Are heat of 1st law(q) and enthalpy of reaction (h) same

Question

I have been studying some basic chapter of thermodynamics in chemistry and physics when I came across these two statements: $$\Delta Q=\Delta U+\Delta W$$ $$\Delta H=\Delta U+P\Delta V$$ So is it true $\Delta Q=\Delta H$ (since $\Delta W=P\Delta V$)

Answer 1

To supplement Aaron's comment, we can always write Q=ΔU+W, where Q is heat transfer to the system, U is the internal energy of the system, and W is the work done by the system. (It doesn't make much sense to add Δ to Q or W because they're not state functions whose differences are relevant.)

We can also always write H=U+PV for all states by definition, where H is the enthalpy. If the pressure is constant, then we can integrate the differential form dH=dU+PdV+VdP=dU+PdV between two states to obtain ΔH=ΔU+PΔV. In such a case, the heat transfer Q into the system indeed corresponds to ΔH.