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I'm learning about rate law and equilibrium, and the textbook really hammers in that the exponents for the rate law must be determined experimentally - you can't just use the coefficients from the balanced equation. In a generic reaction aA + bB -> cC + dD:

rf = kf[A]w[B]x

rr = kr[C]y[D]z

It is not always the case that w = a, x = b etc.

Equilibrium is then described as the point that the forward reaction is occurring at the same rate as the reverse reaction. For the same reaction as above, we set the two rates to equal each other.

rf = rr

kf[A]w[B]x = kr[C]y[D]z

Kc = kf / kr = [C]y[D]z / [A]w[B]x

I'm told that the exponents here are based on the coefficients of the balanced equation, such that w = a, x = b, y = c, and z = d. This seems to be in direct opposition to what I read earlier about the rate law, as equilibrium is just built from two rates.

Why is this?

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    $\begingroup$ Interesting question, but it has been addressed nicely chemistry.stackexchange.com/questions/6243/…. Basically, you have to distinguish something called as the elementary reaction where the rate constant is indeed based on the equation as written and the overall reaction. $\endgroup$
    – ACR
    Commented Aug 17, 2019 at 19:53
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    $\begingroup$ Equilibrium is not necessarily "just built from two rates"; you don't actually need the concept of a rate to define an equilibrium. The approach you're taught is the simplest and most intuitive, which is why it's taught first, but it has its limitations. Refer to any physical chemistry / thermodynamics textbook. $\endgroup$ Commented Aug 17, 2019 at 20:13
  • $\begingroup$ @orthocresol Thanks. So a higher level definition of equilibrium is NOT "the point at which the forward and reverse rates are equal?" This is just something that is intuitive but already falls apart when trying to use rates equations to define equilibrium equations. $\endgroup$ Commented Aug 17, 2019 at 20:17
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    $\begingroup$ It only works when your forward and reverse reactions are elementary reactions, in which case the coefficients in the rate law are indeed the stoichiometric coefficients. However, the equilibrium constant can be proven to have this form (involving the stoichiometric coefficients), even when the forward and reverse reactions are not elementary. I am trying to find the correct links, it has probably been asked/answered here a few times. $\endgroup$ Commented Aug 17, 2019 at 20:18
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    $\begingroup$ I suggest reading chemistry.stackexchange.com/questions/75274/… which also (perhaps more directly) addresses your question. $\endgroup$ Commented Aug 17, 2019 at 20:25

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