How is "always" spontaneous reconciled with equilibrium?

[MichaelM95]

Many introductory (high school chemistry) texts speak of reactions that are “always” or “never” spontaneous (in the former case they reference reactions where ∆H is - and ∆S is +, for example). I struggle with this concept when considering reactions in closed systems.

Premise: All reactions in a closed system will reach a point of equilibrium (even those that appear to go to completion with ridiculously large K’s).

So how can a reaction that reaches equilibrium be “always” spontaneous. Mathematically (again, sticking to algebra here risks losing a lot of the subtlety laid out in thermodynamics) ∆G=∆Go+RT ln Q=∆H-T∆S . Won’t there “always” be a Q for which ∆G changes sign for a given reaction (again, perhaps it is ridiculously large)? And does this mean that ∆H and ∆S depend on the reaction quotient to such an extent that they will actually change signs? Or is it that when Q>K the reaction IS proceeding in the reverse and the signs on ∆H and ∆S are reversed accordingly to match the reverse process? Finally, is the “always spontaneous” language laid out in introductory texts a false construct? If not, what does it mean in a real example: calling the decomposition of hydrogen peroxide “always spontaneous” (∆H is - and ∆S is + in this case) seems to imply that there is no Q for which the reverse reaction is spontaneous, but that is not true…

I suspect I am missing something fundamental, or that it is a small matter of semantics that is eluding me. Either way, I am hoping to clear this up. Any help would be appreciated.Michael

 

[Borek]

Interesting point. My take is that it is just a lousy language, with "always" meaning "always when starting from substances in their standard state" (as opposed to "when staring from an equilibrium mixture").

But then perhaps there exist some more precise definition that I am not aware of, thermodynamics is not my forte.

 

[Ygggdrasil]

IIRC, the correct set of equations would be:
∆G = ∆Go+RT ln Q
∆Go = ∆Ho - T∆So

But, yes, you are correct in your understanding. For a reaction at equilibrium, ∆G = 0. Therefore, ∆Go = –RT ln Q_eq and the reaction quotient at equilibrium (aka the equilibrium constant) is given by K = Q_eq = exp(–∆Go/RT).

Therefore, when Q < K, ∆G < 0 and the reaction proceeds in the forward direction; when Q > K, ∆G > 0 and the reaction proceeds in the reverse direction; and when Q = K, ∆G = 0 and the reaction is at equilibrium.

The actual change in entropy (∆S) of a reaction will depend on the reaction quotient (a mixture of product and reactant gives additional entropy compared to pure product or pure reactant). This entropy of mixing is not accounted for in ∆So, which considers the change in entropy of changing one molecule of reactant into one molecule of product.

 

[Anindya Mondal]

Gibbs free energy is the only criterion for the spontaneity of a chemical reaction.

 

[Chestermiller]

Many introductory (high school chemistry) texts speak of reactions that are “always” or “never” spontaneous (in the former case they reference reactions where ∆H is - and ∆S is +, for example). I struggle with this concept when considering reactions in closed systems.

Premise: All reactions in a closed system will reach a point of equilibrium (even those that appear to go to completion with ridiculously large K’s).

So how can a reaction that reaches equilibrium be “always” spontaneous. Mathematically (again, sticking to algebra here risks losing a lot of the subtlety laid out in thermodynamics) ∆G=∆Go+RT ln Q=∆H-T∆S . Won’t there “always” be a Q for which ∆G changes sign for a given reaction (again, perhaps it is ridiculously large)? And does this mean that ∆H and ∆S depend on the reaction quotient to such an extent that they will actually change signs? Or is it that when Q>K the reaction IS proceeding in the reverse and the signs on ∆H and ∆S are reversed accordingly to match the reverse process? Finally, is the “always spontaneous” language laid out in introductory texts a false construct? If not, what does it mean in a real example: calling the decomposition of hydrogen peroxide “always spontaneous” (∆H is - and ∆S is + in this case) seems to imply that there is no Q for which the reverse reaction is spontaneous, but that is not true…

I suspect I am missing something fundamental, or that it is a small matter of semantics that is eluding me. Either way, I am hoping to clear this up. Any help would be appreciated.Michael

It's just a rule of thumb to give you a feel for how far the reaction will go before it reaches equilibrium. Your assessment was completely correct.