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 ∆ H r x n 0 using the preferred structure from part (b).

Question

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Answer 1

Question

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 $∆Hrxn0$ using the preferred structure from part (b).

Expert Solution & Answer

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

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 $[NO]$ and $[ONOO-]$ 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 $ONOO-,$ 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 $∆Hrxn0$ .

2) Given:

We are given the table for the values of initial concentrations and initial rates of $NO$ and $ONOO-$ in the four experiments.

Experiment	$[NO]0(M)$	$[ONOO-]0(M)$	Initial Rate $(M/s)$
$1$	$1.25×10-4$	$1.25×10-4$	$2.03×10-11$
$2$	$1.25×10-4$	$0.625×10-4$	$1.02×10-11$
$3$	$0.625×10-4$	$2.50×10-4$	$2.03×10-11$
$4$	$0.625×10-4$	$3.75×10-4$	$3.05×10-11$

3) Formula:

$∆Hrxn0=Bond broken-bonds formed$

4) Calculations:

a) The general rate law equation can be written as

$rate=kNOx[ONOO-]y$

We can write rate equations for the all experiments as

$2.03×10-11=k[1.25×10-4]x1.25×10-4y$ ------------ (1) Experiment 1

$1.02×10-11=k[1.25×10-4]x0.625×10-4y$ ------------ (2) Experiment 2

$2.03×10-11=k[0.625×10-4]x2.50×10-4y$ ------------- (3) Experiment 3

$3.05×10-11=k[0.625×10-4]x3.75×10-4y$ ------------- (4) Experiment 4

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

$2.03×10-111.02×10-11=k[1.25×10-4]x1.25×10-4yk[1.25×10-4]x0.625×10-4y$

$2.03×10-111.02×10-11=1.25×10-4y0.625×10-4y$

$2=[2]y$

$y=1$

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

$1.02×10-112.03×10-11=k[1.25×10-4]x0.625×10-41k[0.625×10-4]x2.50×10-41$

$12=[1.25×10-4]x[0.625×10-4]x×4$

$2=[2]x$

$x=1$

So, we can write the rate law equation as

$rate=kNOONOO-$

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

$2.03×10-11=k[1.25×10-4][1.25×10-4]$

$k=2.03×10-11[1.25×10-4][1.25×10-4]$

Rate constant, $k=1.3×10-3M-1s-1$ .

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $ONOO-$

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 $ONOO-$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip 3

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

c) To find the value of $∆Hrxn0$ :

We got the values of average bond energies in $kJ/mol,$ which was given in Appendix $A 4.1$ .

$Bond$	$Bond energy(kJ/mol)$
$O-O$	$146$
$N-O$	$201$
$N=O$	$607$

In the reaction mechanism, one $O-O$ bond has been broken and one $N-O$ bond has been formed. So $∆Hrxn0$ would be,

$∆Hrxn0=Bond broken-bonds formed$

$∆Hrxn0=(146kJ/mol)-(201kJ/mol)$

$∆Hrxn0=-55kJ/mol$

The estimated value of $∆Hrxn0$ is $55kJ/mol$ .

Conclusion:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

d)

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Answer 2

Question

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 $∆Hrxn0$ using the preferred structure from part (b).

Expert Solution & Answer

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

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 $[NO]$ and $[ONOO-]$ 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 $ONOO-,$ 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 $∆Hrxn0$ .

2) Given:

We are given the table for the values of initial concentrations and initial rates of $NO$ and $ONOO-$ in the four experiments.

Experiment	$[NO]0(M)$	$[ONOO-]0(M)$	Initial Rate $(M/s)$
$1$	$1.25×10-4$	$1.25×10-4$	$2.03×10-11$
$2$	$1.25×10-4$	$0.625×10-4$	$1.02×10-11$
$3$	$0.625×10-4$	$2.50×10-4$	$2.03×10-11$
$4$	$0.625×10-4$	$3.75×10-4$	$3.05×10-11$

3) Formula:

