Lecture 1: Physical Equilibria—The Temperature Dependence of Vapor PressureOur first foray into equilibria is to examine phenomena associated with two phases of matter achieving equilibrium in which thefree energy in each phase is the same and there is no change in the overall values of system state functions.The areas of physical equilibria we will investigate are:- The interface between phases: gas liquids solid. Solution properties like freezing and boiling will be given a thorough thermodynamic treatment.- Mixtures formed when phases are soluble or miscible in one another. Dissolving solids or gases in liquids or mixing two liquids.- Colligative properties that derive from mixing of two phases. These are essential properties of biological and chemical systems.We will draw on the following theoretical treatments from CH 301:- intermolecular force theory - thermodynamicsSo review these concepts!!Specifically, review the intermolecular force and thermodynamic lecture notes and worksheets. Pay special attention to the introductory treatment of phases and phase transitions that are presented at various times during these lectures.Vapor PressureVapor pressure was introduced in the fall semester as a solution property and we learned that vapor pressure rankings could be qualitatively determined from a ranking of intermolecular forces.Definition of vapor pressure: the pressure exerted by the vapor of a substance that exists in a condensed (liquid or solid) phase. IMF Theory and vapor phase ranking: It is intuitive that the stronger the intermolecular force holding a substance together, the less likely it is to enter the vapor phase. Similarly, the smaller the IMF force, the greater the vapor pressure.Just looking at this picture suggests vapor pressure is a surface phenomenon. The more surface the more vapor. The depth of the condensed phase doesn’t matter. The reason is that the particles in the condensed phase with the greatest ability to enter the vapor phase are on the surface.Here is the ranking and trend, contrasting IMF and assorted compounds:As the IMF increases Instantaneous dipoles (<1 kJ/mole) < dipole-dipole (5 kJ/mole) < H-bonding (20 kJ/mole) < ionic (200 kJ/mole)The vapor pressure decreases He > CH4 > C3H8 > CCl4 > CHCl3 . CH3OH > H2O > NaCl > CaOIt is pretty easy to generate these qualitative rankings of vapor pressure by first ranking IMF. Look at the table of values for different compounds at 25oC to see the correlation:Now can we create a quantitative measure of vapor pressure?First recognize that vapor pressure is an equilibrium phenomenon that is temperature dependent. Can we derive an expression that predicts this relationship? First note that at equilibrium, Gvap = Ggas – Gliquid = 0Gliquid = Go liquid because it is pressure independentIn the table of water vapor pressures, note that as the temperature increases, the vapor pressure increases until the boiling point is reached—100 degrees at 1 atmosphere. Also note what appears to be an exponential-like increase in P as T increases. note that at equilibrium, as vapor encounters the surface it condenses and theenergy released promotes a different particle into the vapor. also note that at the higher T more particles achieve separation energy to enter vapor.Ggas = Gogas + RTlnP where RTlnP = correction for change in pressure, so: Gvap = Gogas + RTlnP - Goliquid = Gogas - Goliquid + RTlnP Go vap = RTlnP (the standard free energy of vaporization)But Go vap = Ho vap - TSo vapSo: lnP = - G ovap = - H ovap + S ovap A constant, assuming ideal gas behavior, RT RT R because a mole of any kind of gas has about the same entropy.Or multiplying through by ex P = Ke -H vap/RT This is a very famous mathematical relationship that shows up all the time in nature.What does the above equation suggest for a function? An exponential increase in P with T.Note this functional relationship not just for water but for other compounds.And now for something really useful that scientists do: Combining equations to eliminate a variable – or how to create the famous Clausius Clapeyron equation.Given that Sovap is assumed to be constant in the expression:ln P = - H ovap + S ovap R RWe can rearrange the equation to put everything in terms of Svap0R, identify two states of the system P1, T1 and P2, V2, set them equal and do mad algebra. When it is all over,ln P2P1 = vap0R 1T11T2Note that as delta Hovap decreases with increasing IMF, the boiling point increases and that this correlates with IMF.This is the famous Clausius-Clapeyron equation that allows us to use one value of P1 and T1 to predict the vapor pressure anywhere else on the graph.So let’s test this by using the equation to find the boiling point of water (hey, at least we’ll know if our answer is right since at 100 oC degrees the vapor pressure should be 760 torr). Selecting a P and T from our table of H2O vapor pressures. Like 20 oC and 17.54 torr, and given Ho vap for H2O = 40.7 kJ/mole. We need to work in the correct units so convert 17.54 torr into Pa. 1 torr = 133.3 Pa, so 17.54 torr (133.3Pa/1torr) = 2338.1 ParearranginglnP2P1 = vap0R 1T11T2 1T11T2Rvap0 ln P2P1And substituting with 1 atm = 101,325 Pa1T11(20 2 73) + 8.3145J /kmole40,700J /mole ln 2338.1Pa101,325PaAnswer: T1 = 378K or 105oC Of course the answer should be 100 oC but the answer get get is pretty good considering all the approximations. So there we have it, an equation that models the vapor pressure curve for any compound as a function of
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