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A "van't Hoff apparatus" is mentioned as a means of performing a reaction involving gases in a reversible manner, for instance here.

Unfortunately I have been unable to find a description of such a device using a web search (including on this site). It may be that the device has other names. I intuit from the post above that the apparatus is similar to the device used to measure the Joule-Thompson coefficient of a gas.

Could someone provide a description of the nature and operation of such an apparatus (even if it is only a theoretical device)?

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  • $\begingroup$ Van't Hoff Equilibrium Box $\endgroup$ – Chet Miller Feb 15 '20 at 15:40
  • $\begingroup$ I got here through Enrico Fermi Circa 1936 Title: thermodynamics, pp101-109, Chapter VI Gaseous Reactions, section 22. Per Fermi this is a gedunken featuring hypothetical ‘perfect semipermeable membranes’ that only allow a certain gas to pass through. Maybe helpful? $\endgroup$ – B.L. Jul 11 '20 at 4:49
  • $\begingroup$ Thanks for the reference. I think you mean "gedanken experiment" $\endgroup$ – Buck Thorn Jul 11 '20 at 9:19
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The Van't Hoff equilibrium box is a chamber containing a mixture of reactants and products at equilibrium. It is connected to an array of cylinders, each containing one of the pure reactants or products. The chamber itself is operated at constant total pressure and temperature. The cylinders are each connected to the equilibrium box through a semipermeable membrane that only passes that species into the chamber. Pure species are introduced in stoichiometric proportions to the chamber through these membranes at the same partial pressures as in the reaction mixture, and pure products are removed in stoichiometric proportions at the same partial pressures as in the reaction mixture. So the operation of the equilibrium box is reversible, and the delta G in going from pure reactants in to pure products out is zero.

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  • $\begingroup$ Ok, thanks! A search for Van't Hoff equilibrium box did turn up something substantial, such as this: eoht.info/page/Van%27t+Hoff+equilibrium+box . I suppose that, in order to separate the products, the pistons must completely evacuate the reaction chamber. $\endgroup$ – Buck Thorn Feb 15 '20 at 16:44
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    $\begingroup$ No. Please see my answer. The reaction mixture in the chamber does not change. The products are removed at the same rate that the reactants are introduced. $\endgroup$ – Chet Miller Feb 15 '20 at 16:46

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