How do coefficients affect the rate law?

1 Answer
May 30, 2015

Do you mean stoichiometric coefficients? Let's take a classic example.

#N_2O_4 rightleftharpoons 2NO_2#

Each coefficient is worked into the rate of appearance/disappearance like so:

#-1/(nu)(d[A])/(dt) = 1/(nu)(d[B])/(dt) = r(t) = k[A]^("order")#

where #nu# is the stoichiometric coefficient, #A# is a reactant, and #B# is a product. Of course, as reactants are used up, products are generated. This reaction only involves #N_2O_4# as a decomposing reactant, so it is first order.

So, we get:

#-(d[N_2O_4])/(dt) = 1/2(d[NO_2])/(dt) = r(t) = k[N_2O_4]^1#

Therefore, the stoichiometric coefficients do not affect how the rate law is written, but they do affect the value of the rate constant #k#.

Also, the reaction order does not correspond to the stoichiometric coefficients; it's only a coincidence here. The reaction order only says how many molecules per mole participate in the reaction (molecularity is usually a good indication), not how many moles of each molecule there are (stoichiometric coefficients).