Fred wrote:
...
>> Controversy solved?
>>
> Yes, but to show how the dH values vary check out this URL
> on Hess' Law.
>
> http://wine1.sb.fsu.edu/chm1045/notes/Energy/HessLaw/Energy04.htm
Yes indeed! Enthalpy variation is path-independent, which makes perfect
sense since it results from bonds (intramolecular AND intermolecular) broken
and made: what matters is the net energy balance between the final bonds in
the final products and the initial bonds in the initial reactants (chemical
bond energy is nothing but coulombic force potential energy, and heat is
kinetic energy, and the sum of them must stay constant if some of the bond
energy turns into heat like in the reaction we have been discussing)
There is only so much bond energy in 2H2(g) and O2(g), and so much bond
energy in H2O(l), net energy is the difference, whatever has happened in
between (besides many things DO happen in between as you pointed out Fred).
This makes me think of a nice exercise I wrote a few weeks ago to convince
someone else of path-independence of produced (or absorbed) energy in a net
reaction. Here it is it's quite relevant to what is discussed here, the
lucky few here who have my spreadsheet (approved by Michael MacKubre, mind
you) can do it without any table lookup nor calculations:
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1/ In the water evaporation reaction H2O(l)->H2O(g), how much energy
produced (or absorbed if negative) per g of water?
2/ Then try the same net reaction but with an intermediate step in
the path:
2a/ Dissociate the liquid water: H2O(l)->H2(g) + 0.5 O2(g)
(by electrolysis or whatever, it's irrelevant)
How many J per g of H2O(l)?
2b/ Now recombine the gases to get vapor: H2(g) + 0.5 O2(g) -> H2O(g)
(explosively, catalytically, it's irrelevant)
How many J per g of H2O(g)?
3/ How many J net for 2a/ + 2b/ per g of water, how does it compare
with the result for 1/?
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Cheers,
Michel
