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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