1. 20 B Week V Chapters 11 and 18 Colligative Properties and Chemical Kinetics Dissolution reactions and Arrhenius type Acid/Base rxns Colligative properties, Vapor Pressure Lowering, Boiling Point Elevation, Freezing Point Depression and Osmotic Pressure, The Rates of Chemical Reactions A-->B Rate=d[A]/dt= -d[B]/dt. units molL -1 s -1 A is the Reactant and B is the product of the reaction Rate Laws at early times =k[A] n k is the rate constant and n is the reaction order Elementary reactions: single step reactions Unimolecular(1 st ), Bimolecular(2 nd ), Termolecular

2. Fig. 10-21, p. 463

3. Fig. 10-16, p. 459 The Rate of evaporation vs time as the vapor pressure approaches Equilibrium Evaporation rate equals the condensation Rate All the Macroscopic Properties, P, V, and T are only defined at Equilibrium. Which means PV=nRT and the vdW Eq. can only be use under Equilibrium conditions C 2 H 4 ( l )  C 2 H 4 (g) Evaporation  Equilibrium P,V and T well defined

4. Table 10-3, p. 460 H 2 O P-T Phase Diagram Super heated H 2 O liquid Will spontaneously vaporize Super cooled H 2 O liquid will Spontaneously freeze Equilibrium Vapor Pressure In both spontaneous processes the system will go to the Equilibrium State (Phase) and Pressure

5. Fructose C 6 H 12 O 6 Hydrated Fructose C 6 H 12 O 6 H-bonds Dissolution of a Nonvolatile((zero partial pressure) solute, e.g., sugars or salts. Heterogeneous Phase equilibrium of a 2 component solution

6. The dissolution reaction with water as a solvent: A(s)  A( aq ) A is a molecular solid and the molecular units, or monomers, do not dissociate in solution. Like fructose with lots of OH ( hydroxyl groups) for forming H-bonds and which makes the monomer more stable in aq soln Electro Static potential

7. Table 10-3, p. 460 H 2 O P-T Phase Diagram Super heated H 2 O liquid Will spontaneously vaporize Super cooled H 2 O liquid will Spontaneously freeze Equilibrium Vapor Pressure In both spontaneous processes the system will go to the Equilibrium State (Phase) and Pressure

8. Fig. 11-10, p. 491 Solvent vapor pressure versus Mole fraction X 1 =n 1 /(n 1 + n 2 ) Ideal Solution Raoult’s Law Ignores intermolecular Forces (interactions) Nonideal or Real solns Where intermolecular Forces are always present Raoult’s Law P 1 =X 1 P ° 1 P ° 1 =vapor pressure of pure solvent P 1 = solvent vapor Pressure

9. Fig. 10-16, p. 459 The Rate of evaporation vs time as the vapor pressure approaches Equilibrium Evaporation rate equals the condensation Rate All the Macroscopic Properties, P, V, and T are only defined at Equilibrium. Which means PV=nRT and the vdW Eq. can only be use under Equilibrium conditions C 2 H 4 ( l )  C 2 H 4 (g) Evaporation  Equilibrium P,V and T well defined

10. Fig. 11-10, p. 491 Solvent vapor pressure versus Mole fraction X 1 =n 1 /(n 1 + n 2 ) Raoult’s Law P 1 =X 1 P ° 1 basis of all 4 colligative Properties Vapor Pressure Lowering Boling Point Elevation Freezing point depression And Osmotic pressure

11. Table 10-3, p. 460 H 2 O P-T Phase Diagram Super heated H 2 O liquid Will spontaneously vaporize Super cooled H 2 O liquid will Spontaneously freeze Equilibrium Vapor Pressure In both spontaneous processes the system will go to the Equilibrium State (Phase) and Pressure

12. Vapor Pressure Lowering in a two component heterogeneous soln Raoult’s Law P 1 =X 1 P ° 1 Can also be written as  P 1 =P 1 - P ° 1 = X 1 P ° 1 - P ° 1 = -(1- X 1 ) P ° 1 =- X 2 P ° 1 Implies Vapor Pressure Lowering since  P 1 < 0 When the solute is added.

