# Problem: 03-02-2

## Question

The rate law for a particular reaction is rate $=k[\mathrm{XY}]^2$. In an experiment, the initial rate of the reaction is determined to be $0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})$ when the initial concentration of $\mathrm{XY}$ is $0.40 \mathrm{~mol} / \mathrm{L}$.

-   What is the value of the rate constant, $k$, for the reaction?
    -   [ ] $0.10 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
    -   [ ] $0.40 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
    -   [x] $1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
    -   [ ] $2.5 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$

## Solution

lo tind the value of the rate constant for the reaction, let's first solve the rate law for $k$ :

$$
k=\frac{\text { rate }}{[\mathrm{XY}]^2}
$$

Next, let's plug in the initial rate and concentration given in the text:

$$
\begin{aligned}
k & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{(0.40 \mathrm{~mol} / \mathrm{L})^2} \\
& =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{0.16 \mathrm{~mol}^2 / \mathrm{L}^2} \\
& =1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
\end{aligned}
$$

So, the value of the rate constant for the reaction is
$1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$