Consider the acid catalyzed reaction;AH + H+ ----> AH,+rate = k1[AH][H*] (fast)AH2+----> AH + H+rate =k2[АН2*] (fast)AH,+ + B ----> BH+ + AHrate = 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 therate 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 =

Steady state approximation is a technique to obtain the rate law. In this method we assume that the…

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

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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