# For the reaction "N"_2"O"_3(g) -> "NO"(g) + "NO"_2(g), given the following data, calculate the rate constant?

## ${\text{N"_2"O"_3(g) -> "NO"(g) + "NO}}_{2} \left(g\right)$ ["N"_2"O"_3]" "" Initial Rate" $\text{0.1 M" " "" } r \left(t\right) = 0.66$ $\text{M/s}$ $\text{0.2 M" " "" } r \left(t\right) = 1.32$ $\text{M/s}$ $\text{0.3 M" " "" } r \left(t\right) = 1.98$ $\text{M/s}$

Mar 2, 2016

So, we have:

${\text{N"_2"O"_3(g) -> "NO"(g) + "NO}}_{2} \left(g\right)$

["N"_2"O"_3]" "" Initial Rate"
$\text{0.1 M" " "" } r \left(t\right) = 0.66$ $\text{M/s}$
$\text{0.2 M" " "" } r \left(t\right) = 1.32$ $\text{M/s}$
$\text{0.3 M" " "" } r \left(t\right) = 1.98$ $\text{M/s}$

Since you were given only the concentrations of the reactant, we are of course focusing on what happens when changing the reactant concentration.

Notice how doubling $\left[{\text{N"_2"O}}_{3}\right]$ doubles the rate, and tripling $\left[{\text{N"_2"O}}_{3}\right]$ triples the rate. i.e.

$2 \times \left[{\text{N"_2"O}}_{3}\right] \leftrightarrow 2 \times 0.66$ $\text{M/s}$
$3 \times \left[{\text{N"_2"O}}_{3}\right] \leftrightarrow 3 \times 0.66$ $\text{M/s}$

We can make the relationship saying that:

((["N"_2"O"_3])/(["N"_2"O"_3]_0))^m = (r(t))/(r_0(t)

where $\left({\left[{\text{N"_2"O"_3])/(["N"_2"O}}_{3}\right]}_{0}\right) = 2$ for the ratio of the new concentration to the old concentration of the reactant and $\frac{r \left(t\right)}{{r}_{0} \left(t\right)} = 2$ for the ratio of the new rate to the old rate of reaction. $m$ is the order of the reactant.

This is to compare how changing the concentration affects the rate. In examining trials 1 and 2, we see:

${2}^{m} = 2 \to m = 1$

So the reactant is first order, and since it is the only reactant, the overall reaction is first order.

Now we can write the rate law as:

$\setminus m a t h b f \left(r \left(t\right) = k \left[{\text{N"_2"O}}_{3}\right]\right)$

Thus, to get $k$, simply pick a value of $r \left(t\right)$ and $\left[{\text{N"_2"O}}_{3}\right]$, both of which are in the same trial as each other, and solve for $k$:

$0.66$ $\text{M/s" = kxx"0.1 M}$

$\textcolor{b l u e}{k = 6.6}$ $\textcolor{b l u e}{{\text{s}}^{- 1}}$

And if you pick trials 2 or 3 for the calculation of $k$, you will get the same thing.