Question

Derivation of spontaneous change has \(\Delta G \le 0\).

Answer 1

When doing a reaction in chemistry, it's usually in an open beaker sitting on the table. The temperature of the room is constant, just room temperature, and the pressure is constant, since it's just the atmospheric pressure outside that day. Because of this, it'll be useful to us to immediately get this out of the way that for our chemical process we have the change in temperature and change in pressure equal to zero, in fancy calculus symbols:
\[dP=0\]\[dT=0\]

(If this scares you, just pretend I have written \(\Delta P = 0\) and \(\Delta T = 0\) instead and throughout, you'll get pretty much the same idea out of this it will just be less general)

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Now we have three unavoidable things we learn in thermodynamics, the first and second law along with a little formula for work.

First off, that the change in internal energy is the sum of the change in work and change in heat \[ dU = dq+dw\]
And we have that the entropy is always greater than or equal to the change in heat divided by the temperature,
\[dS \ge \frac{dq}{T} \implies T*dS \ge dq\]
And finally we have that the change in work is the pressure times the change in volume(I can derive this for you if you ask, this is fun and simple to show surprisingly!),
\[dw = -p *dV\]

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That above is all we need to understand Gibb's free energy. The rest is following the consequences of algebraically moving stuff around and defining what Gibb's free energy is to just be something handy for us to call it. So let's jump into the actual algebra:

Take these two, and add them together to get the first equation:
\[dq \le TdS \]\[dw =-pdV \]
\[dq+dw \le TdS -pdV\]\[dU \le TdS -pdV \]
Just move the two terms on the right to the left side by adding and subtracting:
\[dU +pdV-TdS \le 0\]
Here we get tricky and add a fancy form of zero, since recall from the beginning of the discussion, \(dT=0\) and \(dp=0\) so we can add zero to anything, we add it to the left side after multiplying them in a fancy way:

\[dU +Vdp+pdV -SdT-TdS \le 0\]

Why did we do this? Well we are doing this to reverse the product rule from calculus:
\[d(fg)=gdf+fdg\]
Let's apply it to these two terms here:
\[dU+d(pV)-d(TS) \le 0\]
another calculus rule to apply here, is that the derivative is a linear operator. All that means is we can do this:
\[d(f+g)=df+dg\]
So again we reverse this step to get:
\[d(U+pV-TS) \le 0\]
Since \(H=U+pV\) we plug that in:
\[d(H-TS) \le 0\]

And then let's just go ahead and call this thing here G by definition!
\[G \equiv H-Ts\]
\[dG \le 0\]
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That concludes the derivation but remember, this was all just algebra (ok and some calculus) but we didn't introduce any new things along the way after introducing the chemistry concepts, we just used the math to combine them into one tiny line.

Answer 2

Thank you.
There's a related question what factors in cell potential at physiological environment which relates to Gibbs Free and pH.

Answer 3

fundemental maxwell relations i think from my thermodynamic days?

Answer 4

Closely related! The Maxwell relations are what hinge upon the idea that order of taking second derivatives doesn't matter, so for instance:

\[dU=TdS-pdV\] can be looked at as being a total differential:
\[dU = \frac{\partial U}{ \partial S} dS+ \frac{\partial U}{\partial V} dV\]
Then from here we equate the partials to the corresponding parts in the first equation, and then take their second derivatives to get:
\[\frac{\partial T}{\partial V} = - \frac{\partial p}{\partial S}\]

And of course you gotta keep a lot of things constant during these processes.

Answer 5

Thank you for the explanation woodward

Answer 6

I need to chew on this for a while