Question

The formation of propanol on a catalytic surface is believed to proceed by the following mechanism

O_2 + 2S 2 O * S

C_3H_6 + O * S

C_3H_5OH * S C_3H_5OH * S

C_3H_5OH + S

Suggest a rate-limiting step and derive a rate law.

Answer 1

The rate law for the formation of propanol in a catalytic hydrogenation process is derived considering the mechanism of the reaction and identifying the rate-limiting step. The slowest step of the mechanism dictates the overall rate, but without more details, an accurate rate law cannot be determined.

Step-by-step explanation:

The question pertains to the mechanism of catalytic hydrogenation of alkenes, specifically the formation of propanol, and deriving a rate law based on the proposed mechanism. When determining the rate-limiting step in a reaction, we look for the slowest step that determines the overall rate of the reaction. The middle step in the given mechanism is assumed to be the fastest and thus not rate-limiting. To derive the rate law, we analyze the stoichiometry and kinetics of the relevant steps, which usually involve steps before the rate-limiting step. However, without the full context of the mechanism and the actual slow step, it is not possible to accurately determine the rate-limiting step or to derive a rate law. Typically, we would use the stoichiometry of the reaction and knowledge of the reaction mechanism to write rate equations for each step and deduce the overall rate law based on the rate-limiting step.

Answer 2

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)

Step-by-step 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