What does the $-T \Delta S$ term of the Gibb's free energy equation represent?

Question

So as we know, Gibb's free energy for a constant pressure reaction is given as

$ \Delta G = \Delta H - T\Delta S$

The enthalpy term represents the change in the internal energy of the system. What does the entropy term represent? Does it act as some extra "reservoir" of energy stored as potential energy or some other quantity?

I know that the entropy of a system is the integral of $1/T$ of the surroundings against heat absorbed by the system, but then does that mean that this entropy term is just energy absorbed by the system throughout the process? I'm thinking of a reaction that produces gas doing mechanical work on the surroundings. Would that work be what is represented by the entropy term?

Answer 1

The Gibbs free energy is defined (in difference form) as $$\Delta G\equiv\Delta H-T\Delta S=\Delta U+P\Delta V-T\Delta S,$$

where $H$ is the enthalpy, $T$ is the temperature, $S$ is the entropy, $U$ is the internal energy, $P$ is pressure, and $V$ is volume. Note that two terms have been added to the internal energy. The first, $P\Delta V$, represents the work done on a surrounding constant-pressure atmosphere; it acknowledges the fact that our system takes up space and needs to do work on its surroundings simply to exist at a finite volume. This is useful because we often consider as systems the materials and processes around us, which are surrounded by air at 1 atm.

The second term, $T\Delta S$, represents energy provided by the surrounding constant-temperature environment; it acknowledges the fact that at nonzero temperature, there's always a nonzero likelihood that some microscale process can occur (such as an atom evaporating from the surface of condensed matter). You can think of the surroundings as providing energy for free, even fluctuatingly large energies, simply by acting as a large temperature reservoir. This concept is also useful because again, the materials and processes around us are at room temperature, perhaps, or some other well-regulated temperature. As a result, we often observe phenomena that require energy and therefore would be nonsensical if we operated under the equilibrium assumption of $\Delta U=0$.

The additional $T\Delta S$ implies that Nature—at least constant-temperature Nature—favors processes that release energy (such as strong bonding, which puts molecules in low-enthalpy states) but also favors processes that increase entropy (such as a formerly fixed atom that can now bounce around in the gas phase with an endless possibility of locations, speeds, and orientations). Which factor dominates depends on the coefficient of $\Delta S$, namely, the temperature $T$. At higher temperature, the equilibrium phase is always the higher-entropy phase, for instance. Oxidation reactions may reverse because of the benefit of creating a gas product. Condensed matter sublimates and evaporates and is cooled in the processes. None of this can be understood without considering the $T\Delta S$ term and by modifying our criterion of equilibrium (at constant pressure and temperature) from $\Delta U=0$ to $\Delta G=0$.

Answer 2

Most of transformations that we observe in nature do not involve isolated systems, but systems in thermal contact with the environment. In such cases we can consider the environment as a thermostat at $T$. For a system exchanging heat with the environment we can state:

$$\Delta S\ge \int_{A}^{B}\frac{dQ}{T}$$ This inequality holds for every generic transformation between two states $A$ and $B$ of a thermodynamic system. Since $T$ is a constant, we can just say $Q\le T\Delta S$. We can imagine the generic transformation between $A$ and $B$ as a constant-pressure or a constant-volume transformation (with or without mechanical work), in each case we can write $W\le T\Delta S- \Delta U$ as a consequence of the first principle. By analogy with mechanical potential energy, whose minima identify the equilibrium points, one can define a "thermodynamic potential" such that its minima are the equilibrium thermodynamic states.

Let us consider a physical situation in which constant-pressure transformations with fixed $T$ are involved: the changes of state (or phase changes of a certain substance) represent some physical examples in which temperature remains constant even in the presence of a heat exchange. When a substance is in equilibrium with two coexisting phases at an assigned pressure, the temperature of the system is uniquely determined by the properties of the substance regardless of the quantity of substance involved in each phase. Since the different phases of the same substance generally have different densities, the transfer of a fraction of the substance from one phase to another implies a change in volume and therefore a work $W=P\Delta V$. The first law of thermodynamics then reads $Q=\Delta U+P\Delta V=\Delta H$. Left to itself, the system can only evolve towards a state of maximum entropy, therefore, the function $G=H-TS$ is always such that $\Delta G\le 0$ in the evolution towards equilibrium (remember that $T\Delta S\ge \Delta H$). The equality $\Delta G=0$ holds in the case of a reversible transformation.