CCL: Regarding the Gibbs-Helmholtz equation

 Hi Bob and Rob,
I don't think it's necessarily true that internal energy (and, thus, enthalpy) has no microscopic correlates. To go back to Bob's breakdown, change in internal energy is going to in some sense correspond to the change in distribution of atoms and electrons: these are the intramolecular and intermolecular potential energy terms, including the electron distribution term. I don't think the concept of absolute internal energy is particularly meaningful, though, since we don't have a meaningful internal energy zero. Relative internal energy difference between reactants and products does sort of make sense.
A volume change, if any, is work done by the system, or alternatively work received by the system from the surroundings.
But H is not, in a microscopic sense, composed of G and S. All the energy terms are incorporated into H, by defining it as E (or U, which is how I learned it) + PV. So those energy terms end up forming part of G, but only because they were part of H first.
As for entropy and heat capacity, there's a mathematical relationship between those two for sure, but the relationship isn't necessarily straightforward. You'd need to look at the equipartition theorem for more details. In particular, if heat is pumped into a system without changing its temperature, that heat is all absorbed as increased entropy; but to the extent that the heat does change the temperature, it's absorbed as enthalpy/internal energy. Then again, increasing enthalpy/internal energy usually increases entropy in bulk, because there's a greater number of ways to build up the state of the system, but not always. To use a very imperfect analogy, entropy is thus like a shadow: when a person or object moves (increase in internal energy), the shadow also moves (change in entropy), but you don't have to put extra effort into moving the shadow.
This also implies that in general, the higher a substance's heat capacity across a range of temperatures, the richer the ensemble of states that substance can adopt, and the higher entropy it has, throughout that temperature range. So temperature is a "measure" of internal energy, or of enthalpy; difficulty in heating or cooling is a "measure" of entropy.
In an electrochemical cell, this relationship doesn't go away. The cell as a whole, like any other reaction, could release heat; this means delta-H for the cell reaction is negative, and the released heat is part of what increases the entropy of the whole universe, and meanwhile the particles making up the cell are adopting, collectively, a lower-energy ensemble of states. Or it could absorb heat, in which case delta-H is positive; the particles making up the cell are now adopting, collectively, a higher-energy ensemble of states, but one that is in some sense more likely to occur, or less highly organised, so they steal heat from the environment in order to get there.
Delta-G in an electrochemical cell absolutely corresponds to the electrochemical potential generated by that cell; but that's because that potential is really just another way of expressing the non-equilibrium nature of the system.
I don't know if that all makes sense. Increasing the temperature of a real-world system would seem to increase both its internal energy (and thus, its enthalpy) and its entropy. But you can't point to an isolated molecule and say that the entropy of that molecule is such and such. Entropy is a bulk property, and as I understand things, entropy calculations on a single molecule are done to work out the possible states that molecule can adopt and their relative energies, and so they feed into statistical mechanical calculations to work out the ensemble of possible states of a bulk sample of that substance at some temperature T. If one were dealing with reactions of isolated molecules in what would otherwise be a ridiculously hard vacuum, I'm not even convinced a thermodynamic analysis would make sense, any more than it would make sense to expect a single roll of a fair die to turn up 3.5 because that's the average number you would expect from rolling a million fair dice. 
 On 8/5/2023 3:49 PM, Robert Molt r.molt.chemical.physics:+:gmail.com wrote:
 
 I feel this is a much simpler way to understand this.
 E = sum of all "internal" energies
 I think this one is intuitive.
 H=E+PV
The concept of enthalpy was created because, in a realistic situation on Earth, there is always an atmosphere. If we want to talk about the total energy to make/do something, you need the energy of the thing (E) and you need to do some work to push the air out of the way to have the object there (PV work). Enthalpy is the energy plus making room for it. I said "air," but you could imagine the same reasoning if you wanted to put this in the ocean (just push the water out); the point is, you have to move the existing particles out of the way (unless we are in a vacuum). 
 G=H-TS=E+PV-TS
Now we build a step further. On Earth, we also realistically do not need to do THAT much energy; we can let the thermal energy around us "help" us (hence the minus sign). In other words, let us subtract off what the environment can do for us thermally; any realistic process on Earth is going to have the effects of some kind of thermal bath.
Gibbs energy (or Gibbs free energy) is just like any other energy, it's just accounting for the effects of NOT being in a vacuum.
The discussion about how all of G and H relate to more microscopic notions (like rotations, translations, etc.) is erroneous, in my opinion. These are all just energies accounting for differing environmental effects of not being in a vacuum. 
 In terms of "why" is
 "d(delta Grxn)/dT  =  - (delta Srxn)  constant P>"
a true statement, this is an irrelevant question. Josiah Gibbs /defined/ the Gibbs free energy this way (G=H-TS), therefore the aforementioned expression follows from calculus.
I've always had a fantasy that I would write a thermodynamics book the way Griffiths has written "the" book on undergraduate electrodynamics and quantum mechanics. 
 On 8/5/23 7:04 AM, Robert T Hanlon robertthanlon,gmail.com wrote:
Ben - thank you very much for engaging in this discussion.  I want to focus my response here on my belief that delta-G does correspond to a physical quantity.  And this is where I"m getting stuck.  See what you think.
Gibbs created the following equation to quantify the maximum external work that a chemical reaction could produce.
Maximum external work = -delta-Grxn = -(delta-Hrxn – T delta-Srxn)      (constant T,P)
He then applied this equation to analyze an electrochemical cell.  He assigned delta-Grxn to cell voltage and T delta-Srxn to the heating/cooling requirement to maintain the cell at constant temperature.
