We will assume an understanding of the
postulates of the
Kinetic Molecular Theory and of the energetics of
chemical reactions. We will also assume an understanding of phase
equilibrium and reaction equilibrium, including the temperature
dependence of equilibrium constants.
We have carefully examined the observation
that chemical reactions come to equilibrium. Depending on the
reaction, the equilibrium conditions can be such that there is a
mixture of reactants and products, or virtually all products, or
virtually all reactants. We have not considered the time scale for
the reaction to achieve these conditions, however. In many cases,
the speed of the reaction might be of more interest than the final
equilibrium conditions of the reaction. Some reactions proceed so
slowly towards equilibrium as to appear not to occur at all. For
example, metallic iron will eventually oxidize in the presence of
aqueous salt solutions, but the time is sufficiently long for this
process that we can reasonably expect to build a boat out of iron.
On the other hand, some reactions may be so rapid as to pose a
hazard. For example, hydrogen gas will react with oxygen gas so
rapidly as to cause an explosion. In addition, the time scale for a
reaction can depend very strongly on the amounts of reactants and
their temperature.
In this concept development study, we seek an
understanding of the rates of chemical reactions. We will define
and measure reaction rates and develop a quantitative analysis of
the dependence of the reaction rates on the conditions of the
reaction, including concentration of reactants and temperature.
This quantitative analysis will provide us insight into the process
of a chemical reaction and thus lead us to develop a model to
provide an understanding of the significance of reactant
concentration and temperature.
We will find that many reactions proceed quite
simply, with reactant molecules colliding and exchanging atoms. In
other cases, we will find that the process of reaction can be quite
complicated, involving many molecular collisions and rearrangements
leading from reactant molecules to product molecules. The rate of
the chemical reaction is determined by these steps.
We begin by considering a fairly simple
reaction on a rather elegant molecule. One oxidized form of
buckminsterfullerene
C60C60
is
C60O3C60O3,
with a three oxygen bridge as shown in Figure 1.
C60O3C60O3
is prepared from
C60C60
dissolved in toluene solution at temperatures of
0°C0°C
or below. When the solution is warmed,
C60O3C60O3
decomposes, releasing
O2O2
and creating
C60OC60O
in a reaction which goes essentially to completion. We can actually
watch this process happen in time by measuring the amount of light
of a specific frequency absorbed by the
C60O3C60O3
molecules, called the
absorbance. The absorbance is proportional to the
concentration of the
C60O3C60O3
in the toluene solution, so observing the absorbance as a function
of time is essentially the same as observing the concentration as a
function of time. One such set of data is given in Table 1, and is shown in the graph in
Figure 2.
Oxidized Buckminsterfullerene Absorbance during Thermal Decomposition at 23°C
| time (minutes) |
C60O3C60O3
absorbance |
| 3 |
0.04241 |
| 9 |
0.03634 |
| 15 |
0.03121 |
| 21 |
0.02680 |
| 27 |
0.02311 |
| 33 |
0.01992 |
| 39 |
0.01721 |
| 45 |
0.01484 |
| 51 |
0.01286 |
| 57 |
0.01106 |
| 63 |
0.00955 |
| 69 |
0.00827 |
| 75 |
0.00710 |
| 81 |
0.00616 |
| 87 |
0.00534 |
| 93 |
0.00461 |
| 99 |
0.00395 |
The rate at which the decomposition reaction
is occurring is clearly related to the rate of change of the
concentration
[C60O3][C60O3],
which is proportional to the slope of the graph in Figure 2. Therefore, we define the rate of
this reaction as
Rate=-ddt[C60O3]≈-Δ[C60O3]ΔtRate
t
[C60O3]
Δ[C60O3]Δt
(1)
We want the rate of reaction to be positive,
since the reaction is proceeding forward. However, because we are
measuring the rate of disappearance of the reactant in this case,
that rate is negative. We include a negative sign in this
definition of rate so that the rate in
Equation 1 is a positive number. Note also that
the slope of the graph in
Figure 2
should be taken as the derivative of the graph, since the graph is
not a straight line. We will approximate that derivative by
estimating the slope at each time in the data, taking the change in
the absorbance of the
C60O3C60O3
divided by the change in time at each time step. The rate,
calculated in this way, is plotted as a function of time in
Figure 3.
