Chapter 13, Problem 13.135QA

Interpretation Introduction

To find:

a) State whether the rate law is consistent with the observed order of the reaction for $\mathrm{N}\mathrm{O}$ and ${\mathrm{O}}_3$ if the given reaction were to occur in a single step.

b) Find the value activation energy of the reaction.

c) Find the rate of the reaction at ${25}^{\mathrm{o}}\mathrm{C}$ when $\left[\mathrm{N}\mathrm{O}\right]={3\times 10}^{-6\mathrm{ }}\mathrm{M}$ and $\left[{\mathrm{O}}_3\right]={5\times 10}^{-9\mathrm{ }}\mathrm{M}$

d) Find the values of the rate constant at ${10}^{\mathrm{o}}\mathrm{C}$ and ${35}^{\mathrm{o}}\mathrm{C}$.

Answer to Problem 13.135QA

Solution:

a) Yes, if the given reaction occur in a single step the rate law would be consistent with the observed order of the reaction for $\mathrm{N}\mathrm{O}$ and ${\mathrm{O}}_3$

b) The value activation energy of the reaction is $62.5\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

c) The rate of the reaction at ${25}^{\mathrm{o}}\mathrm{C}$ when $\left[\mathrm{N}\mathrm{O}\right]={3\times 10}^{-6\mathrm{ }}\mathrm{M}$ and $\left[{\mathrm{O}}_3\right]={5\times 10}^{-9\mathrm{ }}\mathrm{M}$ is ${1.2\times 10}^{-12\mathrm{ }}\mathrm{M}/\mathrm{s}$

d) The value of the rate constant at ${10}^{\mathrm{o}}\mathrm{C}$ is $21\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$ and at ${35}^{\mathrm{o}}\mathrm{C}$ it is

${1.8\times 10}^{-2}{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

Explanation of Solution

1) Concept:

From the given reaction, we will check, how the rate of reaction depends on concentration of ${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}$ and ${\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}$. We could use the given data to calculate activation energy of the reaction. We could use the rate law and data to determine the rate of reaction.

We have to use the formula for calculating activation energy.

$\mathrm{l}\mathrm{n}\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

2) Given:

We have given the reaction

${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}\to \mathrm{ }{\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}$

This reaction is first order with respect to $\mathrm{N}\mathrm{O}$ and ${\mathrm{O}}_3$.

Temperature ${(}^{\mathrm{o}}\mathrm{C})$	Temperature ${(}^{\mathrm{o}}\mathrm{K})$	Rate constant ${(\mathrm{M}}^{-1}{\mathrm{s}}^{-1})$
$25$	$298$	$80$
$75$	$348$	$3000$

At ${25}^{\mathrm{o}}\mathrm{C}$, $\left[\mathrm{N}\mathrm{O}\right]={3\times 10}^{-6\mathrm{ }}\mathrm{M}$ and $\left[{\mathrm{O}}_3\right]={5\times 10}^{-9\mathrm{ }}\mathrm{M}$

3) Formula:

$\mathrm{l}\mathrm{n}\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

Where,

${\mathrm{k}}_1$ and ${\mathrm{k}}_2$= rate constant

${\mathrm{E}}_{\mathrm{a}}$= Activation energy

$\mathrm{R}$= Gas constant = $8.314\mathrm{J}/{\mathrm{m}\mathrm{o}\mathrm{l}}^{\mathrm{o}}\mathrm{C}$

${\mathrm{T}}_1$ and ${\mathrm{T}}_2$= Temperature

4) Calculations:

a) The given reaction is,

${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}\to \mathrm{ }{\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}$

Here one molecule of  ${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}$ collides with one molecule of ${\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}$. The molecules must collide to exchange atoms. Therefore the rate of collision depends on the number of collisions per unit time. Assume a closed system in which there is one molecule of ${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}$ and more number of ${\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}$ molecules. Here one molecule of ${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}$ collides with different ${\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}$ molecules and we get the product. If the concentration of ${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}$ molecules is doubled (that means two molecules), then those ${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}$ molecules will collide with different ${\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}$ molecules. Therefore doubling the concentration of ${\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}$ doubles the rate of collision. The rate of collision also doubles, if the concentration of ${\mathrm{O}}_{3\mathrm{ }\left(\mathrm{g}\right)}$ is doubled. Thus the rate law for the reaction is,

$Rate=k \left[NO\right]\left[O_3\right]$

Therefore, if the given reaction occurs in a single step, the rate law would be consistent with the observed order of the reaction for $\mathrm{N}\mathrm{O}$ and ${\mathrm{O}}_3$.

b) To calculate activation energy, we have formula,

$\mathrm{l}\mathrm{n}\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

So, by putting the values,

$\mathrm{l}\mathrm{n}\frac{\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}{\left(3000\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.314\mathrm{J}/{\mathrm{m}\mathrm{o}\mathrm{l}}^{\mathrm{o}}\mathrm{C}}\left(\frac{1}{{348}^{\mathrm{o}}\mathrm{K}}-\frac{1}{{298}^{\mathrm{o}}\mathrm{K}}\right)$

$-3.6243={\mathrm{E}}_{\mathrm{a}}\times {(-5.799\times 10}^{-5})\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

${\mathrm{E}}_{\mathrm{a}}=62498\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

