Free enthalpyGibbs 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 (combustion of gasoline)

Free-energy change during a nonspontaneous reaction

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 upto 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

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 liberatedwhen 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 Go

f value of the stablest natural allotrope of an element is regarded zero.G = G (products) – G (reactants) andGo = Go

f (products) – Gof (reactants)

Standard FreeEnergies ofFormation( at 25 oC)

Calculation of the standard free enthalpy of formation: Gf

o = Hfo – T• So , where

Gfo = the standard free enthalpy of formation of the chemical

compound studiedHf

o = the standard enthalpy of formation of the chemical compound studiedSo = 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 Hf

o value (which is in the order of magnitude of kilojoules) the difference between the values of Gf

o and Hfo is

generally moderate.

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) Gf

o (kJ/mol)H2 130.7 0 0O2 205.1 0 0H2O 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/molGf

o = Hfo – T • So and after substitution :

Gfo = –285.8 – (– 48.7) = – 237.1 kJ/mol

Dependence of free enthalpy (Gibbs free energy) content on concentration:G = Gf

o + RT ln c (J/mol), at 25 oC:G = Gf

o + 8.3 • 298 • 2.3 • lg c (J/mol))G = Gf

o + 5.71 • lg c (kJ/mol)Free enthalpy change of reversible chemical reactions:A + B C + DG = G (products) – G (reactants)G = GC + GD – (GA + GB)G = [Gf

o(C) + 5.71 • log cC ] + [Gf

o(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 are 1 M each:G = Go , and because in equilibrium: G = 0Go = – 5.71 • log Ke (kJ/mol)

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

Free enthalpy content of the concentration gradient

Free enthalpy change of redox reactions: G = – n • F • E and at standard conditions: Go = – n • F • EkJ/mol, ifR

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

Various ways for the determination of free energy changes of chemical reactions:

1. Go = Gfo (products) – Gf

o (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

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 )

The chemiluminescent reaction of a firefly

Luciferin (organic peroxide) + ATP

in the presence of luciferase

light

Relationship between equilibrium constant (Ke) and temperature At T1 temperature:

G1o = Ho – T1 • So at 25-37 oC the change of Ho

and So with T is negligible,G1

o = – 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)

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

Determination of Ho : from log Ke versus 1/T plotDetermination of Go : from Ke or Eo

Determination of So : from (Ho– Go)/T expression

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