Chapter 15, Problem 17E

(a)

Interpretation Introduction

Interpretation:

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ between 0 and $2.16\times {10}^4\text{s}$ should be calculated. The average rate of production of ${\text{O}}_2$ over the same time period should be predicted by using the rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ .

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 the reduce amount of reactant in per unit time or increase the amount of product in per unit time.

Answer to Problem 17E

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ and production of ${\text{O}}_2$ between 0 and $2.16\times {10}^4\text{s}$ is $2.31\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ and $1.16\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ respectively.

Explanation of Solution

The reaction is given as follows:

  $2{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)$

The given data is shown below:

  

The rate of the reaction is given as follows:

  $-\frac{1}{2}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_2\right]}{\Delta t}=+\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}$

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ between 0 and $2.16\times {10}^4\text{s}$ is calculated as follows:

  $\begin{array}{c}\text{Average}\text{rate}\text{of}\text{decomposition}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}=\frac{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_f-{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_i}{t_f-t_i}\\ =\left(\frac{0.500-1.000}{2.16\times {10}^4-0}\right)\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\\ =-2.31\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

Therefore, the average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ is $2.31\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ .

Similarly, the average rate of production of ${\text{O}}_2$ over the same time period is calculated as follows:

  $\begin{array}{c}+\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}=-\frac{1}{2}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_2\right]}{\Delta t}\\ =\frac{1}{2}\left(2.31\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\right)\\ =1.16\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

Therefore, the average rate of production of ${\text{O}}_2$ over the same time period is $1.16\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ .

(b)

Interpretation Introduction

Interpretation:

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ and average rate of production of ${\text{O}}_2$ between $2.16\times {10}^4\text{s}$ and $4.32\times {10}^4\text{s}$ 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 the reduce amount of reactant in per unit time or increase the amount of product in per unit time.

Answer to Problem 17E

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ and average rate of production of ${\text{O}}_2$ between $2.16\times {10}^4\text{s}$ and $4.32\times {10}^4\text{s}$ is $1.16\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ and $5.80\times {10}^{-6}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ respectively.

Explanation of Solution

The reaction is given as follows:

  $2{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)$

The given data is shown below:

  

The rate of the reaction is given as follows:

  $-\frac{1}{2}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_2\right]}{\Delta t}=+\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}$

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ between $2.16\times {10}^4\text{s}$ and $4.32\times {10}^4\text{s}$ is calculated as follows:

  $\begin{array}{c}\text{Average}\text{rate}\text{of}\text{decomposition}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}=\frac{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_f-{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_i}{t_f-t_i}\\ =\left(\frac{0.250-0.500}{4.32\times {10}^4-2.16\times {10}^4}\right)\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\\ =-1.16\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

Therefore, the average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_2$ is $1.16\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ .

Similarly, the average rate of production of ${\text{O}}_2$ over the same time period is calculated as follows:

  $\begin{array}{c}+\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}=-\frac{1}{2}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_2\right]}{\Delta t}\\ =\frac{1}{2}\left(1.16\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\right)\\ =5.80\times {10}^{-6}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

Therefore, the average rate of production of ${\text{O}}_2$ over the same time period is $5.80\times {10}^{-6}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ .