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