Pages

Driving reactions with equilibrium constants


In the reaction above, K1 is about equal to K2, and they are both in favor of the carbonyls. So we would expect a 50:50 mixture.

The reason that doesn't happen is due to an additional equilibrium - the deprotonation of the acid:


This drives it over to the right.

The process can also be represented as an energy diagram:


Below is another example of a reaction which can be driven left or right, depending on pH:

No comments:

Post a Comment