chapter_14au.ppt

Chemical
Kinetics
Chapter 14
Chemical Kinetics
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten

Chemical
Kinetics
Kinetics
• Studies the rate at which a chemical
process occurs.
• Besides information about the speed at
which reactions occur, kinetics also
sheds light on the reaction mechanism
(exactly how the reaction occurs).

Chemical
Kinetics
Factors That Affect Reaction Rates
• Physical State of the Reactants
 In order to react, molecules must come in
contact with each other.
 The more homogeneous the mixture of
reactants, the faster the molecules can
react.

Chemical
Kinetics
• Concentration of Reactants
 As the concentration of reactants increases,
so does the likelihood that reactant
molecules will collide.

Chemical
Kinetics
• Temperature
 At higher temperatures, reactant
molecules have more kinetic energy,
move faster, and collide more often and
with greater energy.

Chemical
Kinetics
• Presence of a Catalyst SHOW MOVIE
 Catalysts speed up reactions by
changing the mechanism of the
reaction.
 Catalysts are not consumed during
the course of the reaction.

Chemical
Kinetics
Reaction Rates
Rates of reactions can be determined by
monitoring the change in concentration of
either reactants or products as a function of
time.

Chemical
Kinetics
Reaction Rates
In this reaction, the
concentration of
butyl chloride,
C4H9Cl, was
measured at various
times.
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

Chemical
Kinetics
Reaction Rates
The average rate of
the reaction over
each interval is the
change in
concentration divided
by the change in time:
Average rate =
[C4H9Cl]
t

Chemical
Kinetics
Reaction Rates
• Note that the average
rate decreases as the
reaction proceeds.
• This is because as the
reaction goes forward,
there are fewer
collisions between
reactant molecules.

Chemical
Kinetics
Reaction Rates
• A plot of concentration
vs. time for this reaction
yields a curve like this.
• The slope of a line
tangent to the curve at
any point is the
instantaneous rate at
that time.

Chemical
Kinetics
Reaction Rates
• All reactions slow down
over time.
• Therefore, the best
indicator of the rate of a
reaction is the
instantaneous rate near
the beginning.

Chemical
Kinetics
Reaction Rates and Stoichiometry
• In this reaction, the ratio
of C4H9Cl to C4H9OH is
1:1.
• Thus, the rate of
disappearance of
C4H9Cl is the same as
the rate of appearance
of C4H9OH.
Rate =
-[C4H9Cl]
t
=
[C4H9OH]
t

Chemical
Kinetics
• What if the ratio is not 1:1?
2 HI(g)  H2(g) + I2(g)
•Therefore,
Rate = − 1
2
[HI]
t
=
[I2]
t

Chemical
Kinetics
• To generalize, then, for the reaction
aA + bB cC + dD
Rate = −
1
a
[A]
t
= −
1
b
[B]
t
=
1
c
[C]
t
1
d
[D]
t
=

Chemical
Kinetics
Concentration and Rate
One can gain information about the rate
of a reaction by seeing how the rate
changes with changes in concentration.

Chemical
Kinetics
Comparing Experiments 1 and 2, when [NH4
+]
doubles, the initial rate doubles.
NH4
+(aq) + NO2
−(aq) N2(g) + 2 H2O(l)

Chemical
Kinetics
Likewise, comparing Experiments 5 and 6,
when [NO2
−] doubles, the initial rate doubles.
NH4
+(aq) + NO2
−(aq) N2(g) + 2 H2O(l)

Chemical
Kinetics
• This means
Rate  [NH4
+]
Rate  [NO2
−]
Rate  [NH+] [NO2
−]
or
Rate = k [NH4
+] [NO2
−]
• This equation is called the rate law, and
k is the rate constant.

Chemical
Kinetics
SAMPLE EXERCISE 14.1 Calculating an Average Rate of Reaction
From the data given in the caption of Figure 14.3, calculate the average rate at which A disappears over the
time interval from 20 s to 40 s.
PRACTICE EXERCISE
For the reaction pictured in Figure 14.3, calculate the average rate of appearance of B over the time interval from
0 to 40 s. (The necessary data are given in the figure caption.)
Answer: 1.8  10 –2 M/s
Solution
Analyze: We are given the concentration of A at 20 s (0.54 M) and at 40 s (0.30 M) and asked to calculate the
average rate of reaction over this time interval.
Plan: The average rate is given by the change in concentration, [A], divided by the corresponding change in
time, t.Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity.
Solve:

