# If the units of a gas-phase rate constant are "atm"^(-1)cdot"s"^(-1), then what is the order of the overall reaction?

May 26, 2017

Assuming you mean a reaction such as this one:

${\text{N"_2"O"_4(g) rightleftharpoons 2"NO}}_{2} \left(g\right)$

with rate law

$r \left(t\right) = k {\left({P}_{{N}_{2} {O}_{4}}\right)}^{n}$

where ${P}_{{N}_{2} {O}_{4}}$ is the partial pressure of ${N}_{2} {O}_{4}$ in $\text{atm}$, the rate $r \left(t\right)$ then could have units of $\text{atm/s}$, let's say. $n$ is the order of the reaction (since ${N}_{2} {O}_{4}$ is our only reactant in this example).

If that is the case, then:

${\text{atm"/"s" = [???]cdot"atm}}^{n}$

We claim that the units of $k$ are ${\text{atm"^(-1)"s}}^{- 1}$, so:

${\text{atm"^(color(red)(1))/"s" = 1/("atm"cdot"s")cdot"atm}}^{n}$

= "atm"^(color(red)(n-1))/("s")

But for this to be true, we must have:

$n - 1 = 1$,

or

$\textcolor{b l u e}{n = 2}$

Therefore, the order of the overall reaction is $2$.