Question

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

Answer 1

Assuming you mean a reaction such as this one:

`"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)`

with rate law

`r(t) = k(P_(N_2O_4))^n`

where `P_(N_2O_4)` is the partial pressure of `N_2O_4` in `"atm"`, the rate `r(t)` then could have units of `"atm/s"`, let's say. `n` is the order of the reaction (since `N_2O_4` is our only reactant in this example).

If that is the case, then:

`"atm"/"s" = [???]cdot"atm"^n`

We claim that the units of `k` are `"atm"^(-1)"s"^(-1)`, so:

`"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

`color(blue)(n = 2)`

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