6th Edition Homework 6.19

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Clarissa Cabil 1I
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Joined: Fri Sep 28, 2018 12:19 am

6th Edition Homework 6.19

Postby Clarissa Cabil 1I » Tue Nov 27, 2018 5:36 pm

For part c of homework question 19 in chapter 6 of the 6th edition, the question asks to "Account for the following observations in terms of the type and strength of intermolecular forces. (c) The boiling point of pentane, CH3(CH2)3CH3, is 36.1ºC, whereas that of 2,2-dimethylpropane (also known as neopentane), C(CH3)4, is 9.5ºC."

The solution manual says that "Both molecules have the same molar mass, but pentane is a linear molecule compared to dimethylpropane, which is a compact, spherical molecule. The compactness of the dimethypropane gives it a lower surface area. That means that the intermolecular attractive forces, which are the same type (London forces) for both molecules, will have a larger effect for pentane."

Wouldn't the London forces be stronger in dimethylpropane since the molecule is more compact compared to pentane?

Christopher Tran 1J
Posts: 77
Joined: Fri Sep 28, 2018 12:15 am

Re: 6th Edition Homework 6.19

Postby Christopher Tran 1J » Tue Nov 27, 2018 6:05 pm

London dispersion forces increase as the size of the molecule increases, particularly surface area by which the interactions can take place.
Pentane is linear and not as compact, and therefore should have stronger London forces.


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