What is the rate constant of a reaction if rate 1.5 (mol/L)s, [A] is 1 M, [B] is 3 M, m=2, and n=1?

For a reaction's rate law, in general the relevant portion is:

#\mathbf(r(t) = k[A]^m [B]^n)#

where:

• #r(t)# is the rate of the reaction. We know this to be in #"M/s"#, or #"mol"/("L"cdot"s")#, i.e. #"mol"cdot"L"^(-1)"s"^(-1)# or #"M"cdot"s"^(-1)#.
• #k# is the rate constant, in units that are based on the rest of the rate law. We'll need to determine these units.
• #["A"]# and #["B"]# are the concentrations of #"A"# and #"B"#, respectively, in #"M"#. Recall that #"M"# #=# #"mol/L"#.
• #m# is the order of reactant #"A"# and #n# is the order of reactant #"B"#. These contribute to the overall order of this reaction.

We already have all the information we need to solve this.

#"1.5 M"/"s" = k("1 M")^2("3 M")^1#

This bimolecular reaction is observed to be second order with respect to #"A"#, first order with respect to #"B"#, and third order overall.

Naturally, in the original rate law, since we are multiplying the rate constant of units #"???"# by the concentrations of overall units #"M"^3# to get a rate in #"M"/"s"#, we ought to use #1/("M"^2cdot"s")# for the rate constant to cancel out the units properly.

#color(blue)(k) = (("1.5 M")/"s")/(("1 M"^2)("3 M")) = ("1.5" cancel"M")/"s"xx1/("3 M"^(cancel(3)^(2)))#

#= color(blue)(0.5)# #color(blue)("M"^(-2)"s"^(-1))# or #color(blue)(1/("M"^2cdot"s"))#