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.
Table 1: 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.
Table 2: 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.
Table 3: 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.
Table 4: 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]0
k
t
(15)
Note that, as
tt increases,
1[A]1[A]
increases, so
[A][A]
decreases.
Equation 15 reveals that,
for a reaction which is second order in the reactant
AA, we can
plot
1[A]1[A]
as a function of time to get a straight line with slope equal to
kk. Again, if
we know the rate constant and the initial concentration, we can
find the concentration
[A][A]
at any time of interest during the reaction.
The half-life of a second order reaction
differs from the half-life of a first order reaction. From Equation 15, if we take
[A]=12[A]0[A]12[A]0,
we get
t12=1k[A]0
t12
1
k
[A]0
(16)
This shows that, unlike a first order
reaction, the half-life for a second order reaction depends on how
much material we start with. From
Equation 16, the more concentrated the reactant
is, the shorter the half-life.
It is a common observation that reactions tend
to proceed more rapidly with increasing temperature. Similarly,
cooling reactants can have the effect of slowing a reaction to a
near halt. How is this change in rate reflected in the rate law
equation,
e.g. Equation 9? One
possibility is that there is a slight dependence on temperature of
the concentrations, since volumes do vary with temperature.
However, this is insufficient to account for the dramatic changes
in rate typically observed. Therefore, the temperature dependence
of reaction rate is primarily found in the rate constant,
kk.
Consider for example the reaction of hydrogen
gas with iodine gas at high temperatures, as given in Equation 6. The rate constant of this reaction
at each temperature can be found using the method of initial rates,
as discussed above, and we find in Table 5 that the rate constant increases
dramatically as the temperature increases.
Table 5: Rate Constant for Hydrogen Gas and Iodine Gas
| T (K) |
k
(1Ms1Ms) |
| 667 |
6.80×10-36.80-3 |
| 675 |
9.87×10-39.87-3 |
| 700 |
3.00×10-23.00-2 |
| 725 |
8.43×10-28.43-2 |
| 750 |
2.21×10-12.21-1 |
| 775 |
5.46×10-15.46-1 |
| 800 |
1.27 |
As shown in Figure 6, the rate constant appears to
increase exponentially with temperature. After a little
experimentation with the data, we find in Figure 7 that there is a simple linear
relationship between
lnkk
and
1T1T.
From Figure 7,
we can see that the data in Table 4 fit the equation
lnk=a1T+b
k
a
1
T
b
(17)
where
aa and
bb are
constant for this reaction. It turns out that, for our purposes,
all reactions have rate constants which fit
Equation 17, but with different constants
aa and
bb for each
reaction.
Figure 7 is referred to as
an
Arrhenius plot, after Svante Arrhenius.
It is very important to note that the form of
Equation 17 and the appearance of
Figure 7 are both the same as the
equations and graphs for the temperature dependence of the
equilibrium constant for an endothermic reaction. This suggests a
model to account for the temperature dependence of the rate
constant, based on the energetics of the reaction. In particular,
it appears that the reaction rate is related to the amount of
energy required for the reaction to occur. We will develop this
further in the next section.
At this point, we have only observed the
dependence of reaction rates on concentration of reactants and on
temperature, and we have fit these data to equations called rate
laws. Although this is very convenient, it does not provide us
insight into why a particular reaction has a specific rate law or
why the temperature dependence should obey Equation 17. Nor does it provide any physical
insights into the order of the reaction or the meaning of the
constants aa
and bb in
Equation 17.
We begin by asking why the reaction rate
should depend on the concentration of the reactants. To answer
this, we consider a simple reaction between two molecules in which
atoms are transferred between the molecules during the reaction.
For example, a reaction important in the decomposition of ozone
O3O3 by
aerosols is
O3
(g)+
Cl(g)
→O2
(g)+
ClO(g
)
O3
(g)+
Cl(g)
→O2
(g)+
ClO(g
)
What must happen for a reaction to occur
between an
O3O3
molecule and a
ClCl atom?
