The thermal decomposition of bis-pentafluorosufurtrioxide (SF 5 O 3 SF 5 ) is po

Question

The thermal decomposition of bis-pentafluorosufurtrioxide (SF5O3SF5)

is postulated to proceed by the following mechanism for oxygen partial pressures greater than 100 torr.

with k1 and k-1 reaction rate coefficients

with k2 reaction rate coefficient

with k3 reaction rate coefficient

a) Assume that the first reaction is not at equilibrium, that SF5O and SF5O2 are unstable radicals, and that there is no rate determining step. Develop an expression for the rate of bispentafluorosufurtrioxide decomposition in terms of stable molecules and elementary rate constants above.

b)   Now assume that reaction (3) is actually the rate determining step. Is there any difference in the resulting rate expression?

Explanation / Answer

Since SF5O2 and SF5O are intermediates, their net rate is zero.

d/dt[(SF5O2)= K1[SF5O3SF5]- K-1[ SF5O] [SF5O2] – K2[ SF5O2]2 =0 (1)

d/dt[(SF5O)= K1[SF5O3SF5]- K-1[ SF5O] [SF5O2] + K2[ SF5O2]2 - K3[SF5O]2 =0 (2)

from Eq.1, Eq.2 becomes 2K2[SF5O2]2 = K3[SF5O]2

[SF5O2]2 = (K3/2K2)* [SF5O]2

[SF5O2] = sqrt((K3/2K2)* [SF5O] (3)

Let Sqrt( K3/2K2)= K’

From Eq.1   K1[SF5O3SF5] –K-1*Sqrt(K3/2K2)*[SF5O]* [SF5O] = K2[SF5O]2

K1 [SF5O3SF5] = [SF5O]2 { K-1*Sqrt( K3/2K2)+ K2} = [SF5O]2{ K-1*K’ +K2}= [SF5O]2 * K’’

Where K’’ = { K-1*K’ +K2)/K1

[SF5O3SF5]= [SF5O]2*K’’

[SF5O] = 1/Sqrt(K’’) * [SF5O3SF5] (4)

Hence [SF5O2] = K’/Sqrt(K’’)* [SF5O3SF5]

-d/dt[(SF5O2)= K1[SF5O3SF5]- K-1[ SF5O] [SF5O2] = -K2[SF5O2]2 =-K2*K’’/sqrt(K)* [SF5O3SF5]2 = K’’’[SF5O3SF5]2 where K’’’= -K2K’’/Sqrt(K)

3. when the 3rd reaction is rate determining steap

-d/dt[SF5O3SF5]= K3*[SF5O]2 = K3* {1/Sqrt(K’’) * [SF5O3SF5] }2 = K’’’’[SF5O3SF5]2

K’’’’= K3*{1/Sqrt(K’’)2

The rate constant is different but the order is same.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.