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Topic: Steady-state approximation  (Read 6834 times)

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Offline Big-Daddy

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Steady-state approximation
« on: January 20, 2014, 04:53:05 PM »
In the following problem (see the question), why is it that when writing the differential rate laws d[I1]/dt and d[I2]/dt, the equilibrium between I1 and I2 is omitted?

e.g. my original equation was

d[I1]/dt = k1[A][M]-k-1[I1]-k2[I1]+k-3[I2]-k3[I1] ≈ 0

but the solutions gives

d[I1]/dt = k1[A][M]-k-1[I1]-k2[I1] ≈ 0

omitting the terms concerning the equilibrium between the intermediates I1 and I2. Is this a direct part of applying the steady-state approximation? If not, then is it justified in this case?

Offline curiouscat

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Re: Steady-state approximation
« Reply #1 on: January 24, 2014, 06:34:44 AM »
I tend to like your expression.

Offline Babcock_Hall

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Re: Steady-state approximation
« Reply #2 on: January 24, 2014, 09:31:05 AM »
Do we know with certainty whether or not I1 and I2 are in rapid equilibrium?

Offline Big-Daddy

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Re: Steady-state approximation
« Reply #3 on: January 24, 2014, 12:11:45 PM »
I had only the information given in the question as shown. Is there any good reason to assume rapid equilibrium between the intermediates, from the information given?

Also, if we do assume rapid equilibrium of I2 and I1, why does this mean that the rate terms for this equilibrium will disappear? If k3 and k-3 >> k1, k-1, and k2, shouldn't we set all rate terms except that for the rapid equilibrium to 0? (rather than setting only the fastest terms to 0)

Offline Babcock_Hall

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Re: Steady-state approximation
« Reply #4 on: January 24, 2014, 04:36:34 PM »
Can you clarify something?  In the top figure k1 is a forward rate constant and k-1 is a rate constant for a reverse reaction.  However, in the bottom figure k-1 is the rate contstant for the forward reaction, and k1 is the rate constant for the reverse reaction.  Why is there a difference?

Offline Big-Daddy

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Re: Steady-state approximation
« Reply #5 on: January 24, 2014, 07:22:28 PM »
Yes, sorry about the insipid notation of whoever has been scribbling on the paper (wasn't me). Same thing goes for k3 and k-3. I'd say take kx to be rate constant of forward reversible-reaction x and k-x to be rate constant for the reverse direction of reversible-reaction x. In this question this applies for x=3 and x=1 (the two reversible reactions in the system).

The difference appears because it seems that whoever was doing the writing has learnt all of his kinetics theory correctly except for how to draw reversible-reaction symbols in the correct direction.  :P

Offline Babcock_Hall

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Re: Steady-state approximation
« Reply #6 on: January 25, 2014, 09:48:49 AM »
The only suggestion that I have at this point is to try to get rid of [I2] in your expression, possibly by writing it in terms of [I1], but I am not sure how to do that without making one additional assumption.

Offline Big-Daddy

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Re: Steady-state approximation
« Reply #7 on: January 25, 2014, 10:59:42 PM »
Ah I think I just misread the original question - when it says assume the concentrations of the intermediates are 0, it actually means concentrations, not just rates of change of concentration. Therefore the two extra terms in my original question drop out and we get the version used in the solutions.

I would be interested to know under what conditions it would be valid to approximate that [I1]≈[I2]≈0, in addition to the steady-state approximation.

Offline curiouscat

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Re: Steady-state approximation
« Reply #8 on: January 27, 2014, 03:29:16 AM »
I would be interested to know under what conditions it would be valid to approximate that [I1]≈[I2]≈0, in addition to the steady-state approximation.

I can point you to the answer. Look up the book / slides on Chemical Kinetics / Reaction Engineering by Jim Rawlings of UW-Madison.

If memory serves me right, he gives a beautiful exposition of this nuanced approximation.

I'd say if in doubt, put the derivatives to zero but never the rate.

Offline Babcock_Hall

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Re: Steady-state approximation
« Reply #9 on: January 27, 2014, 07:00:42 PM »
I looked up the original paper, and I think that there may be some pedagogical issues with respect to the steady-state approximation that might be worth exploring.  For the reaction A  ::equil:: B :rarrow: C, Sidney Miller (J. Chem Ed., p. 490) sets both the time derivative of B and [B ] itself to approximately zero in his definition of the steady-state for the reaction above.*  However, suppose that we are looking at the steady state of a simple enzyme reaction mechanism, E + S  ::equil:: E•S  :rarrow: E + P.  Typically, E and E•S are orders of magnitude lower in concentration than S, but not necessarily lower than the concentration of P.  In other words, I wonder whether Miller's definition of the steady-state is too restrictive.

Miller wrote, "The pedagogic message is that kss may vary depending on how certain key concentrations are treated. That is, after one sets the conditions of concentration and time dependence under which the steady state applies to a mechanism ( 1 , 7 , 9 ) one may obtain apparently different rate laws."  This may be pertinent to the homework problem given to the OP.

*As an aside, I recall an old textbook treatment (Atkins, Physical Chemistry, p. 870) of the steady-state approximation applied to a simpler system, A  :rarrow::rarrow: C .  In Atkins' treatment, the concentration of B remains low and does not change much with time (for most of the course of the reaction) when the second rate constant when the rate constant for the second step is much larger that the rate constant for the first step, and he defines this as the steady state.

(mod edit by sjb to fix format)
« Last Edit: January 28, 2014, 02:23:38 AM by sjb »

Offline curiouscat

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Re: Steady-state approximation
« Reply #10 on: January 28, 2014, 03:59:06 AM »
Can you think of a case where it makes sense to put a conc. to zero but not the corresponding derivative?

My opinion is set the derivative to zero but never the conc.

Offline Babcock_Hall

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Re: Steady-state approximation
« Reply #11 on: January 29, 2014, 11:18:48 AM »
curiouscat,

Not offhand, but I am not certain.  I looked around for the presentations that you mentioned and found this link.  There is some discussion of the steady-state here.
http://jbrwww.che.wisc.edu/home/jbraw/chemreacfun/ch5/slides-kinetics-2up.pdf

Offline curiouscat

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Re: Steady-state approximation
« Reply #12 on: January 29, 2014, 01:12:43 PM »
Yes, that's the slides I had in mind. The text has more.

I still maintain it only makes sense to put the derivative to zero & not the conc.

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