Chemical
Kinetics
SAMPLE EXERCISE 14.2 Calculating an Instantaneous Rate of Reaction
Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 (the initial rate).
PRACTICE EXERCISE
Using Figure 14.4, determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s.
Answer: 1.1  10 –4 M/s
Solution
Analyze: We are asked to determine an instantaneous rate from a graph of concentration versus time.
Plan: To obtain the instantaneous rate at t = 0 we must determine the slope of the curve at t = 0. The tangent is
drawn on the graph. The slope of this straight line equals the change in the vertical axis divided by the
corresponding change in the horizontal axis (that is, change in molarity over change in time).
Solve: The straight line falls from [C4H9Cl] = 0.100 M to 0.060 M in the time change from 0 s to 200 s, as
indicated by the tan triangle shown in Figure 14.4. Thus, the initial rate is

Chemical
Kinetics
(b) If the rate at which O2 appears, [O2] t, is 6.0  10–5 M/s at a particular instant, at what rate is O3
disappearing at this same time, –[O3] t?
SAMPLE EXERCISE 14.3 Relating Rates at Which Products Appear and
Reactants Disappear
(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction
Solution
Analyze: We are given a balanced chemical equation and asked to relate the rate of appearance of the product
to the rate of disappearance of the reactant.
Plan: We can use the coefficients in the chemical equation as shown in Equation 14.4 to express the relative
rates of reactions.
Solve: (a) Using the coefficients in the balanced equation and the relationship given by Equation 14.4, we
have:
(b) Solving the equation from part (a) for the rate at which O3 disappears, –[O3] t we have:
Check: We can directly apply a stoichiometric factor to convert the O2 formation rate to the rate at which
the O3 disappears:

Chemical
Kinetics
Answers: (a) 8.4  10 –7 M/s, (b) 2.1  10 –7 M/s
SAMPLE EXERCISE 14.3 continued
PRACTICE EXERCISE
The decomposition of N2O5 proceeds according to the following equation:
If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2  10–7 M/s, what is the rate
of appearance of (a) NO2, (b) O2?

Chemical
Kinetics
Rate Laws
• A rate law shows the relationship
between the reaction rate and the
concentrations of reactants.
• The exponents tell the order of the
reaction with respect to each reactant.
• This reaction is
First-order in [NH4
+]
First-order in [NO2
−]

Chemical
Kinetics
Rate Laws
• The overall reaction order can be found
by adding the exponents on the
reactants in the rate law.
• This reaction is second-order overall.

Chemical
Kinetics
Integrated Rate Laws
Using calculus to integrate the rate law
for a first-order process gives us
ln
[A]t
[A]0
= −kt
Where
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time, t,
during the course of the reaction.

Chemical
Kinetics
Integrated Rate Laws
Manipulating this equation produces…
ln
[A]t
[A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y = mx + b

Chemical
Kinetics
First-Order Processes
Therefore, if a reaction is first-order, a
plot of ln [A] vs. t will yield a straight
line, and the slope of the line will be -k.
ln [A]t = -kt + ln [A]0

Chemical
Kinetics
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC CH3CN

Chemical
Kinetics
This data was
collected for this
reaction at
198.9°C.
CH3NC CH3CN

Chemical
Kinetics
• When ln P is plotted as a function of time, a
straight line results.
• Therefore,
The process is first-order.
k is the negative slope: 5.1  10-5 s−1.

Chemical
Kinetics
Second-Order Processes
Similarly, integrating the rate law for a
process that is second-order in reactant
A, we get
1
[A]t
= kt +
1
[A]0
also in the form
y = mx + b

Chemical
Kinetics
So if a process is second-order in A, a
plot of 1/[A] vs. t will yield a straight line,
and the slope of that line is k.
1
[A]t
= kt +
1
[A]0

Chemical
Kinetics
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g) NO (g) + 1/2 O2 (g)
and yields data comparable to this:
Time (s) [NO2], M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380

Chemical
Kinetics
• Graphing ln [NO2] vs. t
yields:
Time (s) [NO2], M ln [NO2]
0.0 0.01000 −4.610
50.0 0.00787 −4.845
100.0 0.00649 −5.038
200.0 0.00481 −5.337
300.0 0.00380 −5.573
• The plot is not a straight
line, so the process is not
first-order in [A].

Chemical
Kinetics
• Graphing ln
1/[NO2] vs. t,
however, gives this
plot.
Time (s) [NO2], M 1/[NO2]
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263
• Because this is a
straight line, the
process is second-
order in [A].

