Attractions and Boiling-Point Temperatures
The temperature at which a liquid boils differs for different substances because it depends on the force of attractions between a substance’s particles. The stronger the interparticle attractions, the higher the temperature at which the substance will boil. Therefore, if you could predict the relative strengths of different attractions, you could predict which of two substances should have a higher boiling point temperature.
The task of predicting relative boiling point temperatures includes several of the tasks covered in chemistry classes. The figure below shows the connections between these skills. The skills it lists are all interrelated: to do any of them successfully, you need to be able to do the ones that lie above it.
Connections Between Skills This figure shows the connections between the skills covered in earlier chapters and the skills presented in this chapter. To master the new skills, you need to have mastered the old ones. Start at the top of the sequence and work your way down, convincing yourself as you go along that you can do each task.
If you can predict the types of attractions between particles in two different substances, you can also predict the relative strengths of those attractions, and then the relative boiling point temperatures for the substances. Increased strength of attractions leads to decreased rate of evaporation, decreased rate of condensation at equilibrium, decreased concentration of vapor, and decreased vapor pressure at a given temperature. As a result, higher temperatures are necessary to reach the vapor pressure required for boiling. In summary, stronger attractions between particles lead to lower equilibrium vapor pressures and higher boiling point temperatures.
Sample Study Sheet: Predicting Types of Attractions and Relative Strengths of Attractions Between Particles
Tip-off – You are asked to predict the relative strengths of attractions between particles of two substances, or you are asked a question that cannot be answered unless you know those relative strengths. (For example, you are asked to compare certain properties of substances, such as their relative boiling point temperatures.)
General Steps -
Determine the type of attraction between the particles using the following steps (summarized in a Figure below).
Step 1: Classify each substance as either a metallic element, carbon in the diamond form, another nonmetallic element, an ionic compound, or a molecular compound.
Metallic elements have metallic bonds. Go to Step 5.
Carbon in the diamond form–C(dia)–has covalent bonds between its atoms. Go to Step 5.
The atoms of the noble gases and the molecules of other nonmetallic elements are held together by London forces. Go to Step 5.
The ions in ionic compounds are held together by ionic bonds. Go to Step 5.
Continue to Step 2 for molecular compounds.
Step 2: For molecular compounds, draw the Lewis structure for the molecule.
If the Lewis structure contains an O-H, N-H, or H-F bond, the attractions that are broken when the substance boils are hydrogen bonds enhanced by London forces. Go to Step 5.
For other molecular compounds, go to Step 3.
Step 3: If there are no O-H, N-H, or H-F bonds, determine the polarity of the bonds.
If there are no polar bonds, the molecules are nonpolar and London forces are broken when they boil. Go to Step 5.
If there is at least one polar bond, go to Step 4(d).
Step 4: Predict whether the polar bonds are symmetrically or asymmetrically arranged.
If the distribution of polar bonds is symmetrical and their dipoles equal, the molecules are nonpolar and London forces are broken when they boil.
If the distribution of polar bonds is asymmetrical, or symmetrical with unequal dipoles, the molecules are polar and dipole-dipole attractions enhanced by London forces are broken when they boil.
Step 5: Although we cannot predict the relative strengths of attractions between all particles, we can apply one of the following guidelines to predict the relative strengths of attractions between some particles.
For substances that contain particles of about the same size, the substances with chemical bonds (ionic, covalent, or metallic) have stronger attractions between particles than substances with intermolecular attractions (hydrogen bonds, dipole-dipole attractions, or London forces). Chemical bonds are generally stronger than intermolecular attractions.
For molecular substances that contain molecules of about the same size, substances with hydrogen bonds have stronger attractions between the particles than substances with either dipole-dipole attractions or London forces, and substances with dipole-dipole attractions have stronger attractions between the particles than London forces. Hydrogen bonds are generally stronger than dipole-dipole attractions, which are generally stronger than London forces.
For molecular substances that have the same type of intermolecular attraction, larger molecules form stronger mutual attractions. Larger molecules tend to have stronger attractions.
EXAMPLE – Predicting Types and Strengths of Attractions Between Particles:
For each of the following substances, write the name for the type of particle that forms its basic structure and the name of the primary type of attraction between these particles. From each pair of substances, choose the one that you expect to have the stronger interparticle attractions.
a. ethylene, C2H4 (used to make polyethylene plastic)
and propylene, C3H6 (used to make polypropylene plastics for toys and other uses)
b. sodium fluoride, NaF (used to fluoridate municipal water)
and nitrogen trifluoride, NF3 (oxidizer for high-energy fuels)
ammonia, NH3 (used to make fertilizers)
and methane, CH4 (major component in natural gas)
methanol, CH3OH (used as a solvent and a fuel)
and potassium, K (used with sodium in heat-exchange alloys)
e. 1-propanol, CH3CH2CH2OH
waxes and vegetable oils)
and methanol, CH3OH (used to make formaldehyde for embalming fluids)
in the diamond form, C(dia) (used
for windows in space probes)
and fluorine, F2 (used to make compounds added to toothpaste)
Ethylene, C2H4 and propylene, C3H6 are both hydrocarbons, so they are nonpolar
molecular substances with London forces operating between the molecules. Larger
molecules exert stronger London forces, so the attractions between C3H6
molecules are stronger than those between C2H4 molecules.
Sodium is a metal, and fluorine is a nonmetal, so we predict that NaF is an
ionic compound, held together with ionic bonds. The compound NF3 is a molecular
substance with an asymmetrical distribution of polar bonds, so it is a polar
molecular compound with dipole-dipole attractions between the molecules
(enhanced by London forces). Ionic
bonds are stronger than dipole-dipole attractions, so NaF has the stronger
attractions between particles.
Ionic bonds are stronger than dipole-dipole attractions, so NaF has the stronger attractions between particles.
formula NH3 represents a molecular, substance possessing N-H bonds, so it is a
polar molecular compound with hydrogen bonds between the molecules (enhanced by
London forces). Methane is a hydrocarbon, so it is a nonpolar molecular
substance with London forces operating between the particles. For molecules of
about the same size, hydrogen bonds are stronger than London forces, so NH3 has
the stronger interparticle attractions.
is an alcohol, and alcohols are polar molecular compounds with hydrogen bonds
that link the molecules together (enhanced by London forces). Potassium atoms
are held together by metallic bonds. Metallic bonds are stronger than hydrogen
bonds, so potassium has the stronger interparticle attractions.
substances are alcohols, which are polar molecular compounds. The primary
attractions between alcohol molecules are hydrogen bonds, but London forces also
play a part. The larger 1-propanol molecules have stronger London forces between
and fluorine are both nonmetallic elements. Carbon in the diamond form is
composed of atoms held together by covalent bonds. Fluorine, F2, is composed of
molecules held together by London forces. The covalent bonds in a diamond are
much stronger than the London forces between F2 molecules.
EXAMPLE – Predicting Relative Equilibrium Vapor Pressures and Boiling Point Temperatures:
Consider the following pairs of molecules. Which substance in each pair would you expect to have the higher equilibrium vapor pressure (at equal temperatures), and which do you think would have the higher normal boiling point?
a. The ketone 2-heptanone has a clove-like odor and is found in oil of cloves. It also contributes to the odor of bleu cheese. Pentanoic acid, or valeric acid, is also known for its odor and in fact is used for flavorings and perfumes.
b. Butane gas, C4H10, can be used as fuel to cook a steak. Part of the smell as the meat cooks is due to the formation of acrolein, CH2CHCHO.
a. The ketone 2-heptanone is a polar molecular compound with dipole-dipole attractions between the molecules. Pentanoic acid, like all carboxylic acids, has molecules mutually attracted by hydrogen bonds. For molecules of about the same size, hydrogen bonds are stronger than dipole-dipole attractions, and so the attractions are stronger between pentanoic acid molecules. Stronger attractions lead to a lower equilibrium vapor pressure and a higher normal boiling point temperature. With an external pressure of 1 atm, pentanoic acid boils at 185.4 °C, and 2-heptanone boils at 151.5 °C.
b. Butane is a hydrocarbon, so it is a nonpolar molecular compound with London forces between the molecules. The C-O bond in acrolein is polar, so acrolein is a polar molecular substance with dipole-dipole attractions between the molecules. For molecules of about the same size, dipole-dipole attractions are stronger than London forces, so we expect acrolein to have a lower equilibrium vapor pressure and a higher normal boiling point temperature than butane. At one atmosphere of pressure, acrolein boils at 52.5 °C, and butane boils at –0.5 °C.