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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?

Nicholas Hassan
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    Interesting question, but it has been addressed nicely https://chemistry.stackexchange.com/questions/6243/how-is-it-that-the-equilibrium-constant-does-not-depend-on-the-mechanism. 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. – AChem Aug 17 '19 at 19:53
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    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. – orthocresol Aug 17 '19 at 20:13
  • @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. – Nicholas Hassan Aug 17 '19 at 20:17
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    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. – orthocresol Aug 17 '19 at 20:18
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    I suggest reading https://chemistry.stackexchange.com/questions/75274/relation-between-chemical-kinetics-and-chemical-equilibrium which also (perhaps more directly) addresses your question. – orthocresol Aug 17 '19 at 20:25

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