The Gibbs Energy (G) and the Helmholtz Energy (A)
• Previously, We determined all the properties of
Entropy and obtained A and G from Entropy ( and
P,V and T)
• Now use A and G for reactions and
transformations
• Energy type terms: U, H, TS, PV. All are
extensive. (T and P by themselves are intensive).
What combinations of these terms give
meaningful expressions and why?
U
A
PV 
 PV
H
TS

G
Four distinct
energies are
U,H,A,G
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Helmholtz Energies
Spontaneity condition:
S  Ssurroundings  0
wrev  w 
w   Pexternal dV  wnonexpansion
qrev  TdS  q
dU  TdS  wrev  TdS  w
dU  TdS  w
Isothermal:
dT  0
TdS = d (TS )
dA  w
dU  TdS  d U  TS   dA
dA  w
For an isothermal process, dA is the maximum work one can get
from the process.
If you don’t want work, w=0, then dA<0 determines whether the
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process will occur at all (if kept isothermal).
Spontaneity Criterions
For processes occurring at constant T and P (and this implies a chemical
reaction or other complicated multi-component process): The criterion for the
maximum non-PdV (i.e. electrical) work one can have is dG
for dT  dP = 0,
dG  wnonexpansion
At constant T and P
Proof:
dA  w   PdV  wnonexpansion
d  A  PV   dG  wnonexpansion
For a process at constant T and P
G  H  TS
dG  dH  TdS
G  H  T S
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Example Problem
Calculate the maximum non-expansion work that can be produced
by the fuel cell oxidation reaction CH4 (g) + 2O2 (g) → CO2 (g) +
H combustion  CH 4 , g   890kJ
2H2O (l) . [At constant T and P]
Gcomb ,rxn  H combustion  CH 4 , g   T  S  CO 2 , g   2S  H 2O, l   S  CH 4 , g   2S  O 2 , g  
 890kJ
 298.15  103   213.74  2 x  69.61  186.26  3 x 205.138  kJ
Gcomb ,rxn  818 kJ
One mole of methane is worth 2 to 3 cups of yogurt in energy.
Compare the available work done by a reversible heat engine and that
done by an electrochemical fuel cell for methane combustion. For
Thot = 600K and Tcold = 300K, e= 0.50. set qhot = H and w= e qhot.
The maximum work available from the heat engine is 59% of that
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available in the electrochemical fuel cell.
Direction of Spontaneous Change for a constant T,P Process
G  H  T S
Spontaneous when
dG  0
There are two contributions to G that determine if an isothermal,
isobaric chemical transformation is spontaneous: the energetic
contribution, H, and the entropic contribution, TS.
•The entropic contribution to G is larger at higher temperatures.
• A chemical transformation is always spontaneous if H < 0 (an
exothermic reaction) and S > 0.
•A chemical transformation is never spontaneous if H > 0 (an
endothermic reaction) and S < 0.
•For all other cases, (when H and TS have the same sign) the
relative magnitudes and signs of H and TS determine if the
chemical transformation is spontaneous.
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Phase Transition
G  H  T S
H
 Tm  0
S
G  S Tm  T 
• Consider a phase transition from the more stable form to the
less stable form (e.g. from solid to liquid, like ice to water at
0C). DH is positive, because Heat must go into the material
to break the bonds. But heat in under constant P (and T) is qreversible so DS = DH/Tm and is also positive. Notice that
DG=0 at this point. So DG=0 is the transition between
spontaneous and not spontaneous.
• Assume that DS and DH are insensitive to T (and you did this
calculation for the ice to water transition), then when T>0
the TDS term dominates and DG < 0 so ice melts above the
transition temperature. Conversely it freezes below it
because the DH term is larger (and it is positive), so DG >0
when T< Tm.
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The Dependence of the Gibbs and Helmholtz
Energies on P, V, and T
 G 
 G 
dG   SdT  VdP  
   S and 
 V
 T  P
 P T
For a change in P at constant T
 G 
G   dG   
 dP   VdP
P T
Go
P 
P
G
P
P
For liquids and solids
p
G  G T , P   G T , P    VdP  V  P  P   V P
p
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By contrast, the volume of a gaseous system changes appreciably
with pressure.
P
G  G T , P   G T    VdP 
P
P

