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Free enthalpy Gibbs free energy

Free enthalpy Gibbs free energy. Dr. Miklós Tóth, professor of medical chemistry and biochemistry, Department of Medical Chemistry Molecular Biology and Pathobiochemistry, Semmelweis University, Budapest. Free-energy change during a spontaneous reaction

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Free enthalpy Gibbs free energy

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  1. Free enthalpy Gibbs free energy Dr. Miklós Tóth, professor of medical chemistry and biochemistry, Department of Medical Chemistry Molecular Biology and Pathobiochemistry, Semmelweis University, Budapest

  2. Free-energy change during a spontaneous reaction (combustion of gasoline)

  3. Free-energy change during a nonspontaneous reaction

  4. What is the meaning of Gibbs free energy (free enthalpy) ? It is the maximum potential of a system (chemical reaction) which is free to do useful work This potential or capacity to do useful work is an idealization: it never can be recovered completely in any spontaneous process (chemical reaction) because some entropy is always created and a corresponding amount of free energy is used up to create „bound” heat, an energy that is unable to do useful work If there is no possibility at all for a spontaneous chemical reaction to do useful work, all energy changes occur in the form of heat (reaction heat, Ho) Devices suitable to exploit free energy: car engine, battery, muscle

  5. Gibbs free energy (free enthalpy) is a state function: its molar value depends only on state properties. Standard Gibbs free energy of formation (Gof) is the quantity of free energy that is used up or liberated when 1 mol of a compound is produced from the stablest natural allotropic forms of its component elements under thermodynamic standard conditions (25 oC, 101.3 kPa). Since the absolute value of the Gibbs free energy content of a chemical system is not measurable, we use an arbitrary scale where the Gof value of the stablest natural allotrope of an element is regarded zero. G =  G (products) –  G (reactants) and Go =  Gof (products) –  Gof (reactants)

  6. Standard Free Energies of Formation ( at 25 oC)

  7. Calculation of the standard free enthalpy of formation: Gfo = Hfo – T• So , where Gfo = the standard free enthalpy of formation of the chemical compound studied Hfo = the standard enthalpy of formation of the chemical compound studied So = the change of standard entropies in the formation reaction of the chemical compound studied Because under standard conditions the T• So value is markedly less ( it is in the order of magnitude of joules) than the Hfo value (which is in the order of magnitude of kilojoules) the difference between the values of Gfo and Hfo is generally moderate.

  8. Calculation of the standard free enthalpy of formation (Gfo) from the standard enthalpy of formation (Hfo) and the standard entropy change (So) of the „formation reaction”. An example. H2 + 1/2 O2  H2O So(J/K) Hfo (kJ/mol) Gfo (kJ/mol) H2 130.7 0 0 O2 205.1 0 0 H2O 69.9 –285.8 – 237.1  So = 69.9 – (130.7 + 102.5) = – 163.3 J/K – 163.3 J/K (: 1000 and •298)  T•  So = – 48.7 kJ/mol Gfo = Hfo – T • Soandafter substitution: Gfo = –285.8 – (– 48.7) = – 237.1 kJ/mol

  9. Dependence of free enthalpy (Gibbs free energy) content on concentration: G = Gfo + RT ln c (J/mol), at 25 oC: G = Gfo + 8.3 • 298 • 2.3 • lg c (J/mol)) G = Gfo + 5.71 • lg c (kJ/mol) Free enthalpy change of reversible chemical reactions: A + B  C + D G =  G (products) –  G (reactants) G = GC + GD – (GA + GB) G = [Gfo(C) + 5.71 • log cC ] + [Gfo(D) + 5.71 • log cD ] – [Gfo(A) + 5.71 • log cA ] – [Gfo(B) + 5.71 • log cB] G = Go + 5.71 • (log cC + log cD – log cA – log cB ) G = Go + 5.71 log (cC • cD) / (cA • cB) G = Go + 5.71 • log Q, if cA, cB, cC, cD values are1 M each: G = Go , and because in equilibrium: G = 0 Go = – 5.71 • log Ke (kJ/mol)

