Definition of Gibbs Free Energy and its interpretation [closed]

Question

RELATIONSHIP BETWEEN $\Delta G$ & $W_\mathrm{non-PV}$
$dU = dq + dw_\mathrm{py} + dw_\mathrm{non-PV}$
for reversible process at constant T & P $dU+pdV-TdS = dw_\mathrm{non-PV}$
$dH-TdS = dw_\mathrm{non-PV}$
$(dG_\mathrm{system})_\mathrm{T,P} = dw_\mathrm{non-PV}$
$(dG_\mathrm{system})_\mathrm{T,P} = (dw_\mathrm{non-py})_\mathrm{by system}$
Means free energy change for a process is equal to the maximum possible work that can be derived from the process i.e.,
$\Delta G^\circ= W_\mathrm{max}$ (for a reversible change at constant pressure and temperature)

The derivation shown above considers $$dU = dq + dw_\mathrm{pv} + dw_\mathrm{non-pv}$$

(1) Is the dq used here different from the general q (as below) used in thermodynamics at definition level - where W stands for mechanical work?

$$ dU = dq - dW $$

(2) Assuming that the term $dw_\mathrm{pv}$ includes $Vdp + pdV$, is $Vdp$ also considered to be mechanical work generally?

(3) What can be an interpretation of $dw_\mathrm{non-pv}$, and, is there a mathematical way to calculate it? Example involving this would be helpful.

Before reading the above text, I was looking at it the following way:

$dG = dH - TdS \\ dS = dq_\mathrm{rev}/T \\ dG = dH - dq ~~\mathrm{[assuming~ the~ process~ to~ be~ reversible]}\\ = dU + d(PV) - dU - PdV \mathrm{[assuming~statement~2}~\\ \mathrm{~to~be~wrong~ and~work~to ~only~ mean}~ PdV~\mathrm{ by ~definition}\\ = Vdp $

I realize that I am completely wrong. Now, I am confused as well. Please help with the same.

Answer 1

Denbigh, Principles of Chemical Equilibrium, analyzes all this in a more precise way. He considers a closed system initially at temperature T and pressure P, and a process for which all heat transfer takes place with a constant temperature reservoir also at T, and with the system in contact with a surroundings at constant external pressure $P_{ext}=P$. In addition, in the final thermodynamic equilibrium state, the system temperature and pressure are also T and P respectively.

For this situation, the first law of thermodynamics tells us that $$\Delta U=Q-P\Delta V-W_{non\ PV}$$And, for this situation, the second law of thermodynamics tells us that $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the (positive definite) entropy generation due to irreversibility. If we combine these two equations by eliminating Q, we obtain:$$\Delta U=T\Delta S-P\Delta V-W_{non\ PV}-T\sigma$$Rearranging this equation then gives:$$W_{non PV}=-\Delta G-T\sigma$$So, for this situation, for the same initial and final end states, the maximum non-PV work that the system can do is $-\Delta G$ if the process is reversible and less than that if the process is irreversible.

Answer 2

I would like to answer the question of the title : Definition of the Gibbs Free Energy and its interpretation.

The free energy is a concept created by Gibbs in the $19$th century. At that times Gibbs was trying to apply in chemistry the concepts developed in physics and mechanics for spontaneous phenomena. In physics, a spontaneous process happen each time an object has the possibility to fall from a high position to a lower position. And never the contrary. An object cannot spontaneously pass from a lower position to a higher level.

This is related to the concept of potential energy. Each object has a mechanical potential energy $mgh$ which depends on its level $h$ measured above an arbitrary reference level. This reference level which may be the level of the sea (in which case $h$ is called altitude). But, main result for Gibbs is that $mgh$ is a mechanical potential energy. This energy has the property that when an object with a high mechanical potential energy has the possibility to go to a lower level, the transformation is spontaneous. The passage from a low level to a higher level is never spontaneous. Never !

