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I am familiar with the idea that rate law must be determined experimentally. I am also familiar with the idea that there is no formulated way for coming up with the rate law besides just experimentation.

My question is: why isn't the rate law of:

$$ aA + bB \rightarrow \mbox{something}$$

just:

$$r = k[A]^a[B]^b$$

?

I mean, the equilibrium constant expression involves something to that effect, and the equilibrium constant expression should be precisely talking about when the rates for each direction are equal. So if the rate law is not the above, then why does the equilibrium have that expression?

EDIT: A bit off topic, but how is it possible for something to be zeroth order?

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  • $\begingroup$ Zeroth order in this schema: If a or b = 0, the rate is no longer proportional to the concentration of that reactant as any number to the power of zero is 1. The reaction is then zeroth order with respect to that reactant. $\endgroup$ – J. LS Apr 21 '15 at 8:43
  • $\begingroup$ These two past questions may be of some use. $\endgroup$ – Nicolau Saker Neto Apr 21 '15 at 11:22
  • $\begingroup$ @NicolauSakerNeto Sneaky linking it on these and two, I nearly missed that. And yes, I did also write something related to the topic, but it is more an example than an actual explanation. Shameless self promotion. !( ^_^ )! $\endgroup$ – Martin - マーチン Apr 21 '15 at 14:10
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Since rate of a reaction depends actually upon the way reactants react, there is not surety that the quoted stoichiometric coefficients may be the same for rate of a reaction, the rate depends on the way or mechanism of the reaction, and since all reactions have a bit different mechanism no strict formula could be quoted. It is only experimentally derived and can be theoretically approached if we know that it is an elementary (simple step) reaction or it falls under a already classified or studied reaction.

Zero Order, what if a reactant is in very large excess, now the rate is not limited or dependent on its concentration, also it's concentration is almost constant, so it is also merged into the constant and hence we have zero order, some adsorption reactions are maybe zero order too, it is because it actually depends on the surface area of the substance and it's constant.

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The order of reaction with respect to each of the reactants, sometimes called to partial order is not equal to the stoichiometric coefficients of the reaction. They are instead dependent on how the reaction proceeds, this is why it must be determined experimentally. So basically it depends on the reaction mechanism.

The equilibrium constant has nothing to do with time as it is measured once the equilibrium has been reached using the concentrations of the reactants and products at that time.

Your rate could be:

$$rate=k[A]^x[B]^y$$

Where $x$ and $y$ are the orders with respect to A and B respectively rather than the stoichiometric coefficients.

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