## Steady-State Approach vs. Pre-Equilibrium Approach [ENDORSED]

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

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abram_wassily_1G
Posts: 10
Joined: Wed Sep 21, 2016 2:56 pm

### Steady-State Approach vs. Pre-Equilibrium Approach

Okay so I am still not entirely sure what separates the Steady-State Approach from the Pre-Equilibrium Approach. Is it correct to assume that if the first elementary step in a reaction mechanism is slow, we use the Steady-State Approach? And, if the first elementary step in a reaction mechanism is fast, we use the Pre-Equilibrium Approach? Thank you.

Jeffreyho97
Posts: 10
Joined: Fri Jul 15, 2016 3:00 am

### Re: Steady-State Approach vs. Pre-Equilibrium Approach  [ENDORSED]

I don't think it matters whether the first or second step is the slow step. Lavelle said that it probably is better to just use the pre-equilibrium step because it is more straightforward. Remember they both give the same result.

Marissa Petchpradub 2F
Posts: 15
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Steady-State Approach vs. Pre-Equilibrium Approach

Yes, I believe it is correct to assume we use the Pre-Equilibrium approach when the first elementary step in a reaction mechanism is fast.

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