Question

How do coefficients affect the rate law?

Answer 1

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).