[galpinj]

Hello guys,

I have a few conceptual questions regarding entropy. I've learned that S = Q / T, and that entropy is zero at absolute zero; however, if the temperature were zero in the above equation, wouldn't S equal infinity? How do chemists consolidate these opposing facts?

Please refer to this link: http://www.eoht.info/m/page/Absolute+zero

He states that at absolute zero, entropy approaches infinity, which is in direct opposition to the third law of thermodynamics.

Secondly, I've had difficulty understanding those tables that give values for ΔH, ΔG, and ΔS. I understand that ΔH is the heat of formation, but how did they calculate the ΔS?

Lastly, I'm having issues conceptually understanding ΔG = ΔH - TΔS. If entropy can be measured by Q/T (where Q is heat input), doesn't it, in essence, measure the same thing as enthalpy? If an endothermic reaction needs heat (ΔH), then ΔS will be the heat that was put in, no?

Please help as this is confusing me so much

[Borek]

I've learned that S = Q / T

Nope, it is dS=dQ/T - we can calculate entropy change, but not an absolute value. Assuming it is zero at T=0 is just setting a reference point.

[mikasaur]

In your linked article the author states that the value is "infinite or undefined". Undefined is a better term for this. There's actually a really interesting YouTube video about zero and what happens when you divide by zero: https://www.youtube.com/watch?v=BRRolKTlF6Q

I think you may be confusing S with ΔS. For a table like this they define H⁺(aq) as being S = 0 and everything else is measured around it. The difference in absolute entropy of the products and reactants represents ΔS for the reaction.

[Irlanur]

https://en.wikipedia.org/wiki/Third_law_of_thermodynamics

[galpinj]

https://en.wikipedia.org/wiki/Third_law_of_thermodynamics

I thought it might be ΔS = ΔQ /T. Nonetheless, when T is equal to zero S will be equal so some undefined/infinite number. Mathematically, what equation are they using to prove that S is zero when T = 0K?

[galpinj]

In your linked article the author states that the value is "infinite or undefined". Undefined is a better term for this. There's actually a really interesting YouTube video about zero and what happens when you divide by zero: https://www.youtube.com/watch?v=BRRolKTlF6Q

I think you may be confusing S with ΔS. For a table like this they define H⁺(aq) as being S = 0 and everything else is measured around it. The difference in absolute entropy of the products and reactants represents ΔS for the reaction.

That was a really interesting link, and I've now spent a great deal of time trying to better understand when/why zero is undefined. I'm still having some trouble relating it to entropy, however...

[mikasaur]

From the wikipedia article:

The entropy of a perfect crystal lattice as defined by Nernst's theorem is zero provided that its ground state is unique, because ln(1) = 0. If the system is composed of one-billion atoms, all alike, and lie within the matrix of a perfect crystal, the number of permutations of one-billion identical things taken one-billion at a time is Ω = 1. Hence:

[itex]S-S_0=k_B\ln\Omega=k_B\ln1=0[/itex]

The difference is zero, hence the initial entropy S₀ can be any selected value so long as all other such calculations include that as the initial entropy. As a result the initial entropy value of zero is selected S₀ = 0 is used for convenience.

There's also a mathematical formulation section in the wikipedia article for another explanation. So if you're really looking for a mathematical equation, there you go.

Your desire to understand is commendable, but I think you might be thinking a little too hard on this. The details and the "whys" of thermodynamics can be incredibly complicated. If you think of entropy as a measure of disorder where gases are more disordered than liquids which are more disordered than solids, then a really, really solid solid at a really low temperature (the lowest temperature possible) has 0 disorder. This is absolute 0, where S = 0.

[Borek]

what equation are they using to prove that S is zero when T = 0K?

It is not something you can prove. S=0 at T=0 is just a defined reference point.

[galpinj]

Yeah, perhaps I'm trying to understand this on too deep of a level. Unfortunately, math isn't my greatest strength, so I have trouble understanding the equations that some people have outlined. I was more hoping for a conceptual reasoning for the difference. I'll just take it at face value that S=O when T=0

[Irlanur]

The Problem with Thermo or Physics in general is that there is no such thing as an accepted set of axioms. One way to handle your question is that you simply postulate the reference point S=0 at T=0, which is the case in Phenomenological Thermodynamics.
Also your Mathematical problem isn't really one. You have an term f(x)/x this can easily converge to 0 at x=0, you have to look at the mathematical limit/limes, and you might need to use the rule of l'hôpital. You can't just say it goes to infinity because x=0.

In a statistical thermo approach, the entropy at equilibrium is defined as the logarithm of the number of accessible microstates. At absolute zero and without degeneracies you only have a single ground state populated and ln(1)=0.

[mjc123]

To add to what others have said, dS = dQ_rev/T - it is only equal to dQ/T when the process is reversible (i.e. is in equilibrium at all times).
At equilibrium dG = dH -TdS = dQ_rev - T*dQ_rev/T = 0
For a non-reversible process (and that includes a "reversible" reaction approaching equilibrium from a non-equilibrium condition) dG ≠ 0 and dS ≠ dQ/T.

[galpinj]

To add to what others have said, dS = dQ_rev/T - it is only equal to dQ/T when the process is reversible (i.e. is in equilibrium at all times).
At equilibrium dG = dH -TdS = dQ_rev - T*dQ_rev/T = 0
For a non-reversible process (and that includes a "reversible" reaction approaching equilibrium from a non-equilibrium condition) dG ≠ 0 and dS ≠ dQ/T.

Oh I see! So because the reaction is not reversible when the process is out of equilibrium, the ΔH value will not be the same as the ΔS. So at equilibrium, the minute changes are occurring at a constant temperature, and ΔG will obviously be zero. I think I get it...