P
We will use this relation many times.
nRT
P
dP  nRT ln
P
P
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Chemical Reactions
• Review: How H and S are written for chemical reactions,
when T and P are fixed.
• In a Chemical Reaction, the concentration of all species are
tied together.
N
0
 i i


0 
 2 NH3  1N2  3H 2
i 1
ni  n   i x
o
i
dni   i dx
X, or dX is the extent of
advancement (or retardation if X is
negative). Get X from any species
concentration change specified.
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Reaction H and S (at constant T and P)
For a reaction, the change in the number of moles is controlled by the
value of X, the advancement parameter.
 H 
 H 
dH  
dT  
dP


 T  P ,n1 ,n2 ...
 P T ,n1 ,n2 ...
 H 
 H 

dn1  
dn1  ......


 n1 T , P ,n2 ...
 n2 T , P ,n1 ...
N
 H 
 H 
fixed Tand P : dH   dni 
 dx  i 



n

n
i 1
i 1
 i T , P ,n j
 i T , P ,n j .
N
N
 S 
o
dS  dX   i 

dX


S


i i i 
i 1
i 1
 ni T , P ,n j .
N
Grxn  H rxn  T Srxn
dH  dH rxn dX
d H  H rxn dX
d S  S rxn dX
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The Gibbs Energy of a Reaction Mixture
 G 
 G 
 G 
 G 
dG  
dT

dP

dn

dn1  ......
1







 T  P ,n1 ,n2 ...
 P T ,n1 ,n2 ...
 n1 T ,P ,n2 ...
 n2 T ,P ,n1 ...
define the chemical potential, the molar Gibbs energy mi, as
 G 
mi  


n
 i  P ,T ,n  n
j
i
 G 
 G 
dG  
dT

dP   mi dni



 T  P ,n1 ,n2 ...
 P T ,n1 ,n2 ...
i
dG   mi dni  dx  m
i i
i
i
dGrxn   i mi
i
 When T and P are fixed:
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Why is mi called the chemical potential of species i? This can be
understood by assuming that the chemical potential for species i has
the values miI in region I, and miII in region II of a given mixture
with miI  miII . If dni moles of species i are transported from region
I to region II, at constant T and p, the change in G is given by
dG   miI dni  miII dni   miII  miI  dni  0
Matter flows from a region of high chemical potential to one of low
chemical potential
At equilibrium, the chemical potential of each individual species is
the same throughout a mixture.
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Ideal gas mixture at constant temperature
After equilibrium has been reached,
m Hpure T , PH   m Hmixture T , PH
2
2
pure
H2
P
m
mixture
H2
T , P 
H2
The chemical potential of
each gas only depends on
its own partial pressure.
This comes from the
mixing entropy
2
P
mixture
H2
2

 xH2 Ptotal
 PH 2 
 m H 2 (T , P )  RT ln  o 
P 


 P 
o
o
 m H 2 (T , P )  RT ln  o   RT ln  H 2
P 
o


 m Ho 2 (T , P)  RT ln  H 2

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
In a mixture of A and B: Compare Gibbs energies due to mixing
• Integrate up the equation for dG in terms of n holding the
chemical potential fixed:
Pi
o
from mi  mi  P   RT ln o  mi  P   RT ln xi
P
dG   mi dni
o
i
G   mi ni   mio ni  RT  ni ln xi
i
i
i
G pure   mio ni and G mix  RT  ni ln xi
i
i
Apply to a two component system: n  nA  nB
G pure  n  xA m Ao  xB m Bo  and G mix  nRT  xA ln xA  xB ln xB 
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In a Chemical Reaction for an Ideal Gas Mixture in
Terms of the mi, the Gibbs Energy is:
0   vi  i
i
dG   mi dni
Use X: the extent of reaction
i
ni  niinitial  vi X
dni  vi dX


dG    vi mi  dX  Grxn dX
 i

 G 
Grxn  
   vi mi
 X T , P
i
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The direction of spontaneous change is that for which G
is negative.
THEREFORE
If
 G 

 < 0, the reaction proceeds spontaneously as written.
 X T , P
 G 
•If 
 > 0, the reaction proceeds spontaneously in the

X

T , P
opposite direction, and a negative dX makes the reaction
spontaneous in the opposite direction (i.e. make reactants).
 G 
•If  X  = 0, the reaction system is at equilibrium, and

T , P
there is no direction of spontaneous change.
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