Chemical
Kinetics
SAMPLE EXERCISE 14.4 Relating a Rate Law to the Effect of Concentration on Rate
Consider a reaction for which rate = k[A][B]2. Each of the following boxes represents a
reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of
increasing rate of reaction.
Solution
Analyze: We are given three boxes containing different numbers of spheres representing mixtures containing
different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to
rank the mixtures in order of increasing reaction rates.
Plan: Because all three boxes have the same volume, we can put the number of spheres of each kind into the
rate law and calculate the rate for each box.
Solve: Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate:
Box 2 contains 7 red spheres and 3 purple spheres:

Chemical
Kinetics
PRACTICE EXERCISE
Assuming that rate = k[A][B], rank the mixtures represented in this Sample Exercise in order of increasing rate.
Answer: 2 = 3 < 1
Box 3 contains 3 red spheres and 7 purple spheres:
The slowest rate is 63k (box 2), and the highest is 147k (box 3). Thus, the rates vary in the order 2 < 1 < 3.
Check: Each box contains 10 spheres. The rate law indicates that in this case [B] has a greater influence on
rate than [A] because B has a higher reaction order. Hence, the mixture with the highest concentration of B
(most purple spheres) should react fastest. This analysis confirms the order 2 < 1 < 3.

Chemical
Kinetics
SAMPLE EXERCISE 14.5 Determining Reaction Orders and Units for Rate Constants
(a)What are the overall reaction orders for the rate laws described in Equations 14.9 and 14.10? (b) What are
the units of the rate constant for the rate law for Equation 14.9?
PRACTICE EXERCISE 14.11H2 + I2  2 HI Rate = k [H2] [I2]
(a) What is the reaction order of the reactant H2 in Equation 14.11? (b) What are the units of the rate constant
for Equation 14.11?
Answers: (a) 1, (b) M–1 s–1
Solution 14.10 CHCl3 + Cl2  CCl4 + HCl Rate = k[CHCl3] [Cl2]^.5
14.9 2N2O5  4 NO2 + O2 Rate = k [N2O5]
Analyze: We are given two rate laws and asked to express (a) the overall reaction order for each and (b) the
units for the rate constant for the first reaction.
Plan: The overall reaction order is the sum of the exponents in the rate law. The units for the rate constant, k,
are found by using the normal units for rate (M/s) and concentration (M) in the rate law and applying algebra to
solve for k.
Solve: (a) The rate of the reaction in Equation 14.9 is first order in N2O5 and first order overall. The reaction
in Equation 14.10 is first order in CHCl3 and one-half order in Cl2. The overall reaction order is three halves.
(b) For the rate law for Equation 14.9, we have
So
Notice that the units of the rate constant change as the overall order of the reaction changes.

Chemical
Kinetics
SAMPLE EXERCISE 14.6 Determining a Rate Law from Initial Rate Data
The initial rate of a reaction was measured for several different starting concentrations of A and
B, and the results are as follows:
Using these data, determine (a) the rate law for the reaction, (b) the magnitude of the rate constant, (c) the rate
of the reaction when [A] = 0.050 M and [B] = 0.100 M.
Solution
Analyze: We are given a table of data that relates concentrations of reactants with initial rates of reaction and
asked to determine
(a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentrations not listed in the table.
Plan: (a) We assume that the rate law has the following form: Rate = k[A]m[B]n, so we must use the given data
to deduce the reaction orders m and n. We do so by determining how changes in the concentration change the
rate. (b) Once we know m and n, we can use the rate law and one of the sets of data to determine the rate
constant k. (c) Now that we know both the rate constant and the reaction orders, we can use the rate law with the
given concentrations to calculate rate.
Solve: (a) As we move from experiment 1 to experiment 2, [A] is held constant and [B] is doubled. Thus, this
pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with respect
to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the
reaction rate. The rate law is therefore zero order in B (that is, n = 0).

Chemical
Kinetics
In experiments 1 and 3, [B] is held constant so these data show how [A] affects rate. Holding [B] constant while
doubling [A] increases the rate fourfold. This result indicates that rate is proportional to [A]2 (that is, the reaction
is second order in A). Hence, the rate law is
This rate law could be reached in a more formal way by taking the ratio of the rates from two experiments:
Using the rate law, we have
2n equals 1 under only one condition:
We can deduce the value of m in a similar fashion
Using the rate law gives

Chemical
Kinetics
Because 2m = 4, we conclude that
(b) Using the rate law and the data from experiment 1, we have
(c) Using the rate law from part (a) and the rate constant from part (b), we have
Because [B] is not part of the rate law, it is irrelevant to the rate, provided that there is at least some B present to
react with A.
Check: A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see
if we can correctly calculate the rate. Using data from experiment 3, we have
Thus, the rate law correctly reproduces the data, giving both the correct number and the correct
units for the rate.

