[cas124]

So for a reaction like A + B => AB , I learned that at equilibrium the forward rate constant (k on) is equal to the reverse rate constant (k off). Then I learned that the dissociation constant Kd is represented by: (k off / k on) at equilibrium. If both of these are true, why isn't Kd always equal to 1?  Am I misunderstanding something? I've seen Kd have multiple values other than 1 and I don't understand how at equilibrium k off can be unequal to k on.

[chenbeier]

I learned that at equilibrium the forward rate constant (k on) is equal to the reverse rate constant (k off)

This is wrong. They are different, depending which side is the equilibrium. And so if you have koff/kon = Kd, it is not 1.

[mjc123]

At equilibrium, the forward rate is equal to the reverse rate, i.e.
kf[A][B ] = kr[AB]
so Keq = [AB]/[A][B ] = kf/kr
(Pedantically, in this case Keq is an association constant, not a dissociation constant.)