2. Review Standard Gibbs Free Energy  G o =  G o < 0  G o > 0  G o = 0 spontaneous non-spontaneous equilibrium HoHo - T  S o glucose  G o = + 16.7 kJ/mol ATP  G o = - 30.5 kJ/mol glucose  G o = - 13.8 kJ/mol + phosphate  glucose-6-phosphate+ H 2 O + H 2 O  ADP + phosphate + ATP  ADP + glucose-6-phosphate

3. Q = K =  G = K > Q K < Q reaction quotient = [products] equilibrium constant = [products]  G < 0 K = Q m [reactants] initial n m equilibrium [reactants] n equilibrium - RTln ( )  G > 0  G = 0 K /Q/Q spontaneous non-spontaneous equilibrium Non-standard conditions

4. GoGo  G = - RT ln(K/Q) Impose Standard conditions: Q =  G = [reactants] initial = [products] initial = 1(M or atm) 1 o - RT ln(K) Standard Free Energy

5.  G o = - RT ln K ln (a/b) =  G = - RT ln (K/Q) =  G = - RT ln (K/Q) -RT ln K + RT ln Q ln a- ln b Non-Standard Conditions  G = GoGoGoGo + RT ln Q

6.  G o = - RT ln(K) glucose + phosphate  glucose-6-phosphate+ H 2 O  G o = 16.7 kJ/mol non-spontaneous = -RT ln K -RT K =  G = - RTln ( )K /Q/Q Q<1.2 x 10 -3 GG < 0 spontaneousstart with no product Q = [products] [reactants] initial standard conditions non-standard conditions 1.2 x 10 -3 < K

7. Initially Will more products or reactants be formed? Q =0.5 /0.3 2 = 5.6 Non-Standard Conditions  G = GoGoGoGo + RT ln Q [NO 2 ] = 0.3 M[N 2 O 4 ] = 0.5 M 2NO 2 (g) N 2 O 4 (g)

8.  G o rxn =  G o f products -  G o f reactants = 97.8 -  G rxn = = -0.5 kJ/mol More will be formed - 4.8 kJ/mol -4.8 +(8.314 x 10 -3 )(298)ln 5.6 a) product b) reactant 2(51.3) = Non-Standard Conditions  G = GoGoGoGo + RT ln Q 2NO 2 (g) N 2 O 4 (g) Initially [NO 2 ] = 0.3 M[N 2 O 4 ] = 0.5 M

9. A K = [A] e [B] e Le Chatelier’s Principle Add A Remove A [C]i[C]i [A] i [B] i  G = -RT ln (K/Q) QK/QK/Q [C]e[C]e a) increase b) decrease > 1 increase< 1 +BC  Q = Add B decreaseQK/QK/Q GG < 0>

10. Temperature dependence of K exothermic reactionheat = endothermic reactionheat = at low T favor reaction at high T at low T favor reaction reverse at high T favor reaction forward a) forward b) reverse favor reaction reverse product reactant

11. Temperature dependence of K  G o = -RT ln K  G o =  H o - T  S o -RT ln K = ln K = -  H o R 1T1T +  S o R exothermic  H o < 0 y m x b increase T K + ln K 1/T T  H o - T  S o decrease

12. Temperature dependence of K  G o = -RT ln K  G o =  H o - T  S o -RT ln K =  H o - T  S o ln K = -  H o R 1T1T +  S o R endothermic  H > 0 y m x b increase Tincrease K - ln K 1/T T