2. Termodinamica Chimica dario.bressanini@uninsubria.it http://www.unico.it/~dario/thermo Miscele e Soluzioni Universita’ degli Studi dell’Insubria Corsi di Laurea in Scienze Chimiche e Chimica Industriale

3. © Dario Bressanini2 At 20  C the partial molar volumes of acetone and chloroform in a mixture containing 354.4 g of acetone and 645.2 g of chloroform are 74.17 cm 3 mol -1 and 80.23 cm 3 mol -1 respectively. What is the volume of the solution. What would be the result if the same calculation was performed using densities? At 20  C, the density of acetone is 0.787 g/cm 3 and for chloroform it is 1.499 g/cm 3.

4. © Dario Bressanini3 Change in Gibbs Energy for an Open System Under conditions of constant temperature and pressure For a binary mixture at constant temperature and pressure This equation may be expressed in integrated form as follows. Let the system increase in size but keep the relative proportions of the components constant. Temperature and pressure are also held constant. Under these conditions the chemical potentials are also constant. This is the Gibbs energy at fixed proportion of n A relative to n B.

5. © Dario Bressanini4 But we also know that under the same conditions In general the chemical potentials do vary with composition. Therefore at constant temperature and pressure Therefore it must be true that This expression is a special case of the Gibbs-Duhem equation. What this means is that the chemical potentials each component in a mixture are coupled. In other words, they can not vary independently. The results presented here are general to all partial molar quantities.

6. © Dario Bressanini5 Non-Ideal Solutions Most solutions are not ideal:  Case 1: If the intermolecular forces between A and B molecules are weaker than those between A molecules and between B molecules, then there is a greater tendency for these molecules to leave the solution than in the case of an ideal solution.  P T > P A + P B  Case 2: If A molecules attract B molecules more strongly than the do their own kind, the vapor pressure of the solution is less than the sum of the vapor pressures.  P T < P A + P B

7. © Dario Bressanini6 Positive Deviations from Raoult’s Law Observed pressure higher than than predicted by Raolut’s Law. Observed pressure higher than predicted by Raolut’s Law. Obtained when A  A and B  B interactions are stronger than A  B interactions. Obtained when A  A and B  B interactions are stronger than A  B interactions. Characterized by  H soln > 0 Characterized by  H soln > 0 Example: ethanol and hexane Example: ethanol and hexane

8. © Dario Bressanini7 Negative Deviations from Raoult’s Law Observed pressure lower than than predicted by Raolut’s Law. Observed pressure lower than predicted by Raolut’s Law. Obtained when A  A and B  B interactions are weaker than A  B interactions. Obtained when A  A and B  B interactions are weaker than A  B interactions. Characterized by  H soln > 0 Characterized by  H soln > 0 Example: acetone and water (these can hydrogen-bond to each other) Example: acetone and water (these can hydrogen-bond to each other)

9. © Dario Bressanini8

10. 9 Disaster: (1700 dead) from Gas Solubility In the African nation of Cameroon in 1986 a huge bubble of CO 2 gas escaped from Lake Nyos and moved down a river valley at 20 m/s (about 45 mph). Because CO 2 is denser than air, it hugged the ground and displaced the air in its path. More than 1700 people suffocated. The CO 2 came from springs of carbonated groundwater at the bottom of the lake. Because the lake is so deep, the CO 2 mixed little with the upper layers of water, and the bottom layer became supersaturated with CO 2. When this delicate situation was changed, perhaps because of an earth-quake or landslide, the CO 2 came out of the lake water just like it does when a can of soda is opened. In the African nation of Cameroon in 1986 a huge bubble of CO 2 gas escaped from Lake Nyos and moved down a river valley at 20 m/s (about 45 mph). Because CO 2 is denser than air, it hugged the ground and displaced the air in its path. More than 1700 people suffocated. The CO 2 came from springs of carbonated groundwater at the bottom of the lake. Because the lake is so deep, the CO 2 mixed little with the upper layers of water, and the bottom layer became supersaturated with CO 2. When this delicate situation was changed, perhaps because of an earth-quake or landslide, the CO 2 came out of the lake water just like it does when a can of soda is opened. Lake Nyos in Cameroon, the site of a natural disaster. In 1986 a huge bubble of CO 2 escaped from the lake and asphyxiated more than 1700 people.

11. © Dario Bressanini10 P T Adding a solute to a pure liquid elevates its T boil by lowering its vapor pressure. Adding a solute to a pure liquid elevates its T boil by lowering its vapor pressure.  (Raoult’s Law)  It also stabilizes liquid against solid (lowers T fusion ) »Lower P wins, remember? Click to see the new liquid regions and Click to see the new liquid regions and  2 colligative properties in 1! Cambiamenti del Diagramma di Fase

12. © Dario Bressanini11 Variazione della Pressione di Vapore

13. © Dario Bressanini12

14. © Dario Bressanini13

15. © Dario Bressanini14

16. © Dario Bressanini15 Dialysis Process of cleaning blood Process of cleaning blood

17. © Dario Bressanini16 OsmosisOsmosis

18. © Dario Bressanini17 The phenomenon of freezing-point de-pression can be turned around to apply to the precipitation of a solute from solution. Just by thinking of the solute as the substance that “freezes out” we get - ln x B = (  fus H/R)(1/T - 1/T*) The phenomenon of freezing-point de-pression can be turned around to apply to the precipitation of a solute from solution. Just by thinking of the solute as the substance that “freezes out” we get - ln x B = (  fus H/R)(1/T - 1/T*)  Here  fus H is the enthalpy of fusion of the SOLUTE (the component that solidifies first) and T* is the freezing (melting) point of the pure SOLUTE. Solubilita’

19. © Dario Bressanini18 Interpretation of  H soln  H soln =  H 1 +  H 2 +  H 3 where CONCLUSION: CONCLUSION: For good solubility (  H soln small or negative),  H 3 must compensate for  H 1 and  H 2. For good solubility (  H soln small or negative),  H 3 must compensate for  H 1 and  H 2.  H 1 =solute-solute energy (> 0)  H 1 =solute-solute energy (> 0)  H 2 =solvent-solvent energy (> 0)  H 2 =solvent-solvent energy (> 0)  H 3 =solute-solvent energy (< 0)  H 3 =solute-solvent energy (< 0)