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4. Reaction Order and Rate Constants
Separating the variables and integrating, using the initial condition that for t = 0, [A] = [A]0,
gives
−∫
[A]
[A]0
t
d[ A]
= k1 ∫ dt
0
[ A]
(4.2)
from which
− ln [ A ] + ln [ A ]0 = k1t
(4.3)
or
log[ A] = −
k1 t
+ log[ A]0
2.303
(4.4)
These equations show that for a first-order reaction, the logarithm of the concentration is
directly proportional to time, and that the proportionality constant is the rate constant.
These equations are often given in the exponential form
[A] = [A]0 exp ( − k1t )
(4.5)
[ A ]0 = [ A ] + [ B]
(4.6)
Using the relationship
we can describe the variation of product concentration with time
[B] = [A]0 ⎡1 − exp ( − k1t )⎤
⎣
⎦
(4.7)
The confirmation of the reaction order is obtained by noting that for a first-order process
the rate is directly proportional to the concentration. This dependence can be shown using
the initial velocity for various well-defined concentrations. The initial rate for each concentration is obtained from the tangent at time t = 0 of the experimental concentration–time
curve. For the solvolysis of 2-chloro-2-methylpropane in water, the initial velocities
obtained for different concentrations show a linear dependence on the initial concentration
that passes through the origin, with a slope k1, confirming that the reaction is first order
(see Figure 3.2).
The analysis of the kinetic data following the differential law of eq. (4.1) requires the
initial determination of the tangents to the curve of time dependence versus concentration,
which can be somewhat subjective, followed by the graph of initial velocity as a function
of the initial concentration. The combination of these steps can lead to errors.
Alternatively, the same kinetic data can be analysed more quickly and securely using the
integrated rate law of eq. (4.3). Figure 4.1 shows that, provided that one is reasonably far
from the end of the reaction, the logarithm of the concentration is a linear function of the
reaction time. The slope of these plots (using natural logarithms) gives the rate constant.
If logarithms to the base 10 are used, they need to be multiplied by ln (10) = 2.303. For
the set of data shown in Figure 4.1 the slope for the lowest concentration of 2-chloro-2methylpropane is seen to be much less than those for the other concentrations. This is
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79
Figure 4.1 Dependence of the logarithm of the concentration of reactant upon time for the solvolysis of 2-chloro-2-methylpropane.
likely to be caused by experimental error in the measurement of small changes in conductivity at these low concentrations.
Another parameter commonly used to characterise the rates of a first-order reaction is
the half-life, t½. For any kinetic process, this is defined as the time taken for the concentration of reactant to be reduced to half its initial value. For a first-order process, we can
write from eq. (4.5)
[ A]0 / 2
= exp(−k1t1 2 )
[ A]0
(4.8)
from which
t1 2 =
ln 2 0.693
=
k1
k1
(4.9)
We can see from the above equation that for a first-order process the half-life does not
depend upon the initial concentration. This is seen with the data on the solvolysis of
2-chloro-2-methylpropane in water, where from Table 3.1, the first half-life is about 210
sec, giving a rate constant k1 = 3.3ϫ10Ϫ3 secϪ1 , while for the second half-life, corresponding to the time necessary for the concentration of the reactant to be reduced from [A]0/2
to [A]0/4, the value is virtually identical. Similarly, the half-lives for all steps of this reaction are the same within experimental error. This provides an excellent technique for confirming the reaction order, and, as a working definition, it is normally accepted that if the
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4. Reaction Order and Rate Constants
integrated rate plot of the logarithm of change in concentration as a function of time is linear for three half-lives (which corresponds to 87.5% of the reaction, and implies that the
first three half-lives are identical), the kinetics can be considered to follow good first-order
behaviour. One area where the concept of half-lives is commonly used is in the decay of
radioactive nuclei. All these reactions follow first-order kinetics.
4.1.2
Second-order reactions
Returning to the dimerisation of 2,5-dimethyl-3,4-diphenylcyclopentadienone, discussed
in Chapter 3, the second-order rate law in this case is of the type
−
d[ A]
= k2 [ A]2
dt
(4.10)
Separating the variables, and integrating with the initial conditions that for t = 0,
[A] = [A]0, we get
t
d[ A]
= k2 ∫ dt
2
[ A ]0 [ A ]
0
−∫
[A]
(4.11)
from which
1
1
= k2 t +
[ A]
[ A]0
(4.12)
The experimental data presented in Chapter 3 allow us to analyse the kinetics of this reaction using the differential law of equation (4.10), and to obtain the dependence of the
dienone concentration upon time. To confirm the reaction order, we can take the logarithms
of eq. (4.10) to give
⎛ d[ A] ⎞
log ⎜ −
= log(k2 ) + 2 log([ A])
⎝ dt ⎟
⎠
(4.13)
This predicts that a plot of logarithm of rate as a function of logarithm of monomer concentration will give a straight line whose slope gives the order of the reaction (Figure 3.4),
while the intercept gives the rate constant. However, since this value is derived from
extrapolation over a wide range, which is likely to reduce the precision of the value, it is
normally better to obtain the rate constant by plotting rate as a function of [dienone]2.