  10. A + B  C + D Go = – 5.71• log Ke (kJ/mol) Ke  1, Go  0 : the reaction is exergonic if it starts from 1 M concentrations of A,B, C and D Ke  1, Go  0 : the reaction is endergonic when it starts from 1 M concentrations of A, B, C and D Ke = 1, Go = 0 : the reaction is in equilibrium at 1 M concentrations of A, B, C and D

  11. Free enthalpy content of the concentration gradient

  12. Free enthalpy change of redox reactions: G = – n • F • E and at standard conditions: Go = – n • F • Eo (kJ/mol, ifR = 96.5)

  13. Relationship between the equilibrium constant (Ke) and the standard electromotive force (Emf o = Eo ) generated by reversible redox reactions ox1 + red2 red1 + ox2 For a reaction in equilibrium: Go = – 5.71 • log Ke (kJ/mol) For a redox reaction: Go = – n • F • Eo (kJ/mol) – n • F • Eo = – 5.71 • log Ke n • F • Eo = 5.71 • log Ke F =96 500 coulomb Eo = (5.71 / 96 500) • (log Ke / n) Eo = 0.059 • log Ke / n or log Ke = n • Eo / 0.059

  14. Various ways for the determination of free energy changes of chemical reactions: 1. Go =  Gfo (products) –  Gfo (reactants) kJ/mol G =  G (products) –  G (reactants) 2. Go = Ho – T • So kJ/mol G = H – T • S 3. G = Go + 5.71 • log Q kJ/mol Go = – 5.71 • log Ke 4. Go = – n • F • Eo kJ/mol, if F = 96.5 kilocoulomb G = – n • F • E

  15. Kinetic interpretation of the equilibrium constant(Ke) v1 A + B  C + D v2 v1 = k1 • [A] • [B] and v2 = k2 • [C] • [D] in equilibrium: v1 = v2 concentrations of A,B,C and D are constant k1 • [A] • [B] = k2 • [C] • [D] k1 / k2 = ( [C] • [D] ) / ( [A] • [B] ) = Ke for any concentrations of A, B, C and D: ( [C] • [D] ) / ( [A] • [B] ) = Q (quotient) from indefinite number of Q there is only one value that equals Ke (in equilibrium Q = Ke)

  16. The chemiluminescent reaction of a firefly Luciferin (organic peroxide) + ATP in the presence of luciferase  light

  17. Relationship between equilibrium constant (Ke) and temperature At T1 temperature: G1o = Ho – T1 • So at 25-37 oCthe change of Ho and So with T is negligible, G1o = – R • T1 • ln Ke1 the change of Ke and Go is more significant Ho – (T1 • So) = – R • T1 • ln Ke1 / :T1 (Ho / T1) – So = – R • ln Ke1 / • – 1 R • ln Ke1 = – (Ho / T1) + So / :R ln Ke1 = – (1/ T1)• (Ho/R) + (So/ R) plot: lg Ke = – (1/ T)• (Ho/2.3•R) + (So/2.3• R)

  18. ln Ke = ( – 1/ T)• (Ho/R) + (So/ R) log Ke = (– 1/ T)• (Ho/2.3 • R) + (So/ 2.3 • R)

  19. Determination of Ho : from log Ke versus 1/T plot Determination of Go :from Ke or Eo Determination of So :from (Ho– Go)/T expression

  20. Free enthalpy change of hydrolysis of the terminal phosphate of ATP ATP + H2O  ADP + Pi Q = ([ADP] • [Pi]) / [ATP], in equilibrium Ke : 2.24 • 105 if [ADP] = [Pi] = [ATP] = 1 M G = G o + 5.71 • log Q (T = 25 oC) if G = O ( in equil.) G o = – 5.71 • log Ke = 30.5 kJ/mol [since -(5.71 • log 2.24 • 10 5) = – (5.71 • 5.35) = – 30.5] In the living cell : [ATP]  [ADP], pH  7.4, T = 37 oC, all of which influence the value of G G = G o + 2.3 • R • T• log ([ADP] • [Pi]) / [ATP] (Q  1, log Q  negative) G = – 52.3 kJ/mol

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