Chapter 13 : Chemical Kinetics

Terms in this set (17)

Kids (including my own) who live in my neighborhood have a unique way of catching
lizards. Armed with cups of ice water, they chase the cold-blooded reptiles into a corner,
and then take aim and fire—or pour, actually. They pour the cold water directly onto the
lizard's body. The lizard's body temperature drops, and it becomes virtually immobilized—
easy prey for little hands. The kids scoop up the lizard and place it in a tub filled with
sand and leaves. They then watch as the lizard warms back up and becomes active again.
They usually release the lizard back into the yard within hours. I guess you could call
them catch-and-release lizard hunters.
Unlike mammals, which actively regulate their body temperature through metabolic
activity, lizards are ectotherms—their body temperature depends on their surroundings.
When splashed with cold water, a lizard's body simply gets colder. The drop in body
temperature slows down the lizard because its movement depends on chemical reactions
that occur within its muscles, and the rates of those reactions—how fast they occur—are
highly sensitive to temperature. When the temperature drops, the reactions that produce
movement occur more slowly; therefore, the movement itself slows down. Cold reptiles
are lethargic, unable to move very quickly. For this reason, reptiles move between sun
and shade to regulate their body temperature.
The rates of chemical reactions, and especially the ability to control those rates, are
important not just in reptile movement but in many other phenomena as well. For example, the launching of a rocket depends on controlling the rate at which fuel burns—too
quickly and the rocket can explode, too slowly and it will not leave the ground. Chemists
must always consider reaction rates when synthesizing compounds. No matter how stable
a compound might be, its synthesis is impossible if the rate at which it forms is too slow.
As we have seen with reptiles, reaction rates are important to life. In fact, the human
body's ability to switch a specific reaction on or off at a specific time is achieved largely
by controlling the rate of that reaction through the use of enzymes (biological molecules
that we explore more fully in Section 13.7).
The knowledge of reaction rates is not only practically important—giving us the
ability to control how fast a reaction occurs—but also theoretically important. As we will
discuss in Section 13.6, knowledge of the rate of a reaction can tell us much about how
the reaction occurs on the molecular scale.
The rate of a chemical reaction is a measure of how fast the reaction occurs. If a chemical
reaction has a slow rate, only a relatively small fraction of molecules react to form products in a given period of time. If a chemical reaction has a fast rate, a large fraction of
molecules react to form products in a given period of time.
When we measure how fast something occurs, or more specifically the rate at which
it occurs, we usually express the measurement as a change in some quantity per unit of
time. For example, we measure the speed of a car—the rate at which it travels—in miles
per hour, and we measure how quickly (or slowly) people lose weight in pounds per
week. Notice that both of these rates are reported in units that represent the change in
what we are measuring (distance or weight) divided by the change in time.
Speed = change in distance
change in time = ∆x
∆t
Weight loss = change in weight
change in time = ∆ weight
∆t
Similarly, the rate of a chemical reaction is measured as a change in the amounts of reactants or products (usually in terms of concentration) divided by the change in time.
Consider the following gas-phase reaction between H2(g) and I2(g):
H2(g) + I2(g) S 2 HI(g)
We define the rate of this reaction in the time interval t1 to t2 as follows:
Rate = - ∆[H2]
∆t = -
[H2]t2 - [H2]t1
t2 - t1
In this expression, [H2]t2
is the hydrogen concentration at time t2 and [H2]t1
is the hydrogen concentration at time t1. Notice that the reaction rate is defined as the negative of the
change in concentration of a reactant divided by the change in time. The negative sign is
part of the definition when we define the reaction rate with respect to a reactant because
reactant concentrations decrease as a reaction proceeds; therefore, the change in the concentration of a reactant is negative. The negative sign thus makes the overall rate positive. (By convention, we report reaction rates as positive quantities.)
We can also define the reaction rate with respect to the other reactant:
Rate = - ∆[I2]
∆t
Since 1 mol of H2 reacts with 1 mol of I2, we define the rate in the same way. We can also
define the rate with respect to the product of the reaction:
Rate = +
1
2
∆[HI]
∆t
Notice that, because product concentrations increase as the reaction proceeds, the change
in concentration of a product is positive. Therefore, when we define the rate with respect
to a product, we do not include a negative sign in the definition—the rate is naturally
positive. Notice also the factor of 1
2 in this definition. This factor is related to the stoichiometry of the reaction. In order to have a single rate for the entire reaction, the definition
of the rate with respect to each reactant and product must reflect the stoichiometric coefficients of the reaction. For this particular reaction, 2 mol of HI are produced from 1 mol
of H2 and 1 mol of I2.
Therefore the concentration of HI increases at twice the rate that the concentration of H2
or I2 decreases. If 100 I2 molecules react per second, then 200 HI molecules form per
second. In order for the overall rate to have the same value when defined with respect to
any of the reactants or products, we must multiply the change in HI concentration by a
factor of one-half.
Consider the graph shown in FiguRe 13.1▲, which represents the changes in concentration for H2 (one of the reactants) and HI (the product) versus time.
Let's examine several features of this graph individually.
Change in Reactant and Product Concentrations
The reactant concentration, as expected, decreases with time because reactants are consumed in a reaction. The product concentration increases with time because products are
formed in a reaction. The increase in HI concentration occurs at exactly twice the rate of
the decrease in H2 concentration because of the stoichiometry of the reaction—2 mol of
HI is formed for every 1 mol of H2 consumed.
The Average Rate of the Reaction
We can calculate the average rate of the reaction for any time interval using Equation
13.1. The table shown here lists H2 concentration ([H2]) at various times, the change in
H2 concentration for each interval (∆[H2]), the change in time for each interval (∆t), and
the rate for each interval (-∆[H2]>∆t).
The rate calculated in this way represents the average rate within the given time interval.