Chemical
Kinetics
PRACTICE EXERCISE
The following data were measured for the reaction of nitric oxide with hydrogen:
Answers: (a) rate = k[NO]2[H2]; (b) k =1.2 M–2s–1; (c) rate = 4.5  10–4 M/s
(a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when
[NO] = 0.050 M and [H2] = 0.150 M.

Chemical
Kinetics
SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1
at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0  10–7
g/cm3. Assume that the average temperature of the lake is 12°C. (a) What is the concentration of the insecticide
on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to drop to
3.0  10–7 g/cm3?
Solution
Analyze: We are given the rate constant for a reaction that obeys first-order kinetics, as well as information
about concentrations and times, and asked to calculate how much reactant (insecticide) remains after one year.
We must also determine the time interval needed to reach a particular insecticide concentration. Because the
exercise gives time in (a) and asks for time in (b), we know that the integrated rate law, Equation 14.13, is
required.
Plan: (a) We are given k = 1.45 yr–1, t = 1.00 yr, and [insecticide]0 = 5.0  10–7 g/cm3, and so Equation 14.13
can be solved for 1n[insecticide]t. (b) We have k = 1.45yr–1, [insecticide]0 = 5.0  10–7 g/cm3, and [insecticide]t
= 3.0  10–7 g/cm3, and so we can solve Equation 14.13 for t.
Solve: (a) Substituting the known quantities into Equation 14.13, we have
We use the ln function on a calculator to evaluate the second term on the right, giving
To obtain [insecticide]t = 1 yr, we use the inverse natural logarithm, or ex, function on the calculator:
Note that the concentration units for [A]t and [A]0 must be the same.

Chemical
Kinetics
Answer: 51 torr
Check: In part (a) the concentration remaining after 1.00 yr (that is,1.2  10–7 g/cm3) is less than the original
concentration (5.0  10–7 g/cm3), as it should be. In (b) the given concentration (3.0  10–7 g/cm3) is greater
than that remaining after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is a
reasonable answer.
(b) Again substituting into Equation 14.13, with [insecticide]t = 3.0  10–7 g/cm3, gives
Solving for t gives
PRACTICE EXERCISE
The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of
6.8  10–4s–1:
If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s?

Chemical
Kinetics
Solution
Analyze: We are given the concentrations of a reactant at various times during a reaction and asked to
determine whether the reaction is first or second order.
Plan: We can plot ln[NO2] and 1/[NO2] against time. One or the other will be linear, indicating whether the
reaction is first or second order.
Is the reaction first or second order in NO2?
SAMPLE EXERCISE 14.8 Determining Reaction Order from the Integrated Rate Law
The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300°C,

Chemical
Kinetics
Solve: In order to graph ln[NO2] and 1/[NO2] against time, we will first prepare the following table from
the data given:

Chemical
Kinetics
PRACTICE EXERCISE
Consider again the decomposition of NO2 discussed in the Sample Exercise. The reaction is second order in NO2
with k = 0.543 M–1s–1. If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining
concentration after 0.500 h?
Answer: Using Equation 14.14, we find [NO2] = 1.00  10–3 M
As Figure 14.8 shows, only the plot of 1/[NO2] versus time is linear. Thus, the reaction obeys a second-order rate
law: Rate = k[NO2]2. From the slope of this straight-line graph, we determine that k = 0.543 M–1 s–1 for the
disappearance of NO2.
Figure 14.8 Kinetic
data for
decomposition of NO2.
The reaction is NO2(g)
NO(g)
+ 1/2O2(g), and the data
were collected at
300°C. (a) A plot of
[NO2] versus time is not
linear, indicating that the
reaction is not first order
in NO2. (b) A plot of
1/[NO2] versus time is
linear, indicating that the
reaction is second order
in NO2.

Chemical
Kinetics
Half-Life
• Half-life is defined
as the time required
for one-half of a
reactant to react.
• Because [A] at t1/2 is
one-half of the
original [A],
[A]t = 0.5 [A]0.

Chemical
Kinetics
Half-Life
For a first-order process, this becomes
0.5 [A]0
[A]0
ln = −kt1/2
ln 0.5 = −kt1/2
−0.693 = −kt1/2
= t1/2
0.693
k
NOTE: For a first-order
process, the half-life
does not depend on [A]0.

