Equilibrium constant expression; why exponents?

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Hello all.

I am wondering why we raise the concentration to the exponent of the stochiometric coefficient. I understand why we take the products of reactants or products.

For instance, let me give an example to show where I am getting mixed up.

Let's say 2A +3B = 8C as a reversible reaction.

Say the equilibrium concentration of A, B, and C are all 2 moles per liter.

then the equation is 2^8 / (2^2)(2^3) = 16

But the concentration of C is the same as the concentration of A and of B...so why is the equation saying the forward reaction is way more favored? If the forward reaction is so heavily favored, why is the concentration the same as for the reactants?
 

Charles Link

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If the forward reaction were heavily favored, the equilibrium constant would be much larger than 16 for this case. It might be 100,000 or more.
 
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I thought being over one meant forward was favored and under one meant backward was favored. But wouldn’t the concentration alone tell you which was favored?
 

Charles Link

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I thought being over one meant forward was favored and under one meant backward was favored. But wouldn’t the concentration alone tell you which was favored?
If equilibrium is reached with nearly equal concentrations of the components, the reaction is not really favored to go in either direction. They basically coexist in that case, as opposed to being a combustible mixture that goes almost entirely into the final product(s). For this type of thing "1" is not the determining factor. Whether it is a 1 or a 16 makes little difference. The final exponent of 8 is larger than normal, but basically, if the equilibrium constant were a ## 10^{-5} ##, (or ## 10^{-8} ## etc.), it would go in the reverse direction, and a ## 10^{+5} ## would mean the forward direction would be favored. ## \\ ## An example of a case where the equilibrium constant would be on the order of unity would be if diatomic molecules like ## O_2 ## or ## N_2 ## were heated to several thousand degrees. Then you could have nearly equal amounts of ## O ## and ## O_2 ##, etc.
 
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Then why even have the exponents at all? Seems the concentration alone tells you enough
 

Charles Link

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Please read the edited additions of post 4. Meanwhile, the exponents are necessary in doing any quantitative calculations using the equilibrium constant. That is how the equilibrium concentrations are computed. I believe it is the Law of Mass Action that says the coefficients in the reaction equation are the exponents in the expression with the equilibrium constant. ## \\ ## Note: In the example I gave at the end of post 4, the equilibrium constant is a function of the temperature. ## \\ ## A very well known equilibrium constant is ## [H^+][OH^-]=1.0 \cdot 10^{-14} ## for water solutions. If you know ## [H^+] ##, you can compute ## [OH^-] ##, etc.
 
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Borek

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I have a feeling you you may be mistaking reaction quotient with equilibrium constant. They both take the same form (products over reactants, all in appropriate powers), but Q is at any moment and can take any value, while K describes equilibrium and has a single value. At equilibrium Q=K, at other points they can differ.
 

Charles Link

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I have a feeling you you may be mistaking reaction quotient with equilibrium constant. They both take the same form (products over reactants, all in appropriate powers), but Q is at any moment and can take any value, while K describes equilibrium and has a single value. At equilibrium Q=K, at other points they can differ.
With the equilibrium constant, it tells what the final results will be, but it doesn't give any information on how long it will take to reach equilibrium. If the components are not at an equilibrium concentration, I think you can always tell what direction the reaction will proceed, but you don't have any information on how quickly the process will take place. I think the OP is presently mostly learning the equilibrium part. That is usually presented before they learn about reaction rates.
 

Borek

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With the equilibrium constant, it tells what the final results will be, but it doesn't give any information on how long it will take to reach equilibrium. If the components are not at an equilibrium concentration, I think you can always tell what direction the reaction will proceed, but you don't have any information on how quickly the process will take place. I think the OP is presently mostly learning the equilibrium part. That is usually presented before they learn about reaction rates.
All true, my statement was addressed to the OP. Sorry for the possible confusion.

I believe the answer to the original question ("why exponents") can be derived with thermodynamics. It requires a bit of knowledge about chemical potentials, Gibbs free energy and the [itex]\Delta G^0 = -RT ln(K)[/itex] dependence. Basically if

[tex]K = \frac {[A]^a[ B]^b}{[C]^c[D]^d}[/tex]

then

[tex]\log(K) = a \log([A]) + b \log([ B]) - c \log([C]) - d \log([D])[/tex]

and with some assumptions the right side can be converted into the change of an intensive property dependent on the amount of substance (yes, it is quite handwavy at this moment, as concentration is an extensive property, so you need to put a standard volume somewhere there; I don't remember details and thermodynamics is not my forte, but I am quite sure this is a general line of thinking).
 

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