It explicitly says only for the mechanically resversible

What we can do is the following. Bring me a book that states clearly that the equation is valid for irreversible process, and we can discuss further.

Up to now, we have one source directly state that what you say is not correct, and this book is pretty much part of the syllabus of a lot of chemical engineering degrees across the world. Thus, it is a good source also.

Ping at me when you have a book that says the contrary so we can discuss further

@MetalStorm But that book only talks about $P_{int}$, and $P_{int}$ is irrelevant. So what book says is irrelevant to my claim. It seems you have forgotten to distinguish between $P_{int}$ and $P_{ext}$ again. So therefore, that book does not refute my claim and in perfect agreement with $\Delta H = Q$ for an irreversible isobaric process because you can ignore $P_{int}$.

@MetalStorm Also does that mean you disagree with Chet Miller? Because his last message perfectly agrees with my claim.

@MetalStorm In short, what your book states is only true if we use $P_{int}$. But am I using $P_{ext}$ and ignoring $P_{int}$, so there is no contradiction between my claim and what the book says, since $P_{int}$ is irrelevant.

@MetalStorm Also, I think you have to reread your own book very closely again too. What the book says is for a "mechanically reversible, constant-pressure process", meaning that $P_{int} = P_{ext} = $ constant throughout the whole process. It is possible for an irreversible isobaric process to be "mechanically reversible" as well.

@MetalStorm So, a "mechanically reversible but thermally irreversible" isobaric process can easily have $\Delta H = Q$. Just like my example of the the piston earlier, without any contradiction from your book. This means an irreversible isobaric process can have $\Delta H = Q$, since your book requires only "mechanical" but "not thermal" reversibility, but a "fully reversible" has to have "both reversibilities"...

So, your book is in perfect agreement with me. An irreversible process that is "mechanically reversible" and at constant pressure can certainly have $\Delta H = Q$. Just a matter of reading closely -- your book requires it to be "only mechanically" reversible, but the process can still be "thermally irreversible" and therefore still an irreversible process overall.

Your book is written very carefully, so you cannot assume "mechanically reversible" means "also completely reversible in every possible way", because the latter would have to add "also thermally reversible" as well...

@MetalStorm So I guess your whole confusion was simply assuming that "mechanically reversible" means a process is "wholly reversible", when it does not. It is possible for an irreversible process where $\Delta S_{univ} > 0$ to also be "mechanically reversible" the whole entire time, just like in Chet's example.

But, it is still an "overall irreversible" process because $\Delta S_{univ} > 0$...

@MetalStorm Also, "mechanically reversible" also seems to follow from the requirement that "non-PV work is zero". If all work is PV-only, then we are obviously in a "mechanically reversible" process anyway since "non-PV work equals zero" means that the piston is also massless and friction-less as well, which already implies $P_{int} = P_{ext}$ aka "mechanical reversibility".

@MetalStorm So therefore, an irreversible isobaric process "with zero non-PV work" will always have $\Delta H = Q$ because "zero non-PV work" always means "mechanically reversible" anyway just like your book requires. Which means my claim is correct. A process can be "mechanically reversible" but still "overall irreversible" because it instead can just be thermally irreversible instead.

@MetalStorm So, in summary --> this is exactly what I am claiming here : chat.stackexchange.com/transcript/message/65169631#65169631 and your book is in full agreement with my statement too because "zero non-PV work" automatically means "mechanically reversible" even for an "irreversible isobaric process" where $\Delta S_{univ} > 0$ anyway...

@ChetMiller It seems that the book linked by @MetalStorm only requires a "mechanically reversible" isobaric process for $\Delta H = Q$. But, it is possible for even an "irreversible" isobaric process to still be "mechanically reversible" (have $P_{int} = P_{ext}$ throughout the whole process) as your earlier example clearly demonstrates. Does this mean that my claim that $\Delta H_{irr} = Q$ for irreversible isobaric processes with zero non-PV work is true?

@ChetMiller (I'm asking for clarification as I believe MetalStorm has "misinterpreted" your example, what you wrote earlier.)

@ManRow Hello! You are asking a different question. This discussion initiated because you claimed that $\Delta H = Q$ regardless of reversibility. I link it here

It seems to me that now you moved the goalpost, or being just plain dishonest. Do you accept the need for the process to be mechanically reversible? If you do, then why didn't you agree with the book like 10 days ago?

@MetalStorm But how I am being dishonest? My claim this whole entire time is that it is possible for an irreversible, isobaric process to have $\Delta H = Q$. That is what it has always been from the beginning. Does this mean you finally agree with me?

An "irreversible" process has only one condition -- that $\Delta S_{sys} + \Delta S_{surr} = \Delta S_{univ} > 0$. That is, it increases the total entropy of the universe. There are many ways to do irreversible processes.

An "isobaric" only has only one condition -- that $P_{ext}$ is constant throughout the whole process, regardless of reversibility.

So, when you have something where both $\Delta S_{univ} > 0$ and $P_{ext}$ is constant, then you have an irreversible isobaric process. Correct? My claim is that it is possible for an irreversible isobaric process to have $\Delta H = Q$.

In chat.stackexchange.com/transcript/message/65169603#65169603 I clearly emphasized and explicitly clarified that this requires that "non-PV work is zero".

So, does this mean you finally agree with me then? That an irreversible, isobaric process will have $\Delta H = Q$ if there is no non-PV work? Remember, the process is still both "irreversible" and "isobaric", even if non-PV work is zero! Because $\Delta S_{univ} > 0$ and $P_{ext}$ is constant the whole entire time.