Fast step at equilibrium [ENDORSED]

[Felicia Wei 1B]

In the lecture, it was explained that for the pre equilibrium approach for the first step we assume the first step is at approximately at equilibrium. Can someone explain why because I'm a bit confused still?

[Janelle Gokim 3B]

Since the first step is occurring at a fast rate, there is going to be a buildup of product being formed. With this buildup, some of this product has the potential to be consumed and drive the reverse reaction, therefore both the forward and reverse reactions will be taking place. We assume that this is approximately at equilibrium so we can make use of the equilibrium constant K and create substitution for the intermediate found in the slow step.

[Chem_Mod]

As I discussed in class step 2 is slow. Therefore product formed in step 1 builds up. Product can go back to reactant in step 1. Since forward and reverse reactions are occurring and step 1 is treated as approximately at equilibrium.

[Gigi Elizarraras 2C]

Because the first step is fast there will be a build up of products that eventually o back to being reactants in a reversible chem equation. The process will be assumed to be at equilibrium in the forward and reverse processes:)

hope this helps!

[Chinmayi Mutyala 3H]

The first step was a fast step and right after it was the slow step. Since the slow step takes longer to happen, in that time, for the fast step, there is a buildup of products which will go in the reverse and form reactants. This is the assumption we're making in the pre-equilibrium approach. In the time that it takes for the slow step to occur, we're saying the first step is in equilibrium.