Chapter 11, Problem 25E

At 40°C, H₂O₂ (aq) will decompose according to the following reaction:

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

The following data were collected for the concentration of H₂O₂ at various times.

Times(s)	[H2O2](mol/L)
0	1.000
2.16 × 104	0.500
4.32 × 104	0.250

a. Calculate the average rate of decomposition of H₂O₂ between 0 and 2.16 × 10⁴ s. Use this rate to calculate the average rate of production of O₂(g) over the same time period.

b. What are these rates for the time period 2.16 × 10⁴ s to 4.32 × 10⁴ s?

Interpretation Introduction

Interpretation: The decomposition reaction of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{-2.31\times {10}^{-5}mol/L\cdot s}$. The average rate of production of ${\text{O}}_{\text{2}}\left(g\right)$ between $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{1.16\times 10^{-5}mol/L\cdot s}$.

Explanation of Solution

Explanation

Given

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ at $\text{0}\text{s}$ is $\text{1}\text{.000}\text{mol/L}$.

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ at $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is $\text{0}\text{.500}\text{mol/L}$.

The reaction that takes place is,

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

The rate law for the above equation is written as,

$\text{Rate}=-\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{\Delta t}=\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{H}}_{\text{2}}\text{O}\right]}{\Delta t}=\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}$ (1)

Where,

• $\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$ is the average rate of consumption of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.
• $\Delta \left[{\text{H}}_{\text{2}}\text{O}\right]$ is the average rate of production of ${\text{H}}_{\text{2}}\text{O}$.
• $\Delta \left[{\text{O}}_{\text{2}}\right]$ is the average rate of production of ${\text{O}}_{\text{2}}$.
• $\Delta t$ is the average time.

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is calculated as,

$\text{Average}\text{Rate}=\frac{\text{Final}\text{concentration}-\text{Initial}\text{concentration}}{\text{Final}\text{time}-\text{Initial}\text{time}}$

Substitute the values of concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ at $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ in the above equation.

$\begin{array}{c}\text{Average}\text{Rate}=\frac{\text{Final}\text{concentration}-\text{Initial}\text{concentration}}{\text{Final}\text{time}-\text{Initial}\text{time}}\\ =\frac{\left(\text{0}\text{.500}-\text{1}\text{.00}\right)\text{mol/L}}{\left(\text{2}\text{.16}\times {\text{10}}^{\text{4}}-\text{0}\right)\text{s}}\\ =\underset{\_}{-2.31\times {10}^{-5}mol/L\cdot s}\end{array}$

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is $-\text{2}{\text{.31×10}}^{-\text{5}}\text{mol/L}\cdot \text{s}$.

Substitute the above value in equation (1).

$\begin{array}{c}-\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{\Delta t}=\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}\\ \frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}=-\frac{\text{1}}{\text{2}}\times \left(-\text{2}\text{.31}\times {\text{10}}^{-\text{5}}\text{mol/L}\cdot \text{s}\right)\\ =\underset{\_}{1.16\times 10^{-5}mol/L\cdot s}\end{array}$

Conclusion

Conclusion

The calculated average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{-2.31\times {10}^{-5}mol/L\cdot s}$. The average rate of production of ${\text{O}}_{\text{2}}\left(g\right)$ between $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{1.16\times 10^{-5}mol/L\cdot s}$.

Interpretation Introduction

Interpretation: The decomposition reaction of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ and $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{-1.15\times {10}^{-5}mol/L\cdot s}$. The average rate of production of ${\text{O}}_{\text{2}}\left(g\right)$ between $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ and $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{5.78\times 10^{-6}mol/L\cdot s}$.

Explanation of Solution

Given:

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ at $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is $\text{0}\text{.500}\text{mol/L}$.

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ at $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ is $\text{0}\text{.250}\text{mol/L}$.

The reaction that takes place is,

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

The rate law for the above equation is written as,

$\text{Rate}=-\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{\Delta t}=\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{H}}_{\text{2}}\text{O}\right]}{\Delta t}=\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}$ (2)

Where,

• $\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$ is the average rate of consumption of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.
• $\Delta \left[{\text{H}}_{\text{2}}\text{O}\right]$ is the average rate of production of ${\text{H}}_{\text{2}}\text{O}$.
• $\Delta \left[{\text{O}}_{\text{2}}\right]$ is the average rate of production of ${\text{O}}_{\text{2}}$.
• $\Delta t$ is the average time.

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{0}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ is calculated as,

$\text{Average}\text{Rate}=\frac{\text{Final}\text{concentration}-\text{Initial}\text{concentration}}{\text{Final}\text{time}-\text{Initial}\text{time}}$

Substitute the values of concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ at $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ and $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ in the above equation.

$\begin{array}{c}\text{Average}\text{Rate}=\frac{\text{Final}\text{concentration}-\text{Initial}\text{concentration}}{\text{Final}\text{time}-\text{Initial}\text{time}}\\ =\frac{\left(\text{0}\text{.25}-\text{0}\text{.500}\right)\text{mol/L}}{\left(\text{4}\text{.32}\times {\text{10}}^{\text{4}}-\text{2}\text{.16}\times {\text{10}}^{\text{4}}\right)\text{s}}\\ =\underset{\_}{-1.15\times {10}^{-5}mol/L\cdot s}\end{array}$

The average rate of production of ${\text{O}}_{\text{2}}\left(g\right)$ between $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ and $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{5.78\times 10^{-6}mol/L\cdot s}$.

The average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ and $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ is $-\text{1}\text{.15}\times {\text{10}}^{-\text{5}}\text{mol/L}\cdot \text{s}$.

Substitute the above value in equation (2).

$\begin{array}{c}-\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{\Delta t}=\frac{\text{1}}{\text{2}}\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}\\ \frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}=-\frac{\text{1}}{\text{2}}\times \left(-\text{1}\text{.15}\times {\text{10}}^{-\text{5}}\text{mol/L}\cdot \text{s}\right)\\ =\underset{\_}{5.78\times 10^{-6}mol/L\cdot s}\end{array}$

Conclusion

Conclusion

The calculated average rate of decomposition of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ between $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ and $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{-1.15\times {10}^{-5}mol/L\cdot s}$. The average rate of production of ${\text{O}}_{\text{2}}\left(g\right)$ between $\text{2}\text{.16}\times {\text{10}}^{\text{4}}\text{s}$ and $\text{4}\text{.32}\times {\text{10}}^{\text{4}}\text{s}$ is $\underset{\_}{5.78\times 10^{-6}mol/L\cdot s}$.