Intermediates cannot be part of an overall rate law and must be written in terms of reactants actually consumed. This is done by solving the pre-reversible steps for the intermediate. Step 1: Fast, reversible HA = H+ + A- Step 2: Fast, reversible X + H+ = HX+ Step 3: Slow HX+ = productsWhich of the following is the correct expression solving for [HX+] in terms of the species in Step 2.
Intermediates cannot be part of an overall rate law and must be written in terms of reactants actually consumed. This is done by solving the pre-reversible steps for the intermediate. Step 1: Fast, reversible HA = H+ + A- Step 2: Fast, reversible X + H+ = HX+ Step 3: Slow HX+ = productsWhich of the following is the correct expression solving for [HX+] in terms of the species in Step 2.
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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Intermediates cannot be part of an overall rate law and must be written in terms of reactants actually consumed. This is done by solving the pre-reversible steps for the intermediate.
Step 1: Fast, reversible HA = H+ + A-
Step 2: Fast, reversible X + H+ = HX+
Step 3: Slow HX+ = products
Which of the following is the correct expression solving for [HX+] in terms of the species in Step 2.
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