An alternative treatment uses the integrated law, eq. (4.12), to study the kinetics of this
reaction. The graph of the reciprocal of concentration as a function of time gives a straight
line whose slope is k2 (Figure 4.2). Again, the integrated rate method is quicker and less
subjective than the differential form.
The half-life can also be calculated from eq. (4.12), using
2
1
= k2 t1 2 +
[ A]0
[ A]0
(4.14)
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Figure 4.2 Relationship between the reciprocal of concentration of dienone and time for the reaction of dimerisation of a dienone.
from which
t1 2 =
1
k2 [ A ]0
(4.15)
The half-life of a reaction of this type depends upon the initial concentration of the
reagents.
Another type of second-order reaction involves two different reactants:
2
A + B ⎯k⎯ products
→
(4.I)
from which
−
d[ A]
d[ B]
=−
= k2 [ A][ B]
dt
dt
(4.16)
Using the condition of mass balance, we obtain
[A ] − [A ]0 = [B] − [B]0
[A ] − [B] = [A ]0 − [B]0
(4.17)
Using, separately, the first and third members, and the second and third members of
eq. (4.16), we obtain the two equations [1]:
d[ A]
= − k2 [ B]dt
[ A]
d[ B]
= − k2 [ A]dt
[ B]
(4.18)
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4. Reaction Order and Rate Constants
and subtraction of these then yields
d[ A] d[ B]
−
= k2 ([ A] − [ B]) dt
[ A] [ B]
(4.19)
Using the mass balance eq. (4.17), we can write
d[ A] d[ B]
−
= k2 ([ A]0 − [ B]0 ) dt
[ A] [ B]
(4.20)
which can be integrated from time t = 0 to time t, to give
⎛ [ A] ⎞
⎛ [ B] ⎞
ln ⎜
⎟ − ln ⎜ [ B] ⎟ = k2 t ([ B]0 − [ A]0 )
⎝ [ A]0 ⎠
⎝ 0⎠
(4.21)
⎧[ B] [ A] ⎫
1
ln ⎨ 0
⎬ = k2 t
[ A]0 − [ B]0 ⎩[ A]0 [ B] ⎪
⎭
(4.22)
or, rearranging
A plot of log ([B]/[A]) as a function of the time will have a slope of ([B]0 Ϫ[A]0)k2/2.3.
For the case of second-order reactions in which the initial concentration can be expressed
in terms of a single variable, [A] = [B] = [A]0 – x = [B]0 – x. When x = 0 for t = 0, the integration of eq. (4.16), after substitution for this variable, gives,
x
[A ]0 ([A ]0 − x )
4.1.3
= k2 t
(4.23)
Zero-order reactions
Even though concentration of reactants has been given as one of the factors on which reaction rates depend, there are also cases of reactions that are independent of reactant concentration. In this case, the differential rate law is constant
−
d[ A]
= k0
dt
(4.24)
On integrating, we obtain
[ A] = [ A]0 − k0 t
(4.25)
Thus, a zero-order reaction gives a linear plot of change in concentration as a function of
time, with slope equal to –k0. The units of this rate constant are identical to those of the
rate (e.g. litres per second). As with other reaction orders, we can determine the half-life,
which is given by
[ A]0
= [ A]0 − k0 t1 2
2
(4.26)
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where
t1 2 =
[A ]0
2 k0
(4.27)
Zero-order reactions appear in heterogeneous systems when the surface of a solid phase is
saturated with a reactant, but are also seen in certain homogeneous systems.
4.1.4
Third-order reactions
In the same way that differential and integral rates can be defined for first- and second-order reactions, we can also obtain rate laws for third-order reactions of the general type
3
A + B + C ⎯⎯ products
→
k
(4.II)
The differential rate law for this reaction is
−
d[ A]
= k3 [ A][ B][C]
dt
(4.28)
The integration of this equation can be made using the method of partial fractions, which
leads to
⎛ ⎛ [ A] ⎞ [ B]0 −[ C ]0 ⎛ [ B] ⎞ [ C ]0 −[ A ]0 ⎛ [C] ⎞ [ A ]0 −[ B]0 ⎞
1
ln ⎜
⎟
⎜ [ B] ⎟
⎜ [ C] ⎟
⎟
⎠
⎝ 0⎠
⎝ 0⎠
([A]0 − [B]0 ) ([B]0 − [C]0 ) ([C]0 − [A]0 ) ⎜ ⎜ [A]0 ⎟
⎝⎝
⎠
= k3 t
(4.29)
The use of the expressions (4.28) and (4.29) in the determination of the partial orders of
these reactions and their rate constants is not very practicable. As we will discuss later in
this chapter, an alternative is to use the so-called isolation method.