Chemical
Kinetics
Half-Life
For a second-order process,
1
0.5 [A]0
= kt1/2 +
1
[A]0
2
[A]0
= kt1/2 +
1
[A]0
2 − 1
[A]0
= kt1/2
1
[A]0
=
= t1/2
1
k[A]0

Chemical
Kinetics
SAMPLE EXERCISE 14.9 Determining the Half-life of a First-Order Reaction
The reaction of C4H9Cl with water is a first-order reaction. Figure 14.4 shows how the concentration of C4H9Cl
changes with time at a particular temperature. (a) From that graph, estimate the half-life for this reaction. (b)
Use the half-life from (a) to calculate the rate constant.
PRACTICE EXERCISE
(a) Using Equation 14.15, calculate t1/2 for the decomposition of the insecticide described in Sample Exercise
14.7. (b) How long does it take for the concentration of the insecticide to reach one-quarter of the initial value?
Answers: (a) 0.478 yr = 1.51  10–7 s; (b) it takes two half-lives, 2(0.478 yr) = 0.956 yr
Check: At the end of the second half-life, which should occur at 680 s, the concentration should have
decreased by yet another factor of 2, to 0.025 M. Inspection of the graph shows that this is indeed the case.
Solution
Analyze: We are asked to estimate the half-life of a reaction from a graph of concentration versus time and
then to use the half-life to calculate the rate constant for the reaction.
Plan: (a) To estimate a half-life, we can select a concentration and then determine the time required for the
concentration to decrease to half of that value. (b) Equation 14.15 is used to calculate the rate constant from
the half-life.
Solve: (a) From the graph, we see that the initial value of [C4H9Cl] is 0.100 M. The half-life for this first-order
reaction is the time required for [C4H9Cl] to decrease to 0.050 M, which we can read off the graph. This point
occurs at approximately 340 s.
(b) Solving Equation 14.15 for k, we have

Chemical
Kinetics
Temperature and Rate
• Generally, as temperature
increases, so does the
reaction rate.
• This is because k is
temperature dependent.

Chemical
Kinetics
The Collision Model
• In a chemical reaction, bonds are
broken and new bonds are formed.
• Molecules can only react if they collide
with each other.

Chemical
Kinetics
The Collision Model
Furthermore, molecules must collide with the
correct orientation and with enough energy to
cause bond breakage and formation.

Chemical
Kinetics
Activation Energy
• In other words, there is a minimum amount of energy
required for reaction: the activation energy, Ea.
• Just as a ball cannot get over a hill if it does not roll
up the hill with enough energy, a reaction cannot
occur unless the molecules possess sufficient energy
to get over the activation energy barrier.

Chemical
Kinetics
Reaction Coordinate Diagrams
It is helpful to
visualize energy
changes
throughout a
process on a
reaction coordinate
diagram like this
one for the
rearrangement of
methyl isonitrile.

Chemical
Kinetics
Reaction Coordinate Diagrams
• It shows the energy of
the reactants and
products (and,
therefore, E).
• The high point on the
diagram is the transition
state.
• The species present at the transition state is
called the activated complex.
• The energy gap between the reactants and the
activated complex is the activation energy
barrier.

Chemical
Kinetics
Maxwell–Boltzmann Distributions
• Temperature is
defined as a
measure of the
average kinetic
energy of the
molecules in a
sample.
• At any temperature there is a wide
distribution of kinetic energies.

Chemical
Kinetics
• As the temperature
increases, the curve
flattens and
broadens.
• Thus at higher
temperatures, a
larger population of
molecules has
higher energy.

Chemical
Kinetics
• If the dotted line represents the activation
energy, as the temperature increases, so does
the fraction of molecules that can overcome
the activation energy barrier.
• As a result, the
reaction rate
increases.

Chemical
Kinetics
This fraction of molecules can be found through the
expression
where R is the gas constant and T is the Kelvin
temperature.
f = e−Ea/RT

Chemical
Kinetics
Solution The lower the activation energy, the faster the reaction. The value of  does not affect the rate.
Hence the order is (2) < (3) < (1).
PRACTICE EXERCISE
Imagine that these reactions are reversed. Rank these reverse reactions from slowest to fastest.
Answer: (2) < (1) < (3) because Ea values are 40, 25, and 15 kJ/mol, respectively
SAMPLE EXERCISE 14.10 Relating Energy Profiles to Activation Energies and
Speeds of Reaction
Consider a series of reactions having the following energy profiles:
Assuming that all three reactions have nearly the same frequency factors, rank the reactions from slowest to
fastest.

Chemical
Kinetics
Arrhenius Equation
Svante Arrhenius developed a mathematical
relationship between k and Ea:
k = A e−Ea/RT
where A is the frequency factor, a number that
represents the likelihood that collisions would
occur with the proper orientation for reaction.

