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.

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

its mg hope this

the answer is d.) potassium