For example, the average rate of the reaction in the time interval between 10 and
20 seconds is 0.0149 M>s, while the average rate in the time interval between 20 and
30 seconds is 0.0121 M>s. Notice that the average rate decreases as the reaction progresses. In other words, the reaction slows down as it proceeds. We discuss this further in
the next section where we will see that, for most reactions, the rate depends on the concentrations of the reactants. As the reactants are consumed, their concentrations decrease,
and the reaction slows down.
The Instantaneous Rate of the Reaction
The instantaneous rate of the reaction is the rate at any one point in time, represented by the
instantaneous slope of the curve at that point. We can obtain the instantaneous rate from
the slope of the tangent to the curve at the point of interest. In our graph (Figure 13.1 on
the previous page), we have drawn the tangent lines for both [H2] and [HI] at
50 seconds.
We calculate the instantaneous rate at 50 seconds as follows:
Instantaneous rate (at 50 s) = - ∆[H2]
∆t = - -0.28 M
40 s = 0.0070 M>s
Instantaneous rate (at 50 s) = +
1
2
∆[HI]
∆t = +
1
2
(0.56 M)
40 s = 0.0070 M>s
Notice that, as expected, the rate is the same whether we use one of the reactants or the
product for the calculation. Notice also that the instantaneous rate at 50 seconds
(0.0070 M>s) lies between the average rates calculated for the 10-second intervals just
before and just after 50 seconds.
We can generalize our definition of reaction rates for the generic reaction:
aA + bB S cC + dD
where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients. We then define the rate of the reaction as follows:
Rate = - 1
a
∆[A]
∆t = - 1
b
∆[B]
∆t = +
1
c
∆[C]
∆t = +
1
d
∆[D]
∆t
Knowing the rate of change in the concentration of any one reactant or product at a
point in time allows us to determine the rate of change in the concentration of any
other reactant or product at that point in time (from the balanced equation). However,
predicting the rate at some future time is not possible from just the balanced
equation.
ExAmPlE 13.1 Expressing Reaction Rates
Consider the balanced chemical equation.
H2O2(aq) + 3 I-(aq) + 2 H+(aq) S I3
-(aq) + 2 H2O(l)
In the first 10.0 seconds of the reaction, the concentration of I
- drops from 1.000 M to 0.868 M.
(a) Calculate the average rate of this reaction in this time interval.
(b) Predict the rate of change in the concentration of H+ (that is, ∆[H+]>∆t) during this time interval.
SOLUTION
(a) Use Equation 13.5 to calculate the average rate of the reaction.
(b) Use Equation 13.5 again to determine the relationship between the
rate of the reaction and ∆[H+]>∆t. After solving for ∆[H+]>∆t, substitute the calculated rate from part a and calculate ∆[H+]>∆t.
The rate of a reaction often depends on the concentration of one or more of the reactants.
In 1850, chemist Ludwig Wilhelmy (1812-1864) noticed this effect for the hydrolysis of
sucrose. For simplicity, consider a reaction in which a single reactant, A, decomposes
into products:
As long as the rate of the reverse reaction (the reaction in which the products return to
reactants) is negligibly slow, we can express the relationship—called the rate law—
between the rate of the reaction and the concentration of the reactant as follows:
Rate = k[A]n
where k is a constant of proportionality called the rate constant and n is a number called
the reaction order. The value of n determines how the rate depends on the concentration
of the reactant.
- If n = 0, the reaction is zero order, and the rate is independent of the concentration
of A.
- If n = 1, the reaction is first order, and the rate is directly proportional to the concentration of A
- If n = 2, the reaction is second order, and the rate is proportional to the square of
the concentration of A.
Although other orders are possible, including noninteger (or fractional) orders, these
three are the most common
Zero-Order Reaction
In a zero-order reaction, the rate of the reaction is independent of the concentration of the
reactant.
Rate = k[A]0 = k
Consequently, for a zero-order reaction, the concentration of the reactant decreases linearly with time, as shown in Figure 13.2. Notice the constant slope of the line labeled
zero order in the plot—a constant slope indicates a constant rate. The rate is constant
because the reaction does not slow down as the concentration of A decreases. The graph
in Figure 13.3 shows that the rate of a zero-order reaction is the same at any concentration of A. Zero-order reactions occur under conditions where the amount of reactant actually available for reaction is unaffected by changes in the overall quantity of reactant. For example, sublimation is normally zero order because only molecules at the surface of a substance can sublime, and their concentration does not change as the amount of
subliming substance decreases
First-Order Reaction
In a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant.
Rate = k[A]1
Consequently, for a first-order reaction, the rate slows down as the reaction proceeds
because the concentration of the reactant decreases. You can see this in Figure 13.2
because the slope of the curve (the rate) becomes less steep (slower) with time. Figure
13.3 shows the rate as a function of the concentration of A. Notice the linear relationship
for the first-order reaction—the rate is directly proportional to the concentration.
Second-Order Reaction
In a second-order reaction, the rate of the reaction is proportional to the square of the
concentration of the reactant.
Rate = k[A]2
Consequently, for a second-order reaction, the rate is even more sensitive to the reactant
concentration. You can see this in Figure 13.2 because the slope of the curve (the rate)
flattens out more quickly than it does for a first-order reaction. Figure 13.3 shows the rate
as a function of the concentration of A. Notice the quadratic relationship—the rate of a
second-order reaction is proportional to the square of the concentration.
The order of a reaction can be determined only by experiment. A common way to determine reaction order is the method of initial rates. In this method, the initial rate—the rate
for a short period of time at the beginning of the reaction—is measured at several different initial reactant concentrations to determine the effect of the concentration on the rate.
For example, let's return to our simple reaction in which a single reactant, A, decomposes
into products.
