Gibbs Free Energy Change of a Reaction

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Amok
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So, in my thermo book, it says that the gibbs free energy change of a reaction is the free energy received by the system at constant T,p and constant chemical potential when the extent of the reaction varies by one mol. The part that is confusing me is the "constant chemical potentials"statement. If the extent of the reaction is changing, aren't the chemical potentials changing as well? Can the change in free energy of a reaction be used to predict how much work one can get out of a chemical reaction?
 
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Think of a reaction taking place in a continuous stirred tank reactor. The composition of the mixture in the tank remains constant and so do all the chemical potentials. Nevertheless, as both educts are continuously added and products removed, a chemical reaction is constantly taking place.
 
But that means you never reach an equilibrium state (unless you started at one). It's funny because people always seem to forget this when they talk about G free energy, they usually only mention constant T and P.
 
Yes, in equilibrium Delta G is zero. However, a chemical reaction can also occur (arbitrarily near) equilibrium with constant chemical potentials of the reactants.
Consider a reaction chamber which where H2 O2 and H2O are in equilibrium and H2 and O2 can diffuse into the chamber via semi-permeable membranes from their respective reservoirs and H2O can be removed by diffusion through a similar membrane semipermeable for water only.
By increasing infinitesimally the pressure (i.e. the chemical potentials) of H2 and O2 and reducing the pressure of H2O in the respective reservoirs you can synthesise an arbitrary amount of water reversibly with Delta G =0 and the chemical potentials of the components hold constant.
 
Yes, that's ture. I'm just saying that those conditions are not always present, and people still use delta-g to estimate how much useful work one can get out of chemical reaction. Having constant p and T is easy, having constant chemical potential is harder.
 
Yes, you are right. When you are burning a mixture of hydrogen and oxygen the chemical potentials change during that process and you have to integrate delta G over the reaction coordinate to get the maximal amount of work you can extract.