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The below longish quote is from Rosenberg's Some Aspects on Brønsted's Energetic Theory. It concerns two chemical species, $A$ and $B$, in a stationary flow enclosed by a pair of semi-permeable membranes.

As a definite and simple example we shall consider a tube containing a liquid mixture of uniform temperature and pressure, and consisting of two compounds $A$ and $B$. A stationary state is maintained by continuous supply of $A$ at one end of the tube and withdrawal of it at the same rate at the other end. If such a process has been going on for some time, the system will presumably attain a state of a certain stability compared to other states which comply with the same conditions of mass transport and of temperature and pressure homogeneity. In the stationary state the system will contain gradients in chemical potential for the diffusing as well as for the non-diffusing component, and the latter must have the same tendency to be transported in all directions whether it be along, against or orthogonal to the gradient in its potential. If the supply and withdrawal of $A$ is stopped, all gradients will disappear, the free energy of the system decreasing to a minimum characteristic of the stability of the equilibrium state.


The maintenance of a certain stationary state will be accompanied by the same external 'work effect', i.e. the same loss of work per unit time in the surroundings, irrespective of the nature of the external phenomena attached to the maintenance of stationarity.

For the sake of illustration we shall apply the principle to the abovementioned system: $s_A$ is the amount of matter flowing through any cross section of the system in unit time, and $\mu_{A1}$ and $\mu_{A2}$ are the chemical potentials of A at the two end points I and II of the tube. The loss of work in unit time, $w_A$ is then: $$w_A=(\mu_{A1}-\mu_{A2})s_A$$ If now the two end walls of the tube are arranged to be permeable to B and not to A, it will be possible to establish the same stationary state in the tube as before by letting B flow in the direction opposite to that of A's flow in the first case. Then, the loss of work in unit time is:$$w_B=(\mu_{B2}-\mu_{B1})s_B$$ which, according to the above principle, will equal $w_A$. This loss of work is an entity which is measurable by the changes in the surroundings. Brønsted, rather jestingly, mentioned this principle as the fourth law of energetics to indicate that according to his view it could not be derived from the other laws. enter image description here My questions are (1) if and when it is true that $w_A=w_B$, (2) has this been confirmed experimentally, and (3) if yes, how can it be derived from other thermodynamic principles?

Rosenberg, T. Some Aspects on Brønsted's Energetic Theory. Acta Chem. Scand. 1949, 3, 1215-1219. DOI 10.3891/acta.chem.scand.03-1215 (open access).

  • $\begingroup$ The condition $w_A=w_B$ is equivalent to $d G=0$, which is Eq (1) in Rosenberg. This system as described is in steady state, not in equilibrium, so it is unclear whether one can appeal to Gibbs minimization. Moreover, without a schematic to fix ideas it is difficult to visualize the problem. $\endgroup$
    – Themis
    Mar 25 at 12:09
  • $\begingroup$ @Themis Thank you, I will make a schematic shortly. I do not understand how the $dG=0$ could be equivalent to $w_A=w_B$ without some other implicit assumptions, for example, a kind of a "relativity" or "reciprocity" principle is needed such as it does not matter for the work done (lost) whether the species $A$ moves to the "left" or species "B" moves to the "right" relative to the container between the semi-permeable membranes. Also the $dG=0$ equation itself is a statement between two equilibrium states not describing steady flow as Bronsted's claim that his equation $w_A=w_B$ does. $\endgroup$
    – hyportnex
    Mar 25 at 13:13
  • $\begingroup$ I was not quite right. Looking more carefully at the condition $w_A=w_B$: with $w_A=(\mu_{A1}-\mu_{A2})s_A = -d\mu_A s_A$ and $w_B=(\mu_{B2}-\mu_{B1})s_B = + d\mu_B s_B $, equating the two we get $s_A d\mu_A + s_B d\mu_B = 0$. This is the Gibbs-Duhem equation (set $s_A=dn_A/dt$, $s_B=dn_B/dt$). Still, not sure why we are applying equilibrium conditions in this non equilibrium problem. (Notice that Rosenberg's equation for $w_B$ is the negative of what you wrote.) $\endgroup$
    – Themis
    Mar 25 at 13:54
  • $\begingroup$ @Themis Thanks for noticing the index error. I have also attempted a drawing, please see above. I think Bronsted's implicit intent was to extend the Gibbs-Duhem equation from being applicable to infinitesimal potential differences, and thus reversible changes, to finite potential differences maintained by eternal constraints producing a stationary and explicitly non-equilibrium process. In this "4th law" the difference between the chemical potentials $\mu_{A1}$ and $\mu_{A2}$ is finite, not an infinitesimal as would be in Gibbs-Duhem, and is maintained by the external constraint in the flow. $\endgroup$
    – hyportnex
    Mar 26 at 1:37
  • $\begingroup$ @Themis Also, in Bronsted's view, the Gibbs-Duhem is not really about an actual process but rather the relationship between two neighboring equilibrium points whose internal energy is a 1st order Euler homogeneous function. His work equation (work axiom), $\sum_k (P_{k1}-P_{k2})\delta K_k=0$ ($K_k$ are the conserved quantities moving between potentials $P_{k1}$ and $P_{k2}$), is a process equation from which Gibbs-Duhem follows trivially but it is still restricted to reversible processes. Bronsted only assumes that the equilibrium potential $\mu$ is still meaningful in the steady state flow. $\endgroup$
    – hyportnex
    Mar 26 at 1:49


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