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If we have a multi-step reaction (which involves more than one step to reach the products; a complex reaction, one may say), then is the following equation valid?

$$K_\mathrm{eq} = \frac{K_\mathrm{fwd}}{K_\mathrm{rev}}$$ $K_\mathrm{eq}$ – equilibrium constant of the reaction; $K_\mathrm{fwd}$ – rate constant of the forward reaction; $K_\mathrm{rev}$ – rate constant of the reverse reaction.

I know that it is valid for an elementary or single step reaction, where the expression for the rate involves raising the concentrations of the reactants to their coefficients in the balanced equation.

But is it so for a multi step reaction too? (Where the law of mass action is no longer valid?)

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    $\begingroup$ The case for non-elementary reactions is interesting. In fact, equilibrium constants arise from thermodynamics, with no reference to kinetics whatsoever. In the case of an elementary reaction, the strategy of equalling the forward and reverse produces the same equation, but why does or why doesn't that strategy extend to non-elementary reactions? Does microscopic reversibility guarantee the kinetics of even a very complex network of reactions generate a simple equilibrium constant between the stating materials and final products? I wonder if anyone can provide an answer touching on that. $\endgroup$
    – Nicolau Saker Neto
    Jan 4, 2018 at 7:37
  • $\begingroup$ If each 'step' in a sequence of N steps is an equilibrium and all steps are at equilibrium, there are N equilibrium constants and 2 N rate constants. Write out the differential rate law for each step (e.g. for $\ce{A<=>B}$, where $A$ does not appear in any other reaction, $\frac {d[A]}{dt}=-k_1 \cdot [A] + k_2 \cdot [B]$), and equate each to $0$. Depending on where reactants and products appear in each step, you will find whether the equation you mentioned holds or not, and in some cases you may be able to find a single expression linking the initial reactant and final product. $\endgroup$
    – user6376297
    Jan 4, 2018 at 7:49
  • $\begingroup$ @user6376297 I don’t quite get it. Anyway, what I asked wouldn’t be valid in that case, yes? $\endgroup$
    – Senthil Arihant
    Jan 4, 2018 at 8:35
  • $\begingroup$ I edited your question a bit for clarity. Note that notation $K_\mathrm{b}$ have another meaning, where b stands for "base"; also, the term "reverse" is preferred to "backward". Notations are updated according to IUPAC recommendations. $\endgroup$
    – andselisk ♦
    Jan 4, 2018 at 8:45
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    $\begingroup$ @SenthilArihant : I don't think your question can be answered as such. As I wrote, you need to specify what the individual steps are, to know which concentrations are linked at equilibrium, and how. And BTW, if your overall transformation is composed by multiple reversible steps, what do the single forward and backward rates you mentioned in your post mean? Which reaction do they refer to? You should at least have the situation completely clear in you mind before you question its details. $\endgroup$
    – user6376297
    Jan 4, 2018 at 9:17

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