Can ATP do work at equilibrium? (Pentcho Valev)
fujimoto at u.washington.edu
Fri Nov 7 01:20:43 EST 1997
<RUMYM at BGEARN.ACAD.BG> writes:
>>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?
You set up the conditions of your example so that in equation /6/,
Delta G < 0. There is nothing in the second law which requires Delta G
to be less than zero for any arbitrary reaction, so /2/ is not a
consequence of the second law. In order for /2/ to be a consequence of
the second law, you would have to start from the second law and derive
/2/ from it, which can't be done. On the other hand, the prediction
that Delta G decreases for a spontaneous process at constant T and P is
derivable from the laws of thermodynamics. So, you still have to deal
with the fact, that if you accept that you can use (Delta G) to
determine the direction of a reaction, then you have accepted the
second law of thermodynamics.
> 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 told you for a reversible process at T and P, any chemical work is in
Delta G. Part of my reason for not trying to define chemical work is
that you can't use it to test the second law. The second law speaks of
work, not chemical work. So even if you had the chemical work, you would
have to add to it all the other forms of work in the system to get the
total work before you could test the second law.
However, since you seem to want this so badly that you start getting
nasty when you can't get it...
At constant T and P if there is only chemical and expansion work being
dG = sum_i mu_i dn_i
where mu_i is the chemical potential and n_i the number of moles of
species i, and the sum extends over all the chemical species in the
system. You can try thinking about this as the chemical work, though
if there is any electrical or deformational work involved, they must
also be included, so you will have to stick to uncharged reactants
and products, and stay away from solids.
>>> 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?
No I am not contradicting myself. The system does the maximum work when
the reaction is reversible. However, since all naturally occuring processes
are irreversible, the details of the reaction determine the actual
efficiency of the reaction. (Do you know what reversible means in the
context of the second law?)
>>> 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.
The substitution you make in /7/ [T(Delta S) = Qr] only holds for
a reversible process. I get the impression that you don't know what
reversible means in the context of the second law (hint: it doesn't
simply mean that the reaction has both forwards and backwards reactions,
its more than that). For an irreversible process, T(Delta S) is
greater than Qr, so if Qr=0, then T(Delta S) = 30 kJ/mole > 0 certainly
works. The free energy comes from the fact that the ATP <-> ADP + P
reaction is not at equilibrium.
fujimoto at u.washington.edu
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