fast equilibrium approach v steady state equilibrium

$K = \frac{k_{forward}}{k_{reverse}}$

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

fast equilibrium approach v steady state equilibrium

I have tried to understand thins concept and how it is different from steady state equilibrium, but it is very difficult for me to conceptualize. Can someone try to describe the difference and importance of each?

Noah Fox 1E
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

Re: fast equilibrium approach v steady state equilibrium

Honestly, steady-state is so math intensive and so time consuming that it is most likely best to avoid it at all circumstances.