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Apr 25, 2017 at 11:35 history edited Pritt says Reinstate Monica CC BY-SA 3.0
added 9 characters in body
Apr 24, 2017 at 15:56 comment added Pritt says Reinstate Monica No, rate of reaction is defined as $$\text{Rate}=\frac{d\text{[Reactant]}}{dx}$$
Apr 24, 2017 at 15:30 comment added Teoc @PrittBalagopal Wouldn't the rate be the negative of what you wrote, because we are measuring the rate of formation of product, not reactant?
Apr 24, 2017 at 7:36 comment added Pritt says Reinstate Monica @Display Name, for reactions at equilibrium, consider it has two reactions, one forward and one backward. Suppose for the reaction: $$\ce{mA + nB <=>[K_1][K_2] pC + qD}$$ We consider the rate of reaction as difference between the forward and backward rates. In that way, you will get the rate law to be: $$Rate=K_2[C]^p[D]^q-K_1[A]^m[B]^n$$
Apr 24, 2017 at 5:44 vote accept Teoc
Apr 24, 2017 at 5:43 comment added ringo You should really take a look at this so you can correctly format your future posts.
Apr 24, 2017 at 5:37 comment added Pritt says Reinstate Monica Sorry, my bad, that plus was not meant to be. (I cant edit that comment now, but I think you get it)
Apr 24, 2017 at 5:32 comment added Teoc According to my book rate would be $k[A]^m [B]^n$ is that just a typo on your part? Additionally, how did you derive the rate law for reactions in equilibrium?
Apr 24, 2017 at 5:22 comment added Pritt says Reinstate Monica Those rate laws are directly derived from the reaction. For example, in an elementary reaction: $$\ce{mA + nB -> C}$$ The rate law would be $$Rate=[A]^m+[B]^n$$
Apr 24, 2017 at 5:18 comment added Teoc Sorry, but how did you get your first rate law?
Apr 24, 2017 at 5:17 history answered Pritt says Reinstate Monica CC BY-SA 3.0