1. Gibbs Free Energy Enthalpy changes (  H) and entropy changes (  S) both have a “say” in whether or not a rxn is spontaneous. Spontaneity is determined using the equation for Gibbs free energy… Josiah Willard Gibbs 1839–1903  G (o) =  H (o) – T  S (o) If  G < 0… If  G > 0… If  G = 0… rxn. is spontaneous (i.e., as written) rxn. is AT equilibrium rxn. is nonspontaneous (i.e., spont. ) (THIS INCLUDES PHASE CHANGES AT NBP OR NFP) (o) = std. conditions are optional

2. For the Haber process… N 2 (g) + 3 H 2 (g) 2 NH 3 (g) spontaneous pure N 2 + H 2 pure NH 3 equilibrium mixture (Q = K,  G = 0) Q < K  G < 0 Q > K  G > 0

3. standard free energies of formation,  G f o -- are tabulated for pure solids, pure liquids, gases at ~1 atm pressure, and 1 M solutions -- For elements in their standard states…  G f o = 0 -- For a reaction, the standard free-energy change is found by…  G o =  nG f o P –  mG f o R  G says WHICH WAY a reaction will proceed, but it says NOTHING about the reaction rate.

4. Calculate the standard free-energy change for… PCl 3 (g) + Cl 2 (g)  PCl 5 (g) –286.30–324.6 tabulated  G f o ’s in kJ/mol  G o =  nG f o P –  mG f o R  G o = –324.6 – –286.3 = –38.3 kJ rxn. is spontaneous as written (i.e., left to right)

5. Free Energy and Temperature From  G =  H – T  S, we see that  G varies with temperature. -- When T changes, so does  G. --  H and  S change little with temperature. EX. (a) Calculate  H o,  G o, and  S o for… 2 NO(g) + O 2 (g)  2 NO 2 (g) 90.3033.2 86.7051.8 210.7205240  H f o (kJ/mol)  G f o (kJ/mol) S o (J/mol-K)

6.  H o = 2(33.2) – [2(90.3)]  G o = 2(51.8) – [2(86.7)]  S o = 2(240.0) – [205.0 + 2(210.7)]  H o = –114.2 kJ  G o = –69.8 kJ  S o = –146.4 J/K (b) Estimate  G at 400 K.  G =  H – T  S = –114,200 J – [400 K(–146.4 J/K)] = –55,600 J = –55.6 kJ (Remember:  H and  S vary very, VERY little w /temp.)

7. Estimate the normal boiling point of ethanol. At the NBP… CH 3 CH 2 OH(l) CH 3 CH 2 OH(g) Strategy: Realize that  G = 0. Find  S and  H. Solve for T in  G =  H – T  S.  H f o (kJ/mol) S o (J/mol-K) –277.7 160.7 –235.1 282.7  S =  S o = ~  H =  H o = ~ 122 J/K 42.6 kJ = 349 K (76 o C) LG From  G =  H – T  S… 0

8. From  G =  H – T  S, we see that spontaneity (i.e.,  G) depends on T. We estimated 76 o C; the actual NBP of ethanol is 78.4 o C. As T increases,  G becomes (–) Assume we start at 78.4 o C, where  G = 0… As T decreases,  G becomes (+) -- spontaneous, liquid to gas -- spontaneous, gas to liquid  H = 42,600 J;  S = 122 J/K ++ 0 351.4