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### Topic: Equilibrium yield  (Read 10595 times)

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#### magician4

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##### Re: Equilibrium yield
« Reply #15 on: September 12, 2013, 08:42:31 PM »
Quote
If we didn't include the stoichiometric factors in the definition of the yield, we could easily end up with a 200% yield.(...)
we need to come up with a sensible definition of "yield" here, and, based on this,  need to invent respective "factors" (or functions) , I agree
however, i doubt that this might be as simple as just having the plain stoichiometric factors involved in a manner as proposed by you

for example:
if we'd keep on working on my example given earlier, and include the respective stoichiometric factors a,b,d with their respective c0's ,  we'd find that the respective expression for [Ptotal] should look something like this (if I didn't miscalculate):

$$[P_{total}] = [P_{(I)}] + [P_{(II)}] =$$
$$= \left( 1 - \frac {1}{1 + K_{(II)} } \right) \cdot \left( - \frac {1 - (K_{(I)}c_0(A)^a + K_{(II)}c_0(B)^b)(1 + K_{(II)})} {2 \cdot (1 + K_{(II)} )} + \sqrt { \left( - \frac {1 - (K_{(I)}c_0(A)^a + K_{(II)}c_0(B)^b)(1 + K_{(II)})} {2 \cdot (1 + K_{(II)} )} \right)^2 - \left( K_{(I)}c_0(A)^a + K_{(II)}c_0(B)^b - \frac {K_{(II)}c_0(D)^d }{1 + K_{(II)} } \right) } \right) + \frac {K_{(II)}c_0(D)^d }{1 + K_{(II)} }$$

I fail to see how you could possibly "extrac" the stoichiometric factors a, b, d from this in a simple fashion as proposed by you
.. and this still is a pretty easy situation, with those very very simple reactions considered: no "squares" (or higher orders) in the LMA expressions whatsoever

hence, yes, you might come up with a generalized mathematical expression for "P" even for the most complex situation of interacting equilibria we possibly could think of, and maybe even find a way to extract the stoichiometric factors in sort of a function from this.
Furthermore, you might find a generalized expression of "yield" this could be introduced to.

however, from the practical side, I can't see how the potential result even probably could be worth this effort

... but maybe I'm just being ignorant to an important question

So, if you have fun gnawing at such type of problems: have fun
Speaking for myself however, I hope that it's ok that I'm outa here by now, as I think I've given all I possibly could contribute to this very problem

regards

Ingo
« Last Edit: September 12, 2013, 09:29:33 PM by magician4 »
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