## pre equilibrium

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Posts: 21
Joined: Fri Jul 15, 2016 3:00 am

### pre equilibrium

Hi,
What is the general purpose of the pre equilibrium approach? When do we use it?

Thanks

Anh Khoa 3E
Posts: 14
Joined: Fri Jul 15, 2016 3:00 am

### Re: pre equilibrium

Pre eq is used to determine the rate of the overall reaction when the first step is really fast and the second step is slow. Since the first step is really fast, there is generally a buildup which leads to an equilibrium like mechanism.

Emily_Lenh2A
Posts: 17
Joined: Sat Jul 09, 2016 3:00 am

### Re: pre equilibrium

I'm still very confused by pre-equilibrium; how do we know when to use it(for example, if it were given on the final, in what way would it be worded or what specific words should we look for as cues to use it)?

Hue_Vo_1D
Posts: 19
Joined: Wed Sep 21, 2016 2:56 pm

### Re: pre equilibrium

In any reaction systems, the slow step will always determine the rate of reaction, thus the rate law.
The purpose of using Pre-Equilirbrium Approach is to substitute other terms with intermediate, which is not supposed to be in the Rate Law.
Suppose you have two-step reaction and
the rate limiting step is Step 1: Rate=k(1)[R], in this case you don't have intermediates in rate law and Rate=k[R] as itself is reaction rate law.
the rate limiting step is Step 2: Rate=K(2)[R], in which [R] is product of step 1, meaning [R] in step 2 is intermediate. This time, you use Pre-Equil. Approach to get rid of intermediate.

*Derive rate law using pre-equil approach if intermediate is in the rate law.

Sarah_Kremer_1A
Posts: 21
Joined: Wed Sep 21, 2016 2:56 pm

### Re: pre equilibrium

The normal approach is used when the first step is the slow step (normally given). The pre-equilibrium approach is used when the second step is the slow step (also normally given).