# Problem: 03-02-2 | Toshiki's Note
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# Problem: 03-02-2 ​

Author:Anda Toshiki
Updated:a minute ago
Words:205
Reading:1 min

## Question ​

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

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

## Solution ​

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

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

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

k=0.16 mol/(L⋅s)(0.40 mol/L)2=0.16 mol/(L⋅s)0.16 mol 2/L 2=1.0 L/(mol⋅s)\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} k​=(0.40 mol/L)20.16 mol/(L⋅s)​=0.16 mol2/L20.16 mol/(L⋅s)​=1.0 L/(mol⋅s)​

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