It is clear that the slope of the graph in
Figure 2 changes over the course of
time. Correspondingly, Figure 3
shows that the rate of the reaction decreases as the reaction
proceeds. The reaction is at first very fast but then slows
considerably as the reactant
C60O3C60O3
is depleted.
The shape of the graph for rate versus time
(Figure 3) is very similar to the
shape of the graph for concentration versus time (Figure 2). This suggests that the rate of the
reaction is related to the concentration of
C60O3C60O3
at each time. Therefore, in Figure 4, we plot the rate of the reaction,
defined in Equation 1 and shown in
Figure 3, versus the absorbance of
the
C60O3C60O3.
We find that there is a very simple
proportional relationship between the rate of the reaction and the
concentration of the reactant. Therefore, we can write
Rate=-ddt[C60O3]=k[C60O3]Rate
t[C60O3]k[C60O3]
(2)
where
kk is a
proportionality constant. This equation shows that, early in the
reaction when
[C60O3][C60O3]
is large, the reaction proceeds rapidly, and that as
C60O3C60O3
is consumed, the reaction slows down.
Equation 2 is an example of a
rate law, expressing the relationship between the rate
of a reaction and the concentrations of the reactant or reactants.
Rate laws are expressions of the relationship between
experimentally observed rates and concentrations.
As a second example of a reaction rate, we
consider the dimerization reaction of butadiene gas,
CH2CH2=CHCH-CHCH=CH2CH2.
Two butadiene molecules can combine to form vinylcyclohexene, shown
in Figure 5.
Table 2
provides experimental data on the gas phase concentration of
butadiene
[C4H6][C4H6]
as a function of time at
T=250°CT250°C.
Dimerization of Butadiene at 250°C
| Time (s) |
[C4H6][C4H6] (M) |
Rate (M/s) |
Rate[C4H6]Rate[C4H6] |
Rate[C4H6]2Rate[C4H6]2 |
| 0 |
0.0917 |
9.48×10-69.48-6 |
1.03×10-41.03-4 |
1.13×10-31.13-3 |
| 500 |
0.0870 |
8.55×10-68.55-6 |
9.84×10-59.84-5 |
1.13×10-31.13-3 |
| 1000 |
0.0827 |
7.75×10-67.75-6 |
9.37×10-59.37-5 |
1.13×10-31.13-3 |
| 1500 |
0.0788 |
7.05×10-67.05-6 |
8.95×10-58.95-5 |
1.14×10-31.14-3 |
| 2000 |
0.0753 |
6.45×10-66.45-6 |
8.57×10-58.57-5 |
1.14×10-31.14-3 |
| 2500 |
0.0720 |
5.92×10-65.92-6 |
8.22×10-58.22-5 |
1.14×10-31.14-3 |
| 3000 |
0.0691 |
5.45×10-65.45-6 |
7.90×10-57.90-5 |
1.14×10-31.14-3 |
| 3500 |
0.0664 |
5.04×10-65.04-6 |
7.60×10-57.60-5 |
1.14×10-31.14-3 |
| 4000 |
0.0638 |
4.67×10-64.67-6 |
7.32×10-57.32-5 |
1.15×10-31.15-3 |
We can estimate the rate of reaction at each
time step as in Equation 1, and these
data are presented in Table 2 as
well. Again we see that the rate of reaction decreases as the
concentration of butadiene decreases. This suggests that the rate
is given by an expression like Equation 2. To test this, we calculate
Rate[C4H6]Rate[C4H6]
in Table 2 for each time step. We
note that this is
not a constant, so Equation 2 does not describe the relationship
between the rate of reaction and the concentration of butadiene.
Instead we calculate
Rate[C4H6]2Rate[C4H6]2
in Table 2. We discover that this
ratio is a constant throughout the reaction. Therefore, the
relationship between the rate of the reaction and the concentration
of the reactant in this case is given by
Rate=-ddt[C4H6]=k[C4H6]2Rate
t[C4H6]k[C4H6]2
(3)
which is the rate law for the reaction in
Figure 5. This is a very interesting
result when compared to
Equation 2. In
both cases, the results demonstrate that the rate of reaction
depends on the concentration of the reactant. However, we now also
know that the way in which the rate varies with the concentration
depends on what the reaction is. Each reaction has its own rate
law, observed experimentally.
