1. Vapor Pressure

2. Pressure of the vapor present when equilibrium is achieved between the rate of vaporization and the rate of condensation. At the boiling point, the P atm = P vapor As the vapor pressure on a pot of water is reduced, the energy needed to boil that water is also reduced. Pressure Cooker: By increasing the vapor pressure, additional energy is needed for the water to boil, therefore the water can boil at temperatures above 100C.

3. Effect of Pressure on Boiling Point

4. Vapor Pressure vs. Temperature As the temperature increases, a greater number of molecules have sufficient kinetic energy to convert from the liquid to the vapor phase. There is a nonlinear relationship between the vapor pressure of a liquid and temperature.

5. Vapor Pressure vs. Temperature

6. The Clausius – Clapeyron Equation A mathematical expression which relates the variation of vapor pressure to temperature ln P = (-  H vap /RT) + C where C is a constant IMPORTANCE: When the ln P is plotted vs (1/T) you create a line where the slope is equal to the –  H vap /R Which means you can calculate the enthalpy of vaporization from the slope of the line. R = 8.314 J/Kmol Convert all Temps to Kelvin

7. Vapor Pressure of Solutions A nonvolatile solute lowers the vapor pressure of the solution. The molecules of the solvent must overcome the force of both the other solvent molecules and the solute molecules.

8. Raoult’s Law: P soln =  solvent x P solvent Vapor pressure of the solution = m ole fraction of solvent x vapor pressure of the pure solvent Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

9. Aqueous Solution Pure water Water has a higher vapor pressure than a solution

10. Aqueous Solution Pure water Water evaporates faster from for water than solution

11. The water condenses faster in the solution so it should all end up there. Aqueous Solution Pure water

12. Practice Problem A solution of cyclopentane with a nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane? P soln = X cp P cp 211 torr = X cp (313 torr).674

13. Try one on your own Determine the vapor pressure of a solution at 25  C that has 45 grams of C 6 H 12 O 6, glucose, dissolved in 72 grams of H 2 O. The vapor pressure of pure water at 25  C is 23.8 torr. P solution = X solvent P solvent P solution =.941(23.8 torr) P solution = 22.4 torr

14. Liquid-liquid solutions where both are volatile. Modify Raoult’s Law to P total = P A + P B =  A P A 0 +  B P B 0 P total = vapor pressure of mixture If this equation works then the solution is ideal. Ideal solutions

15. Vapor Pressure of solution

16. Deviations If solvent has a strong affinity for solute (H bonding). Lowers solvent’s ability to escape. Lower vapor pressure than expected. Negative deviation from Raoult’s law.  H soln is large and negative exothermic. Endothermic  H soln indicates positive deviation.

17. χbχb χAχA Vapor Pressure Positive deviations- Weak attraction between solute and solvent Positive ΔH soln

18. χbχb χAχA Vapor Pressure Negative deviations- Strong attraction between solute and solvent Negative ΔH soln

19. Problem #1 The vapor pressure of a solution containing 53.6g of glycerin C 3 H 8 O 3 in133.7g ethanol C 2 H 5 OH is 113 torr at 40C. Calculate the vapor pressure of pure ethanol at 40C assuming that the glycerin is a non volatile, nonelectrolyte solute in ethanol.

20. Answer to #1 P soln = X eth P eth 113torr = 2.90mol/3.48mol (P eth ) 135.6 torr = P eth

21. Problem #2 At a certain temperature, the vapor pressure of pure benzene C 6 H 6 is 0.930atm. A solution was prepared by dissolving 10.0g of a nondissociating, nonvolatile solute in 78.11g of benzene at that temperature. The vapor pressure was found to be 0.900atm. Assuming the solution behave ideally, determine the molar mass of the solute.

22. Answer #2 P soln = X benzene P benzene.900atm = X benzene (.930atm) X benzene =.9677 X solute =.0323 MM = 10.0g/.0323mol = 310g/mol

23. Problem #3 A solution of NaCl in water has a vapor pressure of 19.6 torr at 25C. What is the mol fraction of solute particle in this solution if the vapor pressure of water is 23.8 torr at 25C?

24. Answer #3 P soln = X water P water 19.6torr = X water (23.8torr).824 = X water therefore X solute =.176

25. Problem #3 For the same problem as #3: What is the vapor pressure of the solution at 45C if the vapor pressure of water is 71.9 torr at 45C?

26. Answer #3 P soln =.824(71.9torr) P soln = 59.2 torr

27. Problem #4 A solution is made from 0.0300mol CH 2 Cl 2 and 0.0500mol CH 2 Br 2 at 25C. Assuming that the solution is ideal, calculate the % composition of the vapor at 25C. P CH2Cl2 = 133 torr P CH2Br2 = 11.4 torr

28. Answer #4 P soln = X CH2Cl2 P + X CH2Br2 P P soln = (.03/.08)(133torr) + (.05/.08)(11.4 torr) P soln = 57.0 torr X CH2Cl2 = 49.9 / 57 =.875 = 87.5% X CH2Br2 = 7.13 / 57 =.125 = 12.5%