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 b)
Use tabulated values of free energies of
formation, Gfo.
formation, Gorxn =
Gorxn = Gffo ((products) G o products) Gffo ((reactants)
G o reactants) Gibbs Free Energy, G
Gibbs Free Energy, G 2 Ho  T So Go = Gibbs f ree energy change =
total energy change for system
 energy lost in disordering the system
energy
If reaction is exothermic ( Ho negative) and
entropy increases ( So is +), then Go
must be negative and reaction productfavored.
If reaction is endothermic ( Ho is +), and
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
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(+) decrease() + Reactfavored exo() decrease() ? T dependent 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
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
1209
Reaction is productfavored in spite of negative
Reaction
productfavored
Sorxn.
Reaction is “enthalpy driven”
Reaction “enthalpy Page 1 Ho  T So exo() 5 3 increase(+) ? T dependent Calculating
Calculating
Calculating Gorxn
Gorxn NH4NO3(s) + heat > NH 4NO3 (aq)
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy or entropydriven? 6 Calculating
Calculating
Calculating Gorxn
Gorxn 7 Gorxn =
Gorxn = NH4NO3(s) + heat > NH 4NO3 (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 J/K)
= 6.7 kJ
6.7
Reaction is productfavored in spite of negative
Reaction
productfavored
Horxn.
Reaction is “entropy driven”
Reaction “entropy 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
Gorxn = +300.8 kJ
Reaction is reactantfavored at 298 K
Reaction reactantfavored
At what T does Gorxn just change from being
At
(+) to being ()?
When Gorxn = 0 = Horxn  T Sorxn
When
T= Calculating
Calculating
Calculating
Gffo ((products) G o products) Gorxn
Gorxn 8 Gffo ((reactants)
G o 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]
Note that free energy of formation of an
element in its standard state is 0.
Gorxn = 394.4 kJ
Reaction is productfavored as expected.
Reaction productfavored 10 More thermo?
More thermo Later! Hrxn
467.9 kJ
=
= 835.1 K
0.5603 kJ/K
Srxn Page 2 Free Energy and Temperature 9 2 Fe2O3(s) + 3 C(s) > 4 Fe(s) + 3 CO 2(g)
Horxn = +467.9 kJ
Sorxn = +560.3 J/K
Gorxn = +300.8 kJ
Reaction is reactantfavored at 298 K
Reaction reactantfavored
At what T does Gorxn just change from being
At
(+) to being ()?
When Gorxn = 0 = Horxn  T Sorxn
When Thermodynamics and Keq
Keq is related to reaction favorability.
When Gorxn < 0, reaction moves
When
energetically “downhill”
Gorxn is the change in free energy as
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
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
Here Grxn is less than
Here
Gorxn , so the state
with both reactants
and products
present is more
stable than complete
conversion. Thermodynamics and Keq Thermodynamics and Keq 14 Reactantfavored
reaction.
N2O4 >2 NO2
Gorxn = +4.8 kJ
Here Gorxn is greater
Here
than Grxn , so the
than
state with both
reactants and
products present is
more stable than
complete conversion. 16 Gorxn =  RT lnK
Calculate K for the reaction
N2O4 >2 NO2
Gorxn = +4.8 kJ
Gorxn = +4800 J =  (8.31 J/K)(298 K) ln K
4800 J
lnK = = 1.94
(8.31 J/K)(298K)
K = 0.14
When Gorxn > 0, then K < 1
When Page 3 Thermodynamics and Keq
Keq is related to reaction favorability and so
to Gorxn.
to
The larger the value of Gorxn the larger the
The
value of K. Gorxn =  RT lnK
where R = 8.31 J/K• mol 15 ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.
 Fall '01
 Reynolds
 Organic chemistry

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