Relationship between thermodynamic reversibility and reactions
The process does not have to be considered reversible, and, if a reaction is occurring in the system, it is, of course, not reversible.  We are trying to find the change in enthalpy between the following two states:

State 1:  Reactants at temperature $T_0$ and pressure $P_0$

State 2:  Products at temperature $T_0$ and pressure $P_0$

We would like to measure the $\Delta H$ in such a way that, for the process path we select, all we need to do is measure the amount of heat transferred to the system Q.  The process path we select involves keeping the system in contact with a constant temperature bath at $T_0$ and controlling the motion of the portion of the system boundary at which the volume change occurs in such a way that the force per unit area is always constant and equal to $P_0$.  Since, in general, the work done by the system on its surroundings is determined by the integral of $dW=P_{ext}dV$ (where $P_{ext}$ is the force per unit area acting on the portion of the boundary at which the volume change is occurring), we have selected $P_{ext}$ such that $P_{ext}=P_0$; and the total cumulative work done by the system when it reaches its final equilibrium state will be just $W=P_0\Delta V$.

From the first law, it then follows that, for this irreversible path:
$\Delta U=Q-P_0\Delta V$.  So $\Delta U+P_0\Delta V=\Delta H =Q$