Question to Lukasz Salwinski about work done by ATP (Pentcho Valev)

RUMYM at BGEARN.ACAD.BG RUMYM at BGEARN.ACAD.BG
Mon Oct 20 12:34:07 EST 1997


Lukasz, as you seem to be perfectly sure that thermodynamics works, let me
ask you a question which, if answered correctly, may change your opinion.
   Let us assume that the ATP-ADP-P system does work. If this work is done
(almost) reversibly, thermodynamics says that it is equal to the negative of
the free energy change

    delta G = delta Go  +  RTln((ADP)(P)/(ATP))

Let us also assume that the concentrations of ATP, ADP and P are such that
delta G = -50 KJ/mole. This means that, as one mole ATP is hydrolysed, the
work done is 50 KJ; as one molecule ATP is hydrolysed, the work done is
50 KJ divided by the Avogadro number.

Now if one mole ATP is hydrolysed without work production, the heat released
(the enthalpy of the reaction) is 20 KJ/mole. This is an internal energy
which is obviously used in work production. But if the work produced is
50 KJ/mole, where do the other 30 KJ/mole come from?

Except for Bill Tivol, people in other groups were reluctant to discuss
this problem. Some even said that chemical work is a useless concept.
I hope that this attitude will change - one cannot believe in the
second law which is in fact a statement about the work done in a cyclical
process, and at the same time claim that this work is meaningless.

Pentcho



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