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What is Gibbs free energy? As my book explains:

Gibbs energy is the energy of a system available for work.

So, what does it want to tell? Why is it free? Energy means ability to do work. What is so special about this energy? Can anyone simply explain?

I just want a math-free intuition.

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First, you have system with some energy, named $U$ by physicists. You think you have all the information you need to characterize the system but then some guy comes near and says:

"Whoa, that's bad, the volume of your system can change."

You say:

"No problem, we just add here $pV$. Our new energy is $H=U+pV$."

"But hey," they say, "your temperature can change by external heat, you have to count that also."

"No problem, we subtract $TS$ from our energy, and rename it. $G=U+pV-TS$"

There you are.

People name it "available" energy, because if your system increases in size, it creates work: $V$ increases, so $U$ decreases. But energy is always bounded from below. When you drain everything you can, your system goes to $G=0$.

You cannot drain all $U$ because you would need an external system with $T=0$.

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    $\begingroup$ There is something missing, however. Ok, you explained why $H$ is important in the sentence V increases, so U decreases, (without input heat). But what about $G$? In which scenario does it remain constant? I think the core of the question was almost hit but nearly missed. $\endgroup$ – André Chalella Nov 29 '14 at 23:51
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    $\begingroup$ Note that the Gibbs free energy is used when describing systems at constant temperature; so the TS term describes entropy changes in the environment, not temperature changes in the system. $\endgroup$ – jabirali Nov 30 '14 at 1:31
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Sanaris's answer is a great, succinct list of what each term in the free energy expression stands for: I'm going to concentrate on the $T\,S$ term (which you likely find the most mysterious) and hopefully give a little more physical intuition. Let's also think of a chemical or other reaction, so that we can concretely talk about a system changing and thus making some of its internal energy $H=U+p\,V$ available for work.

The $T S$ term arises roughly from the energy that is needed to "fill up" the rotational, vibrational, translational and otherwise distractional thermal energies of the constituents of a system. Simplistically, you can kind of think of its being related to the idea that you must use some of the energy released to make sure that the reaction products are "filled up" with heat so that they are at the same temperature as the reactants. So the $T S$ term is related to, but not the same as, the notion of heat capacity: let's look at this a bit further.

Why can't we get at all the energy $\Delta H$? Well, actually we can in certain contrived circumstances. It's just that these circumstances are not useful for calculating how much energy we can practically get to. Let's think of the burning of hydrogen:

$$\rm H_2+\frac{1}{2} O_2\to H_2O;\quad\Delta H \approx 143{\rm kJ\,mol^{-1}}\tag{1}$$

This is a highly exothermic one, and also one of the reactions of choice if you want to throw three astronauts, fifty tons of kit and about a twentieth of the most advanced-economy-in-the-world’s-1960s GDP at the Moon.

The thing about one mole of $H_2O$ is that it can soak up less heat than the mole of $H_2$ and half a mole of $O_2$; naively this would seem to say that we can get more heat than the enthalpy change $\Delta H$, but this is not so. We imagine a thought experiment, where we have a gigantic array of enormous heat pads (all individually like an equilibrium “outside world") representing all temperatures between absolute zero and $T_0$ with a very fine temperature spacing $\Delta T$ between them. On my darker days I find myself imagining an experimental kit that looks eerily like a huge pallet on wheels of mortuary shelves, sliding in and out as needed by the experimenter! We bring the reactants into contact with the first heat pad, which is at a temperature $T_1 = T_0 - \Delta T$ a teeny-tiny bit cooler than $T_0$ thus reversibly drawing some heat $\Delta Q(T_1)$ off into the heat pad. Next, we bring the reactants into contact with the second heat pad at temperature $T_2 = T_0 - 2\,\Delta T$, thus reversibly drawing heat $\Delta Q(T_2)$ off into that heat pad. We keep cooling then shifting to the next lowest heat pad until we have visited all the heat pads and thus sucked all the heat off into our heat pads: see my sketch below:

Cooling Down Reactants

Now the reactants are at absolute zero. There is no heat needed to "fill them up" to their temperature, so we can extract all the enthalpy $\Delta H$ from the reaction as useful work. Let's imagine we can put this work aside in some ultrafuturistic perfect capacitor, or some such lossless storage for the time being.

