HW Prolem (4G.4) 7th Edition

Boltzmann Equation for Entropy:

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Helen Mejia 1I
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Joined: Fri Sep 28, 2018 12:24 am

HW Prolem (4G.4) 7th Edition

Postby Helen Mejia 1I » Sun Feb 10, 2019 3:22 pm

Can someone explain how to identify which of BF3 or COF2 will have a higher molar entropy?

Xuan Kuang 2L
Posts: 31
Joined: Wed Nov 14, 2018 12:23 am

Re: HW Prolem (4G.4) 7th Edition

Postby Xuan Kuang 2L » Sun Feb 10, 2019 4:02 pm

COF2 would have a higher molar entropy than BF3.

When you take a look at the molecular structures, BF3 is a trigonal planar with the same B-F bond on each end. This means that is it very ordered, compared to COF2, which has two F's and an O at the ends. This means that for some molecules, it could possibly have an F or an O at a specific end, so it is less ordered because not every end is identical. Thus, COF2 has a higher molar entropy.

Laura Gong 3H
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Re: HW Prolem (4G.4) 7th Edition

Postby Laura Gong 3H » Mon Feb 11, 2019 2:13 pm

COF2 has greater molecular complexity compared to BF3 which means it has a greater number of possible positions/orientations which makes its entropy greater than BF3 which has lesser number of possible positions.


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