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Does entropy of the universe always increase in one of the forward or backward reactions of a reversible reaction and decrease in the other?

If no then how do you explain the entropy of the universe increasing in both directions or decreasing in both directions?!?

If yes, do we get more of the reaction that increases the entropy of the universe? I would presume yes as all real processes increase the entropy of the universe.

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  • $\begingroup$ I need to think more, but usually you define a system and surrounding, and these may make up your universe. The entropy of the universe will always increase. The system may increase or decrease, but the surrounding will more than compensate this to ensure the entropy of the universe increases.( I have been teaching thermo to mechanical engineers for too long and they don't do reactions... I will get back to you) $\endgroup$ – Charlie Crown Dec 1 '19 at 2:48
  • $\begingroup$ Like I said in my comment in your other thread, there is a terminology difference between a reversible reaction and a reversible process. It is possible to devise a reversible process that takes a reversible reaction mixture from its initial state to its final state without increasing the entropy of the universe. $\endgroup$ – Chet Miller Dec 1 '19 at 13:07
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First, a defintion of terms: The meaning of reversible in "reversible reaction" is different from the thermodynamic meaning of reversible. A reversible reaction is one that is kinetically reversible. I.e., one where you can adjust the relative concentration of products and reactants back and forth as you change conditions. By contrast, an example of an irreversible reaction would be cooking an egg, which causes the proteins to denature. The proteins can't be readily undenatured, even if you change the conditions to thermodynamically favor denaturation because, kinetically, the proteins can't find their way back to their undenatured state.

Anyways, let's suppose we have a "reversible reaction", $$\ce{A <=> B},$$ in which the equilbrium constant is, under certain conditions (e.g., at a certain temperature and pressure):

$$K_{eq} = \frac{[B]}{[A]} = 10,$$ i.e., at equilibrium, we will have 10x as much B as A.

This represents the state at which the entropy of the universe is a maximum under those conditions. Thus if, under those conditions, $\frac{[B]}{[A]} <10$, the reaction will proceed in the forward direction until it reaches equilibrium, increasing the entropy of the universe in the process. Likewise if, under those condtions, $\frac{[B]}{[A]} > 10$, the reaction will proceed in the reverse direction until it reaches equilibrium, also increasing the entropy of the universe in the process.

So it is not the case that if, say, $\frac{[B]}{[A]} = 9$, then the entropy of the universe would increase in both directions. Rather, it would increase in the forward direction (until it reached equilibrium), and (theoretically) decrease in the reverse direction. If we instead had $\frac{[B]}{[A]} = 11$, the opposite would happen: decreased entropy of the universe in the forward direction, increased entropy of the universe in the reverse direction (again, until it reached equilibrium).

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  • $\begingroup$ If I understand you correctly, a 'reversible reaction' is one that is kinetically reversible. I.e., one where you can adjust the relative concentration of products and reactants back and forth as you change conditions. But what is your definition of 'thermodynamic reversibility'? $\endgroup$ – TSA Dec 1 '19 at 8:44
  • $\begingroup$ @ETS That's an entirely separate question, and there are entire answers on Chemistry SE devoted to this (I'd recommend a search). But, briefly: a thermodynamically reversible process is an idealized, infinitely slow process (consisting of an infinite no. of infinitesimal steps) in which the system is in continous equilibrium with the surroundings, and for which $\Delta S_{universe} = 0$. [continued] $\endgroup$ – theorist Dec 1 '19 at 16:47
  • $\begingroup$ ....It is called reversible because each step creates only an infinitesimal change, and thus one can do a backwards infinitesimal step and restore both the system and the surrondings to their original states. Since all real processes require $\Delta S_{universe} > 0$, a reversible process is a limiting idealization that is not possible in the real world. $\endgroup$ – theorist Dec 1 '19 at 16:48
  • $\begingroup$ will these do as definitions: an irreversible reaction is one where activation energy in one direction can be achieved without constant human intervention and where activation energy in the other direction cannot be achieved without constant human intervention; a reversible reaction is one where the activation energy in both directions can be achieved without constant human intervention? $\endgroup$ – TSA Dec 1 '19 at 19:44
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    $\begingroup$ @BuckThorn Yes, that's a good point --- I don't think his argument quite passes muster now, but if it were reframed from activation energy to Gibbs energy of activation, which would include an entropic prefactor to account for the difficulty of refolding the proteins, then it might work. But this is why I suggested he give it more thought and then, if he wished, post it as a separate question. Regarding your last sentence, since we're talking about kinetic reversibility now, how you get between the states of course does matter. $\endgroup$ – theorist Dec 4 '19 at 4:07

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