Can ATP do work at equilibrium? (Pentcho Valev)

RUMYM at BGEARN.ACAD.BG RUMYM at BGEARN.ACAD.BG
Wed Nov 5 07:37:49 EST 1997


Bryant Fujimoto wrote:
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.         <

Don't be so sure. Let us consider the simplest reaction

        A <-> B                                             /1/

going from left to right, i.e.

        Vf > Vr                                            /2/

where Vf and Vr are the velocities of the forward and reverse reaction.
Therefore

      Kf(A)  >  Kr(B)                                        /3/

where Kf and Kr are the respective rate constants and ( ) is concentration.
Further

       Kf/Kr  >  (B)/(A)                                    /4/

By taking ln and multiplying by RT we obtain

      0  >  -RTlnK  +  RTln((B)/(A))                      /5/

where K = Kf/Kr is the equilibrium constant, i.e. we obtain

      0  >  delta Go  +  RTln((B)/(A)  =  delta G            /6/

As you can see, /6/ follows from and is equivalent to /2/.  Would you
claim that /2/ is a direct consequence of the second law?
   To be honest, I must admit that delta G is not only that. There is a
special type of reactions (which comprise  transport from  one
macroscopic compartment to another) for which delta G is indeed work.
I tried to discuss this problem a few times, but it proved to be
the most uninteresting one.

>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 ask how much the work is, you answer that it is not one of expansion.
(Question: How many apples are there on the tree?  Answer: There are no
oranges on the tree).


>   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.                                                           <

Don't you contradict yourself here? Does the system really do the maximum
work WHEN THE REACTION IS REVERSIBLE, or the details of the reaction can
decrease that amount?

>   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.     <

   delta G = delta H - T(delta S) = delta H - Qr                   /7/

where Qr is the heat absorbed as reversible work is done. If delta G is
-50 KJ/mole, and as delta H for the ATP system is nearly  -20 KJ/mole,
Qr is 30 KJ/mole, i.e. 30 KJ/mole heat are ABSORBED FROM THE ENVIRONMENT.

Best regards,
Pentcho



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