0.16 Phase equilibrium and intermolecular forces  (Page 7/7)

It turns out that other measurements and calculations reveal that, when present, hydrogen bonding attractions are approximately ten times stronger than dipole-dipole attractions or dispersion forces, for comparably sized molecules. Thus, hydrogen bonding dominates intermolecular attractions for those molecules which are capable of hydrogen bonding.

Of course, these conclusions are based on the set of data in Table 1 and Figure 5. These sixteen molecules are all somewhat comparable, consisting of no more than 5 atoms, and no more than one atom other than hydrogen. The boiling points of other molecules can reveal other trends in the strengths of intermolecular attractions. We will take one example to illustrate this. Let’s compare the normal boiling point of H ₂ O, 100 ºC, to that of octane (C ₈ H ₁₈ ), which is 125 ºC. Octane is symmetric, has no dipole moment, and has no N, O, or F atoms that could hydrogen bond. Therefore, octane molecules attract each other entirely through dispersion forces. And yet, the strength of the attractions between octane molecules is greater than that between water molecules. This reveals that the magnitude of the dispersion force can be dominant in comparing molecules of very different sizes. Dispersion forces can dominate both dipole-dipole interactions in polar molecules, and even hydrogen bonding forces.

Therefore, in attempting to predict which of two molecules might have the stronger intermolecular forces, it is important first to consider first whether the molecules are of comparable sizes or of very different sizes. Provided that the molecules are of comparable size, the dispersion forces should not be too very different. In this case, polar molecules will have stronger intermolecular forces than non-polar molecules, and molecules which exhibit hydrogen bonding will have even stronger intermolecular forces.

Review and discussion questions

1. In the phase diagram for water in Figure 1, start at the point where T = 60 ºC and P = 400 torr. Slowly increase the temperature with constant pressure until T = 100 ºC. State what happens physically to the water during this heating process.
2. In the phase diagram for water in Figure 1, start at the point where T = 60 ºC and P = 400 torr. Slowly lower the pressure at constant temperature until P = 80 torr. State what happens physically to the water during this process.
3. Explain why Figure 1 is both a graph of the boiling point of liquid water as a function of applied pressure and a graph of the vapor pressure of liquid water as a function of temperature.
4. Using arguments from the Kinetic Molecular Theory and the concept of dynamic equilibrium, explain why, at a given applied pressure, there can be one and only one temperature, the boiling point, at which a specific liquid and its vapor can be in equilibrium.
5. Using dynamic equilibrium arguments, explain why a substance with weaker intermolecular forces has a greater vapor pressure than one with stronger intermolecular forces.
6. The vapor pressure of phenol is 400 torr at about 160 ºC, whereas the vapor pressure of dimethyl ether is 400 torr at about -40 ºC. Which of these substances has the greater intermolecular attractions? Which substance has the higher boiling point? Explain the difference in the intermolecular attractions in terms of molecular structure.
7. In Table 4 and Figure 5, the boiling point of stannane (SnH ₄ ) is -52 ºC and the boiling point of phosphine (PH ₃ ) is -87.7 ºC. SnH ₄ is non-polar and PH ₃ is polar. Explain why the boiling point of SnH ₄ is nevertheless higher than the boiling point of PH ₃ .
8. Figure 5 shows that the boiling points of the hydrides in the first period are all unexpectedly high, except for methane (CH ₄ ). Explain why CH ₄ is an exception to this trend.