study of a reaction
the rate of
a reaction REACTION MECHANISMS 535 Postulating
the rate law When we study a reaction that has more than one elementary step, the rate law
for the overall process is given by the rate-determining step, which is the slowest step
in the sequence of steps leading to product formation.
An analogy for the rate-determining step is the flow of traffic along a narrow road.
Assuming the cars cannot pass one another on the road, the rate at which the cars travel
is governed by the slowest-moving car.
Experimental studies of reaction mechanisms begin with the collection of data
(rate measurements). Next, we analyze the data to determine the rate constant and order of the reaction, and we write the rate law. Finally, we suggest a plausible mechanism for the reaction in terms of elementary steps (Figure 13.18). The elementary steps
must satisfy two requirements:
The sum of the elementary steps must give the overall balanced equation for the
• The rate-determining step should predict the same rate law as is determined experimentally.
• Remember that for a proposed reaction scheme, we must be able to detect the presence of any intermediate(s) formed in one or more elementary steps.
The decomposition of hydrogen peroxide and the formation of hydrogen iodide
from molecular hydrogen and molecular iodine illustrate the elucidation of reaction
mechanisms by experimental studies.
Hydrogen Peroxide Decomposition The decomposition of hydrogen peroxide is facilitated by iodide ions (Figure 13.19).
The overall reaction is
2H2O2(aq) 88n 2H2O(l ) O2(g) By experiment, the rate law is found to be
rate FIGURE 13.19 The decomposition of hydrogen peroxide is
catalyzed by the iodide ion. A
few drops of liquid soap have
been added to the solution to
dramatize the evolution of oxygen gas. (Some of the iodide
ions are oxidized to molecular
iodine, which then reacts with
iodide ions to form the brown
triiodide ion, I3 .) Back Forward Main Menu k[H2O2][I ] Thus the reaction is first order with respect to both H2O2 and I .
You can see that H2O2 decomposition does not occur in a single elementary step
corresponding to the overall balanced equation. If it did, the reaction would be second
order in H2O2 (as a result of the collision of two H2O2 molecules). What’s more, the
I ion, which is not even part of the overall equation, appears in the rate law expression. How can we reconcile these facts? First, we can account for the observed rate
law by assuming that the reaction takes place in two separate elementary steps, each
of which is bimolecular:
Step 2: TOC H2O2
I 88n H2O IO k2
IO 88n H2O O2 Study Guide TOC I Textbook Website MHHE Website 536 CHEMICAL KINETICS If we further assume that step 1 is the rate-determining step, then the rate of the reaction can be determined from the first step alone:
rate k1[H2O2][I ] Note that the IO ion is an intermediate because it does not appear in the overall balanced equation. Although the I ion also does not appear in the overall equation, I
differs from IO in that the former is present at the start of the reaction and at its completion. The function of I is to speed up the reaction — that is, it is a catalyst. We will
discuss catalysis in the next section.
The Hydrogen Iodide Reaction A common reaction mechanism is one that involves at least two elementary steps, the
first of which is very rapid in both the forward and reverse directions compared with
the second step. An example is the reaction between molecular hydrogen and molecular iodine to produce hydrogen iodide:
I2(g) 88n 2HI(g) H2(g) Experimentally, the rate law is found to be
rate k[H2][I2] For many years it was thought that the reaction occurred just as written; it was thought
to be a bimolecular reaction involving a hydrogen molecule and an iodine molecule,
as shown above. However, in the 1960s chemists found that the actual mechanism is
more complicated. A two-step mechanism was proposed:
I2 3:4 2I
k1 Step 1:
Step 2: Chemical equilibrium will be
discussed in the next chapter. H2 k2
2I 88n 2HI where k1, k 1, and k2 are the rate constants for the reactions. The I atoms are the intermediate in this reaction.
When the reaction begins, there are very few I atoms present. But as I2 dissociates, the concentration of I2 decreases while that of I increases. Therefore, the forward
rate of step 1 decreases and the reverse rate increases. Soon the two rates become equal,
and a chemical equilibrium is established. Because the elementary reactions in step 1
are much faster than the one in step 2, equilibrium is reached before any significant
reaction with hydrogen occurs, and it persists throughout the reaction.
In the equilibrium condition of step 1 the forward rate is equal to the reverse rate;
k 1[I]2 k1[I2]
[I]2 or k1
k1 2 The rate of the reaction is given by the slow, rate-determining step, which is step 2:
rate k2[H2][I]2 Substituting the expression for [I]2 into this rate law, we obtain Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.5 REACTION MECHANISMS 537 k1k2
[H ][I ]
k1 2 2 rate k[H2][I2] where k k1k2/k 1. As you can see, this two-step mechanism also gives the correct
rate law for the reaction. This agreement along with the observation of intermediate I
atoms provides strong evidence that the mechanism is correct.
Finally we note that not all reactions have a single rate-determining step. A reaction may have two or more comparably slow steps. The kinetic analysis of such reactions is generally more involved.
The following example concerns the mechanistic study of a relatively simple reaction.
EXAMPLE 13.9 The gas-phase decomposition of nitrous oxide (N2O) is believed to occur via two
N2O 88n N2 Step 1:
Step 2: N2O O k2
O 88n N2 O2 Experimentally the rate law is found to be rate k[N2O]. (a) Write the equation for
the overall reaction. (b) Identify the intermediates. (c) What can you say about the
relative rates of steps 1 and 2?
(a) Adding the equations for steps 1 and 2 gives the overall reaction Answer 2N2O 88n 2N2 O2 (b) Since the O atom is produced in the first elementary step and it does not appear
in the overall balanced equation, it is an intermediate.
(c) If we assume that step 1 is the rate-determining step (that is, if k2 k1), then
the rate of the overall reaction is given by
Similar problem: 13.49. and k k1[N2O] k1. PRACTICE EXERCISE The reaction between NO2 and CO to produce NO and CO2 is believed to occur via
Step 2: NO2
NO3 NO2 88n NO
CO 88n NO2 NO3
CO2 The experimental rate law is rate k[NO2]2. (a) Write the equation for the overall
reaction. (b) Identify the intermediate. (c) What can you say about the relative rates
of steps 1 and 2? EXPERIMENTAL SUPPORT FOR REACTION MECHANISMS How can we find out whether the proposed mechanism for a particular reaction is correct? In the case of hydrogen peroxide decomposition we might try to detect the pres- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 538 CHEMICAL KINETICS ence of the IO ions by spectroscopic means. Evidence of their presence would support the reaction scheme. Similarly, for the hydrogen iodide reaction, detection of iodine atoms would lend support to the two-step mechanism. For example, I2 dissociates
into atoms when it is irradiated with visible light. Thus we might predict that the formation of HI from H2 and I2 would speed up as the intensity of light is increased because that should increase the concentration of I atoms. Indeed, this is just what is observed.
In one case, chemists wanted to know which COO bond is broken in the reaction
between methyl acetate and water in order to better understand the reaction mechanism
CH3 O C O O O CH3 + H2O
methyl acetate O
CH3 O C O OH + CH3OH
methanol The two possibilities are
CH3 O C O O O CH3
CH3 O C O O O CH3
(b) To distinguish between schemes (a) and (b), chemists used water containing the oxygen-18 isotope instead of ordinary water (which contains the oxygen-16 isotope). When
the oxygen-18 water was used, only the acetic acid formed contained oxygen-18:
CH3 O C O 18O O H Thus, the reaction must have occurred via bond-breaking scheme (a), because the product formed via scheme (b) would retain both of its original oxygen atoms.
