# 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?

##### 1 Answer

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

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

#r(t)# is therate 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 therate constant, in units that are based on the rest of the rate law. We'll need to determine these units.#["A"]# and#["B"]# are theconcentrationsof#"A"# and#"B"# , respectively, in#"M"# . Recall that#"M"# #=# #"mol/L"# .#m# is theorderof reactant#"A"# and#n# is theorderof 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 **third order** overall.

Naturally, in the original rate law, since we are multiplying the **rate constant** of units **concentrations** of overall units **rate** in *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"))#