[JugemuJugemu]

4.      SSA   and   Mechanistic   analysis
Write   down   bullet   points   that   summarise   the   steady   state   approximation   and   
when   it   may   be   correctly   applied.
For   the   overall   reaction   involved   in   the   pyrolysis   of   ethanal   (acetaldehyde)   the   
following   kinetics   were   observed:

-d[CH3CHO]/dt=   kexp[CH3CHO]3/2
where   kexp is   the   experimentally   observed   rate   constant.Show   that   the   following   chain   mechanism   is   consistent   with   these   observations   
assuming   that   steady   state   may   be   applied   to   the   radical   (=   non-stable   molecule)   
species,   CH3,   CHO,   H   and   CH3CO,   and   assuming   that   the   rates   of   reactions   (1)   and   
(6) are   much   smaller   than   that   of   the   propagation   step   (4). Hence   also   express   
kexp in   terms   of   rate   constants   for   the   elementary   reactions   in   the   mechanism.

(1) CH3CHO → CH3+CHO
(2) CHO → CO+H
(3) H+CH3CHO → H2+CH3CO
(4) CH3+CH3CHO → CH4+CH3CO
(5) CH3CO → CH3+CO
(6) CH3+CH3 → C2H6

The question wants me to find the rate of loss of Ethanal, so I assume I write out:

-d[CH₃CHO]/dt = k₁[CH₃CHO]+k₃[H][CH₃CHO]+k₄[CH₃][CH₃CHO]

Then I find the rates of removal of [H] radical and [CH₃] radicical and apply the steady state correct then solve and plug them back in? H is consumed in reaction 3 and formed in reaction 2. CH₃ is consumed in 6 and 4 and produced in 1 and 5.

My main problems are I am not sure if to ignore reaction 6 when writing out the rate of loss of CH₃? and should I remove thek₁[CH₃CHO] from the equation for -d[CH₃CHO]/dt that I wrote out?

Edit: Just wrote out

-d[H]/dt = k3[H][CH3CHO]-k2[CHO]

and

-d[CH3]/dt = k4[CH3][CH3CHO]+2k6[CH3]-k1[CH3CHO]-k5[CH3CO]

Now nothing is cancelling, I think I should be writing some of these equations backwards as inhibition processes but I am not sure how to do this.

[mjc123]

You have other radical species in your rate expressions that you need to account for, e.g. CHO and CH₃CO. Try and find expressions for these using the steady state approximation, e.g. CHO is made in step 1 and consumed in step 2, so
d[CHO]/dt = k1[CH₃CHO] - k2[CHO] ≈ 0
[CHO] = k1/k2*[CH₃CHO]
Do likewise with the other species and see if you can eliminate them from your rate equations.

[JugemuJugemu]

ok that makes sense. I am not sure where I am going with this though, I could rearange and substitute forever. Am I trying to get everything in -d[H]/dt and -d[CH3]/dt in terms of either [CH3], [H] or [CH3CHO]?

Edit:

Ok so I used your [CHO] part and also wrote out radical [CH3CO] = k3/k5[H][CH3CHO].

Then I set my -d[H]/dt and -d[CH3]/dt equal to each other and subbed in [CHO] and [CH3CO] and quite a lot cancelled.

I ended up with 2k3[H][CH3CHO]-k4[CH3CHO]=2k6[CH3]

Problem being now I have [H] and [CH3] in the same equation.

Edit again ok so from

-d[H]/dt = k3[H][CH3CHO]-k2[CHO]

i get by subbing in [CHO] = k1/k2 [CH3CHO]

 -d[H]/dt=0=k3[H][CH3CHO]-k1[CH3CHO] one can get [H]=k1/k3?