Is the fact that enthalpy equals the heat of reaction at constant pressure only a consequence of the fact that pressure-volume work is expressed as PV only at constant pressure?

In a reaction where PV work is the only kind possible:

Change in H = Change in U (internal energy) + PV

Change in U = q (heat) - w(work) = q - PV (constant pressure is inherently assumed here)

So:

Change in H = q - PV + PV = q

If the pressure-volume work occurs at a non-constant pressure, then it is defined as Integral(PdV). (If the pressure is constant, this simplifies to PV). So technically couldn't you write:

Change in H = q - Integral(PdV) + Intergral(PdV) = q

But I guess if you previously define change in U as q - PV, then it is *already assumed* that the PV is done at a constant pressure (since its not written with an integral), and consequently enthalpy is defined in terms of constant pressure as well.

So my question is really, is it just mathematical convenience that enthalpy = q needs to occur at constant pressure, or is there actually some physical phenomenon that necessitates this definition? Sorry if this is hard to understand. Thanks for any responses.