Now we must heat our reaction products back up to standard temperature, so that we know what we can get out of our reaction if the conditions do not change. So, we simply do the reverse, as sketched below:

Heating Products Up

Notice that I said that $H_2O$ soaks up less heat than the reactants. This means that, as we heat the products back up to standard temperature, we take from the heat pads less heat in warming up the water than we put into them in cooling the reactants down.

So far, so good. We have gotten all the work $\Delta H$ out into our ultracapacitor without losing any! And we're back to our beginning conditions, or so it seems! What's the catch?

The experimental apparatus that let us pull this trick off is NOT back at its beginning state. We have left heat in the heat pads. We have thus degraded them: they have warmed up ever so slightly and so cannot be used indefinitely to repeatedly do this trick. If we tried to do the trick too many times, eventually the heat pads would be at ambient temperature and would not work any more.

So we haven’t reckoned the free energy at the standard conditions, rather we have simply calculated the free energy $\Delta H$ available in the presence of our unrealistic heat sink array. To restore the system to its beginning state and calculate what work we could get if there were no heat sink array here, we must take away the nett heat flows we added to all the heat pads and send them into the outside World at temperature $T_0$. This is the only "fair" measure, because it represents something that we could do with arbitrarily large quantities of reactants.

But the outside World at $T_0$ is warmer than any of the heat pads, so of course this heat transfer can’t happen spontaneously, simply by dent of Carnot’s statement of the second law!

We must therefore bring in a reversible heat pump and use some of our work $\Delta H$ to pump this heat into the outside world to restore standard conditions: we would connect an ideal reversible heat pump to each of the heat pads in turn and restore them to their beginning conditions, as sketched below:

Packing our Playthings Up

This part of the work that we use to run the heat pump and restore all the heat pads, if you do all the maths, is exactly the $T\,\Delta S$ term.

The above is a mechanism whereby the following statement in Jabirali's Answer holds:

Processes that increase the Gibbs free energy can be shown to increase the entropy of the system plus its surroundings, and will therefore be prevented by the second law of thermodynamics.

The nice thing about the above is that it is a great way to look at endothermic reactions. In an endothermic reaction, we imagine having an energy bank that we can borrow from temporarily. After we have brought the products back up to temperature, we find we have both borrowed $-\Delta H$ from the energy bank and put less heat back into the heat pads than we took from them. So heat can now flow spontaneously from the environment to the heat pads to restore their beginning state, because the heat pads are all at a lower temperature than the environment. As this heat flows, we can use a reversible heat engine to extract work from the heat flowing down the gradient. This work is, again, $-T\,\Delta S$, which is a positive work gotten from the heat flowing down the temperature gradient. The $-T\,\Delta S$ can be so positive that we can pay back the $\Delta H$ we borrowed and have some left over. If so, we have an endothermic reaction, and a nett free energy: this energy coming from the heat flowing spontaneously inwards from the environment to fill the higher entropy products (higher than the entropy of the reactants).

Take heed that, in the above, I have implicitly assumed the Nernst Heat Postulate -the not quite correct third law of thermodynamics - see my answer here for more details. For the present discussion, this approximate law is well good enough.

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Short answer: Gibbs free energy $G = U + PV - TS$ combines internal energy $U$, pressure $P$, volume $V$, temperature $T$, and entropy $S$ into a single quantity that measures spontaneity. With that I mean that processes lowering the Gibbs free energy of your system will spontaneously occur, and equilibrium is reached when the Gibbs free energy reaches the lowest possible value.

Processes that increase the Gibbs free energy can be shown to decrease the entropy of the system plus its surroundings, and therefore will be prevented by the second law of thermodynamics. That's why it measures useful energy - your system may contain more energy, but entropy considerations will prevent you from spending it.

