Chapter 15, Problem 36E

(a)

Interpretation Introduction

Interpretation:

The order of the reaction should be determined with respect to oxygen atoms.

  ${\text{O(g)+NO}}_2\text{(g)}\to {\text{NO(g)+O}}_2(\text{g})$

Concept Introduction:

Rate Law can be expressed as an integrated rate law and a differential rate law.

Differential Rate Law: This describes the change in the concentrations of reactant as a function of time.

Integrated Rate Law: This describes the initial concentrations and the measured concentration of one or more reactants as a function of time.

Answer to Problem 36E

The reaction is first order reaction with respect to oxygen.

Explanation of Solution

Given information:

Data is given as:

Time (s)	$\left[O\right]$ atoms/cm3
0	$5.0\times {10}^9$
$\text{1}\text{.0}\times {\text{10}}^{-2}$	$1.9\times {10}^9$
$\text{2}\text{.0}\times {\text{10}}^{-2}$	$6.8\times {10}^8$
$\text{3}\text{.0}\times {\text{10}}^{-2}$	$2.5\times {10}^8$

The order of reaction can be determined by plotting the graph between $\ln \left[\text{O}\right]$ and time or $\frac{1}{\left[\text{O}\right]}$ and time or $\left[\text{O}\right]$ and time.

For first order reaction: $\ln \left[\text{O}\right]$ Vs time, straight line.

For second order reaction: $\frac{1}{\left[\text{O}\right]}$ Vs time, straight line.

For zero order reaction: $\left[\text{O}\right]$ Vs time, straight line.

The given reaction is:

  ${\text{O(g)+NO}}_2\text{(g)}\to {\text{NO(g)+O}}_2(\text{g})$

Now, if the reaction is first order with respect to oxygen, then the integrated law is expressed as:

  $\ln \left[\text{O}\right]=-k^{\prime }t+\ln {\left[\text{O}\right]}_{\text{o}}$

If the reaction is second order with respect to oxygen, then the integrated law is expressed as:

  $\frac{1}{\left[\text{O}\right]}=k^{\prime }t+\frac{1}{{\left[\text{O}\right]}_o}$

If the reaction is zero order with respect to oxygen, then the integrated law is expressed as:

  $\left[\text{O}\right]=-k^{\prime }t+{\left[\text{O}\right]}_{\text{o}}$

Time (s)	$\ln \left[\text{O}\right]$	$\frac{1}{\left[\text{O}\right]}$
0	22.3327	$2.00\times {10}^{-10}$
$\text{1}\text{.0}\times {\text{10}}^{-2}$	21.36512	$5.26\times {10}^{-10}$
$\text{2}\text{.0}\times {\text{10}}^{-2}$	20.3376	$1.47\times {10}^{-9}$
$\text{3}\text{.0}\times {\text{10}}^{-2}$	19.33697	$4.00\times {10}^{-9}$

The graph between $\left[\text{O}\right]$ and time is:

   The graph between $\frac{1}{\left[\text{O}\right]}$ and time is:

  

The graph between $\ln \left[\text{O}\right]$ and time is:

  

From the above graphs, it is clear that the reaction is first order reaction with respect to oxygen as the graph is straight line graph.

(b)

Interpretation Introduction

Interpretation:

The overall rate law and value of rate constant should be calculated.

  ${\text{O(g)+NO}}_2\text{(g)}\to {\text{NO(g)+O}}_2(\text{g})$

Concept Introduction:

Rate Law can be expressed as an integrated rate law and a differential rate law.

Differential Rate Law: This describes the change in the concentrations of reactant as a function of time.

Integrated Rate Law: This describes the initial concentrations and the measured concentration of one or more reactants as a function of time.

Answer to Problem 36E

Rate law is expressed as:

  $\text{Rate}=k\left[\text{O}\right]\left[{\text{NO}}_{\text{2}}\right]$

Since, concentration of nitrogen dioxide is more in comparison to oxygen, thus, rate law is written as:

  $\text{Rate}=k^{\prime }\left[\text{O}\right]$

Rate constant for first order reaction is $1.0\times {10}^{-11}{\text{cm}}^{\text{3}}\text{/molecules}\cdot \text{s}$ .

Explanation of Solution

Given information:

Data is given as:

Time (s)	$\left[O\right]$ atoms/cm3
0	$5.0\times {10}^9$
$\text{1}\text{.0}\times {\text{10}}^{-2}$	$1.9\times {10}^9$
$\text{2}\text{.0}\times {\text{10}}^{-2}$	$6.8\times {10}^8$
$\text{3}\text{.0}\times {\text{10}}^{-2}$	$2.5\times {10}^8$

The order of reaction can be determined by plotting the graph between $\ln \left[\text{O}\right]$ and time or $\frac{1}{\left[\text{O}\right]}$ and time or $\left[\text{O}\right]$ and time.

For first order reaction: $\ln \left[\text{O}\right]$ Vs time, straight line.

For second order reaction: $\frac{1}{\left[\text{O}\right]}$ Vs time, straight line.

For zero order reaction: $\left[\text{O}\right]$ Vs time, straight line.

The given reaction is:

  ${\text{O(g)+NO}}_2\text{(g)}\to {\text{NO(g)+O}}_2(\text{g})$

Rate Law for first is expressed as:

  $\text{Rate}=k\left[\text{O}\right]\left[{\text{NO}}_{\text{2}}\right]$

Since, it is given that nitrogen dioxide is present in large amount in comparison to oxygen.

Thus, rate law is expressed as:

  $\text{Rate}=k^{\prime }\left[\text{O}\right]$

Where, $k^{\prime }$ = pseudo rate constant and it is equal to $k\left[{\text{NO}}_{\text{2}}\right]$

Now, from the graph slope is given as:

Slope = $\frac{\Delta \ln \left[\text{O}\right]}{\Delta t}$

Put the values from graph,

Slope = $\frac{(21.36512-19.33697)}{(1.0\times {10}^{-2}\text{s}-3.0\times {10}^{-2}\text{s})}$

Slope = $-101.5{\text{ s}}^{-1}$

Thus, value of $k^{\prime }=101.5{\text{ s}}^{-1}$

Now,

  $k^{\prime }=k\left[{\text{NO}}_{\text{2}}\right]$

  $\frac{k^{\prime }}{\left[{\text{NO}}_{\text{2}}\right]}=k$

Put the values,

  $k=\frac{101.5{\text{ s}}^{-1}}{\text{1}\text{.0}\times {\text{10}}^{13}{\text{ molecules/cm}}^3}$

  $k=1.0\times {10}^{-11}{\text{cm}}^{\text{3}}\text{/molecules}\cdot \text{s}$

Thus, rate constant for first order reaction is $1.0\times {10}^{-11}{\text{cm}}^{\text{3}}\text{/molecules}\cdot \text{s}$ .