## 15.3 partb ?

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

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### 15.3 partb ?

Hello this is a stupid question but (hey it'll be easy points for someone to answer!)
Ok for 15.3 part b when you are finding the rate of formation of O2, the solutions manual has O2= (6.2 x -3) (1 mol O2/ 2 mol NO2)
So is the general structure for that formula a rate law ? Why do they multiply the rate of reaction of NO2 with the mole ratio?

Alvin Tran 2E
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Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.3 partb ?

That isn't the rate law but the unique rate of the reaction. In part A, you should have calculated the rate of reaction of NO2.
In general for all reactions, aA -> bB + cC
$rate = \frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt}$
The reactants and products in a reaction will have the same unique rate and because there is only 1 mole of O2 compared to 2 moles of NO2, you divide by 2 to get the rate of formation of O2.

Jessica Lutz 2E
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Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.3 partb ?

I thought that the general structure for a rate law looked something like this: Rate=k[A]^n [B]^m (if the reaction had reactants A and B with orders n and m).
Secondly, you multiply by the mole ratio because 6.2 x -3 itself gives the rate of reaction for NO2, not O2. Since 2 moles of NO2 make 1 mole of O2, you divide the rate of NO2 by 2 to get the production rate of O2.

Jenny Cheng 2K
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### Re: 15.3 partb ?

When you are not looking at the unique rate of the reaction, you must specify the species to which the rate refers. The species consumed or produced have rates that are related to the stoichiometry of the reaction.