1. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure  P = - P  x  sol higher boiling point  T = + k b x m sol  lower freezing point  T = - k f x m sol osmotic pressure Colligative Properties

2. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure  P = - P  x  solute higher boiling point  T = + k b x m solute lower freezing point  T = - k f x m solute  osmotic pressure Colligative Properties

3. © University of South Carolina Board of Trustees ●Solvent flow initially in balance. ●Adding solute blocks flow right  left.

4. © University of South Carolina Board of Trustees ●The solvent level on the right rises due to osmotic pressure.

5. © University of South Carolina Board of Trustees ●External pressure can restore equilibrium. 

6. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure  P = - P  x  sol higher boiling point  T = + k b x m sol k f x m sol lower freezing point  T = - k f x m sol  osmotic pressure  = RT x M sol Colligative Properties

7. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: -P  sol lower vapor pressure  P = - P  x  sol k b m sol higher boiling point  T = + k b x m sol k f m sol lower freezing point  T = - k f x m sol RTM sol osmotic pressure  = RT x M sol Colligative Properties moles of solute Solvent properties Each can measure molecular mass

8. © University of South Carolina Board of Trustees Last Lecture Find the molar mass of a solute, if a solution of 1.33 g of the compound dissolved in 25.0 g of benzene has a boiling point of 81.22 o C.

9. © University of South Carolina Board of Trustees Student Example: Molar Mass by Osmotic Pressure A 5.70 mg sample of protein is dissolved in water to give 1.00 mL of solution. Calculate the molar mass of the protein, if the solution has an osmotic pressure of 6.52 Torr at 20  C. R = 0.0821 L-atm/K-mol 1 atm = 760 Torr

10. © University of South Carolina Board of Trustees Chapt. 12 Solutions Sec. 5 Colligative Properties with Ions

11. © University of South Carolina Board of Trustees Ions and Colligative Properties NaCl (s)  Na + (aq) + Cl - (aq) Ca(NO 3 ) 2(s)  Ca 2+ (aq) + 2 NO 3 - (aq) n solute = i  n formula found by colligative properties usually desired van’t Hoff factor 1 1 mole by formula wt. 2 2 moles of solute i = 2 1 1 mole by formula wt. 3 3 moles of solute i = 3

12. © University of South Carolina Board of Trustees Student Example Arrange the following aqueous solutions in order of increasing boiling points: 0.03 molal urea(s) (a nonelectrolyte) 0.01 molal NaOH 0.02 molal BaCl 2 0.01 molal Fe(NO 3 ) 3

13. © University of South Carolina Board of Trustees Student Example Arrange the following aqueous solutions in order of increasing boiling points: 0.03 molal urea(s)  urea(aq)0.03 m solute 0.01 molal NaOH 0.02 molal BaCl 2 0.01 molal Fe(NO 3 ) 3

14. © University of South Carolina Board of Trustees Student Example Arrange the following aqueous solutions in order of increasing boiling points: 0.03 molal urea(s)  urea(aq)0.03 m solute 0.01 molal NaOH  Na + + OH - 0.02 m ions 0.02 molal BaCl 2 0.01 molal Fe(NO 3 ) 3

15. © University of South Carolina Board of Trustees Student Example molal Arrange the following aqueous solutions in order of increasing boiling points: 0.03 molal urea(s)  urea(aq)0.03 m solute 0.01 molal NaOH  Na + + OH - 0.02 m ions 0.02 molal BaCl 2  Ba 2+ + 2 Cl - 0.06 m ions 0.01 molal Fe(NO 3 ) 3

16. © University of South Carolina Board of Trustees Student Example Arrange the following aqueous solutions in order of increasing boiling points: 0.03 molal urea(s)  urea(aq)0.03 m solute 0.01 molal NaOH  Na + + OH - 0.02 m ions 0.02 molal BaCl 2  Ba 2+ + 2 Cl - 0.06 m ions 0.01 molal Fe(NO 3 ) 3  Fe 3+ + 3 NO 3 - 0.04 m ions

