Gibbs energy stuff.

Question

Gibbs energy stuff.

kainui · Answer

Well I was gonna just start deriving it but I think I actually derived it here before so I kinda stopped haha what are we doing about gibbs stuff?

kainui · Answer

The main things are "ok what do I want to do and under what conditions"

kainui · Answer

Well what you'd like to do is just be able to look at just the current state of the beaker with molecules in it and say "yeah this reaction happens" or "nah it's not gonna happen".

So we can take this as a fact of the universe, that (I'm kinda debating if this is a good starting point or not):

$$dS \ge \frac{\delta q }{T}$$

from here, we can throw in $dU = \delta q + \delta w$ to rearrange things around a bit:

$$TdS \ge \delta q$$
$$TdS + \delta w \ge dU$$
Now if we only allow ourselves to look at volume changes as contributing to the work done, such as a the system making gas or expanding or contracting, then $\delta w = -PdV$

$$TdS -PdV \ge dU$$

So now what's next? Well this is chemistry so we have to really think about what our system is and what it's doing. 

We usually will have it on a hot plate in the open air. So when the gas is released, it'll end up going into the atmosphere where the pressure remains the same. If it's at the boiling point of the solvent, like say 100 degrees C of water then we can maintain a constant temperature quite easily. Or -320 degrees C for the boiling point of liquid nitrogen, or -78 C for the sublimation point of dry ice.

That all is to day, $dP=0$ and $dT=0$. There's no change in pressure or temperature during many chemistry reactions cause they're controlled fairly well. You can even have a beaker that's got an inner filling like a thermos that flushes water through at a temperature to keep your system the same temperature -- I've seen this and done this before.

OK so back to the derivation...!

kainui · Answer

Here we are now, work out the product rule to see that it all ends up back to the equation above. $$0 \ge d(U+PV-TS)$$

There's another state function already defined here though, $U+PV= H$ so just throw that in,

$$0 \ge d(H-TS)$$

Just like H is an arbitrary place holder for some convenient stuff all in one place, we do the same and call this quantity 
$$G = H-TS$$

Since it obeys the cute rule that

$$0 \ge dG$$

that's true if the process is spontaneous. How do we know it's spontaneous? Cause we just got to it by merely algebraically rearranging the condition for entropy to increase, but now it's packaged in a nice form that's a bit more convenient to use.

kainui · Answer

Now that's just one aspect of it, we can look at G in and of itself to see whatever other sorta things it has, let's say we follow G along some reversible path at constant temperature and pressure, then what do we have?

$$\Delta G = \Delta H-T\Delta S$$$$\Delta G=\Delta U+P\Delta V-T\Delta S$$

Using the fact that for reversible processes at all the reasonable assumptions mentioned a second ago, we can also look at the non volume work as well:

$$\Delta U = T\Delta S -P \Delta V + w_{rev}$$
Plug this in:

$$\Delta G =  w_{rev}$$

At constant temperature and pressure. Helmholtz energy is similar but ends up being at constant temperature and volume.

There's a cute relationship between the two, and the run through for A is similar to G actually in a lot of ways, their creation is basically "hey let's hijack spontaneity of entropy to try to focus in on specific cases of stuff happening"

$$G = A + PV$$

kainui · Answer

which line of the last post does the skip in steps lose you

parthkohli · Answer

Using the fact that for reversible processes at all the reasonable assumptions mentioned a second ago, we can also look at the non volume work as well:
$$\Delta U = T\Delta S -P \Delta V + w_{rev}$$

parthkohli · Answer

I mean how'd you get that $w_{rev}$ into the equation?

kainui · Answer

Oh ok, that's just extra work, so the total work is:

$$w_{tot} = w_{volume} + w_{non-volume}$$

So $$w_{volume} = -PdV$$
and
$$w_{non-volume} = w_{rev}$$

It's just the extra non volume work I shoulda labelled it better

kainui · Answer

So that work could be used to drive electrical current

parthkohli · Answer

Nah, I do get that much, but still, what makes non-volume work a part of the equation, and in particular, also equal to the change in Gibbs energy?

kainui · Answer

Yeah it's subtle cause at constant temperature and pressure this stands on its own:

$$\Delta G = \Delta U + P\Delta V - T\Delta S$$

But an entirely different formula is:

$$\Delta U = q + w$$

So now usually we consider $q = T \Delta S$ and $w = -P \Delta V$ but in general there is actually more work than this that can happen. So we write $w = -P \Delta V + w_{ex}$ for the extra non volume work.

Plug this all in and everything cancels except the $w_{ex}$. Yeah I know how you are confused but not sure exactly how to get you through this since thermo is kind of a weird web to walk.

kainui · Answer

Yeah it's a bit strange to me, cause on the one hand you end up working up some stuff and say "oh hey this quantity ends up being true for spontaneous reactions, let's just call it G" then you turn around and use it on its own to see what other different things it implies. Tricky, I am pretty sure that Gibbs and all them were on a whole nother level cause it's just way too hardcore for me.

Supposedly Gibbs is the guy who invented the dot and cross product too. wew

parthkohli · Answer

Whattt

Dayum, that's genius.

frostbite · Answer

@Kainui @ParthKohli you can also try do as Gibbs were trying to investigate: How to get the maximum useful work at constant pressure and temperature $dw_{add,max}=dG_{t,p}$

So basically to discover how much useful energy a system contain when you have removed the energy lost as thermal energy. This is also why the Arbeitz funktion has the look it does: The internal energy subtracted the energy lost at the specific temperature of the system.

I do not know if Gibbs was actually aware of the relationship he actually found in terms of spontaneity, that is:
$$\Large \Delta S_{total}=\frac{ - \Delta G }{ T } ~~ ~~dT=0, ~~~dP=0$$

frostbite · Answer

Or if gibbs went through the molecular interpretation using molecular mechanics:
$$\Large S=\frac{ U(T)-U(0) }{ T }+ k \ln(Q)$$

frostbite · Answer

statical mechanics*