/sci/ - Science & Math

[keq v rate law No.5418868]

For an equation like this: aA + bB <--> cC + dD

The rate laws for the forward (f) and reverse (r) reactions would be, respectively:

Rate(f) = kf[A]^n^m, and Rate(r) = kr[C]^x[D]^y, where n, m, x, and y are the exponents which are determined experimentally, not from the stoichiometric coefficients.

The relationship between kinetics and equilibrium in this scenario is Rate(f) = Rate(r), therefore kf/kr, which equals Keq. However I was always taught that the equilibrium expression was:

Keq = [C]^c[D]^d / [A]^a^b, where c, d, a, and b are the stoichiometric coefficients.

My question is if kf/kr = Keq, why are the exponents different? I can't seem to find an explanation for this, but the only thing I can think of is that the above equilibrium expression exponents are actually the same as the experimentally determined rate law exponents, but only for single elementary step reactions are they the same as the stoichiometric coefficients.

[Anonymous No.5419141]

I guess you guess right, mhm..
http://en.wikipedia.org/wiki/Rate_law#Equilibrium_reactions_or_opposed_reactions
For elementary reactions, I would say that the rate law can be motivated by looking at cross sections, as in the statistical theory of reactions.
There the m,n,x,y are the stoichiometric coefficients.
I, too, would guess that if you've got different exponents, it should be because the reaction is not elementary.

[Anonymous No.5419209]

the laws in the form rate=constant*concentration^coefficient are indeed only valid for elementary reactions. It should be the first thing you were told when you heard or read about this. (because a lot of mistakes come from the false asumption that it al<ways works)