Unique rate coefficients

[madisondesilva1c]

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]

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]

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.