Derivation of the Equilibrium Constant


by gsingh2011
Tags: constant, derivation, equilibrium
gsingh2011
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#1
Jan30-11, 03:05 PM
P: 113
I'm trying to understand this: http://en.wikipedia.org/wiki/Equilib...bs_Free_Energy
And there are few steps that I don't get. My first question is probably a stupid one, but where does the [tex]\Sigma[/tex]v[tex]_{j}[/tex]A[tex]_{j}[/tex]=0 come from? What exactly is it summing up, the number of atoms? Are you assuming that the products side is negative? (Because I don't see how else you would get zero) And why does it only go up to n?

I'll ask my other questions after I get some replies because it's possible I'll figure them out after I read the replies.
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Borek
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Jan30-11, 03:23 PM
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Quote Quote by gsingh2011 View Post
where does the [tex]\Sigma[/tex]v[tex]_{j}[/tex]A[tex]_{j}[/tex]=0 come from? What exactly is it summing up, the number of atoms? Are you assuming that the products side is negative? (Because I don't see how else you would get zero) And why does it only go up to n?
My guess is it is just a mass conservation, and yes, products are negative.
gsingh2011
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Jan30-11, 05:26 PM
P: 113
Quote Quote by Borek View Post
My guess is it is just a mass conservation, and yes, products are negative.
Ok, but why does it only sum up to n? The first equation in the derivation shows the first product being n but then shows more products after that. If it was mass conservation, wouldn't it have to sum passed n?

Borek
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Jan31-11, 01:59 AM
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Derivation of the Equilibrium Constant


Where does it sum only to n? It sums for all j.


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