I have read mannaia's answer on the question of 'rate order and confusion', as well as Nicolau Saker Neto's answer to a related question. They have both been very helpful.
As I understand, the equilibrium is actually derived from the rate constants of a reaction. At dynamic equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction, hence in a hypothetical reaction
$$ \ce{pA + qB <=> rC + sD} $$
with the rate equations $Rate_{fwd} = k_{fwd}[A]^{a}[B]^{b}$ and $Rate_{bck} = k_{bck}[C]^c[D]^d$, with $a, b, c,$ and $d$ not related to $p, q, r$ and $s$.
The equilibrium is derived as such: \begin{align} R_{fwd} &= R_{bck}\\ k_{fwd}[A]^{a}[B]^{b}&=k_{bck}[C]^c[D]^d\\ \frac{k_{fwd}}{k_{bck}} &= \frac{[C]^c[D]^d}{[A]^a[B]^b}\\ \therefore k&=\frac {k_{fwd}} {k_{bck}}\\ \end{align}
Based on the derivation, and the fact that the stoichiometric coefficients of the reactants in the rate-determining step (RDS) of the reaction is equal to the order of the reactants in the rate equation, is it correct to state that the reaction equations given as the equilibrium equation is actually the rate-determining step of two reactions?
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