If you are still having a problem with enthalpy vs. internal energy, please read:
ΔE = q + w (ΔE is change in internal energy) eqn (1)
Enthalpy (H) accounts for heat flow in processes occurring at constant pressure when no forms of work are performed other than P-V work.
H = E + PV eqn (2)
H is a state function because E, P, and V are all state functions.
Δ H = Δ (E + P V) eqn (3)
When a change occurs at constant pressure,
Δ H = Δ E + P Δ V (notice that V Δ P = 0, since Δ P = 0 at constant P), eqn (4)
Recalling ΔE = q + w, eqn (1),
The work involved in the expansion or compression of gases is w = – P Δ V,
Substitute –w for P Δ V and q + w for ΔE in eqn (4): Δ H = Δ E + P Δ V = (qp + w) – w = qp , eqn (5)
Δ H = qp, eqn (6)
qp is the heat flow in the process at constant pressure.
Therefore, Δ H (change in enthalpy) equals the heat gained or lost at constant pressure.
Because we can either measure or readily calculate qp, and because so many physical and chemical changes of interest to us occur at constant pressure, enthalpy H is a more useful function than internal energy E.
For most reactions the difference in Δ H and Δ E is small because P Δ V is small.
Last note: relationship between Δ H and heat qp has specific limitations that only P-V work is involved and the pressure is constant.
