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From: Andrew Usher on 25 Dec 2008 20:06
On Dec 25, 5:41 pm, Edward Green <spamspamsp...(a)netzero.com> wrote:

> But is the tendency of paraffin to decompose into graphite and methane
> (?) important for your observation? I haven't noticed that even when
> I find very old candles that they've evaporated, leaving only a stick
> of graphite with a wick in it.

Not for the observation itself, but see below.

> > > Entropy may always cause _some_ mixing, but it's possible that even at
> > > RT the true thermodynamic equilibrium consists of segregated phases of
> > > various chain lengths, with minor intermixing. After all, we have, in
> > > general, multiple phases in mixtures at room temperature: that's some
> > > kind of compromise between entropy of mixing and ... just to make a
> > > rounded phrase ... enthalpy of mixing.
>
> > Yes, that's how it works. But the enthalpic effects here are very
> > small.
>
> Which, for what it's worth, would favor mixing, and increase the
> probability that the mixed state is actually at or near equilibrium.

I believe the enthalpy does favor mixing in the liquid state (if
that's what
you meant), but probably not in the solid.

> > > > My hypothesis says that any solid that does not form one or more
> > > > crystalline phases must be unstable, in the sense that there is some
> > > > other molecular arrangement with lower free energy.
>
> > > I wonder. I'd have to say "no opinion" for now. Just for the sake of
> > > argument, why couldn't a partially ordered pseudo-crystal be the true
> > > equilibrium?
>
> > Actually, I meant stable in the first sense when I said this. That is,
> > hydrocarbons except methane would count as unstable.
>
> Well, I still think you're mixing apples and oranges here. At first
> the question was about the stability and thermodynamic equilibrium of
> solidified paraffin. I think that's at least a somewhat interesting
> question, for which the obvious simplifying assumption would be "the
> chains are all stable".

It is an interesting question, but neither of us have data on the
actual
structure of wax so I don't know how far I could go in that direction.

> Saying the solid paraffin is unstable because
> the molecules are unstable is like saying a perfect crystal of a
> single isomer of hexane is unstable at any temperature because hexane
> is unstable wrt decomposition into graphite and methane.

No. I don't _usually_ use the word 'unstable' in this sense, which is
just
why I qualified it above - 'in the sense ...'.

> > > I see now your original hypothesis was "at sufficiently low
> > > temperatures"... that makes it weaker (hence more plausible). At the
> > > limit of absolute zero we are guaranteed a lowest energy state as the
> > > thermodynamic equilibrium state: note, just to be cute, this doesn't
> > > immediately exclude mixing or disorder -- so long as the mixed or
> > > disordered state happened to be lower (or at least as low) in energy
> > > than any fully ordered state. Do you know any reason this possibility
> > > is excluded?
>
> > No, I don't. It's just that in every case we know, the crystalline
> > state
> > has lower energy, so I make the generalisation.
>
> Heh... if we _did_ find a system with an equilibrium disordered phase
> at 0K we would violate one of the "laws of thermodynamics" -- not one
> of the main ones, mind you, but a kind of add on law which essentially
> says "that doesn't happen". Well, possibly (it's the third law). I'm
> not really clear on this.

The third law states that the entropy of an equilibrium system must
reach
zero (that is, the lowest possible) at the absolute zero. As only a
perfect
crystal has no disorder, I get my statement.

By the way, this is why the decomposition into graphite and methane
_does_ matter: we agree that paraffin wax is not going to unmix and
can
not form perfect crystals. But this doesn't violate the third law
since there
is a lower-energy state: graphite and methane, which do from perfect
crystals.

> I remember the statement as involving
> crystallization at 0K, but that's not the one I find immediately.

Yes, it is commonly (but mistakenly) stated as about perfect
crystals,
and is then tautological.

Andrew Usher
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