# Kinetics of Consecutive Reactions

## Consecutive Reactions

The reactions in which the reactant forms an intermediate and the intermediate forms the product in one or many subsequent reactions are
called consecutive or sequential reactions. Every stage has its own reactant and rate constant. In such reactions the product is not formed directly from the reactant.

Let us consider a simple reaction-

where, A = reactant, B = intermediatea and C = product.

Initially only the reactant 'A' is present. As the reaction starts, concentration of reactant 'A' decreases and produces an intermediate 'B' through k_{1} rate constant. When 'B' is formed, it produces the product 'C' through k_{2} rate constant. After the completion of reaction only 'C' is present and concentrations of A and B will be zero.

The overall rate of reaction depends upon the magnitude of these two (i.e. k_{1} and k_{2}) rate constants.

During the course of the reaction the concentration of 'A', 'B' and 'C' vary as shown in figure-

It can be clearly seen that the concentration of 'A' decreases exponentially, while the concentration of 'B' increases first and then declines. The concentration of 'C' increases continuously and finally becomes equal to the concentration of 'A'.

Now suppose that the initial concentrations of reactant 'A' is [A]_{o}; while the concentrations of 'A', 'B' and 'C' after time 't' are [A], [B] and [C], respectively. So, we can say that

[A]_{o} = [A] + [B] + [C] -----Eq.-X

**1. Rate law in terms of [A]**

The rate of disappearance of reactant 'A' in the given reaction can be given by the following relation-

−d[A]/dt = k_{1} [A] -----Eq.-1

or, −d[A]/[A] = dt.k_{1} -----Eq.-2

Integrating the above equation on bothe side we get-

−ln[A] = k_{1}t + C -----Eq.-3

When 't' = 0, then [A] = [A]_{o}

Now putting these values in equation-3 we get-

C = −ln[A]_{o}

Putting the value of 'C' in equation-3 we get-

−ln[A] = k_{1}.t −ln[A]_{o} -----Eq.-4

or, ln[A]_{o} −ln[A] = k_{1}.t -----Eq.-5

or, −ln[A]/[A]_{o} = k_{1}.t -----Eq.-6

or, ln[A]/[A]_{o} = −k_{1}.t -----Eq.-7

or, [A]/[A]_{o} = e^{−k1.t} -----Eq.-9

or, [A] = [A]_{o}.e^{−k1.t} -----Eq.-10

**2. Rate law in terms of [B]**

The rate of formation of intermediate B can be given by the following relation-

d[B]/dt] = −k_{2} [B] + k_{1} [A]

d[B]/dt] = k_{1} [A] − k_{2} [B]

After putting the value of [A] from Eq.-10 in the above equation, we get a linear differential equation of first order i.e.

d[B]/dt] = k_{1} [A]_{o}.e^{−k1.t} − k_{2} [B]

Integrating and then rearranging the above equation on both side, we get-

[B] = [A]_{o}(k_{1}/(k_{2} − k_{1}))(e^{−k1.t} − e^{−k1.t}) -----Eq.-11

**3. Rate law in terms of [c]**

The overall rate of formation of the product 'C' in the given reaction can be given by the following relation-

d[C]/dt = k_{2} [B]

After putting the value of [A] and [B] into Eq.-X, we get the following result-

[C] = [A]_{o} − [A]_{o}.e^{−k1.t} − [A]_{o}(k_{1}/(k_{2} − k_{1})) (e^{−k1.t} − e^{−k1.t}) -----Eq.-12

or, [C] = [A]_{o}{1 − e^{−k1.t} − (k_{1}/(k_{2} − k_{1})) (e^{−k1.t} − e^{−k1.t}) -----Eq.-13

or, [C] = [A]_{o}{1 − e^{−k1.t} −
(k_{1}.e^{−k1.t}/(k_{2} − k_{1})) + (k_{1}.e^{−k1.t}/(k_{2} − k_{1})) -----Eq.-14

or, [C] = [A]_{o}{1 − e^{k1.t} +
(k_{1}.e^{−k1.t}/(k_{2} − k_{1})) − (k_{1}.e^{−k1.t}/(k_{2} − k_{1})) -----Eq.-15

or, [C] = [A]_{o}{1 − [(k_{2} − k_{1})e^{−k1.t} + k_{1}e^{−k1.t} − e^{−k2.t})]/(k_{2} − k_{1})} -----Eq.-15

or, [C] = [A]_{o}{1 −[(k_{2}e^{−k1t} − k_{1}e^{−k1t} + k_{1}e^{−k1t} − k_{1}e^{−k2t})/(k_{2} − k_{1})]} -----Eq.-16

or, [C] = [A]_{o}{1 −[(k_{2}e^{−k1t} − k_{1}e^{−k2t})/(k_{2} − k_{1})]} -----Eq.-17

or, [C] = [A]_{o}[1 − (1/(k_{2} − k_{1})(k_{2}e^{−k1t} − k_{1}e^{−k2t})] -----Eq.-18

**Maximum Concentration of B**

At the maximum concentration of B-

d[B]/dt = 0

Now differentiating Eq.-11 w.r.t time 't' we get-

d[B]/dt = [A]_{o}(k_{1}/(k_{2} − k_{1}))(−k_{1}e^{−k1.t} + k_{2} e^{−k2.t}) -----Eq.-19

Equating the above equation we get-

k_{2} − k_{1}))(−k_{1}e^{−k1.tmax} + k_{2} e^{−k2.tmax}) = 0 -----Eq.-20

or, k_{1}/k_{2} = e^{(k1 − k2)tmax} -----Eq.-21

or, ln(k_{1}/k_{2}) = (k_{1} − k_{2})t_{max} -----Eq.-22

or, t_{max} = 1/(k_{1} − k_{2}) ln(k_{1}/k_{2}) -----Eq.-23

substituting Eq.-23 in Eq.-11, we get, we get-

[B]_{max} = [A]_{o} (k_{2}/k_{1})^{k2/(k1 − k2}) -----Eq.-24

**Rate Law in Special Case**

*1. When k _{2} >>> k_{1}*

In this case, the value of k

_{1}can be neglected. So, the Eq.-18 takes the new form-

[C] = [A]

_{o}(1 − e

^{k1t})

It can be clearly seen that the concentration of the intermediate practically remains constant, and So, the steady-state approximation can be applied in this case.

*2. When k*

_{1}>>> k_{2}In this case, the value of k

_{2}can be neglected. So, the Eq.-18 takes the new form-

[C] = [A]

_{o}(1 − e

^{k2t})

Example of consecutive reactions-

Saponification of a diester in presence of an alkali