Chemical Thermodynamics 2.8 - Rotational Partition Function

[@mikusxlr]

This subject is the hardest I have had to endure, and I'm in my third and final quarter of it. Thank you so much for these explanations!!! They have been extraordinarily helpful.

[@TMPChem]

Hi Mike. Physical chemistry is certainly the most challenging course in an undergraduate chemistry degree for a large fraction of students. Luckily it's not impossible, and far less difficult than might be perceived on first impression if you've got someone to help guide you through the forest. Glad to be of service. Good luck with the rest of the course.

[@uniquelyleira]

I'm wondering why the q_rot in this video has pi inside the square root (beside T^3) but in the first video (2.0), it wasn't there.

[@RodrigoLemosSilva-hp9tl]

Nice video, I liked so much of your excellent explanation!!!! Can you give more explanation about the symmetry number, or give a good reference about it (a book maybe)? Thank you very much again for this video! I subscribed myself in your channel!

[@TMPChem]

Hi Rodrigo. The symmetry number in general represents the number of unique, indistinguishable ways an object can be rotated to perfectly overlap with its original self. In a symmetry / group theory context, this number is equivalent to the number of proper rotation *operations* for a molecule, plus one. For example, benzene is a D6h molecule. It has a C6 principle axis, and 6 C2 axes which are perpendicular to the C6. This gives a total of 5 (from C6) + 6 * 1 (from each C2) + 1 (the extra 1) = 12, so sigma = 12 for benzene. Linear molecules can have at most 1 unique C2 operation (and only if they have an inversion center), thus for linear molecules sigma is either 1 (C_infV) or 2 (D_infH). Textbook sources on this topic are fairly scant in my experience, but two recommendations would be McQuarrie Physical Chemistry chapter 18 section 6, and the "nonlinear molecules" section of the Wikipedia page for rotational partition function. Good luck!

[@RodrigoLemosSilva-hp9tl]

Thank you very much for your explanation...you have helped me soo much!