Another example is photosynthesis, the process by which green plants produce
glucose from carbon dioxide and water
6CO2 6H2O 88n C6H12O6 6O2 A question that arose early in studies of photosynthesis was whether the molecular oxygen was derived from water, from carbon dioxide, or from both. By using water containing the oxygen-18 isotope, it was demonstrated that the evolved oxygen came from
water, and none came from carbon dioxide, because the O2 contained only the 18O isotopes. This result supported the mechanism in which water molecules are “split” by
2H2O h 88n O2 4H 4e where h represents the energy of a photon. The protons and electrons are used to drive
energetically unfavorable reactions that are necessary for plant growth and function.
These examples give some idea of how inventive chemists must be in studying
reaction mechanisms. For complex reactions, however, it is virtually impossible to
prove the uniqueness of any particular mechanism. 13.6 CATALYSIS For the decomposition of hydrogen peroxide we saw that the reaction rate depends on
the concentration of iodide ions even though I does not appear in the overall equation. We noted that I acts as a catalyst for that reaction. A catalyst is a substance that Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 13.6 A rise in temperature also
increases the rate of a reaction.
However, at high temperatures
the products formed may undergo
other reactions, thereby
reducing the yield. 539 increases the rate of a chemical reaction without itself being consumed. The catalyst
may react to form an intermediate, but it is regenerated in a subsequent step of the reaction.
In the laboratory preparation of molecular oxygen, a sample of potassium chlorate is heated, as shown in Figure 4.12(b). The reaction is
2KClO3(s) 88n 2KCl(s) To extend the traffic analogy used
in this chapter, adding a catalyst
can be compared with building a
tunnel through a mountain to connect two towns that were previously linked by a winding road
over the mountain. CATALYSIS 3O2(g) However, this thermal decomposition process is very slow in the absence of a catalyst.
The rate of decomposition can be increased dramatically by adding a small amount of
the catalyst manganese(II) dioxide (MnO2), a black powdery substance. All of the MnO2
can be recovered at the end of the reaction, just as all the I ions remain following
A catalyst speeds up a reaction by providing a set of elementary steps with more
favorable kinetics than those that exist in its absence. From Equation (13.8) we know
that the rate constant k (and hence the rate) of a reaction depends on the frequency factor A and the activation energy Ea — the larger A or the smaller Ea, the greater the rate.
In many cases, a catalyst increases the rate by lowering the activation energy for the
Let us assume that the following reaction has a certain rate constant k and an activation energy Ea.
B 88n C D In the presence of a catalyst, however, the rate constant is kc (called the catalytic rate
B 88n C D By the definition of a catalyst,
ratecatalyzed rateuncatalyzed Figure 13.20 shows the potential energy profiles for both reactions. Note that the total energies of the reactants (A and B) and those of the products (C and D) are unaffected by the catalyst; the only difference between the two is a lowering of the acti- Potential energy Ea
Potential energy FIGURE 13.20 Comparison of
the activation energy barriers of
an uncatalyzed reaction and the
same reaction with a catalyst.
The catalyst lowers the energy
barrier but does not affect the
actual energies of the reactants
or products. Although the reactants and products are the same
in both cases, the reaction mechanisms and rate laws are different in (a) and (b). A+B E'
A+B C+D C+D Reaction progress
(a) Back Forward Main Menu TOC Reaction progress
(b) Study Guide TOC Textbook Website MHHE Website 540 CHEMICAL KINETICS A catalyst lowers the activation
energy for both the forward and
reverse reactions. vation energy from Ea to Ea. Because the activation energy for the reverse reaction is
also lowered, a catalyst enhances the rates of the reverse and forward reactions equally.
There are three general types of catalysis, depending on the nature of the rateincreasing substance: heterogeneous catalysis, homogeneous catalysis, and enzyme
catalysis. HETEROGENEOUS CATALYSIS In heterogeneous catalysis, the reactants and the catalyst are in different phases. Usually
the catalyst is a solid and the reactants are either gases or liquids. Heterogeneous catalysis is by far the most important type of catalysis in industrial chemistry, especially in
the synthesis of many key chemicals. Here we describe three specific examples of heterogeneous catalysis that account for millions of tons of chemicals produced annually
on an industrial scale.
The Haber Synthesis of Ammonia Ammonia is an extremely valuable inorganic substance used in the fertilizer industry,
the manufacture of explosives, and many other applications. Around the turn of the
century, many chemists strove to synthesize ammonia from nitrogen and hydrogen. The
supply of atmospheric nitrogen is virtually inexhaustible, and hydrogen gas can be produced readily by passing steam over heated coal:
C(s) 88n CO(g) H2O(g) H2(g) Hydrogen is also a by-product of petroleum refining.
The formation of NH3 from N2 and H2 is exothermic:
3H2(g) 88n 2NH3(g) N2(g) H° 92.6 kJ But the reaction rate is extremely slow at room temperature. To be practical on a large
scale, a reaction must occur at an appreciable rate and it must have a high yield of the
desired product. Raising the temperature does accelerate the above reaction, but at the
same time it promotes the decomposition of NH3 molecules into N2 and H2, thus lowering the yield of NH3.
In 1905, after testing literally hundreds of compounds at various temperatures and
pressures, Fritz Haber discovered that iron plus a few percent of oxides of potassium
and aluminum catalyze the reaction of hydrogen with nitrogen to yield ammonia at
about 500°C. This procedure is known as the Haber process.
In heterogeneous catalysis, the surface of the solid catalyst is usually the site of
the reaction. The initial step in the Haber process involves the dissociation of N2 and
H2 on the metal surface (Figure 13.21). Although the dissociated species are not truly
free atoms because they are bonded to the metal surface, they are highly reactive. The
two reactant molecules behave very differently on the catalyst surface. Studies show
that H2 dissociates into atomic hydrogen at temperatures as low as 196°C (the boiling point of liquid nitrogen). Nitrogen molecules, on the other hand, dissociate at about
500°C. The highly reactive N and H atoms combine rapidly at high temperatures to
produce the desired NH3 molecules:
N Back Forward Main Menu TOC 3H 88n NH3 Study Guide TOC Textbook Website MHHE Website 13.6 FIGURE 13.21 The catalytic
action in the synthesis of ammonia. First the H2 and N2 molecules bind to the surface of the
catalyst. This interaction weakens
the covalent bonds within the
molecules and eventually causes
the molecules to dissociate. The
highly reactive H and N atoms
combine to form NH3 molecules,
which then leave the surface. CATALYSIS 541 H2
NH3 Surface of catalyst The Manufacture of Nitric Acid Nitric acid is one of the most important inorganic acids. It is used in the production of
fertilizers, dyes, drugs, and explosives. The major industrial method of producing nitric acid is the Ostwald † process. The starting materials, ammonia and molecular oxygen, are heated in the presence of a platinum-rhodium catalyst (Figure 13.22) to about
4NH3(g) 5O2(g) 88n 4NO(g) 6H2O(g) The nitric oxide readily oxidizes (without catalysis) to nitrogen dioxide:
2NO(g) O2(g) 88n 2NO2(g) When dissolved in water, NO2 forms both nitrous acid and nitric acid:
2NO2(g) H2O(l ) 88n HNO2(aq) HNO3(aq) On heating, nitrous acid is converted to nitric acid as follows:
3HNO2(aq) 88n HNO3(aq)
A hot platinum wire glows when
held over a concentrated ammonia solution. The oxidation of ammonia to produce nitric oxide,
catalyzed by platinum, is highly
exothermic. Back Forward Main Menu H2O(l ) 2NO(g) The NO generated can be recycled to produce NO2 in the second step.
