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A follow up question from this answer:

Compare:
$\ce{A(aq) +B(aq)->AB(aq)}$
$\ce{2A(aq) +B(aq)->A2B(aq)}$

We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants. Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.

Is there a way to "normalize" the $K_c$ of reactions like that shown above, such that a fair comparison can be made, regarding whether one reaction is more product favoured than the other?

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Since the equilibrium constant depends on the concentration of the species involved, there is no way to normalize it in the way I think you imagine it. A fair comparison of equilibrium constants is already given. The equilibrium constant is, as its name gives away already, constant. As such there is no normalization to be made.

Also, the quote is slightly off the point where it says that

the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants

because the equilibrium constant might actually be the same (except for the unit, of course). The way you calculate the constant changes, but inferring from that that the value changes drastically is simply overreaching.

If $K > 1$ then the equilibrium lies on the side of the products, if $K<1$ then the equilibrium lies on the side of the reactants.

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  • $\begingroup$ Is there any scenario where chemist would even compare equilibrium constants of different reactions like I described, or is K simply used to infer the product-reactant relationship for some single reaction? I have a feeling that my question may not even be practically valid. $\endgroup$ – Yandle Oct 31 '14 at 21:55
  • $\begingroup$ I have never compared the equilibrium constants of different reactions. Now rate constants are quite the different matter. $K$ here is just used to calculate amounts of substance at equilibrium. $\endgroup$ – tschoppi Oct 31 '14 at 22:55
  • $\begingroup$ The equilibrium constant is also the ratio of rate constants, when using the standard definitions (activity, etc.) it is possible to compare these. However it is much easier to compare reaction energies directly to compare reactions of different types. $\endgroup$ – Martin - マーチン Nov 26 '14 at 5:21

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