11
VAPORIZATION Experiment AND INTERMOLECULAR FORCES
I. Learning Objectives...
To understand how heats of vaporization vary with the strength of the intermolecular forces present in various solvents. To relate enthalpy of vaporization to intermolecular forces.
II.
Background Information
Intermolecular Forces
Intermolecular forces are involved when molecules associate with one another in the liquid or solid states. Intermolecular forces include ion-dipole interactions, dipole-dipole interactions, and induced-dipole interactions. Since the compounds used in this Dipole-dipole
experiment are covalent, no ion-dipole interactions are involved.
interactions are weaker than ionic interactions and arise from interactions between partial charges. Hydrogen bonding is a particular type of dipole-dipole interaction that involves a link between the highly electronegative N, O, and F atoms and an electropositive H atom (which is bonded to a N, O, or F). Normally, bond energies associated with hydrogen bonds are in the range of 20-30 kJ/mol. Many molecular interactions involve induced dipoles (dispersion forces). As two nonpolar molecules approach one another, repulsions and/or attractions between their electrons and nuclei lead to distortions in their electron clouds. These slight distortions result in momentarily induced dipoles that lead to intermolecular attraction. Generally, a larger molecule (with more electrons) produces greater fluctuation in charge distribution and thereby a larger induced dipole. Dispersion forces generally increase with molecular weight. Dispersion forces are also affected by the arrangement in 3-D space of the atoms in a molecule. Unbranched (or straight chain) compounds tend to have stronger The atoms of neighboring branched
dispersion forces than their branched isomers.
molecules are unable to position themselves as close together as their unbranched
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counterparts. When compared to hydrogen bonding, typical bond energies for dispersion forces are considerably lower and typically near 2 kJ/mol. The strength of these forces, however, lies in the cumulative effect of the large number of interactions possible as molecular complexity increases.
Vaporization and Intermolecular Forces
When moving from the world of the atom to the world around us, it becomes clear that the numbers of atoms and molecules in the universe are extremely large. Statistics are necessary when working with these huge numbers since there is no way to calculate the behavior and interactions of all of the molecules in even a very small collection. Ludwig Boltzmann and James Clerk Maxwell contributed to the science of statistical thermodynamics in which the properties of very large numbers of molecules is examined. The Maxwell-Boltzmann probability distribution function is commonly used to describe kinetic energies in a large collection of atoms or molecules. Kinetic energy is the energy of motion and is calculated by the formula
KE = 1 2 mv 2
where m is the mass of the particle and v is its velocity. In any sample of a gas or liquid, some molecules have very low energies, others have very high energies, but most have some intermediate energy, as described by the Maxwell-Boltzmann Distribution illustrated in Figure 9.1. Based on this distribution of kinetic energies, temperature is defined as the average kinetic energy of the system. This definition implies that all liquid or gas samples at the same temperature will have the same average kinetic energy. Vaporization is the conversion of a substance from the liquid state to the gaseous state. Any of the molecules in a liquid that have more kinetic energy than the intermolecular forces holding them in the liquid state move into the gas phase. Therefore, at any given temperature only the fraction of molecules with kinetic energies greater than the intermolecular forces (as shown in Figure 9.1) will be able to break free. The molecules remaining in the liquid phase have a lower average kinetic energy and therefore a lower temperature. This change in temperature explains why perspiration evaporating from the skin provides a cooling effect.
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If all liquids have the same kinetic energy at a given temperature, why does rubbing alcohol (isopropyl alcohol) feel cooler than water when it is applied to the skin? The reason for this difference in cooling is due to the strength of the intermolecular forces, as illustrated in Figure 9.2. The intermolecular attractions between water molecules are much stronger than the attractions between the molecules of isopropyl alcohol. Therefore, there are fewer water molecules that have enough kinetic energy to move into the gas phase. Because a larger number of high-energy water molecules remain in the liquid phase after evaporation, the average kinetic energy (or temperature) is higher for water relative to isopropyl alcohol. In other words, water will exhibit a smaller change in temperature during vaporization than isopropyl alcohol because it has stronger intermolecular forces.
