Edit: Why is the work lower when a reaction becomes irreversible?

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Summary:

The actual Gibbs free energy of a reversible reaction is lower if circumstances makes the reaction irreversible. How can this be explained?

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According to a trusted source (a textbook, if you must know), the actual Gibbs free energy of an otherwise reversible reaction, becomes lower than expected if the local circumstances makes the reaction effectively irreversible. The circumstances are that neither enough energy, nor a reaction path is available. Specifically, the reaction in question is the reduction of molecular oxygen to water, done by an enzyme. The reverse reaction, water oxidation to oxygen and protons, is not possible in that enzyme and under normal physiological conditions. So, the idea is that the delta-G for oxygen reduction is less than what it is "supposed to" be, because of the fact that an equilibrium cannot be defined? The book only stated this matter-of-factly, and did not offer any explanation.
I am no expert in thermodynamics, so I was hoping that someone here could explain. Thanks in advance!

Edit: I just checked the exact wording in that book and realized that it doesn't say explicitly that the delta-G is lower, that was my own interpretation. Shoot. What it does say is that the full "potential" of the delta-G of the reaction cannot be used, I guess you can interpret it as saying that the work is less than maximum. But I still don't understand why this is a consequence of irreversibility. Sorry to mess things up.
 
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epenguin
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You would have to give us a sufficiently extensive quotation of the text for us to know or guess what they mean. You do not even tell us what this oxygen is being reduced by.

It may just be that an equilibrium constant gives no idea of what relative concentrations are in the living cell.

Example - isomerases. Simple because one reactant and one product only and of their nature the equilibrium constant is often not far from 1. but if the product is being removed efficiently by other enzymes, their concentration ratio in vivo can be far from 1.

It is not unknown for formulations of physical principles in biochemistry textbooks to be a bit shaky.
 
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All right!
First a brief description of the enzymatic reaction: The enzyme in question is cytochrome oxidase (a.k.a. Complex IV), more specifically the caa3-type oxidase in mammal mitochondrial membranes. Oxygen is reduced to water by electron donation from cytochrome-c, plus uptake of protons by cytochrome oxidase from one side (the "inside") of the mitochondrial membrane. The potential for oxygen reduction at pH7 (close enough to physiological pH) is +820 mV. The total Gibbs energy for the reaction can be calculated taking into account that the reduction potential at which electrons are donated to cytochrome oxidase is +290 mV, or thereabouts (physiological conditions), and also that there is an electrochemical membrane potential difference of 180 mV (it can be measured!). This leads to the grand total of about 350 mV per four electrons, that is "left" and can be used to do work. The maximum work given by ΔG= -nFE would then be about -135 kJ/mol. The actual work done by cytochrome oxidase as a result, is the translocation of four protons from the "inside" to the "outside" of the membrane, resulting in a proton concentration gradient across the membrane.

Now comes the part that confuses me: The textbook states that
"Four electrons falling through this potential would be sufficient to translocate up to six protons across the membrane against a [potential difference] of about 200 mV. However, unlike the remainder of the respiratory chain, Complex IV is irreversible. The actual H+/O stoichiometry for the proton-pumping activity alone is lower, 4H+/4e-, reflecting this lack of reversibility."

My confusion is about that last bit, that although calculations would have it that there is enough energy for translocation of six protons,"only" four protons can be translocated because of irreversibility. I hope that was clear and detailed enough.

Edit: And as I wrote in my original post, it is really an irreversible reaction, *in that enzyme*. Of course, water can be oxidized back to oxygen and protons, it will just never, ever happen in a mitochondrion.
 
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There is a difference between a reversible process and a reversible reaction. A reversible process is one in which the final state for the system plus surroundings at equilibrium cannot be returned to the initial state of these entities without causing a change in some other external entity. A reversible reaction is one in which a reaction can proceed forward or backward depending on the concentrations of the reactants and products in the initial state. In the context you are describing, this process, thermodynamically, is reversible, but, cannot proceed to final equilibrium in a practical amount of time (or cannot reverse) because an enzyme is required to enhance the kinetics. But, if the enzyme is not present, the reaction slows down to an imperceptible rate, That means that the reaction is not able to reach equilibrium in practice, and the final concentrations of reactants and products is the same as if the magnitude of the change in Gibbs free energy were less than it would be at equilibrium. So the entire argument in your reference is related not to the true change in Gibbs free energy that would occur if the reaction could proceed to final equilibrium, but to the change that would occur up to the point where the kinetics has slowed (because of removal of enzyme) to a negligible rate.
 
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Oooooh! That really explains it! Thank you so much for the answer Chestermiller!:smile:
 
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epenguin
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I'm sorry I could not reply earlier. Partly it is very hot where I am right now and I have no concentration. but also what I could understand I could not make any sense out of what your textbook authors are saying. and even suspect it may be nonsense. Can you tell us what the book you are using is? I tho pught it would be better to get the people that there are here who are more used to this kind of bioenergetics and enzyme system, and how they are talked about who might pick up on is is better. So I shall ask for this to be put in another place where it gets more their attention.

It seems to me that although the mechanism is singular and complicated it is not that astounding that the catalytic cycle involving the transfer of four electrons to oxidised cyt c involve also taking up four protons from one side of the membrane and discharging four on the other side. A larger number than four seems harder to imagine.

I'm not sure where irreversibility comes into it. Theoretically no reaction is irreversible, standard free energy changes are finite.I far as I can see all the redox potentials mentioned are finite.
 
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Dear epenguin thank you for replying anyway. No actually, the textbook makes perfect sense to me now. I think Chestermiller explained it well. However, it may also be the case that there is no mechanism for transferring more than four protons per four electrons. Like you say, the proton translocation is complex and seems to be coupled to redox reactions at the catalytic site - exactly how is a matter of debate, but most likely there is no way to achieve proton transfer without coupling it to a change in redox state, within a relevant amount of time anyway. Nevertheless, the statement in the textbook intrigued me, and since I trust the authors I wanted to understand what they meant. The book in question is Bioenergetics 4th ed., by David G. Nicholls and Stuart J. Ferguson.
 

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