## When to use pre-equilibrium approach?

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

Lizzie Zhang 2C
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### When to use pre-equilibrium approach?

There are basically two approaches that the textbook talked about: pre-equilibrium and steady state approximation. My understanding right now is that if no one tells me which step is the slowest, then I use the steady state; if there is a rate determining step, I use the pre-equilibrium.

I just want to make sure that this understanding is correct.
Thanks!

Priyanka Bhakta 1L
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: When to use pre-equilibrium approach?

From my understanding, you are correct.

Michelle Nguyen 2L
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

### Re: When to use pre-equilibrium approach?

You are correct that the pre-equilibrium assumption depends on information given on which step is the slow step. However, you shouldn't think about the two assumptions as mutually exclusive. That is, rather than thinking you should use either one or the other, it is more like you would often need to use both in order to show that a proposed reaction mechanism is plausible. In the textbook example, the steady-state approximation was first used to set the rate of formation of the intermediate to 0 (assume there is a relatively low and constant concentration of intermediate throughout the reaction, hence the rate of intermediate formation is close to 0). After using the steady state to write the rate law without including intermediate concentration, the book then uses the pre-equilibrium condition that the rate of consumption of intermediate in the slow step is negligible to its rates of formation and consumption in the forward and reverse reactions of the faster first step in order to simplify the rate law. Thus, both approaches are used in the same problem.