† Wilhelm Ostwald (1853–1932). German chemist. Ostwald made important contributions to chemical kinetics, thermodynamics, and electrochemistry. He developed the industrial process for preparing nitric acid that now bears his name. He received the Nobel Prize in Chemistry in 1909. TOC Study Guide TOC Textbook Website MHHE Website 542 CHEMICAL KINETICS FIGURE 13.22 Platinumrhodium catalyst used in the
Ostwald process. Catalytic Converters At high temperatures inside a running car ’s engine, nitrogen and oxygen gases react
to form nitric oxide:
N2(g) A hydrocarbon is a compound
containing only carbon and
hydrogen atoms. Hydrocarbons
are the main constituents of
natural gas and gasoline. O2(g) 88n 2NO(g) When released into the atmosphere, NO rapidly combines with O2 to form NO2.
Nitrogen dioxide and other gases emitted by an automobile, such as carbon monoxide
(CO) and various unburned hydrocarbons, make automobile exhaust a major source of
Most new cars are equipped with catalytic converters (Figure 13.23). An efficient
catalytic converter serves two purposes: It oxidizes CO and unburned hydrocarbons to
CO2 and H2O, and it reduces NO and NO2 to N2 and O2. Hot exhaust gases into which
air has been injected are passed through the first chamber of one converter to accelerate the complete burning of hydrocarbons and to decrease CO emission. (A cross section of the catalytic converter is shown in Figure 13.24.) However, since high temperatures increase NO production, a second chamber containing a different catalyst (a
transition metal or a transition metal oxide such as CuO or Cr2O3) and operating at a
lower temperature are required to dissociate NO into N2 and O2 before the exhaust is
discharged through the tailpipe. FIGURE 13.23 A two-stage
catalytic converter for an automobile. Exhaust manifold Exhaust pipe
source of secondary air Back Forward Main Menu TOC Catalytic converters Study Guide TOC Textbook Website MHHE Website 13.6 CATALYSIS 543 FIGURE 13.24 A crosssectional view of a catalytic converter. The beads contain platinum, palladium, and rhodium,
which catalyze the combustion of
CO and hydrocarbons. HOMOGENEOUS CATALYSIS In homogeneous catalysis the reactants and catalyst are dispersed in a single phase,
usually liquid. Acid and base catalyses are the most important types of homogeneous
catalysis in liquid solution. For example, the reaction of ethyl acetate with water to
form acetic acid and ethanol normally occurs too slowly to be measured.
CH3 O C O O O C2H5 + H2O
ethyl acetate O
CH3 O C O OH + C2H5OH
ethanol In the absence of the catalyst, the rate law is given by
rate k[CH3COOC2H5] However, the reaction can be catalyzed by an acid. In the presence of hydrochloric
acid, the rate is faster and the rate law is given by
rate kc[CH3COOC2H5][H ] Note that because kc k, the rate is determined solely by the catalyzed portion of the
Homogeneous catalysis can also take place in the gas phase. A well-known example of catalyzed gas-phase reactions is the lead chamber process, which for many
years was the primary method of manufacturing sulfuric acid. Starting with sulfur, we
would expect the production of sulfuric acid to occur in the following steps:
H2O(l ) O2(g) 88n SO2(g)
O2(g) 88n 2SO3(g)
SO3(g) 88n H2SO4(aq) In reality, however, sulfur dioxide is not converted directly to sulfur trioxide; rather,
the oxidation is more efficiently carried out in the presence of the catalyst nitrogen
2SO2(g) 2NO2(g) 88n 2SO3(g) 2NO(g) Overall reaction: 2SO2(g) 2NO(g) O2(g) 88n 2NO2(g)
O2(g) 88n 2SO3(g) Note that there is no net loss of NO2 in the overall reaction, so that NO2 meets the criteria for a catalyst. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 544 CHEMICAL KINETICS In recent years chemists have devoted much effort to developing a class of metallic compounds to serve as homogeneous catalysts. These compounds are soluble in
various organic solvents and therefore can catalyze reactions in the same phase as the
dissolved reactants. Many of the processes they catalyze are organic. For example,
a red-violet compound of rhodium, [(C6H5)3P]3RhCl, catalyzes the conversion of a
carbon-carbon double bond to a carbon-carbon single bond as follows:
This reaction is important in the
food industry. It converts “unsaturated fats” (compounds containing
many CPC bonds) to “saturated
fats” (compounds containing few
or no CPC bonds). AA
C P C + H2
OC OC O
HH Homogeneous catalysis has several advantages over heterogeneous catalysis. For one
thing, the reactions can often be carried out under atmospheric conditions, thus reducing production costs and minimizing the decomposition of products at high temperatures. In addition, homogeneous catalysts can be designed to function selectively for a
particular type of reaction, and homogeneous catalysts cost less than the precious metals (for example, platinum and gold) used in heterogeneous catalysis.
ENZYME CATALYSIS Of all the intricate processes that have evolved in living systems, none is more striking or more essential than enzyme catalysis. Enzymes are biological catalysts. The
amazing fact about enzymes is that not only can they increase the rate of biochemical
reactions by factors ranging from 106 to 1012, but they are also highly specific. An enzyme acts only on certain molecules, called substrates (that is, reactants), while leaving the rest of the system unaffected. It has been estimated that an average living cell
may contain some 3000 different enzymes, each of them catalyzing a specific reaction
in which a substrate is converted into the appropriate products. Enzyme catalysis is homogeneous because the substrate and enzyme are present in aqueous solution.
An enzyme is typically a large protein molecule that contains one or more active
sites where interactions with substrates take place. These sites are structurally compatible with specific substrate molecules, in much the same way as a key fits a particular lock. In fact, the notion of a rigid enzyme structure that binds only to molecules
whose shape exactly matches that of the active site was the basis of an early theory of
enzyme catalysis, the so-called lock-and-key theory developed by the German chemist
Emil Fischer† in 1894 (Figure 13.25). Fischer ’s hypothesis accounts for the specificity
† Emil Fischer (1852–1919). German chemist. Regarded by many as the greatest organic chemist of the nineteenth century,
Fischer made many significant contributions in the synthesis of sugars and other important molecules. He was awarded the
Nobel Prize in Chemistry in 1902. FIGURE 13.25 The lock-andkey model of an enzyme’s specificity for substrate molecules.
+ Enzyme Back Forward Main Menu TOC Products
complex Study Guide TOC Enzyme Textbook Website MHHE Website Potential energy Potential energy 13.6 S CATALYSIS 545 ES
E+S P E+P Reaction progress Reaction progress (a) (b) FIGURE 13.26 Comparison of (a) an uncatalyzed reaction and (b) the same reaction catalyzed by an enzyme. The plot in (b) assumes that the catalyzed reaction has a two-step mechanism, in which the second step (ES 88n E P) is rate-determining. of enzymes, but it contradicts research evidence that a single enzyme binds to substrates of different sizes and shapes. Chemists now know that an enzyme molecule (or
at least its active site) has a fair amount of structural flexibility and can modify its
shape to accommodate more than one type of substrate.
The mathematical treatment of enzyme kinetics is quite complex, even when we
know the basic steps involved in the reaction. A simplified scheme is given by the following elementary steps:
S 3:4 ES
k1 Rate of product formation k2
ES 7778n E All active sites
at and beyond
FIGURE 13.27 Plot of the rate
of product formation versus substrate concentration in an enzyme-catalyzed reaction. Back Forward Main Menu P where E, S, and P represent enzyme, substrate, and product, and ES is the enzymesubstrate intermediate. It is often assumed that the formation of ES and its decomposition back to enzyme and substrate molecules occur rapidly and that the ratedetermining step is the formation of product. Figure 13.26 shows the potential energy
profile for the reaction.
