1. Free Energy and Equilibrium  G of a reaction (run at standard conditions,  G  ) is the change in free energy accompanying the chemical reaction in which the reactants are converted completely to products. However, the Free Energy at Equilibrium (when both products and reactants are present) is always lower than the Free Energy of the pure reactants or the pure products.

2. Gibbs Free Energy & Equilibrium For a Thermodynamically Favored Process, -  G

3. Gibbs Free Energy & Equilibrium 1.If  G is negative, the forward reaction is spontaneous and K > 1 2.If  G is 0, the system is at equilibrium. 3.If  G is positive, the reaction is spontaneous in the reverse direction and K < 1. The “curve” is also different as the pure products have more Free Energy than the pure reactants

4. Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT lnQ  G = Gibbs Free Energy change under any other condition than standard conditions.

5. Free Energy and Equilibrium At equilibrium, Q = K, and  G = 0. The equation becomes 0 =  G  + RT lnK Rearranging, this becomes  G  =  RT lnK or, solving for K, K = e  G  /RT

6. Sample Problem The reaction of nitrogen gas and hydrogen gas to form one mole of ammonia has a  G o of - 16,370 J/mol at 25 o C. What is the Kp at standard conditions? Use:  G  =  RT lnK, and solve for K. Notice that your units need to be in Joules for  G  (not kJ as given in reference tables) because the gas constant, R, is in Joules. Answer: Kp = 738

7. How Temperature Affects “K” 0 =  G  + RT lnK Consider the above equation, ask yourself, what would happen to “K” if you raise the temperature (all else being constant)? What would happen to “K” if you lowered the temperature? –Remember, once K drops below 1, the lnK increases in negative magnitude with smaller values of K.

8. Implications & Summary Free Energy at Equilibrium is always lower than either the pure reactants or pure products. When ∆ r G<0, the reaction is proceeding spontaneously toward equilibrium. When ∆ r G>0, the reaction is beyond the equilibrium point and is no longer spontaneous in the forward direction, but will be in the reverse direction.

9. Summary & Application When ∆ r G=0, the reaction is at equilibrium and K=Q Application: Fe 3+ + SCN - ↔ FeSCN 2+ –First, recall mixing the yellow iron solution and colorless thiocyanate solutions together, what immediately happened? –What happens to the entropy of the system as products are formed? Is this favorable thermodynamically?

10. Application Continued… –Entropy decreases as you form product, so, what must be true of the change in enthalpy as the reaction moves forward making more product? –Recall placing into hot and cold water. What happen? Consider both the change in entropy and enthalpy of the forward and reverse reactions. What caused the reaction to shift to the right? To the left? What role does temperature play in all this?