From owner-biophysics@net.bio.net Sun Jan 11 22:00:00 1998
Path: biosci!daresbury!not-for-mail
From: Pentcho Valev <RUMYM@BGEARN.ACAD.BG>
Newsgroups: bionet.biophysics
Subject: Entropy change for a chemical reaction
Date: 12 Jan 1998 09:26:02 -0000
Lines: 34
Sender: lpddist@mserv1.dl.ac.uk
Distribution: bionet
Message-ID: <69cnja$2ue@mserv1.dl.ac.uk>
Original-To: biophys@dl.ac.uk

I wrote:
>For a normal closed system dS is not zero. One
>can easily imagine this in the particular case in which a solute reaction
>releases a gas, e.g. A + B <-> C + D  where D is a gas. Then we can carry out
>the reaction reversibly in both directions with the help of a piston. The
>reaction can be either endo- or exothermic, and, in addition, some work is
>done as the piston is pushed. The sum of the endothermic heat dQend and the
>work done dW gives dS:

>       dQabsorbed = dQend + dW = TdS                                  /1/

Bryant Fujimoto replied:>>>>>>>>>>>>>>>
  Where did you get this equation?  Assuming you define your system
correctly, dQabsorbed = TdS for a reversible reaction.  However, I
see no reason why they should necessarily equal dQend + dW.<<<<<<<<<

Just consider a liquid-vapour system. As it expands isothermally and
reversibly, heat is absorbed as the liquid turns into vapour (the
process is endothermic) and, in addition, work dW = PdV is done by the
vapour. dQabsorbed is to compensate for both heat-consumming processes.

Theoretically much more interesting is the case in which dQend is
negative (i.e. the chemical reaction is exothermic). Then dQend and dW
have opposite signs and we can easily imagine concentrations of the
reagents such that  dQend = -dW. Then, accoring to /1/,

            dS = 0                                           /2/

for a reversible and, therefore, for the respective SPONTANEOUS chemical
chemical process. This is even more than an absurdity and can be regarded
as a counterexample against thermodynamics.

Best regards,
Pentcho

.