Series Reactions & Steady State Approximation | Physical Chemistry PDF Download

Series Reaction:

Consider the following series reaction scheme  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 an elementary first-order reaction.
  Then the rate law expression is :

                …(1)
        …(2)
                         …(3)

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

then the rate law expression is 

…(4)
The expression for [A] is substituted into the rate law of I resulting in

This differential equation has a standard form and after setting

[I]₀ = 0, the solution is

The expression for [P] is

[A]₀ = [A] + [I] + [P]

[P] = [A]₀ – [A] – [I]
So

Case I. Let                    k_A >> k_I

then 

 k_I – k_A ≈ –k_A
and e^-k_At ≈ 0
∴[P] = [A]₀ – [A] – [I]

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

[A] = [A]₀ e^-k_At                                                          ............(1)

 

                             ...........(2)
                          ............(3)
 

The graph representation for case I i.e. when k_A >> k_I.

Case II. k_A >> k_I

 k_I – k_A ≈ k_I

∴

The graph representation of case II i.e. when k_A >> k_I

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.

 

Q.1. Consider the following reaction

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:

Applying the steady state for I we get

and

then

This is expression for [P].

Q.2. Using steady-state approximation find the rate law for    for the following given equation.

Sol.

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

and 

Q.3. 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. 

Sol. The intermediate are NO & NO3.
 The rate law are:

= k_b[NO₂[[NO₃] – k_c [NO][N₂O₅] = 0
= k_a[N₂O₅] – k_a’ [NO₂][NO₃] – k_b[NO₂][NO₃] = 0
 = –_ka[N₂O₅] + k_a’[NO₂][NO₃] – k_c[NO][N₂O₅]

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