## Course reader pg 73 example

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Michael Lonsway 3O
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### Course reader pg 73 example

For the pre-equilibrium approach example why is there a 2 in front of the reaction constant? Is it because 2 products (2NO2 ) are formed? Also, is the intermediate only present in the rate law if the first reaction is fast and the second reaction is slow?

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### Re: Course reader pg 73 example

Yes you are correct! The 2 in front of the reaction constant is there because there are 2 NO2 formed. And yes, the intermediate is the product of the reaction for a fast step and is the reactant of the slow step.

Chem_Mod
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### Re: Course reader pg 73 example

Also see the end of the kinetics section in the course reader where I have additional discussion on this.