## Stoichiometric coefficients and rates

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

Cameron Sasmor 1G
Posts: 11
Joined: Fri Sep 26, 2014 2:02 pm

### Stoichiometric coefficients and rates

For a chemical reaction, why are the opposites of the chemical equation's stoichiometric coefficients the diving factors used to calculate the rates of the reaction?

For example for the equation: 2NO2(g) ---> 2NO(g) + O2(g)

The rate of reaction can be calculated as -1/2*(delta NO2)/(delta t)

Why isn't the -1/2 a -2?

Chem_Mod
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### Re: Stoichiometric coefficients and rates

If the coefficient of the reactant is 2, then it will be consumed twice as fast, so dA/dT = -2k[A] (instead of the usual -k[A], assuming first order). So -1/2 dA/dT = -k[A]

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Stoichiometric coefficients and rates

If you look at the bottom of page 53 in the course reader (sorry, I have the 2014 course reader) or the second page in the kinetics section, you can see why it is better to use the fraction and to put the coefficient in the denominator. Otherwise, you would have to use improper fraction or a least common denominator to set up the rate equation.