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.
Best Answer
1) Why is −ΔGsystem not a true energy as written above?To convince yourself that the Gibbs function is not a true energy consider for example the free expansion of an ideal gas.
In this process there is no change in temperature, enthalpy or kinetic energy of the gas. An indeed $\Delta U = 0$; i.e. $U$ is a true energy. From the perspective of $G$, the expansion is accompanied by a decrease of $\Delta G = -T-\Delta S$. But it is not appropriate to state that this is an amount of energy which has been transferred to the environment. As long as we assume the law of conservation of energy there must be somewhere an energy increase to exactly match each observed energy decrease. Therefore $G$ does not qualify as a true energy.
Regarding the quantity that is changing $-T\Delta S$, it corresponds to the energy that would have been transferred as heat if the expansion had been conducted reversibly. I will clarify this aspect answering your next question.
2) If −ΔGsystem is used to increase the entropy of the universe, then how could it be used for work as commonly told?You should devise a reversible process between states 1 and 2 at P=K and T=K, in this case: $\Delta G_{sys} = G_2 − G_1 = W_{nonP-V}$. For the irr process ΔGsystem consists of degraded potential work which accounts for the increase in the entropy of the universe.
Again consider again the isothermal expansion of an ideal gas under reversible and irreversible conditions (free expansion) in both cases $\Delta U = 0$, $w = q$.
When the ideal gas is isothermally and reversibly expanded from state 1 to state 2, we have:
$w_{rev} = q_{rev} = n RT \ln \frac{V_{2}}{V_{1}}$
$q$ is reversibly transferred from the environment to the gas, and the increase in the entropy of the gas is $S_2 − S_1 = n R \ln \frac{V_{2}}{V_{1}}$. The entropy of the environment decreases by an equal amount and, therefore, entropy is not created; that is, $ΔS_{tot} = 0$
However, if the gas is allowed to expand freely from $V_1$ to $V_2$, then, since the gas performs no work, no thermal energy is transferred from the reservoir to the gas, and there is no change in the entropy of the environment. Since entropy is a state function, the value of $S_2 − S_1$ is independent of the process path, and hence, the entropy created, $ΔS_{tot}$, equals $S_B − S_A$, which equals $n R \ln \frac{V_{2}}{V_{1}}$. This entropy is created as a result of the degradation of the work which would have been performed by the gas had the expansion not been carried out against zero pressure. This degraded work equals $w_{max}$ as well as $q_{rev}$.
I'd like to clarify the background out of which $(G_2 − G_1) + T \Delta S_{irr} = 0$ is derived.
$G \equiv H -TS$
for a transformation betweeb 1 and 2:
$(G_2 − G_1) = (H_2 − H_1) − (T_2 S_2 − T_1 S_1) = (U_2 − U_1) + (P_2 V_2 − P_1V_1) − (T_2 S_2 − T_1 S_1)$
For a closed simple system, the First Law gives
$(U_2 − U_1 ) = q − w$
thus
$\Delta G = q − w + (P_2 V_2 − P_1V_1) − (T_2 S_2 − T_1 S_1)$
if the process is carried out a $T=K$ and $P=K$
$\Delta G = q − w + P (V_2 − V_1) − T (S_2 − S_1)$
$w$ can be expressed as
$w = w' + P (V_2 − V_1)$
where:
Substituting into the previous Eqns
$\Delta G = q − w' − T \Delta S$
from Clausius inequality
$q ≤ T \Delta S$
Then $w' ≤ − \Delta G$
$dw'_{max} = dG$
if you integrate between 1 and 2 and choose a reversible path you can extract the maximum work out the system.
$−w' = \Delta G + T \Delta S_{irr}$
Thus, for an isothermal, isobaric process, during which no form of work other than P -V work is performed (i.e., w' = 0),
$\Delta G + T \Delta S_{irr} = 0$
Such a process can only occur spontaneously (with a consequent increase in entropy) if the Gibbs free energy decreases.