1. Lecture 354/29/05 TSC 006 Tuesday, May 3 from 2:30 - 3:30 p.m.

2. QUIZ 1.Calculate ∆H rxn for 2NOCl (g)  N 2 (g) + O 2 (g) + Cl 2 (g) given the following set of reactions: ½ N 2 (g) + ½ O 2 (g)  NO (g) ∆H= 90.3 kJ NO (g) + ½ Cl 2 (g)  NOCl (g) ∆H= -38.6 kJ 2.Give an example of an endothermic and spontaneous process.

3. 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

4. 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)

5. Is the reaction spontaneous? ∆H° rxn = Σn ∆H° f (products) - Σn ∆H° f (reactants) ∆S° = ΣS°(products) - ΣS°(reactants)

6. Gibbs Free Energy ( ∆G) Developed by Josiah Gibbs Combines enthalpy and entropy ∆G < 0  spontaneous reaction ∆G > 0  spontaneous in other direction ∆G = 0  equilibrium ∆G° = ∆H° - T∆S° ∆G = ∆H - T∆S ∆G° = Σn∆G f ° (products) - Σn∆G f ° (reactants)

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

8. Gibbs Free Energy ∆G = ∆H - T∆S ∆H∆S-T∆S∆G spontaneous? example -+- always negative always 2O 3 (g)  3O 2 (g) +-+ always positive never 3O 2 (g)  2O 3 (g) --+ negative at low T positive at high T at low T H 2 O (l)  H 2 O (s) ++- positive at low T negative at high T at high T H 2 O (s)  H 2 O (l)

9. Gibbs Free Energy ( ∆G) R = 8.314 J/mol-K at equilibrium ∆G = 0 and Q = K ∆G° = 0 K = 1 (because of standard conditions) ∆G° > 0 K < 1 ∆G° 1

11. 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