There is a paradox - this group is "thermodynamic" by definition and yet
it is reluctant to discuss thermodynamic problems. Perhaps the way I set
them some time ago was not perfect, but was that important? Does
biothermodynamics contain no problems? Why has nobody else started
a more fruitful discussion?
Let me try again - I hope I can be more convincing now.
The most basic definition of chemical (useful) work is given by the first
law of thermodynamics:
delta H = Qabs - Wu /1/
where delta H is the enthalpy of a chemical reaction, Qabs is the heat
absorbed and Wu is the chemical work extracted from the reaction.
Let us consider the EXOTHERMIC reaction
A -> B /2/
If it is not coupled to another reaction, conversion of one mole A into B
would just release heat equal to -(delta H). However if /2/ is coupled to
another reaction, some of the energy released (-delta H) could be transfered
to the other chemical system in the form of Wu, and the rest would still be
dissipated as heat (Qabs, which is negative in this case). This is what /1/
says: the energy released by one mole of A is either entirely dissipated as
heat (delta H) or, if the reaction is coupled, some of it is transfered as
chemical work (Wu).
Please let me know if this balance sounds plausible. If yes, we would go
further to show that the other definition of chemical work:
-Wu = delta G = delta Go + RTln((B)/(A)) /3/
is incompatible with it. What I was not able to explain before is that
/3/ is in principle correct but it corresponds to a different PHYSICAL
course of the reaction - in the first case Wu comes from the
INTRAMOLECULAR energy of A, whereas /3/ corresponds to a special course
in which the intramolecular energy does not matter - heat is absorbed from
the environment and converted into Wu.
For the ATP system, /3/ remains a definition of potential useful work done
under reversible conditions, but only if this special course of the
reaction takes place. It never does. The implications are incredible.
Best regards,
Pentcho Valev
