Chapter 14, Problem 14.14QP

(a)

Interpretation Introduction

Interpretation: The rate law for the given reaction has to be deduced.

Concept Introduction:

Rate law: An expression relating the rate of a reaction to the rate constant and concentrations of the reactants that are involved in the reaction.

For a given type of reaction:

$\text{aA}\text{+}\text{bB}\stackrel{}{\to }\text{cC+}\text{dD}$

The rate law takes the form as:

$\text{Rate}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}$

$\text{k}$= Rate constant. It is the constant of proportionality between the rate of a reaction and the concentration of the reactants. Its unit is $s^{-1}$.

$\left[\text{A}\right]$-Concentration of the reactant A in molarity.

$\left[\text{B}\right]$-Concentration of the reactant B in molarity.

It is important to note that x and y are not related to a and b.

Rather, they are determined experimentally.

Explanation of Solution

The reaction to be considered is:

${\text{F}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2ClO}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{2FClO}}_{\text{2}}\left(\text{g}\right)$

Figure 1

In this table:

The data entries in 1^st and 3^rd columns show the following observations:

• While holding $\left[{\text{ClO}}_{\text{2}}\right]$ as constant, if $\left[{\text{F}}_{\text{2}}\right]$ is doubled then the rate of the reaction is also doubled. Therefore, the rate of the reaction is directly proportional to the concentration of ${\text{F}}_{\text{2}}$.
• While holding $\left[{\text{F}}_{\text{2}}\right]$ as constant, if $\left[{\text{ClO}}_{\text{2}}\right]$ is quadruple then the rate of the reaction is increased by four times. Therefore, the rate of the reaction is also directly proportional to the concentration of ${\text{ClO}}_{\text{2}}$.

The experimental observations tell that the rate of the reaction is dependent on the concentrations of the reactants such as ${\text{F}}_{\text{2}}$ and ${\text{ClO}}_{\text{2}}$.

So, the rate law for the reaction takes the form as:

$\begin{array}{l}\text{Rate}\text{α}\left[{\text{F}}_{\text{2}}\right]\left[{\text{ClO}}_{\text{2}}\right]\\ \text{Rate}\text{=}\text{k}\left[{\text{F}}_{\text{2}}\right]\left[{\text{ClO}}_{\text{2}}\right]\end{array}$

(b)

Interpretation Introduction

Interpretation:  The rate constant of the considered reaction has to be calculated.

Concept Introduction:

Rate constant:

$\text{k}$= Rate constant. It is the constant of proportionality between the rate of a reaction and the concentration of the reactants. Its unit is $s^{-1}$.

For a given type of reaction:

$\text{aA}\text{+}\text{bB}\stackrel{}{\to }\text{cC+}\text{dD}$

The rate law takes the form as:

$\text{Rate}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}$

The rate constant takes the form as:

$\begin{array}{c}\text{Rate}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}\\ \text{k}=\frac{\text{Rate}}{{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}}\end{array}$

It is important to note that x and y are not related to a and b.

Rather, they are determined experimentally.

Answer to Problem 14.14QP

The rate constant of the reaction is $\text{k}=1.2{\text{M}}^{-1}{\text{s}}^{\text{-1}}$.

Explanation of Solution

The reaction to be considered is:

${\text{F}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2ClO}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{2FClO}}_{\text{2}}\left(\text{g}\right)$

Figure 1

Among the three experiments in the table, data of any one experiment can be considered to calculate the rate constant for the reaction.

The rate law takes the form as:

$\text{Rate}\text{=}\text{k}\left[{\text{F}}_{\text{2}}\right]\left[{\text{ClO}}_{\text{2}}\right]$

The rate constant takes the form as:

$\begin{array}{c}\text{Rate}\text{=}\text{k}\left[{\text{F}}_{\text{2}}\right]\left[{\text{ClO}}_{\text{2}}\right]\\ \text{k}=\frac{\text{Rate}}{\left[{\text{F}}_{\text{2}}\right]\left[{\text{ClO}}_{\text{2}}\right]}\end{array}$

Considering the 3^rd experiment:

$\begin{array}{l}\left[{\text{F}}_{\text{2}}\right]\text{=}\text{0}\text{.20}\text{M}\\ \left[{\text{ClO}}_{\text{2}}\right]\text{=}\text{0}\text{.010}\text{M}\\ \text{Rate}\text{=}\text{2}\text{.4}\times {\text{10}}^{-3}\text{M/s}\end{array}$

$\begin{array}{c}\text{k}=\frac{\text{Rate}}{\left[{\text{F}}_{\text{2}}\right]\left[{\text{ClO}}_{\text{2}}\right]}\\ =\frac{\text{2}\text{.4}\times {\text{10}}^{-3}\text{M/s}}{\left(\text{0}\text{.20}\text{M}\right)\times \left(\text{0}\text{.010}\text{M}\right)}\\ =1.2{\text{M}}^{-1}{\text{s}}^{\text{-1}}\end{array}$

Therefore, the rate constant of the reaction is $\text{k}=1.2{\text{M}}^{-1}{\text{s}}^{\text{-1}}$

Conclusion

The rate constant of the considered reaction has been calculated.

(c)

Interpretation Introduction

Interpretation: From the given data, the rate of the reaction has to be calculated.

Concept Introduction:

The rate of a reaction is the measure of how fast a reactant is consumed or a product is formed. It can be calculated using the rate law expression.

For a given type of reaction:

$\text{aA}\text{+}\text{bB}\stackrel{}{\to }\text{cC+}\text{dD}$

The rate law takes the form as:

$\text{Rate}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}$

$\text{k}$= Rate constant. It is the constant of proportionality between the rate of a reaction and the concentration of the reactants. Its unit is $s^{-1}$.

$\left[\text{A}\right]$-Concentration of the reactant A in molarity.

$\left[\text{B}\right]$-Concentration of the reactant B in molarity.

It is important to note that x and y are not related to a and b.

Rather, they are determined experimentally.

Answer to Problem 14.14QP

The rate of the reaction is $2.4\times {10}^{-4}{\text{Ms}}^{\text{-1}}$.

Explanation of Solution

The reaction to be considered is:

${\text{F}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2ClO}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{2FClO}}_{\text{2}}\left(\text{g}\right)$

Given data:

$\begin{array}{l}\left[{\text{F}}_{\text{2}}\right]\text{=}\text{0}\text{.010}\text{M}\\ \left[{\text{ClO}}_{\text{2}}\right]\text{=}\text{0}\text{.020}\text{M}\end{array}$

It is known that $\text{k}=1.2{\text{M}}^{-1}{\text{s}}^{\text{-1}}$

$\begin{array}{c}\text{Rate}\text{=}\text{k}\left[{\text{F}}_{\text{2}}\right]\left[{\text{ClO}}_{\text{2}}\right]\\ =\left(1.2{\text{M}}^{-1}{\text{s}}^{\text{-1}}\right)\times \left(\text{0}\text{.010}\text{M}\right)\times \left(\text{0}\text{.020}\text{M}\right)\\ =2.4\times {10}^{-4}{\text{Ms}}^{\text{-1}}\end{array}$

Therefore, the rate of the reaction is $2.4\times {10}^{-4}{\text{Ms}}^{\text{-1}}$.

Conclusion

From the given data, the rate of the reaction has been calculated.