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# Gibbs Free Energy

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```Gibbs Free Energy, G
в€†Suniv = в€†Ssurr + в€†Ssys
пЃ„ S univ
=
пЂ­пЃ„ H sys
T
+ пЃ„ S sys
Multiply through by -T
-Tв€†Suniv = в€†Hsys - Tв€†Ssys
-Tв€†Suniv = change in Gibbs free energy
for the system = в€†Gsystem
Under standard conditions вЂ”
o
в€†G
sys
=
o
в€†H
sys
-
o
Tв€†S
sys
1
o
в€†G
Gibbs free
=
o
в€†H
-
o
Tв€†S
2
energy change =
total energy change for system
- energy lost in disordering the system
If reaction is
вЂў exothermic (negative в€† Ho) (energy dispersed)
вЂў and entropy increases (positive в€†So)
(matter dispersed)
вЂў then в€†Go must be NEGATIVE
вЂў reaction is spontaneous (and productfavored).
o
в€†G
=
o
в€†H
-
o
Tв€†S
3
Gibbs free energy change =
total energy change for system
- energy lost in disordering the system
If reaction is
вЂў endothermic (positive в€†Ho)
вЂў and entropy decreases (negative в€†So)
вЂў then в€†Go must be POSITIVE
вЂў reaction is not spontaneous (and is reactant-
favored).
Gibbs Free Energy, G
o
в€†G
=
o
в€†H
-
o
Tв€†S
в€†Ho
в€†So
в€†Go Reaction
exo(вЂ“)
increase(+)
вЂ“
Prod-favored
endo(+)
decrease(-)
+
React-favored
exo(вЂ“)
decrease(-)
?
T dependent
endo(+)
increase(+)
?
T dependent
4
Gibbs Free Energy, G
o
в€†G
=
o
в€†H
-
5
o
Tв€†S
Two methods of calculating в€†Go
a)
Determine в€†Horxn and в€†Sorxn and use
GIbbs equation.
b)
Use tabulated values of
free energies
of formation, в€†Gfo.
в€†Gorxn = пЃ“ в€†Gfo (products) - пЃ“ в€†Gfo (reactants)
Free Energies of Formation
Note that в€†GЛљf for an element = 0
6
Calculating в€†Gorxn
7
Combustion of acetylene
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
в€†Horxn = -1238 kJ
Use standard molar entropies to calculate
в€†Sorxn = -97.4 J/K or -0.0974 kJ/K
в€†Gorxn = -1238 kJ - (298 K)(-0.0974 J/K)
= -1209 kJ
Reaction is product-favored in spite of negative
в€†Sorxn.
Reaction is вЂњenthalpy drivenвЂќ
Calculating в€†Gorxn
NH4NO3(s) + heat ---> NH4NO3(aq)
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy- or entropy-driven?
8
Calculating в€†Gorxn
9
NH4NO3(s) + heat ---> NH4NO3(aq)
From tables of thermodynamic data we find
в€†Horxn = +25.7 kJ
в€†Sorxn = +108.7 J/K or +0.1087 kJ/K
в€†Gorxn = +25.7 kJ - (298 K)(+0.1087 J/K)
= -6.7 kJ
Reaction is product-favored in spite of negative
в€†Horxn.
Reaction is вЂњentropy drivenвЂќ
Gibbs Free Energy, G
o
в€†G
=
o
в€†H
-
10
o
Tв€†S
Two methods of calculating в€†Go
a)
Determine в€†Horxn and в€†Sorxn and use
GIbbs equation.
b)
Use tabulated values of
free energies
of formation, в€†Gfo.
в€†Gorxn = пЃ“ в€†Gfo (products) - пЃ“ в€†Gfo (reactants)
Calculating в€†Gorxn
в€†Gorxn = пЃ“ в€†Gfo (products) - пЃ“ в€†Gfo (reactants)
11
Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
в€†Gorxn = в€†Gfo(CO2) - [в€†Gfo(graph) + в€†Gfo(O2)]
в€†Gorxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an
element in its standard state is 0.
в€†Gorxn = -394.4 kJ
Reaction is product-favored as expected.
Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
в€†Horxn = +467.9 kJ
в€†Sorxn = +560.3 J/K
в€†Gorxn = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does в€†Gorxn just change from
being (+) to being (-)?
When в€†Gorxn = 0 = в€†Horxn - Tв€†Sorxn
T =
пЃ„ H rxn
пЃ„ S rxn
=
467.9 kJ
0.5603 kJ/K
= 835.1 K
12
13
More thermo?
You betcha!
Thermodynamics and Keq
пЃµ
FACT: в€†Gorxn is the change in free energy
when pure reactants convert COMPLETELY
to pure products.
пЃµ
FACT: Product-favored systems have
Keq > 1.
пЃµ
Therefore, both в€†GЛљrxn and Keq are related
to reaction favorability.
14
Thermodynamics and Keq
Keq is related to reaction favorability and so
to в€†Gorxn.
The larger the value of K the more negative
the value of в€†Gorxn
o
в€†G rxn
= - RT lnK
where R = 8.31 J/KвЂўmol
15
Thermodynamics and Keq
в€†Gorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
в€†Gorxn = +4.8 kJ
в€†Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K
ln K =
-
4800 J
(8.31 J/K)(298
K)
=
K = 0.14
When в€†Gorxn > 0, then K < 1
- 1.94
16
в€†G, в€†GЛљ, and Keq
вЂў в€†G is change in free energy at nonstandard conditions.
вЂў в€†G is related to в€†GЛљ
вЂў в€†G = в€†GЛљ + RT ln Q
where Q = reaction quotient
вЂў When Q < K or Q > K, reaction is
spontaneous.
вЂў When Q = K reaction is at equilibrium
вЂў When в€†G = 0 reaction is at equilibrium
вЂў Therefore, в€†GЛљ = - RT ln K
17
в€†G, в€†GЛљ, and Keq
Figure 19.10
18
в€†G, в€†GЛљ, and Keq
вЂў Product favored
reaction
вЂў вЂ“в€†Go and K > 1
вЂў In this case в€†Grxn is
< в€†Gorxn , so state with
both reactants and
products present is
MORE STABLE than
complete conversion.
19
в€†G, в€†GЛљ, and Keq
20
Product-favored
reaction.
2 NO2 ---> N2O4
в€†Gorxn = вЂ“ 4.8 kJ
Here в€†Grxn is less than
в€†Gorxn , so the state
with both reactants
and products
present is more
stable than complete
conversion.
в€†G, в€†GЛљ, and Keq
21
Reactant-favored
reaction.
N2O4 --->2 NO2
в€†Gorxn = +4.8 kJ
Here в€†Gorxn is greater
than в€†Grxn , so the
state with both
reactants and
products present is
more stable than
complete conversion.
Thermodynamics and Keq
пЃµ
Keq is related to reaction favorability.
пЃµ
When в€†Gorxn < 0, reaction moves
energetically вЂњdownhillвЂќ
пЃµ
в€†Gorxn is the change in free energy when
reactants convert COMPLETELY to
products.
22
```
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