06/14/2004
(K. Stone)
Vapor Pressure, Enthalpy of Vaporization,
and Intermolecular Forces
Mike Perona, California State University, Stanislaus
Purpose: Measure the equilibrium vapor pressure of several liquids at room temperature and explain the results in terms of intermolecular forces. By measuring the equilibrium vapor pressure of a liquid at two temperatures, it will be possible to calculate the molar enthalpy of vaporization for that liquid.
Introduction
You have probably observed that if water is left in an open container at room temperature it will eventually evaporate completely, even though its temperature is never raised to the boiling point. The explanation for this behavior is that some of the molecules in the liquid have enough kinetic energy to overcome the intermolecular forces which hold the molecules of the liquid together, and are able to escape into the vapor phase. The molecules in the vapor phase then diffuse away, and a net loss of liquid occurs.
The evaporative process results in a momentary loss of the most energetic molecules from the liquid sample. Heat energy is continuously absorbed from the surroundings and so the supply of energetic molecules is maintained. This continues until all of the liquid has evaporated.
If the liquid water is placed in a closed container at constant temperature, however, the amount of liquid water will at first decrease and then remain constant. It will appear as if evaporation has stopped. In fact, evaporation does not stop. What happens is that since the vapor phase molecules cannot escape, their number initially increases. This results in an increase in the frequency of their collisions with each other and with the liquid surface. Collisions with the liquid surface lead to condensation. Ultimately the rate of condensation becomes equal to the rate of evaporation, and the relative amounts of water in the two phases remains constant. This corresponds to a condition of dynamic equilibrium. The word "dynamic" emphasizes the fact that both evaporation and condensation are still occurring, but they are occurring at the same rate. The water molecules in the vapor phase exert a pressure on the container walls. This pressure is referred to as the equilibrium vapor pressure of water or just the vapor pressure.
The vapor pressure of a liquid depends upon the molecular structure of the liquid and upon the temperature. Increasing the temperature increases the average kinetic energy of the liquid molecules, and hence the fraction of molecules with sufficient energy to escape from the liquid phase. Thus, the vapor pressure increases with temperature.
The variation of the vapor pressure, P, with the absolute or Kelvin temperature, T, is given by
where DHvap is the molar enthalpy of vaporization in units of J/mol and R is the gas constant, also in units of J/mol*K. (R = 8.314 J/mol*K).
This equation can be solved for DHvap:
Thus, by measuring the vapor pressure at two different temperatures, the molar enthalpy of vaporization can be determined. As you might expect, DHvap increases with the strength of the intermolecular force.
The structure of the molecules in the liquid determines the strength of the intermolecular forces holding the molecules together. If the intermolecular force is strong, the fraction of molecules which have enough energy to escape into the vapor is small and the vapor pressure will be small. On the other hand, if the intermolecular force is relatively weak, a large fraction of the molecules will be able to escape into the vapor and the vapor pressure will be large. By comparing the vapor pressures of various liquids at constant temperature, we can determine the relative strengths of the intermolecular forces in the liquids.
Intermolecular Forces:
The weakest intermolecular force is the induced dipole-induced dipole or London force. It arises from the fact that the shape of the electronic cloud in a molecule fluctuates with time. This fluctuation results in an instantaneous dipole moment in the molecule which induces momentary dipole moments in neighboring molecules. The attraction between these momentary dipoles results in the London force. This force exists between all types of molecules whether polar or nonpolar since all molecules contain electrons. It is the primary force responsible for holding together the molecules in nonpolar liquids and solids.
The strength of the London force increases with the number of electrons in the molecule. Since the number of electrons increases with molar mass, we generally find the strength of the London force increases with molar mass.
In the case of polar substances, London forces exist, but in addition, the presence of a permanent dipole moment in the molecules results in a much stronger dipole-dipole intermolecular force. Thus, polar substances will generally have lower vapor pressures than nonpolar substances with the same molar mass. For example, consider the substances butane, C4H10, and acetone, C3H6O, both of which have a molar mass of 58, and hence equally strong London forces. The vapor pressures of butane and acetone at 25C, respectively are: 1735 mm Hg and 200. mm Hg. Why is the vapor pressure of butane so much higher than that of acetone? The answer is that in butane, which contains only hydrogen and carbon, the bonds are nonpolar since hydrogen and carbon have nearly equal electronegativities. Hence the molecule as a whole is nonpolar, and the only significant intermolecular force is the London type. On the other hand, acetone is a polar molecule as shown below.
