1. Equilibrium – Where? vs. Why? The value of K is an indication of WHERE the equilibrium rests. We haven’t addressed WHY the equilibrium exists where it does. ΔGΔG

2. A ball rolling down two different hills. Potential energy “drives” the ball to the point of minimum potential energy.

3. In the same way, free energy (ΔG) of a chemical reaction decreases until it reaches a minimum value. “Chemical Potential” A ball rolling down two different hills.

4. The ball will come to rest at the position of lowest potential energy. A reaction will proceed to the point where free energy is lowest. This drawing is called a free-energy curve.

6. Free Energy of the Mixture (G) becomes more negative during the course of any natural process. as a chemical reaction takes place, G only falls and will never become more positive. Eventually a point is reached where any further transformation of reactants into products would cause G to increase. At this point G is at a minimum, and no further net change can take place; the reaction is at equilibrium.

7.  G vs  G o   G  applies only when the reactants and products are in their standard states. – Their normal state at that temperature – Partial pressure of gas = 1 atm – Concentration = 1 M  G calculated for given (current) conditions. Only one value for  G  for a given reaction. Many possible values for  G.

8. N 2 O 4 2NO 2

9. ΔGΔG ΔG = ΔG  + RT ln Q R = 8.31 J/mol. K

10. Consider the reaction at 298 K: 2H 2 S(g) + SO 2 (g) 3S(s, rhombic) + 2H 2 O(g)  G o rxn = -102 kJ Calculate  G rxn under these conditions: P H2S = 2.00 atm P SO2 = 1.50 atm P H2O = 0.100 atm

11. ΔGΔG ΔG = ΔG  + RT ln Q At equilibrium, ΔG = 0 ΔG  = -RT ln K and Q = K

12. Calculate K p for the reaction 2H 2 O(l) 2H 2 (g) + O 2 (g) Given that ΔG  [H 2 O(l)] = -237.1 kJ/mol