I know I'm wrong, but do Le Chateliers and Rate determing step contradict?

[deleted74029]

For instance if 2A + B---------><--------3AB

Increasing A should produce more 3AB right? But if no more B is produce than usual wouldn't it still only allow for the same amount of 3AB to be produced?

---

Members don't see this ad.

 

[161927]

For instance if 2A + B---------><--------3AB

Increasing A should produce more 3AB right? But if no more B is produce than usual wouldn't it still only allow for the same amount of 3AB to be produced?

If you change the concentrations of any of the reactants, the system will respond to this stress by producing more product. The fact that you have kept B constant is not important.

 

[deleted74029]

Yeah I think I was confusing speed with production.

 

[HawkeyePostOp]

For instance if 2A + B---------><--------3AB

Increasing A should produce more 3AB right? But if no more B is produce than usual wouldn't it still only allow for the same amount of 3AB to be produced?

It still shifts right without additional B, because as long as the reaction doesn't go to completion, B isn't totally consumed. Adding A reacts increasing portions of B.

 

[Sachiel]

I could be wrong about this, but you're getting two different things confused.
And this is a very good topic that the MCAT LOVES to screw people on.

Le Chatelier's talks about big K, as in the equilbrium situation of a reaction, what is FAVORABLE, meaning what will be happening more.

Rate determining step talks about how FAST something goes.

Just because something is favorable to happen, does not mean it will happen within any amount of time.