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equilibrium constant

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GCT:

--- Quote from: ksr985 on May 15, 2005, 11:30:44 AM ---I thought of something while studying eqbm, havent managed to locate the answer yet. please see if you can help:

If i have a reversible reaction of the type:

A(s) gives B(g)

The eqbm constt. Kp for this reaction will be equal to the partial pressure of B at eqbm.

Now if i take 100 moles of A, and at eqbm, i have 5 moles of B, then Kp is 5.

Suppose i take only 2 moles of A. then the maximum number of moles of B that can ever be present is 2. So, will the eqbm never be reached? and if it is, what is the correct value of Kp for the rxn? If eqbm is not reached, but A gets exhausted, how will i know that this is not the eqbm point in the lab, considering the amount of B will be constant thereafter?

--- End quote ---

Kp is in pressure, atm, you made the proposal with moles which is incorrect.  Your query should now be resolved.

ksr985:
oops, stupid mistake, change all the moles to atmospheres of partial pressure. query remains.

GCT:

--- Quote ---query remains
--- End quote ---

really?  With any adequate container (not too large), you'll have a partial pressure due to decomposition, doesn't this seem practical?

GCT:

--- Quote from: ksr985 on May 15, 2005, 02:24:06 PM ---oops, stupid mistake, change all the moles to atmospheres of partial pressure. query remains.

--- End quote ---

Alright, I think I see where your confusion is...with adequate amounts of A, there will the pressure will always be, say...5atm and this does not depend on the original amount of A.  If you don't have enough A, all of the A will simply decompose and you'll have less than 5 atm at equilibrium.

ksr985:
ahh.. exactly. i dont even know if all the A will decompose, that is also uncertain. all i know is that eqbm cannot be acieved. who's to say what position the reaction will occupy at any given time.

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