Intuition on Gibbs Free Energy In many places, I read that Gibbs Free Energy is called "free energy" because some of the enthalpy that comes from the chemical reaction becomes "waste heat" due to a change in the entropy of the system, and only the rest is usable as work. However, I cannot imagine how this is the case. Take an exothermic reaction with a positive change in entropy for example. Because $\Delta G = \Delta H - T \Delta S$, then we find that $\Delta G$ is more negative than $\Delta H.$ That is, more energy is released by the system than the enthalpy of reaction. How does this fit in with the "waste heat" explanation, that not all of the energy released is usable? There is more energy released now.
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 $\Delta S\;.$