$∆Hrxn0=Bond broken-bonds formed$

4) Calculations:

a) The general rate law equation can be written as

$rate=kNOx[ONOO-]y$

We can write rate equations for the all experiments as

$2.03×10-11=k[1.25×10-4]x1.25×10-4y$ ------------ (1) Experiment 1

$1.02×10-11=k[1.25×10-4]x0.625×10-4y$ ------------ (2) Experiment 2

$2.03×10-11=k[0.625×10-4]x2.50×10-4y$ ------------- (3) Experiment 3

$3.05×10-11=k[0.625×10-4]x3.75×10-4y$ ------------- (4) Experiment 4

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

$2.03×10-111.02×10-11=k[1.25×10-4]x1.25×10-4yk[1.25×10-4]x0.625×10-4y$

$2.03×10-111.02×10-11=1.25×10-4y0.625×10-4y$

$2=[2]y$

$y=1$

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

$1.02×10-112.03×10-11=k[1.25×10-4]x0.625×10-41k[0.625×10-4]x2.50×10-41$

$12=[1.25×10-4]x[0.625×10-4]x×4$

$2=[2]x$

$x=1$

So, we can write the rate law equation as

$rate=kNOONOO-$

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

$2.03×10-11=k[1.25×10-4][1.25×10-4]$

$k=2.03×10-11[1.25×10-4][1.25×10-4]$

Rate constant, $k=1.3×10-3M-1s-1$ .

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $ONOO-$

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 $ONOO-$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip 3

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

c) To find the value of $∆Hrxn0$ :

We got the values of average bond energies in $kJ/mol,$ which was given in Appendix $A 4.1$ .

$Bond$	$Bond energy(kJ/mol)$
$O-O$	$146$
$N-O$	$201$
$N=O$	$607$

In the reaction mechanism, one $O-O$ bond has been broken and one $N-O$ bond has been formed. So $∆Hrxn0$ would be,

$∆Hrxn0=Bond broken-bonds formed$

$∆Hrxn0=(146kJ/mol)-(201kJ/mol)$

$∆Hrxn0=-55kJ/mol$

The estimated value of $∆Hrxn0$ is $55kJ/mol$ .

Conclusion:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

d)

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Answer 3

Question

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 $∆Hrxn0$ using the preferred structure from part (b).

Expert Solution & Answer

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

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 $[NO]$ and $[ONOO-]$ 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 $ONOO-,$ 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 $∆Hrxn0$ .

2) Given:

We are given the table for the values of initial concentrations and initial rates of $NO$ and $ONOO-$ in the four experiments.

Experiment	$[NO]0(M)$	$[ONOO-]0(M)$	Initial Rate $(M/s)$
$1$	$1.25×10-4$	$1.25×10-4$	$2.03×10-11$
$2$	$1.25×10-4$	$0.625×10-4$	$1.02×10-11$
$3$	$0.625×10-4$	$2.50×10-4$	$2.03×10-11$
$4$	$0.625×10-4$	$3.75×10-4$	$3.05×10-11$

3) Formula:

$∆Hrxn0=Bond broken-bonds formed$

4) Calculations:

a) The general rate law equation can be written as

$rate=kNOx[ONOO-]y$

We can write rate equations for the all experiments as

$2.03×10-11=k[1.25×10-4]x1.25×10-4y$ ------------ (1) Experiment 1

$1.02×10-11=k[1.25×10-4]x0.625×10-4y$ ------------ (2) Experiment 2

$2.03×10-11=k[0.625×10-4]x2.50×10-4y$ ------------- (3) Experiment 3

$3.05×10-11=k[0.625×10-4]x3.75×10-4y$ ------------- (4) Experiment 4

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

$2.03×10-111.02×10-11=k[1.25×10-4]x1.25×10-4yk[1.25×10-4]x0.625×10-4y$

$2.03×10-111.02×10-11=1.25×10-4y0.625×10-4y$

$2=[2]y$

$y=1$

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

$1.02×10-112.03×10-11=k[1.25×10-4]x0.625×10-41k[0.625×10-4]x2.50×10-41$

$12=[1.25×10-4]x[0.625×10-4]x×4$

$2=[2]x$

$x=1$

So, we can write the rate law equation as

$rate=kNOONOO-$

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

$2.03×10-11=k[1.25×10-4][1.25×10-4]$

$k=2.03×10-11[1.25×10-4][1.25×10-4]$

Rate constant, $k=1.3×10-3M-1s-1$ .

