Chapter 13, Problem 13.38QP

Interpretation Introduction

Interpretation: The decomposition of ${\text{N}}_2{\text{O}}_5$ and data based on the given reaction is given. The average reaction rate between the consecutive measurement times is to be calculated.

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 average reaction rate between the consecutive measurement times for the given reaction.

Answer to Problem 13.38QP

Solution

The average reaction rate between the time period $0.0\text{s}$ and $1.45\text{s}$ is $\underset{\_}{-0.0986\times 10^{-12}molecules/c{m}^{3}s}$.

The average reaction rate between the time period $1.45\text{s}$ and $2.90\text{s}$ is $\underset{\_}{-0.0890\times 10^{12}molecules/c{m}^{3}s}$.

The average reaction rate between the time period $2.90\text{s}$ and $4.35\text{s}$ is $\underset{\_}{-0.0807\times 10^{12}molecules/c{m}^{3}s}$.

The average reaction rate between the time period $4.35\text{s}$ and $5.80\text{s}$ is $\underset{\_}{-0.0731\times 10^{12}molecules/c{m}^{3}s}$.

Explanation of Solution

Explanation

Given

The given data is,

Time $\left(\text{s}\right)$	$\left[{\text{N}}_2{\text{O}}_5\right]\left(\text{molecules}/{\text{cm}}^3\right)$
$0.0$	$1.500\times {10}^{12}$
$1.45$	$1.357\times {10}^{12}$
$2.90$	$1.228\times {10}^{12}$
$4.35$	$1.111\times {10}^{12}$
$5.80$	$1.005\times {10}^{12}$

Table 1

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

$\text{Rate of concentration change}=\frac{\left(\begin{array}{l}{\text{Final concentration of N}}_2{\text{O}}_5-\\ {\text{Initial concentration of N}}_2{\text{O}}_5\end{array}\right)}{\text{Final time}-\text{Initial time}}$ (1)

Substitute the value of initial and final concentration of ${\text{N}}_2{\text{O}}_5$ and initial and final time as $0.0\text{s}$ and $1.45\text{s}$ in the above equation,

$\begin{array}{c}\text{Rate of concentration change}=\frac{\left(\begin{array}{l}\text{1}\text{.357}\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)-\\ 1.\text{500}\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}\right)}{1.45\text{s}-0.\text{0}\text{s}}\\ =\underset{\_}{-0.0986\times 10^{12}molecules/c{m}^{3}s}\end{array}$

The average reaction rate between the time period $0.0\text{s}$ and $1.45\text{s}$ is $\underset{\_}{-0.0986\times 10^{12}molecules/c{m}^{3}s}$.

Substitute the value of initial and final concentration of ${\text{N}}_2{\text{O}}_5$ and initial and final time as $1.45\text{s}$ and $2.90\text{s}$ in the equation (1),

$\begin{array}{c}\text{Rate of concentration change}=\frac{\left(\begin{array}{l}\text{1}\text{.228}\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)-\\ 1.\text{357}\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}\right)}{2.90\text{s}-1.45\text{s}}\\ =\underset{\_}{-0.0890\times 10^{12}molecules/c{m}^{3}s}\end{array}$

The average reaction rate between the time period $1.45\text{s}$ and $2.90\text{s}$ is $\underset{\_}{-0.0890\times 10^{12}molecules/c{m}^{3}s}$.

Substitute the value of initial and final concentration of ${\text{N}}_2{\text{O}}_5$ and initial and final time as $2.90\text{s}$ and $4.35\text{s}$ in the equation (1),

$\begin{array}{c}\text{Rate of concentration change}=\frac{\left(\begin{array}{l}1.111\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)-\\ \text{1}\text{.228}\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}\right)}{4.35\text{s}-2.90\text{s}}\\ =\underset{\_}{-0.0807\times 10^{12}molecules/c{m}^{3}s}\end{array}$

The average reaction rate between the time period $2.90\text{s}$ and $4.35\text{s}$ is $\underset{\_}{-0.0807\times 10^{12}molecules/c{m}^{3}s}$.

Substitute the value of initial and final concentration of ${\text{N}}_2{\text{O}}_5$ and initial and final time as $4.35\text{s}$ and $5.80\text{s}$ in the equation (1),

$\begin{array}{c}\text{Rate of concentration change}=\frac{\left(\begin{array}{l}1.005\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)-\\ \text{1}\text{.111}\times {\text{10}}^{12}\left(\text{molecules}/{\text{cm}}^3\right)\end{array}\right)}{5.80\text{s}-4.35\text{s}}\\ =\underset{\_}{-0.0731\times 10^{12}molecules/c{m}^{3}s}\end{array}$

The average reaction rate between the time period $4.35\text{s}$ and $5.80\text{s}$ is $\underset{\_}{-0.0731\times 10^{12}molecules/c{m}^{3}s}$.

Conclusion

The average rate of reaction is defined for a long interval of time and the negative sign shows that the reactant is getting consumed in the reaction.