Respuesta :
Answer:
the rate limiting step could be the fist forward reaction or the second one. the rate laws are
-dCch/dt= k₃ *Cch
-dCo/dt= √(k₆*k₃/k₁) * (CS/Co)^(1/2) *Cch^(3/2)
Explanation:
For the following reactions
O₂ + 2 S → 2 O*S ,
C₃H₆ + O*S → C₃H₅OH * S
C₃H₅OH * S → C₃H₅OH + S
where k₁ , k₂ forward and backward reaction rates for the first equation. 3,4 for the second one and 5 and 6 for the third one respectively
Assuming for the reaction rates , elementary rate laws and pseudo stationary hypothesis ( meaning that the S-complexes are unstable enough to react fast and do not accumulate, so the net rate of reaction is 0).
denoting
[O₂] = Co ,[S] = Cs , [O*S ] = Cos ,[C₃H₆] = Cch , [C₃H₅OH] = Cco , [C₃H₅OH * S} = Ccos
then for the S complexes
-dCs/dt = k₁* Co*Cs² - k₂ * Cos² - k₆*Cch*Cs = 0
-dCos/dt = k₁ * Cos² + k₃ *Cch *Cos - k₄*Ccos = 0
-dCcos/dt = k₄*Ccos + k₅*Ccos - k₆*Cch*Cs = 0
then the total concentration of occupied sites CS is
CS= Cos + Ccos + Cs
from the third equation
k₄*Ccos + k₅*Ccos - k₆*Cch*Cs = 0
Ccos = k₆/(k₄+k₅)*Cch*Cs
from the first equation + second equation
k₁* Co*Cs² - k₂ * Cos² - k₆*Cch*Cs + k₁ * Cos² + k₃ *Cch *Cos - k₄*Ccos = 0
k₁* Co*Cs² - k₆*Cch*Cs + k₃ *Cch *Cos - k₄*Ccos = 0
k₁* Co*Cs² - k₆*Cch*Cs + k₃ *Cch *Cos - k₄*k₆/(k₄+k₅)*Cch*Cs = 0
Cos = k₁/k₃* Co*Cs²/Cch - [k₆+k₄*k₆/(k₄+k₅)] *Cs
then
CS= Cos + Ccos + Cs
CS = k₁/k₃* Co*Cs²/Cch - [k₆+k₄*k₆/(k₄+k₅)] *Cs + k₆/(k₄+k₅)*Cch*Cs + Cs
k₁/k₃* Co*Cs²/Cch - [k₆+k₄*k₆/(k₄+k₅)] *Cs + k₆/(k₄+k₅)*Cch*Cs + Cs - CS = 0
denoting K1=k₁/k₃ , K2 = k₆+k₄*k₆/(k₄+k₅) , K3=k₆/(k₄+k₅)
K1*Co/Cch * Cs² + (K2 + K3*Cch+1)* Cs - CS = 0
Cs = [-(K2 + K3*Cch+1) + √ (K2 + K3*Cch+1)² +4*K1*Co/Cch*CS)]/(2*K1*Co/Cch )
since the final expression would be too complex, we can do some assumptions in order to obtain a rate that is easier to handle.
assuming that
k₅ >> k₆ ( meaning that C₃H₅OH * S decomposes to propanol and the original active site of the surface, rather than generating another high energy S-complex)
then K3≈0 , K2≈ k₆
Cs = -k₆ ( √(1+4*K1/k₆*Co/Cch*CS) - 1 ) / (2*K1*Co/Cch)
for big Oxygen and total sites concentrations ( reaction with air or oxygen excess) → K1/k₆*Co/Cch*CS >> 1
Cs= 2* √(K1/k₆*Co/Cch*CS)/[(2*K1*Co/Cch)] = √[CS/(K1*k₆*Co/Cch)]
=√[CS*Cch/(K1*k₆*Co)]
then the reaction rate of propane is
-dCch/dt= k₃ *Cch *Cos - k₄*Ccos = k₃ *Cch * (1+k₄*Ccos)
since the concentration of complexes is small
-dCch/dt= k₃ *Cch
for the oxygen
-dCo/dt= k₁* Co*Cs² - k₂ * Cos² = k₆*Cch*Cs = k₆ *Cch*√[CS*Cch/(K1*k₆*Co)]
= √(k₆*k₃/k₁) * √(CS/Co)* Cch^(3/2)
-dCo/dt= √(k₆*k₃/k₁) * (CS/Co)^(1/2) *Cch^(3/2)
therefore
-dCch/dt= k₃ *Cch
-dCo/dt= √(k₆*k₃/k₁) * (CS/Co)^(1/2) *Cch^(3/2)
from our assumptions that k₆ is small, then the rate limiting step is the first forward reaction ( in k₁) because if k₁ >> k₃, then the O*S concentration grows faster than its consumption , stopping the forward reaction in the first equation and limiting the oxygen reaction to make propanol. On the other hand, if k₃ is small , the rate limiting step would be the second forward reaction
