Re: gibbs free energy equation

On Jan 24, 8:11 pm, Craig <cager...@xxxxxxx> wrote:
On Jan 24, 3:04 pm, renai <bense...@xxxxxxxxxxx> wrote:

I'm visualizing sold iron being attacked by oxygen gas molecules which
together are transforming into iron oxide solid molecules.  Of course
I understand why the entropy of the products is lower and that thermal
energy is being released via the bond breakage of the reactants vs.
those formed by the products, but why couldn't in principle all of
that thermal energy be soaked up as mechanical work by the
surroundings via say expansion without letting any leak out as heat.
As I said, I know the 2nd law says this can't work but I'm not
intuiting why when visualizing the process step by step.

I'll try to answer your question on the level you're asking for.

Your main question seems to reduce to why the efficiency of converting
heat to work is limited.  What does it look like, on a microscopic
scale, when something gets hot?  Atoms start vibrating and moving
around in every direction possible.  What is work?  Physics defines it
as displacement against an opposing force.  In other words, the work
needed to, say, lift an object, requires a concerted application of
energy up, against gravity.  Intuitively, think of converting heat to
work as like herding cats.  You're trying to get all of that random
thermal energy to "line up" in a particular direction.

Does that do it for you?

- Craig

I understand what your saying (I think), but of course for an
exothermic reaction whereby the entropy of the products is greater
than that of the reactants, gibbs puts no limitations on efficiency,
i.e. quasistatically you can get 100% of the chemical energy released
converted into work since DS of universe would be positive. Do you
see what I mean, or am I just not quite understanding what you mean?
.

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