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MichaelM95
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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
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