Does the change in Gibbs free energy depend on initial and final concentrations?
$$\mathrm dG = -S\mathrm dT+ V\mathrm dP + \sum \mu_i \mathrm dN_i$$
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 $\ce{AB -&gt; B + A}$ this is $$(\mu_\ce{A}\Delta N_\ce{A} + \mu_\ce{B}\Delta N_\ce{B}) - \mu_\ce{AB}\Delta N_\ce{AB}$$