Obviously, for these two particles to react, they must come into
close proximity to one another so that an
OO atom can be
transferred from one to the other. In general, two molecules cannot
trade atoms to produce new product molecules unless they are close
enough together for the atoms of the two molecules to interact.
This requires a collision between molecules.
The rate of collisions depends on the
concentrations of the reactants, since the more molecules there are
in a confined space, the more likely they are to run into each
other. To write this relationship in an equation, we can think in
terms of probability, and we consider the reaction above. The
probability for an
O3O3
molecule to be near a specific point increases with the number of
O3O3
molecules, and therefore increases with the concentration of
O3O3
molecules. The probability for a
ClCl atom to be
near that specific point is also proportional to the concentration
of ClCl atoms.
Therefore, the probability for an
O3O3
molecule and a
ClCl atom to be
in close proximity to the same specific point at the same time is
proportional to the
[O3][O3]
times
[Cl][Cl].
It is important to remember that not all
collisions between
O3O3
molecules and
ClCl atoms will
result in a reaction. There are other factors to consider including
how the molecules approach one another. The atoms may not be
positioned properly to exchange between molecules, in which case
the molecules will simply bounce off of one another without
reacting. For example, if the
ClCl atom
approaches the center
OO atom of the
O3O3
molecule, that
OO atom will
not transfer. Another factor is energy associated with the
reaction. Clearly, though, a collision must occur for the reaction
to occur, and therefore there rate of the reaction can be no faster
than the rate of collisions between the reactant molecules.
Therefore, we can say that, in a
bimolecular reaction, where two molecules
collide and react, the rate of the reaction will be proportional to
the product of the concentrations of the reactants. For the
reaction of
O3O3
with ClCl, the
rate must therefore be proportional to
[O3][Cl][O3][Cl],
and we observe this in the experimental rate law in Table 4. Thus, it appears that we can
understand the rate law by understanding the collisions which must
occur for the reaction to take place.
We also need our model to account for the
temperature dependence of the rate constant. As noted at the end of
the last
section, the temperature dependence of the rate
constant in Equation 17 is the same as
the temperature dependence of the equilibrium constant for an
endothermic reaction. This suggests that the temperature dependence
is due to an energetic factor required for the reaction to occur.
However, we find experimentally that Equation 17 describes the rate constant
temperature dependence regardless of whether the reaction is
endothermic or exothermic. Therefore, whatever the energetic factor
is that is required for the reaction to occur, it is not just the
endothermicity of the reaction. It must be that all reactions,
regardless of the overall change in energy, require energy to
occur.
A model to account for this is the concept of
activation energy. For a reaction to occur, at least
some bonds in the reactant molecule must be broken, so that atoms
can rearrange and new bonds can be created. At the time of
collision, bonds are stretched and broken as new bonds are made.
Breaking these bonds and rearranging the atoms during the collision
requires the input of energy. The minimum amount of energy required
for the reaction to occur is called the activation energy,
EaEa.
This is illustrated in Figure 8,
showing conceptually how the energy of the reactants varies as the
reaction proceeds. In Figure 8(a), the
energy is low early in the reaction, when the molecules are still
arranged as reactants. As the molecules approach and begin to
rearrange, the energy rises sharply, rising to a maximum in the
middle of the reaction. This sharp rise in energy is the activation
energy, as illustrated. After the middle of the reaction has passed
and the molecules are arranged more as products than reactants, the
energy begins to fall again. However, the energy does not fall to
its original value, so this is an endothermic reaction.
Figure 8(b) shows
the analogous situation for an exothermic reaction. Again, as the
reactants approach one another, the energy rises as the atoms begin
to rearrange. At the middle of the collision, the energy maximizes
and then falls as the product molecules form. In an exothermic
reaction, the product energy is lower than the reactant
energy.