Chemical
Kinetics
Arrhenius Equation
Taking the natural
logarithm of both
sides, the equation
becomes
ln k = -Ea ( ) + ln A
1
RT
y = mx + b
Therefore, if k is determined experimentally at
several temperatures, Ea can be calculated
from the slope of a plot of ln k vs. 1/T.

Chemical
Kinetics
Solution
Analyze: We are given rate constants, k, measured at several temperatures and asked to determine the
activation energy, Ea, and the rate constant, k, at a particular temperature.
Plan: We can obtain Ea from the slope of a graph of ln k versus 1/T. Once we know Ea, we can use Equation
4.21 together with the given rate data to calculate the rate constant at 430.0 K.
SAMPLE EXERCISE 14.11 Determining the Energy of Activation
The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures
(these are the data in Figure 14.12):
(a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at
430.0 K?

Chemical
Kinetics
Solve: (a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of
each temperature, 1/T, and the natural log of each rate constant, ln k. This gives us the table shown at the right:
A graph of ln k versus 1/T results in a straight line, as shown in Figure 14.17.
Figure 14.17 Graphical
determination of activation energy.
The natural logarithm of the rate
constant for the rearrangement of
methyl isonitrile is plotted as a
function of 1/T. The linear relationship
is predicted by the Arrhenius equation
giving a slope equal to – Ea/R.

Chemical
Kinetics
We report the activation energy to only two significant figures because we are limited by the precision with
which we can read the graph in Figure 14.17.
The slope of the line is obtained by choosing two well-separated points, as shown, and using the coordinates of
each:
Because logarithms have no units, the numerator in this equation is dimensionless. The denominator has the
units of 1/T, namely, K–1. Thus, the overall units for the slope are K. The slope equals –Ea/R. We use the value
for the molar gas constant R in units of J/mol-K (Table 10.2). We thus obtain
(b) To determine the rate constant, k1, at T = 430.0 K, we can use Equation 14.21 with Ea = 160 kJ/ mol, and one
of the rate constants and temperatures from the given data, such as k2 = 2.52  10–5s–1 and T2 = 462.9 

Chemical
Kinetics
PRACTICE EXERCISE
Using the data in Sample Exercise 14.11, calculate the rate constant for the rearrangement of methyl isonitrile
at 280°C.
Answer: 2.2  10–2s–1
Thus,
Note that the units of k1 are the same as those of k2.

Chemical
Kinetics
Reaction Mechanisms
The sequence of events that describes
the actual process by which reactants
become products is called the reaction
mechanism.

Chemical
Kinetics
Reaction Mechanisms
• Reactions may occur all at once or
through several discrete steps.
• Each of these processes is known as an
elementary reaction or elementary
process.

Chemical
Kinetics
Reaction Mechanisms
The molecularity of a process tells how many
molecules are involved in the process.

Chemical
Kinetics
Multistep Mechanisms
• In a multistep process, one of the steps will
be slower than all others.
• The overall reaction cannot occur faster than
this slowest, rate-determining step.

Chemical
Kinetics
Slow Initial Step
• The rate law for this reaction is found
experimentally to be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but the
rate of the reaction does not depend on its
concentration.
• This suggests the reaction occurs in two steps.
NO2 (g) + CO (g)  NO (g) + CO2 (g)

Chemical
Kinetics
Slow Initial Step
• A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
• The NO3 intermediate is consumed in the second
step.
• As CO is not involved in the slow, rate-determining
step, it does not appear in the rate law.

Chemical
Kinetics
Fast Initial Step
• The rate law for this reaction is found to
be
Rate = k [NO]2 [Br2]
• Because termolecular processes are
rare, this rate law suggests a two-step
mechanism.
2 NO (g) + Br2 (g)  2 NOBr (g)

Chemical
Kinetics
Fast Initial Step
• A proposed mechanism is
Step 2: NOBr2 + NO  2 NOBr (slow)
Step 1 includes the forward and reverse reactions.
Step 1: NO + Br2 NOBr2 (fast)

Chemical
Kinetics
Fast Initial Step
• The rate of the overall reaction depends
upon the rate of the slow step.
• The rate law for that step would be
Rate = k2 [NOBr2] [NO]
• But how can we find [NOBr2]?

