[MrTeo]

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 k₃ is the last reaction's kinetic constant (k₁ and k₂ 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!

[MrTeo]

*Ignore me, I am impatient*... anyone out there? I just need a quick review of my work, c'mon!