Transcript Entropy and Free Energy - Manasquan Public Schools

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Under standard conditions —
o
∆G
sys
=
o
∆H
sys
-
o
T∆S
sys
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 ENTROPY
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”
∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)
Use tabulated values of
o
∆Gf ,
free energies of formation.
Note that ∆G˚f for an element = 0
free energy of a standard state element is 0.
∆Gorxn =∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)]
∆Gorxn = -394.4 kJ - [ 0 + 0]
∆Go
rxn
= 394.4 kJ
Reaction is product-favored as expected.
∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)
More thermo?
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 ∆Gorxn.
o
∆G rxn
= - RT lnK
where R = 8.31 J/K•mol
The larger the value of K the more
negative the value of ∆Gorxn