1919

Under standard conditions —

∆Gosys = ∆Ho

sys - T∆Sosys

free energy = total energy change for system

- energy change in disordering the system

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

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

then ∆Go must be NEGATIVE reaction is spontaneous (and

product-favored).

If reaction is

endothermic (positive ∆Ho)and entropy decreases

(negative ∆So)then ∆Go must be POSITIVE

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

ENTHALPY and ENTROPYENTHALPY and ENTROPY

use Gibb’s equation

∆Go = ∆Ho - T∆So

Determine ∆Horxn and ∆So

rxn

Combustion of acetyleneCombustion of acetyleneCC22HH22(g) + 5/2 O(g) + 5/2 O22(g) --> 2 CO(g) --> 2 CO22(g) + H(g) + H22O(g)O(g)

Use enthalpies of formation to calculate ∆Ho

rxn = -1238 kJ

Use standard molar entropies to calculate ∆So

rxn = -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 ∆So

rxn.

Reaction is “enthalpy driven”

Is the dissolution of ammonium nitrate product-favored?

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

NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)

From tables of thermodynamic data we find ∆Ho

rxn = +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 kJReaction is product-favored in spite of

positive ∆Horxn.

Reaction is “entropy driven”

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

Use tabulated values of ∆Gfo,

free energies of formation.

∆∆GGoorxnrxn = = ∆G ∆Gff

oo (products) - (products) - ∆G ∆Gffoo (reactants) (reactants)∆∆GGoo

rxnrxn = = ∆G ∆Gffoo (products) - (products) - ∆G ∆Gff

oo (reactants) (reactants)

Note that ∆G˚f for an element = 0

free energy of a standard state element is 0.

∆Gorxn =∆Gf

o(CO2) - [∆Gfo(graph) + ∆Gf

o(O2)]

∆Gorxn = -394.4 kJ - [ 0 + 0]

∆Gorxn = -394.4 kJ

Reaction is product-favored as expected.

∆∆GGoorxnrxn = = ∆G ∆Gff

oo (products) - (products) - ∆G ∆Gffoo (reactants) (reactants)∆∆GGoo

rxnrxn = = ∆G ∆Gffoo (products) - (products) - ∆G ∆Gff

oo (reactants) (reactants)

More thermo?More thermo? You betcha!You betcha!

∆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.

Keq is related to reaction favorability and so to ∆Go

rxn.

∆Gorxn = - RT lnK

where R = 8.31 J/K•mol

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

rxn