Chemical
Kinetics
Fast Initial Step
• NOBr2 can react two ways:
With NO to form NOBr
By decomposition to reform NO and Br2
• The reactants and products of the first
step are in equilibrium with each other.
• Therefore,
Ratef = Rater

Chemical
Kinetics
Fast Initial Step
• Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
• Solving for [NOBr2] gives us
k1
k−1
[NO] [Br2] = [NOBr2]

Chemical
Kinetics
Fast Initial Step
Substituting this expression for [NOBr2]
in the rate law for the rate-determining
step gives
k2k1
k−1
Rate = [NO] [Br2] [NO]
= k [NO]2 [Br2]

Chemical
Kinetics
SAMPLE EXERCISE 14.12 Determining Molecularity and Identifying Intermediates
It has been proposed that the conversion of ozone into O2 proceeds by a two-step mechanism:
(a) Describe the molecularity of each elementary reaction in this mechanism. (b) Write the equation for the
overall reaction. (c) Identify the intermediate(s).
Solution
Analyze: We are given a two-step mechanism and asked for (a) the molecularities of each of the two
elementary reactions, (b) the equation for the overall process, and (c) the intermediate.
Plan: The molecularity of each elementary reaction depends on the number of reactant molecules in the
equation for that reaction. The overall equation is the sum of the equations for the elementary reactions. The
intermediate is a substance formed in one step of the mechanism and used in another and therefore not part of
the equation for the overall reaction.
Solve: (a) The first elementary reaction involves a single reactant and is consequently unimolecular. The
second reaction, which involves two reactant molecules, is bimolecular.
(b) Adding the two elementary reactions gives
Because O(g) appears in equal amounts on both sides of the equation, it can be eliminated to give the net
equation for the chemical process:
(c) The intermediate is O(g). It is neither an original reactant nor a final product, but is formed in the first
step of the mechanism and consumed in the second.

Chemical
Kinetics
Answers: (a) Yes, the two equations add to yield the equation for the reaction. (b) The first elementary reaction
is unimolecular, and the second one is bimolecular. (c) Mo(CO)5
(a) Is the proposed mechanism consistent with the equation for the overall reaction? (b) What is the molecularity
of each step of the mechanism? (c) Identify the intermediate(s).
PRACTICE EXERCISE
For the reaction
the proposed mechanism is

Chemical
Kinetics
SAMPLE EXERCISE 14.13 Predicting the Rate Law for an Elementary Reaction
If the following reaction occurs in a single elementary reaction, predict the rate law:
Comment: Experimental studies of this reaction show that the reaction actually has a very different rate law:
Rate = k[H2][Br2]1/2
Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we
can conclude that the mechanism must involve two or more elementary steps.
Answers: (a) Rate = k[NO]2[Br2] (b) No, because termolecular reactions are very rare
PRACTICE EXERCISE
Consider the following reaction: (a) Write the rate law for the reaction,
assuming it involves a single elementary reaction. (b) Is a single-step mechanism likely for this reaction?
Solution
Analyze: We are given the equation and asked for its rate law, assuming that it is an elementary process.
Plan: Because we are assuming that the reaction occurs as a single elementary reaction, we are able to write
the rate law using the coefficients for the reactants in the equation as the reaction orders.
Solve: The reaction is bimolecular, involving one molecule of H2 with one molecule of Br2. Thus, the rate law
is first order in each reactant and second order overall:
Rate = k[H2][Br2]

Chemical
Kinetics
(a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction.
(b) The rate law for the overall reaction is just the rate law for the slow, rate-determining elementary reaction.
Because that slow step is a unimolecular elementary reaction, the rate law is first order:
Rate = k[N2O]
SAMPLE EXERCISE 14.14 Determining the Rate Law for a Multistep Mechanism
The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism:
Solution
Analyze: Given a multistep mechanism with the relative speeds of the steps, we are asked to write the overall
reaction and the rate law for that overall reaction.
Plan: (a) The overall reaction is found by adding the elementary steps and eliminating the intermediates.
(b) The rate law for the overall reaction will be that of the slow, rate-determining step.
Solve: (a) Adding the two elementary reactions gives
Omitting the intermediate, O(g), which occurs on both sides of the equation, gives the overall reaction:

Chemical
Kinetics
Answer: Because the rate law conforms to the molecularity of the first step, that must be the rate-determining
step. The second step must be much faster than the first one.
The experimental rate law is rate = k[O3][NO2]. What can you say about the relative rates of the two steps of the
mechanism?
PRACTICE EXERCISE
Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen:
The reaction is believed to occur in two steps

