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CHAPTER. 19. Gibbs Free Energy , ∆ G. Under standard conditions — ∆ G o sys = ∆ H o sys - T∆S o sys. free energy = total energy change for system - energy change in disordering the system. ∆G o = ∆H o - T ∆S o.

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Gibbs free energy g
Gibbs Free Energy, ∆G

Under standard conditions —

∆Gosys = ∆Hosys - T∆Sosys

free energy = total energy change for system

- energy change in disordering the system


G o h o t s o
∆Go = ∆Ho- T∆So

If reaction is exothermic (negative ∆ Ho) (energy dispersed)

and entropy increases (positive ∆So) (matter dispersed)

then∆Gomust beNEGATIVE

reaction is spontaneous (and product-favored).


G o h o t s o1
∆Go = ∆Ho- T∆So

If reaction is

endothermic (positive ∆Ho)

and entropy decreases

(negative ∆So)

then ∆Gomust be POSITIVE

reaction is NOT spontaneous (and is reactant-favored).



1st method calculating g o
1st method:calculating ∆Go

use Gibb’s equation

∆Go = ∆Ho - T∆So

Determine ∆Horxn and ∆Sorxn


Combustion of acetylene

C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)

Use enthalpies of formation to calculate

∆Horxn= -1238 kJ

Use standard molar entropies to calculate

∆Sorxn= -97.4 J/K or -0.0974 kJ/K

∆Gorxn= -1238 kJ - (298 K)(-0.0974 J/K)

= -1209 Kj

Reaction is product-favored in spite of negative ∆Sorxn.

Reaction is “enthalpy driven”


NH4NO3(s) + heat ---> NH4NO3(aq)

Is the dissolution of ammonium nitrate product-favored?

If so, is it enthalpy- or entropy-driven?


NH4NO3(s) + heat ---> NH4NO3(aq)

From tables of thermodynamic data we find

∆Horxn = +25.7 kJ

∆Sorxn = +108.7 J/K or +0.1087 kJ/K

∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)

= -6.7 kJ

Reaction is product-favored in spite of positive ∆Horxn.

Reaction is “entropy driven”


2nd method of calculating g o
2nd method of calculating ∆Go

Use tabulated values of ∆Gfo,

free energies of formation.

∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)


Free energies of formation
Free Energies of Formation

Note that ∆G˚f for an element = 0


C graphite o 2 g co 2 g
C(graphite) + O2(g) --> CO2(g)

free energy of a standard state element is 0.

∆Gorxn =∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)]

∆Gorxn = -394.4 kJ - [ 0 + 0]

∆Gorxn = -394.4 kJ

Reaction is product-favored as expected.

∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)


More thermo?

You betcha!


K eq and thermodynamics
Keqand Thermodynamics

∆Gorxn is the change in free energy when pure reactants convert COMPLETELY to pure products.

Product-favored systems have Keq > 1.

Therefore, both ∆G˚rxn and Keq are related to reaction favorability.


Thermodynamics and k eq
Thermodynamics and Keq

Keq is related to reaction favorability and so to ∆Gorxn.

∆Gorxn = - RT lnK

where R = 8.31 J/K•mol

The larger the value of K the more negative the value of ∆Gorxn


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