Chapter 13, Problem 13.129AP

(a)

Interpretation Introduction

Interpretation: The role of Nitrogen dioxide radical in neurotransmission and human immune system and one of the reactions involving Peroxynitrite ion is given. Various questions based on the rate law and rate constant have to be answered.

Concept introduction: According to the rate law the rate of reaction depends on the concentration of reactants that are involved in the reaction. Therefore, for a reaction like $\text{A}\to \text{B}$, the rate law equation is given as,

$\text{Rate}=k{\left[\text{A}\right]}^m$

Where,

• $k$ is the rate constant.
• $m$ is the order of the reaction.

To determine: The rate law and the value of rate constant for the given reaction.

Answer to Problem 13.129AP

Solution

The rate law expression for the given reaction is,

$\text{Rate}=k{\left[\text{NO}\right]}^1{\left[{\text{ONOO}}^{-}\right]}^1$

The value of rate constant is $\underset{\_}{1.30\times 10^{-4}{M}^{-1}{s}^{-1}}$.

Explanation of Solution

Explanation

The rate law expression for the given reaction is,

$\text{Rate}=k{\left[\text{NO}\right]}^a{\left[{\text{ONOO}}^{-}\right]}^b$ (1)

The table for the given reaction is given as,

Experiment	${\left[\text{NO}\right]}_0\left(\text{M}\right)$	${\left[{\text{ONOO}}^{-}\right]}_0\left(\text{M}\right)$	Rate $\left(\text{M}/\text{s}\right)$
$1$	$1.25\times {10}^{-4}$	$1.25\times {10}^{-4}$	$2.03\times {10}^{-11}$
$2$	$1.25\times {10}^{-4}$	$0.625\times {10}^{-4}$	$1.02\times {10}^{-11}$
$3$	$0.625\times {10}^{-4}$	$2.50\times {10}^{-4}$	$2.03\times {10}^{-11}$
$4$	$0.625\times {10}^{-4}$	$3.75\times {10}^{-4}$	$3.05\times {10}^{-11}$

Table 1

The rate law equation for the four experiments is given as,

$2.03\times {10}^{-11}=k{\left[1.25\times {10}^{-4}\right]}^a{\left[1.25\times {10}^{-4}\right]}^b$ (2)

$1.02\times {10}^{-11}=k{\left[1.25\times {10}^{-4}\right]}^a{\left[0.625\times {10}^{-4}\right]}^b$ (3)

$2.03\times {10}^{-11}=k{\left[0.625\times {10}^{-4}\right]}^a{\left[2.50\times {10}^{-4}\right]}^b$ (4)

$3.05\times {10}^{-11}=k{\left[0.625\times {10}^{-4}\right]}^a{\left[3.75\times {10}^{-4}\right]}^b$ (5)

Divide equation (2) by equation (3),

$\begin{array}{c}\frac{2.03\times {10}^{-11}}{1.02\times {10}^{-11}}=\frac{k{\left[1.25\times {10}^{-4}\right]}^a{\left[1.25\times {10}^{-4}\right]}^b}{k{\left[1.25\times {10}^{-4}\right]}^a{\left[0.625\times {10}^{-4}\right]}^b}\\ 2.0={\left(2.0\right)}^a\end{array}$

Take log on both the sides of the above equation,

$\begin{array}{c}\log 2.0=b\log 2.0\\ 0.3010=b\times \left(0.3010\right)\\ b=1.0\end{array}$

Divide equation (2) by equation (4),

$\begin{array}{c}\frac{2.03\times {10}^{-11}}{2.03\times {10}^{-11}}=\frac{k{\left[1.25\times {10}^{-4}\right]}^a{\left[1.25\times {10}^{-4}\right]}^1}{k{\left[0.625\times {10}^{-4}\right]}^a{\left[2.50\times {10}^{-4}\right]}^1}\\ 1.0={\left(2.0\right)}^a\times 0.5\\ 2.0={\left(2.0\right)}^a\end{array}$

Take log on both the sides of the above equation,

$\begin{array}{c}\log 2.0=a\log 2.0\\ 0.3010=a\times \left(0.3010\right)\\ a=1.0\end{array}$

Substitute the values of $a$ and $b$ in equation (1),

$\text{Rate}=k{\left[\text{NO}\right]}^1{\left[{\text{ONOO}}^{-}\right]}^1$

Substitute the value of rate and concentration of all species from experiment (1) in above equation,

$\begin{array}{c}\text{Rate}=k{\left[\text{NO}\right]}^1{\left[{\text{ONOO}}^{-}\right]}^1\\ 2.03\times {10}^{-11}\left(\text{M}/\text{s}\right)=k{\left[1.25\times {10}^{-4}\text{M}\right]}^1{\left[1.25\times {10}^{-4}\text{M}\right]}^1\\ 2.03\times {10}^{-11}\left(\text{M}/\text{s}\right)=k\times 1.5625\times {10}^{-8}{\text{M}}^2\\ k=\underset{\_}{1.30\times 10^{-4}{M}^{-1}{s}^{-1}}\end{array}$

(b)

Interpretation Introduction

To determine: The Lewis structure of Peroxynitrite ion and formal charges and the most preferred form.

Answer to Problem 13.129AP

Solution

The final Lewis structures are obtained as,

Figure 1

Figure (IV) has the least formal charge and is therefore, the most stable structure.