Again, I'm now seeking a physical understand of these terms:  delta-Grxn, delta-Hrxn, and T delta-Srxn
My assumption is that the following physical/chemical changes occur during a chemical reaction at constant T,P: 
 1)  Orbital electron redistribution
2)  Change in intramolecular potential energies - e.g., vibration, rotation 3)  Change in intermolecular potential energies - e.g., attraction, repulsion 
 4)  Change in volume for given pressure
Each of these can contribute to heat effects, and the sum of all are captured in the value of delta-Hrxn as measured in a reaction calorimeter.
Thus, the value of delta-Hrxn is composed of two heat-effect terms, delta-Grxn and T delta-Srxn, and it is delta-Grxn that determines reaction spontaneity at constant T,P.
So another way to look at my question is:  How do the changes in (1), (2), (3), and (4) line up with these terms: delta-Hrxn, delta-Grxn, and T delta-Srxn?
The entropy values involved in delta-Srxn are based on the absolute entropies of the reactants and the products.  These values take into account both heat capacity and volume. Thus, to me, these values align with (2), (3), and (4). These values do not align with (1).
If the above is true, then delta-Grxn must align with (1), as this is the only effect remaining.  Hence my thought that delta-Grxn quantifies the change in orbital electron energies (and also the voltage in an electrochemical cell) and it is this quantification that determines whether or not a reaction is spontaneous.
It's this logical progression of thought that leads me to having a hard time understanding how the delta-Grxn in (1) is related to the T delta-Srxn in (2), (3), and (4).... especially with regards to the Gibbs-Helmholtz equation: 
 d(delta-Grxn)/dT = - delta-Srxn  (constant P)
In light of the above, I don't understand this equation on a physical/chemical basis.  Hence why I'm stuck. 
 regards,
 Bob
 rthanlon%a%mit.edu
On Fri, Aug 4, 2023 at 3:42 PM Ben Roberts ben^roberts.geek.nz <http://roberts.geek.nz>; <owner-chemistry%a%ccl.net> wrote: 
     Sent to CCL by: Ben Roberts [ben++roberts.geek.nz
     <http://roberts.geek.nz>;]
     Hi Bob,
     To clarify, you're looking for a statistical-mechanical
     explanation for
     how delta-G changes as the temperature changes? So, are you
     looking at
     the same chemical reaction, but carried out at a variety of
     different
     temperatures?
     Delta-G, of course, doesn't correspond to any physical quantity,
     except
     for the relative concentrations (properly, "activities", if I
 recall
     correctly) of reactants and products at equilibrium. So this
     reflects
     the principle that changing the temperature at which a reaction
     takes
     place changes the reaction's delta-G, and thus the position of the
     equilibrium.
     To explain this in statistical-mechanical terms, I'd look at
     Boltzmann's
     equation, S = kB ln W, where W measures the number of ways of
     building
     up the macrostate; it's sort of like a binomial probability. The
     image
     of fifty red balls and fifty blue balls comes to mind; assuming no
     differences in physical and chemical properties between the balls
     other
     than their colours, such that there is no difference between
     attraction
     to the same colour ball and attraction to its opposite, the
     highest-entropy configurations are those where the balls are more or
     less evenly mixed throughout the container, and the
     lowest-entropy ones
     are those where the balls congregate together by colour.
     When a reaction is spontaneous, the entropy of the whole universe
     must
     increase; so a decrease in entropy of the system must be
     outweighed by
     an increase in the entropy of the surroundings. In an exothermic
     (delta-H less than zero) process, this is achieved by giving
     energy to
     the surroundings in the form of heat; the heat helps the
     particles in
     the surroundings to adopt a higher-entropy (faster moving and more
     evenly mixed, so to speak) configuration. Of course, a process
     can be
     both exothermic and increasing the entropy of the system.
     But the higher the temperature of the system and surroundings,
     the less
     marginal difference to the surroundings the temperature increase
     will
     make in terms of the higher-entropy state they can adopt. This is
     reflected in delta-S(surroundings) being inversely proportional to
     temperature. At high temperatures, the system entropy change
     dominates
     the question of whether the reaction will be spontaneous.
     If the reaction causes the system to become higher entropy but is
     endothermic (delta-H greater than zero), we know this because the
     sample
     cools down, so it borrows heat from the rest of the universe. This
     causes the rest of the universe to become slightly lower entropy
     as the
     particles cool down and become less evenly mixed, but not enough to
     offset the increasing entropy of the system. I've seen this with
     dissolution of some salts (I forget the names now; it was in high
     school
     25 years ago) where the act of dissolution causes chilling.
     Does that help at all?
     Regards,
     Ben Roberts
     On 8/4/2023 10:18 AM, Robert T Hanlon rthanlon_._mit.edu
     <http://mit.edu>;
 wrote:
     > Sent to CCL by: "Robert T Hanlon" [rthanlon^mit.edu
     <http://mit.edu>;]
     > In the message I recently sent out concerning my interest in
     the physical
     > explanation of the Gibbs-Helmholtz equation, the delta symbol
     that I used in the
     > equation didn't work in this message box.  So the equation
     shown was incorrect.
     > Here is a re-write of the equation using the word "delta"
     rather than the delta
     > symbol:
     >
     > d(delta Grxn)/dT  =  - (delta Srxn)  constant P>
     >
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