Free Energy = Max Useful Work

As we know, energy can be written as the sum of heat energy and work energy, $$dU=đQ+đW$$ We can split this up further by specifying the type of work. $$dU=đQ+đW_{expansion}+đW_{non-expansion}$$ At this point we're going to put one restriction on the system. We're going to say that the system is at STP, and remains there. Now for the hard part. First we're going to change our equation for energy. We know that $đQ=TdS$, and that $đW_{expansion}=-PdV$, energy can be rewritten as follows, $$dU=TdS-PdV+đW_{non-expansion}$$ At this point, we recall and combine the definitions of enthalpy, and free energy, $$dH=dU+d(VP)$$ $$dG=dH-d(TS)$$ Thus, $$dG=dU+d(VP)-d(TS)$$ $$dG=TdS-PdV+đW_{non-expansion}+d(VP)-d(TS)$$ Now we expand the differentials and cancel terms, $$dG=\cancel{TdS}-\cancel{PdV}+đW_{non-expansion}+VdP+\cancel{PdV}-\cancel{TdS}-SdT$$ $$dG=đW_{non-expansion}+VdP-SdT$$ Recall at this point that the system is at STP. This means that both the pressure and temperature are constant, and thus, those two differentials are 0. Therefore, $$dG=đW_{non-expansion}$$

So what exactly does this non-expansion, or "useful," work mean? Non-expansion work is the intrinsic energy of the molecules. It is the ability of the molecules to do work due to their composition, shape, and chemical activity, rather than their thermodynamic state. You may recall that Gibbs energy can be written as it is below, $$dG=-SdT+VdP+\mu dN$$ The "useful" work is that associated with $\mu dN$. Since many chemical processes occur at constant temperature and pressure, this chemical contribution is the only nonzero component.