# Given the reaction A -> B, if the rate multiplies by 100 when the concentration of A is multiplied by 10, what is the order of the reaction?

## I know it's 2 but why?

Jul 24, 2016

Pretend you don't know the answer yet. The reaction given is:

$\stackrel{\text{Reactant(s)")overbrace(A) -> stackrel("Product(s)}}{\overbrace{B}}$

When we write the rate law, or the equation that relates the rate $r \left(t\right)$ (as a function of time) with the rate constant $k$ and the concentration $\left[\text{ }\right]$ of all reactants (in this case, $A$), we have:

$\setminus m a t h b f \left(r \left(t\right) = k \stackrel{\text{Reactant(s) Only")overbrace([A]^m) = stackrel("Reactant(s)")overbrace(-1/nu_A(d[A])/(dt)) = stackrel("Product(s)}}{\overbrace{+ \frac{1}{\nu} _ B \frac{d \left[B\right]}{\mathrm{dt}}}}\right)$

where ${\nu}_{A}$ is the stoichiometric coefficient of $A$.

NOTE: Stoichiometric coefficients do not necessarily correspond with the order.

$\frac{d \left[X\right]}{\mathrm{dt}}$ is the rate of change in concentration of $X$. If $X$ is a reactant, it is a rate of disappearance. If $X$ is a product, it is a rate of appearance.

Now, we know that $k$ remains constant no matter what, as long as it's the same reaction. Therefore, when $\left[A\right]$ changes, $r \left(t\right)$ must change, because $k$ will stay the same.

But, there's also the order of the reactant $A$, labeled as $m$, which is just the exponent in the rate law. When we change $\left[A\right]$, the value of $m$ determines how much $r \left(t\right)$ changes.

The question asks us to multiply $\setminus m a t h b f \left(\left[A\right]\right)$ by $\setminus m a t h b f \left(10\right)$ and tells us how much $\setminus m a t h b f \left(r \left(t\right)\right)$ changes, so let's do that.

$\underline{\underline{\textcolor{red}{x}}} \cdot r \left(t\right) = k {\left(10 \left[A\right]\right)}^{m} = \underline{\underline{\textcolor{red}{{10}^{m}}}} \cdot k {\left[A\right]}^{m}$

But whatever we do to one side, we have to do to the other, so we have to multiply by $x$, a multiplier that is dependent on $m$. We are given that $x = 100$, so:

$100 = {10}^{m}$

Since ${10}^{2} = 100$, the order of reactant $A$ is $\setminus m a t h b f \left(2\right)$, i.e. $m = 2$. So, the resultant rate law has the exponent $\textcolor{red}{2}$ for $\left[A\right]$:

$\textcolor{b l u e}{r \left(t\right) = k {\left[A\right]}^{\textcolor{red}{2}}}$

NOTE:

It's a little different when a second reactant comes into play, but in that case, we rig different experiment trials so that the concentration of other reactants stay the same as well.

That way, the influence of changing concentrations gives a clear relationship with the rate.