Acetone molecules experience both the London Force
and the much stronger dipole-dipole force. This explanation is
supported by the fact that the molar enthalpy of vaporization of
acetone is greater than that of butane (32.0 kJ/mole vs. 24.3
kJ/mole). That is, more energy is required to vaporize one mole
of acetone than one mole of butane.
An especially strong dipole-dipole intermolecular force, called a hydrogen bond, results when a molecule contains a hydrogen atom attached to either of the highly electronegative atoms O, N or F. This results in highly polar covalent bonds. For example, in water the O-H bond is highly polar due to the large electronegativity difference between the O and H atoms. This H atom bears a large positive partial charge and the O atom bears a large negative partial charge. The attraction between the charged ends of the water molecules results in hydrogen bond formation.
A class of compounds in which hydrogen bonding is important are alcohols. These are compounds containing the group -O-H. For example, ethyl alcohol has the structure
In this experiment you will measure the vapor pressure of the compounds heptane
(C7H16), hexane (C6H14 ), and ethyl alchol (C2H6O), at room temperature. You will also measure the vapor pressure of heptane at OșC.
Procedure
To measure the vapor pressure of a liquid at room temperature set up your apparatus as shown in Figure 1. With the pinch clamp closed, the flask is connected to an aspirator, and the aspirator turned ON. The aspirator pumps air out of the flask and the pressure measured by the manometer drops. The height of the mercury column on the left, hL , increases and the height of the column on the right, hR, decreases. The aspirator does not remove all of the air in the flask, and a certain pressure of air, (Pair)rt will remain in the flask. The subscript "rt" stands for "room temperature." The value of (Pair)rt is measured with the manometer. The manometer measures the difference between the pressure inside the flask and the pressure of the atmosphere, Patm.
The value of (Pair)rt is given by equation 1.
(Pair)rt = Patm - Dh1, where Dh1 = (hL - hR)air (1)
Once (Pair)rt is measured, the liquid whose vapor pressure is to be measured is placed in the funnel, the screw clamp is closed, and some of the liquid is admitted to the flask by opening the pinch clamp. The pressure now rises due to the pressure exerted by the liquid's vapor pressure, (Pvap)rt. The total pressure in the flask at room temperature, (PTOT)rt according to Dalton's Law of Partial Pressures, is equal to the sum of the partial pressures exerted by the remaining air and by the liquid's vapor. That is,
(PTOT)rt = (Pair)rt + (Pvap)rt (2)
(PTOT)rt is measured with the manometer, and is given by
(PTOT)rt = Patm - Dh2 (3)
where Dh2 = (hL - hR)vap and is now the difference in the heights of the mercury columns when both air and vapor are present in the flask.
The vapor pressure of the liquid is obtained by solving Eq.(2) for (Pvap)rt
(Pvap)rt = (PTOT)rt - (Pair)rt (4)
In summary, the method involves measuring (Pair)rt
by evacuating the flask and measuring the pressure with no liquid
present, and then admitting the liquid and measuring PTOT,
the sum of the pressures of the remaining air, and the liquid's
vapor pressure. The difference between (PTOT)rt
and (Pair)rt is the vapor pressure.
In measuring the vapor pressure of a liquid near C the flask is placed in an ice water bath. The pressure in the system with no liquid present, (Pair)iw, and the pressure with the liquid present (PTOT)iw, must both be measured at the temperature of the ice water bath. The subscript "iw" stands for "ice water." The vapor pressure at the ice water temperature given by
(Pvap)iw = (PTOT)iw - (Pair)iw (5)
Eye protection must be worn when performing this experiment.
Part 1. Vapor pressure at room temperature
Record the barometric pressure. [Barometer Instructions] Measure room temperature. Half fill two 1. L beakers with tap water. Add hot or cold tap water, and stir. Add ice cubes to the second beaker to prepare an ice bath.