We would like to understand what determines
the specific dependence of the reaction rate on concentration in
each reaction. In the first case considered above, the rate depends
on the concentration of the reactant to the first power. We refer
to this as a
first order reaction. In the second case above, the
rate depends on the concentration of the reactant to the second
power, so this is called a
second order reaction. There are also
third order reactions, and even
zeroth order reactions whose rates do not depend on the
amount of the reactant. We need more observations of rate laws for
different reactions.
The approach used in the previous section to
determine a reaction's rate law is fairly clumsy and at this
point difficult to apply. We consider here a more systematic
approach. First, consider the decomposition of
N2O5(g)N2O5(g).
2N2O5(g)→4
NO2(g)+O2(g)2N2O5(g)→4
NO2(g)+O2(g)
We can create an initial concentration of
N2O5N2O5
in a flask and measure the rate at which the
N2O5N2O5
first decomposes. We can then create a different initial
concentration of
N2O5N2O5
and measure the new rate at which the
N2O5N2O5
decomposes. By comparing these rates, we can find the order of the
decomposition reaction. The rate law for decomposition of
N2O5(g)N2O5(g)
is of the general form:
Rate=k[N2O5]m
Rate
k
[N2O5]
m
(4)
so we need to determine the exponent
mm. For
example, at
25°C25°C
we observe that the rate of decomposition is
1.4×10-3Ms1.4-3Ms
when the concentration of
N2O5N2O5
is
0.020M0.020M.
If instead we begin we
[N2O5]=0.010M[N2O5]0.010M,
we observe that the rate of decomposition is
7.0×10-4Ms7.0-4Ms.
We can compare the rate from the first measurement
Rate1Rate1 to
the rate from the second measurement
Rate2Rate2.
From
Equation 4, we can write
that
Rate1Rate2=k[N2O5]1
mk[N2O5]2
m=1.4×10-3Ms7.0×10-4Ms=k0.020Mmk0.010Mm
Rate
1
Rate
2
k
[N2O5]1
m
k
[N2O5]2
m
1.4-3
M
s
7.0-4
M
s
k
0.020
M
m
k
0.010
M
m
(5)
This can be simplified on both sides of the
equation to give
2.0=2.0m2.02.0m
Clearly, then
m=1m1,
and the decomposition is a first order reaction. We can also then
find the first order rate constant
kk for this
reaction by simply plugging in one of the initial rate measurements
to
Equation 4. We find that
k=0.070s-1k0.070s-1.
This approach to finding reaction order is
called the method of initial rates, since it relies on fixing the
concentration at specific initial values and measuring the initial
rate associated with each concentration.
So far we have considered only reactions which
have a single reactant. Consider a second example of the method of
initial rates involving the reaction of hydrogen gas and iodine
gas:
H2(g)+I2(g)→2HI(g)H2(g)+I2(g)→2HI(g)(6)
In this case, we expect to find that the rate
of the reaction depends on the concentrations for both reactants.
As such, we need more initial rate observations to determine the
rate law. In
Table 3,
observations are reported for the initial rate for three sets of
initial concentrations of
H2H2
and
I2I2.
Hydrogen Gas and Iodine Gas Initial Rate Data at 700K
| Experiment |
[H2]0[H2]0
(M) |
[I2]0[I2]0
(M) |
Rate (M/sec) |
| 1 |
0.10 |
0.10 |
3.00×10-43.00-4 |
| 2 |
0.20 |
0.10 |
6.00×10-46.00-4 |
| 3 |
0.20 |
0.20 |
1.19×10-31.19-3 |
Following the same process we used in the
N2O5N2O5
example, we write the general rate law for the reaction as
Rate=k[H2]n[I2]m
Rate
k
[H2]
n
[I2]
m
(7)
By comparing experiment 1 to experiment 2, we
can write
Rate1Rate2=k[H2]1
n[I2]1
mk[H2]2
n[I2]2
m=3.00×10-4Ms6.00×10-4Ms=k0.10Mm0.10Mnk0.20Mm0.10Mn
Rate
1
Rate
2
k
[H2]1
n
[I2]1
m
k
[H2]2
n
[I2]2
m
3.00-4
M
s
6.00-4
M
s
k
0.10
M
m
0.10
M
n
k
0.20
M
m
0.10
M
n
(8)
This simplifies to
0.50=0.50m1.00n0.500.50m1.00n
from which it is clear that
m=1m1.
Similarly, we can find that
n=1n1.
The reaction is therefore first order in each reactant and is
second order overall.
