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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?

Michael
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  • @KarstenTheis I saw that question; however, the answer didn't help me very much because to me that explanation seems more like "lucky" algebraic manipulation, instead of explaining how to reconcile the equation for the rate constant with the idea that in chemical equilibrium the rates of the forward and reverse reactions are equal. If the reaction is multistage, then how do you reconcile the equation for the equilibrium constant with the idea of equal forward/reverse reaction rates? – Michael Mar 30 '19 at 20:56
  • @KarstenTheis I guess another way of asking my question is this: why when you multiply out the rate constants for each step, as Philipp did, does this always equal (D^dE^e)/(A^aB^b)? At my current level of understanding, Philipp's math seems just "lucky" in the specific explanation he provided, and I fail to see how to generalizes to any equation. Sorry if I am missing something obvious here – Michael Mar 30 '19 at 20:59
  • You are already convinced that the equilibrium constant expression can be derived from the rate laws for elementary reactions. A multistage reaction is a combination of elementary reactions where all the intermediates cancel out of the net equation. Its equilibrium constant is the product of all the elementary steps, and the intermediates will cancel out of that product as well, leaving you with the equilibrium expression as given in textbooks. The cancelling is not luck, it always happens that way. – Karsten Mar 30 '19 at 21:01
  • Also, you can derive the equilibrium constant expression without invoking any kinetics by asking at which concentrations the Gibbs free energy of reaction is zero. Again, you get the stoichiometric coefficients as the exponents in the equilibrium constant expression. So lucky, elegant, or wonderful, the kinetic derivation and the thermodynamic derivation give the same result. – Karsten Mar 30 '19 at 21:02
  • I wrote up the general argument as answer for the previous, similar question. – Karsten Mar 30 '19 at 21:43
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    @KarstenTheis This is exactly what I needed. Thanks so much! – Michael Mar 30 '19 at 21:50
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    @KarstenTheis - Thanks for the correction. I deleted my answer which was wrong. – MaxW Mar 30 '19 at 23:51

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