What is enthalpy theoretically? Why must ΔH equal q at constant P?

[MonDie]

First, I will give my understanding of enthalpy so far, and you can correct and/or advance my understanding where it is fitting to do so.

H = E + PV

A system's enthalpy (H) is its kinetic and potential energy (E) plus the energy (work?) required for it to displace its surroundings, exressed as pressure multiplied by volume (PV).

It seems the canonical example of something displacing its surroundings (via pressure-volume work) is combustion pushing a piston up through a cylinder. Work is force acting through a distance (w = F · D). The heavier the piston is (i.e. the more pressure it exerts), the more force will be needed to move it upward a given distance. Hypothetically, if force (F) never went above zero, there would never be any work to displace the piston, thus the gas couldn't occupy any volume of space beneath the piston. In combustion, potential energy is released from the molecules, and the energy either becomes heat or performs work. More work will mean a greater distance, thus a larger volume of space occupied beneath the piston.

It is said that the enthalpy of a system is not necessarily contained within the system. Does that mean the PV in the equation actually represents the potential energy of the displaced surroundings (e.g. the raised piston)?

At constant pressure, enthalpy change is equal to heat (q), where heat is the exchange of thermal energy. I don't see how that is consistent with the above concepts or the following equation.

ΔH = ΔE + P · ΔV

If neither pressure nor volume change, we would be left with ΔH = ΔE. I know ΔE = q + w, so ΔE will be equal to heat (q) as long as the system isn't doing or recieving any work (w). Thus, ΔH should equal q only when P, V, and w are all constant. However, every source says you only need to keep pressure constant to measure ΔH. My textbook even points out specifically that a bomb calorimeter keeps volume constant and measures ΔE, while the coffe-cup calorimeter keeps pressure constant and measures ΔH. How isn't this nonsense?!

Also, how does a coffee-calorimeter keep the pressure constant without releasing any heat? A rising temperature will cause more water to evaporate. If you hold the water vapor inside, you will increase the pressure. If you release the water vapor, your releasing some of the heat formed by the reaction.

Are these effects insignificant if you only add a miniscule amount of the limiting reactant? Is there not enough time for the water vapor concentration to increase significantly?

Edited April 22, 2013 by Mondays Assignment: Die

[Zeppos10]

first,I think it is better to define H=U+pV, where U is the internal energy.

second, for reasons I cannot explain, there is a habit in thermodynamics to NOT distinguish the inside pressure (notably of gases) from the pressure of the environment (notably the atmosphere).

It can be argued that the p in (U+pV) must be the pressure of the environment, in line with the idea of the displaced surroundings mentioned above.

The whole thing is worked out in:

Mannaerts SH, Energy-balance of the Joule-Thomson experiment: Enthalpy change at decompression. npt-procestechnologie. 2010; 17(4)18-22. (This paper can be retrieved through researchgate.)