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $ONOO-$

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 $ONOO-$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip 3

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

c) To find the value of $∆Hrxn0$ :

We got the values of average bond energies in $kJ/mol,$ which was given in Appendix $A 4.1$ .

$Bond$	$Bond energy(kJ/mol)$
$O-O$	$146$
$N-O$	$201$
$N=O$	$607$

In the reaction mechanism, one $O-O$ bond has been broken and one $N-O$ bond has been formed. So $∆Hrxn0$ would be,

$∆Hrxn0=Bond broken-bonds formed$

$∆Hrxn0=(146kJ/mol)-(201kJ/mol)$

$∆Hrxn0=-55kJ/mol$

The estimated value of $∆Hrxn0$ is $55kJ/mol$ .

Conclusion:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

d)

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Answer 4

Question

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 $∆Hrxn0$ using the preferred structure from part (b).

Expert Solution & Answer

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

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 $[NO]$ and $[ONOO-]$ 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 $ONOO-,$ 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 $∆Hrxn0$ .

2) Given:

We are given the table for the values of initial concentrations and initial rates of $NO$ and $ONOO-$ in the four experiments.

Experiment	$[NO]0(M)$	$[ONOO-]0(M)$	Initial Rate $(M/s)$
$1$	$1.25×10-4$	$1.25×10-4$	$2.03×10-11$
$2$	$1.25×10-4$	$0.625×10-4$	$1.02×10-11$
$3$	$0.625×10-4$	$2.50×10-4$	$2.03×10-11$
$4$	$0.625×10-4$	$3.75×10-4$	$3.05×10-11$

3) Formula:

$∆Hrxn0=Bond broken-bonds formed$

4) Calculations:

a) The general rate law equation can be written as

$rate=kNOx[ONOO-]y$

We can write rate equations for the all experiments as

$2.03×10-11=k[1.25×10-4]x1.25×10-4y$ ------------ (1) Experiment 1

$1.02×10-11=k[1.25×10-4]x0.625×10-4y$ ------------ (2) Experiment 2

$2.03×10-11=k[0.625×10-4]x2.50×10-4y$ ------------- (3) Experiment 3

$3.05×10-11=k[0.625×10-4]x3.75×10-4y$ ------------- (4) Experiment 4

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

$2.03×10-111.02×10-11=k[1.25×10-4]x1.25×10-4yk[1.25×10-4]x0.625×10-4y$

$2.03×10-111.02×10-11=1.25×10-4y0.625×10-4y$

$2=[2]y$

$y=1$

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

$1.02×10-112.03×10-11=k[1.25×10-4]x0.625×10-41k[0.625×10-4]x2.50×10-41$

$12=[1.25×10-4]x[0.625×10-4]x×4$

$2=[2]x$

$x=1$

So, we can write the rate law equation as

$rate=kNOONOO-$

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

$2.03×10-11=k[1.25×10-4][1.25×10-4]$

$k=2.03×10-11[1.25×10-4][1.25×10-4]$

Rate constant, $k=1.3×10-3M-1s-1$ .

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $ONOO-$

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 $ONOO-$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip 3

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

c) To find the value of $∆Hrxn0$ :

We got the values of average bond energies in $kJ/mol,$ which was given in Appendix $A 4.1$ .

$Bond$	$Bond energy(kJ/mol)$
$O-O$	$146$
$N-O$	$201$
$N=O$	$607$

In the reaction mechanism, one $O-O$ bond has been broken and one $N-O$ bond has been formed. So $∆Hrxn0$ would be,

$∆Hrxn0=Bond broken-bonds formed$

$∆Hrxn0=(146kJ/mol)-(201kJ/mol)$

$∆Hrxn0=-55kJ/mol$

The estimated value of $∆Hrxn0$ is $55kJ/mol$ .

Conclusion:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

d)

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Answer 5

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 $∆Hrxn0$ using the preferred structure from part (b).

Expert Solution & Answer

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

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 $[NO]$ and $[ONOO-]$ 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 $ONOO-,$ 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 $∆Hrxn0$ .

2) Given:

We are given the table for the values of initial concentrations and initial rates of $NO$ and $ONOO-$ in the four experiments.

