Timeline for Chemical kinetics with the reaction of tert-butyl bromide with azide ion
Current License: CC BY-SA 3.0
11 events
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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
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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 |