A S products
In an experiment, the initial rate is measured at several different initial concentrations
with the following results:
Notice that, in this data set, when the concentration of A doubles, the rate doubles—the
initial rate is directly proportional to the initial concentration. The reaction is therefore
first order in A, and the rate is equal to the rate constant times the concentration of A:
Rate = k[A]1
Notice that, in this data set, when the concentration of A doubles, the rate doubles—the
initial rate is directly proportional to the initial concentration. The reaction is therefore
first order in A, and the rate is equal to the rate constant times the concentration of A:
Rate = k[A]1
We can determine the value of the rate constant, k, by solving the rate law for k and substituting the concentration and the initial rate from any one of the three measurements.
Here we use the first measurement:
Rate = k[A]1
k = rate
[A] = 0.015 M>s
0.10 M = 0.15 s-1
Notice that the rate constant for a first-order reaction has units of s-1
.
The following data sets show how measured initial rates differ for zero-order and for
second-order reactions:
For a zero-order reaction, the initial rate is independent of the reactant concentration—the
rate is the same at all measured initial concentrations. For a second-order reaction, the initial rate quadruples for a doubling of the reactant concentration—the relationship between
concentration and rate is quadratic. The rate constants for zero- and second-order reactions
have different units than for first-order reactions. The rate constant for a zero-order reaction
has units of M # s-1
, and that for a second-order reaction has units of M-1 # s-1
.
If we are unsure about how the initial rate is changing with the initial reactant concentration, or if the numbers are not as obvious as they are in these examples, we can
substitute any two initial concentrations and the corresponding initial rates into a ratio of
the rate laws to determine the order (n).
rate 2
rate 1 = k[A]n
2
k[A]n
1
For example, we can substitute the last two measurements in the data set we were given
for the second-order reaction to determine that n = 2:
0.240 M>s
0.060 M>s = k(0.40 M)
n
k(0.20 M)
n
4.0 = a
0.40
0.20 b
n
= 2n
log 4.0 = log(2n
)
= n log 2
n = log 4
log 2
= 2
So far, we have considered a simple reaction with only one reactant. How do we define
the rate law for reactions with more than one reactant? Consider the generic reaction:
aA + bB S cC + dD
As long as the reverse reaction is negligibly slow, the rate law is proportional to the concentration of A raised to the m multiplied by the concentration of B raised to the n:
Rate = k[A]m[B]n
where m is the reaction order with respect to A and n is the reaction order with respect to
B. The overall order is the sum of the exponents (m + n). For example, the reaction
between hydrogen and iodine has been experimentally determined to be first order with
respect to hydrogen, first order with respect to iodine, and thus second order overall
H2(g) + I2(g) S 2 HI(g) Rate = k[H2]
1
[I2]
1
Similarly, the reaction between hydrogen and nitrogen monoxide has been experimentally determined to be first order with respect to hydrogen, second order with respect to
nitrogen monoxide, and thus third order overall.
2 H2(g) + 2 NO(g) S N2(g) + 2 H2O(g) Rate = k[H2]
1
[NO]2
The rate law for any reaction must always be determined by experiment, often by the
method of initial rates described previously. We can't simply look at a chemical equation
and determine the rate law for the reaction. When the reaction has two or more reactants, the
concentration of each reactant is usually varied independently of the others to determine the
dependence of the rate on the concentration of that reactant. Example 13.2 shows how to
use the method of initial rates to determine the order of a reaction with multiple reactants.
The rate laws we have examined so far show the relationship between the rate of a reaction and the concentration of a reactant. However, we often want to know the relationship between the concentration of a reactant and time. For example, the presence of
chlorofluorocarbons (CFCs) in the atmosphere threatens the ozone layer. One reason
CFCs pose such a significant threat is that the reactions that consume them are so slow.
Legislation has resulted in reduced CFC emissions, but even at much lower rates of emission, CFC concentrations in the atmosphere have dropped very slowly. Nonetheless, we
would like to be able to predict how their concentration changes with time. How much
will be left in 20 years? In 50 years?
The integrated rate law for a chemical reaction is a relationship between the concentrations of the reactants and time. For simplicity, we return to a single reactant decomposing into products.
A S products
The integrated rate law for this reaction depends on the order of the reaction; let's examine each of the common reaction orders individually.
First-Order Integrated Rate law
If our simple reaction is first order, the rate is equal to the rate constant multiplied by the
concentration of A.
Rate = k[A]
Since Rate = -∆[A]>∆t, we can write:
- ∆[A]
∆t = k[A]
In this form, the rate law is also known as the differential rate law.
We can use calculus (see margin) to integrate the differential rate law to obtain the
first-order integrated rate law:
ln [A]t = -kt + ln [A]0
We can also rearrange the integrated rate law by subtracting ln[A]0 from both sides of the
equation:
ln [A]t = -kt + ln [A]0
ln [A]t - ln [A]0 = -kt
Since ln A - ln B = ln (A>B), we can rearrange this equation to get:
ln [A]t
[A]0
= -kt
where [A]t is the concentration of A at any time t, k is the rate constant, and [A]0 is the
initial concentration of A. Equations 13.12 and 13.13 are equivalent.
Notice that the integrated rate law shown in Equation 13.12 has the form of an equation for a straight line.
Therefore, for a first-order reaction, a plot of the natural log of the reactant concentration
as a function of time yields a straight line with a slope of -k and a y-intercept of ln [A]0,
as shown in FiguRe 13.5 ◀. (Note that the slope is negative but that the rate constant is
always positive.)
ExAmPlE 13.4 The First-Order Integrated Rate Law: Determining the Concentration of a
Reactant at a Given Time
In Example 13.3, you determined that the decomposition of SO2Cl2 (under the given reaction conditions) is first order and has
a rate constant of +2.90 * 10-4 s-1. If the reaction is carried out at the same temperature and the initial concentration of
SO2Cl2 is 0.0225 M, what is the SO2Cl2 concentration after 865 s?
SORT
You are given the rate constant of a first-order reaction and the initial concentration of the reactant. You are
asked to find the concentration at 865 seconds.
