Chapter 13, Problem 13.138QA

Interpretation Introduction

To determine:

a) Calculate the rate constant for the given reaction.

b) Determine the rate of the reaction at ${23}^{\mathrm{o}}\mathrm{C}$ and ${\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}=10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$.

c) Draw a graph showing ${\mathrm{P}}_{\mathrm{P}\mathrm{A}\mathrm{N}}$ as a function of time from $0$ to $200\mathrm{ }\mathrm{h}$.

Answer to Problem 13.138QA

Solution:

d) The rate constant for the given second order reactionis $\mathrm{k}=9.52\times {10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}$.

a) The rate of the reaction at ${23}^{\mathrm{o}}\mathrm{C}$ and ${\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}=10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$ is $0.10\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}/\mathrm{h}$.

b) A graph showing ${\mathrm{P}}_{\mathrm{P}\mathrm{A}\mathrm{N}}$ as a function of time from $0$ to $200\mathrm{ }\mathrm{h}$ is given below.

Explanation of Solution

1) Concept:

The given process is second order in peroxyacetyl nitrate, so write rate law for the reaction. From given data and formula for half-life period for second order reaction, rate constant for the reaction can be calculated.

2) Given:

The given reaction is second order in peroxyacetyl nitrate.

${\mathrm{t}}_{1/2}=100\mathrm{ }\mathrm{h}$

${\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}=10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$

3) Formulae:

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{\mathrm{k}{\left[\mathrm{X}\right]}_0}$

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= half-life period

k = rate constant

${\left[\mathrm{X}\right]}_0$= initial pressure

$\frac{1}{{\left[\mathrm{X}\right]}_{\mathrm{t}}}=\mathrm{k}\mathrm{t}+\mathrm{ }\frac{1}{{\left[\mathrm{X}\right]}_0}$

${\left[\mathrm{X}\right]}_{\mathrm{t}}$= pressure after time t.

4) Calculations:

a) The half-life of the second order reaction, that is,$100\mathrm{ }\mathrm{h}$ and initial pressure of ${\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2$ is $10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$.

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{\mathrm{k}{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}$

$\mathrm{k}=\frac{1}{{\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}$

$\mathrm{k}=\frac{1}{\left(100\mathrm{ }\mathrm{h}\right)\left(10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\right)}$

Rate constant

$\mathrm{k}={9.52\times 10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}$

b) We can write the general rate law equation for the given second order reaction as

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}^2$

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\left({9.52\times 10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}\right){\left(10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\right)}^2$

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=0.10\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}/\mathrm{h}$.

c) To draw the graph, we have to first find the values of ${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{\mathrm{t}}$ at different time intervals from $0\mathrm{ }$ to $200\mathrm{ }\mathrm{h}$. For that, we can use the integrated rate law equation of second order reaction.

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{\mathrm{t}}}=\mathrm{k}\mathrm{t}+\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_0}$

So at $0\mathrm{ }\mathrm{h}{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{\mathrm{t}}=10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$

1. At $50\mathrm{ }\mathrm{h}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{50}}=\left({9.52\times 10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}\right)\left(50\mathrm{ }\mathrm{h}\right)+\frac{1}{10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{50}}=0.1428$

${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{50}=7.000\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$

2. At $100\mathrm{ }\mathrm{h}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{100}}=\left({9.52\times 10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}\right)\left(100\mathrm{ }\mathrm{h}\right)+\frac{1}{10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{100}}=0.1904$

${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{100}=5.2510\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$

3. At $150\mathrm{ }\mathrm{h}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{150}}=\left({9.52\times 10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}\right)\left(150\mathrm{ }\mathrm{h}\right)+\frac{1}{10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{150}}=0.2380$

${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{150}=4.2010\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$

4. At $200\mathrm{ }\mathrm{h}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{200}}=\left({9.52\times 10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}\right)\left(200\mathrm{ }\mathrm{h}\right)+\frac{1}{10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}$

$\frac{1}{{{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{200}}=0.3273$

${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{200}=3.500\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$

Now we have to plot the graph of ${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{\mathrm{t}}$ versus time

$\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{ }\left(\mathrm{h}\right)$	${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{\mathrm{t}}\left(\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\right)$
$0$	$10.5$
$50$	$7.000$
$100$	$5.2510$
$150$	$4.2010$
$200$	$3.500$

We have plotted time on X-axis and ${{[\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}]}_{\mathrm{t}}\left(\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\right)$ on Y-axis.

Conclusion:

a) The rate constant for the given second order reaction is,$\mathrm{k}=9.52\times {10}^{-4}{\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}^{-1}{\mathrm{h}}^{-1}$.

b) The rate of the reaction at ${23}^{\mathrm{o}}\mathrm{C}$ and ${\mathrm{P}}_{{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}}_3{\mathrm{N}\mathrm{O}}_2}=10.5\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$ is,$0.10\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}/\mathrm{h}$.

c) A graph showing ${\mathrm{P}}_{\mathrm{P}\mathrm{A}\mathrm{N}}$ as a function of time from $0$ to $200\mathrm{ }\mathrm{h}$ is plotted.