December 27, 2024, 03:25:37 PM
Forum Rules: Read This Before Posting


Topic: Enthalpy change = 0  (Read 17535 times)

0 Members and 1 Guest are viewing this topic.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Enthalpy change = 0
« Reply #30 on: May 14, 2013, 05:25:31 AM »
Sorry to bring this up again, but I was reading over it and it's thrown up a few more queries.

1) I don't think I understand the point about isotopic exchange reactions having K=4 when we take the assumption that ΔHr°=0. You pointed out that this is because it was "at natural abundance", but natural abundance only shows that the system has reached equilibrium. So then, for any isotopic exchange reaction, the constant is independent of isotopic abundances after all (so long as these abundances are at equilibrium, they just cancel out, so we do not actually need the abundances to derive the constant). I showed this math a few posts ago (sorry it wasn't in LaTeX, it was a multi-step thing).

2) curiouscat wrote a few posts above:

it'd be interesting to try and get the same answer for ΔS using a more direct approach. i.e. counting possible configurations etc.

I assume by "configurations" you mean isotopic combinations (as opposed to microstates or something quantum, which is probably better avoided for me). How would we go about doing this?
« Last Edit: May 14, 2013, 09:11:56 AM by Big-Daddy »

Sponsored Links