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

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answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e