## When to use steady-state vs. pre-equilibrium

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

Haesoo Kim 3O
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm

### When to use steady-state vs. pre-equilibrium

When we are deriving the overall rate law implied by a proposed mechanism, when do we use the steady-state approach and when do we use pre-equilibrium approach, and why? Thank you.

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

### Re: When to use steady-state vs. pre-equilibrium

Both techniques will get you the same answer but in this class, I think we are just going to use the pre-equilibrium approach that Dr. Lavelle went over in class.

Sanmeet Atwal 1D
Posts: 17
Joined: Fri Sep 26, 2014 2:02 pm

### Re: When to use steady-state vs. pre-equilibrium

Justin's right, for this class we use the pre-equilibrium approach to find the overall rate law!

104607508
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

### Re: When to use steady-state vs. pre-equilibrium

when writing the rate laws for elementary reactions the textbook says "The reverse of step 2 is too slow to affect the rate and so is not included." (634) how are we supposed to know when a reaction is to slow to include?

juliana alden 2D
Posts: 12
Joined: Fri Sep 25, 2015 3:00 am

### Re: When to use steady-state vs. pre-equilibrium

We only use the reverse reaction if the step is at equilibrium, because the step is too slow there is no bottleneck or build and therefore the step is not equilibrium and the reverse reaction is not taken into account. The problem should indicate if the step is slow or fast, where fast steps are usually considered to be at equilibrium and therefore have a forward and reverse reaction rate