1. Gibb’s Free Energy Chapter 19

2. GG Gibbs free energy describes the greatest amount of mechanical work which can be obtained from a given quantity of a reactants in a given initial state Also referred to as the chemical potential of a system

3. Gibbs Free Energy  T  S universe is defined as the Gibbs free energy,  G. When  S universe is positive,  G is negative. Therefore, when  G is negative, a process is spontaneous.

4. Gibbs Free Energy 1.If  G is negative, the forward reaction is spontaneous (products are favored at equilibrium, 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 but thermodynamically not favored in forward direction.

5. Spontaneous Means reaction will occur but has no specification as to how long it will take to occur – Spontaneous does not mean quick or that it will occur at a measurable rate, just that it is thermodynamically favored without something outside the system to cause it

6. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation,  G . f  G  =  n  G  (products)   m  G  (reactants) ff where n and m are the stoichiometric coefficients.  G  is for a substance at standard state which is P =1 atm and concentrations of reactants and products that equal 1 M

7. Practice Problem Use the data from Appendix C to calculate the standard free energy change for the following reaction at 298 K:P 4 (g) + 6 Cl 2 (g)  4 PCl 3  G  =  n  G  (products)   m  G  (reactants) = 4  G  (PCl 3 ) -  G  (P 4 ) - 6  G  (Cl 2 ) = 4*(-269.6) – 1(24.4) – 0 = -1102.8 KJ

8. Free Energy Changes At temperatures other than 25°C,  G° =  H   T  S  If  H  = -196.6 KJ and  S  = -189.6 J/K, is the reaction spontaneous at a temperature of 298 K?  G° = -196.6 – 298(-189.6* 1kJ/1000J) = -120.8 kJ (spontaneous)

9. Free Energy and Temperature There are two parts to the free energy equation: –  H  — the enthalpy (internal energy) term – T  S  — the entropy (disorder) term The temperature dependence of free energy, then comes from the entropy term.

10. Free Energy and Temperature Any process that has both  H  0 will always be spontaneous (  G  1) Any process that has both  H  > 0 (endothermic) and  S  0) and favors the reactants (K < 1)

11. Just because a reaction is exothermic, does not mean that it decreases in entropy – Each different type of reaction will have to be looked at individually and both the values for enthalpy and entropy must be considered to determine if the reaction is thermodynamically favored (  G  < 0)

12. Conditions that can cause nonspontaneous reactions to occur Electricity – Charging a battery – Electrolysis Light – Photosynthesis in plants (converting sunlight into food) – Photoionization of an atom (removing additional electrons from an atom) Coupling favorable reactions with unfavorable reactions – Conversion of ATP to ADP releases energy in living things – Series of reactions with common intermediates are grouped together with an overall  G  < 0

13. Thermodynamically Favored but not at a measurable rate Processes that are thermodynamically favorable but do not occur at measurable rates are said to be under kinetic control – Common reason is High activation energy – Note: Just because a thermodynamically favored reaction does not occur at a measurable rate does not mean that it is at equilibrium

14. Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)

15. 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, K = e  G  /RT

16. This means the equilibrium constant is related to free energy – When  G  is large compared to the thermal energy (RT), then K deviates strongly from 1 (away from equilibrium) – When a reaction is at equilibrium  G  is 0 (since there is no change between reactants and products) so K = 1

17.  G  1 – Reaction is classified as “exergonic” Spontaneous  G  > 0, the reaction favors products so K < 1 – Reaction is classified as “endergonic” nonspontaneous

18. Practice Problem Use data from Appendix C to calculate the standard free energy change,  G°and the equilibrium constant, K, at 298 K for the reaction H 2 (g) + Br 2 (l)  2HBr(g)  G°= -106.4 kJ/mol K = 4 x 10 18 Spontaneous, since K > 1 products are favored