Good day.

A notion of chemical potential teaches us, that a chemical reaction

$A+B \to C$

at finite tempereture actually goes in both directions

$A+B \rightleftharpoons C$

and the intensity of a given diection depends on:

1) Temperature

2) Energy difference E(A+B) versus E(C)

For my argument it is enough that it exists in both directions. Let me assume A, B and C are gasses and A has the smallest molecules, B intermadiate and C the biggest ones.

I put A,B,C in a box and an quilibrium settles. Then I use a molecular sieve to extract A, I prevent A from going back to the box by using a vacuum-cleaner. A new equilibrium with excess of B comes into the place. I use (for a differential time where no more A passes) a molecular sieve with bigger holes to extract B. Then alternating sieves frequently I extract A and B, colling down the pool.

Is that possible?

I think it contradicts the second law of thermidynamics... I create A and B by cooling the pool. Then I can combust A and B....

Let us say $A=H_2$, $B=O_2$ and $C=H_20$ - it would lead to "energyless" dissociation of water.

What is wrong with my arguments? :

  • "Bothdirectional" chemical reactions are a fact.

  • Existence of molecular sieves is a fact...



1 Answer 1


Assuming the reverse reaction (the endothermic one) will happen under the conditions of the presumably thermally isolated box, it will reduce the temperature inside the box. As the temperature decreases, the probability of this reverse reaction occurring will decrease, since it is not energetically favourable, so the reaction will become slower and slower as time passes.

Your suggestion is to speed it up by inducing a favourable chemical potential by extracting the left hand molecules using a sieve. This will generate a chemical potential difference across the molecular sieve, and to ensure that you are extracting the correct molecules using your vacuum cleaner (i.e. to stop them from going back through the sieve), you will need to apply a larger chemical potential difference than the one across the sieve. This energy you will need to supply to your vacuum cleaner will increase as the conditions for the reverse reaction become less and less favourable.

These "free energy" problems generally boil down to achieving a level of engineering efficiency that is impossible. Your system needs to be perfectly thermally isolated, and your vacuum cleaner needs to be lossless in order to get a net energy loss of zero (there is no question of gaining energy from the process).

  • $\begingroup$ Are you telling that the energy to power vacuum-cleaner is the energy I would need anyway in any "standard" procedure (e.g. electrolysis)? That this energy compensates for apparent breaking of the second TD law? Well, I am not sure... I just need a huge box (infinite) so that once a molecule goes through the sieve, theres is no wall to bounce back (wall at infinity). I am not telling you are wrong, I am just telling your argument does not seem to me so apparent. $\endgroup$
    – F. Jatpil
    Feb 14, 2017 at 14:52
  • $\begingroup$ Yes, if you had an infinite box then there would always be a favourable chemical potential for extracting molecules from the system. Are you suggesting that it will be possible to make a net gain of energy via this process? $\endgroup$
    – Paraquat
    Feb 14, 2017 at 14:57
  • $\begingroup$ Of course not energy itself, I am talking about second thermodynamical law. The dissociated molecules are those coming from the high-energy tail of the Boltzman distribution: two molecules C with high speed collide, producing AABB (if the speed is low the collision is elastic). If AABB are removed from the system, the energy is removed, the system (after AABB removed) cools down (the kinetic energy of the system is lowerd by the potential (chemical) energy of AABB). Energy is of course conserved. $\endgroup$
    – F. Jatpil
    Feb 14, 2017 at 15:03
  • $\begingroup$ Well you're locally reducing the entropy of the system, but the system as described is not isolated. If you did a detailed calculation, you'd find that the entropy of the universe increases. $\endgroup$
    – Paraquat
    Feb 14, 2017 at 15:08
  • 1
    $\begingroup$ @F.Jatpil it might become more apparent if you think of the question "is there a way to take two boxes at temperatures $T_1 > T_2$ and somehow increase $T_1$ by introducing a box at $T_3 = 0,$ as in absolute zero?". The question is answered clearly as "yes, a heat engine from box 2 to box 3 clearly produces work which can be fed to a heat pump from box 2 to box 1, increasing $T_1.$" This is not a violation of the second law of thermodynamics precisely because box 3 increases its entropy even if it is large enough to not change its temperature substantially. $\endgroup$
    – CR Drost
    Feb 14, 2017 at 16:03

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