STRATEGIZE
Use the first-order integrated rate law to determine the SO2Cl2 concentration at t = 865 s.
SOLVE
Substitute the rate constant, the initial concentration, and the time into the integrated rate law.
Solve the integrated rate law for the concentration of
[SO2Cl2]t.
Second-Order Integrated Rate law
If our simple reaction (A S products) is second order, the rate law is proportional to the
square of the concentration of A.
Rate = k[A]2
Since Rate = -∆[A]>∆t, we can write the differential rate law:
- ∆[A]
∆t = k[A]2
Again, we can use calculus to integrate the second-order differential rate law. We leave
this derivation to an exercise (see Exercise 89 at the end of this chapter). The derivation
results in the second-order integrated rate law:
1
[A]t
= kt +
1
[A]0
The second-order integrated rate law is also in the form of an equation for a straight line.
1
[A]t
= kt +
1
[A]0
y = mx + b
Notice, however, that we must now plot the inverse of the concentration of the reactant as
a function of time. The plot yields a straight line with a slope of k and an intercept of
1>[A]0
Zero-Order Integrated Rate law
If our simple reaction is zero order, the rate law is defined as
Rate = k[A]0
Since Rate = -∆[A]>∆t, we can write:
- ∆[A]
∆t = k
We can integrate the zero-order differential rate law (see margin) to obtain the zero-order
integrated rate law:
[A]t = -kt + [A]0
The zero-order integrated rate law in Equation 13.17 is also in the form of an equation for
a straight line. A plot of the concentration of the reactant as a function of time yields a
straight line with a slope of -k and an intercept of [A]0,
ExAmPlE 13.5 The Second-Order Integrated Rate Law: using
Graphical analysis of Reaction Data
Consider the equation for the decomposition of NO2.
NO2(g) S NO(g) + O(g)
The concentration of NO2 is monitored at a fixed temperature as a function of time during the decomposition reaction, and the data are tabulated in the margin. Show by
graphical analysis that the reaction is not first order and that it is second order. Determine the rate constant for the reaction.
SOLUTION
To show that the reaction
is not first order, prepare a
graph of ln [NO2] versus
time.
The plot is not linear (the
straight line does not fit
the data points), confirming that the reaction is not
first order. In order to show
that the reaction is second
order, prepare a graph of
1>[NO2] versus time.
This graph is linear (the
data points fit well to a
straight line), confirming
that the reaction is indeed
second order. To obtain the
rate constant, determine
the slope of the best fitting line. The slope is
0.255 M-1 # s-1; therefore,
the rate constant is
0.255 M-1 # s-1.
The half-life (t1>2) of a reaction is the time required for the concentration of a reactant to
fall to one-half of its initial value. For example, if a reaction has a half-life of 100 seconds, and if the initial concentration of the reactant is 1.0 M, then the concentration will
fall to 0.50 M in 100 s. The half-life expression—which defines the dependence of halflife on the rate constant and the initial concentration—is different for different reaction
orders.
First-Order Reaction Half-life
From the definition of half-life and from the integrated rate law, we can derive an expression for the half-life. For a first-order reaction, the integrated rate law is:
ln [A]t
[A]0
= -kt
At a time equal to the half-life (t = t1>2), the concentration is exactly half of the initial
concentration: ([A]t = 1
2 [A]0). Therefore, when t = t1>2, we can write the expression:
ln
1
2 [A]0
[A]0
= ln 1
2 = -kt1>2
Solving for t1>2 and substituting -0.693 for ln 1
2, we arrive at the following expression for
the half-life of a first-order reaction:
t1>2 = 0.693/k
Notice that, for a first-order reaction, t1>2 is independent of the initial concentration. For
example, if t1>2 is 100 s, and if the initial concentration is 1.0 M, the concentration falls to
0.50 M in 100 s, then to 0.25 M in another 100 s, then to 0.125 M in another 100 s, and
so on (FiguRe 13.8▼). Even though the concentration changes as the reaction proceeds, the
half-life (how long it takes for the concentration to halve) is constant. A constant half-life
is unique to first-order reactions, making the concept of half-life particularly useful for
first-order reactions.
ExAmPlE 13.6 Half-Life
Molecular iodine dissociates at 625 K with a first-order rate constant of 0.271 s-1.
What is the half-life of this reaction?
SOLUTION
Since the reaction is first order, the half-life is
given by Equation 13.19. Substitute the value of
k into the expression and calculate t1>2.
Second-Order Reaction Half-life
For a second-order reaction, the integrated rate law is:
1
[A]t
= kt +
1
[A]0
At a time equal to the half-life (t = t1>2), the concentration is exactly one-half of the initial concentration ([A]t = 1
2[A]0). We can therefore write the following expression at
t = t1>2:
1
1
2[A]0
= kt1>2 +
1
[A]0
With this expression, we can solve for t1>2:
kt1>2 = 1
1
2[A]0
- 1
[A]0
kt1>2 = 2
[A]0
- 1
[A]0
t1>2 = 1
k[A]0
Notice that, for a second-order reaction, the half-life depends on the initial concentration.
So if the initial concentration of a reactant in a second-order reaction is 1.0 M, and the
half-life is 100 s, the concentration falls to 0.50 M in 100 s. However, the time it takes for
the concentration to fall to 0.25 M is now longer than 100 s because the initial concentration has decreased. The half-life continues to get longer as the concentration decreases.
Zero-Order Reaction Half-life
For a zero-order reaction, the integrated rate law is:
[A]t = -kt + [A]0
making the substitutions (t = t1>2; [A]t = 1
2[A]0). We write the following expression at t = t1>2:
1
2[A]0 = -kt1>2 + [A]0
We then solve for t1>2:
t1>2 = [A]0
2k
Notice that, for a zero-order reaction, the half-life also depends on the initial concentration, but here the half-life gets shorter as the concentration decreases.
