Guy Brown wrote:
I do not think that delta G was ever meant to be traditional work, only it<could
predict the maximal amount of work extractable from a system in certain <
circumstances. <
"Traditional" was not the right term. The work meant is that appearing in
delta G = delta H - T(delta S) = Q - Wu - T(delta S) = -Wu /1/
where Wu is the useful work done by the system, this work is done reversibly,
and all quantities refer to the ATP-ADP-Pi system (not to the system on which
work is done - its delta G is equal but with an opposite sign).
So this work is quite real - the "certain circumstances" consist only in
reversibility of the process - any time the coupled reaction is close to
equilibrium, this maximum work should be expected.
Unfortunately this simple result has quite a misleading expression in
textbooks - it is stated there that delta G is the maximum work that CAN be
done, and this "CAN" is sometimes not explained. So one is lead to think that
CAN means "only in the presence of a special mechanism" - I have seen this
explicitly stated in biochemistry textbooks - I do not remember - maybe
Lehninger.
There is a very simple way to solve the problem with delta G. Were you
never impressed by the fact that the fundamental result
delta G = delta Go + RTln((ADP)(Pi)/(ATP)) /2/
is never accompanied by a physical picture of the process to which this
delta G corresponds? After all, delta G should always describe a real process;
moreover, if this process is reversible, delta G describes also the work done
in it; and yet in no textbook any example is given. Let us do this - describe
a reversible process in which the work done is given by /2/. This will
clarify the whole picture.
Let us assume that ATP, ADP and Pi are ideal solutes. Then let us transfer
reversibly one mole ATP from the original system to a system where the reaction
ATP <-> ADP + Pi is at equilibrium. E.g. the mole is evaporated, expanded
and liquified into the system at equilibrium. Calculations show that the
works at vaporization and liquification cancel, so what remains is the work
of expansion:
Wexp = RTln(ATP) - RTln(ATP)eq /3/
where (ATP) and (ATP)eq are the concentrations of ATP in the original system
and the system at equilibrium respectively. Note that this is PV work, but
it can enter legitimately into delta G since V is not the volume of the
original system (only PV work affecting the volume of the ORIGINAL system does
not enter into delta G). Note also that this work is done AT THE EXPENSE OF
HEAT ABSORBED FROM THE ENVIRONMENT, i.e. it has nothing to do with the
internal energy of the ATP molecule.
In the system at equilibrium, the mole is reversible transformed into
products. At this stage the internal energy is released as heat, i.e. the
process presupposes work done at the expense of heat absorbed from the
environment but dissipation of the internal energy. This is very suspicious -
in a standard chemical reaction we expect just this energy to do the work.
The last stage consists in reversible transpotation of the products (one
mole of ADP and one mole of Pi) from the system at equilibrium to the
original system. Works analogous to /3/ are done.
The sum of all works amounts to the negative of /2/, as should be expected.
But note that this is work done at the expense of heat absorbed from the
environment, whereas the internal energy (the enthalpy) has obviously nothing
to do with this work and is lost (in this particular process) as heat released.
On the contrary, it is this internal energy (the enthalpy) which does the
work in a chemical reaction, whereas the work defining /2/ is simply absent.
The two works cannot be equal - I have not proved this, but have just presented
an example allowing everyone to find a solution in physical (not abstract)
terms.
The "enthaply concept" does not predict that the net reaction will be <
exothermic, <
this depends on how much electrical work is done. A reaction can be more than
100% efficient if we define efficiency as delta H (driven reaction)/-delta <
H (driving <
reaction), indeed it can be infinite. Of course it can not be greater than 100%
efficient if we define efficiency as delta G (driven reaction)/-delta G <
(driving reaction). <
<
We can see this more clearly if the driving reaction has no enthalpy <
change, e.g. <
an isomerisation reaction A > B where A and B have the same internal energy <
(and the same standard molar entropy). <
If this reaction is coupled to the pumping of an ion against an electrical <
gradient, <
then if [A] > [B], ions can be pumped up hill doing work. The enthaply <
efficiency <
of this net reaction is infinite as the delta H of the driving reaction is <
zero, but <
work is done on the driven reaction (transport of the ion). The reaction <
is driven <
forward by the concentration difference of A to B. <
I would be glad to analyse this example. However let us make it real -
the reaction is not isomerization, but movement of H+ down the transmembrane
pH gradient, in the absence of a tansmembrane electrical gradient:
H+out -> H+in /4/
This reaction has the properties you refer to, and it synthesized ATP. How?
It seems that the concentration difference is the only reasonable "force",
but this is not the case. Let me explain this in the next message - this one
has already become too long.
Pentcho