Experiment	$[NO]0(M)$	$[ONOO-]0(M)$	Initial Rate $(M/s)$
$1$	$1.25×10-4$	$1.25×10-4$	$2.03×10-11$
$2$	$1.25×10-4$	$0.625×10-4$	$1.02×10-11$
$3$	$0.625×10-4$	$2.50×10-4$	$2.03×10-11$
$4$	$0.625×10-4$	$3.75×10-4$	$3.05×10-11$

3) Formula:

$∆Hrxn0=Bond broken-bonds formed$

4) Calculations:

a) The general rate law equation can be written as

$rate=kNOx[ONOO-]y$

We can write rate equations for the all experiments as

$2.03×10-11=k[1.25×10-4]x1.25×10-4y$ ------------ (1) Experiment 1

$1.02×10-11=k[1.25×10-4]x0.625×10-4y$ ------------ (2) Experiment 2

$2.03×10-11=k[0.625×10-4]x2.50×10-4y$ ------------- (3) Experiment 3

$3.05×10-11=k[0.625×10-4]x3.75×10-4y$ ------------- (4) Experiment 4

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

$2.03×10-111.02×10-11=k[1.25×10-4]x1.25×10-4yk[1.25×10-4]x0.625×10-4y$

$2.03×10-111.02×10-11=1.25×10-4y0.625×10-4y$

$2=[2]y$

$y=1$

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

$1.02×10-112.03×10-11=k[1.25×10-4]x0.625×10-41k[0.625×10-4]x2.50×10-41$

$12=[1.25×10-4]x[0.625×10-4]x×4$

$2=[2]x$

$x=1$

So, we can write the rate law equation as

$rate=kNOONOO-$

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

$2.03×10-11=k[1.25×10-4][1.25×10-4]$

$k=2.03×10-11[1.25×10-4][1.25×10-4]$

Rate constant, $k=1.3×10-3M-1s-1$ .

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $ONOO-$

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 $ONOO-$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip 3

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

c) To find the value of $∆Hrxn0$ :

We got the values of average bond energies in $kJ/mol,$ which was given in Appendix $A 4.1$ .

$Bond$	$Bond energy(kJ/mol)$
$O-O$	$146$
$N-O$	$201$
$N=O$	$607$

In the reaction mechanism, one $O-O$ bond has been broken and one $N-O$ bond has been formed. So $∆Hrxn0$ would be,

$∆Hrxn0=Bond broken-bonds formed$

$∆Hrxn0=(146kJ/mol)-(201kJ/mol)$

$∆Hrxn0=-55kJ/mol$

The estimated value of $∆Hrxn0$ is $55kJ/mol$ .

Conclusion:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

d)

Answer 6

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 $∆Hrxn0$ using the preferred structure from part (b).

Expert Solution & Answer

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

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 $[NO]$ and $[ONOO-]$ 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 $ONOO-,$ 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 $∆Hrxn0$ .

2) Given:

We are given the table for the values of initial concentrations and initial rates of $NO$ and $ONOO-$ in the four experiments.

Experiment	$[NO]0(M)$	$[ONOO-]0(M)$	Initial Rate $(M/s)$
$1$	$1.25×10-4$	$1.25×10-4$	$2.03×10-11$
$2$	$1.25×10-4$	$0.625×10-4$	$1.02×10-11$
$3$	$0.625×10-4$	$2.50×10-4$	$2.03×10-11$
$4$	$0.625×10-4$	$3.75×10-4$	$3.05×10-11$

3) Formula:

$∆Hrxn0=Bond broken-bonds formed$

4) Calculations:

a) The general rate law equation can be written as

$rate=kNOx[ONOO-]y$

We can write rate equations for the all experiments as

$2.03×10-11=k[1.25×10-4]x1.25×10-4y$ ------------ (1) Experiment 1

$1.02×10-11=k[1.25×10-4]x0.625×10-4y$ ------------ (2) Experiment 2

$2.03×10-11=k[0.625×10-4]x2.50×10-4y$ ------------- (3) Experiment 3

$3.05×10-11=k[0.625×10-4]x3.75×10-4y$ ------------- (4) Experiment 4

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

$2.03×10-111.02×10-11=k[1.25×10-4]x1.25×10-4yk[1.25×10-4]x0.625×10-4y$

$2.03×10-111.02×10-11=1.25×10-4y0.625×10-4y$

$2=[2]y$

$y=1$

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

$1.02×10-112.03×10-11=k[1.25×10-4]x0.625×10-41k[0.625×10-4]x2.50×10-41$

$12=[1.25×10-4]x[0.625×10-4]x×4$

$2=[2]x$

$x=1$

So, we can write the rate law equation as

$rate=kNOONOO-$

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

$2.03×10-11=k[1.25×10-4][1.25×10-4]$

$k=2.03×10-11[1.25×10-4][1.25×10-4]$

Rate constant, $k=1.3×10-3M-1s-1$ .

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $ONOO-$

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 $ONOO-$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip 3

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

c) To find the value of $∆Hrxn0$ :

We got the values of average bond energies in $kJ/mol,$ which was given in Appendix $A 4.1$ .

$Bond$	$Bond energy(kJ/mol)$
$O-O$	$146$
$N-O$	$201$
$N=O$	$607$

In the reaction mechanism, one $O-O$ bond has been broken and one $N-O$ bond has been formed. So $∆Hrxn0$ would be,

$∆Hrxn0=Bond broken-bonds formed$

$∆Hrxn0=(146kJ/mol)-(201kJ/mol)$

$∆Hrxn0=-55kJ/mol$

The estimated value of $∆Hrxn0$ is $55kJ/mol$ .

Conclusion:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

d)

Answer 7

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 $∆Hrxn0$ using the preferred structure from part (b).

Answer 8

Expert Solution & Answer

Answer to Problem 13.139QA

Solution:

a) The rate law for the given reaction is $rate=kNO[ONOO-]$ and rate constant,

$k=1.30×10-3M-1s-1$ .

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

c) The value of $∆Hrxn0=-55 kJ$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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 $[NO]$ and $[ONOO-]$ 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 $ONOO-,$ 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 $∆Hrxn0$ .

2) Given:

We are given the table for the values of initial concentrations and initial rates of $NO$ and $ONOO-$ in the four experiments.

Experiment	$[NO]0(M)$	$[ONOO-]0(M)$	Initial Rate $(M/s)$
$1$	$1.25×10-4$	$1.25×10-4$	$2.03×10-11$
$2$	$1.25×10-4$	$0.625×10-4$	$1.02×10-11$
$3$	$0.625×10-4$	$2.50×10-4$	$2.03×10-11$
$4$	$0.625×10-4$	$3.75×10-4$	$3.05×10-11$

3) Formula:

$∆Hrxn0=Bond broken-bonds formed$

4) Calculations:

a) The general rate law equation can be written as

$rate=kNOx[ONOO-]y$

We can write rate equations for the all experiments as

$2.03×10-11=k[1.25×10-4]x1.25×10-4y$ ------------ (1) Experiment 1

$1.02×10-11=k[1.25×10-4]x0.625×10-4y$ ------------ (2) Experiment 2

$2.03×10-11=k[0.625×10-4]x2.50×10-4y$ ------------- (3) Experiment 3

$3.05×10-11=k[0.625×10-4]x3.75×10-4y$ ------------- (4) Experiment 4

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

$2.03×10-111.02×10-11=k[1.25×10-4]x1.25×10-4yk[1.25×10-4]x0.625×10-4y$

$2.03×10-111.02×10-11=1.25×10-4y0.625×10-4y$

$2=[2]y$

$y=1$

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

$1.02×10-112.03×10-11=k[1.25×10-4]x0.625×10-41k[0.625×10-4]x2.50×10-41$

$12=[1.25×10-4]x[0.625×10-4]x×4$

$2=[2]x$

$x=1$

So, we can write the rate law equation as

$rate=kNOONOO-$

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

$2.03×10-11=k[1.25×10-4][1.25×10-4]$

$k=2.03×10-11[1.25×10-4][1.25×10-4]$

Rate constant, $k=1.3×10-3M-1s-1$ .

b) To draw Lewis diagrams, we have to first find the total number of valence electrons in $ONOO-$

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 $ONOO-$ is drawn by completing the octet of each element present in the molecule.

And the resonance structures are as follows:

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip 3

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