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    $\begingroup$ So processes that increase the Gibbs free energy can be shown to decrease the entropy of the universe, and that's why the second law of thermodynamics prevents it. $\endgroup$ – jabirali Nov 30 '14 at 10:42
  • $\begingroup$ Sir, what does $PV$ mean? To me, $P\Delta{V}$ is the work and it does make sense. Does the former mean the work done against $P$ to make vol. $V$ from $0$ ?? $\endgroup$ – user36790 Nov 30 '14 at 13:29
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    $\begingroup$ And who does work? The system? Or the surroundings?? $\endgroup$ – user36790 Nov 30 '14 at 13:32
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    $\begingroup$ Yeah, that sounds right! For a concrete example: when a chemical reaction produces gas, then the volume of your system increases. In order for this expansion to happen, then your system has to "push away" the atmosphere, which means that the system performs work on the environment. Conversely, if your chemical reaction consumes gas, then $\Delta V$ will be negative, and so the atmosphere will "push together" your system, and the environment performs work on your system. That's how the work $P\Delta V$ enters when you calculate changes in free energy $\Delta G$. $\endgroup$ – jabirali Nov 30 '14 at 15:22
  • $\begingroup$ By the way, the quantity $H = U + PV$ that measures how much energy is left after accounting for volume changes under constant atmospheric pressure, is known as enthalpy. $\endgroup$ – jabirali Nov 30 '14 at 15:24
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In the simplest sense of it, the Free Energy is the heat of the system minus the compulsory heat loss due to entropy. So, in short, it is the amount of "energy" left over in the system, after we consider losses due to entropy. So basically, some amount of heat is wasted, and the remaining amount is useful. And, this remaining amount is the Gibbs Free Energy.

$G=H-TS$, here $H$ is the amount of energy contained in the system, and $TS$ is the amount that will inevitably be wasted. That's how it makes sense to me anyway.

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Gibbs free energy is a thermodynamic potential that measures the "usefulness" or process-initiating work obtainable from a thermodynamic system at a constant temperature and pressure (isothermal, isobaric). Just as in mechanics, where potential energy is defined as capacity to do work, similarly, different potentials have different meanings. The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a closed system; this maximum can be attained only in a completely reversible process. When a system changes from a well-defined initial state to a well-defined final state, the Gibbs free energy $ΔG$ equals the work exchanged by the system with its surroundings, minus the work of the pressure forces, during a reversible transformation of the system from the same initial state to the same final state.

Gibbs free energy is also the chemical potential that is minimized when a system reaches equilibrium at constant pressure and temperature. Its derivative, with respect to the reaction coordinate of the system, vanishes at the equilibrium point. As such, it is a convenient criterion of spontaneity for processes with constant pressure and temperature.

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There are two forces influencing the spontaneity of a reaction:

(1) The tendency of a system to attain a state of minimum energy and maximum order or stability.

(2) The tendency of a system to attain a state of maximum energy and minimum order or entropy.

If a system attains maximum stability, it attains minimum entropy; if it attains maximum entropy, it attains minimum stability. Therefore, the two always oppose each other. If we want to predict the spontaneity of a reaction, we must take into account the effect of both forces. But how?

Since we need both criteria (which are in opposition) in determining whether a given reaction will be spontaneous or not, an equation is needed that takes into account both criteria. And, that equation is the Gibbs Free energy equation. The $\Delta H$ term represents the first criterion (stability) and the $T\Delta S$ term represents the second criterion (entropy). The minus sign before $T\Delta S$ means that the two criteria are in opposition.

Note that equilibrium is achieved when stability ($\Delta H$) and entropy ($T\Delta S$) work out a compromise such that both are satisfied, like when neither side can succeed at defeating the other in a game of tug-of-war. Then $$\Delta H=T\Delta S \\\\\\\\ \implies \boxed {\Delta G=0} \quad (\text{equilibrium condition})$$.

Please feel free to edit the answer and improve it.

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  • $\begingroup$ This answer fails to mention the environment, which, as other answers have pointed out, is central to this discussion. $\endgroup$ – akrasia Nov 30 '14 at 10:44
  • $\begingroup$ @akrasia Thanks for the response. The intention of my answer was to simply explain what the purpose of the Gibbs energy equation is and what it means intuitively. Of course, a further explanation of the sign of $\Delta G$ must include the point you have made. Since you have given the advice with all good intentions, I shall find some time later to extend the explanation and include your point. $\endgroup$ – Gaurav Nov 30 '14 at 11:01
  • $\begingroup$ @user36790 If you want an explanation of why $\Delta G$ cannot be positive for a spontaneous reaction, take a look at the second paragraph of jabirali's answer. It does the job excellently well. $\endgroup$ – Gaurav Nov 30 '14 at 11:50

protected by Qmechanic Nov 30 '14 at 2:16

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