In general, the rate of such a reaction is given by the equation
k[ES] The concentration of the ES intermediate is itself proportional to the amount of the
substrate present, and a plot of the rate versus the concentration of substrate typically
yields a curve like that shown in Figure 13.27. Initially the rate rises rapidly with increasing substrate concentration. However, above a certain concentration all the active
sites are occupied, and the reaction becomes zero order in the substrate. In other words,
the rate remains the same even though the substrate concentration increases. At and
beyond this point, the rate of formation of product depends only on how fast the ES
intermediate breaks down, not on the number of substrate molecules present. TOC Study Guide TOC Textbook Website MHHE Website 546 CHEMICAL KINETICS SUMMARY OF
KEY EQUATIONS k[A]x[B]y • rate
• ln [A]0
[A] • ln [A] kt Ae • ln k
• ln k1
k2 Half-life for a first-order reaction. (13.6) kt Ea/RT Ea
R Relationship between concentration and time for a
Equation for the graphical determination of k for a
first-order reaction. (13.4) y) gives the (13.5) 1
2 •k SUMMARY OF FACTS
AND CONCEPTS ln [A]0 Rate law expressions. The sum (x
overall order of the reaction. (13.3) kt 0.693
•t (13.1) Relationship between concentration and time for a
The Arrhenius equation expressing the dependence of
the rate constant on activation energy and
temperature. (13.8) 1
T ln A (13.9) Ea T1 T2
T1T2 (13.11) Equation for the graphical determination of activation
Relationships of rate constants at two different
temperatures. 1. The rate of a chemical reaction is the change in the concentration of reactants or products
over time. The rate is not constant, but varies continuously as concentrations change.
2. The rate law expresses the relationship of the rate of a reaction to the rate constant and the
concentrations of the reactants raised to appropriate powers. The rate constant k for a given
reaction changes only with temperature.
3. Reaction order is the power to which the concentration of a given reactant is raised in the
rate law. Overall reaction order is the sum of the powers to which reactant concentrations
are raised in the rate law. The rate law and the reaction order cannot be determined from
the stoichiometry of the overall equation for a reaction; they must be determined by experiment. For a zero-order reaction, the reaction rate is equal to the rate constant.
4. The half-life of a reaction (the time it takes for the concentration of a reactant to decrease
by one-half) can be used to determine the rate constant of a first-order reaction.
5. In terms of collision theory, a reaction occurs when molecules collide with sufficient energy, called the activation energy, to break the bonds and initiate the reaction. The rate constant and the activation energy are related by the Arrhenius equation.
6. The overall balanced equation for a reaction may be the sum of a series of simple reactions, called elementary steps. The complete series of elementary steps for a reaction is the
7. If one step in a reaction mechanism is much slower than all other steps, it is the ratedetermining step.
8. A catalyst speeds up a reaction usually by lowering the value of Ea. A catalyst can be recovered unchanged at the end of a reaction.
9. In heterogeneous catalysis, which is of great industrial importance, the catalyst is a solid
and the reactants are gases or liquids. In homogeneous catalysis, the catalyst and the reactants are in the same phase. Enzymes are catalysts in living systems. KEY WORDS
Activated complex, p. 529
Activation energy (Ea), p. 529 Back Forward Bimolecular reaction, p. 534
Catalyst, p. 539 Main Menu TOC Chemical kinetics, p. 508
Elementary step, p. 533 Study Guide TOC Enzyme, p. 544
First-order reaction, p. 518 Textbook Website MHHE Website 547 QUESTIONS AND PROBLEMS Rate constant (k), p. 512
Rate law, p. 514 Half-life (t ), p. 522
Intermediate, p. 534
Molecularity of a reaction,
2 Second-order reaction, p. 524
Termolecular reaction, p. 534
Unimolecular reaction, p. 534 Reaction mechanism,
Reaction order, p. 515
Reaction rate, p. 508 QUESTIONS AND PROBLEMS
THE RATE OF A REACTION THE RATE LAW Review Questions Review Questions 13.1 What is meant by the rate of a chemical reaction?
What are the units of the rate of a reaction?
13.2 Distinguish between average rate and instantaneous
rate. Which of the two rates gives us an unambiguous measurement of reaction rate? Why?
13.3 What are the advantages of measuring the initial rate
of a reaction?
13.4 Can you suggest two reactions that are very slow
(take days or longer to complete) and two reactions
that are very fast (reactions that are over in minutes
or seconds)? 13.9 Explain what is meant by the rate law of a reaction.
13.10 What is meant by the order of a reaction?
13.11 What are the units for the rate constants of firstorder and second-order reactions?
13.12 Consider the zero-order reaction: A 88n product.
(a) Write the rate law for the reaction. (b) What are
the units for the rate constant? (c) Plot the rate of
the reaction versus [A].
13.13 The rate constant of a first-order reaction is 66 s 1.
What is the rate constant in units of minutes?
13.14 On which of the following properties does the rate
constant of a reaction depend? (a) reactant concentrations, (b) nature of reactants, (c) temperature Problems 13.5 Write the reaction rate expressions for the following reactions in terms of the disappearance of the
reactants and the appearance of products:
(a) H2(g) I2(g) 88n 2HI(g)
(b) 5Br (aq) BrO3 (aq) 6H (aq) 88n
13.6 Write the reaction rate expressions for the following reactions in terms of the disappearance of the
reactants and the appearance of products:
(a) 2H2(g) O2(g) 88n 2H2O(g)
(b) 4NH3(g) 5O2(g) 88n 4NO(g) 6H2O(g)
13.7 Consider the reaction
2NO(g) Forward NH4 (aq) TOC 2H2O(l ) is given by rate k[NH4 ][NO2 ]. At 25°C, the rate
constant is 3.0 10 4/M s. Calculate the rate of
the reaction at this temperature if [NH4 ] 0.26 M
and [NO2 ] 0.080 M.
13.16 Use the data in Table 13.2 to calculate the rate of
the reaction at the time when [F2] 0.010 M and
[ClO2] 0.020 M.
13.17 Consider the reaction
B 88n products From the following data obtained at a certain temperature, determine the order of the reaction and calculate the rate constant:
[A] [B] RATE (M/s) 1.50
3.00 3H2(g) 88n 2NH3(g) Main Menu NO2 (aq) 88n N2(g) A Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of
0.074 M/s. (a) At what rate is ammonia being
formed? (b) At what rate is molecular nitrogen reacting? Back 13.15 The rate law for the reaction O2(g) 88n 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of
0.066 M/s. (a) At what rate is NO2 being formed?
(b) At what rate is molecular oxygen reacting?
13.8 Consider the reaction
N2(g) Problems 1.50
1 13.18 Consider the reaction Study Guide TOC X Y 88n Z Textbook Website MHHE Website 548 CHEMICAL KINETICS From the following data, obtained at 360 K, (a) determine the order of the reaction, and (b) determine
the initial rate of disappearance of X when the concentration of X is 0.30 M and that of Y is 0.40 M.
INITIAL RATE OF
DISAPPEARANCE OF X (M/s) [X] [Y] 0.053
0.30 A 88n B The rate of the reaction is 1.6 10 2 M/s when the
concentration of A is 0.35 M. Calculate the rate constant if the reaction is (a) first order in A and (b) second order in A.
13.21 Cyclobutane decomposes to ethylene according to
C4H8(g) 88n 2C2H4(g) Determine the order of the reaction and the rate constant based on the following pressures, which were
recorded when the reaction was carried out at 430°C
in a constant-volume vessel. 400
122 Determine the order of the reaction and the rate constant based on the following data: Main Menu TOC 1
2 Problems 13.27 What is the half-life of a compound if 75 percent of
a given sample of the compound decomposes in
60 min? Assume first-order kinetics.
