Does the change in Gibbs free energy depend on initial and final concentrations?

Suppose there is a decomposition reaction, and by varying different factors, the equilibrium can be disturbed to alter the final concentration of compounds. So, does the value of ∆G here depend upon the initial and final concentration? The basic rule of ∆G = G(products) - G(reactants) seems unsatisfactory, because, it seems pretty absurd to me, that the value of ∆G is the same when 0.01% of a gas decomposes, and when 85% of it decomposes.




$$\mathrm dG = -S\mathrm dT+ V\mathrm dP + \sum \mu\_i \mathrm dN\_i$$
If (products) and (reactants) are at the same temperature and pressure then $\mathrm dT$ and $\mathrm dP$ are both $0$, and $\mathrm dG = \sum \mu\_i\mathrm dN\_i$. Integration gives $\Delta G = \sum \mu\_i\Delta N\_i$. $\mu$ is the chemical potential which is just the specific molar Gibbs free energy. For the decomposition of $\ce{AB -> B + A}$ this is $$(\mu\_\ce{A}\Delta N\_\ce{A} + \mu\_\ce{B}\Delta N\_\ce{B}) - \mu\_\ce{AB}\Delta N\_\ce{AB}$$


 1

The equation $\Delta G = G(products) - G(reactants)$ refers to a change of state in which you start out with pure reactants and end up with pure products. Of course, if you start out in thermodynamic equilibrium state A (containing an equilibrium mixture of reactants and products) and you make a change so that you end up in thermodynamic equilibrium state B (containing a different equilibrium mixture of reactants and products), the change in free energy is not given by that equation. However, you should be able to calculate what that change in free energy actually is (between states A and B).


 0