Explanation of Solution

Explanation

The Lewis structure for Peroxynitrite ion is given as,

Figure 1

The formal charge is calculated by the formula,

$\text{Formal}\text{charge}=\left[\begin{array}{r}\left(\text{Number}\text{of}\text{valence}\text{electrons}\right)-\\ \left(\begin{array}{l}\text{Number}\text{of}\text{non}\text{bonding}\text{electrons}\\ \text{around}\text{the}\text{atom}\end{array}\right)-\\ \frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{bonding}\text{electrons}\\ \text{around}\text{the}\text{atom}\end{array}\right)\end{array}\right]$

In figure (I), the formal charge of doubly bonded Oxygen at the left hand-side is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(\text{4}\right)-\frac{1}{2}\left(4\right)\right]\\ =2-2\\ =\underset{\_}{0}\end{array}$

In figure (I), the formal charge of Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(5\right)-\left(2\right)-\frac{1}{2}\left(6\right)\right]\\ =3-3\\ =\underset{\_}{0}\end{array}$

In figure (I), the formal charge of first singly bonded Oxygen adjacent to Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(\text{4}\right)-\frac{1}{2}\left(4\right)\right]\\ =2-2\\ =\underset{\_}{0}\end{array}$

In figure (I), the formal charge of second bonded Oxygen adjacent to first Oxygen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(6\right)-\frac{1}{2}\left(2\right)\right]\\ =0-1\\ =\underset{\_}{\underset{\_}{-1}}\end{array}$

In figure (II), the formal charge of first singly bonded Oxygen at the left hand-side adjacent to the Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(6\right)-\frac{1}{2}\left(2\right)\right]\\ =0-1\\ =\underset{\_}{-1}\end{array}$

In figure (II), the formal charge of Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(5\right)-\left(2\right)-\frac{1}{2}\left(6\right)\right]\\ =3-3\\ =\underset{\_}{0}\end{array}$

In figure (II), the formal charge of first doubly bonded Oxygen adjacent to Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(2\right)-\frac{1}{2}\left(6\right)\right]\\ =4-3\\ =\underset{\_}{1}\end{array}$

In figure (II), the formal charge of second bonded Oxygen adjacent to doubly bonded Oxygen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(6\right)-\frac{1}{2}\left(2\right)\right]\\ =0-1\\ =\underset{\_}{-1}\end{array}$

In figure (III), the formal charge of first singly bonded Oxygen at the left hand-side adjacent to the Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(6\right)-\frac{1}{2}\left(2\right)\right]\\ =0-1\\ =\underset{\_}{-1}\end{array}$

In figure (III), the formal charge of Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(5\right)-\left(4\right)-\frac{1}{2}\left(4\right)\right]\\ =1-2\\ =\underset{\_}{-1}\end{array}$

In figure (III), the formal charge of first singly bonded Oxygen adjacent to Nitrogen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(2\right)-\frac{1}{2}\left(6\right)\right]\\ =4-3\\ =\underset{\_}{1}\end{array}$

In figure (III), the formal charge of second doubly bonded Oxygen adjacent to singly bonded Oxygen is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(4\right)-\frac{1}{2}\left(4\right)\right]\\ =2-2\\ =\underset{\_}{0}\end{array}$

The final Lewis structures are obtained as,

Figure 2

The structure that carries the least formal charge is the most stable structure and it is seen that figure (IV) has the least formal charge and is therefore, the most stable structure.

(c)

Interpretation Introduction

To determine: The value of $\Delta H{°}_{\text{rxn}}$ for the given reaction.

Answer to Problem 13.129AP

Solution

The value of $\Delta H{°}_{\text{rxn}}$ is $\underset{\_}{-55kJ/mol}$.

Explanation of Solution

Explanation

The preferred structure is Figure (IV). The bond energy for different bonds present in the figure as referred from Table A4.1 is tabulated as,

Type of bond	Bond energy $\left(\text{kJ}/\text{mol}\right)$
$\text{N}-\text{O}$	$201$
$\text{N=O}$	$607$
$\text{O=O}$	$146$

Table 1

The given reaction is,

$\text{NO}\left(aq\right)+{\text{ONOO}}^{-}\left(aq\right)\to {\text{NO}}_2\left(aq\right)+{\text{NO}}_2{}^{-}\left(aq\right)$

The heat of reaction is given as,

$\Delta H{°}_{\text{rxn}}=\sum m\Delta H{°}_{\text{Bond breaking}}+\sum n\Delta H{°}_{\text{Bond making}}$

Where,

• $m$ is the coefficient of reactants.
• $n$ is the coefficients of products.
• $\Delta H{°}_{\text{Bond breaking}}$ is the energy absorbed in breaking of bond.
• $\Delta H{°}_{\text{Bond making}}$ is the energy released in breaking of bond.

Substitute the value of all the bonds at their suitable place in above equation,

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\sum m\Delta H{°}_{\text{Bond breaking}}+\sum n\Delta H{°}_{\text{Bond making}}\\ =\left(\begin{array}{l}\left(607\text{kJ}/\text{mol}+607\text{kJ}/\text{mol}+\text{201}\text{kJ}/\text{mol}+146\text{kJ}/\text{mol}\right)\\ +\left(-607\text{kJ}/\text{mol}-201\text{kJ}/\text{mol}-607\text{kJ}/\text{mol}-201\text{kJ}/\text{mol}\right)\end{array}\right)\\ =\underset{\_}{-55kJ/mol}\end{array}$

Conclusion

1. a. The rate law expression for the given reaction is,

   $\text{Rate}=k{\left[\text{NO}\right]}^1{\left[{\text{ONOO}}^{-}\right]}^1$

The value of rate constant is $\underset{\_}{0.12\times 10^{10}{M}^{-1}{s}^{-1}}$.

1. b. The final Lewis structures are obtained as,

Figure 2

Figure (IV) has the least formal charge and is therefore, the most stable structure.

1. c. The value of $\Delta H{°}_{\text{rxn}}$ is $\underset{\_}{-55kJ/mol}$.