How does enthalpy help us to find the state of equilibrium? For an adiabatic system with constant external (generalized) forces, in a state out of equilibrium we have:
$$\delta H\leq 0 $$
If I understand correctly, derivation of this inequality does not rely on the second law of thermodynamics, which states how systems evolve toward equilibrium. Here is the derivation:
Having constant external forces $\mathbf{J}$, the work input to the system is $\delta W \leq \mathbf{J}.\delta \mathbf{x}$  , where $\mathbf{J}$ and $\mathbf{x}$ are generalized forces and displacements respectively. Equality is achieved for a quasi-static change with $\mathbf{J_{internal}}=\mathbf{J}$, but generally there is some loss of the external work to dissipation.
Now the first law gives $\delta E\leq \mathbf{J}.\delta \mathbf{x}$, or $\delta H\leq 0$, where $H=E-\mathbf{J}.\mathbf{x}$  .
My question is what is the underlying physical principle behind this inequality, and how can we deduce the direction of evolution toward equilibrium without drawing on the second law?
 A: I've been thinking more about it, and I think I can finally see a reason for his inequality, as well as its connection to the second law (which I believe is behind every approach to equilibrium). I'll work the example of pressure, but it can be trivially generalized to a sum of generalized works.
First, the reason for his inequality, I think, is that out of equilibrium (which is the equilibrium of interest for enthalpy, as there's no heat transfer), the pressure in our system need not be equal to the pressure of the outside reservoir (which is also assumed constant). Therefore, let us first define the actual work done on the system:
$$\delta W = - p_{sys} dV_{sys}$$
Where $p_{sys}$ is the pressure inside our system. Now, an important point is that as there's no transfer of heat, for spontaneous processes the entropy must increase (or stay constant) both for the system and for the reservoir, separately. Let us see, in particular, what happens with the entropy of the reservoir, $S_{R}$. If we take the reservoir and the system as making up our microcanonical ensemble, with a given fixed energy and volume, $E_{tot}$ and $V_{tot}$, then 
$$S_{R}(E_{R} , V_{R}) = S_{R}(E_{tot}-E_{sys}, V_{tot}-V_{sys}) \approx S_{R}(E_{tot}, V_{tot})-E_{sys} \frac{\partial S_{R}}{\partial E_{R}} \bigg|_{E_{R}=E_{tot}} -V_{sys} \frac{\partial S_{R}}{\partial V_{R}} \bigg|_{V_{R}=V_{tot}} = S_{R}(E_{tot}, V_{tot})-\frac{E_{sys}}{T_{R}}-\frac{p_{R} V_{sys}}{T_{R}}$$
Sorry if it is a bit cluttered, but I just wanted to be precise. The $R$ and $sys$ underscripts refer to the reservoir and the system, respectively. This approximation works, because the system is assumed to be much smaller than the reservoir. From the above, you can immediately derive that enthalpy should go to a minimum. But to answer your question, let's take a small change:
$$\delta S_{R}(E_{R} , V_{R})=-\frac{\delta E_{sys}}{T_{R}}-\frac{p_{R} \delta V_{sys}}{T_{R}}=-\frac{\delta Q+\delta W}{T_{R}}-\frac{p_{R} \delta V_{sys}}{T_{R}}=-\frac{p_{R} \delta V_{sys}+\delta W}{T_{R}}=\frac{(p_{sys}-p_{R})\delta V_{sys}}{T_{R}}$$
Where we have used the fact that $\delta Q=0$. Now, we know that $\delta S_{R}(E_{R} , V_{R}) \geq 0$ which implies (for non-zero and positive $T_{R}$) that $p_{R} \delta V_{sys} \geq \delta W$ Which is the inequality your derivation uses.
What we find from here, is that your inequality is indeed an expression of the second law. Furthermore, we find this implies that in this case in which we only allow the system to do work and not transfer heat, we have $(p_{sys}-p_{R})\delta V_{sys} \geq 0$, which is an expression of mechanical equilibrium (and Le Chatelier's principle): the system will decrease in volume if the pressure of the reservoir is higher, and viceversa, and equilibrium is reached when the pressures are equal (as there is where a small change in volume doesn't change entropy, so it must be at an extremum).