However, as previously discussed, third-order reactions involving the simultaneous collision of three molecules are very unlikely in comparison with alternative reaction mechanisms involving a series of elementary second-order steps. A good example of a reaction
of this type is the oxidation of vitamin C by hexacyanoferrate(III) discussed in the previous chapter. Although the stoichiometry of this reaction shows the consumption of three
molecules of reactants, in reaction (3.X), the mechanism involves a series of simpler steps
in which the reaction order is never greater than two, reactions (3.XI) and (3.XII). This is
a typical example of a complex reaction whose mechanism involves a sequence of elementary steps. The sum of these steps gives the overall stoichiometry of the reaction. The
effect of ionic strength on the kinetics suggests a mechanism with a step involving the
reaction between two ions of the same charge. This excludes reaction (3.X) as the ratedetermining step in the chemical process.
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4.2
RATES OF COMPLEX REACTIONS
As seen above, the majority of kinetic processes involves more than one elementary
step. However, the fact that a mechanism is in agreement with the kinetic law does not
mean that it is the correct mechanism for the reaction. The same kinetic law can, and frequently does, correspond to more than one possible mechanism. A mechanism is always a
theoretical hypothesis of how a reaction occurs. We can never prove a mechanism from the
kinetic behaviour, but can only eliminate certain hypotheses. A good illustration is given
by the gas-phase reaction at 400 ЊC
H 2 + I 2 ⎯⎯ 2 HI
→
(4.III)
studied by Bodenstein in 1894, which was considered to be an elementary process.
However, even at this time he recognised that it was difficult to find chemically simple reactions in the gas phase, i.e. reactions whose rates are proportional to the product of the concentrations of reactants raised to integral powers. Reaction (4.III) was presented as an
example of an elementary kinetic process in all text books until Sullivan in 1967 detected
the presence of free radicals in the system at 350 ЊC, and proposed an alternative mechanism
I 2 ⎯energy → 2I
⎯⎯
H 2 + 2I ⎯⎯ 2HI
→
(4.IV)
2I + M ⎯⎯ I 2 + M
→
where M is an inert gas. In summary, there are a number of important questions needed to
test this mechanism. Mechanisms of the reactions are proposed on the basis of the
observed kinetic behaviour. Once proposed, a very detailed physical chemical study is
necessary before any mechanism can be considered as accepted. It is necessary to confirm
the feasibility of all the steps proposed in addition to demonstrating the possible existence
of the different intermediates. Because of various experimental difficulties, these tasks are
difficult to carry out for the majority of the chemical reactions. In addition to kinetic studies, involving, for example, predictions of the effects of various factors on rates and product distributions, these studies must be complemented by other chemical and physical
studies, in particular attempts to try to trap and characterise the intermediates.
Complex reaction mechanisms can conveniently be grouped within the following classification: consecutive reactions, parallel reactions and reversible reactions. Parallel reactions are those in which the same species participates in two or more competitive steps.
Consecutive reactions are characterised by the product of the first reaction being a reactant
in a subsequent process, leading to formation of the final product. Reversible reactions are
those in which the products of the initial reaction can recombine to regenerate the reactant.
As complex reactions follow a reaction mechanism involving various elementary steps, the
determination of the corresponding kinetic law involves the solution of a system of differential equations, and the complete analytical solution of these systems is only possible for the
simplest cases. In slightly more complicated cases it may still be possible to resolve the system of corresponding differential equations using methods such as Laplace transforms or
matrix methods. However, there are systems which cannot be resolved analytically, or whose
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analytical solution is so complex that it is not easily applied. In the absence of information on
the orders of magnitude of the rate constants involved, the treatment of these kinetic systems
is made using numerical methods. These methods allow us to obtain concentrations of the
reactants for discrete time intervals, which can then be represented graphically and compared
with experimental data to provide an insight into the changes occurring. Further, although the
results obtained numerically are inherently approximations, we can estimate and allow for the
errors involved, which can be expressed as error limits when compared with experimental
data. In some cases, it may also be possible to introduce some changes of a chemical nature
in the system to help simplify and analyse these complex systems. For example, if we know
the relative magnitudes of some of the rate constants, it may be possible to simplify the system of complex differential equations involved by modifying the initial concentrations of
reactants and obtaining a rate equation that can be solved analytically. Finally, when these
procedures cannot be used to resolve complex systems of differential equations, it is still possible to use stochastic treatments such as Markov chains or the Monte Carlo method.