13.28 The thermal decomposition of phosphine (PH3) into
phosphorus and molecular hydrogen is a first-order
4PH3(g) 88n P4(g) 2NOBr(g) 88n 2NO(g) ClCO2CCl3(g) 88n 2COCl2(g) Forward 13.23 Write an equation relating the concentration of a reactant A at t 0 to that at t t for a first-order reaction. Define all the terms and give their units. Do
the same for a second-order reaction.
13.24 Define half-life. Write the equation relating the halflife of a first-order reaction to the rate constant.
13.25 Write the equations relating the half-life of a second-order reaction to the rate constant. How does it
differ from the equation for a first-order reaction?
13.26 For a first-order reaction, how long will it take for
the concentration of reactant to fall to one-eighth its
original value? Express your answer in terms of the
half-life (t ) and in terms of the rate constant k. 6H2(g) The half-life of the reaction is 35.0 s at 680°C.
Calculate (a) the first-order rate constant for the reaction and (b) the time required for 95 percent of
the phosphine to decompose.
13.29 The rate constant for the second-order reaction PC4H8 (mmHg) 13.22 The following gas-phase reaction was studied at
290°C by observing the change in pressure as a function of time in a constant-volume vessel: Back 15.76
4.44 Review Questions 1
1164 PClCO2CCl3 (mmHg) THE RELATION BETWEEN REACTANT CONCENTRATION
AND TIME 13.19 Determine the overall orders of the reactions to
which the following rate laws apply: (a) rate
k[NO2]2, (b) rate k, (c) rate k[H2][Br2] ,
(d) rate k[NO]2[O2].
13.20 Consider the reaction TIME (s) TIME (s) Br2(g) is 0.80/M s at 10°C. (a) Starting wth a concentration of 0.086 M, calculate the concentration of
NOBr after 22 s. (b) Calculate the half-lives when
[NOBr]0 0.072 M and [NOBr]0 0.054 M.
13.30 The rate constant for the second-order reaction
2NO2(g) 88n 2NO(g) O2(g) is 0.54/M s at 300°C. How long (in seconds) would
it take for the concentration of NO2 to decrease from
0.62 M to 0.28 M? Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS ACTIVATION ENERGY
Review Questions 13.31 Define activation energy. What role does activation
energy play in chemical kinetics?
13.32 Write the Arrhenius equation and define all terms.
13.33 Use the Arrhenius equation to show why the rate
constant of a reaction (a) decreases with increasing
activation energy and (b) increases with increasing
13.34 The burning of methane in oxygen is a highly
exothermic reaction. Yet a mixture of methane and
oxygen gas can be kept indefinitely without any apparent change. Explain.
13.35 Sketch a potential energy versus reaction progress
plot for the following reactions:
(a) S(s) O2(g) 88n SO2(g) H o
H o 242.7 kJ
(b) Cl2(g) 88n Cl(g) Cl(g)
13.36 The reaction H H2 88n H2 H has been studied for many years. Sketch a potential energy versus reaction progress diagram for this reaction.
Problems 13.37 Variation of the rate constant with temperature for
the first-order reaction
2N2O5(g) 88n 2N2O4(g) O2(g) is given in the following table. Determine graphically the activation energy for the reaction.
T (K) k (s 1) 298
4 CO(g) Cl2(g) 88n COCl2(g) O3(g) 88n NO2(g) Forward Main Menu Review Questions 13.43 What do we mean by the mechanism of a reaction?
13.44 What is an elementary step? What is the molecularity of a reaction? Classify the following elementary reactions as unimolecular, bimolecular, and termolecular:
(a) 2NO Br2 88n 2NOBr
(b) CH3NC 88n CH3CN
(c) SO O2 88n SO2 O
13.45 Reactions can be classified as unimolecular, bimolecular, and so on. Why are there no zero-molecular
reactions? Explain why termolecular reactions are
13.46 What is the rate-determining step of a reaction? Give
an everyday analogy to illustrate the meaning of
13.47 The equation for the combustion of ethane (C2H6)
is TOC 7O2(g) 88n 4CO2(g) 6H2O(l) Explain why it is unlikely that this equation also represents the elementary step for the reaction.
13.48 Specify which of the following species cannot be
isolated in a reaction: activated complex, product,
Problems 13.49 The rate law for the reaction
2NO(g) O2(g) the frequency factor A is 8.7 1012 s 1 and the activation energy is 63 kJ/mol. What is the rate constant for the reaction at 75°C? Back REACTION MECHANISMS 2C2H6(g) at 250°C is 1.50 103 times as fast as the same reaction at 150°C. Calculate the activation energy for
this reaction. Assume that the frequency factor is
13.39 For the reaction
NO(g) 13.40 The rate constant of a first-order reaction is 4.60
10 4 s 1 at 350°C. If the activation energy is 104
kJ/mol, calculate the temperature at which its rate
constant is 8.80 10 4 s 1.
13.41 The rate constants of some reactions double with
every 10-degree rise in temperature. Assume that a
reaction takes place at 295 K and 305 K. What must
the activation energy be for the rate constant to double as described?
13.42 The rate at which tree crickets chirp is 2.0 102
per minute at 27°C but only 39.6 per minute at 5°C.
From these data, calculate the “activation energy”
for the chirping process. (Hint: The ratio of rates is
equal to the ratio of rate constants.) 3 13.38 Given the same reactant concentrations, the reaction 549 Cl2(g) 88n 2NOCl(g) is given by rate k[NO][Cl2]. (a) What is the order of the reaction? (b) A mechanism involving the
following steps has been proposed for the reaction: Study Guide TOC Textbook Website MHHE Website 550 CHEMICAL KINETICS NO(g) Cl2(g) 88n NOCl2(g) CATALYSIS NOCl2(g) NO(g) 88n 2NOCl(g) Review Questions If this mechanism is correct, what does it imply
about the relative rates of these two steps?
13.50 For the reaction X2 Y Z 88n XY XZ it is
found that doubling the concentration of X2 doubles
the reaction rate, tripling the concentration of Y
triples the rate, and doubling the concentration of Z
has no effect. (a) What is the rate law for this reaction? (b) Why is it that the change in the concentration of Z has no effect on the rate? (c) Suggest a
mechanism for the reaction that is consistent with
the rate law.
13.51 The rate law for the decomposition of ozone to molecular oxygen
2O3(g) 88n 3O2(g) is
rate Problems [O3]2
[O2] k 13.59 Most reactions, including enzyme-catalyzed reactions, proceed faster at higher temperatures. However,
for a given enzyme, the rate drops off abruptly at a
certain temperature. Account for this behavior.
13.60 Consider the following mechanism for the enzymecatalyzed reaction: The mechanism proposed for this process is
O3 3:4 O
O O2 k2
O3 88n 2O2 E Derive the rate law from these elementary steps.
Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing O2 concentration.
13.52 The rate law for the reaction
2H2(g) 2NO(g) 88n N2(g) 2H2O(g) is rate k[H2][NO]2. Which of the following mechanisms can be ruled out on the basis of the observed
H2 NO 88n H2O
O N NO 88n N2 O H2 88n H2O H2 88n N2 H2O
H2O Mechanism III
2NO 34 N2O2
N2O Back Forward H2 88n N2O
H2 88n N2 k1
S 3:4 ES
ES 88n E (fast) (slow)
(fast) H2O Main Menu (slow)
(fast) TOC (slow) Derive an expression for the rate law of the reaction
in terms of the concentrations of E and S. (Hint: To
solve for [ES], make use of the fact that, at equilibrium, the rate of forward reaction is equal to the rate
of the reverse reaction.)