Figure 8 thus
shows that an energy barrier must be surmounted for the reaction to
occur, regardless of whether the energy of the products is greater
than (Figure 8(a)) or less than
(Figure 8(b)) the energy of the
reactants. This barrier accounts for the temperature dependence of
the reaction rate. We know from the kinetic molecular theory that
as temperature increases the average energy of the molecules in a
sample increases. Therefore, as temperature increases, the fraction
of molecules with sufficient energy to surmount the reaction
activation barrier increases.
Although we will not show it here, kinetic
molecular theory shows that the fraction of molecules with energy
greater than
EaEa
at temperature
TT is
proportional to
ⅇ-EaRTEaRT.
This means that the reaction rate and therefore also the rate
constant must be proportional to
ⅇ-EaRTEaRT.
Therefore we can write
kT=Aⅇ-Ea
RT
k
T
A
Ea
R
T
(18)
where
AA is a
proportionality constant. If we take the logarithm of both sides of
Equation 18, we find that
lnkT=-Ea
RT+lnA
k
T
Ea
R
T
A
(19)
This equation matches the experimentally
observed
Equation 17. We recall that a
graph of
lnkk
versus
1T1T
is observed to be linear. Now we can see that the slope of that
graph is equal to
-EaREaR.
As a final note on Equation 19, the constant
AA must have
some physical significant. We have accounted for the probability of
collision between two molecules and we have accounted for the
energetic requirement for a successful reactive collision. We have
not accounted for the probability that a collision will have the
appropriate orientation of reactant molecules during the collision.
Moreover, not every collision which occurs with proper orientation
and sufficient energy will actually result in a reaction. There are
other random factors relating to the internal structure of each
molecule at the instant of collision. The factor
AA takes
account for all of these factors, and is essentially the
probability that a collision with sufficient energy for reaction
will indeed lead to reaction.
AA is commonly
called the
frequency factor.
Our collision model in the previous section
accounts for the concentration and temperature dependence of the
reaction rate, as expressed by the rate law. The concentration
dependence arises from calculating the probability of the reactant
molecules being in the same vicinity at the same instant.
Therefore, we should be able to predict the rate law for any
reaction by simply multiplying together the concentrations of all
reactant molecules in the balanced stoichiometric equation. The
order of the reaction should therefore be simply related to the
stoichiometric coefficients in the reaction. However, Table 4 shows that this is incorrect for
many reactions.
Consider for example the apparently simple
reaction
2ICl(g)+H2(g)→2HCl(g)+I2(g)2ICl(g)+H2(g)→2HCl(g)+I2(g)
(20)
Based on the collision model, we would assume
that the reaction occurs by
2ICl2ICl
molecules colliding with a single
H2H2 molecule. The probability for
such a collision should be proportional to
[ICl]2[H2][ICl]2[H2].
However, experimentally we observe (see
Table 4) that the rate law for this
reaction is
Rate=k[ICl][H2]Ratek [ICl][H2](21)
As a second example, consider the
reaction
NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g)
(22)
It would seem reasonable to assume that this
reaction occurs as a single collision in which an oxygen atom is
exchanged between the two molecules. However, the experimentally
observed rate law for this reaction is
Rate=k[NO2]2Ratek[NO2]2(23)
In this case, the
[CO][CO]
concentration does not affect the rate of the reaction at all, and
the
[NO2][NO2]
concentration is squared. These examples demonstrate that the rate
law for a reaction cannot be predicted from the stoichiometric
coefficients and therefore that the collision model does not
account for the rate of the reaction. There must be something
seriously incomplete with the collision model.
The key assumption of the collision model is
that the reaction occurs by a single collision. Since this
assumption leads to incorrect predictions of rate laws in some
cases, the assumption must be invalid in at least those cases. It
may well be that reactions require more than a single collision to
occur, even in reactions involving just two types of molecules as
in Equation 22. Moreover, if more than
two molecules are involved as in Equation 20, the chance of a single collision
involving all of the reactive molecules becomes very small. We
conclude that many reactions, including those in Equation 20 and Equation 22, must occur as a result of several
collisions occurring in sequence, rather than a single collision.
The rate of the chemical reaction must be determined by the rates
of the individual steps in the reaction.
