Consider the acid catalyzed reaction; АН + H+ > AH2* rate = k1[AH][H*] (fast) ---- AH2+ ----> AH + H+ rate = k2[AH2+] (fast) %3D AH2+ + B ----> BH+ + AH k3[AH2+][B] (slow) Using the pre-equilibrium approximation derive the rate law. rate = (kık3/k2)[H+][B] rate = (kık2/k3)[HA][H+][B] rate = (kık3/k2)[HA][H+][B] rate = (kık3/k2)[HA][B]
Consider the acid catalyzed reaction; АН + H+ > AH2* rate = k1[AH][H*] (fast) ---- AH2+ ----> AH + H+ rate = k2[AH2+] (fast) %3D AH2+ + B ----> BH+ + AH k3[AH2+][B] (slow) Using the pre-equilibrium approximation derive the rate law. rate = (kık3/k2)[H+][B] rate = (kık2/k3)[HA][H+][B] rate = (kık3/k2)[HA][H+][B] rate = (kık3/k2)[HA][B]
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter11: Chemical Kinetics
Section: Chapter Questions
Problem 11.96PAE: The following statements relate to the reaction for the formation of HI: H2(g) + I2(g) -* 2 HI(g)...
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![Consider the acid catalyzed reaction;
AH + H+ ----> AH,+
rate = k1[AH][H*] (fast)
AH2+
----> AH + H+
rate =
k2[АН2*] (fast)
AH,+ + B ----> BH+ + AH
rate = k3[AH2*[B] (slow)
Using the pre-equilibrium approximation derive the rate law.
rate =
(kık3/k2)[H*][B]
rate =
(kık2/k3)[HA][H*[B]
rate =
(kık3/k2)[HA][H+][B]
rate =
(kık3/k2)[HA][B]
If the only source of H+ is from HA show that it is possible to derive a rate law that is independent of H+. What is the
rate law?
Hint: Add the fast equilibrium HA <------> H+ + A¯ to the proposed mechanism in part a).
(kık3Ka/2/k2)[HA]!/2[B]
(kık3Ka1/2/k2)[HA]3/2[B]
rate = (kık3/k2K1/2)[HA][B]
rate = (kık3/k2Ka1/2)[HA]3/2[B]
rate =
rate =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa567ba2-54db-4ff8-80b0-e6bd2304d289%2Fd321af2f-c625-461e-9f41-a7e1441045e0%2Fy0xp38_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the acid catalyzed reaction;
AH + H+ ----> AH,+
rate = k1[AH][H*] (fast)
AH2+
----> AH + H+
rate =
k2[АН2*] (fast)
AH,+ + B ----> BH+ + AH
rate = k3[AH2*[B] (slow)
Using the pre-equilibrium approximation derive the rate law.
rate =
(kık3/k2)[H*][B]
rate =
(kık2/k3)[HA][H*[B]
rate =
(kık3/k2)[HA][H+][B]
rate =
(kık3/k2)[HA][B]
If the only source of H+ is from HA show that it is possible to derive a rate law that is independent of H+. What is the
rate law?
Hint: Add the fast equilibrium HA <------> H+ + A¯ to the proposed mechanism in part a).
(kık3Ka/2/k2)[HA]!/2[B]
(kık3Ka1/2/k2)[HA]3/2[B]
rate = (kık3/k2K1/2)[HA][B]
rate = (kık3/k2Ka1/2)[HA]3/2[B]
rate =
rate =
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