Part B Which species is a reaction intermediate? > View Available Hint(s) O X O E O F O M ON Submit • Part C What is the rate law for step 1 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][c}* type k*[A]*[C]*3. • View Available Hint(s) Rate = Submit • Part D What is the rate law for step 2 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A]|c* type k*[A]*[CJ*3. • View Available Hint(s) Rate = Submit
Part B Which species is a reaction intermediate? > View Available Hint(s) O X O E O F O M ON Submit • Part C What is the rate law for step 1 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][c}* type k*[A]*[C]*3. • View Available Hint(s) Rate = Submit • Part D What is the rate law for step 2 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A]|c* type k*[A]*[CJ*3. • View Available Hint(s) Rate = Submit
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
Related questions
Question
100%
![• Part B
Which species is a reaction intermediate?
• View Available Hint(s)
O E
O F
O M
ON
Submit
Part C
What is the rate law for step 1 of this reaction?
Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C type k*[A]*[C]*3.
> View Available Hint(s)
Rate =
Submit
Part D
What is the rate law for step 2 of this reaction?
Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][c]* type k*[A]*[C]*3.
• View Available Hint(s)
Rate =
Submit](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa1934326-e7b1-4bbf-8a9f-e6656915b5c9%2F70b76460-4e98-4567-a3dd-4a5c80aa273b%2Fha34y8x_processed.png&w=3840&q=75)
Transcribed Image Text:• Part B
Which species is a reaction intermediate?
• View Available Hint(s)
O E
O F
O M
ON
Submit
Part C
What is the rate law for step 1 of this reaction?
Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C type k*[A]*[C]*3.
> View Available Hint(s)
Rate =
Submit
Part D
What is the rate law for step 2 of this reaction?
Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][c]* type k*[A]*[C]*3.
• View Available Hint(s)
Rate =
Submit
![Leaming Goal:
To understand how elementary steps make up a mechanism and how the rate law for an elementary step can
be determined.
Order and rate law of a reaction
Very often, a reaction does not tell us the whole story. For instance, the reaction
The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an
elementary step. For example, the rate law for step 1 is
NO2(9) + CO(g)–→NO(g) + CO2 (9)
rate = k[NO2]*
does not involve a collision between an NO, molecule and a CO molecule. Based on experimental data at
moderate temperatures, this reaction is thought to occur in the following two steps:
The exponent "2" is used because the reaction involves two NOz molecules. The rate law for step 2 is
1. NO2(9) + NO2(g)→NO3 (9) + NO(g)
2. NO:(g) + CO(g)→CO2(g) + NO2(g)
rate = k[NO3]"(COj' = k[NO:]|CO]
because the reaction involves only one molecule of each reactant the exponents are omitted.
Each individual step is called an elementary step. Together, these elementary steps are called the reaction
mechanism.
Overall, the resulting reaction is
Analyzing a new reaction
NO,(9) + CO(g)-+NO(9) + CO, (9)
Consider the following elementary steps that make up the mechanism of a certain reaction:
1. 3XE + F
2. E+ M→F + N
Notice that in the elementary steps NO, appears both as a product and then as a reactant; therefore it cancels
out of the final chemical equation. NO; is called a reaction intermediate. Also notice that 2 molecules of NO2
appear in the reactants of the first step and 1 molecule of NOz appears as product of the second step, the net
effect leaves only 1 molecule of NO, as a reactant in the net equation.
Molecularity is the proper tem for "how the molecules collide" in a reaction. For example, step 1 is bimolecular
because it involves the collision of two molecules. Step 2 is also bimolecular for the same reason. Unimolecular
reactions involve only one molecule in the reactants. Though rare, collisions among three molecules can occur.
Such a reaction would be called femolecular.
Part A
What is the overall reaction?
Express your answer as a chemical equation.
• View Available Hint(s)
Template Symbols undo redo Teset keyboard shortcuts help
2x+ 21=m+2z
DA chemical reaction does not occur for this question.
Submit
Prevlous Anawere
Part B
Which species is a reaction intermediate?
• View Available Hint(s)
O X
O E
O F
O M
ON
Submit](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa1934326-e7b1-4bbf-8a9f-e6656915b5c9%2F70b76460-4e98-4567-a3dd-4a5c80aa273b%2Fbqr2qc1h_processed.png&w=3840&q=75)
Transcribed Image Text:Leaming Goal:
To understand how elementary steps make up a mechanism and how the rate law for an elementary step can
be determined.
Order and rate law of a reaction
Very often, a reaction does not tell us the whole story. For instance, the reaction
The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an
elementary step. For example, the rate law for step 1 is
NO2(9) + CO(g)–→NO(g) + CO2 (9)
rate = k[NO2]*
does not involve a collision between an NO, molecule and a CO molecule. Based on experimental data at
moderate temperatures, this reaction is thought to occur in the following two steps:
The exponent "2" is used because the reaction involves two NOz molecules. The rate law for step 2 is
1. NO2(9) + NO2(g)→NO3 (9) + NO(g)
2. NO:(g) + CO(g)→CO2(g) + NO2(g)
rate = k[NO3]"(COj' = k[NO:]|CO]
because the reaction involves only one molecule of each reactant the exponents are omitted.
Each individual step is called an elementary step. Together, these elementary steps are called the reaction
mechanism.
Overall, the resulting reaction is
Analyzing a new reaction
NO,(9) + CO(g)-+NO(9) + CO, (9)
Consider the following elementary steps that make up the mechanism of a certain reaction:
1. 3XE + F
2. E+ M→F + N
Notice that in the elementary steps NO, appears both as a product and then as a reactant; therefore it cancels
out of the final chemical equation. NO; is called a reaction intermediate. Also notice that 2 molecules of NO2
appear in the reactants of the first step and 1 molecule of NOz appears as product of the second step, the net
effect leaves only 1 molecule of NO, as a reactant in the net equation.
Molecularity is the proper tem for "how the molecules collide" in a reaction. For example, step 1 is bimolecular
because it involves the collision of two molecules. Step 2 is also bimolecular for the same reason. Unimolecular
reactions involve only one molecule in the reactants. Though rare, collisions among three molecules can occur.
Such a reaction would be called femolecular.
Part A
What is the overall reaction?
Express your answer as a chemical equation.
• View Available Hint(s)
Template Symbols undo redo Teset keyboard shortcuts help
2x+ 21=m+2z
DA chemical reaction does not occur for this question.
Submit
Prevlous Anawere
Part B
Which species is a reaction intermediate?
• View Available Hint(s)
O X
O E
O F
O M
ON
Submit
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY