Chapter 13, Problem 13.35QP

(a)

Interpretation Introduction

Interpretation: The role of Chlorine monoxide in the creation of Ozone holes in the stratosphere over the Earth’s polar region is given. Various questions based on the rate of reactions are to be answered.

Concept introduction: The change in the concentration of reactant or product with time is known as rate of reaction and chemical kinetics deal with the rate of reaction.

To determine: The rate of change of ${\text{Cl}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$.

Answer to Problem 13.35QP

Solution

The rate of change of ${\text{Cl}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is $\underset{\_}{1.15M/s}$.

Explanation of Solution

Explanation

Given

The rate of consumption of $\text{ClO}$ is $-2.3\times {10}^7\text{}\text{M}/\text{s}$.

The chemical equation for the given reaction is,

$\text{2ClO}\left(g\right)\to {\text{Cl}}_2\left(g\right)+{\text{O}}_2\left(g\right)$

The rate expression is given as,

$\text{Rate}=-\frac{1}{2}\frac{\Delta \left[\text{ClO}\right]}{\Delta t}=\frac{\Delta \left[{\text{Cl}}_{\text{2}}\right]}{\Delta t}=\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}$

The rate of formation of ${\text{Cl}}_{\text{2}}$ is given as,

$-\frac{1}{2}\frac{\Delta \left[\text{ClO}\right]}{\Delta t}=\frac{\Delta \left[{\text{Cl}}_{\text{2}}\right]}{\Delta t}$

Rearrange the above equation to obtain the value of rate of formation of ${\text{Cl}}_{\text{2}}$,

$\frac{\Delta \left[{\text{Cl}}_{\text{2}}\right]}{\Delta t}=-\frac{1}{2}\frac{\Delta \left[\text{ClO}\right]}{\Delta t}$

Substitute the value of rate of change of $\text{ClO}$ in the above equation as,

$\begin{array}{c}\frac{\Delta \left[{\text{Cl}}_{\text{2}}\right]}{\Delta t}=-\frac{1}{2}\left(-2.3\times {10}^7\text{M}/\text{s}\right)\\ =\underset{\_}{1.15M/s}\end{array}$

Therefore, the rate of change of ${\text{Cl}}_{\text{2}}$ is $\underset{\_}{1.15M/s}$.

Rearrange the above equation to obtain the value of rate of formation of ${\text{O}}_{\text{2}}$,

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

Substitute the value of rate of change of $\text{ClO}$ in the above equation as,

$\begin{array}{c}\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}=-\frac{1}{2}\left(-2.3\times {10}^7\text{M}/\text{s}\right)\\ =\underset{\_}{1.15M/s}\end{array}$

Therefore, the rate of change of ${\text{O}}_{\text{2}}$ is $\underset{\_}{1.15M/s}$.

(b)

Interpretation Introduction

Interpretation: The role of Chlorine monoxide in the creation of Ozone holes in the stratosphere over the Earth’s polar region is given. Various questions based on the rate of reactions are to be answered.

Concept introduction: The change in the concentration of reactant or product with time is known as rate of reaction and chemical kinetics deal with the rate of reaction.

To determine: The rate of formation of ${\text{O}}_{\text{2}}$ and ${\text{ClO}}_2$.

Answer to Problem 13.35QP

Solution

The rate of formation of ${\text{O}}_{\text{2}}$ and ${\text{ClO}}_2$ is $\underset{\_}{2.9\times 10^{4}M/s}$.

Explanation of Solution

Explanation

Given

The rate of consumption of $\text{ClO}$ is $-2.9\times {10}^4\text{}\text{M}/\text{s}$.

The given reaction is,

$\text{ClO}\left(g\right)+{\text{O}}_3\left(g\right)\to {\text{Cl}}_2\left(g\right)+{\text{O}}_2\left(g\right)$

The rate expression is given as,

$\text{Rate}=-\frac{\Delta \left[\text{ClO}\right]}{\Delta t}=-\frac{\Delta \left[{\text{O}}_3\right]}{\Delta t}=\frac{\Delta \left[{\text{Cl}}_{\text{2}}\right]}{\Delta t}=\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}$

The rate of formation of ${\text{O}}_{\text{2}}$ is given as,

$-\frac{\Delta \left[\text{ClO}\right]}{\Delta t}=\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}$

Rearrange the above equation to obtain the value of rate of formation of ${\text{O}}_{\text{2}}$,

$\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}=-\frac{\Delta \left[\text{ClO}\right]}{\Delta t}$

Substitute the value of rate of change of ${\text{O}}_{\text{2}}$ in the above equation as,

$\begin{array}{c}\frac{\Delta \left[{\text{O}}_{\text{2}}\right]}{\Delta t}=-\frac{\Delta \left[\text{ClO}\right]}{\Delta t}\\ =-\left(-2.9\times {10}^4\text{M}/\text{s}\right)\\ =\underset{\_}{2.9\times 10^{4}M/s}\end{array}$

Therefore, the rate of formation of ${\text{O}}_{\text{2}}$ is $\underset{\_}{2.9\times 10^{4}M/s}$.

The rate of formation of ${\text{ClO}}_{\text{2}}$ is given as,

$-\frac{\Delta \left[\text{ClO}\right]}{\Delta t}=\frac{\Delta \left[{\text{ClO}}_{\text{2}}\right]}{\Delta t}$

Rearrange the above equation to obtain the value of rate of formation of ${\text{ClO}}_{\text{2}}$,

$\frac{\Delta \left[{\text{ClO}}_{\text{2}}\right]}{\Delta t}=-\frac{\Delta \left[\text{ClO}\right]}{\Delta t}$

Substitute the value of rate of change of $\text{ClO}$ in the above equation as,

$\begin{array}{c}\frac{\Delta \left[{\text{ClO}}_{\text{2}}\right]}{\Delta t}=-\frac{\Delta \left[\text{ClO}\right]}{\Delta t}\\ =-\left(-2.9\times {10}^4\text{M}/\text{s}\right)\\ =\underset{\_}{2.9\times 10^{4}M/s}\end{array}$

Therefore, the rate of formation of ${\text{ClO}}_{\text{2}}$ is $\underset{\_}{2.9\times 10^{4}M/s}$.

Conclusion

The negative sign shows that the reactants are getting consumed in the reaction while the positive sign shows that the products are formed in the reaction.