predicting entropy

Boltzmann Equation for Entropy:

Moderators: Chem_Mod, Chem_Admin

Payton Kammerer 2B
Posts: 49
Joined: Tue May 01, 2018 3:00 am

predicting entropy

Postby Payton Kammerer 2B » Tue Mar 10, 2020 11:25 pm

4G.3 Which would you expect to have a higher molar entropy at T=0, single crystals of BF3 or COF2? Why?

The answer given in the solutions manual says that COF2 is the right answer because it is asymmetrical, with an O and two F instead of three F. I also said it was COF2, but I said that it was because this molecule had less electronegativity difference between its atoms to influence the molecular orientation away from randomness. Is this correct as well? If so, which factor, polarity or asymmetry, takes precedence in answering this sort of question, and why?

Myka G 1l
Posts: 100
Joined: Fri Aug 30, 2019 12:17 am

Re: predicting entropy

Postby Myka G 1l » Wed Mar 11, 2020 12:21 am

Even though BF3 and COF2 are both trigonal planar, COF2 has both oxygen and fluorine atoms occupying the same locations, as opposed to BF3 having only fluorine occupy these locations. It would be more likely for disorganization to occur within the COF2 molecule because of the different atoms rather than the identical fluorine atoms occupying sites around the boron atom. I think the electronegativity is a factor but not as important and the asymmetry.

Ryan Yee 1J
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

Re: predicting entropy

Postby Ryan Yee 1J » Fri Mar 13, 2020 6:57 pm

Because they both have the same geometric shape, the orientation due to asymmetry is the biggest contributor to their different entropies, and the asymmetry just also contributes to polarity.

Posts: 101
Joined: Wed Feb 27, 2019 12:17 am

Re: predicting entropy

Postby Ayushi2011 » Sat Mar 14, 2020 3:09 pm

More the disorganization, more is the entropy which is why COF2 will have more entropy compared to BF3.

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