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AnnM
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- TL;DR Summary
- The actual Gibbs free energy of a reversible reaction is lower if circumstances makes the reaction irreversible. How can this be explained?
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.
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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