I suspect that part of the problem here lies in how the high school level
defines 'entropy' (or has it defined for them by their books). It appears
that JD's definition is a great deal broader than the one we tend to see at
the high school level. In point of fact, having read what he's written, I
have no idea what he means by the term. I'll undoubtedly say something
ELSE that John will object to, but THIS is how the term is used at the high
school level:
The Gibbs Free Energy Change (delta-G) = (delta-H) - T(delta-S), where
delta-H is the change in enthalpy (positive delta-H indicates endothermic -
heat absorbed; negative delta-H indicates exothermic - heat released), T is
temperature on the Kelvin (Absolute) scale (always positive), and delta-S
is the change in entropy. At the high school level, entropy is equated
with the state of matter, where an increase in modes of freedom indicates
an increase in entropy. This is, apparently, something John disagrees
with. Anyway, if the delta-G is negative, a reaction will occur
spontaneously; if positive, the reaction would have to be forced to occur
(you'd have to pump in energy). By this definition, entropy is NOT "always
the boss". At high temperature (Kelvin scale), the entropy term tends to
dominate. At lower temperatures, an exothermic reaction, regardless of
what that may do to the entropy, tends to dominate. In answer to the last
question Paul asked, therefore, the reason that an endothermic reaction
might proceed spontaneously is if it occurs at a high enough temperature
and results in a gain in entropy sufficient to overcome the gain in
enthalpy.
A very simple example (which isn't really even a 'reaction') should suffice
to clarify. Water freezes at low temperature, but melts at higher
temperature. Freezing is an exothermic process (delta-H negative), so by
the first term of the Gibbs Free Energy equation, freezing would occur
spontaneously if that were the only consideration (I referred to this as
driving tendency, and, of course, John took issue with my wording). But
freezing decreases entropy because you lose both translational and
rotational freedoms. So delta-S in this case is negative, and since it is
'subtracted', it would work against spontaneous freezing (This I referred
to as the second driving tendency). So water will freeze if the
temperature is low enough for the enthalpy term to dominate, but melts if
the temperature is high enough for the entropy term to dominate.
As I indicated earlier, John seems to equate delta-G with 'entropy', and
that is contrary to everything I was ever taught in high school, college,
and graduate school. That doesn't mean he's wrong; I certainly wouldn't
bet against him, but I would have thought that in the course of attaining
degrees in physics and chemistry, and teaching both at the high school
level for 30 years, I'd have run across this before now.
With respect to John's RDX example; if the formation occurs spontaneously,
then the products represent the most stable configuration available based
on the particles available and the conditions. It does not mean more
stable than some other combination of particles under some other set of
conditions. I'm not familiar with the reaction specifics for RDX, but
bonded atoms are at some level of stability to start with. Their energy is
raised in order to break existing bonds so that the particles may assume
the configuration which is most favorable in terms of the Gibbs Free Energy
possibilities. The most negative delta-G will be preferred, and, if it is
a different configuration than the reactants, then the new products, under
those conditions, have assumed the most stable configuration available to
them under the prevailing conditions.