Summarizing the Integrated Rate Law (see Table 13.1):
▶ The reaction order and rate law must be determined experimentally.
▶ The differential rate law relates the rate of the reaction to the concentration of the
reactant(s).
▶ The integrated rate law (which is mathematically derived from the differential rate
law) relates the concentration of the reactant(s) to time.
▶ The half-life is the time it takes for the concentration of a reactant to fall to one-half
of its initial value.
▶ The half-life of a first-order reaction is independent of the initial concentration.
▶ The half-lives of zero-order and second-order reactions depend on the initial
concentration.
In the opening section of this chapter, we saw that lizards become lethargic when their
body temperature drops because the chemical reactions that control their muscle movement slow down at lower temperatures. The rates of chemical reactions are, in general,
highly sensitive to temperature. For example, at room temperature, a 10 °C increase in
temperature increases the rate of a typical biological reaction by two or three times. How
do we explain this highly sensitive temperature dependence?
Recall that the rate law for a reaction is Rate = k[A]n
. The temperature dependence of
the reaction rate is contained in the rate constant, k (which is actually a constant only when
the temperature remains constant). An increase in temperature generally results in an increase
in k, which results in a faster rate. In 1889, Swedish chemist Svante Arrhenius wrote a paper
quantifying the temperature dependence of the rate constant. The modern form of the
Arrhenius equation, which relates the rate constant (k) and the temperature in kelvins (T), is:
where R is the gas constant (8.314 J>mol # K), A is a constant called the frequency factor
(or the pre-exponential factor), and Ea is the activation energy (or activation barrier).
The activation energy (Ea) is an energy barrier or hump that must be surmounted for the
reactants to be transformed into products (FiguRe 13.9 ▼). We examine the frequency factor more closely in the next section of this chapter; for now, we can think of the frequency
factor (A) as the number of times that the reactants approach the activation barrier per
unit time.
To understand each of these quantities better, consider the simple reaction in which
CH3NC (methyl isonitrile) rearranges to form CH3CN (acetonitrile):
Let's examine the physical meaning of the activation energy, frequency factor, and exponential factor for this reaction.
The Activation Energy
shows the energy of the molecule as the reaction
proceeds. The x-axis represents the progress of the reaction
from left (reactant) to right (product). To get from the reactant
to the product, the molecule must go through a high-energy
intermediate state called the activated complex, or transition state. Even though the overall reaction is energetically
downhill (exothermic), it must first go uphill to reach the activated complex because energy is required to initially weaken
the H3C¬N bond and allow the NC group to begin to rotate.
The energy required to create the activated complex is the activation energy. The higher
the activation energy, the slower the reaction rate (at a given temperature).
The Frequency Factor
Recall that the frequency factor represents the number of approaches to the activation
barrier per unit time. Any time the NC group begins to rotate, it approaches the activation
barrier. For this reaction, the frequency factor represents the rate at which the NC part of
the molecule wags (vibrates from side to side). With each wag, the reactant approaches
the activation barrier. However, approaching the activation barrier is not equivalent to
surmounting it. Most of the approaches do not have enough total energy to make it over
the activation barrier
The Exponential Factor
The exponential factor is a number between 0 and 1 that represents the
fraction of molecules that have enough energy to make it over the activation barrier on a given approach. In other words, the exponential factor is the fraction of approaches that are actually successful and result
in the product. For example, if the frequency factor is 109>s and the
exponential factor is 10-7
at a certain temperature, then the overall rate
constant at that temperature is 109>s * 10-7 = 102>s. In this case, the
CN group is "wagging" at a rate of 109>s. With each wag, the activation barrier is approached. However, only 1 in 107
molecules have sufficient energy to actually make it over the activation barrier for a given
wag.
The exponential factor depends on both the temperature (T) and the
activation energy (Ea) of the reaction.
Exponential factor = e-Ea>RT
As the temperature increases, the number of molecules having enough
thermal energy to surmount the activation barrier increases. At any
given temperature, a sample of molecules has a distribution of energies,
as shown in FiguRe 13.11◀. Under common circumstances, only a small
fraction of the molecules has enough energy to make it over the activation barrier. Because of the shape of the energy distribution curve, however, a small change in temperature results in a large difference in the
number of molecules having enough energy to surmount the activation
barrier. This explains the sensitivity of reaction rates to temperature.
Summarizing Temperature and Reaction Rate:
▶ The frequency factor is the number of times that the reactants approach the activation
barrier per unit time.
▶ The exponential factor is the fraction of approaches that are successful in surmounting the activation barrier and forming products.
▶ The exponential factor increases with increasing temperature, but decreases with an
increasing value for the activation energy.
The frequency factor and activation energy are important quantities in understanding the
kinetics of any reaction. To see how we can measure these factors in the laboratory, consider again Equation 13.24: k = Ae-Ea>RT. Taking the natural log of both sides of this
equation, we get:
ln k = ln 1Ae-Ea>RT
ln k = ln A + ln e-Ea>RT
ln k = ln A - Ea
RT
y = mx + b
Equation 13.26 is in the form of a straight line. A plot of the natural log of the rate constant (ln k) versus the inverse of the temperature in kelvins (1>T) yields a straight line
with a slope of -Ea >R and a y-intercept of ln A. Such a plot is an Arrhenius plot and is
commonly used in the analysis of kinetic data,
In some cases, where either data are limited or plotting capabilities are absent, we
can calculate the activation energy knowing the rate constant at just two different temperatures. We can apply the Arrhenius expression in Equation 13.25 to the two different
temperatures as follows:
ln k2 = - Ea
R a 1
T2
b + ln A ln k1 = - Ea
R a 1
T1
b + ln A
We then subtract ln k1 from ln k2 as follows:
ln k2 - ln k1 = c - Ea
R a 1
T2
b + ln Ad - c - Ea
R a 1
T1
b + ln Ad
Rearranging, we get the two-point form of the Arrhenius equation:
ln
k2
k1
= Ea
R a 1
T1
- 1
T2
b
Example 13.8 shows how you can use this equation to calculate the activation energy
from experimental measurements of the rate constant at two different temperatures.
