December 22, 2024, 06:46:28 PM
Forum Rules: Read This Before Posting


Topic: Steady State Approximation  (Read 1798 times)

0 Members and 1 Guest are viewing this topic.

Offline XGen

  • Full Member
  • ****
  • Posts: 127
  • Mole Snacks: +9/-4
Steady State Approximation
« on: July 26, 2012, 11:15:52 PM »
I found the following problem online while I was looking for steady-state approximation problems, and after I did it I found I didn't have an answer key. Could anyone tell me if my answers/thinking are correct?

Given the mechanism:
        k1
PdL2  ::equil:: PdL + L
        k-1

               k2
PdL + ArX  :rarrow: PdLArX

1. Use the steady-state approximation to solve for the concentration of the intermediate
species [PdL].
2. Write the rate law for the formation of the oxidative addition product PdLArX.



We can start by realizing the rate of formation for PdL can be represented as k1[PdL2], and the rates of consumption as k2[ArX][PdL] and k-1[PdL][L].

From this, we can write d[PdL]/dt = k1[PdL2] - k2[ArX][PdL] - k-1[PdL][L] = 0, as this is the assumption for steady-state, that the concentrations of the intermediates do not change. After manipulation, [PdL] = k1[PdL2]/(k2[ArX] + k-1[L]).

For the second part, we begin by realizing that the rate of formation for PdLArX can be written as k2[PdL][ArX]. Substituting the result in from the first part, d[PdLArX]/dt = k1k2[PdL2][ArX]/(k2[ArX] + k-1[L]).

Thanks for your time!

Sponsored Links