Account for the following observations in terms of the type and strength of intermolecular forces. (a) The melting point of solid xenon is -112 C and that of solid argon is 2189 C. (b) The vapor pressure of diethyl ether (C2H5OC2H5) is greater than that of water. (c) The boiling point of pentane, CH3(CH2)3CH3, is 36.1 C, whereas that of 2,2-dimethylpropane (also known as neopentane), C(CH3)4, is 9.5 C.
I'm overall confused on the whole problem and don't know how to start
3F.19
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Re: 3F.19
(a) The melting point of solid xenon is -112 C and that of solid argon is 2189 C. (b) The vapor pressure of diethyl ether (C2H5OC2H5) is greater than that of water. (c) The boiling point of pentane, CH3(CH2)3CH3, is 36.1 C, whereas that of 2,2-dimethylpropane (also known as neopentane), C(CH3)4, is 9.5 C.
You can consider this based on types and relative strengths of intermolecular forces; typically, the stronger the IMF, the more resistant it is to changes in state.
(a) Xe is larger than Ar, which means Xe has higher polarizability (a more easily distorted electric field), and thus greater induced dipole-induced dipole forces. Because Xe has greater IMF, more energy is required to break those forces from solid to liquid state, and so it has a higher melting point.
(b) Similarly, water has polar bonds and higher IMFs than diethyl ether. This means that water has stronger interactions that hold water molecules more tightly together, which means it's more difficult to vaporize, which leads to lower vapor pressure of liquid water at a given temperature.
(c) First, note that pentane and neopentane both have induced dipole-induced dipole forces. Though both have the same molar mass, pentane is a linear-shaped molecule in contrast to neopentane's spherical shape. This means that pentane will have higher surface area/greater exposure to IMFs --> greater IMFs --> higher boiling point.
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