# Equilibrium Constant and Reaction Mechanism [duplicate]

How is it possible for the equilibrium constant to not depend on a reaction's mechanism?

For the elementary reaction

$$aA + bB \rightleftharpoons dD + eE$$

I understand that the rate of the forward reaction is $$k_f [A]^a[B]^b$$ and that the rate of the reverse reaction is $$k_r[D]^d[E]^e$$, and from that it follows that the equilibrium constant is $$K_c = \frac{k_f}{k_r} = \frac{[D]^d[E]^e}{[A]^a[B]^b}$$

What bothers me, however, is that my chemistry textbook says that for ANY reaction, elementary or not, of the form

$$aA + bB \rightleftharpoons dD + eE$$

the equilibrium constant will be

$$K_c = \frac{[D]^d[E]^e}{[A]^a[B]^b}$$

just as in the elementary reaction. Wouldn't this imply that the reaction rate of this forward reaction is $$k_f [A]^a[B]^b$$ (just as in the elementary reaction), and that the reaction rate of the reverse reaction is $$k_r[D]^d[E]^e$$ (also the same as the elementary reaction), which might not necessarily be true depending on the reaction mechanism? Or is the math behind calculating the equilibrium constant for a "complex"/multistage reaction different than the math behind a regular equation?