ADDITIONAL PROBLEMS CO(g) 88n NO(g) NO2(g)
H2O P (fast equilibrium) 13.61 Suggest experimental means by which the rates of
the following reactions could be followed:
(a) CaCO3(s) 88n CaO(s) CO2(g)
(b) Cl2(g) 2Br (aq) 88n Br2(aq) 2Cl (aq)
(c) C2H6(g) 88n C2H4(g) H2(g)
(d) C2H5I(g) H2O(l ) 88n C2H5OH(aq)
H (aq) I (aq)
13.62 List four factors that influence the rate of a reaction.
13.63 “The rate constant for the reaction (slow)
(fast) Mechanism II
H2 2NO 88n N2O
N2O 13.53 How does a catalyst increase the rate of a reaction?
13.54 What are the characteristics of a catalyst?
13.55 A certain reaction is known to proceed slowly at
room temperature. Is it possible to make the reaction proceed at a faster rate without changing the
13.56 Distinguish between homogeneous catalysis and
heterogeneous catalysis. Describe three important
industrial processes that utilize heterogeneous catalysis.
13.57 Are enzyme-catalyzed reactions examples of homogeneous or heterogeneous catalysis? Explain.
13.58 The concentrations of enzymes in cells are usually
quite small. What is the biological significance of
this fact? CO2(g) 6 is 1.64 10 /M s.” What is incomplete about
13.64 In a certain industrial process involving a heteroge- Study Guide TOC Textbook Website MHHE Website 551 QUESTIONS AND PROBLEMS neous catalyst, the volume of the catalyst (in the
shape of a sphere) is 10.0 cm3. Calculate the surface area of the catalyst. If the sphere is broken down
into eight spheres, each having a volume of 1.25 cm3,
what is the total surface area of the spheres? Which
of the two geometric configurations of the catalyst
is more effective? (The surface area of a sphere is
4 r 2, where r is the radius of the sphere.) Based on
your analysis here, explain why it is sometimes dangerous to work in grain elevators.
13.65 Use the data in Example 13.4 to determine graphically the half-life of the reaction.
13.66 The following data were collected for the reaction
between hydrogen and nitric oxide at 700°C:
2H2(g) 2NO(g) 88n 2H2O(g) N2(g) 1
3 [H2] [NO] 0.025
0.0125 2.4 10 6
1.2 10 6
0.60 10 6 (a) Determine the order of the reaction. (b) Calculate
the rate constant. (c) Suggest a plausible mechanism
that is consistent with the rate law. (Hint: Assume
that the oxygen atom is the intermediate.)
13.67 When methyl phosphate is heated in acid solution,
it reacts with water:
CH3OPO3H2 H2O 88n CH3OH CH3COOC2H5(aq) H2O(l ) 88n
CH3COOH(aq) C2H5OH(aq) shows first-order characteristics — that is, rate
k[CH3COOC2H5] — even though this is a secondorder reaction (first order in CH3COOC2H5 and first
order in H2O). Explain.
13.69 Explain why most metals used in catalysis are transition metals.
13.70 The reaction 2A 3B 88n C is first order with respect to A and B. When the initial concentrations
are [A] 1.6 10 2 M and [B] 2.4 10 3 M,
the rate is 4.1 10 4 M/s. Calculate the rate constant of the reaction. Forward Main Menu TOC H
Br2 8877777n CH3COCH2Br
catalyst H Br The rate of disappearance of bromine was measured
for several different concentrations of acetone,
bromine, and H ions at a certain temperature: [CH3COCH3] [Br2] [H ] 0.30
0.050 RATE OF DISAPPEARANCE OF Br2 (M/s) 0.050
5 (a) What is the rate law for the reaction?
(b) Determine the rate constant. (c) The following
mechanism has been proposed for the reaction:
CH3 O C O CH3 + H3O+
CH3 O C O CH3 + H2O
CH3 O C P CH2 + Br2 H3PO4 If the reaction is carried out in water enriched with
O, the oxygen-18 isotope is found in the phosphoric acid product but not in the methanol. What
does this tell us about the mechanism of the reaction?
13.68 The rate of the reaction Back CH3COCH3 INITIAL RATE
0.010 EXPERIMENT 13.71 The bromination of acetone is acid-catalyzed: 13.72 13.73 13.74
13.75 +OH B
CH3 O C O CH3 + H2O
CH3 O C P CH2 + H3O+ (slow)
CH3 O C O CH2Br + HBr (fast) Show that the rate law deduced from the mechanism
is consistent with that shown in (a).
The decomposition of N2O to N2 and O2 is a firstorder reaction. At 730°C the half-life of the reaction
is 3.58 103 min. If the initial pressure of N2O is
2.10 atm at 730°C, calculate the total gas pressure
after one half-life. Assume that the volume remains
2I 88n 2SO2
I2 proThe reaction S2O2
ceeds slowly in aqueous solution, but it can be catalyzed by the Fe3 ion. Given that Fe3 can oxidize I and Fe2 can reduce S2O2 , write a
plausible two-step mechanism for this reaction.
Explain why the uncatalyzed reaction is slow.
What are the units of the rate constant for a thirdorder reaction?
The integrated rate law for the zero-order reaction
A 88n B is [A] [A]0 kt. (a) Sketch the following plots: (i) rate versus [A] and (ii) [A] versus
t. (b) Derive an expression for the half-life of the
reaction. (c) Calculate the time in half-lives when Study Guide TOC Textbook Website MHHE Website 552 CHEMICAL KINETICS 13.76 13.77 13.78 13.79 13.80 13.81 13.82 13.83 the integrated rate law is no longer valid, that is,
when [A] 0.
A flask contains a mixture of compounds A and B.
Both compounds decompose by first-order kinetics.
The half-lives are 50.0 min for A and 18.0 min for
B. If the concentrations of A and B are equal initially, how long will it take for the concentration of
A to be four times that of B?
Referring to Example 13.4, explain how you would
measure the partial pressure of azomethane experimentally as a function of time.
The rate law for the reaction 2NO2(g) 88n N2O4(g)
is rate k[NO2]2. Which of the following changes
will change the value of k? (a) The pressure of NO2
is doubled. (b) The reaction is run in an organic solvent. (c) The volume of the container is doubled.
(d) The temperature is decreased. (e) A catalyst is
added to the container.
The reaction of G2 with E2 to form 2EG is exothermic, and the reaction of G2 with X2 to form 2XG is
endothermic. The activation energy of the exothermic reaction is greater than that of the endothermic
reaction. Sketch the potential energy profile diagrams for these two reactions on the same graph.
In the nuclear industry, workers use a rule of thumb
that the radioactivity from any sample will be relatively harmless after ten half-lives. Calculate the
fraction of a radioactive sample that remains after
this time period. (Hint: Radioactive decays obey
Briefly comment on the effect of a catalyst on each
of the following: (a) activation energy, (b) reaction
mechanism, (c) enthalpy of reaction, (d) rate of forward step, (e) rate of reverse step.
When 6 g of granulated Zn is added to a solution of
2 M HCl in a beaker at room temperature, hydrogen gas is generated. For each of the following
changes (at constant volume of the acid) state
whether the rate of hydrogen gas evolution will be
increased, decreased, or unchanged: (a) 6 g of powdered Zn is used; (b) 4 g of granulated Zn is used;
(c) 2 M acetic acid is used instead of 2 M HCl;
(d) temperature is raised to 40°C.
