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?

Feb 23, 2016

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

$\setminus m a t h b f \left(r \left(t\right) = k {\left[A\right]}^{m} {\left[B\right]}^{n}\right)$

where:

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

Naturally, in the original rate law, since we are multiplying the rate constant of units $\text{???}$ by the concentrations of overall units ${\text{M}}^{3}$ to get a rate in $\text{M"/"s}$, we ought to use $\frac{1}{\text{M"^2cdot"s}}$ 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)))

$= \textcolor{b l u e}{0.5}$ $\textcolor{b l u e}{{\text{M"^(-2)"s}}^{- 1}}$ or $\textcolor{b l u e}{\frac{1}{\text{M"^2cdot"s}}}$