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Unformatted text preview: Gibbs Free Energy, G
Gibbs Free Energy, G
Suniv = Ssurr +
Suniv = Hsys
T 1 Ssys
+ Go = Ssys = Ho T So Gibbs Free Energy, G
Gibbs Free Energy, G
Go = 4 Ho  T So Two methods of calculating Go
a)
Determine Horxn and
Determine
GIbbs equation. Sorxn and use
and b)
Use tabulated values of free energies of
formation, Gfo.
formation, Gorxn =
Gorxn = Gffo (products) G o ((products)
products) Gffo (reactants)
G o ((reactants)
reactants) Gibbs Free Energy, G
Gibbs Free Energy, G 2 Ho  T So Go = Gibbs free energy change =
total energy change for system
 energy lost in disordering the system
If reaction is exothermic ( Ho negative) and
negative)
entropy increases ( So is +), then Go
must be negative and reaction productfavored.
If reaction is endothermic ( Ho is +), and
is
entropy decreases ( So is ), then Go
must be + and reaction is reactantfavored. Multiply through by T
T Suniv = Hsys  T Ssys
T Suniv = change in Gibbs free energy
change
for the system = Gsystem
for
Under standard conditions — Go Gibbs Free Energy, G
Gibbs Free Energy, G Calculating
Calculating
Calculating Gorxn
Gorxn Ho So Go Reaction increase(+)  Prodfavored endo(+)
endo(+) decrease() + Reactfavored exo()
exo() decrease() ? T dependent endo(+)
endo(+) Combustion of acetylene
C2H2(g) + 5/2 O 2(g) > 2 CO 2(g) + H 2O(g)
Use enthalpies of formation to calculate
Horxn = 1238 kJ
1238
Use standard molar entropies to calculate
Sorxn = 97.4 J/K or 0.0974 kJ/K
97.4
Gorxn = 1238 kJ  (298 K)(0.0974 J/K)
1238
= 1209 kJ
Reaction is productfavored in spite of negative
Sorxn.
Reaction is “enthalpy driven” Page 1 Ho  T So exo()
exo() 5 3 increase(+) ? T dependent Calculating
Calculating
Calculating Gorxn
Gorxn NH4NO3(s) + heat > NH 4NO3 (aq)
aq)
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy or entropydriven? 6 Calculating
Calculating
Calculating Gorxn
Gorxn Gorxn =
Gorxn = NH4NO3(s) + heat > NH 4NO3 (aq)
aq)
From tables of thermodynamic data we find
Horxn = +25.7 kJ
+25.7
Sorxn = +108.7 J/K or +0.1087 kJ/K
+108.7
Gorxn = +25.7 kJ  (298 K)(+0.1087 J/K)
+25.7
= 6.7 kJ
Reaction is productfavored in spite of negative
Horxn.
Reaction is “entropy driven” Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) > 4 Fe(s) + 3 CO 2(g)
Horxn = +467.9 kJ
Sorxn = +560.3 J/K
+467.9
+560.3
Gorxn = +300.8 kJ
+300.8
Reaction is reactantfavored at 298 K
At what T does Gorxn just change from being
At
just
(+) to being ()?
When Gorxn = 0 = Horxn  T Sorxn
When
T= Calculating
Calculating
Calculating 7 Gffo (products) G o ((products)
products) Gorxn
Gorxn 8 Gffo (reactants)
G o ((reactants)
reactants) Combustion of carbon
C(graphite) + O2(g) > CO 2(g)
Gorxn = Gfo(CO2)  [ Gfo(graph) + Gfo(O2)]
(graph)
Gorxn = 394.4 kJ  [ 0 + 0]
394.4
Note that free energy of formation of an
element in its standard state is 0.
Gorxn = 394.4 kJ
394.4
Reaction is productfavored as expected. 10 Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) > 4 Fe(s) + 3 CO 2(g)
Horxn = +467.9 kJ
Sorxn = +560.3 J/K
+467.9
+560.3
Gorxn = +300.8 kJ
+300.8
Reaction is reactantfavored at 298 K
At what T does Gorxn just change from being
At
just
(+) to being ()?
When Gorxn = 0 = Horxn  T Sorxn
When Thermodynamics and Keq More thermo?
thermo? Later! Hrxn
467.9 kJ
=
= 835.1 K
0.5603 kJ/K
Srxn Page 2 9 Keq is related to reaction favorability.
is
favorability.
When Gorxn < 0, reaction moves
When
0,
energetically “downhill”
Gorxn is the change in free energy as
is
reactants convert completely to products.
But systems often reach a state of
equilibrium in which reactants have not
converted completely to products.
In this case Grxn is < Gorxn , so state with
In
so
both reactants and products present is
more stable than complete conversion. 12 Thermodynamics and Keq 13 Productfavored
reaction.
2 NO2 > N 2O4
>
Gorxn = 4.8 kJ
4.8
Here Grxn is less than
Here
is
Gorxn , so the state
so
with both reactants
and products
present is more
stable than complete
conversion. Thermodynamics and Keq Thermodynamics and Keq 14 Reactantfavored
reaction.
N2O4 >2 NO2
>2
Gorxn = +4.8 kJ
+4.8
Here Gorxn is greater
Here
is
than Grxn , so the
than
so
state with both
reactants and
products present is
more stable than
complete conversion. 16 Gorxn =  RT lnK
RT
Calculate K for the reaction
N2O4 >2 NO2
Gorxn = +4.8 kJ
>2
+4.8
Gorxn = +4800 J =  (8.31 J/K)(298 K) ln K
+4800
4800 J
lnK = = 1.94
(8.31 J/K)(298K)
K = 0.14
When Gorxn > 0, then K < 1
When
0, Page 3 Thermodynamics and Keq
Keq is related to reaction favorability and so
is
to Gorxn.
to
The larger the value of Gorxn the larger the
The
the
value of K. Gorxn =  RT lnK
RT
where R = 8.31 J/K•mol
J/K•mol 15 ...
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This note was uploaded on 11/12/2011 for the course CHEM 101 taught by Professor Staff during the Fall '09 term at North South University.
 Fall '09
 Staff

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