# What is the order of the reaction A -> B if changing the concentration of A from "0.05 M" to "0.1 M" changes the rate from 2 xx 10^(-3) "M/s" to 1.6 xx 10^(-2) "M/s"?

May 15, 2016

The reaction is:

$\text{A" -> "B}$

The trick with this is that the order of the reactant $\text{A}$, given that it is the only reactant, is the order of the reaction.

The order of a reactant is basically the contribution of its concentration to the change of reaction rate.

To put this in a larger context, the reaction rate $r \left(t\right)$ (as a function of time) is related to the concentration $\left[\text{A}\right]$ for reactant $\text{A}$ by a rate constant $k$, as follows:

$\setminus m a t h b f \left(r \left(t\right) = k {\left[\text{A}\right]}^{m} = - \frac{d \left[A\right]}{\mathrm{dt}}\right)$

where $m$ is the order of reactant $A$ and $- \frac{d \left[A\right]}{\mathrm{dt}}$ is the rate of disappearance of $A$ (you don't need the rate of disappearance for this problem, but I thought I'd provide that for a brief review).

The above equation is known as a rate law.

Note that you have two rates and two initial concentrations (you do not need equilibrium concentrations---in fact, those are not practical for kinetics studies, because the reaction is most active when it first starts).

These are for each trial run for a particular kinetics study, and can be used to determine the order of the reaction as follows.

Suppose we compared rate $2$ with rate $1$ relative to the change in initial concentration ${\left[\text{A}\right]}_{i}$ of $\text{A}$. Then what we really do is divide the rate laws for trials $2$ and $1$.

$\frac{{r}_{2} \left(t\right)}{{r}_{1} \left(t\right)} = {\left(\frac{{\left[A\right]}_{i , 2}}{{\left[A\right]}_{i , 1}}\right)}^{m}$

$\frac{1.6 \times {10}^{- 2}}{2 \times {10}^{- 3}} = {\left(\left(\text{0.1 M")/("0.05 M}\right)\right)}^{m}$

${2}^{m} = 8$

$\textcolor{b l u e}{m = 3}$

For this process, we thus describe it by saying it is a third-order unimolecular reaction.

What this order says is that if you double the concentration of reactant $\text{A}$, you make the rate of reaction multiply by $\setminus m a t h b f \left(8\right)$.