Outline:1/26/07 n n Chemistry Seminar – 4pm n n Pick up Quiz #2 – from me n n Today: Chapter 14 (cont’d) Entropy Free Energy Example calculations

Quiz #2 Average: 6.1 (20.5% got 9 or 10….)

On what does entropy depend? n n Temperature Phase (s)  ( )  (g) obvious less obvious n Molar mass n Concentration Generally, as molar mass , intermolecular disorder  Table 14-2: H 2 = J/K mol (page 580) F 2 = J/K mol Cl 2 = J/K mol

On what does entropy depend? n n Temperature Phase (s)  ( )  (g) obvious less obvious n Molar mass n Concentration p.126

Given thermodynamic tables,  S rxn can be calculated:   CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g) (205.1) (188.8) J/K mol J/ K mol J/K mol J/K mol   S rxn =  S o products -  S o reactants  =  5.2 J/K But isn’t it spontaneous ? Need to check with the universe…

How useful is entropy? If you can calculate the  S of the universe you can tell if a reaction is spontaneous...  S universe not always easy to calculate...

The last thermo variable:  G n n Gibb’s Free Energy:    G sys =  H sys  T  S sys If  G < 0 the system spontaneously reacts…. For  G you only need to consider the system (not the system + surroundings !)

Prove it! Since  H sys = q sys (at constant P)   =  q surr   Then  H sys =  T  S surr   Plug into:  G sys =  H sys  T  S sys   G sys =  T  S surr  T  S sys   G sys =  T (  S surr  S sys )  or  G sys =  T (  S universe )

Given thermodynamic tables,  G rxn can be calculated:   CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g)    (  228.6) kJ/mol kJ/mol kJ/mol kJ/mol   G rxn =   G o products -   G o reactants  =  801 kJ/mol very spontaneous at 298 K !

Reality check: Pb 2+ (aq) + SO 4 2- (aq)  PbSO 4(s) Pb 2+ (aq) + SO 4 2- (aq)  PbSO 4(s) Ca 2+ (aq) + SO 4 2- (aq)  CaSO 4(s) Ca 2+ (aq) + SO 4 2- (aq)  CaSO 4(s) Predict which is spontaneous…. Qualitative Chemistry rules revisited!

Which is spontaneous? Pb 2+ (aq) + SO 4 2- (aq)  PbSO 4(s) Pb 2+ (aq) + SO 4 2- (aq)  PbSO 4(s) Ca 2+ (aq) + SO 4 2- (aq)  CaSO 4(s) Ca 2+ (aq) + SO 4 2- (aq)  CaSO 4(s)   24.4   kJ/mol kJ/mol kJ/mol kJ/mol kJ/mol kJ/mol     kJ/mol kJ/mol kJ/mol kJ/mol kJ/mol kJ/mol spontaneous!  G rxn =  14.5 kJ/molspontaneous! non-spontaneous!  G rxn = 6.3 kJ/mol non-spontaneous!

Better yet,  G rxn at different temperatures can be calculated:   H o rxn and  S o rxn are relatively independent of temperature   G T rxn   H o rxn  T  S o rxn

p.126

Example: 4 CuO (s)  2Cu 2 O (s) + O 2 (g) At what temperature will this reaction become spontaneous? Or conversely, at what temperature a reaction will become spontaneous!  The beauty of  G T rxn is that you can now determine whether a reaction is spontaneous at any reasonable temp!

  G T rxn  0 when first spontaneous…. Solve: 0 =  H o rxn  T  S o rxn Example: 4 CuO (s)  2Cu 2 O (s) + O 2(g) For this reaction:   H o rxn  kJ   S o rxn  J/K  (from appendix D)  Or:  kJ  kJ/K)  

Summary to date:   E,  H,  S and  G are defined  First law calculations:  E = q + w   H rxn,  phase change problems   S calculations: T  S = q   G o rxn =  H o rxn  T  S o rxn problems  Since  H o rxn and  S o rxn are relatively independent of temperature:   G T rxn   H o rxn  T  S o rxn

Practice Problems: Chapter 14 n 14.11, 14.15, 14.17, 14.19, 14.23, 14.25, 14.27, 14.31, 14.35, 14.37, 14.38, 14.41, 14.43, 14.49, 14.51, 14.53, 14.55, 14.57, 14.61, 14.65, 14.67, 14.71, 14.75, 14.77, 14.79, 14.81, 14.91, ,