Each step in a complex reaction is a single
collision, often referred to as an
elementary process. In single collision process step,
our collision model should correctly predict the rate of that step.
The sequence of such elementary processes leading to the overall
reaction is referred to as the
reaction mechanism. Determining the mechanism for a
reaction can require gaining substantially more information than
simply the rate data we have considered here. However, we can gain
some progress just from the rate law.
Consider for example the reaction in Equation 22 described by the rate law in
Equation 23. Since the rate law
involved
[NO2]2[NO2]2,
one step in the reaction mechanism must involve the collision of
two
NO2NO2
molecules. Furthermore, this step must determine the rate of the
overall reaction. Why would that be? In any multi-step process, if
one step is considerably slower than all of the other steps, the
rate of the multi-step process is determined entirely by that
slowest step, because the overall process cannot go any faster than
the slowest step. It does not matter how rapidly the rapid steps
occur. Therefore, the slowest step in a multi-step process is thus
called the
rate determining or
rate limiting step.
This argument suggests that the reaction in
Equation 22 proceeds via a slow step
in which two
NO2NO2
molecules collide, followed by at least one other rapid step
leading to the products. A possible mechanism is therefore
NO2+NO2→NO3+NO
NO2+NO2→NO3+NO
(24)
NO3+CO→NO2+CO2
NO3+CO→NO2+CO2
(25)
If
Step
1 is much slower than
Step 2, the rate of the
reaction is entirely determined by the rate of
Step 1. From our
collision model, the rate law for
Step 1 must be
Rate=k[NO2]2Ratek[NO2]2,
which is consistent with the experimentally observed rate law for
the overall reaction. This suggests that the mechanism in
Equation 24 and
Equation 25 is the correct description of the
reaction process for
Equation 22, with
the first step as the rate determining step.
There are a few important notes about the
mechanism. First, one product of the reaction is produced in the
first step, and the other is produced in the second step.
Therefore, the mechanism does lead to the overall reaction,
consuming the correct amount of reactant and producing the correct
amount of reactant. Second, the first reaction produces a new
molecule,
NO3NO3,
which is neither a reactant nor a product. The second step then
consumes that molecule, and
NO3NO3
therefore does not appear in the overall reaction, Equation 22. As such,
NO3NO3
is called a
reaction intermediate. Intermediates play important
roles in the rates of many reactions.
If the first step in a mechanism is rate
determining as in this case, it is easy to find the rate law for
the overall expression from the mechanism. If the second step or
later steps are rate determining, determining the rate law is
slightly more involved. The process for finding the rate law in
such a case is illustrated in Exercise 11.
When
C60O3C60O3
in toluene solution decomposes,
O2O2
is released leaving
C60O3C60O3
in solution.
Based on the data in Figure 2
and Figure 3, plot the concentration
of
C60OC60O
as a function of time.
How would you define the rate of the reaction in terms of the
slope of the graph from above? How is the rate
of appearance of
C60OC60O
related to the rate of disappearance of
C60O3C60O3?
Based on this, plot the rate of appearance of
C60OC60O
as a function of time.
The reaction 2N2O5(g)→4NO2(g)+O2(g)2N2O5(g)→4NO2(g)+O2(g) was found in this
study to have rate law given by
Rate=k[N2O5]Ratek[N2O5]
with
k=0.070s-1k0.070s-1.
How is the rate of appearance of
NO2NO2
related to the rate of disappearance of
N2O5N2O5?
Which rate is larger?
Based on the rate law and rate constant, sketch a plot of
[N2O5][N2O5],
[NO2][NO2],
and
[O2][O2]
versus time all on the same graph.
For which of the reactions listed in
Table 4 can you be certain that
the reaction does not occur as a single step collision? Explain
your reasoning.
Consider two decomposition reactions for
two hypothetical materials, A and B. The decomposition of A is
found to be first order, and the decomposition of B is found to be
second order.
