Intuition on Gibbs Free Energy
From Second Law of Thermodynamics viz. $$\Delta S_\textrm{sys}+ \Delta S_\textrm{res}\geq 0\,,$$ it can be concluded that $$W_\textrm{sys}\leq (U_1- U_2)- T(S_1 -S_2) \tag{1}\;.$$

Now, $$W_\textrm{sys}= \underbrace{W_T}_\textrm{pressure-volume/compression work}+ \underbrace{A_T}_\textrm{non-compression work}\;.$$

Putting the the first term on the RHS of $(1)\,,$ it yields,

\begin{align}A_T &amp;\leq (U_1- U_2)- T(S_1 -S_2) - W_T\\ &amp;\leq (U_1- U_2)- T(S_1 -S_2)+ P(V_1 -V_2)\;.\end{align}

Now, this is equal to the negative change of Gibbs energy $G$ provided the thermodynamic process is done at constant pressure and temperature. 

So, that means,  $$A_T \leq G_1 -G_2\;.$$ Note the $\leq$ sign; this implies $(G_1-G_2)$ is the maximum value $A_T$ can attain.

Since, $\Delta G \lt 0\;,$ this implies $A_T$ is positive. Now, if the process is reversible, then the non-compression work $A_T$ is maximum viz: $$A_T= G_1 -G_2\;.$$

This definitely supports the term free energy as the decrease in $G$ gives the amount of energy "that can be 'freed' and made available for work". 




  That is, more energy is released by the system than the enthalpy of reaction.


Of course, the extra energy came from heat energy absorbed from the environment as the entropy of the products was greater than that of the reactants by $\Delta S\;.$