Is Gibbs free energy really a true energy that can be used to do work?

Gibbs energy came from the relation $$\Delta S_\text{universe}= \Delta S_\text{surroundings}+\Delta S_\text{system}\ge 0.$$

By little algebra, we can arrive at $$\Delta S_\text{universe}= \frac{-\Delta H_\text{system}}{T} + \Delta S_\text{system} \\ \implies -T\Delta S_\text{universe} = \Delta H_\text{system} - T\Delta S_\text{system}$$

From the relation of Gibbs free energy $$\Delta G_\text{system} = \Delta H_\text{system}- T\Delta S_\text{system}\;$$ we get $$\Delta S_\text{universe}= \frac{-\Delta G_\text{system}}{T}.$$

So, that means, $-\Delta G_\text{system}$ is the energy dispersed to the universe from the system to increase the entropy of the universe.

Now, all we know that the energy that has been dispersed to for the sake of entropy cannot be used for work viz: $$d S= \frac{\delta Q}{T}\; ;$$ we can't use $\delta Q$ to do work as it is used up for increasing entropy. Similarly, how could we use $-\Delta G_\text{system}$ for doing work as it has been used for increasing the entropy of the universe?

As written by Frank Lambert in his site:

[...] Strong and Halliwell rightly maintained that $- \Delta G\; ,$ the "free energy", is not a true energy because it is not conserved. Explicitly ascribed in the derivation of the preceding paragraph (or implicitly in similar derivations), $- \Delta G$ is plainly the quantity of energy that can be dispersed to the universe, the kind of entity always associated with entropy increase, and not simply energy in one of its many forms. Therefore, it is not difficult to see that $\Delta G$ is indeed not a true energy. Instead, as dispersible energy, when divided by $T\; ,$ $\frac{\Delta G}{T}$ is an entropy function — the total entropy change associated with a reaction, not simply the entropy change in a reaction, i.e., $S_\text{products} - S_\text{reactants}\; ,$ that is the $\Delta S_\text{system}.$ ...

My questions are:

$\bullet$ Why is $-\Delta G_\text{system}$ not a true energy as written above? Why can't it be conserved? I want to understand what those lines actually mean?

$\bullet$ If $-\Delta G_\text{system}$ is used to increase the entropy of the universe, then how could it be used for work as commonly told$^1$?

$^1$ I'm referring to this statement: Gibbs energy is the energy of the system available to do work.

• Well, I guess one intuitive way of seeing why in general one cannot speak of conservation of free energies, is because they depend on entropy, and we know entropy is not conserved. – user929304 Nov 13 '15 at 14:27
• @user 929304: Why are not they conserved? Could you give me a example to visualise the fact? – user36790 Nov 13 '15 at 14:29
• Take the free energy of a system that evolves irreversibly from a state $G_0$ to $G_1,$ the irreversiblity of the evolution comes with a positive entropy production of the universe, thus free energy given by $G_1$ must be larger. (take the simple case where everyhing else in kept unchanged, internal energy, pressure etc.). Since you cannot bound the entropy, free energies cannot be bounded either, thus tend to decrease or increase as the system evolves, i.e. not conserved unlike total energy. – user929304 Nov 13 '15 at 14:37
• @user 929304: How can it be used as work as it is dispersed to the universe for increase in entropy? I'm not getting the reason; you can't use $\delta Q$ to do work; it is lost; isn't it? – user36790 Nov 13 '15 at 15:49
• $T\Delta S$ is the "unusable" part of the free energy. That's why you can e.g. interpret the Helmholtz free energy ($F=\Delta U -T\Delta S$) as the maximum extractable work from a thermodynamic process at constant temperature and volume. Conversely, the more disordered the state you want to reach is, the higher its entropy, the lesser the amount of necessary work to create that state. Now in going from Helmholtz to Gibbs, an additional amount of work $PV$ is added due to change of volume e.g. Hope this helps! – user929304 Nov 13 '15 at 16:06