Gibbs was looking to find a potential energy valid for chemistry. For a while he thought that the enthalpy $\ce{H}$ was this potential energy, because in nearly all spontaneous chemical reactions, heat is emitted. So that the molecules pass spontaneously from an initially high $\ce{H}$level to an lower one, and the difference is emitted as heat. This is usually right. But it is not valid for endothermic reactions, which are also spontaneous. So enthalpy is not the energy that Gibbs was looking for. What was the potential energy that Gibbs was looking for, so that in all cases, it decreases in spontaneous processes ?

It is not heat content. Another way of looking at this problem is to say : What is the potential energy that is liberated in all chemical reactions, that is in endo- or exothermic reactions ?

For Gibbs, this strange and new potential energy, called "free energy", is the electrical energy liberated in galvanic cells. Galvanic cells are working by chemical reactions. And the galvanic cell is producing spontaneously electrical energy, whatever the endothermic or exothermic character of the internal chemical reaction. A cell will always produce electricity spontaneously, never the contrary. The chemical reaction may be endo- or exothermic, it must be exoergic. A galvanic cell cannot absorb electrical energy : it would be a non-sense.

So, for Gibbs, all reactants in a cell possess an initially high free enthalpy $G$ ($G$ for Gibbs). And, when spontaneously working, the cell emits an electrical energy Delta $G$ as non-pV work, equal to : Delta $G = - zEF$, where $z$ is the number of electrons liberated in the chemical equation of the cell, $E$ is the tension of the cell, and $F$ is the Faraday (= $96500$ $Cb$).

After this definition, Gibbs decided to behave like the chemists in calorimetry. Using all known galvanic cells, he measured their $E$ values and calculated the corresponding values of free enthalpy of formation of all compounds found in cells. These values are published in tables, like the enthalpies of formation.

Answer 3

based on the answers, comments and discussions, I thought of summarizing what I understood:

• For any system, the energy present (requisite for its existence) is called Internal Energy.
• I as part of the surrounding can interact with the system by either giving or gaining Heat or by applying/extracting work.
• While Heat, Internal Energy and Work help me in altering conditions of a system, say to restore conditions after a particular process is done, they do not give me an Intergrated view of how and if the process is of any use to me (in deriving some productive output)
• Why should I do any process? Because it can give me energy which can be used for other applications. This energy is termed as Gibbs Free Energy and also called as usable work (because it can be utilized in any know form, say electrical or magnetic or say to cause movement of something, etc.) Gibbs Free Energy tells me the usable energy I can extract from a process. As long as it is negative (it releases usable Energy, I can consider my process to be spontaneous, feasible and possible for it to occur)
• Now is this Gibbs Free energy same as Internal Energy? Internal Energy just tells us about the energy content a system can hold. If Iam unable to extract this energy for any external use, there is no point in using Internal Energy to denote whether a reaction would happen or not. (when it happens, I should get some change in the environment in the form of the usable energy I mentioned above). Gibbs Free Energy on the other hand is defined to tell me this, whether a process is spontaneous or not.
• But is Gibbs Free Energy same as Work done? Gibbs Free Energy is known to be Usable work. In a process apart from the Usable energy released, the process may also involve a change in volume. Say, I have a gas in a container and through a process, I increase the size of the piston. The volume is changing and for this new state of gas of high volume, the internal energy is different. The change in Internal Energy is compensated by work done on/by gas to carry this process. Such a work is called W_pv and clearly this is not included in Gibbs Free Energy. (Iam not able to use it for any external use, so there is no point in considering it to denote spontaneity of a process)

UPDATE : Another way of seeing it has come to my mind, please correcct if its wrong. The first law of thermodynamics can be rearranged to be written as $W_{non-pv} = \Delta U - Q - W_{pv} $ So, any change in one variable, can causes changes in other variables. Different permutations of distribution of the changes leads to different processes. While $W_{pv} $ indicates the work done by changing P and V conditions, it's not necessary that all of it is converted to useful energy or W$_{non-pv} $, ie. Gibbs Energy.

Thank you BuckThorn, Chet Miller, Maurice for your help through your answers and comments!

I know that topics such as these vary a lot as they are interpreted in different ways. Please let me know if you have a better way of defining this as it might help us all!