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Previously in Chem 104: Solutions: macroscopic & microscopic Deducing Enthalpies of Solution Energetics (Entha

Previously in Chem 104: Solutions: macroscopic & microscopic Deducing Enthalpies of Solution Energetics (Enthalpies) of Dissolution—they can be exo- and endothermic. TODAY QUIZ answer key out - send self-check by Tuesday midnight Another BIG IDEA- Equilibrium

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Previously in Chem 104: Solutions: macroscopic & microscopic Deducing Enthalpies of Solution Energetics (Entha

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  1. Previously in • Chem 104: • Solutions: macroscopic & microscopic • Deducing Enthalpies of Solution • Energetics (Enthalpies) of Dissolution—they can be • exo- and endothermic • TODAY • QUIZ answer key out - send self-check by Tuesday midnight • Another BIG IDEA- Equilibrium • Effects on Dissolution by: • DP • DT • Effects of Dissolution on: • vapor pressure • boiling • freezing

  2. A really BIG IDEA in chemistry EQUILIBRIUM EQUILIBRIUM EQUILIBRIUM EQUILIBRIUM

  3. A really BIG IDEA in chemistry EQUILIBRIUM EQUILIBRIUM EQUILIBRIUM EQUILIBRIUM Closely followed by Le Chatelier’s Principle

  4. We need another word to pair with EQUILIBRIUM Dynamic As applied to a solid in equilibrium with a solvent in solution:

  5. EQUILIBRIUM Combined with Le Chatelier’s Principle “ a system at equilibrium when disturbed will adjust to remove or minimize the effect of the disturbance” Example 1. This explains phase changes in phase diagrams

  6. H2O Phase Diagram solid liquid Pressure, atm gas ice water 1 atm Temperature, deg C 0 deg

  7. H2O Phase Diagram solid liquid Pressure, atm make more water 5 atm ice gas 1 atm Temperature, deg C 0 deg melts lower T

  8. EQUILIBRIUM Combined with Le Chatelier’s Principle This explains almost all Solution behaviors Example 2. How gas pressure affects solubility Sg = kh Pg Sg = kh Pg Gas solubility depends on gas partial pressure Henry’s Law Constant

  9. EQUILIBRIUM Le Chatelier’s Principle Combined with Explains almost all Solution behaviors A related example: How temperature affects gas solubility Gas solution + energy (exothermic) Gas + solvent Increase T and: Gas solution + energy Gas + solvent

  10. EQUILIBRIUM Le Chatelier’s Principle Combined with Explains almost all Solution behaviors Another related example: How temperature affects solid solubility M-(aq) + X-(aq) + energy Case A. MX + water Increase T and: M-(aq) + X-(aq) + energy MX + water Less dissolves M-(aq) + X-(aq) Case B. MX + water + energy Increase T and: MX + water + energy M-(aq) + X-(aq) More dissolves

  11. EQUILIBRIUM Le Chatelier’s Principle Combined with Explains almost all Solution behaviors Example 3. How a solute affects solvent vapor pressure Psolvent = Xsolvent Posolvent

  12. EQUILIBRIUM Le Chatelier’s Principle Combined with Explains almost all Solution behaviors Example 3. How a solute affects gas pressure DPsolvent = - Xsolute Posolvent

  13. EQUILIBRIUM Le Chatelier’s Principle Combined with Explains almost all Solution behaviors Example 4. How a solute affects solvent boiling point When does a liquid boil? What changes if we add a solute? DTbp = Kbp msolute Lower vapor pressure, requires more energy, higher T, to get to atmospheric pressure: Boiling Point Elevation DTbp = Kbp msolute Ebullioscopic constant

  14. EQUILIBRIUM Le Chatelier’s Principle Combined with Explains almost all Solution behaviors By extension: How a solute affects solvent freezing point DTfp = Kfp msolute

  15. EQUILIBRIUM Le Chatelier’s Principle Combined with Explains almost all Solution behaviors DPsolv = Xsolvent Posolvent DPsolv = XsolventPosolvent Three examples of Colligative Properties: colligative – depends only on the number of particles DTbp = Kbp msolute DTbp = Kbp msolute DTfp = Kfp msolute DTfp = Kfp msolute

  16. Calculations based on Colligative Properties: requires different concentration units We need to deal with concentration expressed as mole fraction, molality because collogative properties deal only with number of particles in solution, not identities. M, molarity = #moles solute / liter solvent m, molality = #moles solute / kg solvent Weight % = (mass A/ total mass) x 100% Mole fraction= #moles A / total # moles

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