@MetalStorm I should emphasize the zero non-PV work requirement . If non-PV work is nonzero, i.e, $\delta w_{non-PV} \ne 0$, then $dU = \delta q - p_{ext}dV + \delta w_{non-PV}$ and isobaric $dH = \delta q + \delta w_{non-PV} \ne \delta q$. So, my claim that isobaric $\Delta H_{irr} = Q$ requires that non-PV work is zero, just like as in the "irreversible heat-bath" example.

@ChetMiller I think MetalStorm posted a question in the last comment you replied to, so I'm not sure which claim you are agreeing with. Are you saying $\Delta H = Q$ is only true for reversible isobaric processes? (this is what MetalStorm is claiming) My claim is that $\Delta H = Q$ can also be true for irreversible isobaric processes as well provided that "non-PV work is zero" (i.e, that all "work" is PV-only!)

Yes, of course. We are only considering the "PV work" in $w$.

If you can move the piston (day using a control system with feedback (such that the force per unit area on the inside face of the piston is the same as initially throughout the change) and a very large heat flux is imposed (such the the process is irreversible), then $\Delta H$ will equal Q (provided the final state is a thermodynamic equilibrium state). Of course, it is very difficult to control the force on the inside face of the piston in this throughout an irreversible process like this.