$\Delta U = Q + W$, where the signs have been assumed to be positive (work is done on the system and heat is absorbed). But, also, $Q$ can be written as $Q=\Delta U + W$, since $W=P \Delta V$.
So is $Q=\Delta H$?
This somehow doesn't feel right. So, I want to know, if I am wrong (I believe so) where I have gone wrong?
The definition of enthalpy $H=U+PV$ can be compared to a statement of the first law at constant pressure $Q = \Delta U + P\Delta V$ with $\Delta H = Q$.
It is not a good idea to use @Q and @W because you cannot have a final amount of heat take away an initial amount of heat and similarly for work done. Just use Q and W for heat into system and work done by system. I have used this formulation as this is a Physics forum and prefer it to the convention used by Chemists.
$H=U+PV$ is a state function on the $P$-$V$ diagram because $U$, $P$, and $V$ are state functions. $Q$ is not a state function.
Using $\Delta H=\Delta U+\Delta (PV)$ and $\Delta U=Q-W$ (where $Q$ is the heat added to the system and $W$ is the work done by the system) implies that: $$ \Delta H = Q-W +\Delta (PV) $$
In addition, using $W=P\Delta V$ implies that: $$ \Delta H=Q-P\Delta V +\Delta (PV) $$
Therefore,
$\Delta H=Q$ when pressure is constant.
or
$\Delta H=Q_p$
where $Q_p$ means heat added at constant pressure.
And pressure is very close to constant over many chemical processes.
The first law of thermodynamics is often confusing, because of the "normal" designation of work being positive if the system is doing work on the environment, and heat transfer being positive if heat is being added to the system. This confusion can be easily fixed, as noted below.
All terms in the first law, (e.g., changes in internal energy, heat, and work) are describing energy transfers. In my opinion, all energy is equivalent, so work or heat added to a system raises its internal energy, and such energy transfers should have a positive sign associated with them. Likewise, all forms of energy leaving a system should have a negative sign associated with them, because such energy transfers decrease the internal energy of the system.
In addition to the signs noted above, real world processes are highly variable. Since the first law of thermodynamics is essentially an energy balance, this means that the formulation of the first law should correspond to the process at hand. Thus, if heat is entering a system and work is leaving a system, the "first law" equation for this process would look a bit different than the "first law" equation for a process in which both heat and work were entering a system, and heat was leaving a system. This concept becomes even more important when one notes that some text books designate work done on the environment as positive while other textbooks designate such work as negative.
Regarding the original poster's question, and following the sign conventions noted above, the enthalpy change of a system CAN be equal to the heat entering the system, if no work is done on the environment and the system does not transfer heat to the environment. Usually, this is not the case, but the answer does indeed depend on the particular system that is involved.
The important point is that the first law is an equation considering the total change between two different states . First this process that you describe is a quasistatic one with constant presure and work done by the system changing is volume; second, like the internal energy $\Delta U$, the change of enthalpy $\Delta H$ is a state variable that describe the system (what happens with the energy of atoms, molecules, etc. at the microscale, if you wish), not the process; third, the work done and heat added are process variables, what happen between the environment and the system. Here, the first law only is a numerical equivalence, not a physical equivalence between them, and you can't give a definition of enthalpy based in the heat added or removed from a system in all the cases.
The point you're missing is that $$H=U+PV$$ means $$\Delta H=\Delta U+P\Delta V+\color{red}{V\Delta P}$$ Only at constant pressure, you have $\Delta P=0$ and so $$\Delta H=Q\quad (\text{at constant pressure})$$ or $$\Delta H=Q_p$$
Hope this helps. Ask anything if not clear :)