Assuming that the two reactions have the same rate constant
at the same temperature, sketch
[A][A]
and
[B][B]
versus time on the same graph for the same initial conditions,
i.e. [A]0=[B]0[A]0[B]0.
Compare the half-lives of the two reactions. Under what
conditions will the half-life of B be less than the half-life of A?
Under what conditions will the half-life of B be greater than the
half-life of A?
A graph of the logarithm of the equilibrium
constant for a reaction versus
1T1T
is linear but can have either a negative slope or a positive slope,
depending on the reaction, as was observed here. However, the
graph of the logarithm of the rate constant for a reaction versus
1T1T
has a negative slope for essentially every reaction. Using
equilibrium arguments, explain why the graph for the rate constant
must have a negative slope.
Using Equation 18 and the data in Table 5, determine the activation energy for the reaction H2(g)+I2(g)→2HI(g)H2(g)+I2(g)→2HI(g).
We found that the rate law for the reaction H2(g)+I2(g)→2HI(g)H2(g)+I2(g)→2HI(g) is
Rate=k[H2][I2]
Rate
k
[H2]
[I2]
. Therefore, the
reaction is
second order overall but
first order in
H2H2.
Imagine that we start with
[H2]0=[I2]0[H2]0[I2]0
and we measure
[H2][H2]
versus time. Will a graph of
ln[H2][H2]
versus time be linear or will a graph of
1[H2]1[H2]
versus time be linear? Explain your reasoning.
As a rough estimate, chemists often assume
a
rule of thumb that the rate of any reaction
will double when the temperature is increased by
10°C10°C.
What does this suggest about the activation energies of
reactions?
Using Equation 18, calculate the activation energy of a
reaction whose rate doubles when the temperature is raised from
25°C25°C
to
35°C35°C.
Does this rule of thumb estimate depend on the temperature
range? To find out, calculate the factor by which the rate constant
increases when the temperature is raised from
100°C100°C
to
110°C110°C,
assuming the same activation energy you found above. Does the rate
double in this case?
Consider a very simple hypothetical
reaction
A+B↔2C
A+B↔2C
which comes to
equilibrium.
At equilibrium, what must be the relationship between the
rate of the forward reaction,
A+B→2C
A+B→2C
and the reverse
reaction
2C→A+B
2C→A+B
?
Assume that both the forward and reverse reactions are
elementary processes occurring by a single collision. What is the
rate law for the forward reaction? What is the rate law for the
reverse reaction?
Using the previous results from here and here, show that the
equilibrium constant for this reaction can be calculated from
Kc=kfkrKckfkr, where kfkf is the
rate constant for the forward reaction and krkr is the rate constant
for the reverse reaction.
Consider a very simple hypothetical
reaction
A+B↔C+D
A+B↔C+D
. By examining
Figure 8, provide and explain the
relationship between the activation energy in the forward
direction, Ea,fEa,f, and in the reverse direction, Ea,rEa,r . Does this
relationship depend on whether the reaction is endothermic
(Figure 8(a)) or exothermic (Figure 8(b))? Explain.
For the reaction H2(g)+I2(g)→2HI(g)H2(g)+I2(g)→2HI(g), the rate law is
Rate=k[H2][I2]
Rate
k
[H2]
[I2]
. Although this
suggests that the reaction is a one-step elementary process, there
is evidence that the reaction occurs in two steps, and the second
step is the rate determining step:
I2↔2I
I2↔2I
(26)
H2+2I→2HI
H2+2I→2HI
(27)
Where
Step 1 is fast and
Step2 is
slow.
If the both the forward and reverse reactions in Step 1 are much
faster than Step2, explain why
Step 1 can
be considered to be at equilibrium.
What is the rate law for the rate determining step?
Since the rate law above depends on the
concentration of an intermediate
II, we need to
find that intermediate. Calculate
[I][I]
from Step 1,
assuming that Step1 is at equilibrium.
Substitute
[I][I]
from above into
the rate law found previously to find the overall rate law for the
reaction. Is this result consistent with the experimental
observation?
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