January 06, 2025, 08:35:15 PM
Forum Rules: Read This Before Posting


Topic: Enthalpy - clarification  (Read 3916 times)

0 Members and 1 Guest are viewing this topic.

Offline Schrödinger

  • Chemist
  • Sr. Member
  • *
  • Posts: 1162
  • Mole Snacks: +138/-98
  • Gender: Male
Enthalpy - clarification
« on: April 04, 2010, 02:14:11 PM »
Hey guys

I need some help understanding the concept of enthalpy.
I read that Enthalpy is defined as the heat flow (in or out of the system) during a process at constant pressure.

I also read the following in wikipedia :
As a differential expression, the value of H can be defined as[3]



where

    δ represents the inexact differential,
    U is the internal energy,
    δQ = TdS is the energy added by heating during a reversible process,
    δW = pdV is the work done by the system in a reversible process,
    dS is the increase in entropy (joules per kelvin),
    p is the constant pressure,
    dV is an infinitesimal volume, and
    T is the temperature (kelvins).

But why is this correct?
Remembering that enthalpy is defined at constant pressure, dp=0. So why is the Vdp term still present in the above equation?
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
- William Jennings Bryan

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: Enthalpy - clarification
« Reply #1 on: April 04, 2010, 06:11:37 PM »
Enthalpy is not defined as the amount of heat flow in/out of a system at constant pressure.  Enthalpy is defined as U + pV.  The fact that ΔH = q for a process at constant pressure is just a useful corollary to this definition.

Offline Schrödinger

  • Chemist
  • Sr. Member
  • *
  • Posts: 1162
  • Mole Snacks: +138/-98
  • Gender: Male
Re: Enthalpy - clarification
« Reply #2 on: April 05, 2010, 12:49:28 AM »
You said $$ \Delta H /$$ = U + pV, without any other specifications.
But isn't this just q for a process? How is enthalpy different from q then?
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
- William Jennings Bryan

Offline Schrödinger

  • Chemist
  • Sr. Member
  • *
  • Posts: 1162
  • Mole Snacks: +138/-98
  • Gender: Male
Re: Enthalpy - clarification
« Reply #3 on: April 05, 2010, 03:23:18 AM »
I'm sorry. I just found out that I completely misread the definition.  :-[
I have understood that $$ \Delta H = q_{p} /$$  is just a way to calculate $$ \Delta{H} /$$, but is not defined at constant pressure.

Thanks for helping me out
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
- William Jennings Bryan

Offline Juan R.

  • Chemist
  • Full Member
  • *
  • Posts: 148
  • Mole Snacks: +24/-3
  • Gender: Male
    • The Center for CANONICAL |SCIENCE)
Re: Enthalpy - clarification
« Reply #4 on: April 05, 2010, 08:34:01 AM »
For a simple system U = U(S,V,N) but sometimes we are interested in pressure p as variable. Then we want to introduce a new function enthalpy H = H(S,p,N).

The general procedure to change a variable (e.g. V) by its conjugate (e.g. p) is achieved using a Legendre transformation. For the change V--> p this is defined as

H == U + pV

Some texts give a geometrical interpretation of the transformation but I prefer a more direct form to understand it. Just notice that we want to eliminate a term pdV from dU and add a new term Vdp, then we add to U the term pV because

dH = dU + pdV + Vdp

dH = (TdS + mudN -pdV) + pdV + Vdp

dH = TdS + mudN + Vdp

That is we obtained the function H = H(S,p,N)

The advantage of H is that for constant p

dH = DQ

and thus changes in enthalpy are directly related to heat changes of processes at constant p. For instance, chemical reactions under constant atmospheric pressure.
The first canonical scientist.

Sponsored Links