Review Standard Gibbs Free Energy G o = G o < 0 G o > 0 G o = 0 spontaneous non-spontaneous equilibrium HoHo - T S o glucose G o = kJ/mol ATP G o = kJ/mol glucose G o = kJ/mol + phosphate glucose-6-phosphate+ H 2 O + H 2 O ADP + phosphate + ATP ADP + glucose-6-phosphate
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
GoGo 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
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 = GoGoGoGo + RT ln Q
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 GG < 0 spontaneousstart with no product Q = [products] [reactants] initial standard conditions non-standard conditions 1.2 x < K
Initially Will more products or reactants be formed? Q =0.5 /0.3 2 = 5.6 Non-Standard Conditions G = GoGoGoGo + RT ln Q [NO 2 ] = 0.3 M[N 2 O 4 ] = 0.5 M 2NO 2 (g) N 2 O 4 (g)
G o rxn = G o f products - G o f reactants = G rxn = = -0.5 kJ/mol More will be formed kJ/mol (8.314 x )(298)ln 5.6 a) product b) reactant 2(51.3) = Non-Standard Conditions G = GoGoGoGo + RT ln Q 2NO 2 (g) N 2 O 4 (g) Initially [NO 2 ] = 0.3 M[N 2 O 4 ] = 0.5 M
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 GG < 0>
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
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
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