# Vapor Pressure, Enthalpy of Vaporization, and Intermolecular by variablepitch333

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```									                     Vapor Pressure, Enthalpy of Vaporization,

and Intermolecular Forces

Dr. Mike Perona, California State University, Stanislaus

Procedure modified by Dr. Scott Russell 8/30/2007

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 ΔHvap 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 ΔHvap:
Thus, by measuring the vapor pressure at two different temperatures, the molar
enthalpy of vaporization can be determined. As you might expect, ΔHvap
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 that 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 shown above.
In this experiment you will measure the vapor pressure of the compounds
heptane (C7H16), hexane (C6H14 ), and ethyl alcohol (C2H6O), at room
temperature. You will also measure the vapor pressure of heptane at 0ºC.

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 through a trap to a vacuum line. The vacuum line pumps air out of the
flask. The vacuum line 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 microlabs pressure
sensor. Before the vacuum is pulled on the flask, the pressure should read ~760
Torr. Once the vacuum is open the pressure should drop rapidly to a range of
80-100 Torr.

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 (1)

(PTOT)rt is measured by the pressure sensor after introducing the liquid into the

The vapor pressure of the liquid is obtained by solving Eq.(1) for (Pvap)rt

(Pvap)rt = (PTOT)rt - (Pair)rt (2)
In measuring the vapor pressure of a liquid near 00C 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              (3)

Funnel
Pinch clamp
To microlabs
pressure sensor
Screw
clamp

To lab
vacuum
line

Trap
Figure 1.

Eye protection must be worn when performing this experiment.

Part 1a. Measurement of (Pair)rt:

Turn on the microlabs FS-522 interface (green button on the front). Start the
microlabs software by double clicking on the “Microlab.exe” icon on the desktop.
A window will open; double click on “Microlab experiment”. This opens the
microlabs interface where you can design your own methods. A method for this
lab has already been created. To access it, click file, then open, then double
click on the file named “Vapor Pressure” in the “Saved Experiments” folder. This
experiment will allow you to monitor pressure (Torr) versus time (sec). The data
will be displayed as a graph and a spreadsheet.

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. 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 vacuum trap.
Insert the medicine dropper from the vacuum trap tubing into the thin-walled
tubing. Close the rubber tubing below the funnel with the pinch clamp. Click
“start” in the microlabs software, the pressure should initially read ~760 Torr.
Now open the lab vacuum line fully, and slowly loosen the screw clamp to
evacuate the flask. The pressure reading on the microlabs display will drop
rapidly. Continue evacuating with the vacuum line until pressure readings at one-
minute intervals are within 3 Torr. Close the screw clamp and disconnect the
flask from the trap. You now have an isolated system.

Note the pressure readings on the computer. If there is no leak, the pressure
should be stable over a period of several minutes, and you are ready to proceed.
If there is a leak, it must be stopped. Normally checking and tightening the seals
at the rubber stopper, the screw clamp, and the pinch clamp can fix any leaks.
When you are certain that there are no leaks, record your measured (Pair)rt in
your notebook and proceed to part 1b.

Part 1b: Vapor pressure of hexane:

At this point you should have the screw clamp and pinch clamp closed and the
pressure should be stable (i.e. no leaks, see Part 1a). To obtain (PTOT)rt of
hexane and air, 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.

Monitor the pressure at one-minute intervals until consecutive readings agree
within 3 Torr. Record the final pressure (PTOT)rt in your notebook.

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 blowing
compressed air (fume hood) into the flask to evaporate any remaining liquid.

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 OºC

Thoroughly dry the flask with compressed air. 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 for 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.

You must repeat each measurement twice. That is, measure the vapor pressure
of ethanol, hexane and heptane at room temperature twice and measure the
vapor pressure of heptane near 0ºC twice.

1. Results section

Arrange your data in tabular form. A sample data sheet, which you may want to
use, is shown below. Calculate values 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 that 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.

Sample Spreadsheet for Vapor Pressure Experiment
Trial 1: hexane at room temp.
Temperature =                    Pressure (Torr)