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From what I can ascertain, the formulation of the chemical equilibrium constant is somewhat arbitrary. I don't understand the motivation for putting the stoichiometric coefficients in the exponents. This seems like you are double counting. For instance, if A -> 4B, for every mole of A that decomposes to B produces 4 moles of B. So my reaction would already have a higher amount of B. Then I take those four moles and raise to the 4th power. So if I started with 2 moles of A and 1 decomposed, I would have one mole of A to 4 moles of B. Then I would have a equilibrium constant of 256. How is this superior to not having exponents and just having an equilibrium constant of 4? Or for that matter there are many arbitrary ways I can stick on or not the stoichiometric coefficients. Why not take the 4th root of B over the first root of A?

Edit: Many are saying that this is a duplicate of Law of Mass Action. However, in that question the stoichiometric coefficients just HAPPENED to be the exponents. And as pointed out in the answer the order in the rate law doesn't have to be the same as them. My question is dealing with the Eqilibrium Constant. The Equipibrium Constant is defined such that the stoichiometric coefficients are ALWAYS the exponents. Why?

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  • $\begingroup$ Related: chemistry.stackexchange.com/questions/68195/… $\endgroup$
    – Mockingbird
    Commented Feb 14, 2017 at 4:23
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    $\begingroup$ Sorry, but I don't understand how this could be considered a duplicate of that question, @KlausWarzecha. $\endgroup$
    – Martin - マーチン ♦
    Commented Feb 14, 2017 at 5:54
  • $\begingroup$ It is definitely related @Mockingbird. Your question did not come up on my search though. $\endgroup$
    – user1934868
    Commented Feb 14, 2017 at 15:06
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    $\begingroup$ There are some good questions and answers here on how the form of the equilibrium constant is obtained. See: Is there a reason for the mathematical form of the equilibrium constant? Truthfully, you can define the equilibrium constant any way you wish to. However, if you wish to use the typical thermodynamics equations, you will have to define it in a specific way. If you define $K = [\ce{B}]/[\ce{A}]^2$ for a reaction $\ce{A -> 4B}$ then $\Delta G^\circ = -RT\ln K$ will no longer apply (...) $\endgroup$
    – orthocresol
    Commented Feb 16, 2017 at 12:19
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    $\begingroup$ It is perhaps easier to think of it this way. Let's say you have a chemical reaction $a\ce{A} + b\ce{B} \to c\ce{C} + d\ce{D}$. When you run through all the thermodynamics proofs and whatnot, the result you get out of it is $$\Delta G^\circ = -RT \ln \left(\frac{[\ce{C}]^c_{\mathrm{eq}}[\ce{D}]^d_{\mathrm{eq}}}{[\ce{A}]^a_{\mathrm{eq}}[\ce{B}]^b_{\mathrm{eq}}}\right)$$ If, and only if, you define that entire thing as $K$ will you obtain the familiar relation. $\endgroup$
    – orthocresol
    Commented Feb 16, 2017 at 12:21

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