13. Fig. 11-10, p. 491 Solvent vapor pressure versus Mole fraction X 1 =n 1 /(n 1 + n 2 ) Raoult’s Law P 1 =X 1 P ° 1 basis of all 4 colligative Properties Vapor Pressure Lowering Boling Point Elevation Freezing point depression And Osmotic pressure Attractive Forces

14. Table 10-3, p. 460 Pure H 2 O P-T Phase Diagram Equilibrium Vapor Pressure Vapor Pressure Lowering After dissolution

15. Fig. 11-11, p. 493 Boiling Point Elevation =T’ b = T b + X 2 P ° 1 /S T b = boiling pt T’ b elevated boiling pt Vapor pressure Lowering With added solvent  P 1 = - X 2 P ° 1  T b =T’ b – T b (1/S)X 2 S=-  P/  T b

16. Fig. 11-11, p. 493 Boiling Point Elevation T b = boiling point T’ b = elevated boiling pt. Vapor pressure Lowering With added solvent  P 1 = - X 2 P ° 1 S= -  P/  T b Solve for  T b  T b =T’ b – T b =(1/S)X 2

17. Fig. 10-6, p. 450 In Solutions, for example when NaCl(s) is dissolved in H 2 O( l ). + H 2 O  NaCl(s) + H 2 O( l )  Na + ( aq ) +Cl - ( aq ) ( aq ) means an aqueous solution, where water is the solvent, major component. The solute is NaCl, which is dissolved, minor component Water molecules solvates the ions the Cation (Na + ) and the Anion (Cl - ). The forces at play here are Ion dipole forces Dissolution of a polar solid by a polar solid by a polar liquid A non-polar liquid e.g., benzene, would not dissolve NaCl?

18. Table 11-2, p. 494 Boiling Point Elevation= T b ’ = T b + X 2 P° 1 /S In dilute a solutions n 1 >>n 2 X 2 = n 2 /(n 1 + n 2 )~ n 2 /n 1 =(m 2 / M 2 )/(m 1 / M 1 ) T b ’ = T b + X 2 P° 1 /S T b ’-T b =(1/S) (m 2 / M 2 )/(m 1 / M 1 )  T=K b (m 2 / M 2 )/(m 1 [1000gkg -1 )  T=K b m ( m=molality of soln) m=(m 2 / M 2 )/(m 1 [1000gkg -1 ]) M 1 (gmol -1 ) is the molar mass of the solvent and M 2 (gmol -1 ) of the solute ad Since S and M 1 are properties of the solvent then we can define K b = M 1 /S(1000 gKg -1 )  T b = K b m ( m=molality of soln) m 1 is the mass of the solute m 2 is the mass of the solvent

19. Boiling Point Elevation (  T b = T’ b -T b )  T b = K b m ( m=molality of soln)

20. As an example: NaCl(s) dissolves completely in water. NaCl(s) + H 2 O( l )  Na + ( aq ) +Cl - ( aq ) 1.0 mole NaCl(s) produces 2 moles of ions in soln Given 0.058 grams of NaCl(s) is dissolved in 10 grams of H 2 0( l ) What is the boiling point(T’ b ) at p=1 atm? T b =100°C for pure water  T b = K b m ( m=molality of soln) K b (H 2 O)= 0.512 K kg mol -1 The molality m=(m 2 / M 2 )/(m 1 /[1000gkg -1 ]) m 2 mass of solute; m 1 mass od solvent

