# Why are the stoichiometric coefficients used as exponents in the chemical equilibrium constant?

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?

• – Mockingbird Feb 14 '17 at 4:23
• Sorry, but I don't understand how this could be considered a duplicate of that question, @KlausWarzecha. – Martin - マーチン Feb 14 '17 at 5:54
• It is definitely related @Mockingbird. Your question did not come up on my search though. – user1934868 Feb 14 '17 at 15:06
• @klaus warzecha. That question has to do with rate laws, which do not use stoichiometric coefficients. The equilibrium constant does. – user1934868 Feb 14 '17 at 15:09
• 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 (...) – orthocresol Feb 16 '17 at 12:19