Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

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Consecutive elementary reaction (Series reaction): 

Consider the following series reaction scheme Series Reaction - Chemical Kinetics Chemistry Notes | EduRev In this, the reactant A decays to four intermediate I, and this intermediate undergoes subsequent decay resulting in the formation of product P. The above series is elementary first order reaction.
Then the rate law expression is :

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev               …(1)
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev        …(2)
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev                         …(3)

Let only the reactant A is present at t = 0 such that

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
then the rate law expression is 

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev…(4)
The expression for [A] is substituted into the rate law of I resulting in

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

This differential equation has a standard form and after setting

[I]0 = 0, the solution is

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

The expression for [P] is

[A]0 = [A] + [I] + [P]

[P] = [A]0 – [A] – [I]
So

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Case I. Let                    kA >> kI

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

then 

 kI – kA ≈ –kA
and e-kA≈ 0
∴[P] = [A]0 – [A] – [I]

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

 

The rate of formation of product can be determined by slowest step.

[A] = [A]0 e-kAt                                                          ............(1)

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev                             ...........(2)
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev                          ............(3)
 

The graph representation for case I i.e. when kA >> kI.

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Case II. kA >> kI

 kI – k≈ kI
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
The graph representation of case II i.e. when kA >> kI

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

The Steady-State Approximation.
The steady-state approximation assume that, after an initial induction period, an interval during which the concentration of intermediate ‘I’ rise from zero, and during the major part of the reaction, the rates of change of concentration of all reaction intermediate are negligibly small.

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Problem.  Consider the following reaction

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

assuming that only reactant A is present at t = 0, what is the expected time dependence of [P] using the steady state approximation ?
 Sol.
The differential rate expression for this reaction are:

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Applying the steady state for I we get

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

and

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

then

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
This is expression for [P].

Problem. Using steady state approximation find the rate law for Series Reaction - Chemical Kinetics Chemistry Notes | EduRev   for the following given equation.

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Sol.

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

I1 & I2 are intermediate & apply steady state approximation on intermediate, we get

 Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

and 

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

Problem.  Using steady state approximation, derive the rate law for the decomposition of 

N2O5. 2N2O5(g) → 4NO2(g) + O2(g) 

On the basis of following mechanism. 

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev
Sol. The intermediate are NO & NO3.
 The rate law are:

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev= kb[NO2[[NO3] – kc [NO][N2O5] = 0
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev= ka[N2O5] – ka’ [NO2][NO3] – kb[NO2][NO3] = 0
Series Reaction - Chemical Kinetics Chemistry Notes | EduRev = –ka[N2O5] + ka’[NO2][NO3] – kc[NO][N2O5]

and replacing the concentration of intermediate by using the equation above gives 

Series Reaction - Chemical Kinetics Chemistry Notes | EduRev

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