Phase Equilibrium and Intermolecular Forces 1
17.4 Observation 3: Boiling Points and Intermolecular Forces
Earlier in this study, we determined that the boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the pressure applied externally, e.g. by the atmosphere or by a piston trapping the liquid and gas in a cylinder. When the applied pressure is 1 atm, we refer to this as the normal boiling point. If we compare the normal boiling points of two liquids, the liquid with the higher normal boiling point clearly requires a higher temperature to reach a vapor pressure of 1 atm. From our work on dynamic equilibrium, we know that this higher temperature is required to provide su cient kinetic energy for the molecules in the liquid to overcome stronger attractions between the molecules. Overcoming these attractions is necessary for a molecule to escape the liquid and join the vapor phase.
This line of reasoning means that, when we compare the normal boiling points of two liquids, we are also indirectly comparing the strengths of the intermolecular attractions in those two liquids. The liquid with a higher boiling point has stronger intermolecular attractions.
What determines the strength of these attractions? To nd out, we can analyze experimental data for the boiling points of many liquids and look at the properties of the corresponding molecules. A useful set of compounds to look at are the covalent compounds formed by combining hydrogen with each of the elements in the main group, Groups IV to VII. For example, in the rst row of the periodic table, these include CH4, NH3, H2O, and HF. Table 1 gives the experimentally observed normal boiling points of the sixteen hydrides from Groups IV to VII in the rst four rows of the periodic table.
Boiling Point (C)
178 CHAPTER 17. PHASE EQUILIBRIUM AND INTERMOLECULAR FORCES At rst glance, the values of the boiling points seem to be all over the place. Any patterns that might exist are not obvious. But there are patterns if we look at the data long enough, and those patterns can reveal to us what determines the intermolecular attractions. First, we can see that, for the compounds in each row of the periodic table, the compound with the lowest boiling point is from Group IV: CH4, SiH4, GeH4, and SnH4. Note that these are not the lowest four boiling points in Table 1. Rather, for each period, they are lowest boiling points of the compounds in each period. Notice also that the boiling points increase as we move down the table in Group IV, so that the heavier mass molecules have higher boiling points. This is easiest to see if we put them together on a chart in Figure 4.
Figure 17.3
These two observations suggest that we might nd patterns if we put all of the sixteen hydrides on a chart together, sorted by the period each are in. The result is shown in Figure 5, which is just Figure 4 expanded to show all four groups in the data set of Table 1. With this chart, we see that the two patterns we described for Group IV work for Groups V to VII, but we also see three dramatic exceptions to those patterns in H2O, NH3, and HF.
179
Figure 17.4
First, as we noted before, the hydrides from Group IV are the lowest boiling point compounds in each period. Second, in each group, the boiling point increases as we move down the periodic table. In fact, this second pattern is perhaps the most pronounced trend in the data. From this, we draw our rst conclusion about the strengths of intermolecular attractions: for similar types of molecules, the molecules with larger atoms (more mass, more protons) have stronger intermolecular attractions.
Why would this be? The answer is not obvious but does make sense once we know it. Remember that these are all neutral molecules. As such, we might have imagined that there would no positive-negative interactions. However, each molecule consists of a large of number of protons from the nuclei of the atoms, and an equal number of electrons, both core electrons and valence electrons, which are either shared or unshared. When two molecules are close to each other, the positive and negative charges in each of the molecules interact with each other. We might again imagine that the attractions of opposite charges would be exactly oset by the repulsions of like charges. This would be true if the charges were uniformly distributed in the molecules, as we would expect for non-polar molecules. But when the molecules are close enough to each other, the attractions and repulsions cause the charges to rearrange such that the attractions become signi cantly more favorable.
Such an arrangement of electrons in two adjacent highly simpli ed molecules is shown in Figure 6. Note that two nonpolar molecules become polarized when they are close to each other, due to the attraction of the negative charges in one molecule to the positive charges in the other molecule, and vice versa. The result is a net attractive force between the two molecules. This type of force is called the dispersion force, sometimes also called the London force after the discoverer.
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180 CHAPTER 17. PHASE EQUILIBRIUM AND INTERMOLECULAR FORCES
Figure 17.5
What makes the dispersion force larger? The data in Table 1 and Figure 5 tell us: molecules with more positive and negative charges, like SnH4, have stronger attractions than molecules with fewer positive and negative charges, like CH4. We say that the molecule with more charges is more polarizable, meaning it is easier for the molecule to become polarized in the presence of other electrical charges. The more polarizable a molecule is, the stronger the intermolecular forces will be.