ExAmPlE 13.8 using the Two-Point Form of the arrhenius Equation
The reaction between nitrogen dioxide and carbon monoxide is:
NO2(g) + CO(g) S NO(g) + CO2(g)
The rate constant at 701 K is measured as 2.57 M-1 # s-1, and that at 895 K is measured as 567 M-1 # s-1. Find the activation
energy for the reaction in kJ>mol.
SORT
You are given the rate constant of a reaction at two different temperatures and asked to find the activation energy
STRATEGIZE
Use the two-point form of the Arrhenius equation, which
relates the activation energy to the given information and R (a constant).
SOLVE
Substitute the two rate constants and the two temperatures into
the equation.
Solve the equation for Ea, the activation energy, and convert to kJ>mol.
We saw previously that the frequency factor in the Arrhenius equation represents the
number of approaches to the activation barrier per unit time. We now refine that idea for
a reaction involving two gas-phase reactants.
A(g) + B(g) S products
In the collision model, a chemical reaction occurs after a sufficiently energetic collision between two reactant molecules (FiguRe 13.12▶). In collision theory, therefore, each
approach to the activation barrier is a collision between the reactant molecules.
Consequently, the value of the frequency factor should simply be the number of collisions that occur per second. However, the frequency factors of most (though not all) gasphase chemical reactions tend to be smaller than the number of collisions that occur per
second. Why?
In the collision model, we separate the frequency factor into two distinct parts, as
shown in the following equations:
where p is the orientation factor and z is the collision frequency. The collision frequency is the number of collisions that occur per unit time, which we can calculate for a
gas-phase reaction from the pressure of the gases and the temperature of the reaction
mixture. Under typical conditions, a single molecule undergoes on the order of 109
collisions every second.
To understand the orientation factor, consider the following reaction:
In order for the reaction to occur, two NOCl molecules must collide with
sufficient energy. However, not all collisions with sufficient energy lead
to products because the reactant molecules must also be properly oriented. Consider each of the possible orientations of the reactant molecules (shown in the margin ) during a collision. The first two collisions,
even if they occur with sufficient energy, do not result in a reaction
because the reactant molecules are not oriented in a way that allows the
chlorine atoms to bond. If two molecules are to react with each other,
they must collide in such a way that allows the necessary bonds to break and form. For
the reaction of NOCl(g), the orientation factor is p = 0.16. This means that only 16 out
of 100 sufficiently energetic collisions are actually successful in forming the products.
Most chemical reactions do not occur in a single step but through several steps. When we
write a chemical equation to represent a chemical reaction, we usually represent the overall reaction, not the series of individual steps by which the reaction occurs. Consider the
reaction in which hydrogen gas reacts with iodine monochloride:
H2(g) + 2 ICl(g) S 2 HCl(g) + I2(g)
The overall equation simply shows the substances present at the beginning of the reaction
and the substances formed by the reaction—it does not show the intermediate steps. A
reaction mechanism is the series of individual chemical steps by which an overall chemical reaction occurs. For example, the reaction between hydrogen and iodine monochloride occurs through the following proposed mechanism:
Step 1 H2(g) + ICl(g) S HI(g) + HCl(g)
Step 2 HI(g) + ICl(g) S HCl(g) + I2(g)
In the first step, an H2 molecule collides with an ICl molecule and forms an HI molecule
and an HCl molecule. In the second step, the HI molecule formed in the first step collides
with a second ICl molecule to form another HCl molecule and an I2 molecule. Each step
in a reaction mechanism is an elementary step. Elementary steps cannot be broken down
into simpler steps—they occur as they are written.
One of the requirements for a valid reaction mechanism is that the individual steps in
the mechanism must add to the overall reaction. For example, the proposed mechanism
sums to the overall reaction as shown here
H2(g) + ICl(g) S Hl(g) + HCl(g)
Hl(g) + ICl(g) S HCl(g) + I2(g)
= H2(g) + 2 ICl(g) S 2 HCl(g) + I2(g)
Species—such as HI—that form in one step of a mechanism and are consumed in another
are reaction intermediates. A reaction intermediate is not found in the balanced equation for the overall reaction but plays a key role in the mechanism. A reaction mechanism
is a complete, detailed description of the reaction at the molecular level—it specifies the
individual collisions and reactions that result in the overall reaction. As such, reaction
mechanisms are highly sought-after pieces of chemical knowledge.
How do we determine reaction mechanisms? Recall from the opening section of this
chapter that chemical kinetics are not only practically important (allowing us to control
the rate of a particular reaction), but also theoretically important because they can help us
determine the mechanism of the reaction. We can piece together a reaction mechanism by
measuring the kinetics of the overall reaction and working backward to write a mechanism consistent with the measured kinetics.
As we have noted, most chemical reactions occur through a series of elementary steps. In
most cases, one of those steps—which we call the rate-determining step—is much
slower than the others. The rate-determining step in a chemical reaction is analogous to
the narrowest section on a freeway. If a freeway narrows from four lanes to two lanes, the
rate at which cars travel along the freeway is limited by the rate at which they can travel
through the narrow section (even though the rate could be much faster along the four-lane
section). Similarly, the rate-determining step in a reaction mechanism limits the overall
rate of the reaction (even though the other steps occur much faster) and therefore determines the rate law for the overall reaction.
As an example, consider the reaction between nitrogen dioxide gas and carbon monoxide gas,
NO2(g) + CO(g) S NO(g) + CO2(g)
The experimentally determined rate law for this reaction is Rate = k[NO2]
2
. We can see
from this rate law that the reaction must not be a single-step reaction; otherwise, the rate
law would be Rate = k[NO2][CO]. A possible mechanism for this reaction is:
NO2(g) + NO2(g) S NO3(g) + NO(g) Slow
NO3(g) + CO(g) S NO2(g) + CO2(g) Fast
(on the next page) shows the energy diagram accompanying this mechanism.
