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Understanding Gibbs Free Energy in Chemical Reactions: Spontaneity and Non-Standard Conditions

This comprehensive overview explores the concept of Gibbs Free Energy (ΔG) and its significance in determining the spontaneity of chemical reactions. We delve into the relationship between ΔG, enthalpy (ΔH), and entropy (ΔS), presenting the equation ΔG = ΔH - TΔS. The analysis includes conditions that render reactions spontaneous (ΔG < 0) versus non-spontaneous (ΔG > 0) and the implications of equilibrium (ΔG = 0). Further, we touch on the effects of concentration (reaction quotient Q) and temperature on Gibbs Free Energy, with practical examples to illustrate these concepts.

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Understanding Gibbs Free Energy in Chemical Reactions: Spontaneity and Non-Standard Conditions

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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 N2O4 will be formed

  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 decrease Q K/Q > 1 Add A Add B > < 0 G Remove A increase Q K/Q < 1

  9. Temperature dependence of K exothermic reaction heat = product favor forward reaction 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 = m x b ln K y 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 ln K = + So R - ln K m b y x endothermic H > 0 1/T T increase T increase K

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