# How do coefficients affect the rate law?

May 30, 2015

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

${N}_{2} {O}_{4} r i g h t \le f t h a r p \infty n s 2 N {O}_{2}$

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

$- \frac{1}{\nu} \frac{d \left[A\right]}{\mathrm{dt}} = \frac{1}{\nu} \frac{d \left[B\right]}{\mathrm{dt}} = r \left(t\right) = k {\left[A\right]}^{\text{order}}$

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 ${N}_{2} {O}_{4}$ as a decomposing reactant, so it is first order.

So, we get:

$- \frac{d \left[{N}_{2} {O}_{4}\right]}{\mathrm{dt}} = \frac{1}{2} \frac{d \left[N {O}_{2}\right]}{\mathrm{dt}} = r \left(t\right) = k {\left[{N}_{2} {O}_{4}\right]}^{1}$

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).