${\mathrm{E}}_{\mathrm{a}}=62.5\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

Therefore, the value of activation energy of the reaction is $62.5\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.

c) This reaction is first order with respect to $\mathrm{N}\mathrm{O}$ and ${\mathrm{O}}_3$. Therefore the rate law of the reaction is

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\left[\mathrm{N}\mathrm{O}\right]\left[{\mathrm{O}}_3\right]$

At ${25}^{\mathrm{o}}\mathrm{C}$, $\left[\mathrm{N}\mathrm{O}\right]={3\times 10}^{-6\mathrm{ }}\mathrm{M}$ and $\left[{\mathrm{O}}_3\right]={5\times 10}^{-9\mathrm{ }}\mathrm{M}$

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)\times \left({3\times 10}^{-6\mathrm{ }}\mathrm{M}\right)\times \left({5\times 10}^{-9\mathrm{ }}\mathrm{M}\right)$

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}={1.2\times 10}^{-12}\mathrm{M}/\mathrm{s}$

Therefore, at ${25}^{\mathrm{o}}\mathrm{C}$, when $\left[\mathrm{N}\mathrm{O}\right]={3\times 10}^{-6\mathrm{ }}\mathrm{M}$ and $\left[{\mathrm{O}}_3\right]={5\times 10}^{-9\mathrm{ }}\mathrm{M}$, the rate of reaction is ${1.2\times 10}^{-12}\mathrm{M}/\mathrm{s}$.

d) The values of the rate constant at ${10}^{\mathrm{o}}\mathrm{C}$ and ${35}^{\mathrm{o}}\mathrm{C}$.

e)

i. Rate constant at ${10}^{\mathrm{o}}\mathrm{C}$:

We are using the formula of activation energy. By putting value of rate constant at ${25}^{\mathrm{o}}\mathrm{C}\mathrm{ }\mathrm{a}\mathrm{s}{\mathrm{K}}_1$ we can find the value of ${\mathrm{K}}_2$ which is the rate constant at ${10}^{\mathrm{o}}\mathrm{C}={283}^{\mathrm{o}}\mathrm{K}$.

$\mathrm{l}\mathrm{n}\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

$\mathrm{l}\mathrm{n}\frac{\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}{{\mathrm{k}}_2}=\frac{(62498\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})}{\left(8.314\mathrm{J}/{\mathrm{m}\mathrm{o}\mathrm{l}}^{\mathrm{o}}\mathrm{C}\right)}\left(\frac{1}{{283}^{\mathrm{o}}\mathrm{K}}-\frac{1}{{298}^{\mathrm{o}}\mathrm{K}}\right)$

$\mathrm{l}\mathrm{n}\frac{\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}{{\mathrm{k}}_2}=\left(1.3370\right)$

$\frac{\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}{{\mathrm{k}}_2}=3.8079$

${\mathrm{k}}_2=21\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$

ii. Rate constant at ${35}^{\mathrm{o}}\mathrm{C}$:

We are using the formula of activation energy. By putting value of rate constant at ${25}^{\mathrm{o}}\mathrm{C}\mathrm{ }\mathrm{a}\mathrm{s}{\mathrm{K}}_1$ we can find the value of ${\mathrm{K}}_2$ which is the rate constant at ${35}^{\mathrm{o}}\mathrm{C}={308}^{\mathrm{o}}\mathrm{K}$.

$\mathrm{l}\mathrm{n}\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

$\mathrm{l}\mathrm{n}\frac{\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}{{\mathrm{k}}_2}=\frac{(62498\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})}{\left(8.314\mathrm{J}/{\mathrm{m}\mathrm{o}\mathrm{l}}^{\mathrm{o}}\mathrm{C}\right)}\left(\frac{1}{{308}^{\mathrm{o}}\mathrm{K}}-\frac{1}{{298}^{\mathrm{o}}\mathrm{K}}\right)$

$\mathrm{l}\mathrm{n}\frac{\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}{{\mathrm{k}}_2}=(-0.8190)$

$\frac{\left(80\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)}{{\mathrm{k}}_2}=0.4480$

${\mathrm{k}}_2={1.8\times 10}^2{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$

At ${10}^{\mathrm{o}}\mathrm{C}$ and ${35}^{\mathrm{o}}\mathrm{C}$, the value of rate constant is $21\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$ and ${1.8\times 10}^2{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$ respectively.

Conclusion:

a) Yes, if the given reaction occurs in a single step the rate law would be consistent with the observed order of the reaction for $\mathrm{N}\mathrm{O}$ and ${\mathrm{O}}_3$

b) The value activation energy of the reaction is $62.5\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

c) The rate of the reaction at ${25}^{\mathrm{o}}\mathrm{C}$ when $\left[\mathrm{N}\mathrm{O}\right]={3\times 10}^{-6\mathrm{ }}\mathrm{M}$ and $\left[{\mathrm{O}}_3\right]={5\times 10}^{-9\mathrm{ }}\mathrm{M}$ is ${1.2\times 10}^{-12\mathrm{ }}\mathrm{M}/\mathrm{s}$

d) The value of the rate constant at ${10}^{\mathrm{o}}\mathrm{C}$ is $21\mathrm{ }{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$ and at ${35}^{\mathrm{o}}\mathrm{C}$ it is ${1.8\times 10}^{-2}{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

e)