When a fully suicidal enzyme follows a second-order equation

A fully suicidal enzyme can follow an effective second-order rate law, but only in the appropriate concentration regime. The binding intermediate matters, as follows.

To be specific, a suicide reaction is a reaction that results in enzyme inactivation. This can be full or partial, as is the case of reactions whose product is reactive and that sometimes reacts as is the case with alanine racemase and chloroalanine. An enzyme with no way to regenerate its reduced or oxidised FAD may also stop after one turnover, but cofactor exhaustion is not itself irreversible enzyme inactivation.
A second-order rate law has a total concentration exponent of two; reaction order is not defined simply by counting the reactants. A normal enzyme under Michaelis–Menten kinetics will go back and react with a substrate over and over.
Given a reaction as follows:

For the scheme above, conversion of ES consumes the active enzyme rather than releasing E again:

$$\frac{d[E]}{dt}=k_{\mathrm{off}}[ES]-k_{\mathrm{on}}[E][S].$$

Under a quasi-steady-state approximation for ES, define K_M = (k_off + k_cat)/k_on and [E]_active = [E] + [ES]. Then:

$$\frac{d[P]}{dt}=\frac{k_{\mathrm{cat}}[E]_{\mathrm{active}}[S]}{K_M+[S]}.$$

This has a Michaelis–Menten-like saturation form, but [E]_active decreases as product forms. At low substrate concentration ([S] ≪ K_M), it reduces to:

$$\frac{d[P]}{dt}\approx\frac{k_{\mathrm{cat}}}{K_M}[E]_{\mathrm{active}}[S].$$

That is the effective second-order limit. At saturating substrate, the rate instead approaches k_cat[E]_active, so a single turnover does not by itself guarantee second-order kinetics. The approximation also requires binding to relax faster than the changing active-enzyme and substrate pools; otherwise the full differential equations are needed.