where K(subscript)On is the rate of Enzyme (E) + Substrate (S) going to Enzyme-Substrate complex (ES).
I think I understand why K(subscript)Off and K(subscript)catalysis have units of time^-1. They're going to be multiplied by a molar amount of ES, so you end up with mol/time, which makes sense as a rate, right? But then why isn't this the case for K(subscript)On? If moles are also in the denominator, they will cancel when multiplied by a molar amount of an input, leaving just time^-1 as an output, which I'm not sure how to interpret.
kon is a bimolecular rate constant. The rate law is rate = kon[E][S]. Since rate is molar/second and [E] and [S] have units of molar, the units of kon have to be molar-1second-1 so that when you multiply by molar2, the units of rate work out to be molar/second.
Ah, it is multiplied by two mole values--apologies for not noticing that, and many thanks for your time, Mr. Shapiro.
rate=k[E][S]
rate=molar/sec
k=?
E=molar
S=molar
dividing both side by [E][S]
k=rate/[E][S]
Therefore the unit of k is molar-¹ sec-¹.
On a point of order: Please write rate constants with a small k, the capital K is reserved for equilibrium constants.
Blaine Dillingham, There are only two constants in the M-M equation: rm (Vm) and KM. The rate constants of the intermediate stages about which you write appear in deriving the M-M equation. Moreover, the units of the rate constants result from the assumed concentration units and the adopted rate lows. Thus, if the concentration is expressed in mol/m3, k1 has the unit m3/(mol·s), not 1/(mol·s) and k-1 has the unit 1/s.
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