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

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?

• Nitpick but if the enthalpy term represented the change in internal energy why wouldn't that term actually be $\Delta U$. Apr 19, 2021 at 8:42
• @jacob1729 Well, I think that's what it represents anyways. I think that the reason it represents the change in internal energy here is that this is a constant pressure process. But I may be wrong. I don't understand thermodynamics too well Apr 19, 2021 at 9:27
• When you say "internal energy", do you mean the energy function of the system (i.e. $U$ in canonical terminology), or the energy of 'internal degrees of freedom' that may be released or stored in exo/endo-thermic reactions (or something else entirely?)
– TLDR
Apr 19, 2021 at 17:23
• Apr 19, 2021 at 18:32

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 $$\Delta G=0$$.

• Hmm, I see what you are saying, but that still begs the question of why any heat absorbed by the system from the environment wouldn't be represented in the internal energy term and instead have a new term. If the system were to absorb heat from the environment, wouldn't that raise its internal energy? Apr 21, 2021 at 0:49
• Could you be more specific? A system can absorb heat from the environment as its internal energy decreases, increases, and stays constant. Apr 21, 2021 at 6:01
• Basically, wouldn't the internal energy of a gas be equal to = any heat energy in the system from the start + any new heat absorbed + any potential energy? So wouldn't any heat energy absorbed by the system would show up in the internal energy of the gas instead of having a separate term associated with it? That's what I'm wondering about. Apr 21, 2021 at 10:58
• I’m not sure if this satisfactorily addresses your question, but make sure not to conflate equilibrium calculations with energy bookkeeping. Both have to be satisfied because they express different thermodynamic laws (specifically, the Second and First). The presence of a new term in an equilibrium calculation doesn’t change the fact that energy added to a system minus energy removed must be energy stored. Apr 21, 2021 at 16:37

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.