The first step has a much larger activation energy than the second step. The greater activation energy results in a much smaller rate constant for the first step compared to the second
step. The first step determines the overall rate of the reaction, and the predicted rate law is
therefore Rate = k[NO2]
2
, which is consistent with the observed experimental rate law.
For a proposed reaction mechanism such as the one shown above to be valid—
mechanisms can only be validated, not proven—two conditions must be met:
1. The elementary steps in the mechanism must sum to the overall reaction.
2. The rate law predicted by the mechanism must be consistent with the experimentally observed rate law.
We have already seen that the rate law predicted by the proposed mechanism is consistent
with the experimentally observed rate law. We can check whether the elementary steps
sum to the overall reaction by adding them together.
NO2(g) + NO2(g) S NO3(g) + NO(g) slow
NO3(g) + CO(g) S NO2(g) + CO2(g fast
=NO2(g) + CO(g) S NO(g) + CO2(g) Overall.
The mechanism fulfills both of the requirements and is therefore valid. A valid mechanism is not necessarily a proven mechanism (because other mechanisms may also fulfill
both of the requirements). We can only say that a given mechanism is consistent with
kinetic observations of the reaction and is therefore possible. Other types of data—such
as the experimental evidence for a proposed intermediate—can further strengthen the
validity of a proposed mechanism.
When the proposed mechanism for a reaction has a slow initial step—such as the one
shown previously for the reaction between NO2 and CO—the rate law predicted by the
mechanism normally contains only reactants involved in the overall reaction. However,
when a mechanism begins with a fast initial step, then some other subsequent step in the
mechanism is the rate-limiting step. In these cases, the rate law predicted by the ratelimiting step may contain reaction intermediates. Since reaction intermediates do not
appear in the overall reaction equation, a rate law containing intermediates cannot correspond directly to the experimental rate law. Fortunately, however, we often can express
the concentration of intermediates in terms of the concentrations of the reactants of the
overall reaction.
In a multistep mechanism where the first step is fast, the products of the first step
build up because the rate at which they are consumed is limited by some slower step further down the line. As those products build up, they begin to react with one another to
re-form the reactants. As long as the first step is fast enough in comparison to the ratelimiting step, the first-step reaction will reach equilibrium. We indicate the equilibrium as
follows:
Reactants L
k1
k-1 =Products
The double arrows indicate that both the forward reaction and the reverse reaction occur.
If equilibrium is reached, the rate of the forward reaction equals the rate of the reverse
reaction.
Consider the reaction by which hydrogen reacts with nitrogen monoxide to form
water and nitrogen gas.
2 H2(g) + 2 NO(g) S 2 H2O(g) + N2(g)
The experimentally observed rate law is Rate = k[H2][NO]2
. The reaction is first order
in hydrogen and second order in nitrogen monoxide. The proposed mechanism is as
follows:
2 NO(g) L
k1
k-1
N2O2(g Fast
H2(g) + N2O2(g) L
k2
H2O(g) + N2O(g) Slow (rate limiting)
N2O(g) + H2(g) L
k3
N2(g) + H2O(g) Fast
2 H2(g) + 2 NO(g) S 2 H2O(g) + N2(g) Overall
To determine whether the mechanism is valid, we must determine whether the two conditions described previously are met. As we can see above, the steps do indeed sum to the
overall reaction, so the first condition is met.
The second condition is that the rate law predicted by the mechanism is consistent
with the experimentally observed rate law. Since the second step is rate limiting, we write
the following expression for the rate law: Rate = k2[H2][N2O2]
This rate law contains an intermediate (N2O2) and can therefore not be immediately reconciled with the experimentally observed rate law (which does not contain intermediates). Because of the equilibrium in the first step, however, we can express the
concentration of the intermediate in terms of the reactants of the overall equation. Since
the first step reaches equilibrium, the rate of the forward reaction in the first step equals
the rate of the reverse reaction. Rate (forward) = Rate (backward)
The rate of the forward reaction is given by:
Rate = k1[NO]2
The rate of the reverse reaction is given by:
Rate = k-1[N2O2]
The rate of the reverse reaction is given by:
Rate = k-1[N2O2]
Since these two rates are equal at equilibrium, we can write the expression:
k1[NO]2 = k-1[N2O2]
Rearranging, we get:
[N2O2] = k1
k-1
[NO]2
We can substitute this expression into Equation 13.28, the rate law obtained from the
slow step.
Rate = k2[H2][N2O2]
= k2[H2]
k1
k-1
[NO]2
= k2k1
k-1
[H2][NO]2
If we combine the individual rate constants into one overall rate constant, we get the predicted rate law.
Rate = k[H2][NO]2
This rate law is consistent with the experimentally observed rate law, so the second condition is met and the proposed mechanism is valid.
EXAMPlE 13.9 Reaction Mechanisms
Ozone naturally decomposes to oxygen by the reaction:
2 O3(g) S 3 O2(g)
The experimentally observed rate law for this reaction is:
Rate = k[O3]
2[O2]
-1
Show that the following proposed mechanism is consistent with the experimentally observed rate law.
O3(g) L
k1
k-1
O2(g) + O(g) Fast
O3(g) + O(g) Lk2
2 O2(g) Slow
SOLUTION
To determine whether the mechanism is valid, you must first determine
whether the steps sum to the overall reaction. Since the steps do indeed
sum to the overall reaction, the first condition is met.
The second condition is that the rate law predicted by the mechanism must
be consistent with the experimentally observed rate law. Since the second
step is rate limiting, write the rate law based on the second step
Because the rate law contains an intermediate (O), you must express the
concentration of the intermediate in terms of the concentrations of the
reactants of the overall reaction. To do this, set the rates of the forward
reaction and the reverse reaction of the first step equal to each other.
