1. Effect of Pressure Copyright © 2011 Pearson Canada Inc. Slide 1 of 46 General Chemistry: Chapter 13 William Henry found that the solubility of a gas increases with increasing pressure. C = kP gas k = C P gas = 23.54 mL 1.00 atm = 23.54 ml N 2 /atm k C P gas = = 100 mL = 4.25 atm 23.54 ml N 2 /atm

2. Copyright © 2011 Pearson Canada Inc. Slide 2 of 46 General Chemistry: Chapter 13

3. Copyright  2011 Pearson Canada Inc. 13 - 3

4. Vapor Pressures of Solutions Raoult, 1880s. – Dissolved solute lowers vapor pressure of solvent. – The partial pressure exerted by solvent vapor above an ideal solution is the product of the mole fraction of solvent in the solution and the vapor pressure of the pure solvent at a given temperature. P A =  A P° Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 4 of 46 A

5. Raoult’s Law – Esters?

6. Raoult’s Law – the Kitchen In many cases our enjoyment of foods and beverages results in part from the presence of volatile components in what we eat/drink. Sometimes warming of food is employed which increases the vapour pressure of all volatile components.

7. Mixture with Two (or more) Volatile Components: Solutions can have two (or more) volatile components. The vapor pressure of the mixture can be written as P total = χ A P A 0 + χ B P B 0 + χ C P C 0 +…….. In the (unlikely!) event that a mixture contained only two components (A and B) with P A = P B the vapor pressure of the mixture would not vary with the composition (χ A etc).

8. Liquid-Vapor Equilibrium: Ideal Solutions Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 8 of 46 Liquid-vapor equilibrium for benzene-toluene mixtures at 25°C

9. Raoult’s Law Example: 1. (a) A benzene/toluene mixture is prepared by mixing 500.0 g benzene, C 6 H 6, and 500.0 g toluene, C 7 H 8, at 25.0 0 C. At this temperature pure benzene and pure toluene have vapor pressures of 12.68 kPa and 3.79 kPa respectively. Calculate the total vapor pressure of the mixture. (b) Find the mole fraction of benzene and toluene in the gas phase.

10. Raoult’s Law – Vapor Pressure Lowering: For a mixture of two substances Raoult’s Law tells us that P total = χ A P A 0 + χ B P B 0 where P A 0 and P B 0 are the vapor pressure of the two pure substances (at a particular T). If A is totally involatile, P A 0 = 0, then a mixture of A and B will have a total vapor pressure less than that of A. We can roughly estimate molar masses using this effect.

11. Raoult’s Law Example: 44.2 g of a simple carbohydrate (known to be either C 12 H 22 O 11 or C 6 H 12 O 6 ) are dissolved in 99.5 g of water at 24.0 o C. The vapor pressures of pure water and the solution at 24.0 o C are 2.986 kPa and 2.859 kPa respectively. Find the molar mass of the carbohydrate and hence its identity.

12. Raoult’s Law As a mnemonic device can you think of a way of thinking about Raoult’s Law that involves “probabilities”?

13. Osmotic Pressure Interesting changes occur when two solutions of different concentrations are placed in close contact. The next slide shows what happens when two aqueous solutions of different concentrations, each containing a nonvolatile solute are placed under a transparent enclosure. Water moves over time from the dilute solution (A) to the concentrated solution. Why? (Vapor pressure H 2 O higher above the left hand container. Why?)

14. Osmotic Pressure Copyright © 2011 Pearson Canada Inc. Slide 14 of 46 FIGURE 13-16 Observing the direction of flow of water vapor General Chemistry: Chapter 13

15. Osmotic Pressure – cont’d: On the previous slide there is a net transfer of water (solvent) molecules between the two containers until both containers have the same concentration or the vapor pressure of H 2 O is the same over both containers. Transfer of solvent need not occur through the gas phase! When two solutions of different concentration are separated by a solvent permeable membrane there is a surprise!

16. Osmosis FIGURE 13-17 Copyright © 2011 Pearson Canada Inc. Slide 16 of 46 General Chemistry: Chapter 13 Does pure water or sugar water have the higher vapor pressure. Mechanical analogy?

17. Osmotic Pressure – cont’d: On the previous slide the movement of solvent molecules generates enough pressure to lift solution through the tube. The size of this osmotic pressure can be calculated with a “familiar looking” equation. Here n is the number of moles of solute (initially a nonelectrolyte!!!!) and V is the solution volume (in Liters).

18. Osmotic Pressure Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 18 of 46 πV = nRT π = RT n V = MRT For dilute solutions of nonelectrolytes: