Chapter 15, Problem 23E

(a)

Interpretation Introduction

Interpretation:

The rate law and the value of the rate constant for the reaction ${\text{2ClO}}_2\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{ClO}}_3{}^{-}\left(aq\right)+{\text{ClO}}_2{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$ should be calculated.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduce amount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 23E

The rate law and the value of the rate constant for the given reaction are $k{\left[{\text{ClO}}_2\right]}^2\left[{\text{OH}}^{-}\right]$ and $2.30\times {10}^2{\text{mol}}^{-2}{\text{L}}^2{\text{s}}^{-1}$ respectively.

Explanation of Solution

The given reaction is ${\text{2ClO}}_2\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{ClO}}_3{}^{-}\left(aq\right)+{\text{ClO}}_2{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The initial concentration and rate constant is given as follows:

  

The rate law for the given reaction is represented as follows:

  $\text{Rate}=k{\left[{\text{ClO}}_2\right]}^a{\left[{\text{OH}}^{-}\right]}^b$ ...(1)

Where,

• $k$ is the rate constant.
• a and b is the order of the reaction with respect to ${\text{ClO}}_2$ and ${\text{OH}}^{-}$.

Substitute the values of 1^st row and 2^nd in equation (1) and divide 1^st row by 2^nd row values as shown below:

  $\begin{array}{c}\frac{{\left(\text{Rate}\right)}_1}{{\left(\text{Rate}\right)}_2}=\frac{{\left(k{\left[{\text{ClO}}_2\right]}^a{\left[{\text{OH}}^{-}\right]}^b\right)}_1}{{\left(k{\left[{\text{ClO}}_2\right]}^a{\left[{\text{OH}}^{-}\right]}^b\right)}_2}\\ \frac{5.75\times {10}^{-2}}{2.30\times {10}^{-1}}=\frac{\cancel{k}{\left(0.0500\right)}^a\cancel{{\left(0.100\right)}^b}}{\cancel{k}{\left(0.100\right)}^a\cancel{{\left(0.100\right)}^b}}\\ 0.25={\left(0.5\right)}^a\\ {\left(0.5\right)}^2={\left(0.5\right)}^a\\ a=2\end{array}$

Therefore, the order of reaction with respect to ${\text{OH}}^{-}$ is 2.

Similarly, substitute the values of 2^nd and 3^rd row in equation (1) and divide 2^nd by 3^rd values as shown below:

  $\begin{array}{c}\frac{{\left(\text{Rate}\right)}_2}{{\left(\text{Rate}\right)}_1}=\frac{{\left(k{\left[{\text{ClO}}_2\right]}^a{\left[{\text{OH}}^{-}\right]}^b\right)}_2}{{\left(k{\left[{\text{ClO}}_2\right]}^a{\left[{\text{OH}}^{-}\right]}^b\right)}_1}\\ \frac{2.30\times {10}^{-1}}{1.15\times {10}^{-1}}=\frac{\cancel{k}\cancel{{\left(0.100\right)}^a}{\left(0.100\right)}^b}{\cancel{k}\cancel{{\left(0.100\right)}^a}{\left(0.0500\right)}^b}\\ 2.0={\left(2.0\right)}^b\\ b=1\end{array}$

Therefore, the order of reaction with respect to ${\text{ClO}}_2$ is 1.

Substitute a = 2 and b = 1 in equation (1).

  $\text{Rate}=k{\left[{\text{ClO}}_2\right]}^2\left[{\text{OH}}^{-}\right]$

Therefore, the rate law for the reaction is $k{\left[{\text{ClO}}_2\right]}^2\left[{\text{OH}}^{-}\right]$.

Substitute 1^st row values of given table in equation (1).

  $\begin{array}{c}5.75\times {10}^{-2}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}=k{\left(0.0500\text{mol}{\text{L}}^{-1}\right)}^2\left(0.100\text{mol}{\text{L}}^{-1}\right)\\ k=\frac{5.75\times {10}^{-2}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}}{{\left(0.0500\text{mol}{\text{L}}^{-1}\right)}^2\left(0.100\text{mol}{\text{L}}^{-1}\right)}\\ =2.30\times {10}^2{\text{mol}}^{-2}{\text{L}}^2{\text{s}}^{-1}\end{array}$

Therefore, the value of rate constant for the given reaction is $2.30\times {10}^2{\text{mol}}^{-2}{\text{L}}^2{\text{s}}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

The initial rate for an experiment where ${\left[{\text{ClO}}_2\right]}_0=0.175\text{mol/L}$ and ${\left[{\text{OH}}^{-}\right]}_0=0.0844\text{mol/L}$ is should be calculated.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduce amount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 23E

The initial rate for an experiment where ${\left[{\text{ClO}}_2\right]}_0=0.175\text{mol/L}$ and ${\left[{\text{OH}}^{-}\right]}_0=0.0844\text{mol/L}$ is $0.594\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$.

Explanation of Solution

The given reaction is ${\text{2ClO}}_2\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{ClO}}_3{}^{-}\left(aq\right)+{\text{ClO}}_2{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$. The rate law for the reaction is shown below:

  $\text{Rate}=k{\left[{\text{ClO}}_2\right]}^2\left[{\text{OH}}^{-}\right]$

Substitute the value of ${\left[{\text{ClO}}_2\right]}_0=0.175\text{mol/L}$ , ${\left[{\text{OH}}^{-}\right]}_0=0.0844\text{mol/L}$ and $k=2.30\times {10}^2{\text{mol}}^{-2}{\text{L}}^2{\text{s}}^{-1}$ in above expression.

  $\begin{array}{c}\text{rate}=2.30\times {10}^2{\text{mol}}^{-2}{\text{L}}^2{\text{s}}^{-1}{\left(0.175\text{mol/L}\right)}^2\left(0.0844\text{mol/L}\right)\\ =0.594\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

Therefore, the initial rate for an experiment where ${\left[{\text{ClO}}_2\right]}_0=0.175\text{mol/L}$ and ${\left[{\text{OH}}^{-}\right]}_0=0.0844\text{mol/L}$ is $0.594\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$.