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Vapor Pressure and Enthalpy of Vaporization
The term "vapor" is applied to the gas of any compound that would normally be found as a liquid at room temperature and pressure. For example, water, gasoline, rubbing alcohol, and finger nail polish remover (ethyl acetate) are all normally liquids, but they all evaporate to form a gas. These gases are referred to as vapors. In contrast, we would not talk about nitrogen, oxygen or methane vapor since these materials are normally gases at room temperature and pressure. As described above, molecules in a liquid that gain enough kinetic energy to break the bonds holding them to the rest of the molecules in the liquid enter the gas phase. This process can also occur in reverse gas molecules can collide with the surface of the liquid and re-enter the liquid. If you place a liquid in a closed container, at first there are very few molecules of the liquid in vapor form, and so the reverse process doesn't occur very often. As more of the liquid converts to vapor, the rate of the reverse process increases until finally there are the same number of molecules entering and leaving the liquid. Once this equilibrium is reached, the number of molecules in vapor form doesn't change, and we can compute the pressure of the vapor. The relationship between vapor pressure and temperature is given by an equation called the Clausius-Clapeyron equation:
ln P =
H vap RT
+C
(Eq. 1)
In this equation P is the vapor pressure, T is the absolute temperature (in Kelvin), R is the gas constant (8.314 J/molK), Hvap is the enthalpy of vaporization per mole of substance, and C is a constant. The Clausius-Clapeyron equation predicts that a graph of ln P versus 1/T should give a straight line with a slope equal to - Hvap/R. Thus, we can use such a plot to determine the enthalpy of vaporization of a substance by finding the slope. As an example of the application of the Clausius-Clapeyron equation, the vapor pressure data for ethanol are graphed in Figure 9.3 as ln P vs. 1/T. Note that the data lies on a straight line with a negative slope. We can use the slope of the line to determine the experimental Hvap for ethanol, as shown:
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slope =
H vap J R J kJ = 38650 = 38.65 mol K mol mol
H vap = slope R = ( 4648.79 K ) 8.314
************************************************************************* Additional Background Reading: See Chemistry: The Central Science 10th Edition, Brown, Lemay, and Bursten Chapter 11 (Intermolecular Forces), Pages 445-453; (Clausius-Clapeyron Equation), Page 462.
III. Experimental Procedures
A. Hardware Setup
1. Start the computer. 2. Check that the Science Workshop 500 interface is connected to the power source. When the interface is properly connected, the green power light is illuminated.
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3. Connect the computer to the interface using the cable provided. Attach the cable to the USB port of the computer. 4. Check that the temperature sensor is connected to analog port A.
B.
Software Setup
1. Open Data Studio. 2. Select CREATE EXPERIMENT. 3. The program looks for the interface to initialize. If the interface is not connected, properly it prompts the user to SCAN or PICK the interface for initialization. Select SCAN. If this does not work, see a TA. 4. In the experimental setup window, click on Port A of the interface box, scroll down to the Temperature Sensor and double click on it. This selects the sensor and automatically connects it to port A (the temperature sensor icon should now be shown to be connected to that port). 5. At the bottom of the Experiment Set-up screen, change the sample rate to 1 reading every 10 seconds. Also change the unit of measure for the temperature sensor to K. 6. Create a table and graph of temperature. 7. Save the activity. SELECT
DEVOID OF PUNCTUATION. A FILE NAME WITH LESS THAN
10
CHARACTERS AND
D. Equipment Setup
1. Gently clamp the temperature sensor to a metal support rod on the bench top, using a three-prong clamp. 2. Use a 400-mL or 600-mL beaker to collect liquid waste under the suspended temperature sensor. 3. Obtain a wash bottle and fill it with distilled water. Rinse the tip of the temperature sensor by squirting it with water thoroughly on all sides. Carefully collect all liquid in the waste beaker under the temperature sensor. Dry the temperature sensor with a Utility Wipe.