Solve the expression from the previous step for [O], the concentration of
the intermediate.
Finally, substitute [O] into the rate law predicted by the slow step.
Throughout this chapter, we have learned ways to control the rates of chemical reactions.
For example, we can speed up the rate of a reaction by increasing the concentration of the
reactants or by increasing the temperature. However, these ways are not always feasible.
There are limits to how concentrated we can make a reaction mixture, and increases in temperature may allow unwanted reactions—such as the decomposition of a reactant—to occur.
Alternatively, reaction rates can be increased by using a catalyst, a substance that
increases the rate of a chemical reaction but is not consumed by the reaction. A catalyst
works by providing an alternative mechanism for the reaction—one in which the rate-determining step has a lower activation energy. For example, consider the noncatalytic destruction of ozone in the upper atmosphere.
O3(g) + O(g) S 2 O2(g)
In this reaction, an ozone molecule collides with an oxygen atom to form two oxygen
molecules in a single elementary step. The reason that Earth has a protective ozone layer
in the upper atmosphere is that the activation energy for this reaction is fairly high and the
reaction, therefore, proceeds at a fairly slow rate. The ozone layer does not rapidly
decompose into O2. However, the addition of Cl atoms (which come from the photodissociation of human-made chlorofluorocarbons) to the upper atmosphere makes available
another pathway by which O3 can be destroyed. The first step in this pathway—called the
catalytic destruction of ozone—is the reaction of Cl with O3 to form ClO and O2.
Cl + O3 S ClO + O2
This is followed by a second step in which ClO reacts with O,
regenerating Cl.
ClO + O S Cl + O2
Notice that if we add the two reactions, the overall reaction is
identical to the noncatalytic reaction.
Cl + O3 S ClO + O2
ClO + O S Cl + O2
= O3 + O S 2 O2
However, the activation energy for the rate-limiting step in this
pathway is much smaller than for the first, uncatalyzed pathway (as shown in FiguRe 13.14▶), and therefore the reaction
occurs at a much faster rate. Note that the Cl is not consumed
in the overall reaction—this is characteristic of a catalyst.
Catalysis can be divided into two types: homogeneous and heterogeneous (FiguRe 13.15▼). In homogeneous catalysis, the catalyst
exists in the same phase as the reactants. The catalytic destruction
of ozone by Cl is an example of homogeneous catalysis—the
chlorine atoms exist in the gas phase with the gas-phase reactants. In heterogeneous catalysis, the catalyst exists in a different phase than the reactants. The use of solid catalysts with
gas-phase or solution-phase reactants is the most common type of heterogeneous catalysis.
Research has shown that heterogeneous catalysis is most likely responsible for the
ozone hole over Antarctica. In 1985, when scientists discovered the dramatic drop in
ozone over Antarctica, they wondered why it existed only there and not over the rest of the planet. After all, the chlorine from chlorofluorocarbons that catalyzes ozone destruction is evenly distributed throughout the entire atmosphere.
As it turns out, most of the chlorine that enters the atmosphere from chlorofluorocarbons gets bound up in chlorine reservoirs, substances such as ClONO2 that hold chlorine
and prevent it from catalyzing ozone destruction. The unique conditions over Antarctica—
especially the cold isolated air mass that exists during the long dark winter—result in
clouds that contain solid ice particles. These unique clouds are called polar stratospheric
clouds (or PSCs), and the surfaces of the ice particles within these clouds appear to catalyze the release of chlorine from their reservoirs:
ClONO2 + HCl ¡PSCs Cl2 + HNO2
When the sun rises in the Antarctic spring, the sunlight dissociates the chlorine molecules
into chlorine atoms.
Cl2 ¡light 2 Cl
When the sun rises in the Antarctic spring, the sunlight dissociates the chlorine molecules
into chlorine atoms.
Cl2 ¡light 2 Cl
The chlorine atoms then catalyze the destruction of ozone by the mechanism discussed
previously. This continues until the sun melts the stratospheric clouds, allowing chlorine
atoms to be reincorporated into their reservoirs. The result is an ozone hole that forms
every spring and lasts about 6-8 weeks.
Perhaps the best examples of chemical catalysis are found in living organisms. Most of
the thousands of reactions that must occur for an organism to survive would be too slow
at normal temperatures. So living organisms rely on enzymes, biological catalysts that
increase the rates of biochemical reactions. Enzymes are usually large protein molecules
with complex three-dimensional structures. Within the enzyme structure is a specific area
called the active site. The properties and shape of the active site are just right to bind the
reactant molecule, usually called the substrate. The substrate fits into the active site in a
manner that is analogous to a key fitting into a lock (FiguRe 13.16▼). When the substrate
binds to the active site—through intermolecular forces such as hydrogen bonding and
dispersion forces, or even covalent bonds—the activation energy of the reaction is greatly
lowered, allowing the reaction to occur at a much faster rate. The general mechanism by
which an enzyme (E) binds a substrate (S) and then reacts to form the products (P) is:
E + S L ES Fast
ES -forced into a geometry that stresses the bond. Weakening of this bond lowers the activation energy for the reaction, increasing the reaction rate. The reaction can then proceed
toward equilibrium—which favors the products—at a much lower temperature. E + P Slow, rate limiting
Sucrase is an enzyme that catalyzes the breaking up of sucrose (table sugar) into
glucose and fructose within the body. At body temperature, sucrose does not break into
glucose and fructose because the activation energy for the reaction is high, resulting in a
slow reaction rate. However, when a sucrose molecule binds to the active site within
sucrase, the bond between the glucose and fructose units weakens because glucose is forced into a geometry that stresses the bond. Weakening of this bond lowers the activation energy for the reaction, increasing the reaction rate. The reaction can then proceed
toward equilibrium—which favors the products—at a much lower temperature.
;