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?
  • $0.10 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
  • $0.40 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
  • $1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
  • $2.5 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$

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 $1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$