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E. Data Collection
Follow the procedure below to observe temperature changes upon evaporation of several known alcohols. 1. Each sample to be tested is in a scintillation vial that is located in the hood. Dip the tip of the temperature sensor in the vial for about 30 seconds to allow the temperature sensor to equilibrate. Click on START and immediately remove the vial. Be sure to recap the vial before placing in the hood. 2. Follow the temperature change on the graph. When a minimum is reached and the temperature begins to go up again click STOP. 3. Select the SCALE TO FIT button in the top left corner of the graph to rescale the yaxis. 5. Rinse the temperature sensor with water and dry thoroughly using a Utility wipe. 6. Rename the run. To do so, go under the data window at the left and click on Run #1, then click and hold on the title for 2-5 seconds. This brings up a box around the title and allows renaming of the data set. Choose an appropriate name for the sample tested (eg. Methanol). 7. Complete this procedure for all of the alkanes and alcohols. Be sure to allow the temperature sensor to reach room temperature before starting the next run. Each time START is selected the program overlays the new set of data on the same graph. Alternatively, select the runs to be included on each graph under the DATA arrow, the runs can simply be toggled on and off. Remember: Do not have greater than 8 active displays open simultaneously. This overloads the memory and may create problems opening the file later.
IV. Data Analysis
A. Determination of Vapor Pressure and Hvap 1. Using the calculate feature in DataStudio, take the inverse of the temperature measurements (1/T). 2. Using the calculate feature, compute the natural log of the vapor pressure (atm) for each solvent using the equations below (enter the equation exactly as
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written).
In the equations, T is the temperature measurement for the
corresponding solvent. heptane: octane: nonane: methanol: ethanol: 1-propanol: 1-butanol: sec-butanol: tert-butanol: ln P = (1/(-2.2473E-4T))+20.982 ln P = (1/(-2.0366E-4T))+22.388 ln P = (1/(-1.9356E-4T))+23.155 ln P = (1/(-2.1253E-4T))+23.816 ln P = (1/(-1.9454E-4T))+24.509 ln P = (1/(-1.7765E-4T))+25.502 ln P = (1/(-1.6769E-4T))+25.202 ln P = (1/(-1.7317E-4T))+25.607 ln P = (1/(-1.7353E-4T))+27.118
3. Plot ln P vs 1/T for each solvent. The data should lie on a straight line with a negative slope. Remember to rescale the graph so that the data fill the entire graph. Determine the slope of the line and record the values in Data Table I. 4. Using the Clausius-Clapeyron equation, determine the experimental Hvap for each sample and record the values in Data Tables I and III. experimental Hvap to the literature values of Compare the
Hvap for each sample by
calculating the percent error and record the values in Data Tables I and III.
B. Comparing molecular structure, Hvap and molar mass of alkanes and alcohols 1. Draw the structural formula for each of the solvents listed in Table II. You can draw the structures on your computer by downloading MDL ISIS/Draw from http://www.mdli.com/downloads/downloadable/index.jsp. 2. Open an editable data table to enter the molar mass and Hvap of each alkane. To do so, click on EXPERIMENT in the top toolbar, select NEW EMPTY TABLE. An empty table opens. Double click on this data set shown in the data display window at the left. This opens the properties for this editable data set. Rename the table and the variables to reflect x-values of Molar Mass (g/mol) and the y-values of Hvap (kJ/mol). Select OK.
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3. Manually enter these data values of molar mass and alkane with the lowest MW.
Hvap starting with the
4. Click and drag this data set atop the graph icon in the display window at the bottom left. This should open a plot of Hvap vs. molar mass. 5. Repeat this procedure to generate a new graph of methanol, ethanol, and 1-propanol. 6. Look for trends in the graphs and explain how intermolecular forces are related to the trends. 7. Compare and contrast the data (structural formula, molar mass, DHvap) for the alkanes and the alcohols. Use intermolecular forces to explain any similarities or differences you see in the data. Hvap vs. molar mass for
C. Effect of Branching on Intermolecular Attraction Each group will analyze three alcohols (n-butanol, sec-butanol, and tert-butanol) to see the effect of molecular shape on intermolecular attraction.
1. Draw the structural formula of each of the alcohols in Data Table III. 2. Record the average Hvap values for each of the alcohols in Data Table III. 3. Based on the Hvap values, rank the alcohols in order of decreasing
intermolecular strength, with 1 representing the strongest intermolecular forces and 3 the weakest. Record these values in Data Table III. 4. Using the results in Data Table III, explain how the molecular shape effects the intermolecular attraction of the molecule. Things to think about: 1) what types of intermolecular forces are present; 2) how would the molecular shape affect how these types of interactions?
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