Let's now compare the boiling points of compounds in the same period from Group IV and from Group VII, e.g. SiH4 versus HCl. The boiling point of HCl is larger. However, if we count the charges in these two molecules, we discover that they have the same number of electrons and the same number of protons.
This suggests that the two molecules should be equally polarizable and therefore should have equal disper-sion forces and therefore should have equal boiling points. But this is not true. Something else must be contributing to the dierence in boiling points than just the dispersion forces.
To spot the dierence between these two molecules, we need to rely on our previously developed knowledge of electronegativities, molecular geometries, and molecular polarity. HCl is a polar linear molecular because the Cl atom is much more electronegative than the H atom. Si atoms are actually slightly less electronegative than H atoms, suggesting that SiH4should also be polar. But, we also recall from our electron domain model that SiH4has a symmetric tetrahedral geometry, like CH4. As such, like CH4, SiH4has no molecular dipole moment. From this comparison, we can conclude that, in comparing two molecules with similar dispersion forces, the molecule with a dipole moment will have stronger intermolecular attractions than the molecule
181 without a dipole moment. This explains why, in each period in Figure 5, the Group IV hydride compound always has the lowest boiling point. Each of these compounds has nonpolar molecules due to their symmetry.
By contrast, two HCl molecules will have stronger attractions as the positive end of one HCl will be attracted to the negative end of the other HCl, and vice versa.
We can now conclude that molecules attract one another via dispersion forces and, if the molecules are polar, via dipole-dipole attractions. Understanding these two types of intermolecular attractions works well to explain the major patterns observed in Figure 5.
This does not explain the exceptions, however. Why are the boiling points of NH3, H2O, and HF so abnormally high? Given that these are small mass molecules, we would not expect them to have larger than average dispersion forces. All three are polar molecules, but there is nothing to suggest that there dipole moments are unusually high. There must be a dierent type of intermolecular attraction that is unique to these three molecules out of this set of molecules.
We need a pattern to analyze. What do these three molecules have in common with each other that the other molecules in Table 1 do not? N, O, and F are all strongly electronegative atoms and are also amongst the smallest atoms in the periodic table. In the Lewis structures for all three molecules, O, N and F all have non-bonded, lone pairs of electrons. These three properties, taken together in a single molecule, must present a uniquely strong intermolecular attraction.
Chemists account for this strong bonding via a model called hydrogen bonding. This is a uniquely strong form of dipole-dipole attraction that only occurs when a molecule contains a hydrogen atom bonded to an N atom, an O atom, or an F atom. Due to the strength of the electronegativity of these atoms, the N-H bond or O-H bond or F-H bond is highly polar, meaning that the H atom is almost a bare, positively charged hydrogen nucleus. This strong positive charge on one molecule is in turn strongly attracted to the negatively charged lone pair electrons on the N, O, or F atom of another molecule. This is illustrated in Figure 7.
Figure 17.6
Note that size seems to be important in hydrogen bonding as well. HF has a much higher boiling point than HCl, indicating that HF has stronger intermolecular attractions. Even though the Cl atom is strongly electronegative and has lone pair electrons in the HCl molecule, the Cl atom is apparently too large to support the uniquely strong dipole-dipole attraction we call hydrogen bonding.
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 com-parably sized molecules. Thus, hydrogen bonding dominates intermolecular attractions for those molecules
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182 CHAPTER 17. PHASE EQUILIBRIUM AND INTERMOLECULAR FORCES 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 H2O, 100 ºC, to that of octane (C8H18), 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 dierent 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 rst to consider rst whether the molecules are of comparable sizes or of very dierent sizes.
Provided that the molecules are of comparable size, the dispersion forces should not be too very dierent. 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.
17.5 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 speci c 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 dierence in the intermolecular attractions in terms of molecular structure.
7. In Table 4 and Figure 5, the boiling point of stannane (SnH4) is -52 ºC and the boiling point of phosphine (PH3) is -87.7 ºC. SnH4 is non-polar and PH3 is polar. Explain why the boiling point of SnH4 is nevertheless higher than the boiling point of PH3.
8. Figure 5 shows that the boiling points of the hydrides in the rst period are all unexpectedly high, except for methane (CH4). Explain why CH4is an exception to this trend.