17. © University of South Carolina Board of Trustees Student Example Arrange the following aqueous solutions in order of increasing boiling points: 0.03 molal urea(s)  urea(aq)0.03 m solute 0.01 molal NaOH  Na + + OH - 0.02 m ions  0.02 molal BaCl 2  Ba 2+ + 2 Cl - 0.06 m ions  0.01 molal Fe(NO 3 ) 3  Fe 3+ + 3 NO 3 - 0.04 m ions 

18. © University of South Carolina Board of Trustees Chapt. 12 Solutions Sec. 6 Vapor Pressure of Liquid-Liquid Solutions (skip)

19. © University of South Carolina Board of Trustees Chapt. 13 Kinetics (Later)

20. © University of South Carolina Board of Trustees Chapt. 14 Chemical Equilibrium Sec. 1 What is Chemical Equilibrium?

21. © University of South Carolina Board of Trustees What is Chemical Equilibrium? ●Both products and reactants remain ●Macroscopically Static: System has finished changing (not just slow)

22. © University of South Carolina Board of Trustees Macroscopically Unchanging 2 NO 2(g) N 2 O 4(g) brown colorless equilibrium Next Day

23. © University of South Carolina Board of Trustees What is Chemical Equilibrium? ●Both products and reactants remain ●Macroscopically Static: System has finished changing (not just slow)

24. © University of South Carolina Board of Trustees What is Chemical Equilibrium? ●Both products and reactants remain ●Macroscopically Static: System has finished changing (not just slow) ●Molecularly Dynamic: Forward and backward rates are in balance

25. © University of South Carolina Board of Trustees Molecules Dynamic 2 NO 2(g)  N 2 O 4(g) brown colorless equilibrium Next Day

26. © University of South Carolina Board of Trustees What is Chemical Equilibrium? ●Both products and reactants remain ●Macroscopically Static: System has finished changing (not just slow) ●Molecularly Dynamic: Forward and backward rates are in balance ●Equilibrium concentrations independent of starting conditions

27. © University of South Carolina Board of Trustees Independent of Starting Conditions final:2 NO 2 (g)  N 2 O 4 (g) 2 NO 2 (g)  N 2 O 4 (g) initial: 100% NO 2 (g) 100 % N 2 O 4 (g) equilibrium concentrations

28. © University of South Carolina Board of Trustees Chapt. 14 Chemical Equilibrium Sec. 1 How are the Equilibrium Concentrations Calculated?

29. © University of South Carolina Board of Trustees Experiments: Equilibrium concentrations from several initial concentrations Initial concentration, M Equilibrium concentration, M [NO 2 ][N 2 O 4 ] 0.02000.00000.01034.86 x 10 -3 0.00000.01000.01034.86 x 10 -3 2 NO 2 (g)  N 2 O 4 (g) junk 0.03000.00000.01348.29 x 10 -30.62 0.04000.00000.016111.9 x 10 -30.74 [N 2 O 4 ][NO 2 ] Ratio I 0.47 0.47 [N 2 O 4 ] [NO 2 ]

30. © University of South Carolina Board of Trustees Experiments: Equilibrium concentrations from several initial concentrations Initial concentration, M Equilibrium concentration, M Ratio I [NO 2 ][N 2 O 4 ] 0.02000.00000.01034.86 x 10 -3 0.4745.8 0.00000.01000.01034.86 x 10 -3 0.4745.8 0.03000.00000.01348.29 x 10 -3 0.6246.2 0.04000.00000.016111.9 x 10 -3 0.7445.9 2 NO 2 (g)  N 2 O 4 (g) Ratio II [N 2 O 4 ][NO 2 ][N 2 O 4 ] [NO 2 ] [NO 2 ] 2 constant !

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32. © University of South Carolina Board of Trustees Equilibrium Constant Expression 2 NO 2 (g)  N 2 O 4 (g) Example: a A + b B  c C + d D In General: products – top reactants – bottom concentration

33. © University of South Carolina Board of Trustees Equilibrium Constant Expression 1 2 NO 2 (g)  1 N 2 O 4 (g) Example: abcd a A + b B  c C + d D In General: products – top reactants – bottom Equilibrium Constant a numbera number independent of starting amountsindependent of starting amounts

34. © University of South Carolina Board of Trustees Examples CO (g) + H 2 O (g)  H 2(g) + CO 2(g) Cl 2(g)  2 Cl (g) 2 O 3(g)  3 O 2(g)