# Why does the $V_{\mathrm{max}}$ of an enzyme not change with competitive inhibition? Shouldn't it decrease since there are fewer active sites?

My textbook states that $V_{\mathrm{max}}$ doesn't change for competitive inhibition. This makes no sense to me, since decreasing the amount of active sites available should lower the rate of an enzyme catalyzed reaction.

My attempt at rationalizing this is that $V_{\mathrm{max}}$ is dependent on the rate at which the enzyme/substrate complex dissociates into free enzyme + product by the equation Kcat * [E]. So, since it only takes into account the enzyme/substrate complex, we basically don't consider any enzymes with inhibited active sites.

Is that reasonable?

You can think of $V_{max}$ as describing the nature of the active site - it's deftness at converting substrates to the product, the rest is just probability. Competitive inhibitor does not change properties of the active site - they just hang there for some amount of time until the E-I complex dissociates hence they don't affect the maximum theoretical conversion rate of that enzyme. Competive inhibitors only decrease the chance of inhibitor binding to the enzyme. Thus you can always raise the concetration of your substrate to the state that probability (now the other way around) of inhibitor binding the enzyme will become negligible with regard to the substrate allowing it to work at his maximum rate.