Analyzing a new reaction Consider the following elementary steps that make up the mechanism of a certain reaction: 1. 3X->E +F 2. E + MF+N Bart 0
Analyzing a new reaction Consider the following elementary steps that make up the mechanism of a certain reaction: 1. 3X->E +F 2. E + MF+N Bart 0
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section: Chapter Questions
Problem 115QRT
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Question
![Mechanisms and Molecularity
7 of 18
<>
I Review | Constants | Periodic Table
Order and rate law of a reaction
Learning Goal:
To understand how elementary steps make up a
mechanism and how the rate law for an elementary
step can be determined.
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
Very often, a reaction does not tell us the whole story.
For instance, the reaction
rate = k[NO2]²
The exponent "2" is used because the reaction involves two NO2 molecules. The rate law for step 2 is
NO2(9) + CO(9)→NO(g)+ CO2(g)
rate = k[NO3]"[coj = k[NO3][C0]
does not involve a collision between an NO2
molecule and a CO molecule. Based on experimental
data at moderate temperatures, this reaction is
thought to occur in the following two steps:
because the reaction involves only one molecule of each reactant the exponents are omitted.
1. NO2(9) + NO2(g)→NO3(g) + NO(g)
2. NO3(9) + C0(g)→CO2(g) + NO2(9)
Analyzing a new reaction
Consider the following elementary steps that make up the mechanism of a certain reaction:
Each individual step is called an elementary step.
Together, these elementary steps are called the
reaction mechanism.
1. 3X-E +F
2. E + M F+ N
Overall, the resulting reaction is
Part A
NO2(g) + CO(9)→NO(g)+CO2(9)
Notice that in the elementary steps NO3 appears](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87faf35d-bdff-4fde-b9fe-6459d170f061%2F686e5313-8fd3-435b-baca-f143996caee0%2F2bi0kkr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Mechanisms and Molecularity
7 of 18
<>
I Review | Constants | Periodic Table
Order and rate law of a reaction
Learning Goal:
To understand how elementary steps make up a
mechanism and how the rate law for an elementary
step can be determined.
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
Very often, a reaction does not tell us the whole story.
For instance, the reaction
rate = k[NO2]²
The exponent "2" is used because the reaction involves two NO2 molecules. The rate law for step 2 is
NO2(9) + CO(9)→NO(g)+ CO2(g)
rate = k[NO3]"[coj = k[NO3][C0]
does not involve a collision between an NO2
molecule and a CO molecule. Based on experimental
data at moderate temperatures, this reaction is
thought to occur in the following two steps:
because the reaction involves only one molecule of each reactant the exponents are omitted.
1. NO2(9) + NO2(g)→NO3(g) + NO(g)
2. NO3(9) + C0(g)→CO2(g) + NO2(9)
Analyzing a new reaction
Consider the following elementary steps that make up the mechanism of a certain reaction:
Each individual step is called an elementary step.
Together, these elementary steps are called the
reaction mechanism.
1. 3X-E +F
2. E + M F+ N
Overall, the resulting reaction is
Part A
NO2(g) + CO(9)→NO(g)+CO2(9)
Notice that in the elementary steps NO3 appears
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