Part 1a. Measurement of (Pair)rt:
Make sure that all of the connections in the apparatus, shown in Figure 1, are tight and no moisture is present between stopper and flask. The manometer is especially fragile, and must be handled carefully. Do not tip it as this may result in spilling mercury.
Clamp the flask in the room temperature water bath so that it is covered by water as much as possible. Close the screw clamp that connects the flask to the aspirator. Insert the medicine dropper from the aspirator tubing into the thin-walled tubing. Close the rubber tubing below the funnel with the pinch clamp. Now turn on the aspirator fully, and slowly loosen the screw clamp to evacuate the flask. The mercury column in the left leg of the manometer will begin to rise. Continue evacuating with the aspirator until the heights of the mercury columns remain constant. Close the screw clamp.
Note the heights of the mercury columns on both sides of the manometer. If there is no leak, the mercury levels will not change over a period of several minutes, and you are ready to proceed. If there is a leak, it must be stopped. Normally this can be accomplished by checking and tightening the seals at the rubber stopper, at the screw clamp, and at the pinch clamp.
When you are certain that there are no leaks, measure the temperature of the water bath and the mercury heights (hL and hR), and record the data in your notebook. (Measurements should be converted to mm.
Part 1b: Vapor pressure of hexane:
To obtain (PTOT)rt of hexane and air, evacuate the flask with the aspirator until the heights of the mercury columns remain constant. Close the screw clamp, and pour 2 mL of hexane into the funnel. Do not inhale the vapor. Slowly open the pinch clamp and admit some of the liquid into the flask. The liquid enters the flask very rapidly; make sure you close the pinch clamp while there is still about 0.5 mL of liquid remaining in the funnel. If any air enters the flask at this point you must start over.
Check the mercury heights at 3-min. intervals until consecutive readings agree within 3 mm Hg. Then record the mercury heights and temperature.
Loosen the screw clamp slowly, and allow the contents of the flask to return to atmospheric pressure. Open the pinch clamp and drain the remaining liquid into the flask. Remove the rubber stopper from the flask, and pour the liquid into the solvent waste container. No water should be used to clean the flasks, but the flask should be dried between trials. This is accomplished by clamping the flask upside down to the ringstand and evaporating any remaining liquid by moving the aspirator tubing and medicine dropper around the inside of the flask while the aspirator is completely ON.
Part 1c: Vapor pressure of heptane and ethanol: Repeat the procedure in Part 1b., with heptane and then with ethanol at room temperature.
Part 2 Vapor pressure of heptane near OșC
Clean and thoroughly dry the flask. Reassemble the apparatus, and place the flask in the ice-water bath. Follow the procedure in Part 1a, to measure (Pair)iw, the pressure in the flask with no liquid present. Again, with the flask in the ice-water bath, repeat the procedure in Part 1b fpr heptane, to obtain (PTOT)iw, the pressure with liquid present.
Upon completion of the experiment, return the apparatus to atmospheric pressure by slowly opening the screw clamp, thereby admitting air into the flask and manometer.
NOTE: Ideally, you will want to repeat each measurement twice. That is, measure the vapor pressure of ethanol, hexane and heptane at room temperatrue twice and measure the vapor pressure of heptane near OșC twice.
Reporting Your Results
1. Results section
Arrange your data in tabular form. Sample data sheets which you may want to copy are on the next pages. Calculate values of Pair, PTOT, and Pvap for each experiment and enter the values in your data sheet. Show all calculations.
2. Conclusions section
The conclusion section should contain sentences that answer the following questions.
a) What are your measured values of the vapor pressures of heptane, hexane, and ethanol at each temperature?
b) Based on your vapor pressure measurements at room temperature, arrange the three compounds in order of increasing strength of the intermolecular forces. Explain your reasoning in terms of types of forces present.
c) How does the vapor pressure vary with molar mass? What conclusions can you draw from this behavior about the kinds of intermolecular forces which are predominant in each of the compounds, C7H16, C6H14 and C2H6O?
d) Using your values for the vapor pressure of heptane obtained at two different temperatures, calculate the molar enthalpy of vaporization of heptane.
