Unique rate coefficients

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Unique rate coefficients

Postby madisondesilva1c » Fri Mar 01, 2019 3:12 pm

The in class examples had coefficients that were 1 and 2, however what is the best way to approach a reaction where the stoichiometric coefficients do not divide out in a neat fashion?

Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

Re: Unique rate coefficients

Postby Michael Novelo 4G » Fri Mar 01, 2019 3:55 pm

For the unique rate we use the form aA -> bB + cC or can be different depending on how many Reactants and products there are. Regardless of the coefficients it will always have the form of 1/b(coefficient) × d [b]/dt for example. The Reactants form will be have a negative sign and products will not. after dividing by the coefficients Reactants and products should equal each other and have the same instantaneous rate for the unique equation.

Kevin ODonnell 2B
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

Re: Unique rate coefficients

Postby Kevin ODonnell 2B » Fri Mar 01, 2019 6:48 pm

I know you are referring to the example where 2NO2 -> 2NO + O2 turned into - d[NO2]/dt = d[NO]/dt = 2d[O2]/dt. However, it is much easier to just use fractions for more complicated problems where this problem and the unique rate would be -(1/2)d[NO2]/dt = (1/2)d[NO]/dt = d[O2]/dt where the 1/2 comes from 1/stoichiometric coefficient unique to that molecule.

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