Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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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
Check Mark

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.

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip  1

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.

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.139QA , additional homework tip  2

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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Chapter 13 Solutions

Chemistry: An Atoms-Focused Approach

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