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Review. Standard Gibbs Free Energy. G o =. H o. - T S o. spontaneous. G o < 0. G o = - 30.5 kJ/mol. non-spontaneous. G o > 0. G o = + 16.7 kJ/mol. G o = 0. equilibrium. G o = - 13.8 kJ/mol. . glucose-6-phosphate. + H 2 O. + phosphate. glucose. + H 2 O . - PowerPoint PPT Presentation
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Review Standard Gibbs Free Energy
Go =
Go < 0Go > 0Go = 0
spontaneousnon-spontaneous
equilibrium
Ho - TSo
glucose
Go = + 16.7 kJ/mol
ATP
Go = - 30.5 kJ/mol
glucose
Go = - 13.8 kJ/mol
+ phosphate glucose-6-phosphate + H2O
+ H2O ADP + phosphate
+ ATP ADP + glucose-6-phosphate
Q =
K =
G =
K > Q
K < Q
reaction quotient = [products]
equilibrium constant = [products]
G < 0
K = Q
m
[reactants]initialn
initialmequilibrium
[reactants]nequilibrium
- RT ln ( )
G > 0G = 0
K /Q
spontaneousnon-spontaneousequilibrium
Non-standard conditions
Go
G = - RT ln(K/Q)
Impose Standard conditions:
Q =
G =
[reactants]initial = [products]initial= 1(M or atm)
1
o - RT ln(K)
Standard Free Energy
GGoo = - RT ln K = - RT ln K
ln (a/b) =
G = - RT ln (K/Q) =
G = - RT ln (K/Q)G = - RT ln (K/Q)
-RT ln K + RT ln Q
ln a - ln b
Non-Standard Conditions
G =G = GGoo + RT ln Q+ RT ln Q
Go = - RT ln(K)
glucose + phosphate glucose-6-phosphate + H2O
Go = 16.7 kJ/mol
non-spontaneous
= -RT ln K-RT -RT
K =
G = - RT ln ( )K /Q
Q < 1.2 x 10-3G < 0
spontaneous start with no product Q = [products][reactants] initial
initial
standard conditions
non-standard conditions
1.2 x 10-3
< K
Initially
Will more products or reactants be formed?Q = 0.5 /0.32 = 5.6
Non-Standard Conditions
G =G = GGoo + RT ln Q+ RT ln Q
[NO2] = 0.3 M [N2O4] = 0.5 M
2NO2 (g) N2O4 (g)
Gorxn = Go
f products - Gof reactants
= 97.8 -
Grxn= = -0.5 kJ/mol
More N2O4 will be formed
- 4.8 kJ/mol
-4.8 +(8.314 x 10-3)(298)ln 5.6
2(51.3) =
Non-Standard Conditions
G =G = GGoo + RT ln Q+ RT ln Q2NO2 (g) N2O4 (g)
Initially [NO2] = 0.3 M [N2O4] = 0.5 M
A K =
[A]e[B]e
Le Chatelier’s Principle
Add A
Remove A
[C]i
[A]i[B]i
G = -RT ln (K/Q)
Q K/Q
[C]e
> 1
increase < 1
+B C Q =
Add Bdecrease Q K/Q
G < 0>
Temperature dependence of K
exothermic reaction heat =
endothermic reaction heat =
at low T favor forward reactionat high T
at low T favor reaction reverseat high T favor reaction forward
favor reaction reverse
product
reactant
Temperature dependence of K
Go = -RT ln K Go = Ho - TSo
-RT ln K =
ln K = - Ho
R1T
+ So
R
Ho < 0
y m x b
increase T K
ln K
1/T
T
Ho - TSo
decrease
Temperature dependence of K
Go = -RT ln K Go = Ho - TSo
-RT ln K = Ho - TSo
ln K = - Ho
R1T
+ So
R
endothermic H > 0
y m x b
increase T increase K
-ln K
1/T
T
