This question already has an answer here:

We know that the rate of a reaction $\ce{aA + bB -> cC + dD}$, the rate of the forward reaction is given by $r_\mathrm f = k_\mathrm f[\ce A]^p[\ce B]^q$ where $ a\neq p$ and $ b\neq q$ according to Chemical Kinetics.

However when studying Chemical Equilibrium, $\ce{aA + bB <=> cC + dD}$ when we write the forward and backward reaction rates and equate them, we write them as $r_\mathrm f = k_\mathrm f[\ce A]^a[\ce B]^b$ where the exponents are equal to the stoichiometric coefficients.

How is this the case?


marked as duplicate by Mithoron, A.K., Community Sep 3 '18 at 12:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


In your example for equilibrium there is a key assumption which you were either not taught, or glossed over. It is the assumption that both forward and backward reactions are elementary, which makes the use of the stoichiometric coefficients valid. For a generic multi-step reaction the exponents are not necessarily equal to the stoichiometric coefficients.

  • $\begingroup$ We are actually taught equilibrium a year before kinetics. So will a non elementary multi step reaction never achieve equilibrium OR will we define the new $K_c as K_c = \frac{[C]^w[D]^x}{[A]^y[B]^z}$ where $w,x,y,z \neq$ stoichiometric coeffecients? $\endgroup$ – harshit54 Sep 2 '18 at 18:11
  • $\begingroup$ (1) Of course it will reach equilibrium. Everything does! (2) the equilibrium constant $K$ is still defined using stoichiometric coefficients. It is just that you can no longer derive it in that fashion, by saying that $K$ is a ratio of rate constants. There are other, more rigorous, ways of deriving it. This definition of the equilibrium constant using stoichiometric coefficients is necessary to fulfill all the other thermodynamic relationships, such as $\Delta G^\circ = -RT \ln K$, see e.g. my comments here. $\endgroup$ – orthocresol Sep 2 '18 at 18:20
  • $\begingroup$ Sir, your comments lead to a lot many pages. Could you give me a link to the direct proof of the above expression. I don't have the Peter Rock's book. $\endgroup$ – harshit54 Sep 2 '18 at 18:33
  • $\begingroup$ You should look in a proper physical chemistry textbook, it can be somewhat involved. chemistry.stackexchange.com/a/42532/16683 is the standard proof but if you are new to the topic then you will find it to be lacking in detail. If you don't know what activities or chemical potentials are, or how they relate to $\Delta G$, then I suggest coming back to it at a later stage in your education. Or, of course, reading a physical chemistry textbook, which should explain all these concepts. $\endgroup$ – orthocresol Sep 2 '18 at 19:00
  • $\begingroup$ Yup I think that you are correct. I don't understand a thing. Thanks for the answer. $\endgroup$ – harshit54 Sep 2 '18 at 19:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.