Chemical
Kinetics
Solution
Analyze: We are given a mechanism with a fast initial step and asked to write the rate law for the overall
reaction.
Plan: The rate law of the slow elementary step in a mechanism determines the rate law for the overall
reaction. Thus, we first write the rate law based on the molecularity of the slow step. In this case the slow step
involves the intermediate N2O2 as a reactant. Experimental rate laws, however, do not contain the concentrations
of intermediates, but are expressed in terms of the concentrations of starting substances. Thus, we must relate the
concentration of N2O2 to the concentration of NO by assuming that an equilibrium is established in the first step.
Solve: The second step is rate determining, so the overall rate is
Rate = k2[N2O2][Br2]
SAMPLE EXERCISE 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step
Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the
experimentally observed one:
We solve for the concentration of the intermediate N2O2 by assuming that an equilibrium is established in step 1;
thus, the rates of the forward and reverse reactions in step 1 are equal:

Chemical
Kinetics
What is the expression relating the concentration of Br(g) to that of Br2(g)?
Substituting this expression into the rate expression gives
Thus, this mechanism also yields a rate law consistent with the experimental one.
PRACTICE EXERCISE
The first step of a mechanism involving the reaction of bromine is

Chemical
Kinetics
Catalysts
• Catalysts increase the rate of a reaction by
decreasing the activation energy of the
reaction.
• Catalysts change the mechanism by which
the process occurs.

Chemical
Kinetics
Catalysts
One way a
catalyst can
speed up a
reaction is by
holding the
reactants together
and helping bonds
to break.

Chemical
Kinetics
Enzymes
• Enzymes are
catalysts in
biological systems.
• The substrate fits
into the active site of
the enzyme much
like a key fits into a
lock.

Chemical
Kinetics
The decomposition reaction is determined to be first order. A graph of the partial pressure of HCOOH versus
time for decomposition at 838 K is shown as the red curve in Figure 14.28. When a small amount of solid
ZnO is added to the reaction chamber, the partial pressure of acid versus time varies as shown by the blue curve
in Figure 14.28.
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
Formic acid (HCOOH) decomposes in the gas phase at elevated temperatures as follows:
(a) Estimate the half-life and first-order rate constant for formic acid decomposition.
(b) What can you conclude from the effect of added ZnO on the decomposition of formic acid?
Figure 14.28 Variation in
pressure of HCOOH(g) as
a function of time at 838
K. The red line
corresponds to
decomposition when only
gaseous HCOOH is
present. The blue line
corresponds to
decomposition in the
presence of added ZnO(s).

Chemical
Kinetics
Solution (a) The initial pressure of HCOOH is 3.00  102 torr. On the graph we move to the level at
which the partial pressure of HCOOH is 150 torr, half the initial value. This corresponds to a time of about
6.60 x 102s, which is therefore the half-life. The first-order rate constant is given by Equation 14.15:
k = 0.693/t1/2 = 0.693/660 s = 1.05  10–3 s–1.
(c) The progress of the reaction was followed by measuring the partial pressure of formic acid vapor at
selected times. Suppose that, instead, we had plotted the concentration of formic acid in units of mol/L. What
effect would this have had on the calculated value of k?
(d) The pressure of formic acid vapor at the start of the reaction is 3.00  10 2 torr. Assuming constant
temperature and ideal-gas behavior, what is the pressure in the system at the end of the reaction? If the volume
of the reaction chamber is 436 cm3, how many moles of gas occupy the reaction chamber at the end of the
reaction?
(e) The standard heat of formation of formic acid vapor is Calculate Hº for the
overall reaction. Assuming that the activation energy (Ea) for the reaction is 184 kJ/mol, sketch an approximate
energy profile for the reaction, and label Ea, Hº, and the transition state.
SAMPLE INTEGRATIVE EXERCISE continued
(b) The reaction proceeds much more rapidly in the presence of solid ZnO, so the surface of the oxide must
be acting as a catalyst for the decomposition of the acid. This is an example of heterogeneous catalysis.
(d) According to the stoichiometry of the reaction, two moles of product are formed for each mole of
reactant. When reaction is completed, therefore, the pressure will be 600 torr, just twice the initial pressure,
assuming ideal-gas behavior. (Because we are working at quite high temperature and fairly low gas pressure,
assuming ideal-gas behavior is reasonable.) The number of moles of gas present can be calculated using the
ideal-gas equation (Section 10.4):
(c) If we had graphed the concentration of formic acid in units of moles per liter, we would still have
determined that the half-life for decomposition is 660 seconds, and we would have computed the same value for
k. Because the units for k are s–1, the value for k is independent of the units used for concentration.

Chemical
Kinetics
From this and the given value for Ea, we can draw an approximate energy profile for the reaction, in analogy to
Figure 14.15.
SAMPLE INTEGRATIVE EXERCISE continued
(e) We first calculate the overall change in energy, Hº (Section 5.7 and Appendix C), as in

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