rates of complex reactions

7/24/2019 RATES OF COMPLEX REACTIONS

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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 fre-

quently 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 C

(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 reac-

tions in the gas phase, i.e. reactions whose rates are proportional to the product of the con-centrations 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 C, and proposed an alternative mechanism

(4.IV)

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 confirmthe 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 stud-

ies, involving, for example, predictions of the effects of various factors on rates and prod-

uct 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 clas-

sification: consecutive reactions, parallel reactions and reversible reactions. Parallel reac-

tions 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, thedetermination of the corresponding kinetic law involves the solution of a system of differen-

tial 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 sys-

tem of corresponding differential equations using methods such as Laplace transforms or

matrix methods. However, there are systems which cannot be resolved analytically, or whose

I 2I

H 2I 2HI

2I M I M

2

energy

2

2

+

+ +

H I 2 HI2 2

+

84 4. Reaction Order and Rate Constants


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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 systemsis 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 sys-

tem 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 pos-

sible 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:

(4.V)

The rate of disappearance of A is given by the differential equation

(4.30)

where kT = k1 + k2. Integrating this equation gives

(4.31)

To determine the rate of formation of B, we have

(4.32)

and, substituting for [A] using expression (4.31), followed by integration with the condi-

tion that for t = 0, [B] = 0, we obtain

(4.33)

We can obtain the variation of [C] with time in the same way:

(4.34)[ ][ ]

1 e0CA

T

T= ( )k

k

k t2

[ ]

[ ]

1 e

0

B

A

T

T

= ( )k

k

k t1

d[ ]

d[ ]

BA

tk=

1

[ ] [ ] [ ] e0 0

A A e AT= = +( )k t k k t1 2

d[ ]

d[ ] [ ] [ ]

AA A A

Tt

k k k= =1 2

A B

A C

1

2

k

k

4.2 Rates of Complex Reactions 85


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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.

(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 + + k

n, 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

(4.VI)

The differential equations describing this sequence of elementary steps are

(4.36)

d[ ]d

[ ]

d[ ]

d[ ] [ ]

d[ ]

d[ ]

A A

BA B

CB

tk

tk k

tk

=

=

=

1

1 2

2

A B C1 2k k

[ ]

[ ]

B

C=

k

k

1

2


Figure 4.3 Variation of the concentrations of reactant and final products for parallel first-order reac-tions. The full lines correspond to the situation in which k1 = k2 and the dashed lines correspond tothe case in which k1 = 2k2.


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As we have seen, the integrated rate equation for the disappearance of A is given by

(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

(4.38)

Multiplying both terms by exp (k2t), we obtain [2]

(4.39)

and as

(4.40)

the equation can be written as the differential equation

(4.41)

which, upon integration, leads to

(4.42)

where the integration constant I is obtained for the simplest case of [B] = 0 at t = 0, as

(4.43)

Substituting this value for Iin eq. (4.42) and dividing by exp (k2t), we obtain the integrated

equation for the dependence of [B] upon time:

(4.44)

which is true for the case in which k1 k2. When k1 = k2, we obtain

(4.45)

whose solution is

(4.46)[ ] [ ] e0

B A= k tk t1

1

d [ ]e [ ] d0

B Ak t

k t21( )=

[ ][ ] e e

0B

A=

( )

k

k k

k t k t

1

2 1

1 2

Ik

k k=

[ ]( )

1

2 1

A0

[ ]e[ ] e

0BA

k t

k k t

k

k kI2

2 1

1

2 1

=

+( )

d [ ]e [ ] e d0

B Ak t k k t

k t22 1

1( )= ( )

e d[ ]

d[ ]e

d [ ]e

d

k t k t

k t

tk

t

2 2

2

2

BB

B+ =

( )

e d[ ]

d[ ]e [ ] e

0

k t k t k k t

tk k2 2

2 1

2 1

BB A+ =

( )

d[ ]

d[ ] [ ] e

0

BB A

tk k

k t+ = 2 1

1

[ ] [ ] e0

A A= k t1

4.2 Rates of Complex Reactions 87


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when [B] = 0 at t = 0. Figure 4.4 shows the time evolution of the concentrations of reac-

tant, 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

(4.VII)

The corresponding differential equations for this mechanism are

(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:

(4.48)

and solving in terms of [B] we obtain

(4.49)B A B A[ ]=[ ] + [ ] [ ]0 0

A B A B[ ] + [ ] =[ ] + [ ]0 0

=

=

d[ ]

d[ ] [ ]

d[ ]

d[ ] [ ]

AA B

BB A

f r

r f

tk k

tk k

A Bf

r

k

k


Figure 4.4 Variation of the concentrations of reactant, intermediate and final product with time forconsecutive first-order reactions. The solid line shows the case in which k1 = k2, while the dashedline is for the condition k1 = 10k2.


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Substituting [B] in the differential expression and rearranging, we can write

(4.50)

where

(4.51)

is a constant. Thus, the integration

(4.52)

leads to

(4.53)

When equilibrium is reached, the forward and reverse reactions have the same rates. This

corresponds to a formulation of the principle of microscopic reversibility. Thus, at equi-

librium, the concentrations of reactant and product do not change with time

(4.54)

i.e.

(4.55)

where the subscript refers explicitly to the equilibrium condition. From this, the equilib-

rium constant is given by

(4.56)

Figure 4.5 shows the evolution of [A] and [B] for the cases in which kf = kr and kf = 3 kr.

The same ideas can be applied to reversible reactions that occur in various steps. The

principle of microscopic reversibility can help resolve more complicated systems of

kinetic equations when the system is in equilibrium.

4.3 METHODS FOR SOLVING KINETIC EQUATIONS

4.3.1 Laplace transforms

Classical methods of integration are used to solve many of the linear differential equations

that describe the kinetic behaviour of chemical reactions. However, the methods of solving

k

kK

f

r

e

e

eq

B

A=

[ ]

[ ] =

[ ]

+ [ ]

=k kf e r e

A B 0

d[ ]

d

d[ ]

d

A Be e

t t= = 0

ln k k

k kk k t

f r

f r

f r

A B

A B

[ ] [ ]

[ ] [ ]

= +( )

0 0

d A

Ad

A

A

f r

[ ]

[ ]

[ ]

[ ]

mk k t

t

= +( )

0 0

+( )+( )

= k

k km

r

f r

A B[ ] [ ]0 0

= +( ) +( )

+( ) +

d[ ]

d

[ ] [ ][ ]

0 0A A BA

f r

r

f rt

k kk

k k

4.3 Methods for Solving Kinetic Equations 89

rates of complex reactions

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