Consider the following reaction mechanism for the addition of HCl to propene in the gas hase. The overall reaction is just HCl + CH;CH=CH2 → CH;CH2CH3, but the experimentally bserved rate law is more complicated than that stoichiometry would suggest. 2 HCI (HCI)2 kz HCl + CH3CH = CH2 2 complex k-2 (НC)2 + compleх + сн,CHCICH, Here (HCl)2 is an HCl dimer and “complex" represents some intermediate species of
Consider the following reaction mechanism for the addition of HCl to propene in the gas hase. The overall reaction is just HCl + CH;CH=CH2 → CH;CH2CH3, but the experimentally bserved rate law is more complicated than that stoichiometry would suggest. 2 HCI (HCI)2 kz HCl + CH3CH = CH2 2 complex k-2 (НC)2 + compleх + сн,CHCICH, Here (HCl)2 is an HCl dimer and “complex" represents some intermediate species of
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...
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![Consider the following reaction mechanism for the addition of HCl to propene in the gas
phase. The overall reaction is just HCI + CH3CH=CH2 → CH;CH2CH3, but the experimentally
observed rate law is more complicated than that stoichiometry would suggest.
2 HCl 2 (HCI),
k_1
k2
HCI + CH3CH = CH, 2 complex
k_2
(HCI)2 + complex → CH3CHCICH3
(Here (HCl)2 is an HCl dimer and “complex" represents some intermediate species of
unknown structure.) Assume that the third step is rate determining, and that the first two
steps both remain in equilibrium throughout the reaction.
d[CH;CHCICH3] Please write
Derive an expression for the rate of formation of CH3CHCICH,'
dt
your final expression in the boxes. HINT: Be sure to use the equilibrium assumption, and
not the steady state approximation. Your final rate law should NOT include the
concentrations of any intermediates.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48e43d24-e797-4c18-89f0-87785ffa58f5%2Fa61ff32f-d7b1-4abb-a5dd-8e0c83cc5fd5%2F8folbaf_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the following reaction mechanism for the addition of HCl to propene in the gas
phase. The overall reaction is just HCI + CH3CH=CH2 → CH;CH2CH3, but the experimentally
observed rate law is more complicated than that stoichiometry would suggest.
2 HCl 2 (HCI),
k_1
k2
HCI + CH3CH = CH, 2 complex
k_2
(HCI)2 + complex → CH3CHCICH3
(Here (HCl)2 is an HCl dimer and “complex" represents some intermediate species of
unknown structure.) Assume that the third step is rate determining, and that the first two
steps both remain in equilibrium throughout the reaction.
d[CH;CHCICH3] Please write
Derive an expression for the rate of formation of CH3CHCICH,'
dt
your final expression in the boxes. HINT: Be sure to use the equilibrium assumption, and
not the steady state approximation. Your final rate law should NOT include the
concentrations of any intermediates.
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