Rate=k[H2][I2]
Rate
k
[H2]
[I2]
(9)
Once we know the rate law, we can use any of
the data from
Table 3 to
determine the rate constant, simply by plugging in concentrations
and rate into
Equation 9. We find that
k=3.00×10-21Msk3.00-21Ms.
This procedure can be applied to any number of
reactions. The challenge is preparing the initial conditions and
measuring the initial change in concentration precisely versus
time. Table 4 provides an
overview of the rate laws for several reactions. A variety of
reaction orders are observed, and they cannot be easily correlated
with the stoichiometry of the reaction.
Rate Laws for Various Reactions
| Reaction |
Rate Law |
| 2NO(g)+O2(g)→2NO2(g)2NO(g)+O2(g)→2NO2(g) |
Rate=k[NO]2[O2]Ratek[NO]2[O2] |
| 2NO(g)+2H2(g)→2N2(g)+2H2O(g)2NO(g)+2H2(g)→2N2(g)+2H2O(g) |
Rate=k[NO]2[H2]Ratek[NO]2[H2] |
| 2ICl(g)+H2(g)→2HCl(g)+I2(g)2ICl(g)+H2(g)→2HCl(g)+I2(g) |
Rate=k[ICl][H2]Ratek [ICl][H2] |
| 2N2O5(g)→4NO2(g)+O2(g)2N2O5(g)→4NO2(g)+O2(g) |
Rate=k[N2O5]Ratek [N2O5] |
| 2NO2(g)+F2(g)→2NO2F(g)2NO2(g)+F2(g)→2NO2F(g) |
Rate=k[NO2][F2]Ratek [NO2][F2] |
| 2H2O2(aq)→2H2O(l)+O2(g)2H2O2(aq)→2H2O(l)+O2(g) |
Rate=k[H2O2]Ratek [H2O2] |
| H2(g)+Br2(g)→2HBr(g)H2(g)+Br2(g)→2HBr(g) |
Rate=k[H2][Br2]12Ratek [H2][Br2]12 |
| O3(g)+Cl(g)→O2(g)+ClO(g)O3(g)+Cl(g)→O2(g)+ClO(g) |
Rate=k[O3][Cl]Ratek [O3][Cl] |
Once we know the rate law for a reaction, we
should be able to predict how fast a reaction will proceed. From
this, we should also be able to predict how much reactant remains
or how much product has been produced at any given time in the
reaction. We will focus on the reactions with a single reactant to
illustrate these ideas.
Consider a first order reaction like
A→productsA→products,
for which the rate law must be
Rate=-ddt[A]
=k[A]
Rate
t
[A]
k
[A]
(10)
From Calculus, it is possible to use
Equation 10 to find the function
[A]t[A]t
which tells us the concentration
[A][A]
as a function of time. The result is
[A]
=[A]0
ⅇ-kt
[A]
[A]0
k
t
(11)
or equivalently
ln[A]
=ln[A]0
-kt
[A]
[A]0
k
t
(12)
Equation 12
reveals that, if a reaction is first order, we can plot
ln[A][A]
versus time and get a straight line with slope equal to
-kk.
Moreover, if we know the rate constant and the initial
concentration, we can predict the concentration at any time during
the reaction.
An interesting point in the reaction is the
time at which exactly half of the original concentration of
AA has been
consumed. We call this time the
half life of the reaction and denote it as
t12t12.
At that time,
[A]=12[A]0[A]12[A]0.
From Equation 12 and using the
properties of logarithms, we find that, for a first order
reaction
t12=ln2k
t12
2
k
(13)
This equation tells us that the half-life of a
first order reaction does not depend on how much material we start
with. It takes exactly the same amount of time for the reaction to
proceed from all of the starting material to half of the starting
material as it does to proceed from half of the starting material
to one-fourth of the starting material. In each case, we halve the
remaining material in a time equal to the constant half-life in
Equation 13.
These conclusions are only valid for first
order reactions. Consider then a second order reaction, such as the
butadiene dimerization discussed above. The general
second order reaction
A→productsA→products
has the rate law
Rate=-ddt[A]
=k[A]
2
Rate
t
[A]
k
[A]
2
(14)
Again, we can use Calculus to find the
function
[A]t[A]t
from
Equation 14. The result is most
easily written as
1[A]
=1[A]0
+kt
1
[A]
1
[A]
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