Chapter 14, Problem 14.41QP

Interpretation Introduction

Interpretation: The data for the dimerization of $\text{ClO}$ to ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is given. The instantaneous rate of change at $1\text{s}$ for $\text{ClO}$ and ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is to be calculated.

Concept introduction: The change in the concentration of reactant or product at a particular instant of time is known as instantaneous rate of reaction.

To determine: The instantaneous rate of change at $1\text{s}$ for $\text{ClO}$ and ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$.

Answer to Problem 14.41QP

Solution

The instantaneous rates of change at $1\text{s}$ for $\text{ClO}$ and ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\underset{\_}{-0.64\times 10^{10}molecules/c{m}^{3}s}$ and $\underset{\_}{0.33\times 10^{11}molecules/c{m}^{3}s}$.

Explanation of Solution

Explanation

Given

The given data is,

Time $\left(\text{s}\right)$	$\left[\text{ClO}\right]\left(\text{molecules}/{\text{cm}}^3\right)$
$0.0$	$2.60\times {10}^{11}$
$1.0$	$1.08\times {10}^{11}$
$2.0$	$6.83\times {10}^{10}$
$3.0$	$4.99\times {10}^{10}$
$4.0$	$3.93\times {10}^{10}$
$5.0$	$3.24\times {10}^{10}$
$6.0$	$2.76\times {10}^{10}$

Table 1

The graph for the above equation is obtained as,

Figure 1

For finding out the instantaneous rate of change a tangent is drawn at $1\text{s}$ as shown above. Then the two points are chosen along the tangent.

The change in the concentration of reactant between the given time period is given as,

$\text{Rate of concentration change}=\frac{\left(\begin{array}{l}\text{Final concentration of ClO}-\\ \text{Initial concentration of ClO}\end{array}\right)}{\text{Final time}-\text{Initial time}}$ (1)

Substitute the value of initial and final concentration of $\text{ClO}$ and initial and final time in the above equation,

$\begin{array}{c}\text{Rate of concentration change}=\frac{\left(\begin{array}{l}6\times {\text{10}}^{10}\left(\text{molecules}/{\text{cm}}^3\right)-\\ 14\times {\text{10}}^{10}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}\right)}{1.75\text{s}-0.\text{5}\text{s}}\\ =\underset{\_}{-0.64\times 10^{10}molecules/c{m}^{3}s}\end{array}$

The instantaneous reaction rate involving change in $\text{ClO}$ is $\underset{\_}{-0.64\times 10^{10}molecules/c{m}^{3}s}$.

The given reaction is,

$2\text{ClO}\to {\text{Cl}}_2{\text{O}}_2$

Initially there was no ${\text{Cl}}_2{\text{O}}_2$, therefore at $0\text{s}$ $\left[{\text{Cl}}_2{\text{O}}_2\right]=0$.

The value of $\frac{\Delta \left[\text{ClO}\right]}{\Delta t}$ is given as,

$\left[{\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}\right]=\frac{\left(\begin{array}{l}\text{Final concentration of ClO}-\\ \text{Initial concentration of ClO}\end{array}\right)}{\text{Final time}-\text{Initial time}}$

Substitute this value of $\frac{\Delta \left[\text{ClO}\right]}{\Delta t}$ in the above equation,

$\left[{\text{Cl}}_2{\text{O}}_2\right]=\frac{1}{2}\Delta \left[\text{ClO}\right]$ (2)

Substitute the value of final concentration and initial concentration of $\text{ClO}$ for the time $t=1\text{s}$,

$\begin{array}{c}\left[{\text{Cl}}_2{\text{O}}_2\right]=\frac{1}{2}\Delta \left[\text{ClO}\right]\\ =\frac{1}{2}\left(\text{1}\text{.08}\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)-2.60\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\right)\\ =0.76\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}$

Substitute the value of final concentration and initial concentration of $\text{ClO}$ for the time $t=2\text{s}$,

$\begin{array}{c}\left[{\text{Cl}}_2{\text{O}}_2\right]=\frac{1}{2}\Delta \left[\text{ClO}\right]\\ =\frac{1}{2}\left(6.83\times {\text{10}}^{10}\left(\text{molecules}/{\text{cm}}^3\right)-2.60\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\right)\\ =0.958\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}$

Substitute the value of final concentration and initial concentration of $\text{ClO}$ for the time $t=3\text{s}$,

$\begin{array}{c}\left[{\text{Cl}}_2{\text{O}}_2\right]=\frac{1}{2}\Delta \left[\text{ClO}\right]\\ =\frac{1}{2}\left(0.499\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)-2.60\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\right)\\ =1.05\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}$

Substitute the value of final concentration and initial concentration of $\text{ClO}$ for the time $t=4\text{s}$,

$\begin{array}{c}\left[{\text{Cl}}_2{\text{O}}_2\right]=\frac{1}{2}\Delta \left[\text{ClO}\right]\\ =\frac{1}{2}\left(0.393\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)-2.60\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\right)\\ =1.10\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}$

Substitute the value of final concentration and initial concentration of $\text{ClO}$ for the time $t=5\text{s}$,

$\begin{array}{c}\frac{\Delta \left[{\text{Cl}}_2{\text{O}}_2\right]}{\Delta t}=\frac{1}{2}\Delta \left[\text{ClO}\right]\\ =\frac{1}{2}\left(0.324\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)-2.60\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\right)\\ =1.14\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}$

Substitute the value of final concentration and initial concentration of $\text{ClO}$ for the time $t=6\text{s}$,

$\begin{array}{c}\left[{\text{Cl}}_2{\text{O}}_2\right]=\frac{1}{2}\Delta \left[\text{ClO}\right]\\ =\frac{1}{2}\left(0.276\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)-2.60\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\right)\\ =1.16\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}$

The obtained data is tabulated as,

Time $\left(\text{s}\right)$	$\left[\text{ClO}\right]\left(\text{molecules}/{\text{cm}}^3\right)$
$0.0$	$0$
$1.0$	$0.76\times {10}^{11}$
$2.0$	$0.958\times {\text{10}}^{11}$
$3.0$	$1.05\times {\text{10}}^{11}$
$4.0$	$1.10\times {\text{10}}^{11}$
$5.0$	$1.14\times {\text{10}}^{11}$
$6.0$	$1.16\times {10}^{10}$

Table 2

The graph from the above table is obtained as,

Figure 2

The change in the concentration of reactant between the given time period is given as,

$\text{Rate of concentration change}=\frac{\left(\begin{array}{l}{\text{Final concentration of Cl}}_{\text{2}}{\text{O}}_{\text{2}}-\\ {\text{Initial concentration of Cl}}_{\text{2}}{\text{O}}_{\text{2}}\end{array}\right)}{\text{Final time}-\text{Initial time}}$ (3)

Substitute the value of initial and final concentration of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ and initial and final time in the above equation,

$\begin{array}{c}\text{Rate of concentration change}=\frac{\left(\begin{array}{l}0.95\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)-\\ 0.6\times {\text{10}}^{11}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}\right)}{1.55\text{s}-0.\text{5}\text{s}}\\ =\underset{\_}{0.33\times 10^{11}molecules/c{m}^{3}s}\end{array}$

The instantaneous reaction rate involving change in ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\underset{\_}{0.33\times 10^{11}molecules/c{m}^{3}s}$.

Conclusion

The instantaneous rates of change at $1\text{s}$ for $\text{ClO}$ and ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\underset{\_}{-0.64\times 10^{10}molecules/c{m}^{3}s}$ and $\underset{\_}{0.33\times 10^{11}molecules/c{m}^{3}s}$.