lab_handout_Experiment_8

# lab_handout_Experiment_8 - Experiment 8 Vapor Pressure of...

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Experiment 8 Vapor Pressure of Liquids

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EXPERIMENT 8: INTERMOLECULAR FORCES VAPOR PRESSURE OF LIQUIDS AND DETERMINATION OF THE ENTHALPY OF VAPORIZATION PRELAB 1. Review the relationship between intermolecular forces, vapor pressure and enthalpy of vaporization. Review the ideal gas law. What does ideal mean? INTRODUCTION Part I: Vapor Pressure and the Determination of the Enthalpy of Vaporization Overview How strong are the intermolecular attractive forces that are present in a given pure substance? These attractive forces represent a form of bonding and so it is best if we enumerate intermolecular forces in terms of ‘Bond Energy’ as illustrated in Figure 1. Intermolecular forces are operating in the liquid Zero intermolecular forces in the gaseous state (assuming ideal gas behavior) Liquid Vapor Energy required to reduce intermolecular attraction to zero => Intermolecular ''bond energy' => H vap Figure 1. Diagram indicating why intermolecular 'bond energy' is equal to H vap The Clausius-Clapeyron equation describes the relationship between vapor pressure (P vap ) and absolute temperature: ln P vap = - H vap RT + B A plot of ln P vap vs 1/T(K) should yield a straight line with a slope of - H vap /R where R = 8.314 J/K . mole.
Consider heating methanol from room temperature to its boiling point of 65 o C. Adding heat increases the kinetic energy of the molecules. Consider what, if anything is happening to the intermolecular forces. It is quite reasonable to assume that as the molecules move faster, the intermolecular forces will decrease. This in turn means that H vap should decrease. When the H vap of methanol is carefully measured in narrow temperature window, H vap values actually decrease (45.0 kJ/mole at 0 o C, 43.4 kJ/mole at 40 o C and 40.7 at 100 o C). We thus write the enthalpy of vaporization of methanol as 40.7 - 45.0 kJ/mole. For any molecule that is polar, the total intermolecular force is a sum the dispersion and polar parts. For n-pentane, diethyl ether, 2-butanone, and 2-butanol: H vap = H vap (dispersion) + H vap (polar). In this experiment, you will estimate H vap of four solvents: n-pentane, diethyl ether, 2-butanone, and 2-butanol. The mass, length and overall size of each of these molecules is determined by five backbone atoms, either carbon or oxygen. The differences in the number of hydrogen atoms on each molecule is an insignificant contribution its general size. Therefore the energy required to disperse the molecules throughout the liquid is approximately the same. When present, the oxygen atom in each molecule is more electronegative than the carbon atoms. It will attract electrons away from the carbon and create a negative pole around itself leaving a positive pole around the carbon atoms. The negative poles of one molecule will attract the positive poles of neighboring molecules causing them to bind together. These polar forces increase the enthalpy needed to separate the molecules so they can enter the gas phase. Since n-

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## This note was uploaded on 11/18/2011 for the course BI 107 taught by Professor Fuller during the Spring '08 term at Montgomery College.

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lab_handout_Experiment_8 - Experiment 8 Vapor Pressure of...

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