Can ATP do work at equilibrium? (Pentcho Valev)

Bryant Fujimoto fujimoto at u.washington.edu
Tue Nov 4 01:56:18 EST 1997


<RUMYM at BGEARN.ACAD.BG> writes:

>Concerning the reaction far from equilibrium

>   ATP + B = ADP + P + C                                    /1/

>Bryant Fujimoto wrote:
>>So what is wrong with Delta G? That is, what is there you would want to<
>>know about the system that can't be described by thermodynamics or     <
>>statistical mechanics (at least in principle).                         <

I'm not exactly sure what the best way to break this to you, but when
you say 

>There is nothing wrong with delta G so far as it shows the direction of
>the reaction. 

you are accepting the second law of thermodynamics.  The above statement
(at constant T and P) is a direct consequence of the second law.
Furthermore, if you assume the properties of the Gibbs free energy 
(instead of deriving them) you can go backwards and derive the second
law.  That is, showing that delta G correctly predicts the direction of
a reaction is a way of checking the second law.

>But I would like to know how much work the system ATP-ADP-P
>does on the system B-C. Would you tell me how to find it (at least in
>principle)? This is not just a curiosity - after all, the second law is a
>statement about the work a system does, so in order to verify the law we
>need to know how much this work is.

Well, Delta G is sometimes called the non-pV-work (at constant T and p).
That is, for a reversible process at constant T and p, it is the total
work minus the part due to any change in the volume of the system. So
any "chemical work" is in there.  

>   I would also like to know how much work the system ATP-ADP-P  does on
>the system B-C when the reaction /1/ is at equilibrium. You would probably
>answer that this work is given by delta G of the reaction  ATP = ADP + P,

-(Delta G) is the maximum non-pV-work which can be done by a system on its
surroundings.  Its only equal to the maximum for a reversible reaction.
If you think of the ATP<->ADP+P reaction as one system, and the B<->C
reaction as another, then -(Delta G) is the maximun amount of work available
for the B<->C reaction, how much is actually done depends on the details
of the reactions.

>but I would paraphrase this answer like this: "The work done is as much as
>the second law says it is". 

You really ought to learn what thermodynamics actually says.

>Moreover, you probably remember that, long time
>ago, Guy Brown declared in the biothermokinetics group that delta G is not
>an energy - some were rather surprised, but then everybody thrust their
>heads in the sand and life became calm again.

Didn't see this. In any event Delta G is not an energy, its a free energy.
If he meant anything else by that statement, then he's wrong.  After all,
G = U + pV - TS, so it must be some sort of energy.

>   So the experiment I proposed aimed at verifying whether or not reversible
>work done by ATP is really equal to (the negative of) delta G. Theoretically
>this is simple - as an ATPase transports, almost reversibly, H+ against an
>electrical gradient, I expect the work done by one mole ATP to be not
>greater than the internal energy (enthalpy) of one mole ATP, i.e. 20 KJ/mole.
>Thermodynamics predicts that the work done can be much greater - e.g.
>50 KJ/mole. This means that, as one mole ATP does work, it uses not only
>its internal energy but also absorbs 30 KJ/mole heat from the environment
>and converts it into work too. It is just this heat absorption that bothers
>me - on one hand, its value must depend on the concentrations of the
>reagents, i.e. it is a statistical phenomenon; on the other, the work
>production consists of one mole independent chemical acts so no  such
>statistics is physically plausible.

The source of the additional energy is in the difference between the
concentrations of ATP and ADP.  It is not absorbed from the environment.

>   I do not think that the second law has been confirmed many times in
>many different experiments. This is simply impossible - in most cases
>the chemical work is not at all defined. The only confirmation I know of
>are galvanic reactions( a very interesting topic which also was found
>uninteresting in the btk group). If you know of other experiments, please
>describe them. If they are more interesting than those which I propose,
>we will discuss them.

See what I wrote at the begining of this post.

>   There is another "uninteresting" way allowing one to verify the second
>law - by analysing the countless implications. For instance, a derivation
>based on the second law leads to to an obviously correct result - the
>osmotic equation. If, as I believe, the second law is not valid, how is
>this possible? I would be happy to discuss such problems, bur have a very
>sad experience in the btk group. (You probably remember also that I found
>a mistake invalidating Prigogine's theorem of minimum entropy production
>at steady state, someone confirmed my conclusion, but then again heads were
>in the sand and my distorted psychology became the most interesting
>problem.)

Pentcho
  You are not owed a response by anyone for your posts.   I remember one
of your posts on Prigogine's work,  I didn't bother to post a correction
because, quite frankly at the time I was getting tired of correcting you.
I obviously cannot prevent you from interpreting a lack of response any
way that pleases you. However, you might want to consider the possiblity
that people are not always willing to take the time to correct you.  
After all, your "head in the sand" comments and others you have posted
make it appear that you feel you are owed a response by others.  It 
appears that you feel that others have an obligation to spend their time
considering your ideas even if they don't think them interesting.

Regards,
Bryant
fujimoto at u.washington.edu



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