The mechanism below represented as follows ;
A + S k₁ AS ( adsorption)
AS k−1 A+S ( desorption)
AS k₂ products (decomposition)
Here,
If, r is the rate of the reaction is proportional to the fraction θ of the surface covered.
r = k₂θ ............................(1)
Assuming steady state approximation for concentration of as,
We have
r= d [AS ] = k₁ [A] [S]−k−₁ [AS]−k₂ [AS] = 0
dt
...............(2)
[AS] = k₁ [A] [S] .....................(3)
k−₁ + k₂
Here, [S] = Cs (1−θ )
[AS] = Cs θ
Therefore, the above equation is,
Csθ = k₁ [A] Cs (1−θ ) .............(4)
k−₁ + k₂
k₁ [A] Cs 1−θ = Csθ (k−₁ + k₂)
1− θ = k−₁ + k₂
θ k₁ [A]
1 − 1 = k−₁ + k₂
θ k₁ [A]
1 = k₁ + k₂ +1
θ k₁ [A]
1 = k−₁ + k₂ + k₁ [A]
θ k₁ [A]
∴ θ = k₁[A] ..................(5)
k−₁ + k₂ + k₁ [A]
Substituting equation (5) in (1)
r = k₂k₁ [ A.] (6)
k₁ [A] + k−₁ + k₂
1 = k₂k₁ [A ]
r k₂ + k₁ [A] + k−₁
1 = k₁ [A] + k−₁ + k₂
r k₂ k₁ [A]
1 = 1 + k−₁ + k₂ (7)
r k₂ k₁k₂ [A]
This is a linear equation of the types below,
Y= mx+c
1 = k₁k₂ ( 1 ) + 1
r k₁k₂ [A] k₂
y m x c
The plot of against 1/r versus 1
[A]
Should give a straight line with slop
k−₁ + k₂
k₁ k₂
and the intercept is 1
K₂
Expressing concentration in terms of partial pressure,
Therefore equation (6) and (7) we get
r = k₁k₂PA
k₁PA + k−₁ + k₂
1 = 1 + k−₁ + k₂ (8)
r k₂ k₁k₂ PA
Special cases:-
Case:1 k₂ >> (k₁[A] + k−₁ )
r = k₁ [A]
Case :2 k₂ >> (k₁ [A] + k−₁)
r = k₁k₂ [A] = k₂k₁ [ A ]
k₁[A] + k−₁ k [A] + 1
= k₂ k' PA
k' PA + 1
Super
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