The proper expression for the rate law using the steady-state approximation is 4, kika 0 What would the rate law be if the concentration of O, was very high? Match the items in the left column to the appropriate blanks in the equation on the right Reset Help If the concentration of Oz is very high then the denominator becomes insignificant when compared to the the rate law would simpilify to Rate = term in the 2k| [NO]? term and ki ĮNO]? 2k NO
The proper expression for the rate law using the steady-state approximation is 4, kika 0 What would the rate law be if the concentration of O, was very high? Match the items in the left column to the appropriate blanks in the equation on the right Reset Help If the concentration of Oz is very high then the denominator becomes insignificant when compared to the the rate law would simpilify to Rate = term in the 2k| [NO]? term and ki ĮNO]? 2k NO
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![k, ka 0|NO
The proper expression for the rate law using the steady-state approximation is:
What would the rate law be if the concentration of O, was very high?
Match the items in the left column to the appropriate blanks in the equation on the right
Reset Help
k1
If the concentration of Og is very high then the
term in the
2k1 [NO]?
denominator becomes insignificant when compared to the
term and
the rate law would simplify to Rate
ki [NO]?
k_1+ kg[O2]
ka [NO]?
kz (O2]
2k: [NO]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00c2a551-ae46-4bbd-9c62-962721a84afa%2Fa07d0738-ecea-4a63-94fb-c2c903c47aee%2F01ism59_processed.png&w=3840&q=75)
Transcribed Image Text:k, ka 0|NO
The proper expression for the rate law using the steady-state approximation is:
What would the rate law be if the concentration of O, was very high?
Match the items in the left column to the appropriate blanks in the equation on the right
Reset Help
k1
If the concentration of Og is very high then the
term in the
2k1 [NO]?
denominator becomes insignificant when compared to the
term and
the rate law would simplify to Rate
ki [NO]?
k_1+ kg[O2]
ka [NO]?
kz (O2]
2k: [NO]
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