The bonding mechanism of a ligand L to the site P goes through an intermediate stadium we'll call PL*:
Express the formation speed of PL using the kinetic constants of these reactions. What would change in the expression found if the reactions had comparable speeds?
≈•≈
This is my attempt to solve it:
Considering that the equilibrum is reached quite fast I think we can neglect the product consumption caused by the second reaction and approximate PL's formation speed as constant and equal to:
where k
3 is the last reaction's kinetic constant (k
1 and k
2 are referred to the direct and inverse reaction of the equilibrium).
After solving the equilibrium expression (with [P]
0≠0, [L]
0≠0, [PL*]
0≠0) I get:
If the reaction speed were comparable, then the expression for [PL] would have been one of the solutions of this differential equation system (as we should have also considered the amount of PL* consumed by the second reaction):
That's all folks
(I don't expect you to revise every passage, especially concerning the gigantic equilibrium expression, but maybe you could throw a glance at the equation system and confirm the ideas on which my answer is based)
Thanks in advance!