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$$\ce{A -> B -> C -> A}$$

In Alberty's Physical Chemistry (7th ed.), the author says that this reaction mechanism is not possible. The reason given is:

Although this mechanism at first sight appears reasonable, no actual process follows this mechanism because it violates the principle of detailed balance. According to this principle, at equilibrium the forward rate of each step is equal to the reverse rate of that step. Thus equilibrium cannot be maintained by this mechanism.

I understand the statement of the principle ("at equilibrium the forward rate of each step is equal to the reverse rate of that step"), but I don't get how it implies that equilibrium cannot be maintained, or that this reaction is not possible.

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  • $\begingroup$ The whole thing looks like this en.wikipedia.org/wiki/… and I am not sure what point they are trying to make. $\endgroup$ – Karl Jul 19 '17 at 19:36
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    $\begingroup$ So, technically speaking, you could imagine $\ce{IBrClC-CH2Cl}$. There are three staggered conformations that could interconvert in exactly the reaction scheme provided. They are all in equilibrium. This essentially refutes the statement from the book. $\endgroup$ – Zhe Jul 19 '17 at 20:04
  • $\begingroup$ @Zhe according to the Wikipedia page linked by Karl, Rudolf Wegscheider proved such a reaction scheme wasn't possible. I doubt the statement is wrong, so my guess is the confusion arises due to the scheme described by Wegscheider not matching what you provided. $\endgroup$ – Tyberius Jul 19 '17 at 20:16
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    $\begingroup$ Whoa, whoa. The steps are supposed to be irreversible? My example does not at all conform to that. @Tyberius I'll think about the reaction again in this context. $\endgroup$ – Zhe Jul 19 '17 at 21:43
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    $\begingroup$ Interesting question. It doesn't make sense to talk about a fully irreversible reaction (which has no reverse rate constants), when also asserting that the forward and reverse rates of each step must be the same. The matter at hand seems to be more subtle than that. Quoting from the Wikipedia page that Karl linked: "... one represents irreversible reactions as limits of reversible steps [...] not all reaction mechanisms with irreversible reactions can be obtained as limits of systems or reversible reactions with detailed balance". $\endgroup$ – orthocresol Jul 20 '17 at 6:17
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In this triangular scheme, taking it literally, it is clear that all the population will end up in species C. If $A \rightarrow B$ is mathematically irreversible (rate constant back is zero) it implies that B is at an energy infinitely below that of A, and the energy of C infinitely below that of B, so clearly C cannot revert to A without the input of an infinite amount of energy. Such an argument hardly makes sense in chemistry, however.

If we now assume that the reaction is reversible but with a very small back rate constant a similar conclusion is obtained, i.e. that the overwhelming (but not total) population ends up in C.

Suppose, without loss of generality and for simplicity, that the activation energy A/B and B/C are the same at $E_A$ and that B is at energy E below A, and C at E below B. Now let the forward rate constant A to B be $k_f=k_0\exp(-E_A/RT) $ and the back rate constant is $k_{BA}=k_0\exp(-E_A/RT)\exp(-E/RT)$ and which is going to be smaller, say it is $10^{-3}$ smaller. The rate constant from C to A is therefore $k_{CA}=k_0\exp(-E_A/RT)\exp(-2E/RT)$ which is now $10^{-6}$ times smaller that the rate constant A to B.

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  • $\begingroup$ I think your anwer is reasonable, I will accept it after waiting for sometime. Thank you. $\endgroup$ – user8277998 Jul 20 '17 at 17:45

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