4.2.1
Parallel first-order reactions
The simplest parallel reactions involve two competitive first-order steps:
1
A ⎯k⎯ B
→
(4.V)
2
A ⎯k⎯ C
→
The rate of disappearance of A is given by the differential equation
d[ A]
= − k1 [ A] − k2 [ A] = kT [ A]
dt
(4.30)
where kT = k1 + k2. Integrating this equation gives
− k +k t
[ A] = [ A]0 e − kT t = [ A]0 e ( 1 2 )
(4.31)
To determine the rate of formation of B, we have
d[ B]
= k1 [ A]
dt
(4.32)
and, substituting for [A] using expression (4.31), followed by integration with the condition that for t = 0, [B] = 0, we obtain
[ B] =
(
k1 [ A]0
1 − e − kT t
kT
)
(4.33)
We can obtain the variation of [C] with time in the same way:
[ C] =
(
k2 [ A]0
1 − e − kT t
kT
)
(4.34)
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Figure 4.3 Variation of the concentrations of reactant and final products for parallel first-order reactions. The full lines correspond to the situation in which k1 = k2 and the dashed lines correspond to
the case in which k1 = 2k2.
From the above two expressions we see immediately that the ratio of the rate constants of
parallel first-order reactions is equal to the ratio of yields of the products.
[ B] k1
=
[ C] k 2
(4.35)
This quotient is called the branching ratio of the reaction.
Figure 4.3 shows the time evolution of [A], [B] and [C] for the cases in which k1 = k2
and k1 = 2k2.
This solution can easily be extended to parallel reactions involving more than two
processes. If there are n parallel reactions, then kT = k1 + k2 + … + kn, and the expression
for the disappearance of A has the same form as eq. (4.30).
4.2.2
Consecutive first-order reactions
Another very common kinetic scheme involves a series of first-order reactions, leading first
to the formation of an intermediate B, which subsequently reacts to give the final product
1
2
A ⎯k⎯ B ⎯k⎯ C
→
→
(4.VI)
The differential equations describing this sequence of elementary steps are
d[ A]
= − k1 [ A]
dt
d[ B]
= k1 [ A] − k2 [ B]
dt
d[C]
= k2 [ B]
dt
(4.36)
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As we have seen, the integrated rate equation for the disappearance of A is given by
[ A] = [ A]0 e − k1t
(4.37)
and can be substituted for [A] in the equation describing the variation of [B] with time.
After rearranging, this can be written as
d[ B]
+ k2 [ B] = k1 [ A]0 e − k1t
dt
(4.38)
Multiplying both terms by exp (k2t), we obtain [2]
e k2 t
d[ B]
k −k t
+ k2 [ B]e k2 t = k1 [ A]0 e ( 2 1 )
dt
(4.39)
and as
e k2 t
(
d [ B]e k2 t
d[ B]
+ k2 [ B]e k2 t =
dt
dt
)
(4.40)
the equation can be written as the differential equation
(
)
d [ B]e k2 t = k1 [ A]0 e ( 2
k − k1 ) t
dt
(4.41)
which, upon integration, leads to
[ B]e k2 t =
k −k t
k1 [ A]0 e ( 2 1 )
+I
k2 − k1
(4.42)
where the integration constant I is obtained for the simplest case of [B] = 0 at t = 0, as
I=−
k1 [ A]0
(4.43)
(k2 − k1 )
Substituting this value for I in eq. (4.42) and dividing by exp (k2t), we obtain the integrated
equation for the dependence of [B] upon time:
[ B] =
which is true for the case in which k1
(
(
k1 [ A]0 e − k1t − e − k2 t
)
k2 − k1
(4.44)
k2. When k1 = k2, we obtain
)
d [ B]e k2 t = k1 [ A]0 dt
(4.45)
[ B] = k1 [ A]0 e − k1t t
(4.46)
whose solution is
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Figure 4.4 Variation of the concentrations of reactant, intermediate and final product with time for
consecutive first-order reactions. The solid line shows the case in which k1 = k2, while the dashed
line is for the condition k1 = 10k2.
when [B] = 0 at t = 0. Figure 4.4 shows the time evolution of the concentrations of reactant, intermediate and final product for the cases in which k1 = k2 and k1 = 10k2.
4.2.3
Reversible first-order reactions
The simplest reversible reactions are of the type
A
kf
kr
B
(4.VII)
The corresponding differential equations for this mechanism are
d[ A]
= kf [ A] − kr [ B]
dt
d[ B]
−
= kr [ B] − kf [ A]
dt
−
(4.47)
Assuming that at time t = 0 the initial concentrations of the two species are [A]0 and [B]0,
we can write the law of conservation of mass for this system:
[A]0 + [B]0 = [A] + [B]
(4.48)
and solving in terms of [B] we obtain
[B] = [A]0 + [B]0 − [A]
(4.49)