1. Chapter 18 Thermodynamics

2. Free Energy and Temperature

3.  G° =  H   T  S   G° =  H  =  S  = + + - at temperatures above 0 o C + at temperatures below 0 o C For ice melting, heat is absorbed and destruction of the crystalline lattice increases disorder, therefore, entropy is on your side but enthalpy is not. Spontaneity is observed only when the temperature is high (above ΔH/ΔS).

4. Free Energy and Temperature  G° =  H   T  S   G° =  H  =  S  = - - + at temperatures above 0 o C - at temperatures below 0 o C For water freezing, heat is given off and formation of the crystalline lattice reduces disorder, therefore, enthalpy is on your side but entropy is not. Spontaneity is observed only when the temperature is low (below ΔH/ΔS).

5. Free Energy and Temperature

6. Free Energy and Equilibrium Under any conditions, 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 and  G =  G  )

7. 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

8.  G 0 =  RT lnK 18.6

9. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12(–394.4) + 6(–237.2 )] – [ 2(124.5 )] = -6405 kJ Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ < 0 spontaneous 18.5

10. CaCO 3 (s) CaO (s) + CO 2 (g)  H 0 = 177.8 kJ  S 0 = 160.5 J/K  G 0 =  H 0 – T  S 0 At 25 0 C,  G 0 = 130.0 kJ  G 0 = 0 at 835 0 C 18.5 Temperature and Spontaneity of Chemical Reactions

11. Gibbs Free Energy and Phase Transitions H 2 O (l) H 2 O (g)  G 0 = 0=  H 0 – T  S 0  S = T HH = 40.79 kJ 373 K = 109 J/K 18.5

12. 18.6

13. 18.7 ATP + H 2 O + Alanine + Glycine ADP + H 3 PO 4 + Alanylglycine Alanine + Glycine Alanylglycine  G 0 = +29 kJ  G 0 = -2 kJ K < 1 K > 1

14. 18.7