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

1 Answer
May 26, 2017

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#.