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Lecture 314/20/07. Section 1 Equilibrium Le Chatelier’s Solubility Section 2 Acid/Base equilibria pH Buffers Titration Section 3 Oxidation numbers Balancing.

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Presentation on theme: "Lecture 314/20/07. Section 1 Equilibrium Le Chatelier’s Solubility Section 2 Acid/Base equilibria pH Buffers Titration Section 3 Oxidation numbers Balancing."— Presentation transcript:

1 Lecture 314/20/07

2 Section 1 Equilibrium Le Chatelier’s Solubility Section 2 Acid/Base equilibria pH Buffers Titration Section 3 Oxidation numbers Balancing Redox reactions Galvanic cells Standard reduction potential table

3 Entropy (Measurement of disorder) Related to number of microstates S = klnW ∆S universe = ∆S system + ∆S surroundings 2 nd Law of Thermodynamics Entropy of the universe increases with spontaneous reactions Reversible reactions ∆S universe = ∆S system + ∆S surroundings = 0 Can be restored to the original state by exactly reversing the change Each step is at equilibrium Irreversible reaction ∆S universe = ∆S system + ∆S surroundings > 0 Original state can not be restored by reversing path spontaneous

4 3 rd Law of thermodynamics S = O at O K S° - entropy gained by converting it from a perfect crystal at 0 K to standard state conditions

5 3 rd Law of thermodynamics S = O at O K S° - entropy gained by converting it from a perfect crystal at 0 K to standard state conditions ∆S° = ΣS°(products) - ΣS°(reactants)

6 Degrees of freedom translational motion molecules in gas > liquid > solid vibrational motion movement of a atom inside a molecule rotational motion rotation of a molecule

7 Entropy trends Entropy increases: with more complex molecules with dissolution of pure gases/liquids/solids with increasing temperature with increasing volume with increasing # moles of gases

8 Which has higher entropy? dry ice orCO 2 liquid water at 25°Corliquid water at 50°C pure Al 2 O 3 (s)orAl 2 O 3 with some Al 2+ replaced with Cr 3+ 1 mole of N 2 at 1 atmor1 mol of N 2 at 10 atm CH 3 CH 2 CH 2 CH 3 (g)or CH 3 CH 3 (g)

9 Is the reaction spontaneous?

10 Gibbs Free Energy ( ∆G) ∆G° = ∆H° - T∆S° ∆G = ∆H - T∆S ∆G° = Σn∆G f ° (products) - Σn∆G f ° (reactants)

11 Gibbs Free Energy ∆G = ∆H - T∆S ∆H∆S-T∆S∆G spontaneous? example -+ 2O 3 (g)  3O 2 (g) +- 3O 2 (g)  2O 3 (g) -- H 2 O (l)  H 2 O (s) ++ H 2 O (s)  H 2 O (l)

12 Gibbs Free Energy (∆G) and equilibrium R = 8.314 J/mol-K

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