Strictly speaking, the rate law derived for the reaction in Problem 13.66 applies only to certain concentrations of H2. The general rate law for the reaction takes the form
rate Forward Main Menu 2N2O5 88n 4NO2 TOC O2 [N2O5] INITIAL RATE (M/s) 0.92
5 Determine graphically the rate law for the reaction
and calculate the rate constant.
13.86 The thermal decomposition of N2O5 obeys first-order kinetics. At 45°C, a plot of ln [N2O5] versus t
gives a slope of 6.18 10 4 min 1. What is the
half-life of the reaction?
13.87 When a mixture of methane and bromine is exposed
to light, the following reaction occurs slowly:
CH4(g) Br2(g) 88n CH3Br(g) HBr(g) Suggest a reasonable mechanism for this reaction.
(Hint: Bromine vapor is deep red; methane is colorless.)
13.88 The rate of the reaction between H2 and I2 to form
HI (discussed on p. 536) increases with the intensity of visible light. (a) Explain why this fact supports the two-step mechanism given. (The color of
I2 vapor is shown on p. 452.) (b) Explain why the
visible light has no effect on the formation of H
13.89 The carbon-14 decay rate of a sample obtained from
a young tree is 0.260 disintegration per second per
gram of the sample. Another wood sample prepared
from an object recovered at an archaeological excavation gives a decay rate of 0.186 disintegration
per second per gram of the sample. What is the age
of the object? (Hint: See Chemistry in Action essay
on p. 527.)
13.90 Consider the following elementary step:
1 k2[H2] where k1 and k2 are constants. Derive rate law expressions under the conditions of very high and very Back low hydrogen concentrations. Does the result from
Problem 13.66 agree with one of the rate expressions here?
13.84 A certain first-order reaction is 35.5 percent complete in 4.90 min at 25°C. What is its rate constant?
13.85 The decomposition of dinitrogen pentoxide has been
studied in carbon tetrachloride solvent (CCl4) at a
certain temperature: 2Y 88n XY2 (a) Write a rate law for this reaction. (b) If the initial rate of formation of XY2 is 3.8 10 3 M/s and
the initial concentrations of X and Y are 0.26 M and
0.88 M, what is the rate constant of the reaction? Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 13.91 In recent years ozone in the stratosphere has been
depleted at an alarmingly fast rate by chlorofluorocarbons (CFCs). A CFC molecule such as CFCl3 is
first decomposed by UV radiation:
CFCl3 88n CFCl2 Cl The chlorine radical then reacts with ozone as follows:
Cl O3 88n ClO ClO O 88n Cl O2
O2 (a) Write the overall reaction for the last two steps.
(b) What are the roles of Cl and ClO? (c) Why is
the fluorine radical not important in this mechanism? (d) One suggestion to reduce the concentration of chlorine radicals is to add hydrocarbons such
as ethane (C2H6) to the stratosphere. How will this
work? (e) Draw potential energy versus reaction
progress diagrams for the uncatalyzed and catalyzed
(by Cl) destruction of ozone: O3 O 88n 2O2. Use
the thermodynamic data in Appendix 3 to determine
whether the reaction is exothermic or endothermic.
13.92 Chlorine oxide (ClO), which plays an important role
in the depletion of ozone (see Problem 13.91), decays rapidly at room temperature according to the
2ClO(g) 88n Cl2(g) TIME (s) 0.12
4.77 Assuming that both steps are first-order, sketch on
the same graph the variations of [A], [B], and [C]
13.96 (a) What can you deduce about the activation energy of a reaction if its rate constant changes significantly with a small change in temperature? (b) If
a bimolecular reaction occurs every time an A and
a B molecule collide, what can you say about the
orientation factor and activation energy of the reaction?
13.97 The rate law for the following reaction
2 Main Menu TOC 6H2(g) It is found that the reaction is independent of [PH3]
as long as phosphine’s pressure is sufficiently high
( 1 atm). Explain.
13.100 Thallium(I) is oxidized by cerium(IV) as follows: 6 Tl 6 2Ce4 88n Tl3 2Ce3 The elementary steps, in the presence of Mn(II), are
as follows: 6
6 Ce4 6 13.93 A compound X undergoes two simultaneous firstorder reactions as follows: X 88n Y with rate constant k1 and X 88n Z with rate constant k2. The ratio of k1/k2 at 40°C is 8.0. What is the ratio at 300°C?
Assume that the frequency factors of the two reactions are the same.
13.94 Consider a car fitted with a catalytic converter. The
first five minutes or so after it is started are the most
13.95 The following scheme in which A is converted to B,
which is then converted to C is known as a consecutive reaction. Forward NO(g) is rate k[NO2]2. Suggest a plausible mechanism
for the reaction, given that the unstable species NO3
is an intermediate.
2.44 105 yr) is
13.98 Radioactive plutonium-239 (t
used in nuclear reactors and atomic bombs. If there
are 5.0 102 g of the isotope in a small atomic
bomb, how long will it take for the substance to decay to 1.0 102 g, too small an amount for an effective bomb?
13.99 Many reactions involving heterogeneous catalysts
are zero order; that is rate k. An example is the
decomposition of phosphine (PH3) over tungsten
(W): Ce Back NO2(g) 88n CO2(g) 4PH3(g) 88n P4(g) [ClO]
3 A 88n B 88n C O2(g) From the following data, determine the reaction order and calculate the rate constant of the reaction 553 4 Tl Mn2 88n Ce3
3 Mn Mn4 3 88n Ce
88n Tl3 Mn3
Mn2 (a) Identify the catalyst, intermediates, and the ratedetermining step if the rate law is rate
k[Ce4 ][Mn2 ]. (b) Explain why the reaction is
slow without the catalyst. (c) Classify the type of
catalysis (homogeneous or heterogeneous).
13.101 Sucrose (C12H22O11), commonly called table sugar,
undergoes hydrolysis (reaction with water) to produce fructose (C6H12O6) and glucose (C6H12O6): Study Guide TOC C12H22O11 H2O 88n C6H12O6
fructose Textbook Website C6H12O6
glucose MHHE Website 554 CHEMICAL KINETICS This reaction is of considerable importance in the
candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose,
called invert sugar, does not crystallize, so the candy
containing this sugar would be chewy rather than
brittle as candy containing sucrose crystals would
be. (a) From the following data determine the order
of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the
rate law does not include [H2O] even though water
is a reactant.
TIME (min) [C12H22O11] 0
0.280 (a) The initiator frequently used in the polymerization of ethylene is benzoyl peroxide [(C6H5COO)2]:
[(C6H5COO)2] 88n 2C6H5COOj This is a first-order reaction. The half-life of benzoyl peroxide at 100°C is 19.8 min. (a) Calculate
the rate constant (in min 1) of the reaction. (b) If
the half-life of benzoyl peroxide is 7.30 h, or
438 min, at 70°C, what is the activation energy (in
kJ/mol) for the decomposition of benzoyl peroxide?
(c) Write the rate laws for the elementary steps in
the above polymerization process, and identify the
reactant, product, and intermediates. (d) What condition would favor the growth of long, high-molarmass polyethylenes?
13.104 Consider the following elementary steps for a consecutive reaction:
k1 13.102 The first-order rate constant for the decomposition
of dimethyl ether
(CH3)2O(g) 88n CH4(g) H2(g) CO(g) is 3.2 10 4 s 1 at 450°C. The reaction is carried
out in a constant-volume flask. Initially only dimethyl ether is present and the pressure is 0.350 atm.
What is the pressure of the system after 8.0 min?
Assume ideal behavior.
13.103 Polyethylene is used in many items, including water pipes, bottles, electrical insulation, toys, and
mailer envelopes. It is a polymer, a molecule with a
very high molar mass made by joining many ethylene molecules together. (Ethylene is the basic unit,
or monomer for polyethylene.) The initiation step is
ki R2 88n 2Rj initiation The Rj species (called a radical) reacts with an ethylene molecule (M) to generate another radical
M 88n M1j Rj Reaction of M1j with another monomer leads to the
growth or propagation of the polymer chain:
M1j kp M 88n M2j propagation This step can be repeated with hundreds of monomer
units. The propagation terminates when two radicals
M′j Back Forward kt M″j 88n M′OM″ Main Menu termination TOC k2 A 88n B 88n C (a) Write an expression for the rate of change of B.
(b) Derive an expression for the concentration of B
under steady-state conditions; that is, when B is decomposing to C at the same rate as it is formed from
13.105 Ethanol is a toxic substance that, when consumed
in excess, can impair respiratory and cardiac functions by interference with the neutrotransmitters
of the nervous system. In the human body, ethanol
is metabolized by the enzyme alcohol dehydrogenase to acetaldehyde, which causes “hangovers.”
(a) Based on your knowledge of enzyme kinetics,
explain why binge drinking (that is, consuming too
much alcohol too fast) can prove fatal. (b) Methanol
is even more toxic than ethanol. It is also metabolized by alcohol dehydrogenase, and the product,
formaldehyde, can cause blindness or death. An antidote to methanol poisoning is ethanol. Explain how
this procedure works.
13.106 The kinetics of the heterogeneous reaction between
magnesium metal and hydrochloric acid
2H (aq) 88n Mg2 (aq) Mg(s) H2(g) can be studied by measuring the increase in hydrogen gas pressure as a function of time ( P/ t). The
rate law of the reaction is given by
rate k[surface area of Mg]a[H ]b The experimental data at 32°C are summarized below: Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS EXPERIMENT [H ]
(M ) SURFACE AREA
OF Mg (mm2) P/ t
0.970 drogen gas from the stoichiometry of the reaction.
[Data taken with permission from J. P. Birk and
D. L. Walters, J. Chem. Educ. 70, 587 (1993).]
13.107 Sketch a potential energy versus reaction progress
diagram for a two-step exothermic reaction in which
the first step is rate-determining.
Answers to Practice Exercises: 13.1 rate (a) Determine the rate law of the reaction (that is,
determine the values a and b and an average value
of k.) (b) In one experiment, 0.0083 g of Mg was
reacted with an excess of HCl. The final pressure
and volume of the hydrogen gas were 93.2 torr and
69.5 mL at 32°C. Calculate the percent error for hy- Back Forward Main Menu TOC 555 13.2 rate [CH4]
k[S2O2 ][I ]; k
8.1 10 /M s. 13.3 66 s. 13.4 2.3 10 4 s 1. 13.5 1.2 103 s. 13.6 (a) 3.2 min.
(b) 2.1 min. 13.7 240 kJ/mol. 13.8 3.13 10 9 s 1. 13.9 (a)
NO2 CO 88n NO CO2. (b) NO3. (c) The first step is ratedetermining. Study Guide TOC Textbook Website MHHE Website C HEMISTRY IN T HREE D IMENSIONS The Molecular Basis of Smell and Taste
bulb To brain Olfactory
Camphor bound to receptor site In mammals, the initial detection
of odors takes place at the rear
of the nose in the cilia, hairlike
sensors attached to nerve fibers
in the olfactory bulb. Receptor
proteins in the cilia bind to odor
molecules that have a complementary shape, stimulating the
nerve fibers to signal the brain,
which identifies the odor. In Chapter 13 we saw that the mechanism of enzyme catalysis is highly specific — an
enzyme can facilitate reactions only of substrates whose shapes fit its receptor site.
Two other biological processes that also require a high degree of specificity are smell
SMELL Camphor-like Floral Ethereal
+ Humans’ sense of smell, or olfaction (from the Latin, olere, to smell facere, to make),
is highly developed. We can recognize approximately 10,000 scents, including mercaptans (disagreeable-smelling compounds contained in skunk spray), which we can
detect at a concentration of one part per billion.
To be smelled, a substance must be volatile and it must be at least slightly soluble in water. Odorous molecules bind to specialized proteins, known as receptors, located on hairlike cilia at the rear of the nose. The binding of odors to these receptors
triggers an electrical signal that travels along nerve fibers from the olfactory bulb to
the brain, where the scent is identified.
Odors are generally divided into seven categories: musky (angelica root oil), camphor-like (moth repellent), floral (roses), pungent (vinegar), peppermint (mint candy),
ethereal (dry-cleaning fluid), and putrid (rotten eggs). A particular smell might fit just
one of these categories, or it might be perceived as a combination of two or more of
According to stereochemical theory, the mechanism of smell depends on the binding of an odor molecule to a specific receptor site, which resembles the lock-and-key
mechanism of enzyme catalysis. If two substances fit the same receptor, they should
have the same odor, even if they differ in chemical composition. There are seven different kinds of olfactory receptors, each of which will accept a molecule that has the
appropriate geometry. For putrid molecules such as hydrogen sulfide (H2S), however,
polarity is more important than shape in linking up with a receptor. In addition, if different portions of a molecule fit different receptors, the molecule should have a mixed
odor. For example, portions of benzaldehyde fit the camphor-like, floral, and peppermint receptors; we recognize the resulting aroma as almond.
Recent studies indicate that the stereochemical model of smell is fundamentally
correct, but reality is much more complicated. There are actually about a thousand odor
receptors, and so all the possible combinations of odorous molecules and receptors enable us to make subtle distinctions between different odors.
556 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Taste
receptor cell Taste
hairs Receptor Bitter Fructose Sour Salt Sensory
nerve Taste bud Sweet To brain Taste receptor cells within taste buds in different areas of the tongue
respond to the sweet, sour, bitter, or salty flavor molecules. One
saporous unit of fructose (C6H12O6), the sweetest kind of sugar, forms
hydrogen bonds with the receptor site on the sweet taste buds.
Saccharin contains a similar saporous unit (shaded area) and therefore
evokes an equivalent response from the sweet taste receptor cells. Saccharin TASTE The human tongue is studded with small conical bumps, or papillae, which house the
taste buds. Each taste bud consists of receptor cells, and extending from each receptor
cell are taste hairs with receptor sites for flavor molecules. The receptor cells can distinguish only four general flavors: sweet, salty, bitter, and sour. The areas of response
to these tastes are located on specific parts of the tongue.
Sour tastes are produced by the hydrogen ions in acids and salty tastes by the anions of salts (for example, chloride ions). Bitterness is due primarily to a class of compounds called alkaloids; examples are quinine, caffeine, and nicotine. Many substances
other than sugar evoke a sweet taste, including ethylene glycol (antifreeze), alcohols,
amino acids, and certain salts of lead and beryllium [for example, lead carbonate hydroxide (white lead), Pb3(OH)2(CO3)2]. (The sweet flavors of ethylene glycol and lead
paint are blamed for the unwitting consumption of these toxic substances by children
The exact mechanism by which a flavor molecule stimulates a taste cell is not
known, but a molecule must be water soluble to reach the taste buds. Molecules that
evoke taste are called sapid (from the Latin, sapere, to taste). Particular tastes are produced by saporous units of the molecules. The saporous unit for sweetness is called a
glucophore. It is believed that the glucophore binds to its receptor protein by forming
hydrogen bonds. Saccharin (C7H5O3NS), a synthetic substitute for sugar, contains a
glucophore similar to the ones in sugar; when the H atom that is bonded to the N atom
in saccharin is replaced with a methyl group (CH3), the sweetness is lost. White lead,
a sweet-tasting pigment used in white paint for many years has two OH groups that
can participate in hydrogen bonding.
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