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Steady-state vs. Pre-equilibrium

Posted: Sat Mar 10, 2018 9:05 am
by Kate Zeile 2D
In lecture, Dr. Lavelle said that the pre-equilibrium approach is simpler than the steady-state approach, but it is also less flexible. What did he mean by "less flexible"?

Re: Steady-state vs. Pre-equilibrium

Posted: Sat Mar 10, 2018 6:50 pm
by Rakhi Ratanjee 1D
During lecture, he mentioned that the steady-state approach assumes a constant intermediate concentration in the rate-limiting step and that the pre-equilibrium approach relies on the reaction before the rate-limiting step is at equilibrium, requiring the equilbrium constant. Therefore, it may be less flexible if the concentration of the intermediate changes due to loss to the surroundings.

Re: Steady-state vs. Pre-equilibrium  [ENDORSED]

Posted: Sat Mar 10, 2018 11:13 pm
by Alexia Joseph 2B
In a steady state approach, a constant intermediate concentration (the d[intermediate]/dt = 0). The pre-equilibrium approach means the reaction before the rate-limiting step is at equilibrium, so you can use the equilibrium constant K. The pre-equilibrium approach is less flexible because it applies to a a reaction in which there is a fast step followed by the slow step.