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.
Kf(A) > Kr(B) /3/
where Kf and Kr are the respective rate constants and ( ) is concentration.
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.
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