21. As an example: NaCl(s) dissolves completely in water. NaCl(s) + H 2 O( l )  Na + ( aq ) +Cl - ( aq ) 1.0 mole NaCl(s) produces 2 moles of ions in soln Given 0.0584 grams of NaCl(s) is dissolved in 10 grams of H 2 0( l ) What is the boiling point(T’ b ) at p=1 atm? T b =100°C for pure water  T b = K b m ( m=molality of soln); K b (H 2 O)= 0.512 K kg mol -1 m=(m 2 / M 2 )/(m 1 [1000gkg -1 ])=n 2 /(m 1 /[1000gkg -1 ]) n 2 =(2)x(0.0584 g/ 58.4 gmol -1 ) = 2.0 x 10 -3 mols m 1 =10 g m =2x10 -3 mols/0.01 kg = 0.2 mols kg -1  T b = (0.512)x(0.2) K = 0.1024 K T b = 100 °C + 0.102 °C= 100.102 °C

22. Fig. 11-11, p. 493 Boiling Point Elevation T b = boiling point T’ b = elevated boiling pt. Vapor pressure Lowering With added solvent  T b =T’ b – T b =(1/S)X 2 0.102 °C

23. Fig. 11-12, p. 496 Freezing Point Depression only consider cases where the pure solvent crystalizes from solution, e.g., ice Crystalizes from salt water and NaCl(s) does not  T f = - K f m ( m=molality of soln)  T f = T’ f - T f Melting point(T f ) lowered to keep the vapor pressure over the pure solid and liquid solution the same at Equilibrium!  P 1 = - X 2 P ° 1 S= -  P/  T f Solved for  T f

24. Table 11-2, p. 494  T f = - K f m ( m=molality of soln)  T f = Example : again for 0.058 gmol -1 of NaCl in 10 g H 2 O( l ) over H 2 O(s) The molality is m = 0.02 gkg -1 (grams of solute per kg of solvent)

25. Table 11-2, p. 494  T f = - K f m ( m=molality of soln)  T f = - (1.86)x(0.02) K= - 0.0372 °C T’ f = 0 +(-0.037)°C=-0.037°C Example : again for 0.058 gmol -1 of NaCl in 10 g H 2 O( l ) over H 2 O(s) The molality is m = 0.02 gkg -1 (grams of solute per kg of solvent)

26. Fig. 11-12, p. 496 Freezing Point Depression only consider cases where the pure solvent crystalizes from solution, e.g., ice Crystalizes from salt water and NaCl(s) does not  T f = - K f m ( m=molality of soln)  T f = T’ f - T f Melting point(T f ) lowered to keep the vapor pressure over the pure solid and liquid solution the same at Equilibrium!  P 1 = - X 2 P ° 1 S= -  P 1 /  T f (pure solvent) Solved for  T f P1P1  T f = 0.037 °C = 0.037 K

27. Fig. 11-13, p. 497 Freezing point depression (  T b ) v.s. Molality of the solution M(s)  M( aq ) MX(s)  M( aq ) +X( aq ) MX 2 (s)  M( aq ) +2X( aq )x MX 3 (s)  M( aq ) +3X( aq )x } Closer to ideal soln

28. p. 499 Osmotic pressure forces water out of a carrot placed I n a salt soln Water doesn’t leave a carrot placed in pure water Osmotic Pressure π

29. Measuring Pressure Hg Barometer mm

30. Pressure= Force/unit area F=mg=  Vg=  hAg F air FHgFHg F Hg = F air P air =F air /A=F Hg /A=  hAg/A=  hg: P air =  hg  =mass/V Mass Density kg/m 3 or kg/cm 3 mm

31. Fig. 11-14, p. 498 Osmotic Pressure π=[solute]RT Solute molar concentration Solute molecules Lowers the rate of Solvent molecules Crossing the Membrane From the solution At Equilibrium the rate of The Solvent molecules Crossing the membrane from solution Is equal to the rate from the solvent Recall that pressure in the tube P=  gh so π=  gh Van’t Hoff proposed π=[solute]RT Which is similar to PV=nRT for an ideal gas. Note that is the solute Concentration but the Solvent mass density  (kg meter -3 )!