Chapter 13, Problem 13.139QA

Interpretation Introduction

To find:

a) Determine the rate law and the rate constant of the reaction at the experimental temperature for which the rate is given.

b) Draw the Lewis structure of peroxynitrite ion including all resonance forms and to assign formal charges on them.

c) Using the average bond energies, estimate the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ using the preferred structure from part (b).

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\left[\mathrm{N}\mathrm{O}\right]\left[\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}\right]$  and rate constant,

$\mathrm{k}=1.30\times {10}^{-3}{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

b) The Lewis structure of peroxynitrite ion including all resonance forms are given below.

c) The value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=-55\mathrm{ }\mathrm{k}\mathrm{J}$

Explanation of Solution

1) Concept:

We are asked to determine the rate law and rate constant for the initial reaction rate data for each four sets of initial reactant concentrations. We could use equations similar to those used to derive a rate law expression from $\left[\mathrm{N}\mathrm{O}\right]$ and $\left[\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}\right]$ data in the given table. From the rate law expression and values of initial reactant concentrations, we can calculate the rate constant. By checking the atoms in the molecule $\left[\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}\right],$ we will draw the Lewis structure and resonating structures. From the formal charge, we could find out most preferred structure. Using the average bond energies, we could estimate the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$.

2) Given:

We are given the table for the values of initial concentrations and initial rates of $\mathrm{N}\mathrm{O}$ and $\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}$ in the four experiments.

Experiment	${\left[\mathrm{N}\mathrm{O}\right]}_0\left(\mathrm{M}\right)$	${\left[\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}\right]}_0\left(\mathrm{M}\right)$	Initial Rate $(\mathrm{M}/\mathrm{s})$
$1$	${1.25\times 10}^{-4}$	${1.25\times 10}^{-4}$	${2.03\times 10}^{-11}$
$2$	${1.25\times 10}^{-4}$	${0.625\times 10}^{-4}$	${1.02\times 10}^{-11}$
$3$	${0.625\times 10}^{-4}$	${2.50\times 10}^{-4}$	${2.03\times 10}^{-11}$
$4$	${0.625\times 10}^{-4}$	${3.75\times 10}^{-4}$	${3.05\times 10}^{-11}$

3) Formula:

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{b}\mathrm{r}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{n}-\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{s}\mathrm{ }\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{e}\mathrm{d}$

4) Calculations:

a) The general rate law equation can be written as

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}{\left[\mathrm{N}\mathrm{O}\right]}^{\mathrm{x}}{\left[\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}\right]}^{\mathrm{y}}$

We can write rate equations for the all experiments as

${2.03\times 10}^{-11}=\mathrm{k}{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{y}}$------------ (1) Experiment 1

${1.02\times 10}^{-11}=\mathrm{k}{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{0.625\times 10}^{-4}\right]}^{\mathrm{y}}$------------ (2) Experiment 2

${2.03\times 10}^{-11}=\mathrm{k}{\left[{0.625\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{2.50\times 10}^{-4}\right]}^{\mathrm{y}}$------------- (3) Experiment 3

${3.05\times 10}^{-11}=\mathrm{k}{\left[{0.625\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{3.75\times 10}^{-4}\right]}^{\mathrm{y}}$------------- (4) Experiment 4

Now, by taking the ratio of equation (1) and (2), we can find value of y.

$\frac{{2.03\times 10}^{-11}}{{1.02\times 10}^{-11}}=\frac{\mathrm{k}{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{y}}}{\mathrm{k}{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{0.625\times 10}^{-4}\right]}^{\mathrm{y}}}$

$\frac{{2.03\times 10}^{-11}}{{1.02\times 10}^{-11}}=\frac{{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{y}}}{{\left[{0.625\times 10}^{-4}\right]}^{\mathrm{y}}}$

$2={\left[2\right]}^{\mathrm{y}}$

$\mathrm{y}=1$

Now to find value of x, we have to put $\mathrm{y}=1$ and take ratio of equation (2) and equation (3).

$\frac{{1.02\times 10}^{-11}}{{2.03\times 10}^{-11}}=\frac{\mathrm{k}{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{0.625\times 10}^{-4}\right]}^1}{\mathrm{k}{\left[{0.625\times 10}^{-4}\right]}^{\mathrm{x}}{\left[{2.50\times 10}^{-4}\right]}^1}$

$\frac{1}{2}=\frac{{\left[{1.25\times 10}^{-4}\right]}^{\mathrm{x}}}{{\left[{0.625\times 10}^{-4}\right]}^{\mathrm{x}}\times 4}$

$2={\left[2\right]}^{\mathrm{x}}$

$\mathrm{x}=1$

So, we can write the rate law equation as

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

Using this rate law and the values of concentrations and initial rates, we can find the rate constant.

So form experiment 1, rate law is

${2.03\times 10}^{-11}=\mathrm{k}{[1.25\times 10}^{-4}\left]\right[{1.25\times 10}^{-4}]$

$\mathrm{k}=\frac{{2.03\times 10}^{-11}}{{[1.25\times 10}^{-4}\left]\right[{1.25\times 10}^{-4}]}$

Rate constant,$\mathrm{k}={1.3\times 10}^{-3}{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}$

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
N	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			1
Valence electrons in NO3-			24

N is the central atom. The Lewis structure of $\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

The first structure for $\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}$ would be preferred because in that, the negative charge is stabilized in more electronegative $\mathrm{O}$ atom, and in other two structures,$\mathrm{O}$ has $+1$ charge, making the structure unstable.

c) To find the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$:

We got the values of average bond energies in $\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l},$ which was given in Appendix $\mathrm{A}\mathrm{ }4.1$.

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}$	$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}(\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$
$\mathrm{O}-\mathrm{O}$	$146$
$\mathrm{N}-\mathrm{O}$	$201$
$\mathrm{N}=\mathrm{O}$	$607$

In the reaction mechanism, one $\mathrm{O}-\mathrm{O}$ bond has been broken and one $\mathrm{N}-\mathrm{O}$ bond has been formed. So $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ would be,

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{b}\mathrm{r}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{n}-\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{s}\mathrm{ }\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{e}\mathrm{d}$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=(146\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})-(201\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=-55\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

The estimated value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ is $55\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.

Conclusion:

a) The rate law for the given reaction is $\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\left[\mathrm{N}\mathrm{O}\right]\left[\mathrm{O}\mathrm{N}\mathrm{O}{\mathrm{O}}^{-}\right]$ and rate constant,

$\mathrm{k}=1.30\times {10}^{-3}{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

b) The Lewis structures of peroxynitrite ion including all resonance forms are given above.

c) The value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=-55\mathrm{ }\mathrm{k}\mathrm{J}$

d)