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### Topic: Why does the equilibrium constant have that form?  (Read 4300 times)

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#### alexmahone

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##### Why does the equilibrium constant have that form?
« on: July 11, 2012, 02:23:55 PM »
For the reaction aA+bB cC+dD,
the equilibrium constant is K=[C]^c[D]^d/[A]^a[B ]^b.

This quantity (K) is the same at equilibrium regardless of the initial concentrations of A, B, C, D. But why does K have that form?

#### Hunter2

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##### Re: Why does the equilibrium constant have that form?
« Reply #1 on: July 12, 2012, 05:01:59 AM »
Quote
This quantity (K) is the same at equilibrium regardless of the initial concentrations of A, B, C, D. But why does K have that form?

This is wrong. The concentrations of A,B,C,D have some relation to each other, they are not independend. For this reason you can do the quotient of Πproducts/educts = K

#### alexmahone

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##### Re: Why does the equilibrium constant have that form?
« Reply #2 on: July 12, 2012, 05:09:46 AM »
Quote
This quantity (K) is the same at equilibrium regardless of the initial concentrations of A, B, C, D. But why does K have that form?

This is wrong. The concentrations of A,B,C,D have some relation to each other, they are not independend. For this reason you can do the quotient of Πproducts/educts = K

But why are they related to each other in that way, and not in some other way?

#### Hunter2

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##### Re: Why does the equilibrium constant have that form?
« Reply #3 on: July 12, 2012, 05:23:01 AM »
Let make it easy A + B => C+D

K = C*D/A*B

If let say 0.5 mol of A reacts with B then also 0.5 B will be consumed, because the factor is 1. On the other hand we get 0.5 C and the same amount D. If you change one concentration of the mixture, the system reacts in this way to replace the new concentration by reacting with the other component or decomposing of the other components.

#### AWK

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##### Re: Why does the equilibrium constant have that form?
« Reply #4 on: July 12, 2012, 06:05:27 AM »
AWK

#### alexmahone

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##### Re: Why does the equilibrium constant have that form?
« Reply #5 on: July 12, 2012, 07:10:24 AM »

#### Borek

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##### Re: Why does the equilibrium constant have that form?
« Reply #6 on: July 12, 2012, 08:29:33 AM »
But why are they related to each other in that way, and not in some other way?

While there are ways of deriving law of mass action from the first principles (for example using statistical mechanics), the simplest answer to this question is - because that's the way it works when you do the experiment. After all, we are dealing with the reality - we don't make these things up, we observe the nature and describe it using best math we can.
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#### alexmahone

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##### Re: Why does the equilibrium constant have that form?
« Reply #7 on: July 12, 2012, 08:30:58 AM »
But why are they related to each other in that way, and not in some other way?

While there are ways of deriving law of mass action from the first principles (for example using statistical mechanics), the simplest answer to this question is - because that's the way it works when you do the experiment. After all, we are dealing with the reality - we don't make these things up, we observe the nature and describe it using best math we can.

Well, my question was whether there is an intrinsic reason why nature works that way.

#### Jorriss

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##### Re: Why does the equilibrium constant have that form?
« Reply #8 on: July 14, 2012, 12:00:16 AM »
Well, my question was whether there is an intrinsic reason why nature works that way.
If you trace it back far enough it shows up as a natural result of the second law of thermodynamics.

#### juanrga

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##### Re: Why does the equilibrium constant have that form?
« Reply #9 on: July 15, 2012, 09:23:15 AM »
For the reaction aA+bB cC+dD,
the equilibrium constant is K=[C]^c[D]^d/[A]^a[B ]^b.

This quantity (K) is the same at equilibrium regardless of the initial concentrations of A, B, C, D. But why does K have that form?

Because at equilibrium, by definition, the forward rate (+) and the backward rate (-) are both equal and constant

k+[A]a[B ]b = k-[C]c[D]d

This can be rewritten as

k+ / k- = [C]c[D]d / [A]a[B ]b

and k+ / k- is what we call K.

Note: The equilibrium constant is really defined using activities, but for many applications it can be approximated by concentrations.

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