HW 9.23

Boltzmann Equation for Entropy:

Moderators: Chem_Mod, Chem_Admin

Paul Adkisson 1D
Posts: 23
Joined: Wed Sep 21, 2016 3:00 pm

HW 9.23

Postby Paul Adkisson 1D » Wed Jan 25, 2017 7:37 pm

Could anyone explain this problem? The Solution manual states that COF2 has greater residual entropy than BF3 because the Fluorine atoms could occupy the same location as the oxygen atom. What does that even mean? I thought that since COF2 is polar it would be more likely to line up in a particular orientation (namely with the oxygen atom of one molecule next to the fluorine atoms of an adjacent molecule), and thus have lower residual entropy.

Posts: 18918
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 745 times

Re: HW 9.23

Postby Chem_Mod » Tue Jan 31, 2017 9:33 am

So both BF3 and COF2 exists as a gas. Comparing how gas and liquid are dispersed, the polar argument could be made for maybe liquid or solid, but the polar interaction is very short ranged that in the gas phase, it is negligible. With this said, imagine placing BF3 in the x,y,z cartesian coordinates, no matter how many 120 degree rotation about the central B atom, the final orientation will always be the same. But COF2, if you try the same thing, will be different for each 120 degree rotation.

Return to “Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy”

Who is online

Users browsing this forum: No registered users and 1 guest