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Review

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

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Review

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  1. Review Standard Gibbs Free Energy Go = Ho - TSo spontaneous Go < 0 Go = - 30.5 kJ/mol non-spontaneous Go > 0 Go = + 16.7 kJ/mol Go = 0 equilibrium Go = - 13.8 kJ/mol  glucose-6-phosphate + H2O + phosphate glucose + H2O  ADP ATP + phosphate  ADP + glucose-6-phosphate glucose + ATP

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

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

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

  5. Go = - RT ln(K) + phosphate  glucose-6-phosphate + H2O glucose -RT -RT /Q G = - RT ln ( ) K G < 0 [products] initial [reactants] initial standard conditions non-spontaneous Go = 16.7 kJ/mol = -RT ln K K = 1.2 x 10-3 Q < K < 1.2 x 10-3 non-standard conditions Q = start with no product spontaneous

  6. Non-Standard Conditions 2NO2 (g) N2O4 (g) G = Go + RT ln Q Initially [NO2] = 0.3 M [N2O4] = 0.5 M Will more products or reactantsbe formed? Q = 0.5 / 0.32 = 5.6

  7. Non-Standard Conditions 2NO2 (g) N2O4 (g) G = Go + RT ln Q Initially [NO2] = 0.3 M [N2O4] = 0.5 M Gorxn= Gof products - Gof reactants = 97.8 - - 4.8 kJ/mol 2(51.3) = Grxn= = -0.5 kJ/mol -4.8 + (8.314 x 10-3)(298) ln 5.6 More will be formed a) product b) reactant

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

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

  10. 1 T Temperature dependence of K Go =  Ho - TSo Go = -RT ln K  Ho - TSo -RT ln K = - Ho R + So R ln K = + ln K m b y x exothermic Ho < 0 1/T increase T decrease K T

  11. 1 T Temperature dependence of K Go =  Ho - TSo Go = -RT ln K -RT ln K =  Ho - TSo - Ho R + So R ln K = - ln K m b y x endothermic H > 0 1/T T increase T increase K

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