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Question about equilibrium constant

by brianparks
Tags: constant, equilibrium
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brianparks
#1
Jan25-06, 07:18 PM
P: 24
I am having serious difficulty understanding the concept of the equilibrium constant in a reversible chemical reaction. Maybe one of you all can help?

Suppose we have the following reversible chemical reaction:

1) aA + bB <---> cC + dD

The equilibrium constant for the reaction is said to be:

Kc1 = [(A^a)(B^b)]/[(c^C)(d^D)]

Now, suppose that we express the above reaction differently, as follows:

2) 2aA + 2bB <---> 2cC + 2dD

My understanding is that these two chemical reactions are the exact same reaction, just expressed with different stoichiometric numbers.

The thing that boggles me, however, is that the second expression of the reaction has a different equilibrium constant than the first.

Kc2 = [(A^2a)(B^2b)]/[(C^2c)(D^2d)]

How can the equilibrium constant (and its associated equation) express or quantify an actual feature of a reaction (i.e., the extent to which it is shifted toward the product or the reactants) if it changes depending on how you express the reaction?

What does the equilbrium constant (and equation) mean? Is it an arbitrary formula used for convenience in solving chemical equilibrium problems that could just as easily be defined differently? Or is it something more?

I have read that you can actually derive the equation using chemical thermodynamics, so it can't just be a arbitrary ratio.

This is really confusing me. Can someone please help?

Thanks,
--Brian
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ksinclair13
#2
Jan25-06, 08:11 PM
P: 102
Quote Quote by brianparks
Suppose we have the following reversible chemical reaction:

1) aA + bB <---> cC + dD

The equilibrium constant for the reaction is said to be:

Kc1 = [(A^a)(B^b)]/[(c^C)(d^D)
It should be: [tex]K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}[/tex]
brianparks
#3
Jan25-06, 09:47 PM
P: 24
Good point. I screwed that up. Anyway, does my question make any sense?

Bystander
#4
Jan26-06, 10:03 AM
Sci Advisor
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PF Gold
P: 1,384
Question about equilibrium constant

What's the relationship between your two "separate" equilibrium constants?
GCT
#5
Jan26-06, 05:01 PM
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P: 1,769
In addition to what others have said, think about the molecular scale of things. The first would be a bimolecular reaction, the second would represent a reaction involving four molecules assuming uniform constants (four molecule ~simultaneous collision)! The two reactions are indeed different. I'm not sure where you got the idea that the two are the same, remember these equations represent what's actually going on.


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